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Traffic Engineering, 4th Edition
《交通工程》，第四版

Roess, R.P., Prassas, E.S., and McShane, W.R.
Roess，R.P.，Prassas，E.S.和 McShane，W.R.

Solutions to Problems in Chapter 2
第二章习题解答

Problem 2-1
习题 2-1

The driver continues to travel at 60 mi/h while taking his foot from the accelerator to the brake. Therefore, the vehicle will travel:
司机在将脚从油门移到刹车踏板的过程中，继续以 60 英里/小时的速度行驶。因此，车辆将行驶：

dr = 1.47 St = 1.47 * 60 * 3.5 = 308.7 ft before the driver’s foot hits the brake.
dr = 1.47 St = 1.47 * 60 * 3.5 = 308.7 英尺，然后司机的脚踩到刹车上。

Problem 2-2
习题 2-2

When the driver first sees the overturned truck, he will continue to move at 65 mi/h during his reaction time. During this time, the vehicle will travel 1.47*65*t feet, or 95.5t ft. Thus, the distance available for braking is 350 – 95.5t feet. This is the distance that is available for deceleration before the vehicle hits the overturned truck. The formula for braking distance is
当司机第一次看到倾覆的卡车时，他将在反应时间内继续以 65 英里/小时的速度行驶。在此期间，车辆将行驶 1.47*65*t 英尺，或 95.5t 英尺。因此，可用于制动的距离为 350 – 95.5t 英尺。这是车辆撞上倾覆的卡车前可用于减速的距离。制动距离的公式为:

In this case, the initial speed (Si) is 65 mi/h. The friction factor, F, is related to the deceleration rate, and is computed by dividing the deceleration rate by the deceleration rate due to gravity, or 10/32.2 = 0.31. The grade is level, i.e., G = 0. The braking distance is 350 – 95.5t. Therefore:
在这种情况下，初始速度（S）为 65 英里/小时。摩擦系数 F 与减速度率有关，其计算方法是将减速度率除以重力引起的减速度率，即 10/32.2 = 0.31。坡度为水平，即 G = 0。制动距离为 350 – 95.5t。因此：

This equation is solved for various values of t from 0.50 to 5.00 s. Note that at the point where the reaction distance becomes more than 350 ft, the final speed is a constant 60
此方程针对从 0.50 到 5.00 秒的不同 t 值求解。请注意，当反应距离超过 350 英尺时，最终速度为恒定的 60

mi/h, and the braking distance is essentially “0.”
英里/小时，制动距离基本上为“0”。

2

Reaction
反应

Time
时间

(s)
(秒)

Reaction
反应

Distance
距离

(ft)
(英尺)

Braking
制动

Distance
距离

(ft)
(英尺)

Final
最终的

Speed (mi/h)
速度 (英里/小时)

0.5

47.8

302.2

37.6

1.0

95.6

254.5

43.1

1.5

143.3

206.7

48.0

2.0

191.1

158.9

52.4

2.5

238.9

111.1

56.5

3.0

286.7

63.4

60.3

3.5

334.4

15.6

63.9

4.0

382.2

0.0

65.0

4.5

430.0

0.0

65.0

5.0

477.8

0.0

65.0

The vehicle is going to hit the overturned truck in any event. For reaction times over approximately 3.75 s, the vehicle will hit the truck at full speed, 60 mi/h.
即使如此，该车辆无论如何都会撞上翻倒的卡车。对于反应时间超过大约 3.75 秒，该车辆将以全速（每小时 60 英里）撞击卡车。

Problem 2-3
问题 2-3

This problem involves only the braking distance, which is assumed to be the same as the length of the measured skid marks. The speed at the collision point is estimated to be 25 mi/h. Working backwards, we can estimate the speed at the beginning of the grass skid, and then the speed at the beginning of the pavement skid, as illustrated below
此问题仅涉及制动距离，该距离假定与测量滑痕长度相同。碰撞点速度估计为 25 英里/小时。通过反向计算，我们可以估计草地滑痕开始时的速度，然后估计路面滑痕开始时的速度，如下所示:

2

3

PAVEMENT F = 0.35
路面摩擦系数 = 0.35

GRASS F = 0.25
草地摩擦系数 = 0.25

Sf2 Sf1 25 mi/h
Sf2 Sf1 25 英里/小时

The braking distance formula has two speeds. Thus, the computation starts with the grass skid, for which we have an estimate of the final speed, 25 mi/h. Then:
制动距离公式包含两个速度。因此，计算从草地滑痕开始，我们对最终速度有 25 英里/小时的估计。然后：

S1 = (250 * 30 * 0.28) + 252 = 2100 + 625 = 2725 Sf 1 = 52.2 mi / h
S1 = (250 * 30 * 0.28) + 252 = 2100 + 625 = 2725 S1 = 52.2 英里 / 小时

This is the speed at which the grass skid started; it is also the speed at which the pavement skid ended. Now, considering the pavement skid:
这是草地滑痕开始时的速度；它也是路面滑痕结束时的速度。现在，考虑路面滑痕：

S2 = (120 * 30 * 0.38) + 52.22 = 1368 + 2725 = 4093 Sf 2 = 64.0 mi / h
S2 = (120 * 30 * 0.38) + 52.22 = 1368 + 2725 = 4093 S2 = 64.0 英里 / 小时

Thus, when the skid began, the vehicle was traveling at 64 mi/h.
因此，当滑痕开始时，车辆行驶速度为 64 英里/小时。

3

4

Problem 2-4
问题 2-4

The total deceleration distance must be evaluated to answer this question. It is the sum of the reaction distance and the braking distance, such that:
必须评估总减速度距离才能回答这个问题。它是反应距离和制动距离之和，因此：

In this case, the reaction time, t, is the AASHTO standard, or 2. 5 s. The friction factor, F, is based upon the standard AASHTO deceleration rate of 11.2 ft/s2 (F = 11.2/32.2 = 0.348) and the deceleration is given as 60 mi’h to 40 mi/h. Then:
在这种情况下，反应时间 t 为 AASHTO 标准，即 2.5 秒。摩擦系数 F 基于 AASHTO 标准减速度率 11.2 ft/s2（F = 11.2/32.2 = 0.348），减速度为从 60 mi/h 到 40 mi/h。然后：

Because the sign maybe seen from 120 ft away, the sign should be placed AT LEAST 412.1-120 = 292.1 ft before the curve.
因为标志可能在 120 英尺外可见，所以标志应至少在弯道前 412.1-120 = 292.1 英尺处放置。

Problem 2-5
例题 2-5

The “yellow” signal must be long enough to allow a vehicle that cannot safely stop before crossing the intersection line to proceed to enter the intersection at their ambient approach speed (40 mi/h). Thus, the last vehicle that should be allowed to enter the intersection during the “yellow” should have been one safe stopping distance away when the “yellow” was initiated. The safe stopping distance for this case is:
“黄灯”信号必须足够长，以允许无法在越过交叉路口线之前安全停车的车辆以其环境通行速度（40 mi/h）驶入交叉路口。因此，在“黄灯”启动时，允许进入交叉路口的最后一辆车应该距离安全停车距离有一个安全停车距离。在这种情况下，安全停车距离为：

At 40 mi/h, it would take a vehicle:
以 40 mi/h 的速度，车辆需要：

to traverse the safe stopping distance. This should be the length of the subject “yellow” interval.
来穿过安全停车距离。这应该是主题“黄灯”间隔的长度。

4

5

Problem 2-6
例题 2-6

The safe stopping distance is:
安全停车距离为：

Note that the standard AASHTO values for reaction time (2.5 s) and friction factor (0.348) are used.
请注意，使用了 AASHTO 标准的反应时间值 (2.5 秒) 和摩擦系数 (0.348)。

Problem 2-7
问题 2-7

The minimum radius of curvature is found as:
最小曲率半径计算如下：

5

Traffic Engineering, 4th Edition
《交通工程》，第四版

Roess, R.P., Prassas, E.S., and McShane, W.R.
Roess，R.P.，Prassas，E.S.和 McShane，W.R.

Solutions to Problems in Chapter 3
第三章习题解答

Problem 3-1
问题 3-1

Information: Horizontal curve, P.I. = 11,500 + 66
信息：水平曲线，交点桩号 = 11,500 + 66

Radius = 1,000 ft
半径 = 1,000 英尺

Angle of deflection = 60o
偏角 = 60°

Find: All relevant characteristics of the curve. Stations of the P.C.
求解：曲线的全部相关特性。曲线起点桩号。

and P.T.
和 P.T.

The degree of curvature is computed using Equation 3-1:
曲率的计算使用公式 3-1：

Then:
然后：

T = R Tan (Δ 2)= 1,000 Tan (602)= 1,000 Tan (30) = 1,000 * 0.5773 = 577.3 ft
T = R Tan (Δ 2)= 1,000 Tan (602)= 1,000 Tan (30) = 1,000 * 0.5773 = 577.3 英尺

LC = 2RSin(Δ 2)= 2 *1,000 * Sin(30) = 2,000 * 0.5000 = 1,000 ft
LC = 2RSin(Δ 2)= 2 *1,000 * Sin(30) = 2,000 * 0.5000 = 1,000 英尺

Stationing:
测站：

P.C. = (11,500+66)-577.3=10,988.7 = 10,900+88.7 P.T. = (10,900+88.7)+1,047=13,035.7=12,000+35.7

1

Problem 3-2
例题 3-2

Information: 3.5o curve
信息：3.5°曲线

60 mi/hdesign speed Two 12-ft lanes
设计速度 60 英里/小时 两条 12 英尺的车道

Use spiral transition curves P.I. = 15,100+26
使用螺旋过渡曲线 P.I. = 15,100+26

Angle of deflection =40o
偏角 = 40°

Find: T.S., S.C., C.S., and S.T.
求解：缓和曲线起终点（T.S.，S.C.）、圆曲线起终点（C.S.，S.T.）。

The design value of superelevation is given by Equation 3-9:
超高设计值由公式 3-9 给出：

From Table 3-3, the design friction factor (fdes) = 0.12. Then:
从表 3-3 中，设计摩擦系数 (fdes) = 0.12。则：

The length of the spiral is typically estimated using Equation 3-13:
螺旋线的长度通常使用公式 3-13 估算：

It may also be computed as the length of the spiral runoff, or the length of the spiral plus tangent runoffs. These are estimated using Equations 3-10 and 3-11, using the following assumptions: Superelevation is achieved by rotating both lanes around the centerline, and the normal drainage cross-slope is 1%. Then:
它也可以计算为螺旋线渐变段长度，或螺旋线长度加上直线渐变段长度。这些长度使用公式 3-10 和 3-11 估算，并基于以下假设：超高是通过绕中心线旋转双车道实现的，正常排水横坡为 1%。则：

e d

2

Where: w = 12 ft
其中：w = 12 英尺

n = 1 lanes rotated around centerline
n = 绕中心线旋转的车道数为 1

eNC = 1%

ed = 2.6%

bw = 1.00 (Seepage 46, 1 lanes rotated around centerline)
bw = 1.00（参见第 461 页图，车道绕中心线旋转）

Δ = 0.45 (Table 3.4, 60 mi/h)
Δ = 0.45（表 3.4，60 英里/小时）

Then:
然后：

Lr + Lt = 69.3 + 26.7 = 96.0 ft
Lr + L = 69.3 + 26.7 = 96.0 英尺

Equation 3-13 is based upon driver comfort. A value of 211 ft will be used. The central angle for the spiral is given by Equation 3-14:
等式 3-13 基于驾驶员舒适度。将使用 211 英尺的值。螺旋线的中心角由等式 3-14 给出：

The central angle of deflection for the circular portion of the curve is given by Equation 3-15:
曲线圆形部分的偏角中心角由等式 3-15 给出：

Δ S = Δ − 2δ = 40.0 − (2 * 3.7) = 32.6o Then:
ΔS = Δ − 2δ = 40.0 − (2 * 3.7) = 32.6° 然后：

TS = ⎢1637Tan

⎞,⎟⎤」⎥ + ⎡⎢L⎝(⎜1637 Cos(3.7) −1637 + ⎜⎝(
⎞,⎟
 * ⎜⎝(Tan
⎞,⎟
 + [211 −1637 Sin (3.7)]

TS = 595.8 + 0.4 + 105.3 = 701.5 ft
TS = 595.8 + 0.4 + 105.3 = 701.5 英尺

3

And:
和：

Then:
然后：

T.S. = P.I. − Ts = (15,100 + 26) − 701.5 = 14,424.5 = 14,400 + 24.5 S.C. = T.S. + Ls = 14,424.5 + 211 = 14,635.5 = 14,600 + 35.5

C.S. = S.C. + Lc = 14,635.5 + 931 = 15,566.5 = 15,500 + 66.5 S.T. = C.S. + Ls = 15,566.5 + 211 = 15,777.5 = 15,700 + 77.5

Problem 3-3
问题 3-3

Information: 5o curve
信息：5 度曲线

65 mi/hdesign speed 2% upgrade
65 英里/小时设计速度 2% 上坡

t = 2.5 s (driver reaction time)
t = 2.5 秒（驾驶员反应时间）

Find: Closest placement of roadside object.
查找：路边物体最近的放置位置。

Placement of roadside objects is based upon the severity of the curve and the safe-stopping distance, computed as:
路边物体的放置基于曲线的严重程度和安全停止距离，计算如下：

ds = 238.9 + 382.7 = 621.6 ft Then: (Equation 3-17)
ds = 238.9 + 382.7 = 621.6 英尺 然后：（方程式 3-17）

M = 1,145.9 *[1 − Cos(15.54)] = 1,145.9 * 0.036551 = 41.9 ft
M = 1,145.9 *[1 − Cos(15.54)] = 1,145.9 * 0.036551 = 41.9 英尺

4

Problem 3-4
例题 3-4

Information: R = 1,200 ft
已知信息：R = 1,200 英尺

60 mi/hdesign speed
设计速度 60 英里/小时

6% maximum superelevation rate Find: Appropriate superelevation rate Using Equation 3-9:
最大超高率 6% 求解：合适的超高率 使用公式 3-9：

Where: fdes = 0.12 (Table 3.3, 60 mi/h)
其中：fdes = 0.12（表 3.3，60 英里/小时）

The maximum design superelevation rate of 6% would be used.
将使用最大设计超高率 6%。

Problem 3-5
例题 3-5

Information: e = 10%
已知信息：e = 10%

70 mi/hdesign speed Three 12-ft lanes
设计速度 70 英里/小时 三条 12 英尺的车道

Rotation around pavement edge Find: Length of superelevation runoff Using Equation 3-10:
路面边缘旋转 求解：加宽过渡段长度 使用公式 3-10：

Where: w = 12 ft (lane width)
式中：w = 12 英尺（车道宽度）

N = 3 (lanes rotated) ed = 10% (given)
N = 3（旋转车道数）ed = 10%（已知）

bw = 0.67 (page 47, 3 lanes rotated) Δ = 0.40 (Table 2.4, 70 mi/h)
bw = 0.67（第 47 页，3 条车道旋转）Δ = 0.40（表 2.4，70 英里/小时）

5

Problem 3-6
例题 3-6

Information:
信息：

Category
类别

Grade 1
一级

Grade 2
二级

Grade 3
三级

Facility Type
设施类型

Rural Freeway
农村高速公路

Rural Arterial
农村干道

Urban Arterial
城市干道

Terrain
地形

Mountainous
山区

Rolling
滚动

Level
水平

Design Speed
设计速度

60 mi/h
60 英里/小时

45 mi/h
45 英里/小时

40 mi/h
40 英里/小时

Find: Maximum grade and critical length of grade for each From Figure 3.15:
查找：每条从图 3.15 中的最大坡度和临界坡长：

Maximum grades are: Grade 1 = 6%
最大坡度为：坡度 1 = 6%

Grade 2 = 5.5% Grade 3 = 7%
坡度 2 = 5.5% 坡度 3 = 7%

Critical lengths of grades are found from Figure 3.18, as shown below. Because all design speeds are less than 70 mi/h, a 10% reduction in speeds is used to determine the critical length.
坡长的临界长度是从图 3.18 中找到的，如下所示。由于所有设计速度都小于 70 英里/小时，因此使用速度降低 10% 来确定临界长度。

6

Grade 1: 800 ft
坡度 1：800 英尺

Grade 2: 850 ft
坡度 2：850 英尺

Grade 3: 700 ft
坡度 3：700 英尺

Problem 3-7
问题 3-7

Information: Vertical curve
信息：竖曲线

+4% grade to -5% grade V.P.I. = 1,500+55
+4% 坡度到 -5% 坡度 V.P.I. = 1,500+55

Elevation of V.P.I. = 500 ft L = 1,000 ft
V.P.I. 的高程 = 500 英尺 L = 1,000 英尺

Find: Stations of the V.P.C. and V.P.T.
求：V.P.C. 和 V.P.T. 的里程桩号

Elevations of the V.P.C. and V.P.T.
V.P.C. 和 V.P.T. 的高程

Elevation points along the curve at 100-ft intervals
曲线沿线每隔 100 英尺的高程点

Location and elevation of the high point Note that the curve described is a crest vertical curve
最高点的坐标和高程 注意，描述的曲线是顶点竖曲线.

An equation for this vertical curve can be constructed in the form of:
对于这条竖曲线，可以构建如下形式的方程：

Y = ax 2 + b + Y
Y = ax² + b + Y

x c o

Where:
其中：

b = G1 = 4

L = 10 (measured in 100' s of ft)
L = 10（以百英尺为单位测量）

Yo = YVPC = 500.00 − (4 * 5) = 480.0 ft Thus:
Yo = YVPC = 500.00 − (4 * 5) = 480.0 英尺 因此：

Yx = −0.45 x 2 + 4x + 480.0
Yx = −0.45 x² + 4x + 480.0

Stations: VPC = (1500+55)-500 = 1,000+55
站点：VPC = (1500+55)-500 = 1,000+55

VPT = (1500+55)+1000 = 2,500+55

7

Elevations:
高程：

Y = −0.45x 2 + 4x + 480.0
Y = −0.45x² + 4x + 480.0

x

Yo = YVPC = 480 ft
Yo = YVPC = 480 英尺

Y1 = −0.45(12 ) + (4 *1) + 480 = 483.55 ft Y2 = −0.45(22 ) + (4 * 2) + 480 = 486.2 ft Y3 = −0.45(32 ) + (4 * 3) + 480 = 487.95 ft Y4 = −0.45(42 ) + (4 * 4) + 480 = 488.80 ft Y5 = −0.45(52 ) + (4 * 5) + 480 = 488.75 ft Y6 = −0.45(62 ) + (4 * 6) + 480 = 487.85 ft Y7 = −0.45(72 ) + (4 * 7) + 480 = 485.95 ft Y8 = −0.45(82 ) + (4 *8) + 480 = 483.20 ft Y9 = −0.45(92 ) + (4 * 9) + 480 = 479.55 ft

Y10 = YPVT = −0.45(102 ) + (4 *10) + 480 = 475.00 ft The high point is found as:
Y10 = YPVT = −0.45(102 ) + (4 *10) + 480 = 475.00 ft 找到最高点：

Y4.44 = −0.45(4.442 ) + (4 * 4.44) + 480 = 488.9 ft

Problem 3-8
问题 3-8

Information:
信息：

Grade
等级

Entry Grade
入口等级

Exit Grade
出口坡度

Design Speed
设计速度

Reaction Time
反应时间

1

3%

8%

45 mi/h
45 英里/小时

2.5 s
2.5 秒

2

-4%

2%

65 mi/h
65 英里/小时

2.5 s
2.5 秒

3

0%

-3%

70 mi/h
70 英里/小时

2.5 s
2.5 秒

Find: Minimum lengths of the above vertical curves.
求解：上述竖曲线最小长度。

To find the minimum lengths of grade, the safe stopping distance for each curve must be computed. To do this, the grade used in the computation will be grade which results in the worst (or highest) safe stopping distance.
为了求得坡度的最小长度，必须计算每条曲线的安全停车距离。为此，计算中使用的坡度将是导致最差（或最大）安全停车距离的坡度。

8

ds1 = (1.47 * 45 * 2.5) + ⎢

 = 165.4 + 178.6 = 344.0 ft ds2 = (1.47 * 65 * 2.5) + ⎢
 = 238.9 + 457.3 = 696.2 ft
= 165.4 + 178.6 = 344.0 英尺 ds2 = (1.47 * 65 * 2.5) + ⎢= 238.9 + 457.3 = 696.2 英尺

It should be noted that Curve 1 is a SAG vertical curve; Curve 2 is a SAG vertical curve; Curve 3 is a CREST vertical curve.
值得注意的是，曲线 1 是下凹竖曲线；曲线 2 是下凹竖曲线；曲线 3 是凸形竖曲线。

We will start each computation assuming that the length of the curve is greater than the safe stopping distance:
我们将假设曲线的长度大于安全停车距离来开始每次计算：

Curve 1 (Use Equation 3-24)
曲线 1（使用公式 3-24）

Curve 2 (Use Equation 3-24)
曲线 2（使用公式 3-24）

Curve 3 (Use Equation 3-22)
曲线 3（使用公式 3-22）

Problem 3-9
问题 3-9

Information: Vertical curve
信息：竖曲线

-4% to +1%
-4% 至 +1%

Minimum length curve t = 2.5 s
最小长度曲线 t = 2.5 秒

70 mi/hdesign speed
设计速度 70 英里/小时

9

VPI = 5100+22

Elevation of the VPI = 1,285 ft
VPI 高程 = 1285 英尺

Find: VPC and VPT
求：VPC 和 VPT

Elevation of points on 100-ft intervals High point and station
100 英尺间隔点的标高 最高点和里程

To begin, we must determine the minimum length of curve. Note that this is a SAG vertical curve.
首先，我们必须确定曲线的最小长度。请注意，这是一个下凹垂直曲线。

Assuming that L > ds, we start using Equation 3-24:
假设 L > ds，我们开始使用公式 3-24：

For convenience of construction, we will round off: L = 985 ft or 9.85 in hundreds of feet.
为方便施工，我们将四舍五入：L = 985 英尺或 9.85 百英尺。

Then:
然后：

b = G1 = −4

Yo = YVPC = 1285 + 4(9.85 / 2) = 1,304.7 Yx = −0.254x2 − 4x + 1,304.7
Yo = YVPC = 1285 + 4(9.85 / 2) = 1,304.7 Yx = −0.254x^2 − 4x + 1,304.7

Elevations are now computed for intervals of x = 1 (100 ft) from “0” to 9.85 (which is the end of the curve). The station of the VPC is (5100+22)- (985/2)=4600+29.4. The station of the VPT is (5100+22)+(985/2) = 5,600+14.6. The spreadsheet table below shows the resulting elevations:
现在计算 x = 1（100 英尺）间隔的标高，从“0”到 9.85（这是曲线末端）。VPC 的里程桩号是 (5100+22) - (985/2) = 4600+29.4。VPT 的里程桩号是 (5100+22) + (985/2) = 5,600+14.6。下表显示了计算出的标高：

10

L

Y

0

1304.7

1

1300.4

2

1295.7

3

1290.4

4

1284.6

5

1278.4

6

1271.6

7

1264.3

8

1256.4

9

1248.1

9.85

1240.7

Because of the two grades involved, the high point of this curve is at its beginning, or 1,304.7 ft.
由于涉及两个坡度，该曲线的最高点位于其起点，即 1,304.7 英尺。

11

1

Traffic Engineering, 4th Edition
《交通工程》，第四版

Roess, R.P., Prassas, E.S., and McShane, W.R.
Roess，R.P.，Prassas，E.S.和 McShane，W.R.

Solutions to Problems in Chapter 5
第 5 章习题解答

Problem 5-1
习题 5-1

The peak rate of flow is computed as v = V/PHF. The table below summarized the results for the information given. A plot of the results is also shown.
峰值流量计算为 v = V/PHF。下表总结了给定信息的计算结果。结果的图示也显示在下方。

Peak Flow Rate vs. PHF
峰值流量率与 PHF

Volume
数量

(veh/h)
(辆/小时)

PHF

Peak
峰值

Flow
流量

Rate
速率

(veh/h)
(辆/小时)

1200

1.00

1200

1200

0.90

1333

1200

0.80

1500

1200

0.70

1714

Flow Rate vs. PHF
流量速率与高峰小时流量系数的关系

Peak Flow Rate (veh/h)
峰值流量速率 (辆/小时)

1800

1600

1400

1200

1000

800

600

400

200

0

0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00

PHF
高峰小时流量系数

Even with the same hourly volume, a small difference in PHF leads to an
即使小时通行量相同，高峰小时流量系数的微小差异也会导致峰值流量速率的巨大差异。交通工程师必须能够定期处理这种峰值特性。

enormous difference in peak flow rates. Traffic engineers must be able to deal with this peaking characteristic on a regular basis.
即使小时通行量相同，高峰小时流量系数的微小差异也会导致峰值流量速率的巨大差异。交通工程师必须能够定期处理这种峰值特性。

1

2

Problem 5-2
问题 5-2

A headway can be converted to a flow rate as follows:
车头时距可以按如下方式转换为流量：

Knowing both flow rate and speed (given), the density may now be computed as:
已知流量和速度（已给出），现在可以计算密度为：

Problem 5-3
问题 5-3

Density is obtained from occupancy as follows:
密度的计算方法如下：

Such a high value is indicative of highly congested conditions within a queue.
如此高的值表明队列内交通极其拥堵。

Problem 5-4
问题 5-4

The table on the next page illustrates the computation of monthly ADT and AWT values.
下一页的表格说明了月度日平均交通量 (ADT) 和日平均旅行时间 (AWT) 值的计算方法。

The AADT is computed as the total annual volume divided by 365 days, or:
年平均日交通量 (AADT) 的计算方法是将全年交通总量除以 365 天，即：

The AAWT is computed as the total weekday volume divided by 260 days, or:
年平均工作日交通量 (AAWT) 的计算方法是将工作日交通总量除以 260 天，即：

2

3

Table: ADT and AWT Computed
表格：ADT 和 AWT 计算结果

1

2

3

4

5

6=4/2

7=5/3

Month
月份

Days
天数

in
在

Month
月份

Weekdays
工作日

in
在

Month
月份

Total
总计

Volume
数量

(vehs)
(车辆)

Weekday
工作日

Volume
数量

(vehs)
(车辆)

ADT for Month (veh/day)
月度 ADT（辆/天）

AWT for Month (veh/day)
月度 AWT（辆/天）

Jan
1 月

31

22

200,000

170,000

6,452

7,727

Feb
2 月

28

20

210,000

171,000

7,500

8,550

Mar
3 月

31

22

215,000

185,000

6,935

8,409

Apr
四月

30

22

205,000

180,000

6,833

8,182

May
五月

31

21

195,000

172,000

6,290

8,190

Jun
六月

30

22

193,000

168,000

6,433

7,636

Jul
七月

31

23

180,000

160,000

5,806

6,957

Aug
八月

31

21

175,000

150,000

5,645

7,143

Sep
九月

30

22

189,000

175,000

6,300

7,955

Oct
十月

31

22

198,000

178,000

6,387

8,091

Nov
十一月

30

21

205,000

182,000

6,833

8,667

Dec
十二月

31

22

200,000

176,000

6,452

8,000

Total
总计

365

260

2,365,000

2,067,000

77,868

95,507

Because the average weekday volume is higher than the total average volume, it is likely that this is a commuter route. The difference is even clearer if the average weekend traffic is computed. The total weekend volume for the year is 2,365,000 – 2,067,000 = 298,000 vehs. There are 365-260 = 105 Saturdays and Sundays in the year. Then, the average weekend traffic is computed as:
由于平均工作日交通量高于平均总交通量，因此这很可能是一条通勤路线。如果计算平均周末交通量，差异将更加明显。全年周末总交通量为 2,365,000 – 2,067,000 = 298,000 辆车。一年中有 365-260 = 105 个星期六和星期日。然后，平均周末交通量计算如下：

This is clearly NOT a recreational route, but one that serves a substantial proportion of regular commuters.
这显然不是一条休闲路线，而是一条服务于大量常客通勤者的路线。

Problem 5-5
问题 5-5

Headway and Spacing can be converted to the macroscopic measures of flow rate and density, as follows:
车头时距和车距可以转换为宏观的流量和密度指标，如下所示：

3

4

Speed is then computed as:
速度计算如下：

Problem 5-6
问题 5-6

The determination of the peak hour is illustrated in the table that follows. Note that the determination is made to the nearest 15 minutes by computing all overlapping hourly volumes for each possible combination of four consecutive 15-minute periods between 4:00 PM and 6:00 PM.
下表说明了高峰小时的确定方法。请注意，通过计算下午 4:00 至下午 6:00 之间每种可能的四个连续 15 分钟时段组合的所有重叠小时交通量，将结果精确到最接近的 15 分钟。

Table: Finding the Peak Hour
表格：寻找高峰小时

Time Period
时间段

Vol (vehs)
交通量 (辆)

Hourly
每小时

Vol
交通量

(vehs)
(车辆)

4:00-4:15

450

NA

4:15-4:30

465

NA

4:30-4:45

490

NA

4:45-5:00

500

1905

5:00-5:15

503

1958

5:15-5:30

506

1999

5:30-5:45

460

1969

5:45-6:00

445

1914

a) The highest hourly volume (within the study period) occurs between 4:30 and 5:30 PM.
a) 研究期间最高小时交通量发生在下午 4 点 30 分至 5 点 30 分之间。

b) The hourly volume is the volume for this hour, or 1,999 vehs/h.
b) 每小时交通量是指该小时的交通量，即 1999 辆/小时。

4

5

c) The highest flow rate is the 15-minute interval within the peak hour
c) 最高流量是指高峰小时内 15 分钟区间内

with the highest 15-minute volume. This is the period between 5:15
交通量最高的 15 分钟时段。该时段介于下午 5:15

and 5:30 PM. The flow rate within this period is 506/0.25 = 2,024 veh/h.
至 5:30 之间，该时段内的流量为 506/0.25 = 2024 辆/小时。

d) The peak hour factor is 1999/2024 = 0.988.
d) 峰时因子为 1999/2024 = 0.988。

Problem 5-7
例题 5-7

The peak flow rate is found as:
峰值流量计算如下：

Problem 5-8
例题 5-8

The density is found as:
密度计算如下：

Problem 5-9
例题 5-9

From textbook Table 5-2, Page 109, for an urban radial facility, K factors range from 0.07 to 0.12. D factors range from 0.55 to 0.60. Then:
根据教材表 5-2，第 109 页，对于城市放射状道路，K 因子范围为 0.07 到 0.12。D 因子范围为 0.55 到 0.60。然后：

DDHV = AADT *K *D
DDHV = AADT * K * D

DDHVlow = 50,000*0.07*0.55 = 1,925 veh / h DDHVhigh = 50,000*0.12*0.60 = 3,600 veh / h
DDHV 低 = 50,000 * 0.07 * 0.55 = 1,925 车/小时 DDHV 高 = 50,000 * 0.12 * 0.60 = 3,600 车/小时

This is a very broad range, and highlights the danger in using such generalized factors for estimating demand.
这是一个非常大的范围，突出了使用这种概括性因素估计需求的危险性。

Problem 5-10
问题 5-10

The TMS is computed as the arithmetic average of individual vehicle speeds. The SMS is a speed computed using the average travel time of the individual vehicles. These computations are shown in the table that follows.
TMS 计算为单个车辆速度的算术平均值。SMS 是使用单个车辆的平均行驶时间计算出的速度。这些计算结果如下表所示。

5

6

Table: TMS and SMS Computed
表：计算的 TMS 和 SMS

Veh
车辆

No.
编号

Travel
行驶

Time
时间

(s)
(秒)

Travel
行驶

Distance
距离

(ft)
(英尺)

Travel
行驶

Speed (ft/s)
速度（英尺/秒）

Travel
行驶

Speed (mi/h)
速度 (英里/小时)

1

20.6

1000

48.54

33.02

2

21.7

1000

46.08

31.35

3

19.8

1000

50.51

34.36

4

20.3

1000

49.26

33.51

5

22.5

1000

44.44

30.23

6

18.5

1000

54.05

36.77

7

19.0

1000

52.63

35.80

8

21.4

1000

46.73

31.79

Total
总计

163.8

392.3

266.8

The TMS is now merely the average of the vehicle speeds, or 266.8/8 = 33.35 mi/h. The SMS is based upon the average travel speed, or 163.8/8 = 20.475 s/veh. Then:
TMS 现在仅仅是车辆速度的平均值，即 266.8/8 = 33.35 英里/小时。SMS 基于平均行程速度，即 163.8/8 = 20.475 秒/车。然后：

6

1

Traffic Engineering, 4th Edition
《交通工程》，第四版

Roess, R.P., Prassas, E.S., and McShane, W.R.
Roess，R.P.，Prassas，E.S.和 McShane，W.R.

Solution to Problems in Chapter 8
第 8 章问题解决方案

Problem 8-1
问题 8-1

The counts shown are for 12 out of 15 minutes. Thus, each count will be multiplied by 15/12 = 1.25 to get the estimated full 15-minute count. Note that all values will be rounded to the nearest vehicle. Missing cells will then be estimated using straight-line interpolation, or extrapolation at the end points. These results are shown in the table below.
显示的计数是 15 分钟中的 12 分钟。因此，每个计数将乘以 15/12 = 1.25，以获得估计的完整 15 分钟计数。请注意，所有值都将四舍五入到最接近的车辆。缺失的单元格将使用直线插值或端点外推来估算。这些结果显示在下表中。

Table: Expanded Count Table
表：扩展计数表

Count
计数

Time
时间

ORIGINAL COUNTS
原计数

EXPANDED COUNTS
扩展计数

East Bound
东行

Westbound
西行

Eastbount
东行

Westbound
西行

Lane 1
车道 1

Lane 2
车道 2

Lane 1
车道 1

Lane 2
车道 2

Lane 1
车道 1

Lane 2
车道 2

Lane 1
车道 1

Lane 2
车道 2

4:00-4:12

360

310

426

450

345

388

4:15-4:27

350

285

438

463

356

401

4:30-4:42

380

330

451

475

366

413

4:45-4:57

370

300

463

469

375

419

5:00-5:12

370

340

447

463

363

425

5:15-5:27

345

280

431

444

350

407

5:30-5:42

340

310

416

425

338

388

5:45-5:57

320

260

400

406

325

369

Italics indicate an interpolated or extrapolated cell.
斜体表示插值或外推单元格。

To answer part b of the question, lane volumes have to be added to get a total directional volume; then, the two directions are added to get a total freeway volume. Each direction is then analyzed to determine the peak hour, the peak hour volume, and the PHF. This is repeated for the highway as a whole. This analysis is shown in the table that follows.
为了回答问题 b，必须将车道流量相加得到总方向流量；然后，将两个方向相加得到总高速公路流量。然后，对每个方向进行分析，以确定高峰小时、高峰小时流量和 PHF。然后对整个高速公路重复此过程。下表显示了这项分析。

1

2

Table: Directional and Freeway Peak Hour
表：方向和高速公路高峰小时

Time Period
时间段

EASTBOUND

WESTBOUND

FREEWAY

Lane 1
车道 1

Lane 2
车道 2

Total
总计

Cum
累计

Lane 1
车道 1

Lane 2
车道 2

Total
总计

Cum
累计

Total
总计

Cum
累计

4:00-4:15

426

450

876

345

388

733

1609

4:15-4:30

438

463

901

356

401

757

1658

4:30-4:45

451

475

926

366

413

779

1705

4:45-5:00

463

469

932

3634

375

419

794

3062

1726

6696

5:00-5:15

447

463

910

3668

363

425

788

3118

1698

6785

5:15-5:30

431

444

875

3642

350

407

757

3118

1632

6760

5:30-5:45

416

425

841

3557

338

388

726

3065

1567

6622

5:45-6:00

400

406

806

3432

325

369

694

2965

1500

6396

The peak hours occur as shown, although for the westbound direction, two different hours produce the same peak hourly flow. In that case, either on can be used.
峰值小时出现的时间如所示，但对于西向方向，两个不同的小时产生了相同的峰值小时流量。在这种情况下，可以使用任一个小时。

As shown, for the EB direction, the peak hour occurs between 4:15 and 5:15 PM, the peak hour volume is 3,668 veh/h, and the PHF is 3668/(4*932) = 0.984. For the WB direction, the peak hour occurs between 4:30 and 5:30 PM (or between 4:15 and 5:15 PM); the peak hour volume is 3,118 veh/h, and the PHF (for 4:30-5:30) is 3118/(4*794) = 0.982. For the freeway as a whole, the peak hour occurs between 4:15 and 5:15 PM, the peak hour volume is 6,785 veh/h, and the PHF is 6785/(4*1726) = 0.983.
如图所示，对于东向（EB）方向，峰值小时出现在下午 4:15 到 5:15 之间，峰值小时交通量为 3668 辆/小时，峰时因子（PHF）为 3668 /（4 * 932）= 0.984。对于西向（WB）方向，峰值小时出现在下午 4:30 到 5:30 之间（或下午 4:15 到 5:15 之间）；峰值小时交通量为 3118 辆/小时，峰时因子（对于 4:30-5:30）为 3118 /（4 * 794）= 0.982。对于整个高速公路，峰值小时出现在下午 4:15 到 5:15 之间，峰值小时交通量为 6785 辆/小时，峰时因子为 6785 /（4 * 1726）= 0.983。

Flow rates are simply the 15-minute volumes divided by 0.25 hours (or multiplied by 4). Flow rates by direction are shown in the table that follows.
流量只是将 15 分钟的交通量除以 0.25 小时（或乘以 4）。按方向划分的流量如下表所示。

Table: Flow Rates by Direction and Period
表：按方向和时间段划分的流量

Time Period
时间段

EASTBOUND

WESTBOUND

Vol
交通量

(veh/h)
(辆/小时)

Flow Rate (veh/h)
流量 (辆/小时)

Vol
交通量

(veh/h)
(辆/小时)

Flow Rate (veh/h)
流量 (辆/小时)

4:00-4:15

876

3504

733

2932

4:15-4:30

901

3604

757

3028

4:30-4:45

926

3704

779

3116

4:45-5:00

932

3728

794

3176

5:00-5:15

910

3640

788

3152

5:15-5:30

875

3500

757

3028

5:30-5:45

841

3364

726

2904

5:45-6:00

806

3224

694

2776

2

3

Problem 8-2
问题 8-2

The sample data must be used to estimate the average number of axles per vehicle, as shown in the following table.
必须使用样本数据来估算每辆车的平均轴数，如以下表格所示。

Table: Axles vs. Vehicles
表格：轴数与车辆

Axle Per Veh
每辆车的轴数

Vehs Obs (veh)
观测车辆数 (辆)

Axles Obs
观测轴数

2

157

314

3

55

165

4

50

200

5

33

165

6

8

48

Total
总计

303

892

The average number of axles per vehicle is 892/303 = 2.944. The estimated number of vehicles in the count period is therefore 11,250/2.944 = 3,821 vehs.
每辆车的平均轴数是 892/303 = 2.944。因此，该计数期间估计的车辆数量为 11,250/2.944 = 3,821 辆。

Problem 8-3
问题 8-3

The manually-measured time is actually recorded of an effective distance because of the sight lines involved. This effective distance is:
手动测量的实际时间是由于涉及的视线而记录的有效距离。这个有效距离是：

deff = d1 *Tanθ = 60 *Tan(80.4) = 60 * 6.013 = 360.78 ft
deff = d1 *Tanθ = 60 *Tan(80.4) = 60 * 6.013 = 360.78 英尺

The observed speed is, therefore, 360.78/8.3 = 43.5 ft/s = 43.5/1.47 = 29.6 mi/h.
因此，观察到的速度为 360.78/8.3 = 43.5 英尺/秒 = 43.5/1.47 = 29.6 英里/小时。

3

1

Traffic Engineering, 4th Edition
《交通工程》，第四版

Roess, R.P., Prassas, E.S., and McShane, W.R.
Roess，R.P.，Prassas，E.S.和 McShane，W.R.

Solutions to Problems in Chapter 9
第 9 章习题解答

Problem 9-1
问题 9-1

The problem calls for estimating a total 12-hour volume for the study data shown. There is one control-count station (Station A, Figure 9.19) and 9 coverage-count stations (Stations 1-9, Figure 9.19). There are several issues that must be addressed in the estimation process:
本题要求根据所示的研究数据估算 12 小时的总交通量。有一个控制计数站（A 站，图 9.19）和 9 个覆盖计数站（1-9 站，图 9.19）。在估算过程中必须解决几个问题：

• Data was taken in three four-hour periods: 8 AM to 12 Noon, 12 Noon to 4 PM, and 4 PM to 8PM. To allow for movement of data crews, however, actual counts were taken for 3.75 hours out of each 4-hour period. All counts, therefore, must be multiplied by 4.00/3.75 = 1.067 to estimate the actual 4-hr counts.
• 数据采集分为三个 4 小时的时间段：上午 8 点到中午 12 点，中午 12 点到下午 4 点，下午 4 点到晚上 8 点。然而，为了方便数据采集人员的移动，实际计数只进行了每个 4 小时时间段中的 3.75 小时。因此，所有计数都必须乘以 4.00/3.75 = 1.067 来估计实际的 4 小时计数。

• Counts were taken using road tubes, and thus represent axle-counts, not vehicle-counts. Sample data on traffic composition (Table 9.16) must be used to estimate the average number of axles per vehicle, which can than be used to convert axle-counts to vehicle-counts.
• 计数是使用道路计数管进行的，因此代表的是轴数计数，而不是车辆计数。必须使用交通构成样本数据（表 9.16）来估算每辆车的平均车轴数，然后才能将轴数计数转换为车辆计数。

• Counts taken during one 4-hour period must be expanded to estimate counts for the 12-hour target period.
• 必须扩展一个 4 小时时间段内的计数以估计 12 小时目标时间段的计数。

• Counts were taken across three days. All counts must, therefore, be adjusted to reflect the average day of the count.
• 计数是在三天内进行的。因此，所有计数都必须进行调整以反映计数的平均日数据。

These conversions can be done in almost any order, and are best done using a spreadsheet. As all results must be rounded to the nearest vehicle, the order of computations and the rounding mechanism used may cause small discrepancies in final answers. In this solution, rounding is done only in the final step, although most of the spreadsheet tables will appear to be rounded at each step.
这些转换可以几乎按任何顺序进行，最好使用电子表格。由于所有结果都必须四舍五入到最近的车辆，计算顺序和使用的舍入机制可能会导致最终答案出现细微差异。在本解决方案中，舍入仅在最终步骤中进行，尽管大多数电子表格表格看起来在每个步骤中都会进行四舍五入。

Table 1, which follows, computes the average number of axles per vehicle from the sample data of Table 9.16. The total number of axles observed is divided by the total number of vehicles observed to determine the conversion factor.
随后的表 1 根据表 9.16 的样本数据计算每辆车的平均轴数。观察到的总轴数除以观察到的总车辆数以确定转换因子。

2

Table 1: Computing the Average Number of Axles Per Vehicle
表 1：计算每辆车的平均轴数

Vehicle Class
车辆类型

Vehicles Observed
观察到的车辆数

Axles
轴数

Observed
观察值

2-axle
2 轴

1,100

2,200

3-axle
3 轴

130

390

4-axle
4 轴

40

160

5-axle
5 轴

6

30

Total
总计

1,276

2,780

Average Axles/Vehicle = 2,780/1,276 = 2.18
平均车轴数/车辆 = 2,780/1,276 = 2.18

The data from the Control Count Station A must now be manipulated to produce conversion values for coverage counts. Two conversions must be conducted: a) from 4-hr counts to 12-hr counts, and b) from 12-hr counts on a particular day to 12-hr counts representing the average of the three days of the study.
现在必须对控制计数站 A 的数据进行处理，以生成覆盖计数的换算值。必须进行两次转换：a) 将 4 小时计数转换为 12 小时计数，以及 b) 将特定日期的 12 小时计数转换为代表研究三天平均值的 12 小时计数。

The first is accomplished by calibrating the percentage of 12-hour volume that occurs in each 4-hour period. For each day of the study, the percentage is computed as (V4/V12)*100. There will be different values for each day of the study. These can be applied separately to coverage counts on the same day, or the average percentages can be applied to all three days.
第一步是通过校准每个 4 小时期间内 12 小时交通量的百分比来完成的。对于研究的每一天，百分比计算为 (V4/V12)*100。研究的每一天的值都不同。这些值可以分别应用于同一天的覆盖计数，或者可以将平均百分比应用于所有三天。

The second conversion is accomplished by calibrating “daily variation factors” for each of the three days of the study. These factors are defined as VAVE/VDAY. The calibration of these values can be based directly on the 3.75-hr axle-counts of Table 9.15. These values could be converted to 4-hr vehicle-counts and used, but the conversions would affect every number equally, and none of the conversion values would be changed. Table 2 illustrates the computation of these conversion values in spreadsheet form.
第二次转换是通过校准研究三天中的每一天的“每日变化因子”来完成的。这些因子定义为 VAVE/VDAY。这些值的校准可以直接基于表 9.15 中的 3.75 小时轴数。这些值可以转换为 4 小时车辆数并使用，但转换会平等地影响每个数字，并且不会更改任何转换值。表 2 以电子表格形式说明了这些转换值的计算。

In terms of expanding counts from 4 hours to 12 hours, the percentages do not vary greatly for each day of the study. Therefore, percentages based upon the average data will be used.
就将计数从 4 小时扩展到 12 小时而言，研究每一天的百分比变化不大。因此，将使用基于平均数据的百分比。

Coverage counts are now expanded to full 12-hour vehicle counts in Table 3, using the following equation:
使用以下等式，覆盖计数现在在表 3 中扩展到完整的 12 小时车辆计数：

V12i = 1.067V3.75i * DFj pk

3

Where:
其中：

V12i

V3.75i 1.067 DFj

pk

= = = = =

12-hour vehicle count for Station i, vehs 3.75-hour axle count for Station i, axles expansion factor, 3.75 hrs to 4 hrs
车站 i 的 12 小时车辆计数，车辆；车站 i 的 3.75 小时轴数，轴；扩展因子，3.75 小时到 4 小时

daily adjustment factor for day j
第 j 天的每日调整因子

percentage of volume occurring during time period k, expressed as a decimal
时间段 k 内发生的体积百分比，以小数表示

Table 2: Calibration of Conversion Values from Control-Count Data
表 2：控制计数数据转换值的校准

Day
星期

Time Period
时间段

Daily Adj Factor
每日调整系数

8:00-11:45

12:00-3:45

4:00-7:45

Total
总计

Axle Counts
车轴计数

Monday
星期一

3,000

2,800

4,100

9,900

1.118

Tuesday
星期二

3,300

3,000

4,400

10,700

1.034

Wednesday
星期三

4,000

3,600

5,000

12,600

0.878

Total
总计

10,300

9,400

13,500

33,200

11,067

Percent of 12-Hour Total
占 12 小时总计的百分比

Monday
星期一

30.30%

28.28%

41.41%

100.00%

NA

Tuesday
星期二

30.84%

28.04%

41.12%

100.00%

NA

Wednesday
星期三

31.75%

28.57%

39.68%

100.00%

NA

Total
总计

31.02%

28.31%

40.66%

100.00%

NA

Table 3: Expansion and Adjustment of Coverage Counts to 12-Hour Vehicle- Counts
表 3：覆盖范围计数扩展和调整到 12 小时车辆计数

Station
站点

Day
星期

Time
时间

Axle Count
车轴数

Exp to 4 Hr
4 小时内预计

Exp to 12-Hr
12 小时内预计

Daily Adj
每日调整

12-Hr Vehs
12 小时车辆

1

Monday
星期一

8:00-11:45

1,900

1.067

0.3102

1.118

7,307

2

Monday
星期一

12:00-3:45

2,600

1.067

0.2831

1.118

10,956

3

Monday
星期一

4:00-7:45

1,500

1.067

0.4066

1.118

4,401

4

Tuesday
星期二

8:00-11:45

3,000

1.067

0.3102

1.034

10,670

5

Tuesday
星期二

12:00-3:45

3,600

1.067

0.2831

1.034

14,030

6

Tuesday
星期二

4:00-7:45

4,800

1.067

0.4066

1.034

13,024

7

Wednesday
星期三

8:00-11:45

3,500

1.067

0.3102

0.878

10,570

8

Wednesday
星期三

12:00-3:45

3,200

1.067

0.2831

0.878

10,589

9

Wednesday
星期三

4:00-7:45

4,400

1.067

0.4066

0.978

11,292

Problem 9-2
问题 9-2

Daily variation factors maybe computed as
每日变化因素可以计算为:

Where:
其中：

VAVE = VDAY =

average daily count for all days of the week, vehs average daily count for each day of the week, vehs
所有工作日的平均日计数，车辆；每周每一天的平均日计数，车辆

These computations are carried out in Table 4.
这些计算在表 4 中进行。

4

Table 3: Calibration of Daily Adjustment Factors
表 3：每日调整系数的校准

Day
星期

Ave Vol
平均交易量

Daily Factor
每日系数

Sunday
星期日

3,500

1.155

Monday
星期一

4,400

0.919

Tuesday
星期二

4,200

0.963

Wednesday
星期三

4,300

0.940

Thursday
星期四

3,900

1.037

Friday
星期五

4,900

0.825

Saturday
星期六

3,100

1.304

TOTAL

AVERAGE

28,300

4,043

Problem 9-3
问题 9-3

a) 5 minutes or 15 minutes. Count 4 of 5 or 13 of 15.
a) 5 分钟或 15 分钟。计算 5 中的 4 或 15 中的 13。

The counting period and the actual count time must be multiples of 1 minute.
计数期间和实际计数时间必须是 1 分钟的倍数。

b) 6 minutes or 18 minutes. Count 4.5 of 6 or 15 of 18. The counting period and the actual count time must be multiples of 90 seconds or 1.5 minutes.
b) 6 分钟或 18 分钟。计算 6 中的 4.5 或 18 中的 15。计数期间和实际计数时间必须是 90 秒或 1.5 分钟的倍数。

c) 6 minutes or 18 minutes. Count 4 of 6 or 16 of 18. The counting period and the actual count time must be multiples of 2 minutes.
c) 6 分钟或 18 分钟。计算 6 中的 4 或 18 中的 16。计数期间和实际计数时间必须是 2 分钟的倍数。

Problem 9-4
问题 9-4

Daily adjustment factors are based upon the data in Table 9.18. The factors, which use the same equation noted in Problem 9-2, are based upon the average of the 4 weeks of data provided.
日常调整因子基于表 9.18 中的数据。这些因子使用与问题 9-2 中相同的公式，基于提供的 4 周数据的平均值。

Monthly adjustment factors are based upon the data in Table 9.19, and are computed using the following equation:
月度调整系数基于表 9.19 中的数据，并使用以下公式计算：

Where:
其中：

MF
乘法因子i

AADT
日平均交通量

ADTi

= =

=

monthly adjustment factor month i
第 i 个月的月度调整系数

average annual daily traffic , vehs/day
平均年日交通量，车辆/日

(estimated as the average of 12 monthly ADTs) average daily traffic for month I, vehs/day
（估计为 12 个月平均 ADT）第 I 个月的平均日交通量，车辆/日

5

Daily adjustment factors are calibrated in Table 4. Monthly adjustment factors are calibrated in Table 5. Monthly variation factors must be themselves “adjusted” to reflect the middle of each month. This is done graphically in Figure 1.
日调整系数在表 4 中进行校准。月调整系数在表 5 中进行校准。月变化系数必须经过自身“调整”以反映每月的中间值，这在图 1 中以图形方式进行。

Table 4: Daily Adjustment Factors Calibrated
表 4：已校准的日调整系数

First Week
第一周

In:
在：

Day of the Week
星期几

Monday
星期一

Tuesday
星期二

Wednesday
星期三

Thursday
星期四

Friday
星期五

Saturday
星期六

Sunday
星期日

TOTAL

January
一月

2,000

2,200

2,250

2,000

1,800

1,500

950

12,700

April
四月

1,900

2,080

2,110

1,890

1,750

1,400

890

12,020

July
七月

1,700

1,850

1,900

1,710

1,580

1,150

800

10,690

October
十月

2,100

2,270

2,300

2,050

1,800

1,550

1,010

13,080

TOTAL

7,700

8,400

8,560

7,650

6,930

5,600

3,650

AVERAGE

1,925

2,100

2,140

1,913

1,733

1,400

913

12,123

DF

0.900

0.825

0.809

0.906

1.000

1.237

1.898

1,732

Table 5: Monthly Adjustment Factors Calibrated
表 5：每月调整系数校准

Third Week In
第三周

Ave 24-Hr Count
24 小时平均值

Monthly Factor
月度系数

January
一月

2,250

0.904

February
二月

2,200

0.924

March
三月

2,000

1.017

April
四月

2,100

0.968

May
五月

1,950

1.043

June
六月

1,850

1.099

July
七月

1,800

1.130

August
八月

1,700

1.196

September
九月

2,000

1.017

October
十月

2,100

0.968

November
十一月

2,150

0.946

December
十二月

2,300

0.884

TOTAL

24,400

AVERAGE

2,033

6

Figure 1: Monthly Adjustment Factors “Adjusted”
图 1：每月调整因子“调整后”

Daily Factor DF
日因子 DF

1.300

1.200

1.100

1.000

0.900

0.800

0 1 2 3 4 5 6 7 8 9 10 11 12

Third Week in Month No.
月份第三周编号

In Figure 1, Monthly Factors are plotted at the end of the 3rd week of each month, when the counts were taken. Ideally, factors should represent the “middle” of the month, which is usually at the end of the 2nd week. The graph approximates four weeks per month (except for Feb, there are actually 4 + a fraction). The end of the 2nd week can, therefore, be approximated as one week earlier than the actual count. The factors for the “middle” of the month are read from the graph, and are entered in Table 6.
在图 1 中，每月因子绘制在每个月的第三周末，当时进行了计数。理想情况下，因子应代表月份的“中间”，通常在第二周末。该图近似为每月四周（二月除外，实际上是 4 周加一小部分）。因此，第二周的末尾可以近似为实际计数前一周。从图表中读取月份“中间”的因子，并输入表 6 中。

Table 6: Monthly Factors Adjusted for the Middle of the Month
表 6：调整至月中月份因子

Month
月份

Adjusted Monthly Factor MF
调整后的月度因子 MF

January
一月

0.900

February
二月

0.920

March
三月

1.005

April
四月

0.975

May
五月

1.020

June
六月

1.090

July
七月

1.120

August
八月

1.200

September
九月

1.060

October
十月

0.975

November
十一月

0.955

December
十二月

0.900

7

Problem 9-5
问题 9-5

The four control count stations of text Table 9.20 are proposed to form a single “group” for the purpose of calibrating Daily Adjustment Factors DF. To be an appropriate grouping, the “average” factor for each day of the week cannot differ from the factors at each station by more than ± 0.10. The grouping is evaluated in Table 7.
文本表 9.20 中的四个控制计数站建议形成一个“组”，以便校准每日调整因子 DF。为了使分组合理，每周每一天的“平均”因子不能偏离每个站点的因子超过 ± 0.10。该分组在表 7 中进行评估。

Table 7: Average Daily Factors for Group and Assessment
表 7：组和评估的平均每日因子

Station
站点

Monday
星期一

Tuesday
星期二

Wednesday
星期三

Thursday
星期四

Friday
星期五

Saturday
星期六

Sunday
星期日

1

1.04

1.00

0.96

1.08

1.17

0.90

0.80

2

1.12

1.07

0.97

1.06

1.02

0.87

0.82

3

0.97

0.99

0.89

1.01

0.86

1.01

1.06

4

1.01

1.00

1.01

1.09

1.10

0.85

0.86

Total
总计

4.14

4.06

3.83

4.24

4.15

3.63

3.54

Average
平均值

1.035

1.015

0.9575

1.06

1.0375

0.9075

0.885

OK Range
正常范围

0.935-1.135

0.915-1.115

0.8575-1.0575

0.96-1.16

0.9375-1.1375

0.8075-1.0075

0.785-0.985

Obviously, three of the factors lie outside the acceptable range. It appears that Station 3 should be eliminated. Assuming that they are still spatially contiguous, Stations 1, 2, and 4 maybe grouped, and must again be tested, as shown in Table 8.
显然，三个因素都超出了可接受的范围。看来应该排除站点 3。假设它们仍然在空间上相邻，站点 1、2 和 4 可能被组合在一起，并且必须再次进行测试，如表 8 所示。

Table 8: Re-Grouped Stations Tested
表 8：重新组合的测试站点

Station
站点

Monday
星期一

Tuesday
星期二

Wednesday
星期三

Thursday
星期四

Friday
星期五

Saturday
星期六

Sunday
星期日

1

1.04

1.00

0.96

1.08

1.17

0.90

0.80

2

1.12

1.07

0.97

1.06

1.02

0.87

0.82

4

1.01

1.00

1.01

1.09

1.10

0.85

0.86

Total
总计

3.17

3.07

2.94

3.23

3.29

2.62

2.48

Average
平均值

1.057

1.023

0.980

1.077

1.097

0.873

0.827

OK Range
正常范围

0.957-1.157

0.923-1.123

0.880-1.080

0.977-1.177

0.997-1.197

0.773-0.973

0.727-0.927

The re-grouping meets the acceptability criteria, and would be used.
重新分组符合可接受性标准，将被采用。

Problem 9-6
问题 9-6

Data from coverage counts within the control area depicted in text Table 9.11 are given. We are asked to estimate the annual VMT for each section counted. To do this, the AADT at each location must be estimated. The following equations are used:
给出了文本表 9.11 中所示控制区域内覆盖范围计数的数据。我们被要求估计每个计数路段的年度车公里数（VMT）。为此，必须估计每个位置的年平均日交通量 (AADT)。使用以下公式：

AADT = Count * DF * MF
AADT = 计数 * 日系数 * 月系数

i ,j i j

Annual VMT = AADT * L
年车公里数 (VMT) = AADT * L

8

Where: AADT = average annual daily traffic, vehs/day
其中：AADT = 年平均日交通量，车辆/天

Counti.j = count taken on day i in month j, vehs/day
Counti.j = 第 i 天第 j 个月测得的计数，车辆/天

DFi = daily factor for day i
DF = 第 i 天的日系数

MFj = daily factor for month j
MF = 第 j 个月的月系数

L = length of the study segment, mi
L = 研究路段长度，英里

These computations are illustrated in Table 9.
表 9 对此计算进行了说明。

Table 9: Estimated AADT and VMT for Stations in Text Table 9.21
表 9：文本表 9.21 中各站点的估计年平均日交通量和车辆英里数

Station
站点

Length (mi)
长度（英里）

Day
星期

Month
月份

DF
自由度

(Table 9.11)
(表 9.11)

MF
乘法因子

(Table 9.11)
(表 9.11)

Daily Vol (vehs/day)
每日车流量 (辆/天)

AADT
日平均交通量

(vehs/day)
(辆/天)

Veh Miles
车辆里程

1

3.0

Wed
周三

March
三月

1.108

1.100

9,120

11,115

33,346

2

2.7

Tue
周二

September
九月

1.121

0.884

10,255

10,162

27,438

3

2.5

Fri
周五

August
八月

1.015

0.882

16,060

14,377

35,943

4

4.6

Sun
太阳

May
五月

0.789

0.949

21,858

16,366

75,286

5

1.8

Thu
周四

December
十二月

1.098

1.114

9,508

11,630

20,934

6

1.6

Fri
周五

January
一月

1.015

1.215

11,344

13,990

22,384

Problem 9-7
问题 9-7

Text table 9.22 gives the data from an origin and destination study. Only sample measurements are made, and an estimate of the actual O-D counts for the period of the study is needed. The O-D matches can be adjusted so that the row totals (origins) are correct, or so that the column totals (destinations) are correct. The accepted methodology is to average these two approaches until all row and column totals are within ±10% of the observed volumes at each origin and destination. This is an iterative process.
表格 9.22 给出了一个起点和终点研究的数据。只进行了样本测量，需要对研究期间实际的 OD 计数进行估计。可以调整 OD 匹配，使行总计（起点）正确，或使列总计（终点）正确。公认的方法是平均这两种方法，直到所有行和列总计都在每个起点和终点的观测量的±10%以内。这是一个迭代过程。

Each row and each column has an adjustment factor that will resolve the row or column totals. The actual adjustment of the O-D matches (cells of the table) are computed as follows:
每行和每列都有一个调整因子，它将解决行或列总计。实际调整 O-D 匹配（表格单元格）的计算如下：

Where:
其中：

ODi+1 ODi Foi

= = =

OD volume for the i+1 iteration, vehs OD volume for the ith iteration, vehs
第 i+1 次迭代的 OD 流量，第 i 次迭代的 OD 流量，veh

adjustment factors based upon resolving origin totals, ith iteration
基于解决原点总计的调整因子，第 i 次迭代

Fdi

=

adjustment factors based upon resolving destination totals, ith iteration
基于解决目的地总计的调整因子，第 i 次迭代

9

In each iteration, the adjustment factors are re-computed as follows:
在每次迭代中，调整因子都会重新计算如下：

Where: Vo
其中：Vo

=

total observed volume at origin “o” (vehs)
“o” 原点处的总观测流量（veh）

Vd

=

total observed volume at destination “d” (vehs)
目的地“d”的总观测体积（车辆）

ODij

=

OD matches for origin “i” and destination “j”,vehs
起点“i”和目的地“j”的 OD 匹配，车辆

These computations are shown in Table 10. Iterations are continued until all origin and destination totals are resolved to ±10%.
这些计算结果显示在表 10 中。 迭代继续，直到所有起点和目的地总数都解决到 ±10%。

Table 10: Origin and Destination Adjustments
表 10：起点和目的地调整

Destination Station
目的地车站

Origin Station
起点车站

Destination Sum
目的地总和

Destination Vol
目的地数量

Fd
终点

1

2

3

4

5

1

50

120

125

210

75

580

1,200

2.069

2

105

80

143

305

100

733

2,040

2.783

3

125

100

128

328

98

779

1,500

1.926

4

82

70

100

125

101

478

985

2.061

5

201

215

180

208

210

1,014

2,690

2.653

Origin Sum
原点总和

563

585

676

1,176

584

3,584

Origin Vol
起源卷

1,820

1,225

1,750

2,510

1,110

8,415

Fo
来源

3.233

2.094

2.589

2.134

1.901

FIRST ITERATION
第一次迭代

Destination Station
目的地车站

Origin Station
起点车站

Destination Sum
目的地总和

Destination Vol
目的地数量

Fd
终点

1

2

3

4

5

1

133

250

291

441

149

1,264

1,200

0.950

2

316

195

384

750

234

1,879

2,040

1.086

3

322

201

289

666

187

1,666

1,500

0.901

4

217

145

232

262

200

1,057

985

0.932

5

591

510

472

498

478

2,550

2,690

1.055

Origin Sum
原点总和

1,579

1,302

1,668

2,617

1,249

8,415

Origin Vol
起源卷

1,820

1,225

1,750

2,510

1,110

8,415

Fo
来源

1.152

0.941

1.049

0.959

0.889

SECOND ITERATION
第二轮迭代

Destination Station
目的地车站

Origin Station
起点车站

Destination Sum
目的地总和

Destination Vol
目的地数量

Fd
终点

1

2

3

4

5

1

139

236

291

421

137

1,224

1,200

0.980

2

353

198

410

767

231

1,959

2,040

1.041

3

331

185

282

619

168

1,584

1,500

0.947

4

226

136

230

248

182

1,023

985

0.963

5

653

509

496

501

465

2,625

2,690

1.025

Origin Sum
原点总和

1,703

1,264

1,709

2,556

1,183

8,415

Origin Vol
起源卷

1,820

1,225

1,750

2,510

1,110

8,415

Fo
来源

1.069

0.969

1.024

0.982

0.939

Note that in each iteration, any origin or destination adjustment factor that is less than 0.900 or more than 1.100 indicates that there is still a discrepancy greater than 10% in origin or destination totals. Iterations are continued until the
请注意，在每次迭代中，任何小于 0.900 或大于 1.100 的起点或终点调整因子都表明起点或终点总额仍存在超过 10% 的差异。迭代将持续到起点和终点的所有调整因子都位于 0.900 到 1.100 的范围内。

10

adjustment factors for both origins and destinations all lie within a range of 0.900 to 1.100.
对于起点和终点，调整因子都位于 0.900 到 1.100 的范围内。

1

Traffic Engineering, 4th Edition
《交通工程》，第四版

Roess, R.P., Prassas, E.P., and McShane. W.R.
罗埃斯、R.P.、普拉萨斯、E.P.和麦克肖恩、W.R.

Solutions to Problems in Chapter 10
第 10 章问题解答

Problem 10-1
问题 10-1

Parts a and b:
a 和 b 部分：

To plot a frequency distribution curve and a cumulative frequency distribution curve, a frequency distribution table must be constructed from the data given. Table 1 provides this. Then:
为了绘制频率分布曲线和累积频率分布曲线，必须根据给定的数据构建频率分布表。表 1 提供了这个表。然后：

• The Frequency Distribution Curve (FDC) is plotted as the % Vehicles in Group vs. the Middle Speed of the group.
• 频率分布曲线 (FDC) 以组内车辆百分比与组的中等速度作图。

• The Cumulative Frequency Distribution Curve (CFDC) is plotted as the Cum % Vehicles in Group vs. the Upper Speed of the speed group.
• 累积频率分布曲线 (CFDC) 以组内车辆累积百分比与速度组的上限速度作图。

• The median speed is the 50th percentile speed from the CFDC.
• 中位数速度是从 CFDC 中获得的第 50 个百分位数速度。

• The modal speed is estimated as the peak of the FDC.
• 众数速度估计为 FDC 的峰值。

• The pace is the 10-mi/h increment in speed that captures the highest percentage of observed speeds compared to any other 10-mi/h increment.
• 步长是 10 英里/小时的速度增量，与任何其他 10 英里/小时增量相比，它捕获了观察到的最高速度百分比。

• The percent vehicles in the pace are found by finding the percentile speed representing the boundaries of the pace, and subtracting their values.
• 步长内的车辆百分比是通过找到表示步长边界的百分位数速度，并减去它们的值来找到的。

Figure 1 shows the two curves, and the solution for the values called for in Part b of the question.
图 1 显示了这两条曲线，以及问题 b 部分中所需值的解决方案。

From Figure 1:
来自图 1：

Mode =
模式 =

42.0 mi/h
42.0 英里/小时

Median =
中位数 =

42.0 mi/h
42.0 英里/小时

Pace =
步调 =

37.5 – 47.5 mi/h
37.5 – 47.5 英里/小时

% Veh in Pace =
步调中车辆百分比 =

72% - 24% = 48%

2

Table 1: Frequency Distribution Table for Speed Data
表 1：速度数据频率分布表

Speed Group (mi/h)
速度组 (英里/小时)

Middle
中等

Speed, S (mi/h)
速度，S（英里/小时）

Observed
观察值

Freq
频数

n

Percent
百分比

Freq (%)
频数（%）

Cum %
累积%

Freq (%)
频数（%）

n*S

n*S2

Low Speed
低速

High Speed
高速

15

20

17.5

0

0.00%

0.00%

0.00

0.00

20

25

22.5

4

2.37%

2.37%

90.00

2,025.00

25

30

27.5

9

5.33%

7.69%

247.50

6,806.25

30

35

32.5

18

10.65%

18.34%

585.00

19,012.50

35

40

37.5

35

20.71%

39.05%

1,312.50

49,218.75

40

45

42.5

42

24.85%

63.91%

1,785.00

75,862.50

45

50

47.5

32

18.93%

82.84%

1,520.00

72,200.00

50

55

52.5

20

11.83%

94.67%

1,050.00

55,125.00

55

60

57.5

9

5.33%

100.00%

517.50

29,756.25

60

65

62.5

0

0.00%

100.00%

0.00

0.00

169

100.00%

7,107.50

310,006.25

30.00%

25.00%

20.00%

15.00%

10.00%

5.00%

0.00% 1

0

20 30

40

Speed, mi/h
速度，英里/小时

7

2%

24%

42

10 20 30 40

Upper Speed of Group, mi/h
Groupmi/h 最高速度

Figure 1: Frequency and Cumulative Frequency Distribution Curves
图 1：频率和累积频率分布曲线

3

Part c:
c 部分：

The mean speed of the distribution is computed using Equation 10-3; the standard deviation is computed using Equation 10-5. Both use column totals from the Frequency Distribution Table.
使用公式 10-3 计算分布的平均速度；使用公式 10-5 计算标准差。两者都使用频率分布表中的列总计。

= 8.11 mi / h
= 8.11 英里/小时

Part d:
d 部分：

The standard error of the mean, E, is computed as:
平均值的标准误差 E 计算如下：

Then:
然后：

95% Confidence : μ = x ± 1.96 E = 42.06 ± (1.96 * 0.624) = 42.06 ± 1.22 (40.84 to 43.28 mi / h) 99.7% Confidence : μ = x ± 3 E = 42.06 ± (3 * 0.624) = 42.06 ± 1.87 (40.19 to 43.93 mi / h)
95%置信区间：μ = x ± 1.96 E = 42.06 ± (1.96 * 0.624) = 42.06 ± 1.22 (40.84 到 43.28 英里/小时) 99.7%置信区间：μ = x ± 3 E = 42.06 ± (3 * 0.624) = 42.06 ± 1.87 (40.19 到 43.93 英里/小时)

Part e:
e 部分：

Sample size is estimated using Equation 10-11 for 95% confidence:
使用公式 10-11 估算 95%置信区间的样本大小：

Part f:
f 部分：

This question is answered by conducting a Chi-Squared Goodness of Fit test. In this test, actual observed frequency values are compared to theoretical frequencies that would have been observed if the distribution were “Normal.” Table 2 shows the test computations.
此问题通过进行卡方拟合优度检验来解答。在这个检验中，实际观察到的频率值与如果分布为“正态”则会观察到的理论频率进行比较。表 2 显示了检验计算。

4

Table 2: Chi-Square Goodness of Fit Test
表 2：卡方拟合优度检验

Speed Group (mi/h)
速度组 (英里/小时)

Observed
观察值

Freq
频数

n

Upper
上限

Limit
限制

"zd"
zd

Prob
概率

z<zd

Tab 7.3
表 7.3

Prob of
概率

of Being in
属于

Group
组

Theoretical
理论

Freq
频数

f

Combined Group
组合组

f n

Group χ2
组χ2

High Speed
高速

Low Speed
低速

。

60

0

。

1.0000

0.0136

2.2984

9.5992

9

0.0374

60

55

9

2.21208385

0.9864

0.0432

7.3008

55

50

20

1.59556104

0.9432

0.1354

22.8826

22.8826

20 0.3631

50

45

32

0.86806412

0.8078

0.1672

28.2568

28.2568

32

0.4959

45

40

42

0.36251541

0.6406

0.2393

40.4417

40.4417

42

0.0600

40

35

35

-0.2540074

0.4013

0.2091

35.3379

35.3379

35

0.0032

35

30

18

-0.8705302

0.1922

0.1241

20.9729

20.9729

18

0.4214

30

25

9

-1.487053

0.0681

0.0493

8.3317

11.5089

13

0.1932

25

20

4

-2.1035758

0.0188

0.0155

2.6195

20

15

0

-2.7200986

0.0033

0.0033

0.5577

TOTAL

169

1.0000

169

169

169

1.5743

After all speed groups are combined to insure that all values of f ≥ 5, the result is a Chi-Square value of 1.5743, with 7 – 3 = 4 degrees of freedom. From text Table 7.11, the probability of a value this high or higher is between 0.75 (for X2 = 1.923) and 0.90 (for X2 = 1.064). Interpolating:
在将所有速度组组合在一起以确保所有 f 值≥5 后，得到卡方值为 1.5743，自由度为 7 – 3 = 4。根据表 7.11，出现此值或更高值的概率在 0.75（对于 X² = 1.923）和 0.90（对于 X² = 1.064）之间。通过插值：

Xd2 Prob (X2 ≥ Xd2)
卡方概率 (X² ≥ Xd²)

1.064 0.90

1.574 ?
1.574？

1.923 0.75

In order to reject the hypothesis that the data and the Normal Distribution are the same, the Prob (X2 ≥ Xd2) would have to be less than 5%. Therefore, the hypothesis is confirmed. The data maybe considered to be normally distributed
为了拒绝数据与正态分布相同的假设，概率 (X² ≥ Xd²) 必须小于 5%。因此，假设成立。数据可以认为服从正态分布。.

Problem 10-2
问题 10-2

a. To determine whether or not the observed reduction in speeds was significant, a Normal Approximation Test must be conducted:
a. 为了确定观察到的速度降低是否显著，必须进行正态近似检验：

From text Table 7.3, the Prob (zd ≤ 4.02) = 0.9999. Thus, the difference is highly significant.
从文本表 7.3 中，概率 (zd ≤ 4.02) = 0.9999。因此，差异非常显著。

5

b. The standard error of the mean for the after sample is:
b. 后样本的均值标准误差为：

Therefore, it is 95% probable that the true mean speed of the distribution lies between 40.8 + (1.96*0.515) and 40.8 – (1.96*0.515) or 39.8 to 41.8 mi/h. As the target speed of 40 mi/h is in this range, it may be considered to have been achieved.
因此，真均速分布介于 40.8 + (1.96*0.515) 和 40.8 – (1.96*0.515) 或 39.8 至 41.8 英里/小时之间，其概率为 95%。由于目标速度 40 英里/小时在此范围内，因此可以认为已达到目标。

Problem 10.3
问题 10.3

Note: This solution assumes that there is one lane being observed.
注意：此解决方案假设只有一个车道正在被观察。

Table 3: Summary of Queue Counts
表 3：队列计数汇总

Clock Time
时钟时间

Cycle Number
周期编号

No. of Vehicles in Queue at:
队列中车辆数量：

+0 s
+0 秒

+15 s
+15 秒

+30 s
+30 秒

+45 s
+45 秒

9:00

1

3

4

2

4

9:01

2

1

2

3

3

9:02

3

4

3

3

4

9:03

4

2

3

3

4

9.04

5

0

1

2

3

9:05

6

2

1

1

2

9:06

7

4

3

3

3

9.07

8

5

5

6

4

9:08

9

2

3

4

3

9:09

10

0

3

2

2

9:10

11

1

2

3

1

9:11

12

1

0

1

0

9:12

13

2

2

1

2

9:13

14

2

3

2

2

9:14

15

4

3

3

3

Sum
总和

33

38

39

40

Total for All Time Periods:
所有时间段的总计：

150

Then:
然后：

6

d = TQ + (FVS * CF )
d = TQ + (FVS * CF)

Where: CF = +2 (Text Table 10.6, FFS = 35 mi/h, VSLC = 20.3)
其中：CF = +2（文本表 10.6，FFS = 35 英里/小时，VSLC = 20.3）

Then: d = 4.66 + (0.701* 2) = 6.062 s / veh
然后：d = 4.66 + (0.701 * 2) = 6.062 s / 车辆

Problem 10-4
问题 10-4

For 95% confidence:
95% 置信度：

e

Table 4 executes this equation for tolerance levels (e) of 2 min, 5 min, and 10 min with base standard deviations (s) of 5 min, 10 min, and 15 min.
表 4 使用该等式计算公差水平 (e) 为 2 分钟、5 分钟和 10 分钟，基本标准差 (s) 为 5 分钟、10 分钟和 15 分钟的情况。

Table 4: Required Number of Samples
表 4：所需样本数量

Tolerance
公差

(min)
(分钟)

Standard Deviations (min)
标准差 (分钟)

5

10

15

2

24

96

216

5

4

15

35

10

1

4

9

Problem 10-5
问题 10-5

Table 5 uses the problem data to determine the average travel time and average running time in each section shown. These values are used to compute the average travel speed and the average running speed.
表 5 使用问题数据确定所示各段的平均行程时间和平均运行时间。这些值用于计算平均行驶速度和平均运行速度。

Note that data for the 1st segment must be added as follows: Cumulative Section Length = 0.50 mi; Cumulative Travel Time = 1.0 min; Delay = 0 s; No. of Stops = 0.
请注意，必须按如下方式添加第一段的数据：累积路段长度 = 0.50 英里；累积行程时间 = 1.0 分钟；延误 = 0 秒；停车次数 = 0。

Then, using the data, Segment Lengths must be established, as well as travel times in each segment (converted to seconds). As an example, select the segment between checkpoints 3 and 4:
然后，使用这些数据，必须确定路段长度以及各路段的行程时间（转换为秒）。例如，选择检查点 3 和 4 之间的路段：

7

Section Length (mi) = 3.50 – 2.25 = 1.25 mi
路段长度（英里）= 3.50 – 2.25 = 1.25 英里

Travel Time = 7 min, 30 s – 4 min, 50 s = 450 s – 290 s = 160 s. Running Time = 160 – 25 = 135 s.
行程时间 = 7 分 30 秒 – 4 分 50 秒 = 450 秒 – 290 秒 = 160 秒。运行时间 = 160 – 25 = 135 秒。

Average Travel Speed = (1.25/160)*3600 = 28.1 mi/h. Average Running Speed = (1.25/135)*3600 = 33.3 mi/h
平均行驶速度 = (1.25/160)*3600 = 28.1 英里/小时。平均运行速度 = (1.25/135)*3600 = 33.3 英里/小时

Table 5: Average Travel Speed and Average Running Speed
表 5：平均行驶速度和平均运行速度

Section
路段

Length (mi)
长度（英里）

Travel
行驶

Time
时间

(s)
(秒)

Running
跑步

Time
时间

(s)
(秒)

Travel Speed (mi/h)
行进速度（英里/小时）

Running
跑步

Speed
速度

(mi/h)
(英里/小时)

1

0.50

60

60

30.0

30.0

2

0.50

65

55

27.7

32.7

3

1.25

165

135

27.3

33.3

4

1.25

160

135

28.1

33.3

5

0.50

100

58

18.0

31.0

6

0.25

77

30

11.7

30.0

7

0.75

87

73

31.0

37.0

40

35

30

25

20

15

10

5

0

Travel Speed
行进速度

o Running Speed
跑步速度

1 2 3 4 5 6 7

Section Number
章节编号

Figure 2: Average Travel Speed and Average Running Speed
图 2：平均旅行速度和平均运行速度

8

The answer to Part b depends upon which measure is to have a tolerance of 3 mi/h – average travel speed or average running speed? We also need an estimate of the standard deviation of travel speed and/or running speed. The standard deviations are computed in Table 6 as:
问题 b 的答案取决于哪个指标的公差为 3 英里/小时——平均旅行速度还是平均运行速度？我们还需要估计旅行速度和/或运行速度的标准差。标准差在表 6 中计算如下：

Table 6: Computation of Standard Deviations
表 6：标准差的计算

Segment
路段

Travel
行驶

Speed (mi/h)
速度 (英里/小时)

Running
跑步

Speed
速度

(mi/h)
(英里/小时)

(TS-Mean)2
(TS-均值)²

(RS-Mean)2
(RS-均值)²

1

30

30

26.72

6.19

2

27.7

32.7

8.19

0.06

3

27.3

33.3

5.96

0.71

4

28.1

33.3

10.85

0.71

5

18.0

31.0

46.65

2.11

6

11.7

30.0

172.71

6.19

7

31.0

37.0

38.49

20.24

Total
总计

173.8

227.4

Mean
均值

24.8

32.5

STD

6.42

3.37

Then:
然后：

e

nATS =

= 17.6, say 18 runs
= 17.6，大约 18 次运行

If each car makes 5 runs, 3 cars would be needed to estimate ATS; only 1 would be needed to estimate ARS.
如果每辆车运行 5 次，则需要 3 辆车来估计 ATS；只需要 1 辆车来估计 ARS。

1

Traffic Engineering, 4th Edition
《交通工程》，第四版

Roess, R.P., Prassas, E.S., and McShane, W.R.
Roess，R.P.，Prassas，E.S.和 McShane，W.R.

Solutions to Problems in Chapter 11
第 11 章习题解答

Problem 11-1
问题 11-1

Deaths and Accidents Per 100,000 Population:
每 10 万人人口中的死亡和事故数量：

Deaths and Accidents Per 10,000 Registered Vehicles
每 1 万辆注册车辆的死亡事故

Acc /10,000 Reg Veh = 360 * (⎜ 10,000 ⎞⎟ = 102.9 per10KReg Veh
每 1 万辆注册车辆的 Acc /1 万 = 360 * (⎜ 10,000 ⎞⎟ = 每 1 万辆注册车辆 102.9 例

⎝ 35,000 ,

Deaths and Accidents Per 100,000,000 Vehicle-Miles Travelled
每 1 亿辆车行驶里程的死亡和事故

National statistics for 2008 are not yet available in final form. 2007 national rates include fatality rates of 1.4 per 100 MVM and 13.64 per 100K Population. The rates in this locality are very high compared to national statistics, and warrant a thorough investigation.
2008 年的国家统计数据目前尚未以最终形式发布。2007 年国家数据包括每 1 亿车行驶里程死亡率为 1.4 例，每 10 万人口死亡率为 13.64 例。本地区的这些比率与国家统计数据相比非常高，需要进行彻底调查。

Problem 11-2
问题 11-2

Because of the small sample size, the Binomial Test should be used. The one- year reduction in accidents at this location is (10/25)*100 = 40%. Then:
由于样本量小，应使用二项检验。该地点一年事故减少了 (10/25)*100 = 40%。然后：

2

The decrease is not statistically significant.
这种下降在统计上不显着。

Problem 11-3
问题 11-3

• Sideswipe and turning accidents may be due to the off-set intersection geometry, which has vehicle paths intersection at unexpected positions within the intersection. Remedy: Install a 3-phase signal with a fully protected LT phase for the NS street.
• 侧撞和转向事故可能是由于错位交叉口几何形状造成的，车辆路径在交叉口内以意想不到的位置相交。补救措施：为 NS 街道安装三相信号，并为其提供完全保护的 LT 相位。

• Some of the rear-end collisions may be related to signal visibility problems. This should be checked, as there are only two pole-mounted signal heads. Remedy: Use overhead signal heads on span wire.
• 一些追尾碰撞可能与信号可见性问题有关。这应该检查，因为只有两个杆安装的信号头。补救措施：使用跨线上的顶置信号头。

3

• Some sideswipe accidents may be related to vehicles approaching in the wrong lane for their movement. Remedy: Better lane-use control markings and signing.
• 一些侧撞事故可能与车辆驶入错误车道有关。补救措施：改进车道使用控制标记和标识。

• Pedestrian accidents may be due to the awkward location of the crosswalks or related reasons. Remedy: Place Pedestrian Signals at proper locations; check signal timing for pedestrians.
• 行人事故可能是由于斑马线的尴尬位置或相关原因造成的。补救措施：将行人信号灯放置在适当的位置；检查行人信号的计时。

Problem 11-4
问题 11-4.

• The “yellow” interval provides a transition allowing a vehicle who cannot stop before the “red” to legally enter the intersection.
• “黄色”间隔提供了一个过渡，允许无法在“红色”之前停车的车辆合法地进入交叉路口。

• The “all-red” interval allows a vehicle who entered the intersection legally on “yellow” to safely clear the intersection before releasing conflicting vehicles.
• “全红”间隔允许在“黄色”上合法进入交叉口的车辆在释放冲突车辆之前安全地离开交叉口。

• Signals remove the major crossing conflicts from the realm of driver judgment to avoid accidents
• 信号将主要交叉冲突从驾驶员判断领域移出，以避免事故发生。.

• Protected LT phases remove LT conflicts with opposing flows from the realm of driver judgment to avoid accidents.
• 受保护的 LT 相位将 LT 冲突与相反方向的流量从驾驶员判断领域移出，以避免事故。

1

TRAFFIC ENGINEERING, 4th Edition
交通工程，第 4 版

Roess, R.P., Prassas, E.S., and McShane, W.R.
Roess，R.P.，Prassas，E.S.和 McShane，W.R.

Solution to Problems in Chapter 13
第 13 章习题解答

Problem 13-1
习题 13-1

Capacity is the maximum flow rate that can reasonably be accommodated by a roadway segment under prevailing traffic, roadway, and control conditions.
容量是在当前交通、道路和控制条件下，一段道路能够合理承受的最大流量。

Service flow rate is the maximum flow rate that can reasonably be accommodated by a roadway segment under prevailing traffic, roadway, and control conditions while maintaining the defined operating characteristics of the designated level of service
服务流量是在当前交通、道路和控制条件下，一段道路能够合理承受的最大流量，同时保持指定服务水平的既定运行特性。.

Problem 13-2
习题 13-2

a. Capacity under ideal conditions is a capacity value (see above) where the traffic, roadway, and control conditions are assumed to be “ideal,” or in conformance with the “base” conditions for the type of segment under consideration.
a. 理想条件下的容量是一个容量值（见上文），其中假设交通、道路和控制条件为“理想”条件，或符合所考虑路段类型的“基础”条件。

b. Capacity is defined in Problem 13-1: It accounts for prevailing conditions that are different from those defined as “base conditions” or “ideal conditions.”
b. 容量在习题 13-1 中定义：它考虑了与定义为“基础条件”或“理想条件”不同的现有条件。

c. Service flow rate is defined above. There is a service flow rate for every LOS from A through E (none for LOS F, which is unstable). The value for LOS C is the maximum flow rate that can reasonably be achieved while maintaining the defined density for LOS C.
c. 服务流量如上定义。从 A 到 E 的每个 LOS 都有服务流量（LOS F 不稳定，因此没有）。LOS C 的值是在保持 LOS C 的定义密度的情况下可以合理达到的最大流量。

Problem 13-3
问题 13-3

The service flow rates for a 6-lane freeway (3 lanes in one direction) are given as follows:
6 车道高速公路（一个方向 3 车道）的服务流量如下：

LOS A LOS B LOS C LOS D LOS E

3,250 veh/h 3,900 veh/h 4,680 veh/h 5,810 veh/h 7,000 veh/h
3,250 车/小时 3,900 车/小时 4,680 车/小时 5,810 车/小时 7,000 车/小时

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2

a. The capacity of the segment is 7,000 veh/h.
a. 该路段的容量为 7,000 车/小时。

b.

SV = SF x PHF. Thus:
SV = SF x PHF。因此：

LOS A 3250*0.95

LOS B 3900*0.95

LOS C 4680*0.95
LOS C 4680 乘以 0.95

LOS D 5810*0.95
LOS D 5810 乘以 0.95

LOS E 7000*0.95
LOS E 7000 乘以 0.95

= = = = =

3,088 veh/h 3,705 veh/h 4,446 veh/h 5,520 veh/h 6,650 veh/h
3,088 辆/小时 3,705 辆/小时 4,446 辆/小时 5,520 辆/小时 6,650 辆/小时

c.

SFIDEAL = SF/(fHV*fp)
SF 理想值 = SF/(fHV 乘以 fp)

LOS A 3250/0.935*1 =
LOS A 3250 除以 0.935 乘以 1 =

LOS B 3900/0.935*1 =
LOS B 3900 除以 0.935 乘以 1 =

LOS C 4680/0.935*1 =
LOS C 4680 除以 0.935 乘以 1 =

LOS D 5810/0.935*1 =
LOS D 5810 除以 0.935 乘以 1 =

LOS E 7000/0.935*1 =

3,476 pc/h 4,171 pc/h 5,005 pc/h 6,214 pc/h 7,487 pc/h
3,476 件/小时 4,171 件/小时 5,005 件/小时 6,214 件/小时 7,487 件/小时

Problem 13-4
问题 13-4

Freeways are divided highways with full control of access. Multilane highways have two or three lanes for traffic in each direction, may be divided or undivided, but do not have full control of access. Signalized and unsignalized intersections and driveways at grade generally exist on multilane highways. Two-lane highways have one lane for each direction of traffic, and are undivided. Passing is accomplished in the opposing lane of traffic.
高速公路是具有完全控制出入的分割公路。多车道公路在每个方向都有两到三条车道，可能分隔或未分隔，但没有完全控制出入。信号灯和无信号灯的交叉路口以及路面上的车道通常存在于多车道公路上。双车道公路每方向只有一条车道，并且未分隔。超车发生在对向车道上。

2

1

TRAFFIC ENGINEERING, 4th Edition
交通工程，第 4 版

Roess, R.P., Prassas, E.S., and McShane, W.R.
Roess，R.P.，Prassas，E.S.和 McShane，W.R.

Solution to Problems in Chapter 14
第 14 章问题解答

Problem 14-1
问题 14-1

The free-flow speed of a multilane highway is estimated using Equation 14-6: FFS = BFFS − fLW − fLC − fM − fA
使用公式 14-6 估计多车道公路的自由行驶速度：FFS = BFFS − fLW − fLC − fM − fA

Where: BFFS = 60 mi/h (given)
其中：BFFS = 60 英里/小时（已知）

fLW = 1.9 mi/h (Table 14.5, 11-ft lanes)
fLW = 1.9 英里/小时（表 14.5，11 英尺车道）

fLC = 0.65 mi/h (Table 14.7, 3 + 6 = 9 ft total lateral clearance)
fLC = 0.65 英里/小时（表 14.7，3 + 6 = 9 英尺总侧向间距）

fM = 1.6 mi/h (Table 14.8, undivided)
fM = 1.6 英里/小时（表 14.8，非分隔）

fA = 3.75 mi/h (Table 14.9, 15 access pts/mi) FFS = 60.00 −1.90 − 0.65 − 3.75 = 53.7 mi / h
fA = 3.75 英里/小时（表 14.9，每英里 15 个入口点）FFS = 60.00 −1.90 − 0.65 − 3.75 = 53.7 英里 / 小时

Problem 14-2
问题 14-2

The free-flow speed of a freeway is estimated using Equation 14-5: FFS = 75.4 − fLW − fLC − 3.22TRD 0.84
自由流速公路速度使用公式 14-5 估计：FFS = 75.4 − fLW − fLC − 3.22TRD^0.84

Where: fLW = 0.0 mi/h (Table 14.5, 12-ft lanes)
其中：fLW = 0.0 英里/小时（表 14.5，12 英尺车道）

fLC = 1.6 mi/h (Table 14.6, 2-ft clearance, 6-lane freeway)
fLC = 1.6 英里/小时（表 14.6，2 英尺间距，6 车道高速公路）

TRD = 3.5 ramps/mi (given)
TRD = 3.5 个坡道/英里（已知）

FFS = 75.4 − 0.0 −1.6 − 3.22 (3.50.84 ) = 75.4 − 0.0 −1.6 − 9.2 = 64.6 mi / h
FFS = 75.4 − 0.0 − 1.6 − 3.22 (3.5^0.84) = 75.4 − 0.0 − 1.6 − 9.2 = 64.6 英里 / 小时

Problem 14-3
问题 14-3

(a) As the total length of the composite grade (2,000+1,000+900 = 3,900 ft) is less than 4,000 ft, the average grade methodology maybe used
(a) 由于复合坡道的总长度 (2,000 + 1,000 + 900 = 3,900 英尺) 小于 4,000 英尺，因此可以使用平均坡度法。.

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© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained
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(b)

(b)
沿着这条坡道再行驶 5,000 英尺，达到 8,800 英尺，卡车重新加速到大约 38 英里/小时的速度，此时它们现在进入了 5% 坡度的最后 3,000 英尺。从沿着 1,800% 的坡度开始，它们又行驶了 3,000 英尺到 4,800 英尺。它们的最终速度约为 28 英里/小时，复合坡度为 5%，长 10,000 英尺。

在 2, 000 英尺的 4% 坡度后，卡车将以大 约 36 英里/小时的速 度行驶。 这是卡车进 入 5, 000 英尺的速度 为 3 %。 就好像卡车 在 3 % 的坡度上行驶 了大约 3, 800 英尺。
在 2,000 英尺的 4% 坡度后，卡车将大约以 36 英里/小时的速度行驶。 这是卡车在 5,000 英尺时的速度为 3%。 就好像卡车在 3% 的坡度上行驶了大约 3,800 英尺。

沿着这个坡度再行驶 5, 000 英尺，达到 8, 800 英尺，卡车重新 加速到大约 38 英里/ 小时的速度，此时它 们现在进入了 5 % 坡 度的最后 3, 000 英尺 。 从沿着 1, 800%的 坡度开始，它们又行 驶了 3, 000 英尺到 4, 800 英尺。 它们的最 终速度约为 28 英里/ 小时，复合坡度为 5 %，长 10, 000 英尺。
沿着这条坡道再行驶 5000 英尺，达到海拔 8800 英尺，卡车重新加速到大约 38 英里/小时，此时它们进入最后的 3000 英尺 5% 的坡道。从 1800%坡道开始，它们又行驶了 3000 英尺达到 4800 英尺。最终速度约为 28 英里/小时，复合坡度为 5%，长度 10000 英尺。

(c)

Rise on 3% Grade: 2,000*0.03
3%坡度上升：2,000*0.03

=

60 ft
60 英尺

Rise on 2% Grade: 1,000*0.02
2%坡度上升：1,000*0.02

=

20 ft
20 英尺

Rise on 4% Grade: 900*0.04
4%坡度上升：900*0.04

Total
总计

=

36 ft
36 英尺

116 ft
116 英尺

Composite Grade = (116/3,900)*100
综合坡度 = (116/3,900)*100

=

2.97%

As this composite grade is longer than 4,000 ft (10,000 ft), and part of the curve has a grade of greater than 4%, this grade must be handled using the graphic composite grade methodology illustrated below.
由于该复合坡度超过 4,000 英尺 (10,000 英尺)，并且曲线部分的坡度大于 4%，因此必须使用下文所示的图形复合坡度法处理该坡度。

After 2,000 ft of 4% grade, trucks will be traveling at approximately 36 mi/h. This is the speed at which trucks enter the 5,000 ft of 3%. It is as if the trucks had been on the 3% grade for approximately 3,800 ft. Traveling another 5,000 ft along this grade, to 8,800 ft, trucks have re-accelerated to an approximate speed of 38 mi/h, at which they now enter the final 3,000 ft of 5% grade. Starting as if they were approximately 1,800 ft along the 5% grade, they travel another 3,000 ft to 4,800 ft. They’re final speed is approximately 28 mi/h, and the composite grade is 5%, 10,000 ft long.
在 4% 坡度 2,000 英尺后，卡车将以大约 36 英里/小时的速度行驶。这是卡车进入 3% 坡度 5,000 英尺的速度。就好像卡车已经行驶了大约 3,800 英尺的 3% 坡度一样。沿着该坡度再行驶 5,000 英尺，到 8,800 英尺，卡车已重新加速到大约 38 英里/小时的速度，此时它们进入最后 3,000 英尺的 5% 坡度。就好像它们已行驶了大约 1,800 英尺的 5% 坡度一样，它们再行驶 3,000 英尺到 4,800 英尺。它们最终速度大约是 28 英里/小时，复合坡度为 5%，长度为 10,000 英尺。

As the initial portion of the grade is the steepest, the composite grade is taken to the end of the first segment: 5%, 4,000 ft.
由于坡度的初始部分是最陡峭的，因此将复合坡度延伸到第一段的末端：5%，4,000 英尺。

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3

Problem 14-4
问题 14-4

From Table 14.11, for rolling terrain, ET = 2.5 and ER = 2.0. Then:
从表 14.11 可以看出，对于起伏地形，ET = 2.5，ER = 2.0。然后：

PC Equivalents for Trucks: 3,200*0.12*2.5 =
卡车的小客车当量：3,200*0.12*2.5 =

960 pc/h
960 辆小客车/小时

PC Equivalents for RVs: 3,200*0.03*2.0 =
房车的小客车当量：3,200*0.03*2.0 =

192 pc/h
192 辆小客车/小时

PC Equivalents for Cars: 3,200*0.85*1.0 =
小轿车的小客车当量：3,200*0.85*1.0 =

2,720 pc/h
2,720 辆小客车/小时

Total Equivalent Volume:
总当量体积：

3,872 pc/h
3,872 辆车/小时

Problem 14-5
例题 14-5

It is necessary to determine the free-flow speed of the subject freeway using Eqn 14-5: FFS = 75.4 − fLW − fLC − 3.22 TRD 0.84
需要根据公式 14-5 确定研究高速公路的自由流速：FFS = 75.4 − fLW − fLC − 3.22 TRD 0.84

Where: fLW = 1.9 mi/h (Table 14-5, 11-ft lanes)
其中：fLW = 1.9 英里/小时（表 14-5，11 英尺车道）

fLC = 0.8 mi/h (Table 14-6, 2-ft clearance, 4 lanes)
fLC = 0.8 英里/小时（表 14-6，2 英尺间隙，4 车道）

TRD = 4.2 ramps/mi (given)
TRD = 4.2 个坡道/英里（已知）

FFS 75.4 1.9 0.8 3.22 (4.20.84 ) 62.0 mi / h
FFS 75.4 1.9 0.8 3.22 (4.20.84 ) 62.0 英里/小时

From 14.10, the 60-mi/h speed-flow relationship is used for this freeway.
从 14.10 开始，该高速公路的 60 英里/小时速度流量关系被使用。

Service flow rate are computed using Eqn 14-2; service volumes are computed using Eqn 14-3:
服务流量使用公式 14-2 计算；服务量使用公式 14-3 计算：

SF = MSF *N *f HV * fp SV = SF * PHF

Maximum service flow rates (MSF) are selected from Table 14.3 for a FFS of 60 mi/h:
对于 60 英里/小时的 FFS，最大服务流量 (MSF) 从表 14.3 中选择：

LOS A – 660 pc/h/ln; LOS B – 1,080 pc/h/ln; LOS C – 1,560 pc/h/ln; LOS D – 2,010 pc/h/ln; LOS E - 2,300 pc/h/ln.
LOS A – 660 辆车/小时/车道；LOS B – 1,080 辆车/小时/车道；LOS C – 1,560 辆车/小时/车道；LOS D – 2,010 辆车/小时/车道；LOS E - 2,300 辆车/小时/车道。

The heavy vehicle factor is based upon passenger car equivalents for trucks on a 4% grade of 1.5 miles. The pce values are different for the upgrade and the downgrade.
重型车辆系数基于 4% 坡度 1.5 英里的卡车乘用车当量。升级和降级 pce 值不同。

ET (upgrade) = 3.75 (Table 14.12, 4% grade, 1.5 mi, 3% trucks interpolated) ET (dngrade) = 1.50 (Table 14.14, 4% grade, < 4mi, 3% trucks extrapolated)
ET（上坡）= 3.75（表 14.12，4% 坡度，1.5 英里，3% 卡车插值）ET（下坡）= 1.50（表 14.14，4% 坡度，< 4 英里，3% 卡车外推）

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4

Then, using Eqn 14.9:
然后，使用公式 14.9：

The PHF is given as 0.92, there are 4 lanes in each direction on the freeway, and the driver population adjustment factor (fp) is 1.00 for a normal driver population. Equations 14-2 and 14-3 are implemented in the spreadsheet table shown below.
PHF 给定为 0.92，高速公路每个方向有 4 条车道，对于正常驾驶员群体，驾驶员人口调整系数 (fp) 为 1.00。公式 14-2 和 14-3 在下表中实现。

Problem 14-6
问题 14-6

To determine the probable LOS for this existing 6-lane multilane highway with FFS = 45 mi/h, the equivalent ideal lane flow must be determined using Eqn 14-1:
为了确定这条现有 6 车道多车道公路的可能 LOS（服务水平），必须使用公式 14-1 确定等效理想车道流量：

Where: V
其中：V

=

4,000 veh/h (given)
4,000 辆/小时（已知）

PHF

=

0.88 (given)
0.88（已知）

N

=

3 lanes (given)
3 车道（已知）

fp

=

1.00 (normal driver population assumed)
1.00（假设为正常驾驶员群体）

4

5

Then: ET = 2.5 (Table 14.11,Rolling Terrain)
则 ET = 2.5（表 14.11，滚动地形）

And: f HV =
以及 fHV =

= 0.847

Then
那么:

= 1,789 pc/ h / ln
= 1,789 台/小时/车道

Comparing this to the MSF values of Table 14.4 for a FFS of 45 mi/h, it is seen that the LOS is E.
与表 14.4 中 45 英里/小时 FFS 的 MSF 值相比，可以看出 LOS 为 E。

Problem 14-7
问题 14-7

This is a design application for a section of freeway that goes from level terrain to a sustained 5%, 2-mile grade. LOS C is the design target. The number of lanes
这是针对一段从平坦地形到持续 5%、2 英里的坡度的免费公路路段的设计应用。设计目标是 LOS C。车道数量

needed to provide this on the (a) upgrade, (b) downgrade, and (c) level terrain is needed. Equation 14-4 is used:
需要在 (a) 升级、(b) 降级和 (c) 水平地形情况下提供此信息。使用公式 14-4：

The FFS of the facility is needed to begin:
开始需要该设施的 FFS：

FFS = 75.4 − fLW − fLC − 3.22TRD0.84

FFS = 75.4 − 0 − 0 − 3.22 (0.50.84 ) = 75.4 −1.8 = 73.6 mi / h SAY 75 mi/ h
FFS = 75.4 − 0 − 0 − 3.22 (0.50.84 ) = 75.4 −1.8 = 73.6 mi / h 约 75 mi/ h

DDHV = 2,500 veh/h (given)
DDHV = 2,500 车/小时（已知）

MSFC = 1,750 pc/h/ln (Table 14.3, FFS = 75 mi/h)
MSFC = 1,750 辆/小时/车道（表 14.3，FFS = 75 英里/小时）

PHF = 0.92 (given)
PHF = 0.92（已知）

fp = 1.00 (normal driver population assumed)
驾驶员因素 fp = 1.00（假设为普通驾驶员群体）

There maybe as many as three different heavy vehicle adjustment factors for the three segments to be analyzed. They are based upon the appropriate passenger car equivalents for trucks and RVs. Level terrain values are selected from Table 14.11; upgrade (5%, 2 mi) values are selected from Table 14.12 for trucks and 14.13 for RVs; downgrade values are selected from Table 14.14 for trucks and Table 14.11 for RVs (level terrain assumed for downgrade). The resulting values are shown in the table that follows:
可能有多达三个不同的重型车辆调整系数需要用于分析的三个路段。这些系数基于卡车和房车的相应乘用车当量。水平路段的值选自表 14.11；上坡路段（5%，2 英里）的值选自表 14.12（卡车）和 14.13（房车）；下坡路段的值选自表 14.14（卡车）和表 14.11（房车）（下坡路段假设为水平路段）。结果值如下表所示：

5

6

Equivalent
当量

Level
水平

Upgrade
升级

Downgrade
降级

ET

1.5

2.5

1.5

ER

1.2

4.5

1.2

Then:
然后：

Then:
然后：

It appears that the provision of a 4-lane freeway will be sufficient to deliver LOS C on all of the defined segments.
看来，提供一条四车道高速公路足以在所有已定义的路段上实现 LOS C 服务水平。

Problem 14-8
问题 14-8

This question concerns an old freeway with projected traffic growth in the future. It asks for an evaluation of LOS at various future time-points. The easiest way to approach this problem is to create a table of service volumes for the freeway which can be matched against future demand levels.
此问题涉及一条旧高速公路，预测未来交通量将会增长。问题要求评估未来不同时间点的服务水平 (LOS)。解决此问题的最简单方法是创建一个高速公路服务量表，以便与未来的需求水平相匹配。

It is first necessary to estimate the FFS of the freeway using Eqn 14-5: FFS = 75.4 − fLW − fLC − 3.22TRD 0.84
首先，需要使用公式 14-5 估算高速公路的自由流速度 (FFS)：FFS = 75.4 − fLW − fLC − 3.22TRD^0.84

Where:
其中：

fLW

fLC

TRD

= = =

1.9 mi/h (Table 14.5, 11-ft lanes)
1.9 英里/小时（表 14.5，11 英尺车道）

3.6 mi/h (Table 14.6, 0-ft clearance, 2 lanes)
3.6 英里/小时（表 14.6，0 英尺净空，2 车道）

4.5 ramps/mi (given)
4.5 个匝道/英里（已知）

FFS = 75.4 −1.9 − 3.6 − 3.22 (4.5084 ) = 58.5 mi / h, SAY 60 mi / h
FFS = 75.4 − 1.9 − 3.6 − 3.22 (4.5)^0.84 = 58.5 英里/小时，约 60 英里/小时

6

7

Then, using Equations 14-2 and 14-3:
然后，使用公式 14-2 和 14-3：

SF = MSF *N *f HV * fp SV = SF * PHF

Where: N = 2 lanes (given)
车道数：N = 2 车道（已知）

PHF = 0.90 (given)
小时车流因子 PHF = 0.90（已知）

fp = 1.00 (normal driver population assumed)
驾驶员因素 fp = 1.00（假设为普通驾驶员群体）

ET = 2.5 (Table 14.11,rolling terrain)
等效小时交通量 ET = 2.5（表 14.11，起伏地形）

fHV = 1/[1+0.07(2.5-1)] = 0.905

and values of MSF are selected for each LOS from Table 14.3 for a FFS or 60 mi/h: LOS A – 660; LOS B – 1080; LOS C – 1560; LOS D – 2010; LOS E = 2300.
并从表 14.3 中为每个 LOS 选择 MSF 值，其自由流速度为 60 英里/小时：LOS A – 660；LOS B – 1080；LOS C – 1560；LOS D – 2010；LOS E = 2300。

Equations 14-3 and 14-4 are implemented in the spreadsheet table shown below:
方程 14-3 和 14-4 已在下面的电子表格中实现：

These values must be compared to the projected demand volumes over the next
这些值必须与未来

20 years to determine the likely LOS that will exist:
20 年的预计需求量进行比较，以确定可能存在的 LOS：

Current Volume = 2,100 = 2,100 veh/h (LOS C)
当前车流量 = 2,100 = 2,100 车/小时（LOS C）

5-Year Forecast = 2,100*1.035 = 2,434 veh/h (LOS C)
5 年预测 = 2,100*1.035 = 2,434 辆/小时 (服务水平 C)

10-Year Forecast = 2,100*1.0310 = 2,822 veh/h (LOS D)
10 年预测 = 2,100*1.0310 = 2,822 辆/小时 (服务水平 D)

15-Year Forecast = 2,100*1.0315 = 3,271 veh/h (LOS D)
15 年预测 = 2,100*1.0315 = 3,271 辆/小时 (服务水平 D)

20-Year Forecast = 2,100*1.0320 = 3,792 veh/h (LOS F)
20 年预测 = 2,100*1.0320 = 3,792 辆/小时 (服务水平 F)

The demand volume will exceed capacity somewhere in the period between 15 and 20 years, near 20 years. Given the lead time for most major re-construction projects, planning should begin no later than year 10.
需求量将在 15 到 20 年之间，接近 20 年时超过容量。鉴于大多数主要重建项目的工期，计划应最迟在第 10 年开始。

7

8

Problem 14-9
问题 14-9

The headway data shown is based upon 160 passenger cars and 40 trucks in the traffic stream. This represents a truck population of (40/200)*100 = 20%. If all headways are considered separately, Equation 14-14 is used to compute the equivalent:
显示的间距数据基于交通流中的 160 辆客车和 40 辆卡车。这代表卡车数量占 (40/200)*100 = 20%。如果所有间距都分别考虑，则使用公式 14-14 计算等效值：

If only the trailing vehicle type matters, average headways for each are as follows:
如果仅考虑后部车辆类型，则每种车辆的平均间距如下：

The equivalent is then found using Equation 14-15:
然后使用公式 14-15 找到等效值：

The two are different precisely because headways clearly depend upon both the lead and trailing vehicle types.
两者之所以不同，正是因为间距明显取决于前部和后部车辆类型。

8

1

TRAFFIC ENGINEERING, 4th Edition
交通工程，第 4 版

Roess, R.P., Prassas, E.S., McShane, W.R.
罗斯，R.P.，普拉斯，E.S.，麦克谢恩，W.R.

Solutions to Problems in Chapter 15
第 15 章问题解答

Problem 15-1
问题 15-1

2,000 ft
2,000 英尺

Freeway
高速公路

3,400 veh/h
3,400 辆/小时

7% trucks
7% 卡车

Ramp 1
坡道 1

800 veh/h
800 辆/小时

3% trucks
3% 卡车

RFFS = 35 mi//h
RFFS = 35 英里/小时

General
一般

Rolling Terrain PHF = 0.95
滚动地形 PHF = 0.95

FFS = 65 mi/h
FFS = 65 英里/小时

ID = 1.6 interchanges/mi
ID = 1.6 个互通立交/英里

Ramp 2
匝道 2

700 veh/h
700 车/小时

5% trucks
5% 卡车

RFFS = 50 mi/h
RFFS = 50 英里/小时

The first step in the analysis is the conversion of all demand volumes to demand flow rates under equivalent base conditions, using Equation 15-1:
分析的第一步是使用公式 15-1 将所有需求量转换为等效基础条件下的需求流量：

From Chapter 14, the passenger car equivalent for trucks in rolling terrain is 2.5. This allows the heavy vehicle factor to be computed for each of the critical volumes shown:
从第 14 章中，滚动地形中卡车的乘用车当量为 2.5。这使得可以计算出所示每个关键车流量的重型车辆系数：

1

2

It is assumed that the driver population is normal, i.e., fp = 1.00 Then:
假设驾驶员群体服从正态分布，即 fp = 1.00。然后：

Analysis of the Merge Segment (Ramp 1)
合并路段（匝道 1）分析

Equation 15-23 is used to estimate the demand flow remaining in lanes 1 and 2 of the freeway immediately upstream of the merge:
公式 15-23 用于估算在合并点上游高速公路 1 号和 2 号车道中剩余的需求流量：

v12 = vF PFM

From Table 15.13, the value of PFM is estimated using Equation 15-25 or 15-27, which includes the effect of the downstream off-ramp. To determine which is appropriate, the equivalence distance is computed using Equation 15-31:
从表 15.13 中，使用公式 15-25 或 15-27 估算 PFM 的值，其中包括下游匝道的的影响。为了确定哪个公式合适，使用公式 15-31 计算等效距离：

Therefore, Equation 15-27 is used. Then:
因此，使用公式 15-27。然后：

P = 0.5487 + 0.2628 (⎜v L ⎞⎟ = 0.5487 + 0.2628 (792 )= 0.653

v12 = 3,955 * 0.653 = 2,587 pc/ h
v12 = 3,955 * 0.653 = 2,587 pcu/h

This result should be reviewed for reasonableness. The flow rate in lane 3 will be 3,955 – 2,587 = 1,368 pc/h. As this is less than the maximum of 2,700 pc/h and 1.5*(2,430/2) = 1,823 pc/h, the result is reasonable.
应该检查此结果的合理性。3 号车道的流量将为 3,955 – 2,587 = 1,368 pcu/h。由于这小于最大值 2,700 pcu/h 和 1.5*(2,430/2) = 1,823 pcu/h，因此结果是合理的。

Using the criteria of Table 15-5, capacity and limiting values should be checked:
使用表 15-5 的标准，应检查容量和限制值：

vFO = vF + vR1 = 3,955 + 880 = 4,835 pc/ h < 7,050 pc/ h OK v12R = 2,430 + 880 = 3,310 pc/ h < 4,600 pc/ h OK
vFO = vF + vR1 = 3,955 + 880 = 4,835 辆车/小时 < 7,050 辆车/小时 OK v12R = 2,430 + 880 = 3,310 辆车/小时 < 4,600 辆车/小时 OK

vR1 = 880 pc/ h < 1,900 pc/ h OK
vR1 = 880 辆车/小时 < 1,900 辆车/小时 OK

2

3

The density in the merge influence area is then estimated using Equation 15-40:
然后，使用公式 15-40 在合并影响区域估算密度：

DR = 5.475 + 0.00734 vR + 0.0078 v12 − 0.00627 La

DR = 5.475 + (0.00734 * 880) + (0.0078 * 2587) − (0.00627 * 500) DR = 5.475 + 6.459 + 20.179 − 3.135 = 29.0 pc/ mi / ln
DR = 5.475 + (0.00734 * 880) + (0.0078 * 2587) − (0.00627 * 500) DR = 5.475 + 6.459 + 20.179 − 3.135 = 29.0 辆车/英里/ln

From Table 15.1, this is LOS D, although very close to the LOS C/Dboundary. Analysis of Diverge Segment (Ramp 2)
根据表 15.1，这是 LOS D，尽管非常接近 LOS C/DboundaryAnalysis of Diverge Segment (Ramp 2)

Equation 15-32 is used to estimate the demand flow in lanes 1 and 2 immediately upstream of the off-ramp:
使用公式 15-32 来估算下坡匝道直接上游车道 1 和 2 的需求流量：

v12 = vR + (vF − vR ) PFD

From Table 15.4, the value of PFD is estimated using either Equation 15-33 or 15- 34, which accounts for the impact of the upstream on-ramp. The choice of which to use is made by computed the equivalence distance using Equation 15-36:
根据表 15.4，使用公式 15-33 或 15-34 来估算 PFD 的值，该值考虑了上游入口匝道的冲击。选择使用哪个公式是通过使用公式 15-36 计算等效距离来确定的：

Note that vF for the of-ramp includes the on-ramp demand flow: 3,955+880 = 4,835 pc/h. Then:
请注意，出口匝道的 vF 包括入口匝道的需求流量：3,955+880 = 4,835 辆车/小时。然后：

880

EQ 0.071 + (0.000023* 4835) − (0.000076 * 792)
0.071 + (0.000023 * 4835) − (0.000076 * 792)

Equation 13-34 is used:
使用公式 13-34：

PFD = 0.717 − (0.000039 * 4835) + (0.604 *880 / 2000) PFD = 0.717 − 0.189 + 0.265 = 0.793

3

4

Then:
然后：

v12 = 792 + 0.793(4835 − 792) = 792 + 3,059 = 3,206 pc/ h
v12 = 792 + 0.793(4835 − 792) = 792 + 3,059 = 3,206 车辆/小时

This result must be checked for reasonableness. The demand flow in lane 3 is 4835 – 3206 = 1,629 pc/h compared to an average flow rate of 3206/2 = 1,603 pc/h. This does not violate either rule for outer lane demand flow (< 2,700 pc/h, <1.5*1603). The analysis moves forward with these values.
必须检查此结果是否合理。车道 3 的需求流量为 4835 – 3206 = 1,629 车辆/小时，与平均流量 3206/2 = 1,603 车辆/小时相比。这并未违反外侧车道需求流量的任何规则（< 2,700 车辆/小时，< 1.5 * 1603）。分析将继续使用这些值。

The capacity of the freeway between the ramps has been checked previously, and is ok. The demand flow in lanes 1 and 2 (3,206 pc/h) is less than the maximum of 4,400 pc/h (Table 15.5) and the off-ramp flow rate of 792 pc/h is less than the capacity of the off-ramp demand flow of 2,100 pc/h (Table 15.5, 50 mi/hRFFS).
之前已检查过匝道之间的高速公路容量，并且符合要求。车道 1 和 2 的需求流量 (3,206 车辆/小时) 小于最大值 4,400 车辆/小时（表 15.5），以及下坡匝道的流量 (792 车辆/小时) 小于下坡匝道需求流量的容量 (2,100 车辆/小时)（表 15.5，50 英里/小时 RFFS）。

The density in the diverge segment is estimated using Equation 15-41:
使用公式 15-41 来估算分流段的密度：

DR = 4.252 + 0.0086 v12 − 0.009 Ld

DR = 4.252 + (0.0086 * 3206) − (0.009 * 300) DR 4.252 + 27.257 − 2.700 = 28.8 pc/ mi / ln
DR = 4.252 + (0.0086 * 3206) − (0.009 * 300) DR 4.252 + 27.257 − 2.700 = 28.8 车辆/英里/车道

From Table 15.1, this is LOS D, but is close to the LOS C/Dboundary. Analysis of a Weaving Configuration
根据表 15.1，这是 D 服务水平，但接近于 C/D 服务水平边界编织配置分析

If a continuous auxiliary lane is built connecting the two ramps, the segment shown below emerges. Demand flows (already in pc/h) assume that there is no ramp-to-ramp traffic.
如果建造一条连接两条匝道的连续辅助车道，则会出现下述路段。需求流量（已经以车辆/小时为单位）假设没有匝道之间的交通。

2,000 ft
2,000 英尺

3955-792 = 3,163

880

792

4

5

By inspection of this diagram, this is a ramp-weave with the following characteristics:
通过检查此图，这是一个斜坡编织，其特征如下：

vw = 880 + 792 = 1,672 pc/ h vnw = 3,163 pc/ h
vw = 880 + 792 = 1,672 件/小时 vnw = 3,163 件/小时

v = 4,835 pc/ h
v = 4,835 件/小时

VR = 1672 / 4835 = 0.346 LCRF = 1

LCFR = 1 NWV = 2

Given the marking of the weaving segment shown, the short length, LS, is equal to the base length, LB, or 2,000 ft. If barrier markings are to be used, the default value of 0.77*LB or 0.77*2000 = 1,540 ft could be used.
鉴于所示编织段的标记，短长度 LS 等于基长 LB 或 2,000 英尺。如果要使用障碍物标记，可以使用默认值 0.77*LB 或 0.77*2000 = 1,540 英尺。

Then, using Equation 15-2:
然后，使用方程式 15-2：

LCmin = (LCFR vFR )+ (LCRFvRF ) = (792 *1)+ (880 *1) = 1,672 lane changes/ h The maximum weaving length is determined using Equation 15-4:
LCmin = (LCFR vFR ) + (LCRFvRF ) = (792 *1) + (880 *1) = 1,672 个车道变化/小时 最大编织长度使用方程式 15-4 确定：

L [5,728(1 VR)1.6 ] [1,566 N ]

LMAX = (5,728 *1.3461.6 ) − (1,566 * 2) = 6,083 ft > 2,000 ft Analysis as a weaving segment is appropriate.
LMAX = (5,728 *1.3461.6 ) − (1,566 * 2) = 6,083 英尺 > 2,000 英尺 将该段作为织入段进行分析是合适的。

The capacity of the weaving segment must now be considered to determine whether stable flow exists or not. Capacity can be limited by either a maximum density of 43 pc/mi/ln, or by the capacity of the configuration for weaving vehicles:
现在需要考虑织入段的容量，以确定稳定流是否存在。容量可能受限于每英里每车道 43 辆车的最大密度，或者受织入车辆配置的容量限制：

Capacity Based Upon Density
基于密度的容量

Equations 15-5 and 15-6 are used:
使用公式 15-5 和 15-6：

cIWL = cIFL − [438.2 (1 + VR)1.6 ]+ [0.0765LS ]+ [119.8 NWV ] cIWL = 2,350 − (438.2 *1.3461.6 ) + (0.0765 * 2000) + (119.8 * 2) cIWL = 2,350 − 704.93 + 153.00 + 239.60 = 2,038 pc/ h / ln

5

6

Then:
然后：

cW1 = cIWL N f HV fp

Where N = 4 and fp is assumed to be 1.00 (for a normal population of drivers). Finding fHV is a bit harder. The demand flow rate within the weaving segment is fed by two entering flows (vF1 and vR1) which have different levels of truck presence. To find fHV, we need the truck percentage for the combined flow. From the problem statement, the demand volume is:
其中 N = 4，fp 假定为 1.00（对于正常的驾驶员群体）。求解 fHV 比较困难。编织段内的需求流量由两个进入流量（vF1 和 vR1）提供，这两个流量的卡车比例不同。为了求解 fHV，我们需要组合流量的卡车百分比。根据题意，需求量为：

3,400 veh/h with 7% trucks: 3,400*0.07 = 238 trucks
3,400 辆/小时，其中 7%为卡车：3,400*0.07 = 238 辆卡车

800 veh/h with 3% trucks: 800*0.03 = 24 trucks
800 辆/小时，其中 3%为卡车：800*0.03 = 24 辆卡车

4,200 veh/h 262 trucks
4,200 辆/小时，262 辆卡车

% trucks = (262/4200)*100 = 6.24%
卡车百分比 = (262/4200)*100 = 6.24%

With ET = 2.5 (rolling terrain), the heavy vehicle adjustment factor is:
当 ET = 2.5（起伏地形）时，重型车辆调整系数为：

and:
以及：

cw1 = 2,038 * 4 * 0.914 *1 = 7,451 veh / h
cw1 = 2,038 * 4 * 0.914 *1 = 7,451 车辆 / 小时

Capacity Based Upon Maximum Weaving Flow
基于最大穿插流量的容量

For a case in which NWV = 2, Equations 15-7 and 15-8 are used to determine capacity:
对于 NWV = 2 的情况，使用方程 15-7 和 15-8 来确定容量：

cW 2 = cIW f HV fp = 6936 * 0.914 *1 = 6,340 veh / h
cW 2 = cIW f HV fp = 6936 * 0.914 *1 = 6,340 车辆 / 小时

The lower of the two capacity values, in this case cW2, controls, and c = 6,340 veh/h. As the demand flow rate is well below this level, the weaving segment is not expected to fail, the analysis continues.
这两种容量值中的较小值，在本例中为 cW2，起控制作用，并且 c = 6,340 车辆/小时。由于需求流量远低于此水平，预计穿插段不会发生故障，分析继续进行。

6

7

It is now necessary to estimate the total lane-changing activity in the weaving segment. Equation 15-11 is used to find the rate of weaving-vehicle lane changes:
现在有必要估算穿插段中的总车道变换活动。使用方程 15-11 来查找穿插车辆车道变换率：

LC LC 0.39 [(L 300)0.5 N2 (1 ID)0.8 ]

LC = 1,6IN+ 0.39 [(2,00 − 300)0.5 42 )(1 +1.6)0.8 ]= 2,027 lane changes/ h
LC = 1,6IN+ 0.39 [(2,00 − 300)0.5 42 )(1 +1.6)0.8 ]= 2,027 车道变换/小时

The lane-changing rate for non-weaving vehicles is a bit more complicated. Equation 15-13 is used to compute the non-weaving lane-change index, INW
非穿插车辆的车道变换率稍复杂一些。使用方程 15-13 计算非穿插车道变换指数 INW:

Equation 15-12 (the one for cases of INW < 1,300) is now used:
现在使用方程 15-12（适用于 INW < 1,300 的情况）：

LCNW1 = LCNW = (0.206 vNW ) + (0.542 LS ) − (192.6 N) LCNW = (0.206 * 3,163) + (0.542 * 2000) − (192.6 * 4) LCNW = 651.6 +1,084.0 − 770.4 = 965 lane changes/ h
LCNW1 = LCNW = (0.206 vNW) + (0.542 LS) − (192.6 N) LCNW = (0.206 * 3163) + (0.542 * 2000) − (192.6 * 4) LCNW = 651.6 + 1084.0 − 770.4 = 965 车道变更/小时

Then: LCALL = 2027 + 965 = 2,992 lane changes/ h
然后 LCALL = 2027 + 965 = 2992 车道变更/小时

The speed of weaving vehicles is now estimated using Equations 15-19 and 15- 20:
编织车辆的速度现在使用公式 15-19 和 15-20 估算：

W = 0.226

0.789 = 0.226 *
0.789 = 0.3105

The speed of non-weaving vehicles is estimated using Equation 15-21:
非编织车辆的速度使用公式 15-21 估算：

SNW = FFS − (0.0072 LCMIN ) − (0.0048 v N) SNW = 65 − (0.0072 *1672) − (0.004848354)

SNW = 65.0 −12.0 − 5.8 = 47.2 mi / h

7

8

The average speed for all vehicles is given by Equation 15-22:
所有车辆的平均速度由公式 15-22 给出：

The average density in the weaving segment is given by Equation 15-23:
编织路段的平均密度由公式 15-23 给出：

From Table 15.1, this is LOS C. Comparing the Results
从表 15.1 可以看出，这是 C 级服务水平。结果比较

The weaving segment design result in LOS C operation, while the ramp design (without a continuous auxiliary lane) result in LOS D operation (although quite close to the LOS C/D boundary). This comparison is slightly in favor of the weaving design, but not significantly so.
编织路段设计导致 C 级服务水平运行，而匝道设计（无连续辅助车道）导致 D 级服务水平运行（尽管非常接近 C/D 级服务水平边界）。这种比较略微有利于编织设计，但并不显著。

We can also compare capacities. Note that capacity is stated as a flow rate for prevailing conditions. For the weaving segment, capacity was computed as 6,340 veh/h. For the ramp design, the capacity between the ramps (the controlling segment) is 7,050 pc/h. When converted to prevailing conditions, this becomes 7050*0.914*1 = 6,434 veh/h. The difference is slightly in favor of the ramp configuration, but not significantly so.
我们还可以比较通行能力。请注意，通行能力表示为在现有条件下的流量。对于编织路段，通行能力计算为 6340 辆/小时。对于匝道设计，匝道之间的通行能力（控制路段）为 7050 辆/小时。换算成现有条件后，则为 7050*0.914*1 = 6434 辆/小时。差异略微有利于匝道配置，但并不显著。

So which should we use? These results do not show great differences, so the decision may rest on non-traffic factors. One factor that may sway a bit towards the side of the ramp design is this: the capacity of the weaving segment was controlled by the amount of weaving vehicles, NOT the overall operation of the segment. This is an indication that the ramp-weave configuration may not be optimal for this situation. Given the demands, it does not seem logical to add a lane to either the on-ramp or off-ramp to create a more favorable configuration. Perhaps, barring unknown non-traffic factors, the ramp design should be favored.
那么我们应该使用哪一个呢？这些结果并没有显示出太大的差异，所以决定可能取决于非交通因素。可能稍微偏向于坡道设计的一个因素是：编织路段的容量是由编织车辆的数量控制的，而不是路段的整体运行情况。这表明坡道编织配置对于这种情况可能并非最佳选择。鉴于需求，在入口匝道或出口匝道增加车道以创建更有利的配置似乎并不合逻辑。也许，除非存在未知的非交通因素，否则应该优先考虑坡道设计。

8

9

Problem 15-2
问题 15-2

FFS = 70 mi/h, ID = 0.9 int/mi
自由流速 (FFS) = 70 英里/小时，车头时距 (ID) = 0.9 车辆/英里

2,500 ft
2500 英尺

2000 pc/h
2000 辆/小时

800 pc/h
800 辆/小时

1,000 pc/h 1,600 pc/h
1000 辆/小时 1600 辆/小时

LCRF = 1
左转车道流量比 (LCRF) = 1

LCFR = 0
右转车道流量比 (LCFR) = 0

The drawing above illustrates the determination of the values of LCRF =1 and LFR = 0. This is a major weave configuration with lane balance at the exit gore area. Note that all demands are shown as flow rates under equivalent base conditions, and require no conversions.
上图说明了 LCRF=1 和 LFR=0 值的确定方法。这是一个主要的编织配置，在出口汇合区车道平衡。请注意，所有需求都显示为等效基准条件下的流量，无需任何转换。

Then:
然后：

vw = 800 + 1,600 = 2,400 pc/ h
vw = 800 + 1600 = 2400 pc/h

vNW = 2,000 + 1,000 = 3,000 pc/ h
vNW = 2000 + 1000 = 3000 pc/h

v = 2,400 + 3,000 = 5,400 pc/ h VR = 2400 / 5400 = 0.444
v = 2400 + 3000 = 5400 pc/h VR = 2400 / 5400 = 0.444

NWV = 3 (by inspection)
NWV = 3（目测）

LCmin = (LCRF * vRF ) + (LCFR *vFR ) = (1* 800) + (0 *1,600) = 800 lane changes/ h
LCmin = (LCRF * vRF) + (LCFR * vFR) = (1 * 800) + (0 * 1600) = 800 车道变更/h

The maximum weaving length is estimated using Equation 15-4: LMAX = [5728 (1 + VR)1.6 ]− [1566 NWV ]
最大编织长度使用公式 15-4 估算：LMAX = [5728 (1 + VR)^1.6] − [1566 NWV]

LMAX = [5728 (1.444)1.6 ]− [1566 * 3] = 5,613 ft > 2,500 ft
LMAX = [5728 (1.444)^1.6] − [1566 * 3] = 5613 英尺 > 2500 英尺

The area will function as a weaving segment, and the analysis continues.
该区域将用作编织段，分析继续进行。

The capacity of the segment, assuming that the overall operation of the facility controls, is estimated using Equation 15-5:
假设设施的整体运行控制，该段的容量使用公式 15-5 估算：

9

10

cIWL = cIFL − [438.2 (1 + VR)1.6 ]+ [0.0765 LS ]+ [119.8 NWV ] cIWL = 2400 − (438.2 *1.4441.6 ] + [0.0765 * 2500] + [119.8 * 3] cIWL = 2400 − 788.8 +191.3 + 359.4 = 2,162 pc/ h / ln
cIWL = cIFL − [438.2 (1 + VR)^1.6] + [0.0765 LS] + [119.8 NWV] cIWL = 2400 − (438.2 * 1.444^1.6) + [0.0765 * 2500] + [119.8 * 3] cIWL = 2400 − 788.8 + 191.3 + 359.4 = 2162 pc/h/ln

cIW = 2162 * 4 = 8,648 pc/ h
cIW = 2162 * 4 = 8648 pc/h

Capacity can also be related to the maximum weaving demand that can be accommodated. This capacity is computed using Equation 15-7, for a case in which NWV = 3:
织机能力也可以与可容纳的最大织造需求相关。当 NWV = 3 时，使用公式 15-7 计算此能力：

The latter value, 7,883 pc/h, is the controlling value of capacity. As the total demand is far less than this value, the analysis continues.
后一个值，7883 pc/h，是能力的控制值。由于总需求远小于此值，因此分析继续进行。

Total lane-changing rates must now be estimated. The weaving lane-change rate is estimated using Equation 15-11:
现在必须估计总车道变更率。使用公式 15-11 估算编织车道变更率：

LCW = 800 + 0.39 * 46.9 *16 *1.67 = 800 + 488.7 = 1,289 lane changes/ h The non-weaving lane-change index is given by Equation 15-13:
LCW = 800 + 0.39 * 46.9 * 16 * 1.67 = 800 + 488.7 = 1289 次车道变更/小时。非编织车道变更指数由公式 15-13 给出：

Therefore, using Equation 15-12:
因此，使用公式 15-12：

LCNW1 = LCNW = (0.206 vNW ) + (0.546 LS ) + (192.6 * N)
LCNW1 = LCNW = (0.206 vNW) + (0.546 LS) + (192.6 * N)

LCNW = (0.206 * 3,000) + (0.546 * 2,500) + (192.6 * 4) = 618 +1,365 + 770 = 2,753 lane changes/ h And: LCALL = 1,289 + 2,753 = 4,042 lane changes/ h
LCNW = (0.206 * 3000) + (0.546 * 2500) + (192.6 * 4) = 618 + 1365 + 770 = 2753 次车道变更/小时。并且：LCALL = 1289 + 2753 = 4042 次车道变更/小时

10

11

Then, using Equations 15-18 and 15-19:
然后，使用公式 15-18 和 15-19：

The non-weaving speed is estimated using Equation 15-21:
使用公式 15-21 估算非编织速度：

SNW = FFS − (0.0072 LCMIN ) − (0.0048 v N) SNW = 70.0 − (0.0072 * 800) − (0.0048 *54004)
SNW = FFS − (0.0072 LCMIN) − (0.0048 v N) SNW = 70.0 − (0.0072 * 800) − (0.0048 * 54004)

SNW = 70.0 − 5.7 − 6.5 = 57.8 mi / h Then:
SNW = 70.0 − 5.7 − 6.5 = 57.8 英里/小时 然后：

From Table 15.1, this is LOS C.
来自表 15.1，这是 LOS C。

The capacity of the weaving segment was computed to be 7,883 pc/h. To convert this to prevailing conditions, it must be multiplied by fHV and fp. The driver population factor can be taken to be 1.00 (for a normal driver population). The heavy vehicle factor must be estimated for 10% trucks in rolling terrain (ET = 2.5). Then:
梭织段的产能计算为 7,883 件/小时。为了换算成现行条件，必须乘以 fHV 和 fp。驾驶员数量因素可以取 1.00（对于正常的驾驶员数量）。必须针对起伏地形中的 10% 卡车估算重型车辆因素 (ET = 2.5)。然后：

c = 7883 * 0.870 *1 = 6,858 veh / h
c = 7883 × 0.870 × 1 = 6858 车/小时

As is the case for all capacity values, this one is stated in terms of the maximum flow rate during a peak 15-minute interval. To obtain a maximum hourly volume for the segment, the result must be multiplied by the PHF:
与所有容量值一样，此值也以 15 分钟高峰时段的最大流量表示。要获得路段的最大小时交通量，必须将结果乘以小时峰值因子 (PHF)：

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12

cHOUR = 6858 * 0.92 = 6.309 veh / h
cHOUR = 6858 × 0.92 = 6309 车/小时

Problem 15-3
问题 15-3

Freeway
高速公路

4,200 veh/h
4200 车/小时

8 % trucks
8% 货车

Level terrain FFS = 65 mi//h PHF = 0.92
平坦路段自由流速度 (FFS) = 65 英里/小时，小时峰值因子 (PHF) = 0.92

 1,000 ft
1000 英尺

Ramp
匝道

700 veh/h
700 车/小时

5% trucks
5% 卡车

Level terrain
平坦路段

RFFS = 40 mi/h PHF = 0.92
RFFS = 40 英里/小时 PHF = 0.92

Equation 15-1 is used to convert the demand volumes to flow rates under equivalent ideal conditions. A normal driving population is assumed (fp = 1.00). The passenger car equivalent for trucks in level terrain is 1.5 (see Chapter 14). Then:
公式 15-1 用于将需求量转换为等效理想条件下的流量。假设为正常的驾驶人群 (fp = 1.00)。平坦路段的卡车乘用车当量为 1.5（参见第 14 章）。然后：

From Table 15.6, Equation 15-25 is used to estimate the flow rate in lanes 1 and 2 immediately upstream of this merge:
根据表 15.6，使用公式 15-25 估算该合并点上游第 1 和第 2 车道的流量：

PFM = 0.5775 + 0.000028 LA = 0.5775 + (0.000028 *1000) = 0.6055 v12 = vF PFM = 4746 * 0.6055 = 2,874 pc/ h
PFM = 0.5775 + 0.000028 LA = 0.5775 + (0.000028 * 1000) = 0.6055 v12 = vF PFM = 4746 * 0.6055 = 2,874 pcu/h

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13

This prediction must be checked for reasonableness. The outer lane (lane 3) will have a demand flow rate of 4,746 – 2,874 = 1,872 pc/h/ln. This is compared to the average flow rate in lanes 1 and 2 (2874/2 = 1,437 pc/h/ln). It is neither more than 2,700 pc/h/ln nor 1.5*1473 = 2,210 pc/h/ln. The predicted distribution is ok.
必须检查此预测的合理性。外车道（第 3 车道）的需求流量将为 4,746 – 2,874 = 1,872 pcu/h/车道。将其与第 1 和第 2 车道的平均流量（2874/2 = 1,437 pcu/h/车道）进行比较。它既不大于 2,700 pcu/h/车道，也不大于 1.5*1473 = 2,210 pcu/h/车道。预测的分布是可以的。

Capacity values must be checked to insure that stable operations are present. The downstream freeway capacity is 7,050 pc/h (Table 15.5, 65 mi/h FFS, 3 lanes in each direction). The downstream freeway demand is 4746+780 = 5,526 pc/h < 7,050 pc/h. vR12 (2,874+780 = 3,654 pc/h) is less than the maximum recommended value of 4,600 pc/h. The ramp demand flow rate of 780 pc/h is less than the capacity of a 40-mi/h ramp (2,000 pc/h). No capacity constraints exist. The analysis continues.
必须检查容量值以确保运行稳定。下游高速公路的容量为 7,050 pcu/h（表 15.5，65 英里/小时 FFS，每个方向 3 车道）。下游高速公路的需求量为 4746 + 780 = 5,526 pcu/h < 7,050 pcu/h。vR12（2,874 + 780 = 3,654 pcu/h）小于推荐最大值 4,600 pcu/h。780 pcu/h 的匝道需求流量小于 40 英里/小时匝道的容量（2,000 pcu/h）。不存在容量限制。分析继续进行。

The density in the merge influence area is estimated by Equation 15-40:
合并影响区域的密度由公式 15-40 估算：

DR = 5.475 + 0.00734 vR + 0.0078v12 − 0.00627 LA

DR = 5.475 + (0.00734 * 780) + (0.0078 * 2874) − (0.00627 *1,000)
DR = 5.475 + (0.00734 * 780) + (0.0078 * 2874) − (0.00627 * 1000)

DR = 5.475 + 7.72 + 22.4 − 6.27 = 29.3 pc/ mi / ln This represents LOS D.
DR = 5.475 + 7.72 + 22.4 − 6.27 = 29.3 pc/mi/ln 这代表 LOS D。

If 1,000 veh/h are added to the ramp demand, making it 1,700 veh/h, a number of things change. Converting to passenger car equivalents under base conditions:
如果在坡道需求中增加 1,000 辆/小时，使其达到 1,700 辆/小时，则会发生一些变化。在基本条件下转换为乘用车当量：

The demand flow in lanes 1 and 2 (v12) does not depend upon the ramp demand, and is unchanged (2,874 pc/h). The downstream freeway flow rate changes to 4746+1893 = 6,639 pc/h, which is still under the capacity of 7,050 pc/h. The ramp capacity is 2,000 pc/h, still larger than the ramp demand of 1,893 pc/h. The volume entering the merge influence area (vR12) is now 2874+1893 = 4,747 pc/h, which is more than the recommended maximum value of 4,600 pc/h. This does not imply that the segment will fail, but may indicate that operating conditions are worse than predicted by the algorithms.
1 号和 2 号车道的需求流量 (v12) 不依赖于坡道需求，并且保持不变 (2,874 pc/h)。下游高速公路的流量变为 4746 + 1893 = 6,639 pc/h，这仍然低于 7,050 pc/h 的容量。坡道容量为 2,000 pc/h，仍然大于 1,893 pc/h 的坡道需求。进入合并影响区域的车辆数量 (vR12) 现在为 2874 + 1893 = 4,747 pc/h，这超过了推荐的最大值 4,600 pc/h。这并不意味着路段会失效，但可能表明运行状况比算法预测的要差。

The density in the merge influence area must be re-computed:
必须重新计算合并影响区域的密度：

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14

DR = 5.475 + (0.00734 *1893) + (0.0078 * 2874) − (0.0062 *1000) DR = 5.475 + 13.895 + 22.417 − 6.200 = 35.6 pc/ h / ln
DR = 5.475 + (0.00734 * 1893) + (0.0078 * 2874) − (0.0062 * 1000) DR = 5.475 + 13.895 + 22.417 − 6.200 = 35.6 pc/h/ln

This is LOS E. The addition of another 1,000 veh/h on the on-ramp would place the merge segment under significant stress. While the demand has not yet reached the capacity of the segment, it is only several hundred vehs/h of doing so. Serious consideration should be given to an improvement project if the additional demand is a realistic expectation.
这是 LOS E。在匝道上增加另外 1000 辆车/小时的交通量将使合并路段承受巨大压力。虽然需求尚未达到路段容量，但也仅差几百辆车/小时就达到饱和。如果额外需求是现实的预期，则应认真考虑改进项目。

Problem 15-4
问题 15-4

FFS = 60 mi/h; ID = 2 int/mi
FFS = 60 英里/小时；ID = 2 个交叉口/英里

3, 100 veh/h
3, 100 辆/小时

1200 ft
1200 英尺

200 ft
200 英尺

700 veh/h
700 车/小时

600 veh/h
600 辆车/小时

RFFS = 45 mi/h
自由流车速 (RFFS) = 45 英里/小时

All Movements
所有车辆

Level terrain 10% trucks PHF = 0.94
平坦路段，10%卡车，PHF = 0.94

RFFS = 45 mi/h
自由流车速 (RFFS) = 45 英里/小时

As in all weaving and ramp analyses, the process begins by converting all demand volumes to flow rates under equivalent base conditions. From Chapter 14, ET = 1.5 for level terrain. Then:
与所有编织和匝道分析一样，该过程首先将所有需求量转换为等效基准条件下的流量。根据第 14 章，平坦路段的 ET = 1.5。然后：

The driver population is assumed to be normal; therefore fp = 1.00. Then:
假设驾驶员群体服从正态分布；因此 fp = 1.00。然后：

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15

Because the merge and diverge segments are on a 4-lane freeway (2 lanes in each direction), much of the analysis is trivial:
由于合并和分流路段位于四车道高速公路（每个方向两车道）上，因此大部分分析都比较简单：

v12 (Ramp 1) = 3,464 pc/ h
v12（匝道 1）= 3,464 pcu/小时

v12 (Ramp2) = 3,464 + 670 = 4,134 pc/ h vR12 (Ramp1) = 3,464 + 670 = 4,134 pc/ h
v12（匝道 2）= 3,464 + 670 = 4,134 pcu/小时 vR12（匝道 1）= 3,464 + 670 = 4,134 pcu/小时

From Table 15.5, capacity values may be checked. The capacity of a 4-lane freeway segment (FFS = 60 mi/h) is 4,600 pc/h. This is greater than the maximum freeway flow between the ramps (4,134 pc/h). Both ramps have a capacity of 2,000 pc/h, which is considerably more than either of the two ramp demand flow rates. An limits on flow rate entering the merge and diverge influence areas (4,600 pc/h for merge; 4,400 pc/h for diverge) are also higher than the demand flow (4,134 pc/h in both cases). No capacity constraints exist.
根据表 15.5，可以检查容量值。四车道高速公路路段（FFS = 60 英里/小时）的容量为 4,600 pcu/小时。这大于匝道之间最大的高速公路流量（4,134 pcu/小时）。两条匝道的容量均为 2,000 pcu/小时，这大大超过了两个匝道的需求流量。进入合并和分流影响区域的流量限制（合并为 4,600 pcu/小时；分流为 4,400 pcu/小时）也高于需求流量（两种情况下均为 4,134 pcu/小时）。不存在容量限制。

Equation 15-40 is used to estimate the density in the merge influence area:
使用公式 15-40 估算合并影响区域的密度：

DR = 5.474 + (0.0073* 670) + (0.0078 * 3464) − (0.00627 *100)
DR = 5.474 + (0.0073 * 670) + (0.0078 * 3464) − (0.00627 * 100)

DR = 5.475 + 4.891 + 27.02 − 0.627 = 36.8 pc/ mi / ln From Table 15.1, this is LOS E.
DR = 5.475 + 4.891 + 27.02 − 0.627 = 36.8 pc/mi/ln 由表 15.1 可知，这是 LOS E。

Equation 15-41 is used to estimate the density in the diverge influence area:
等式 15-41 用于估算分流影响区域的密度：

DR = 4.252 + (0.0086 * 4134) − (0.009 * 200)

DR = 4.252 + 35.552 −1.800 = 38.0 pc/ h From Table 15.1, this is also LOS E.
DR = 4.252 + 35.552 − 1.800 = 38.0 pc/h 由表 15.1 可知，这也是 LOS E。

15

16

The segment clearly operates in LOS E, and is very close to capacity. Several alternatives have been suggested as improvements. Each is analyzed in the sections that follow.
该路段明显运行在 LOS E 等级，并且非常接近饱和状态。已经提出了一些改进方案。以下各节将分析每一种方案。

Analysis of a Weaving Segment
织入式路段分析

Adding a full auxiliary lane creates the following configuration. It is assumed that there is no ramp-to-ramp demand.
增加一条完整的辅助车道会产生以下配置。假设没有匝道间的需求。

FFS = 60 mi/h; ID = 2 int/mi
FFS = 60 英里/小时；ID = 2 个交叉口/英里

1200 ft
1200 英尺

3.464-782=2,682 pc/h

670 pc/h
670 辆/小时

782 pc/h
782 辆/小时

By inspection, this is a ramp-weave segment with LCRF = LCFR = 1 and NWV = 2. Then:
经检验，这是一个坡道编织路段，LCRF = LCFR = 1，NWV = 2。然后：

vW = 670 + 782 = 1,452 pc/ h vNW = 2,682 pc/ h
vW = 670 + 782 = 1,452 辆/小时 vNW = 2,682 辆/小时

v = 1,452 + 2,682 = 4,134 pc/ h
v = 1,452 + 2,682 = 4,134 辆/小时

LCMIN = (670 *1) + (782 *1) = 1,452 lane changes/ h VR = 1452 / 4134 = 0.351
LCMIN = (670 *1) + (782 *1) = 1,452 次/小时 VR = 1452 / 4134 = 0.351

The maximum length for weaving operations to exist is given by Equation 15-4: LMAX = [5,782 (1 + VR)1.6 ] − [1,566 NWV ]
编织作业存在的最大长度由公式 15-4 给出：LMAX = [5,782 (1 + VR)1.6 ] − [1,566 NWV ]

L [5,782 (1 0.351)1.6 ] [1,566 * 2] LMAX = 9,357 − 3,132 = 6,225 ft > 1,200 ft
L [5,782 (1 + 0.351)1.6 ] − [1,566 * 2] LMAX = 9,357 − 3,132 = 6,225 英尺 > 1,200 英尺

The segment will operate as a weaving segment. The analysis continues.
该路段将作为编织路段运行。分析继续。

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17

The capacity of the weaving segment, as determined by overall operations, is estimated using Equation 15-5:
编织路段的容量，由总体运行情况决定，使用公式 15-5 估算：

c = c − [438.2 (1 + VR)1.6 ]+ [0.0765 L ]+ [119.8 N ]
c = c − [438.2 (1 + VR)¹·⁶] + [0.0765 L] + [119.8 N]

c = 2I,00 − [438.2 *1.3511.6 ]+ [0.0765 1200]+ [119 V* 2]
c = 2I₀₀ − [438.2 × 1.35¹·⁶] + [0.0765 × 1200] + [119 V × 2]

cIWL = 2,300 − 709.122 + 91.8 + 239.6 = 1,922 pc/ h / ln
cIWL = 2300 − 709.122 + 91.8 + 239.6 = 1922 pc/h/ln

cIW = 1,922 * 3 = 5,766 pc/ h
cIW = 1922 × 3 = 5766 pc/h

The capacity of the weaving segment, as determined by maximum weaving flow rates, is estimated using Equation 15-7:
根据公式 15-7 估算由最大织机流量确定的织造段容量：

The lower value, 5,766 pc/h, controls. This value is in pc/h. It could be converted to veh/h by multiplying by the heavy vehicle and driver population adjustment factors. That is not needed here. The capacity (in pc/h) is greater than the total demand flow of 4,134 pc/h. There is no breakdown, and the analysis continues.
最低值 5766 pc/h 起控制作用。该值以 pc/h 为单位。可以通过乘以重型车辆和驾驶员数量调整系数将其转换为 veh/h。此处无需进行此操作。容量（以 pc/h 计）大于总需求流量 4134 pc/h。没有发生故障，分析继续进行。

The total lane-changing rate for weaving vehicles is estimated using Equation 15- 11:
使用公式 15-11 估算编织车辆的总换道率：

LCW = 1,452 + 0.39 [30 * 9 * 2.41] = 1,452 + 254 = 1,706 lane changes/ h
LCW = 1452 + 0.39 [30 × 9 × 2.41] = 1452 + 254 = 1706 次换道/小时

The non-weaving vehicle lane-changing index is given by Equation 15-13:
非编织车辆的换道指数由公式 15-13 给出：

Then, using Equation 15-12:
然后，使用公式 15-12：

LCNW1 = LCNW = (0.206 vNW ) + (0.542 LS ) − (192.6 N) LCNW = (0.206 * 2682) + (0.542 *1200) − (192.6 * 3) LCNW = 552.5 + 650.4 − 577.8 = 625 lane changes/ h LCALL = 1706 + 625 = 2,331 lane changes/ h
LCNW1 = LCNW = (0.206 vNW ) + (0.542 LS ) − (192.6 N) LCNW = (0.206 * 2682) + (0.542 *1200) − (192.6 * 3) LCNW = 552.5 + 650.4 − 577.8 = 625 车道变更/小时 LCALL = 1706 + 625 = 2,331 车道变更/小时

17

18

The speed of weaving vehicles is estimated using Equations 15-19 and 15-20:
使用公式 15-19 和 15-20 估计编织车辆的速度：

The speed of non-weaving vehicles is estimated using Equation 15-21:
非编织车辆的速度使用公式 15-21 估算：

SNW = FFS − (0.0072 LCMIN ) − (0.0048v N) SNW = 60 − (0.0072 *1452) − (0.0048 41343)
SNW = FFS − (0.0072 LCMIN ) − (0.0048v N) SNW = 60 − (0.0072 *1452) − (0.0048 * 41343)

SNW = 60 −10.454 − 6.614 = 42.9 mi / h Then:
SNW = 60 − 10.454 − 6.614 = 42.9 英里/小时 然后：

From Table 15.1, this is LOS D. Add a Lane to the Freeway
根据表 15.1，这是 D 级服务水平。在高速公路上增加车道

Adding a lane to the freeway creates the following scenario:
在高速公路上增加车道会产生以下场景：

FFS = 60 mi/h; ID = 2 int/mi
FFS = 60 英里/小时；ID = 2 个交叉口/英里

1200 ft
1200 英尺

3, 100 veh/h
3, 100 辆/小时

300 ft
300 英尺

600 veh/h
600 辆车/小时

RFFS = 45 mi/h
自由流车速 (RFFS) = 45 英里/小时

All Movements
所有车辆

Level terrain 10% trucks PHF = 0.94
平坦路段，10%卡车，PHF = 0.94

RFFS = 45 mi/h
自由流车速 (RFFS) = 45 英里/小时

18

19

This scenario creates a ramp sequence, now attached to a 6-lane freeway instead of a 4-lane freeway. All of the volumes have been converted previously:
此情景创建了一个匝道序列，现在连接到一条 6 车道高速公路，而不是 4 车道高速公路。所有流量数据都已进行过转换：

v F = 3 ,464 pc / h v R1 = 670 pc / h  v R 2 = 782 pc / h
vF = 3,464 辆车/小时，vR1 = 670 辆车/小时，vR2 = 782 辆车/小时

This time, the estimation of v12 values is not trivial. From Table 15.3, PFM (first ramp) is estimated using Equation 15-25 or 15-27. The latter includes the impact of the downstream off-ramp. The decision on which to use is based upon the equivalence distance, given by Equation 15-31:
这一次，v12 值的估计并非简单。根据表 15.3，使用公式 15-25 或 15-27 来估计 PFM（第一个匝道）。后者包括下游出口匝道的冲击。选择使用哪个公式取决于等效距离，由公式 15-31 给出：

Therefore, Equation 15-27 is used, and:
因此，使用公式 15-27，并且：

PFM = 0.5487 + 0.1713 = 0.72

v12 (Ramp1) = vF PFM = 3464 * 0.72 = 2,494 poc/ h
v12 (Ramp1) = vF PFM = 3464 * 0.72 = 2,494 pcu/h

This is checked for reasonableness. The lane 3 flow is 3464 – 2494 = 970 pc/h. Compared to the average flow in lanes 1 and 2 (2494/2 = 1,247 pc/h), neither the limit of 2,700 pc/h/ln or 1.5*1247 = 1,871 pc/h is violated by the lane 3 flow. The predicted distribution is reasonable.
这已进行合理性检查。车道 3 的流量为 3464 – 2494 = 970 pcu/h。与车道 1 和 2 的平均流量 (2494/2 = 1,247 pcu/h) 相比，车道 3 的流量并未超过每车道 2,700 pcu/h 或 1.5*1247 = 1,871 pcu/h 的限制。预测分布是合理的。

The demand is also well within the capacity limitations of Table 15.5. The density in the merge influence area is given by Equation 15-40:
需求也远低于表 15.5 的容量限制。合并影响区域的密度由公式 15-40 给出：

DR = 5.475 + (0.00734 * 670) + (0.0078 * 2494) − (0.0062 * 300) DR = 5.475 + 4.918 + 19.453 −1.860 = 28.0 pc/ mi / ln
DR = 5.475 + (0.00734 * 670) + (0.0078 * 2494) − (0.0062 * 300) DR = 5.475 + 4.918 + 19.453 − 1.860 = 28.0 pcu/mi/ln

From Table 15.1, this is LOS C, but barely. It is right on the boundary between levels of service C and D.
根据表 15.1，服务水平为 C 级，但勉强。它位于 C 级和 D 级服务水平的边界上。

19

20

From Table 15.4, PFD, for the downstream ramp, is computed using either Equation 15-33 or 15-34. The choice is dependent upon the equivalence distance, estimated using Equation 15-36. Also, for Ramp 2, vF = 3464+670 = 4,134 pc/h.
根据表 15.4，下游匝道的 PFD 使用公式 15-33 或 15-34 计算。选择取决于使用公式 15-36 估计的等效距离。此外，对于匝道 2，vF = 3464 + 670 = 4,134 pcu/h。

LEQ = 6,262 ft > 1,200 ft
LEQ = 6,262 英尺 > 1,200 英尺

Therefore, Equation 15-34 is used:
因此，使用公式 15-34：

P = 0.717 − 0.000039 v + 0.604 u

PFD = 0.717 − (0.000039 * 4134) + (0.604 6701200) PFD = 0.717 − 0.161 + 0.337 = 0.893
PFD = 0.717 − (0.000039 * 4134) + (0.604 * 670/1200) PFD = 0.717 − 0.161 + 0.337 = 0.893

v12 (Ramp2) = vR + (vF − vR )PFD = 782 + (4134 − 782) * 0.893 = 3,775 pc/ h
v12（Ramp2）= vR + (vF − vR )PFD = 782 + (4134 − 782) * 0.893 = 3,775 pc/h

This is checked for reasonableness. The lane 3 flow rate is 4134-3775 = 359 pc/h. This is less than the limiting values of 2,700 pc/h/ln or 1.5*(3775/2) = 2,831 pc/h/ln. The lane distribution is deemed reasonable, and the analysis continues:
对此进行了合理性检查。3 号车道的流量为 4134-3775 = 359 pc/h。这小于 2,700 pc/h/ln 或 1.5*(3775/2) = 2,831 pc/h/ln 的限制值。车道分布被认为是合理的，分析继续进行：

None of the capacity limitations of Table 15.5 are violated either, so stable flow is expected and the analysis continues.
表 15.5 中没有任何容量限制被违反，因此预计流量稳定，分析继续进行。

The density in the diverge influence area is estimated using Equation 15-41:
使用公式 15-41 估算分散影响区域的密度：

DR = 4.252 + (0.0086 * 3775) − (0.009* 300) = 4.252 + 32.465 − 2.700 = 34.0 pc/ mi / ln From Table 15.1, this is LOS D, but just barely under the LOS D/Eboundary
DR = 4.252 + (0.0086 * 3775) − (0.009* 300) = 4.252 + 32.465 − 2.700 = 34.0 pc/mi/ln 根据表 15.1，这是 LOS D，但略低于 LOS D/E 边界.

20

21

Another Weaving Segment
另一个编织路段

The third alternative improvement creates another weaving segment:
第三种替代改进方案创建了另一个编织路段：

FFS = 60 mi/h; ID = 2 int/mi
FFS = 60 英里/小时；ID = 2 个交叉口/英里

1200 ft
1200 英尺

3.464-782=2,682 pc/h

670 pc/h
670 辆/小时

782 pc/h
782 辆/小时

Note that in the context of a weaving segment, the 300-ft deceleration lane at the 2nd ramp is not included in the analysis.
请注意，在编织路段的背景下，第二个匝道处的 300 英尺减速车道未包含在分析中。

If the deceleration lane at the off-ramp is ignored, this creates a major weaving configuration with LCRF = 0 and LCFR = 1. NWV for this case is 3 lanes. The volumes shown have already been converted to flow rates under equivalent base conditions.
如果忽略驶离匝道处的减速车道，则会形成一个主要的编织配置，其中 LCRF = 0，LCFR = 1。此情况下 NWV 为 3 车道。所示车流量已根据等效基本条件转换为流量。

vW = 670 + 782 = 1,452 pc/ h vNW = 2,682 pc/ h
vW = 670 + 782 = 1,452 辆/小时 vNW = 2,682 辆/小时

v = 1452 + 2683 = 4,134 pc/ h VR = 1452 / 4134 = 0.351
v = 1452 + 2683 = 4,134 pcu/h VR = 1452 / 4134 = 0.351

LCMIN = (670 * 0) + (782 *1) = 782 lane changes/ h
LCMIN = (670 * 0) + (782 * 1) = 782 车道变更/小时

The maximum length of a weaving segment is given by Equation 15-4:
编织路段的最大长度由公式 15-4 给出：

L [5728(1 0.351)1.6 ] [1566 * 3] 9,269 4,698 4,571 ft 1,200 ft
L [5728(1 - 0.351)^1.6] [1566 * 3] 9,269 4,698 4,571 英尺 1,200 英尺

Therefore, the segment operates as a weaving segment, and the analysis may continue.
因此，该路段作为编织路段运行，可以继续进行分析。

21

22

The capacity of the weaving segment as determined by overall segment operations is given by Equation 15-5:
由整体路段运行确定的编织路段容量由公式 15-5 给出：

c = c − [438.2 (1 +VR)1.6 ]+ [0.0765 L ]+ [119.8 * N ]
c = c − [438.2 (1 + VR)^1.6] + [0.0765 L] + [119.8 * N]

c = 2I,L00 − [438.2 *1.3511.6 ]+ [0.0765 1200]+ [119.8W3]
c = 2IL00 - [438.2 *1.3511.6 ] + [0.0765 * 1200] + [119.8W3]

cIWL = 2,300 − 709.122 + 91.800 + 359.400 = 2,042 pc/ h / ln cIW = 2042 * 3 = 6,126 pc/ h
cIWL = 2,300 − 709.122 + 91.800 + 359.400 = 2,042 pc/h / ln cIW = 2042 * 3 = 6,126 pc/h

The capacity of the segment, as controlled by maximum weaving flow rates, is given by Equation 15-7:
由最大织物流量控制的段容量由等式 15-7 给出：

The smaller value, 6,126 pc/h controls. As this is larger than the total demand of 4,134 pc/h, LOS F will not occur, and the analysis continues.
较小值 6,126 pc/h 控制。由于它大于总需求 4,134 pc/h，因此不会发生 LOS F，分析继续进行。

The number of lane changes made by weaving vehicles is estimated using Equation 15-11:
使用等式 15-11 估算织车道改变次数：

LCW = LCMIN + 0.39 [(LS − 300)0.5 N2 (1 + ID)0.8 ] LCW = 782 + 0.39 [(1200 − 300)0.5 32 (1 + 2)0.8 ]

LCW = 782 + 0.39 [30 * 9 * 2.408] = 782 + 0.39 * 650.16 = 782 + 254 = 1,036 lane change/ h The non-weaving lane-change index is defined by Equation 15-13:
LCW = 782 + 0.39 [30 * 9 * 2.408] = 782 + 0.39 * 650.16 = 782 + 254 = 1,036 车道变换/小时 非编队车道变换指数由公式 15-13 定义：

Then, the number of lane changes made by non-weaving vehicles is estimated using Equation 15-12:
然后，使用公式 15-12 估计非编队车辆的车道变换次数：

LCNW1 = LCNW = (0.206 vNW ) + (0.542 LS ) − (192.6 N)

LCNW = (0.206 * 2682) + (0.542 *1200) − (192.6 * 3) = 552.5 + 650.4 − 577.8 = 625 lane changes/ h LCALL = 1036 + 625 = 1,661 lane changes/ h
LCNW = (0.206 * 2682) + (0.542 *1200) − (192.6 * 3) = 552.5 + 650.4 − 577.8 = 625 车道变换/小时 LCALL = 1036 + 625 = 1,661 车道变换/小时

22

23

The speed of weaving vehicles is estimated using Equations 15-19 and 15-20:
使用公式 15-19 和 15-20 估计编织车辆的速度：

The speed of non-weaving vehicles is estimated using Equation 15-21: SNW = 60 − (0.0072 * 782) − (0.004841343) = 60.0 − 5.6 − 6.6 = 47.8 mi / h Then:
使用公式 15-21 估计非编队车辆的速度：SNW = 60 − (0.0072 * 782) − (0.004841343) = 60.0 − 5.6 − 6.6 = 47.8 英里/小时 然后：

This is LOS D, but very close to the LOS C/Dborder. Comparison
这是 D 级服务水平，但非常接近 C/D 级服务水平边界比较

This is a case in which it is more valuable to look at actual densities than the resulting LOS values. That is because while the existing case operates at LOS E, each of the alternatives operates at LOS D. For ramp cases, the worst of the two LOS results would control. In the last alternative, however, the result is just
这是一个更值得关注实际密度而非结果服务水平值的情况。这是因为，虽然现有情况运行在 E 级服务水平，但每个替代方案都在 D 级服务水平运行。对于匝道情况，两个服务水平结果中的最差者将起控制作用。然而，在最后一个替代方案中，结果刚刚高于 C 级服务水平边界，而在第二个替代方案中，它刚刚低于 E 级服务水平边界——一个更糟糕的结果。在第一个替代方案中，服务水平处于 D 级服务水平范围的中部。查看实际密度：

above the LOS C boundary, while in the second, it is just under the LOS E boundary – a far worse result. In the first alternative, the LOS is in the middle of the LOS D range. Looking at the densities themselves:

Existing: 38 pc/mi/ln
现有：38 pc/mi/ln

Alt 1: 31 pc/mi/ln
备选方案 1：31 pc/mi/ln

Alt 2: 34 pc/mi/ln
备选方案 2：34 pc/mi/ln

Alt 3: 28.4 pc/mi/ln
备选方案 3：28.4 pc/mi/ln

From these results, Alternative 3 appears to be favored.
从这些结果来看，备选方案 3 似乎更受欢迎。

23

24

Problem 15-5
问题 15-5

FFS=70 mi/h; ID = 1 int. per mile.
FFS=70 英里/小时；ID = 每英里 1 个。

1,300 ft
1,300 英尺

3,500 pc/h
3,500 pc/小时

400 pc/h
400 pc/小时

1,100 pc/h
1,100 辆/小时

This problem is greatly simplified by the fact that all of the flow rates have already been converted, and can be used directly.
问题大大简化了，因为所有流量都已换算完毕，可以直接使用。

As illustrated above, LCRF = 2, LCFR = 0, and NWV = 3. Also:
如上所示，LCRF = 2，LCFR = 0，NWV = 3。此外：

vW = 400 + 1,000 = 1,400 pc/ h vNW = 3500 + 1100 = 4,600 pc/ h v = 1400 + 4600 = 6,000 veh / h VR = 1400 / 6000 = 0.233
vW = 400 + 1,000 = 1,400 辆/小时 vNW = 3500 + 1100 = 4,600 辆/小时 v = 1400 + 4600 = 6,000 车/小时 VR = 1400 / 6000 = 0.233

LCMIN = (400 * 2) + (1,000 * 0) = 800 lane changes/ h
LCMIN = (400 * 2) + (1,000 * 0) = 800 车道变更/小时

The maximum weaving length is estimated using Equation 15-4:
最大编织长度使用公式 15-4 估算：

LMAX = [5,728 (1 + 0.233)1.6 ]− [1,566 * 3] = 8008 − 4698 = 3,310 ft > 1,300 ft The segment will operate as a weaving segment.
LMAX = [5,728 (1 + 0.233)1.6] − [1,566 * 3] = 8008 − 4698 = 3,310 英尺 > 1,300 英尺 该路段将作为编织路段运行。

The capacity of the weaving segment based upon overall operation is given by Equation 15-5:
基于整体运行的编织路段容量由公式 15-5 给出：

cIWL = 2,400 − [438.2 (1 + 0.233)1.6 ]+ [0.0765*1300]+ [119.8* 3 ] cIWL = 2,400 − 612.7 + 97.5 + 359.4 = 2,244 pc/ h
cIWL = 2,400 − [438.2 (1 + 0.233)1.6] + [0.0765*1300] + [119.8* 3] cIWL = 2,400 − 612.7 + 97.5 + 359.4 = 2,244 辆/小时

cIW = 2244 * 4 = 8,976 pc/ h
cIW = 2244 * 4 = 8,976 件/小时

24

25

The capacity of the weaving segment based upon maximum weaving flow rate is given by Equation 15-7:
根据最大织造流量，编织段的容量由公式 15-7 给出：

The smaller value, 8,976 pc/h prevails. As it is larger than the actual demand flow rate of 6,000 pc/h, operation is stable and the analysis continues.
较小的值 8,976 件/小时占上风。由于它大于实际需求流量 6,000 件/小时，因此运行是稳定的，并且分析继续进行。

The total lane-changing rate for weaving vehicles is given by Equation 15-11: LCW = 1400 + 0.39 [(1300 − 300)0.5 42 (1 +1)0.8 ]
编织车辆的总车道变更率由公式 15-11 给出：LCW = 1400 + 0.39 [(1300 − 300)0.5 42 (1 +1)0.8 ]

LCW = 1400 + 0.39 * (31.62 *16 *1.74) = 1,743 lane changes/ h
LCW = 1400 + 0.39 * (31.62 *16 *1.74) = 1,743 车道变更/小时

The non-weaving lane-changing index is computed using Equation 15-13:
非编织车道变更指数由公式 15-13 计算：

Therefore, the total rate of non-weaving vehicle lane changes is estimated using Equation 15-12:
因此，使用公式 15-12 估计非编织车辆车道变更的总速率：

LCNW = (0.206 * 4600) + (0.542 *1300) − (192.6 * 4)  LCNW = 947.6 + 704.6 − 770.4 = 882 lane changes/ h LCALL = 1743 + 882 = 2,625 lane changes/ h
LCNW = (0.206 * 4600) + (0.542 *1300) − (192.6 * 4) LCNW = 947.6 + 704.6 − 770.4 = 882 车道变更/小时 LCALL = 1743 + 882 = 2,625 车道变更/小时

The speed of weaving vehicles is estimated using Equations 15-19 and 15-20:
使用公式 15-19 和 15-20 估计编织车辆的速度：

The non-weaving vehicle speed is given by Equation 15-21:
非编织车辆速度由公式 15-21 给出：

SNW = 70.0 − (0.0072 *800) − (0.0048 *60004) = 70.0 − 5.76 − 7.2 = 57.0 mi / h
SNW = 70.0 − (0.0072 *800) − (0.0048 *60004) = 70.0 − 5.76 − 7.2 = 57.0 英里/小时

25

26

Then:
然后：

From Table 15.1, this is LOS C.
来自表 15.1，这是 LOS C。

26

1

TRAFFIC ENGINEERING, 4th Edition
交通工程，第 4 版

Roess, R.P., Prassas, E.S., and McShane, W.R.
Roess，R.P.，Prassas，E.S.和 McShane，W.R.

Solution to Problems in Chapter 16
第 16 章问题的解决方案

Problem 16-1
问题 16-1

Segment Characteristics (given):
段特性（已知）：

Class I two-lane highway Length = 20 miles
一级双车道高速公路长度 = 20 英里

BFFS = 60 mi/h
BFFS = 60 英里/小时

12-ft lanes; 4-ft clear shoulders
12 英尺车道；4 英尺清晰路肩

8 access points/mi
每英里 8 个入口

60% “No Passing” zones 15% trucks; 2% RVs
60% “禁止超车”区域；15% 卡车；2% RV

Rolling terrain
起伏地形

60-40 directional split
60-40 方向分割

V (total both directions) = 400 veh/h PHF = 0.84
V（双向总和）= 400 车/小时 PHF = 0.84

Finding the Current LOS
查找当前服务水平

The free-flow speed (FFS) of the segment must be estimated using Equation 16-6: FFS = BFFS − fLS − fA
必须使用公式 16-6 估算路段的自由流速度（FFS）：FFS = BFFS − fLS − fA

Where: BFFS = 60 mi/h (given)
其中：BFFS = 60 英里/小时（已知）

fLS = 1.3 mi/h (Table 16.5, 12-ft lanes, 4-ft shoulders)
fLS = 1.3 英里/小时 (表 16.5，12 英尺车道，4 英尺路肩)

fA = 2.0 mi/h (Table 16.6, 8 pts/mi – interpolated) FFS = 60.0 −1.3 − 2.0 = 56.7 mi / h
fA = 2.0 英里/小时 (表 16.6，每英里 8 个点 - 插值) FFS = 60.0 −1.3 − 2.0 = 56.7 英里/小时

The demand volume must now be converted to a flow rate under base conditions. There are four such conversions to be made: for each direction, a conversion is necessary for the determination of average highway speed (ATS) and for the determination of percent time spent following (PTSF). By definition, the high-volume direction is Dir 1; the low-volume direction is Dir 2:
现在必须将需求量转换为基准条件下的流量。需要进行四次此类转换：对于每个方向，都需要进行转换以确定平均高速公路速度 (ATS) 和确定跟随时间百分比 (PTSF)。根据定义，高流量方向是 Dir 1；低流量方向是 Dir 2：

1

2

V1

=

400*0.60

=

240 veh/h
240 辆/小时

V2

=

400*0.40

=

160 veh/h
160 辆/小时

All conversions use Equation 16-7:
所有转换都使用公式 16-7：

Where: fG1 = 0.82 for ATS, 0.84 for PTSF
其中：fG1 = 0.82（用于 ATS），0.84（用于 PTSF）

(Table 16.7, for v = 240/0.84 = 286 veh/h, interpolated) fG2 = 0.74 for ATS, 0.79 for PTSF
(表 16.7，对于 v = 240/0.84 = 286 辆/小时，插值) fG2 = 0.74（用于 ATS），0.79（用于 PTSF）

(Table 16.7, for v = 160/0.84 = 190 veh/h, interpolated)
(表 16.7，对于 v = 160/0.84 = 190 辆/小时，插值)

The heavy vehicle adjustment factor is based upon passenger car equivalents for trucks and RVs, using Equation 16-8:
重型车辆调整系数基于卡车和 RV 的小汽车当量，使用公式 16-8：

Where: ET1 = 2.1 for ATS, 1.7 for PTSF
其中：ET1 = 2.1（ATS），1.7（PTSF）

(Table 16.10, for v = 240/0.84 = 286 veh/h, interpolated)
(表 16.10，对于 v = 240/0.84 = 286 辆/小时，插值)

ET2 = 2.3 for ATS, 1.8 for PTSF
ET2 = 2.3（ATS），1.8（PTSF）

(Table 16.10, for v = 160/0.84 = 190 veh/h, interpolated)
(表 16.10，对于 v = 160/0.84 = 190 辆/小时，插值)

ER1 = 1.1 for ATS, 1.0 for PTSF (Table 16.10, ALL)
ER1 = 1.1（ATS），1.0（PTSF）（表 16.10，全部）

ER2 = 1.1 for ATS, 1.0 for PTSF (Table 16.10, ALL)
ER2 = 1.1（ATS），1.0（PTSF）（表 16.10，全部）

Then:
然后：

2

3

And:
和：

The average travel speed for each direction is estimated using Equation 16-10: ATSd = FFS − 0.00776 (vd + vo ) − fnpA
使用公式 16-10 估计每个方向的平均行驶速度：ATSd = FFS − 0.00776（vd + vo ）− fnpA

Where:
其中：

FFS

vd+vo fnpA1

= = =

56.7 mi/h (computed previously) 407+308 = 715 veh/h
56.7 英里/小时（先前计算得出）407+308 = 715 辆/小时

3.0 mi/h
3.0 英里/小时

(Table 16.15, vo = 308, 60% NPZ, 56.7 mi/h, multiple iterations)
(表 16.15，vo = 308，60% NPZ，56.7 英里/小时，多次迭代)

fnpA2

=

2.7 mi/h
2.7 英里/小时

(Table 16.15, vo = 407, 60% NPZ, 56.7 mi/h, multiple iterations)
(表 16.15，vo = 407，60% NPZ，56.7 英里/小时，多次迭代)

Then:
然后：

ATS1 = 56.7 − (0.00776 * 715) − 3.0 = 56.7 − 5.5 − 3.0 = 48.2 mi / h ATS2 = 56.7 − (0.00776 * 715) − 2.7 = 56.7 − 5.5 − 2.7 = 48.5 mi / h
ATS1 = 56.7 − (0.00776 * 715) − 3.0 = 56.7 − 5.5 − 3.0 = 48.2 英里/小时；ATS2 = 56.7 − (0.00776 * 715) − 2.7 = 56.7 − 5.5 − 2.7 = 48.5 英里/小时

The percent time following in each direction is estimated using Equations 16-11:
使用公式 16-11 估算每个方向的跟车百分比时间：

BPTSFd = 100 [1 − exp (a vd )]

Where: a1 = -0.0017 (Table 16.17, vo = 270 pc/h, interpolated)
其中：a1 = -0.0017（表 16.17，vo = 270 pc/h，插值）

b1 = 0.956 (Table 16.17, vo = 270 pc/h, interpolated)
b1 = 0.956（表 16.17，vo = 270 pc/h，插值）

a2 = -0.0021 (Table 16.17, vo = 376 pc/h, interpolated)
a2 = -0.0021 (表 16.17，vo = 376 pc/h，插值)

b2 = 0.929 (Table 16.17, vo = 376 pc/h, interpolated)
b2 = 0.929 (表 16.17，vo = 376 pc/h，插值)

3

4

fnpP V1 V2

Then:
然后：

= = =

48.2%

240 veh/h
240 辆/小时

160 veh/h
160 辆/小时

Direction 1 has an ATS of 48.5 mi/h and a PTSF of 67.8%. Direction 2 has an ATS of 48.5 mi/h and a PTSF of 51.0%
方向 1 的平均行驶速度为 48.5 英里/小时，峰值小时因素为 67.8%。方向 2 的平均行驶速度为 48.5 英里/小时，峰值小时因素为 51.0%。

From Table 16.4, Direction 1 operates at LOS D (based upon PTSF), while Direction 2 operates at LOS C (based upon both ATS and PTSF).
根据表 16.4，方向 1 的服务水平为 D（基于峰值小时因素），而方向 2 的服务水平为 C（基于平均行驶速度和峰值小时因素）。

Analysis of Growth
增长分析

This analysis is best done approximately, and in converted units.
此分析最好以近似值和换算后的单位进行。

The nominal capacity of a 2-lane highway under base conditions is 3,200 pc/h in both directions, or 1,700 in one direction. In this case, we have a 60-40 directional split. Thus, if the 60% direction is limited to 1,700 pc/h, then total 2-way flow is limited to 1700/0.60 = 2,833 pc/h. If the 40% direction is at 1,700 pc/h, then the 2- way flow is limited to 1700/0.40 = 4,250 pc/h > 3,200 pc/h, which is not possible. Obviously, the 60% direction controls, and the 2-way capacity of this facility is 2,823 pc/h. This must be compared to the demand flow rate.
双向两车道高速公路的基本通行能力为每小时 3200 辆次，单向为 1700 辆次。在这种情况下，方向分配比例为 60 比 40。因此，如果 60% 的方向限速为 1700 辆次/小时，则双向总流量限制为 1700/0.60 = 2833 辆次/小时。如果 40% 的方向限速为 1700 辆次/小时，则双向流量限制为 1700/0.40 = 4250 辆次/小时 > 3200 辆次/小时，这是不可能的。显然，60% 的方向控制，该设施的双向通行能力为 2833 辆次/小时。这必须与需求流量进行比较。

The problem is that the demand flow rates were converted from demand volumes using adjustment factors which used directional and opposing flow rates based upon existing operations, NOT capacity operations. Further, there were two sets of conversions: one for ATS, one for PTSF. The conversion for ATS resulted in higher converted flow rates, so that would be the basis of the analysis. Going back to the original tables, and assuming that each of the directional demand flow rates would be higher than 900 pc/h, values of fG for both
问题在于，需求流量是使用调整系数将需求体积转换为的，该调整系数使用了基于现有运营，而不是产能运营的定向和相反的流量。此外，存在两组转换：一组用于 ATS，一组用于 PTSF。ATS 的转换导致更高的转换流量，因此这将是分析的基础。回到原始表格，并假设每个方向的需求流量都将高于 900 pc/小时，则两个方向的 fG 值都将为 1.00（表 16.7）。ET 和 ER 的值分别为 1.3 和 1.1，适用于两个方向（表 16.10）。

4

5

directions would be 1.00 (Table 16.7). Values of ET and ER would be 1.3 and 1.1 respectively for both directions (Table 16.10).
两个方向的值分别为 1.00（表 16.7）。ET 和 ER 的值分别为 1.3 和 1.1，适用于两个方向（表 16.10）。

Then:
然后：

v = 299 + 199 = 498 pc/ h
v = 299 + 199 = 498 pc/小时

The question now is: At a growth rate of 8% per year, how long does it take for the current 2-way demand to reach the capacity of 2,823:
现在的问题是：以每年 8% 的增长率，当前双向需求达到 2,823 的产能需要多长时间？

2823 = 498 *1.08n n = 22 − 23 years
2823 = 498 * 1.08n n = 22 - 23 年

Problem 16-2
问题 16-2

Characteristics of the segment (given):
该段的特征（已知）：

Class I highway
一级公路

Segment length = 10 mi
段长度 = 10 英里

Level terrain
平坦路段

BFFS = 65 mi/h
BFFS = 65 英里/小时

12-ft lanes, 6-ft clear shoulders
12 英尺车道，6 英尺净路肩

15 access pts/mi
每英里 15 个出入点

25% “No Passing” zones 8% trucks, no RVs
25%“禁止超车”区域，8%卡车，禁止房车

Current demand = 800 veh/h 70-30 directional distribution PHF = 0.83
当前需求 = 800 车/小时，70-30 方向分布，PHF = 0.83

The free-flow speed (FFS) of the segment maybe estimated using Equation 16-16: FFS = BFFS − fLS − fA
路段的自由流速度 (FFS) 可以使用公式 16-16 估算：FFS = BFFS − fLS − fA

5

6

where: BFFS = 65 mi/h (given)
其中：BFFS = 65 英里/小时（已知）

fLS = 0.0 mi/h (Table 16.5, 12-ft lanes, 6-ft shoulders)
fLS = 0.0 英里/小时（表 16.5，12 英尺车道，6 英尺路肩）

fA = 3.75 mi/h (Table 16.6, 15 access points/mi,
fA = 3.75 英里/小时（表 16.6，每英里 15 个出入点，

interpolated)
插值）

FFS = 65.0 − 0.0 − 3.75 = 61.25, SAY 61.3 mi / h
FFS = 65.0 − 0.0 − 3.75 = 61.25，约 61.3 英里/小时

Demand volumes must now be converted to demand flow rates under equivalent base conditions, using Equation 16-7.
现在必须将需求量转换为等效基准条件下的需求流率，使用公式 16-7。

There are four conversions to be made: the flow rate in each direction, once each for ATS computations, and once each for PTSF computations.
需要进行四次转换：每个方向的流量，一次用于 ATS 计算，一次用于 PTSF 计算。

V1 = 800*0.70 = 560 veh/h
V1 = 800 * 0.70 = 560 辆/小时

V2 = 800*0.30 = 240 veh/h
V2 = 800 * 0.30 = 240 辆/小时

The grade adjustment factors (fG) are found in Table 16.6 (for both ATS and PTSF determinations). The heavy vehicle adjustment factor (fHV) is based upon passenger car equivalents for trucks. The equivalents (for both ATS and PTSF) determination are found in Table 16.10. The heavy vehicle adjustment factor is computed using Equation 16-8:
坡度调整系数 (fG) 在表 16.6 中找到（用于 ATS 和 PTSF 的确定）。重型车辆调整系数 (fHV) 基于卡车的乘用车当量。等效值（用于 ATS 和 PTSF）的确定在表 16.10 中找到。重型车辆调整系数使用公式 16-8 计算：

The results of these look-ups and computations are summarized in the table below
这些查找和计算的结果总结在下表中:

Item
项目

Direction 1
方向 1

Direction 2
方向 2

ATS

PTSF

ATS

PTSF

V (veh/h)
V（辆/小时）

560

560

240

240

fG (Table 16.7)
fG（表 16.7）

1.00

1.00

1.00

1.00

ET

(Table 16.10)
（表 16.10）

1.1

1.0

1.5

1.1

fHV (Eqn 16-8)
fHV（公式 16-8）

0.992

1.000

0.962

0.992

v (Eqn 16.7)
v（公式 16.7）

680

675

301

291

6

7

ATS is estimated using Equation 16-10: ATSd = FFS − 0.00776 (vd + vo ) − fnpA
ATS 的估算使用公式 16-10：ATSd = FFS − 0.00776 (vd + vo ) − fnpA

Where:
其中：

FFS

vd+vo fnpA1

= = =

61.3 mi/h (previously computed) 680 + 301 = 981 pc/h
61.3 英里/小时（先前计算）680 + 301 = 981 pc/h

1.9 mi/h (Table 16.15, FFS=61.3, vo=301, multiple interpolation.)
1.9 英里/小时（表 16.15，FFS=61.3，vo=301，多次插值。）

fnpA2

=

1.0 mi/h (Table 16.15, FFS=61.3, vo=680, multiple interpolation.)
1.0 英里/小时（表 16.15，FFS=61.3，vo=680，多次插值。）

ATS1 = 61.3 − (0.00776 * 981) −1.9 = 61.3 − 7.6 −1.9 = 51.8 mi / h ATS2 = 61.3 − 7.6 −1.0 = 52.7 mi / h
ATS1 = 61.3 − (0.00776 * 981) −1.9 = 61.3 − 7.6 −1.9 = 51.8 英里/小时 ATS2 = 61.3 − 7.6 −1.0 = 52.7 英里/小时

PTSF is estimated using Equations 16-11:
使用公式 16-11 估算 PTSF：

BPTSFd = 100 [1 − exp (a v )]

Where: a1 = -0.0018 (Table 16.17, vo=291, interpolated)
式中：a1 = -0.0018（表 16.17，vo=291，插值）

b1 = 0.951 (Table 16.17, vo=291, interpolated)
b1 = 0.951（表 16.17，vo=291，插值）

a2 = -0.0038 (Table 16.17, vo=675, interpolated)
a2 = -0.0038（表 16.17，vo=675，插值）

b2 = 0.856 (Table 16.17, vo=675, interpolated)
b2 = 0.856（表 16.17，vo=675，插值）

fnpP = 22.2% (Table 16.16, v=966, 70/30 directional split, 25%
fnpP = 22.2%（表 16.16，v=966，70/30 方向分割，25%

NPZ,multiple iteration.)
NPZ 多次迭代）

V1 = 560

V2 = 240

Direction 1 has an ATS of 51.8 mi/h and a PTSF of 74.1%. Direction 2 has an ATS of 52.7 mi/h and a PTSF of 45.3%
方向 1 的 ATS 为 51.8 英里/小时，PTSF 为 74.1%。方向 2 的 ATS 为 52.7 英里/小时，PTSF 为 45.3%

7

8

From Table 16.4, Direction 1 has LOS B based on ATS and LOS D based on PTSF. LOS D controls. Direction 2 has LOS B based on ATS and LOS B based on PTSF. LOS B prevails.
根据表 16.4，方向 1 根据 ATS 为 LOS B，根据 PTSF 为 LOS D。LOS D 控制。方向 2 根据 ATS 为 LOS B，根据 PTSF 为 LOS B。LOS B 占主导。

Problem 16-3
问题 16-3

Segment characteristics (given):
路段特征（已知）：

Class I highway BFFS = 55 mi/h
一级公路 BFFS = 55 英里/小时

11-ft lanes; 2-ft shoulders
11 英尺车道；2 英尺路肩

5 access points/mi
每英里 5 个入口

80% “No Passing” zones 5% grade, 3 miles long
80%“禁止超车”区域 5%坡度，长 3 英里

70-30 directional split (70% up) 20% trucks; 5% RVs
70-30 方向分流（70% 上行）20% 卡车；5% RV

PHF = 0.85

V = 250 veh/h
V = 250 辆车/小时

The free-flow speed (FFS) is estimated using Equation 16-6: FFS = BFFS − fLC − fA = 55 − 3.0 −1.25 = 50.8 mi / h
自由流速度 (FFS) 使用方程式 16-6 进行估计：FFS = BFFS − fLC − fA = 55 − 3.0 −1.25 = 50.8 英里/小时

The demand volumes must be adjusted to reflect demand flow rates under equivalent base conditions. Equation 16-7 is used. This problem concerns a specific grade, which affects how adjustment factors and passenger car equivalents are determined.
必须调整需求量以反映在等效基准条件下的需求流量。使用方程式 16-7。这个问题涉及特定坡度，它会影响调整因子和乘用车当量是如何确定的。

Once again, four different demand flow rates must be computed: two for each direction of flow (ATS determination; PTSF determination).
再次，必须计算四个不同的需求流量：每个方向流量两个（ATS 确定；PTSF 确定）。

Vup = 250 * 0.70 = 175 veh / h Vdn = 250 * 0.30 = 75 veh / h
Vup = 250 * 0.70 = 175 车/小时 Vdn = 250 * 0.30 = 75 车/小时

8

9

Adjustments for Vup
Vup 调整 :

fG ATS =

0.61 (Table 16.8, 5% grade, 3 mi, v=175/0.85 = 206)
0.61（表 16.8，5%坡度，3 英里，v=175/0.85 = 206）

ET ATS =

13.7 (Table 16.11, 5% grade, 3 mi, v=175/0.85 = 206)
13.7（表 16.11，5%坡度，3 英里，v=175/0.85 = 206）

ER ATS =

1.5 (Table 16.12, 5% grade, 3 mi, v=175/0.85 = 206)
1.5（表 16.12，5%坡度，3 英里，v=175/0.85 = 206）

fG PTSF = 1.00 (Table 16.9, 5% grade, 3 mi, v=175/0.85 = 206)
fG PTSF = 1.00（表 16.9，5%坡度，3 英里，v=175/0.85 = 206）

ET PTSF = 2.2 (Table 16.13, 5% grade, 3mi, v=175/0.85 = 206)
ET PTSF = 2.2（表 16.13，5%坡度，3 英里，v=175/0.85 = 206）

ER PTSF = 1.0 (Table 16.13, ALL)
ER PTSF = 1.0（表 16.13，全部）

Adjustments for Vdn
Vdn 的调整:

fG ATS = 1.00 (Table 16.7, v=75/0.85 = 88)
fG ATS = 1.00 (表 16.7，v=75/0.85 = 88)

ET ATS = 1.9 (Table 16.10, v=75/0.85 = 88)
ET ATS = 1.9 (表 16.10，v=75/0.85 = 88)

ER ATS = 1.0 (Table 16.10, ALL)
ER ATS = 1.0 (表 16.10，全部)

fG PTSF = 1.00 (Table 16.7, v=75/0.85 = 88)
fG PTSF = 1.00 (表 16.7，v=75/0.85 = 88)

ET PTSF = 1.1 (Table 16.10, v=75/0.85 = 88)
ET PTSF = 1.1 (表 16.10，v=75/0.85 = 88)

ER PTSF = 1.0 (Table 16.10, ALL)
ER PTSF = 1.0 (表 16.10，全部)

9

10

Summarizing these results:
总结这些结果：

v

ATS

PTSF

Upgrade
升级

1201

255

Downgrade
降级

104

90

Total
总计

1,305

345

The average travel speed is estimated using Equation 16-10: ATSd = FFS − 0.00776 (vd + vo ) − fnpA
平均行驶速度采用公式 16-10 估算：ATSd = FFS − 0.00776 (vd + vo ) − fnpA

Where:
其中：

FFS =

50.8 mi/h (previously computed)
50.8 英里/小时（先前计算）

vd+vo =

1,305 veh/h
1305 车/小时

fnpA up = 2.5 mi/h (Table 16.15, vo=104, 80% NPZ, 50.8 mi/h)
fnpA 上行 = 2.5 英里/小时（表 16.15，vo=104，80% NPZ，50.8 英里/小时）

fnpB dn = 0.8 mi/h (Table 16.15, vo =1201, 80% NPZ, 50.8 mi/h)
fnpB 下行 = 0.8 英里/小时（表 16.15，vo =1201，80% NPZ，50.8 英里/小时）

ATSup = 50.8 − (0.00776 *1305) − 2.5 = 50.8 −10.1 − 2.5 = 38.2 mi / h ATSdn = 50.8 −10.1 − 0.9 = 39.8 mi / h
ATS 上行 = 50.8 − (0.00776 *1305) − 2.5 = 50.8 −10.1 − 2.5 = 38.2 英里/小时 ATS 下行 = 50.8 −10.1 − 0.9 = 39.8 英里/小时

The percent time spent following is estimated using Equations 16-11:
跟车行驶所占时间的百分比采用公式 16-11 估算：

BPTSFd = 100 [1 − exp (a v )]

Where: aup = -0.0014 (Table 16.17, vo=90)
其中：aup = -0.0014 (表 16.17，vo=90)

bup = 0.973 (Table 16.17, vo=90)
bup = 0.973 (表 16.17，vo=90)

adn = -0.0016 (Table 16.17, vo=255, interpolated)
adn = -0.0016 (表 16.17，vo=255，插值)

bdn = 0.959 (Table 16.17, vo=255, interpolated)
bdn = 0.959 (表 16.17，vo=255，插值)

fnpP = 44.9% (Table 16.16, v=345, 70-30 dir split, 80%NPZ)
fnpP = 44.9% (表 16.16，v=345，70-30 方向分割，80%NPZ)

10

11

Direction 1 has an ATS of 38.2 mi/h and a PTSF of 57.9%. Direction 2 has an ATS of 39.8 mi/h and a PTSF of 24.8%.
第 1 个方向 ATS 为 38.2 英里/小时，PTSF 为 57.9%。第 2 个方向 ATS 为 39.8 英里/小时，PTSF 为 24.8%。

This is LOS E for both directions, based upon speed, and LOS C (Dir 1) and LOS A (Dir 2) based upon PTSF. (Table 16.4). LOS E prevails.
这是两个方向的 LOS E，基于速度，以及基于 PTSF 的 LOS C（方向 1）和 LOS A（方向 2）。（表 16.4）。LOS E 占主导地位。

This example shows the great deleterious effect of severe alignment on the operation of 2-lane highways. This highway is operating at LOS E with only 250 veh/h total demand. Given the obviously severe terrain and low demand volume, it might be more reasonable to classify this highway as Class II. In that case, the ATS values would not determine LOS, and the better LOS ratings determined by PTSF would prevail. This should not be done in isolation, but rather as a review of the classification of all 2-lane highways.
此示例显示了严重对齐对 2 车道高速公路运行的巨大不利影响。这条高速公路仅以 250 辆/小时的总需求运行在 LOS E。鉴于明显严重的地理地形和低需求量，将这条高速公路归类为 II 类可能更合理。在这种情况下，ATS 值将不会确定 LOS，并且由 PTSF 决定的更好 LOS 等级将占主导地位。这不应该孤立地进行，而应作为审查所有 2 车道高速公路的分类。

Problem 16-4
问题 16-4

Segment characteristics (given):
路段特征（已知）：

Class III highway BFFS = 45 mi/h
三类公路，基本自由流速度（BFFS）= 45 英里/小时

11-ft lanes, 4-ft shoulders 100% “No Passing” zones 20 access pts/mi
11 英尺车道，4 英尺路肩，100%“禁止超车”路段，20 个出入点/英里

10% trucks, 3% RVs PHF = 0.81
10%卡车，3%房车，小时峰值因子（PHF）= 0.81

60-40 directional distribution Rolling terrain
60-40 方向分布，起伏地形

Demand = 500 veh/h
交通量 = 500 车/小时

The free-flow speed (FFS) is estimated using Equation 16-6: FFS = BFFS − fLS − fA = 45 −1.7 − 5.0 = 38.3 mi / h
自由流速度（FFS）使用公式 16-6 估算：FFS = BFFS − fLS − fA = 45 −1.7 − 5.0 = 38.3 英里/小时

This is a Class III highway in a developed rural area. As such, the LOS is based upon the percent of free-flow speed maintained (PFFS), which is, in turn, based upon the ATS. Because of this, there is no need to predict the PTSF, as it is not used in estimating the LOS. Thus, only two sets of conversions are needed, one for each direction, both based on ATS.
这是一条位于已开发农村地区的 III 级公路。因此，服务水平 (LOS) 基于保持的自由流速百分比 (PFFS)，而 PFFS 又基于平均行程速度 (ATS)。因此，无需预测峰时行程速度 (PTSF)，因为它不用于估算服务水平 (LOS)。因此，只需要两组换算，每方向一组，均基于平均行程速度 (ATS)。

11

12

V1 = V2 =

500*0.6 = 500*0.4 =

300 veh/h
300 辆车/小时

200 veh/h
200 辆车/小时

Conversions are made using Equation 16-7:
使用公式 16-7 进行转换：

Where:
其中：

PHF = fG1 ATS = fG2 ATS = ET1 ATS = ET2 ATS = ER1 ATS = ER2 ATS =

0.81 (given)
0.81 (已知)

0.88 (Table 16.7, v=300/0.81=370, interpolated) 0.79 (Table 16.7, v=200/0.81=247, interpolated) 2.0 (Table 16.10, v=370)
0.88 (表 16.7，v=300/0.81=370，插值) 0.79 (表 16.7，v=200/0.81=247，插值) 2.0 (表 16.10，v=370)

2.2 (Table 16.10, v=247, interpolated)
2.2 (表 16.10，v=247，插值)

1.1 (Table 16.10, ALL)
1.1 (表 16.10，全部)

1.1 (Table 16.10, ALL)
1.1 (表 16.10，全部)

The average travel speed may now be estimated using Equation 16-10: ATS = FFS − 0.00776 (vo + vd ) − fnpA
平均行驶速度现在可以使用公式 16-10 进行估计：ATS = FFS − 0.00776 (vo + vd ) − fnpA

Where:
其中：

FFS

vo+vd fnpA1

= = =

38.3 mi/h (previously computed) 464 + 351 = 815 pc/h
38.3 英里/小时（先前计算）464 + 351 = 815 英尺/小时

3.0 mi/h (Table 16.15, 100% NPZ, 38.3 mi/h, vo=351, interpolated)
3.0 英里/小时（表 16.15，100% NPZ，38.3 英里/小时，vo=351，插值）

fnpA2

=

2.4 mi/h (Table 16.15, 100% NPZ, 38.3 mi/h, vo=464, interpolated)
2.4 英里/小时（表 16.15，100% NPZ，38.3 英里/小时，vo=464，插值）

12

13

ATS1 = 38.3 − (0.00776 * 815) − 3.0 = 38.3 − 6.3 − 3.0 = 29.0 mi / h ATS2 = 38.3 − (0.00776 * 815) − 3.0 = 38.3 − 6.3 − 2.4 = 29.6 mi / h
ATS1 = 38.3 − (0.00776 * 815) − 3.0 = 38.3 − 6.3 − 3.0 = 29.0 英里/小时 ATS2 = 38.3 − (0.00776 * 815) − 2.4 = 38.3 − 6.3 − 2.4 = 29.6 英里/小时

PFFS1 = 29.0 / 38.3 = 0.757 PFFS2 = 29.6 / 38.3 = 0.773

From Table 16.4, both directions are experiencing LOS C.
根据表 16.4，两个方向都在经历 C 级交通拥堵。

Problem 16-5
问题 16-5

This problem relates to the result of Problem 16-1, restated here for ease of use:
本问题与问题 16-1 的结果相关，为了方便使用，在此重述：

Item
项目

Direction 1
方向 1

Direction 2
方向 2

ATS

PTSF

ATS

PTSF

v

407 pc/h
407 台/小时

376 pc/h
376 台/小时

308 pc/h
308 台/小时

270 pc/h
270 台/小时

ATS

48.2 mi/h
48.2 英里/小时

48.5 mi/h
48.5 英里/小时

PTSF

67.8%

51.0%

The highway in question is a 20-mile segment. A 3-mile passing lane (in each direction) is to be added, beginning 3 miles from the beginning of the segment. Using Figure 16-5 as an example, the following distances can be established:
相关高速公路路段长 20 英里。计划在该路段起点 3 英里处增设 3 英里长的超车道（每个方向各一条）。参照图 16-5，可以确定以下距离：

Direction
方向

Lu
路

Lpl

Lde (Tab 16.21)
Lde（选项卡 16.21）

Ld

ATS

PTSF

ATS

PTSF

1

3 mi
3 英里

3 mi
3 英里

1.7 mi
1.7 英里

8.7 mi
8.7 英里

12.3 mi
12.3 英里

5.3 mi
5.3 英里

2

3 mi
3 英里

3 mi
3 英里

1.7 mi
1.7 英里

11.3 mi
11.3 英里

12.3 mi
12.3 英里

2.7 mi
2.7 英里

Then:
然后：

Where: fpl (both directions) = 1.10 (Table 16.22, v>300≤600)
位置：fpl（双向）= 1.10（表 16.22，v>300≤600）

13

14

Then:
然后：

The adjusted PTSF is estimated using Equation 16-13:
使用公式 16-13 估算调整后的 PTSF：

Where: fpl1 = 0.61 (Table 16.22, v=376 pc/h)
位置：fpl1 = 0.61（表 16.22，v=376 pc/h）

fpl2 = 0.58 (Table 16.22, v=207 pc/h)
fpl2 = 0.58（表 16.22，v=207 pc/h）

Then:
然后：

The effect of the passing lane on ATS is not significant, with speeds increasing about 1 mi/h. The impact on PTSF, however, is much more significant, declining almost 10% in both directions. The LOS based upon speed is still LOS C for both directions. For Direction 1, the LOS based upon PTSF is now C (up from D); for Direction 2, the LOS based upon PTSF is now B (up from C). The prevailing LOS in both directions is now C.
超车道对 ATS 的影响不显著，车速大约提高了 1 英里/小时。然而，对 PTSF 的影响要大得多，双向下降近 10%。基于速度的 LOS 仍然是双向 LOS C。对于方向 1，基于 PTSF 的 LOS 现在为 C（从 D 上升）；对于方向 2，基于 PTSF 的 LOS 现在为 B（从 C 上升）。双向的主要 LOS 现在为 C。

超车道对 ATS 的影响不显著，车速仅提高约 1 英里/小时。然而，对 PTSF 的影响则大得多 ,双向下降近 10%。基于速度的服务水平仍然是双向 C 级。对于方向 1，基于 PTSF 的服 务水平现在是 C 级（从 D 级提升）；对于方向 2，基于 PTSF 的服务水平现在是 B 级（从 C 级提升）。双向的主要服务水平现在都是 C 级。
超车道对 ATS 的影响不显著，车速仅提高约 1 英里/小时。然而，对 PTSF 的影响则大得多，双向下降近 10%。基于速度的服务水平仍然是双向 C 级。对于方向 1，基于 PTSF 的服务水平现在是 C 级（从 D 级提升）；对于方向 2，基于 PTSF 的服务水平现在是 B 级（从 C 级提升）。双向的主要服务水平现在都是 C 级。

14

© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained
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(b)

15

Problem 16-6
问题 16-6

This problem deals with adding a climbing lane to the upgrade of Problem 16-3. The results of that problem (for the upgrade only) are summarized below for convenience:
本问题研究在 问题 16-3 的升级中增加爬坡车道。为了方便起见，以下总结了该问题的结果（仅限升级）：

vu ATS = 1,201 pc/ h vu PTSF = 255 pc/ h ATSu = 38.2 mi / h
vu ATS = 1,201 辆/小时 vu PTSF = 255 辆/小时 ATSu = 38.2 英里/小时

PTSF = 57.9% u
PTSF = 57.9% u

The method for assessing the impact of climbing lanes uses the same equations (16-13, 16-14) as passing lanes, except that:
评估爬坡车道影响的方法使用与超车车道相同的方程式 (16-13、16-14)，不同之处在于：

Lu =Lde = Ld = 0 mi. Lpl = L
Lu =Lde = Ld = 0 英里。Lpl = L

fpl is taken from Table 16.23 instead of 16.22.
fpl 取自表 16.23，而不是 16.22。

Then: fpl ATS =1.14; fpl PTSF = 0.20
然后：fpl ATS =1.14；fpl PTSF = 0.20

And:
和：

The climbing lane will increase the ATS to 43.5 mi/h, which is an increase of 5.3 mi/h, and will decrease the PTSF to 11.6%, a decrease of 46.3%. These are both significant impacts, particularly for passing delays. The upgrade LOS is now LOS D (up from E) for ATS and LOS A (up from C) for PTSF. The prevailing LOS is D (up from E).
攀爬车道将 ATS 提高到每小时 43.5 英里，比原来增加了 5.3 英里/小时，并将 PTSF 降低到 11.6%，下降了 46.3%。这些都是重大影响，尤其会影响超车延误。升级后的服务水平现在是 D 等级（从 E 等级提升），对于 ATS 来说是 D 等级（从 E 等级提升），对于 PTSF 来说是 A 等级（从 C 等级提升）。普遍的服务水平是 D 等级（从 E 等级提升）。

15

1

Traffic Engineering, 4th Edition
《交通工程》，第四版

Roess, R.P., Prassas, E.S., and McShane, W.R.
Roess，R.P.，Prassas，E.S.和 McShane，W.R.

Solutions to Problems in Chapter 18
第 18 章问题解决方案

Problem 18-1
问题 18-1

The solution involves evaluating the four sight triangles of this intersection. As two of the sight triangles are not obstructed, this problem hinges on the corners occupied by the Wheat Field and Barn. Based upon the relative volumes, the NB and SB approaches will be considered to be the minor approaches.
该解决方案涉及评估该路口的四个视距三角形。由于两个视距三角形没有被阻挡，这个问题的关键在于小麦田和谷仓所占据的角落。根据相对的交通量，NB 和 SB 方向将被视为次要方向。

Beginning with the NB approach (Vehicle A will be on this approach), the position of Vehicle A is assumed to be at its safe stopping distance, or:
从 NB 方向开始（车辆 A 将位于该方向），假设车辆 A 的位置在其安全停车距离内：

Using the geometry of the sight triangle, the actual position of Vehicle B (db) when the two drivers first see each other can be determined:
利用视距三角形的几何形状，可以确定两名驾驶员第一次互相看到对方时，车辆 B（db）的实际位置：

In order for the operation to be deemed safe, one of two conditions for db must be less than this value. The first value is based upon the safe stopping distance for Vehicle B, while the second allows the two vehicles to pass the collision point without colliding. Then:
为了使该操作被认为是安全的，db 的两个条件之一必须小于此值。第一个值基于车辆 B 的安全停车距离，而第二个值允许两辆车通过碰撞点而不会发生碰撞。然后：

Neither of these are less than the actual distance of 39.7 ft. Therefore, the intersection cannot be safely operated under basic rules of the road. An analysis of the other obstructed sight triangle could be done, but it is not necessary. As
这两个距离均不小于实际距离 39.7 英尺。因此，根据基本的道路规则，该交叉路口无法安全通行。可以对另一个受阻视线三角形进行分析，但这并非必要。

1

2

long as one sight triangle is unsafe, rules of the road cannot be used as the only form of control.
只要有一个视线三角形不安全，就不能仅依靠道路规则作为控制方式。

Problem 18-2
问题 18-2

In this intersection, there are only two sight triangles, both of which are obstructed. The one that appears to be most restricted is to the left of the vehicle approaching on the one-way street. It will be analyzed first.
在这个路口，只有两个视线三角形，而且都被遮挡了。其中看起来受限程度最大的是位于单行道来车左侧的那个，我们将首先分析它。

The vehicle on the one-way street is Vehicle A. It will be placed one safe- stopping distance from the collision point:
单行道上的车辆为 A 车。我们将 A 车放置在与碰撞点安全停车距离之外：

From the sight triangle, the actual distance of Vehicle B from the collision point when both drivers can see each other is:
从视线三角形来看，当两名驾驶员都能互相看到对方时，B 车与碰撞点的实际距离为：

This must now be compared to the two minimum conditions, as in Problem 18-1 previously:
这现在必须与先前问题 18-1 中的两个最小条件进行比较：

Neither rule for safe operation is met. Therefore, operation under basic rules of the road is not safe. In most cases, a STOP-sign would be recommended.
两种安全操作规则都不满足。因此，按照道路基本规则进行操作是不安全的。在大多数情况下，建议设置停止标志。

As in Problem 18-1, the second sight triangle could be analyzed, but it is not necessary, as the intersection has already failed the safety test for one sight triangle.
与问题 18-1 一样，可以分析第二个视距三角形，但这不是必要的，因为该交叉口已经通过了一个视距三角形的安全测试。

2

3

Problem 18-3
问题 18-3

In this case, the visibility for SB vehicles is not obstructed. Sight distances for NB vehicles approaching the STOP-sign, however, should be checked.
在这种情况下，南向车辆的能见度没有受到阻碍。但是，应该检查北向车辆接近停止标志时的视距。

The distance of Vehicle A from the collision point is found as: dA = 18+ dci
从碰撞点到车辆 A 的距离计算为：dA = 18 + dci

This assumes that the vehicle is stopped at location where the driver’s eye is 18 ft from the curb line. Distance dci is the distance to the center of the nearest conflicting traffic lane. With 14-ft lanes, this is 7 ft for the Vehicle B and 21 ft for Vehicle C. Then:
这假设车辆停在驾驶员眼睛距离路缘线 18 英尺的地方。dci 是到最近冲突车道中心线的距离。如果车道宽 14 英尺，则车辆 B 为 7 英尺，车辆 C 为 21 英尺。然后：

dA (for Veh B) = 18+ 7 = 25 ft dA (for Veh C) = 18+ 21 = 39 ft
dA（对于车辆 B）= 18 + 7 = 25 英尺 dA（对于车辆 C）= 18 + 21 = 39 英尺

The minimum required sight distance for Veh A and B is given by: db (min) = 1.47 Smajtg = 1.47*35*7.5 = 385.9 ft
车辆 A 和 B 的最小所需视距由以下公式给出：db(min) = 1.47 × Smajtg = 1.47 × 35 × 7.5 = 385.9 英尺

This is based upon gap acceptance criteria. The actual visibility distances of both Vehicle A and Vehicle B are computed from the sight triangle. These must be
这基于间隙接受标准。车辆 A 和车辆 B 的实际能见距离是从视三角形计算出来的。这些必须

more than 385.9 ft for safety.
超过 385.9 英尺，以确保安全。

Neither sight distance is safe. There are very few alternatives, however, unless signals are warranted. Sight obstructions might be trimmed back if they are not permanent structures, and/or speed limits reduced on the main street. In practice, drivers will inch forward until they can see approaching vehicles before proceedings.
两个能见距离都不安全。然而，除非有必要的信号，否则可供选择的方案很少。如果视野障碍物不是永久性结构，则可以将其修剪掉，并且/或者可以降低主街道的速度限制。实际上，司机会在能够看到迎面车辆之前，慢慢地向前移动，然后再继续行驶。

3

4

Problem 18-4
问题 18-4

Because the approach speeds are in excess of 40 mi/h, the 70% criteria will be applied for appropriate warrants. Note that the volumes shown are minima over a 10-hour period. Thus, they represent the lowest volume within the 10-hour period. If volumes meet a particular warrant, they will meet it for 10 hours. If they do not meet a particular warrant, it is possible that within the 10 hours, there are hours that do meet the warrant. The information, however, is insufficient to determine this.
由于方法速度超过 40 英里/小时，因此将应用 70% 的标准以获得适当的许可。请注意，显示的交通量是在 10 小时内测得的最小值。因此，它们代表 10 小时内最低的交通量。如果交通量满足特定条件，则它们将在 10 小时内都满足该条件。如果它们不满足特定条件，则可能在 10 小时内存在满足该条件的小时。但是，信息不足以确定这一点。

Warrant 1 Condition A (at 70%) requires 350 veh/h on the major approach, both directions, and 105 veh/h on the highest-volume minor approach. Condition B requirements are 525 veh/h and 53 veh/h respectively. Using the highest volume street as the “major” street (N-S), the actual values are 700 veh/h and 350 veh/h respectively. Both Condition A and B are met. The warrant is met.
许可条件 1 条件 A（70%）要求主要方法（双向）上为 350 车/小时，最高交通量次要方法上为 105 车/小时。条件 B 的要求分别为 525 车/小时和 53 车/小时。使用交通量最大的街道作为“主要”街道（南北向），实际值为 700 车/小时和 350 车/小时。条件 A 和 B 均满足。满足许可条件。

Warrant 2 The figure below shows the 4-hour vehicular warrant (at 70%), with [700,350] point plotted.
许可条件 2 下图显示了 4 小时车辆许可条件（70%），并标注了 [700,350] 点。

The warrant is easily met.
许可条件很容易满足。

Warrant 3 Warrant 3 has two criteria – peak hour delay, and peak hour volume. The problem does not give delay data, so this portion of the warrant cannot be evaluated. The volume warrant (at 70%) is evaluated on the figure that follows:
许可条件 3 许可条件 3 有两个标准——高峰小时延误和高峰小时交通量。问题没有提供延误数据，因此无法评估许可条件的这一部分。交通量许可条件（70%）在接下来的图中进行评估：

4

5

The volume portion of the warrant is met. The warrant is met.
许可条件的交通量部分已满足。满足许可条件。

Warrant 4 This warrant (at 70%) evaluates pedestrian crossings. As the N-S street is the “major” street in this analysis, there are a minimum of 90+70 = 160 peds/h crossing the major street for 10 hours. There are two criteria – one for four hours, one for peak hour only. Both are shown in the figures that follow.
依据 4 本依据（70%）评估人行横道。由于南北向街道在此分析中是“主要”街道，因此在 10 小时内，至少有 90+70 = 160 名行人/小时穿越主要街道。有两个标准——一个为四个小时，一个仅为高峰小时。两个标准都显示在后面的图中。

The 4-hour pedestrian criterion is met.
满足 4 小时行人通行标准。

5

6

The peak-hour pedestrian criterion is not met. Because the 4-hour pedestrian criterion was met, the warrant is met.
未满足高峰小时行人通行标准。由于满足了 4 小时行人通行标准，因此满足了该依据。

Other Warrants Due to lack of information, no other warrant can be evaluated for this intersection.
其他依据 由于缺乏信息，无法评估该交叉路口的其他依据。

Recommendation Obviously, a signal should be placed here. The minimum volumes (most hours will have more than this) meet all volume-based warrants. Ped signals should be used, but there is not enough information given to recommend a specific form of signalization.
建议 很明显，此处应设置信号灯。最小车流量（大部分小时的车流量都超过此值）满足所有基于车流量的条件。应使用人行信号灯，但提供的资料不足以推荐具体的信号灯形式。

Problem 18-5
问题 18-5

Once again, only vehicular volumes are given. Non-volume-based warrants cannot be evaluated. Only Warrants 1 – 3 maybe evaluated with the information given. Note that neither the population nor the approach speeds engage a reduction in criteria, so warrants must be met at 100% in this case. To assist in making this evaluation, the table should be re-arranged to show total 2-way volume on the major street (N-S) and the highest single-direction volume on the minor street (E-W). The table that follows shows this.
同样，这里只给出了车辆通行量。无法评估非基于车流量的条件。根据所提供的信息，只能评估条件 1-3。请注意，人口和通行速度均未导致标准降低，因此在这种情况下必须满足 100% 的条件。为了帮助进行评估，应重新排列表格，显示主要街道（南北向）上的双向总车流量和次要街道（东西向）上最高的单向车流量。下表显示了这一点。

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7

Table: Volumes for Warrant Analysis
表格：条件分析车流量

Hour
小时

Major Street Vol
主要街道车流量

(2-Way)
（双向）

Minor Street Vol
次要街道车流量

(High Dir)
（高方向）

1

50

30

2

100

30

3

175

50

4

300

50

5

450

100

6

700

250

7

850

400

8

850

450

9

750

375

10

400

300

11

300

300

12

300

150

13

300

100

14

350

100

15

350

100

16

450

250

17

600

325

18

700

375

19

800

400

20

800

425

21

400

325

22

200

150

23

100

100

24

100

50

Warrant 1 Warrant 1, Condition A requires minimum volumes of 600 veh/h on the major street (2 ways) and 150 veh/h (one way) of the minor street. Condition B requires 900 veh/h and 75 veh/hrespectively
许可证 1 许可证 1，条件 A 要求主要街道（双向）车流量至少为 600 车/小时，次要街道（单向）车流量至少为 150 车/小时。条件 B 要求分别为 900 车/小时和 75 车/小时.

Hours 6, 7, 8, 9, 17, 18, 19, 20 meet Condition A (8 hours). No hours meet
6、7、8、9、17、18、19、20 小时满足条件 A（8 小时）。没有小时满足

Condition B. The warrant is met.
条件 B。满足许可证条件。

Warrant 2 While all 24 hourly points could be plotted against the 4-hour volume criteria, if the top 4 don’t meet the warrant, no other set will meet the
许可证 2 虽然可以将所有 24 个小时的数据点与 4 小时车流量标准进行对比，但如果前 4 个小时不满足许可证要求，则其他任何组合也不会满足

7

8

warrant. Hours 7, 8, 19 and 20 appear to be the worst periods. These eight points are plotted on the figure below: [850,400], [850,450], [800,400], and [800, 425].
许可证要求。7、8、19 和 20 小时似乎是最差的时段。这八个点的图示如下：[850,400]、[850,450]、[800,400] 和 [800,425]。

As all four points are clearly above the decision line, this warrant is met.
由于所有四个点都明显高于决策线，因此满足此要求。

Warrant 3 Warrant 3 has two parts: peak hour delay, and peak hour volume. There is no delay information given, so the first part cannot be evaluated. The second can be evaluated. The highest volume point [850,450] is plotted. If this hour does not meet the criteria, no other hourly volume pair will.
要求 3 要求 3 有两个部分：高峰时段延误和高峰时段交通量。没有给出延误信息，因此无法评估第一部分。第二部分可以评估。最高交通量点[850,450]已绘制。如果此小时不符合标准，则其他任何小时的交通量对也不会符合。

8

9

As the intersection has 2 lanes (each direction) on the major street and 1 lane (in each direction) on the minor street, the middle decision line is applicable. The warrant is met.
由于主要街道上每个方向有 2 条车道，次要街道上每个方向有 1 条车道，因此适用中间决策线。满足要求。

A signal is warranted by all three of the volume criteria. No particular form of signalization is recommended without additional information.
交通量标准中的所有三个标准都证明需要安装信号灯。没有其他信息，不建议任何具体的信号灯形式。

Problem 18-6
问题 18-6

Because of the 45-mi/h speeds on the major street (E-W), the 70% criteria of the volume warrants apply. Because the minor street is a one-way street, the “total” volume is the “highest directional volume.”
由于主要街道（东西向）的速度为 45 英里/小时，因此适用交通量保证的 70％标准。由于次要街道是单行道，“总”交通量即为“最高方向交通量”。

Warrant 1 Condition A requires minimum volumes of [420, 105]. Condition B requires [630, 53].
要求 1 条件 A 要求最小交通量为[420, 105]。条件 B 要求[630, 53]。

Condition A is met by the following hours: 3-4 PM, 4-5 PM, 5-6 PM, and 6-7 PM. This is only 4 hours, while 8 are required. Condition A is not met.
条件 A 由以下时间段满足：下午 3-4 点，下午 4-5 点，下午 5-6 点和下午 6-7 点。这只有 4 个小时，而需要 8 个小时。条件 A 未满足。

Condition B is met by the following hours: all hours between 1 PM and 11 PM. This is 10 hours. Condition B is met.
条件 B 由以下时间段满足：下午 1 点到晚上 11 点之间所有的小时。这是 10 个小时。条件 B 满足。

The warrant is met.
满足要求。

Warrant 2 The highest four-hour volume period is between 3 PM and 7 PM. These four hours are plotted against the 4-hour vehicular volume warrant criteria. As all of these points are off the volume scale on the 70% criteria for Warrant 2, and the minor street volumes are above the minimums required, the warrant is met.
保证 2 最高四个小时的交通量在下午 3 点至 7 点之间。这四个小时的数据将根据 4 小时的车辆通行量保证标准进行绘制。由于所有这些数据点都在保证 2 的 70%标准的交通量刻度之外，并且支路的交通量超过了最低要求，因此满足了保证条件。

Warrant 3 The highest volume hour [1150, 160] or [1200, 135] are plotted
保证 3 最高交通小时[1150, 160]或[1200, 135]被绘制

against the peak hour vehicular volume warrant criteria (70% level).
根据高峰小时车辆通行量保证标准（70%水平）进行绘制。

9

10

As both points lie above the decision line, the warrant is met.
由于这两个点都位于决策线之上，因此满足了保证条件。

The delay portion of Warrant 3 can also be evaluated for vehicles on the STOP- controlled approach. In the peak hour, 160 vehicle experience 72 s/veh of delay for a total of 72*160 = 11,520 veh-sec, or 3.2 hours of aggregate delay. The warrant requires a minimum of 4.0 hours, so this part of Warrant 3 is not met.
警告 3 的延误部分也可以评估停止控制路口的车辆。在高峰时段，160 辆车辆经历了每车 72 秒的延误，总共为 72*160 = 11,520 车辆-秒，或 3.2 小时的总延误。警告要求至少 4.0 小时，因此警告 3 的这一部分未满足。

The warrant is met, as the volume criterion is met.
警告满足，因为满足了流量标准。

Warrant 4 The highest four hours of pedestrian activity occur between 1 and 4 PM, and between 8 and 9 PM. During these four hours, the major street vehicular volume and pedestrian volumes (crossing the major street) are [800, 200], [855, 210], [1025, 205] and [975,200]. These are plotted against the four-hour pedestrian warrant (70% level). The highest period [1025, 205] is plotted against the one-hour pedestrian warrant (70% level).
警告 4 行人活动最集中的四个小时出现在下午 1 点到 4 点之间以及晚上 8 点到 9 点之间。在这四个小时内，主要街道车辆流量和行人流量（穿过主要街道）分别是 [800, 200]、[855, 210]、[1025, 205] 和 [975, 200]。这些数据与四小时行人警告（70% 水平）进行了比较。最高峰值 [1025, 205] 与一小时行人警告（70% 水平）进行了比较。

10

11

As both criteria are met, this warrant is met.
由于两个标准都满足，该警告满足。

Warrant 5 The School Crossing Warrant is not applicable.
警告 5 学校交通警告不适用。

Warrant 6 The Coordinated Signal Warrant is not applicable.
警告 6 协调信号警告不适用。

Warrant 7 The Crash Experience Warrant can be evaluated. All relevant criteria are met: There is STOP-control in place, there are 8 right turn, 3 left turn, and 4 pedestrian accidents that can be corrected by signalization, and Warrant 1B is met at 100%. The warrant is met.
警告 7 事故经验警告可以评估。所有相关标准都满足：有停止控制，有 8 起右转、3 起左转和 4 起可以通过信号灯纠正的行人事故，并且警告 1B 满足 100%。该警告满足。

Warrant 8 The Roadway Network Warrant is not applicable. Warrant 9 The RR Grade Crossing Warrant is not applicable.
保证 8 道路网保证不适用。保证 9 RR 道口保证不适用。

11

12

A signal is clearly warranted at this location. Given the high pedestrian volumes, use of pedestrian signals is suggested for crossing the major street. If an actuated controller is used, a pedestrian actuator should be provided.
此处显然需要安装信号灯。鉴于行人流量大，建议在穿越主要街道时使用行人信号灯。如果使用自动控制装置，则应提供行人启动装置。

Problem 18-7
问题 18-7

The high speeds on Broad Street allow the use of 70% criteria in all vehicular and pedestrian volume warrants.
布劳德街的高速行驶允许在所有车辆和行人流量保证中使用 70%的标准。

Warrant 1 Condition A requires [420, 105] for 8 hours. Condition B requires [630, 53] for 8 hours. Condition A is met by the following hours: 12-1 PM through 10-11 PM. This condition is met. Condition B is met by the following hours: 11 AM through 10 PM. This condition is met. The warrant is met.
保证 1 条件 A 要求在 8 小时内达到[420, 105]。条件 B 要求在 8 小时内达到[630, 53]。条件 A 在以下时间段内满足：下午 12 点到晚上 10 点到 11 点。满足此条件。条件 B 在以下时间段内满足：上午 11 点到晚上 10 点。满足此条件。满足保证条件。

Warrant 2 The four highest volume hours occur between 1 PM and 5 PM, and are as follows: [875, 150], [930, 165], [930, 170], and [870, 160]. These are plotted
证明 2 四个交易量最高的时段出现在下午 1 点到 5 点之间，具体如下：[875, 150]，[930, 165]，[930, 170]和[870, 160]。这些数据已绘制成图。

on the 4-hour volume criteria (70%).
在 4 小时流量标准 (70%) 中。

This warrant is met.
此保证已满足。

Warrant 3 The highest volume hour occurs between 3 and 4 PM [930, 170]. This is plotted against the peak hour volume criteria (70%).
保证 3 最高流量小时出现在下午 3 点至 4 点之间 [930, 170]。这与高峰小时流量标准 (70%) 相比。

12

13

As the middle decision line is used, the volume portion of the peak-hour warrant is met.
由于使用中间决策线，高峰小时保证的流量部分已满足。

In the worst hour, 170 veh/hr are delayed for 45 s/veh by the STOP-sign. This is 170*45 = 7,650 veh-sec or 2.125 veh-hrs of delay. The warrant requires 4 veh-hrs, so this portion of the warrant is not met.
在最差小时，170 辆车/小时因停止标志而延误 45 秒/辆。这是 170*45 = 7,650 车辆秒或 2.125 车辆小时的延误。保证要求 4 车辆小时，因此此保证部分未满足。

The warrant is met.
满足要求。

Warrant 4 The worst four hours of pedestrian-vehicle conflict in crossing the major street occur between 2 and 6 PM, with the following volumes:
保证 4 在穿越主要街道的人行车辆冲突的最差四个小时发生在下午 2 点至 6 点之间，流量如下：

[930, 122], [930, 135], [870, 140], and [780, 125]. These are plotted vs. the 4-hour pedestrian warrant (70%). The highest of these, [930, 135], is plotted vs. the one- hour pedestrian warrant (70%).
[930, 122]，[930, 135] [870, 140] 和 [780, 125]。这些与 4 小时行人保证 (70%) 相比。其中最高值 [930, 135] 与 1 小时行人保证 (70%) 相比。

13

14

Both the 4-hour and peak hour pedestrian criteria are satisfied. This warrant is met.
4 小时和高峰时段的行人标准都满足。该授权满足。

Warrants 5 and 6 These warrants do not apply.
授权 5 和 6 这些授权不适用。

Warrant 7 The three criteria of this warrant are met. There is already a STOP- sign in place; there are 14 accidents that are susceptible to correction by a signal, and Warrants 1, 2, and 3 are met. This warrant is met.
授权 7 该授权的三个标准都满足。现已设置了停止标志；有 14 起事故可通过信号灯纠正，并且授权 1、2 和 3 都已满足。该授权满足。

Because of the pedestrian accidents, it is imperative that pedestrian signals be used at both crosswalks crossing the major street. If the signal is pretimed, the timing should accommodate pedestrians in every cycle. If the signal is actuated, a pedestrian actuator must be used.
由于行人事故，必须在穿过主干道的两条人行横道上使用行人信号灯。如果信号是预设的，则计时应在每个周期内适应行人。如果信号是活动的，则必须使用行人传感器。

Problem 18-8
问题 18-8

The first part of the solution is to determine the “equivalent” volume crossing the tracks. There are no buses, but there are tractor-trailers. There is also an adjustment for train frequency. For 20 trains per day, from Table 18.11, an adjustment of 1.33 is applied. For 20% tractor-trailers, from Table 18.13, an adjustment of 1.35 is applied. Thus, the equivalent volume crossing the tracks is
解决方案的第一部分是确定穿过铁轨的“等效”流量。没有公共汽车，但有拖车。此外，还有列车频率的调整。对于每天 20 列火车，根据表 18.11，应用 1.33 的调整。对于 20% 的拖车，根据表 18.13，应用 1.35 的调整。因此，穿过铁轨的等效流量为

14

15

150*1.33*1.35 = 269 veh/h. This is plotted against 300 veh/h on the major street on the Figure 18.9 criteria curve:
150*1.33*1.35 = 269 辆/小时。这与图 18.9 标准曲线上的 300 辆/小时相对应：

The warrant is clearly met, and a signal should be placed. It should be
该授权显然满足，应安装信号灯。它应该

coordinated with the RR crossing signals and gates.
与铁路交叉口信号和闸门协调。

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16

1

Traffic Engineering, 4th Edition
《交通工程》，第四版

Roess, R.P., Prassas, E.S., and McShane, W.R.
Roess，R.P.，Prassas，E.S.和 McShane，W.R.

Solutions to Problems in Chapter 20
第 20 章问题解决方案

Problem 20-1
问题 20-1

The easiest way to plot the data is to first compute the average headway for each queue position. The average headway is then plotted against queue position, and the best curve (by eye) is drawn through the data. In computing the average headway for each position, note that all queue positions do not have the same number of observations.
绘制数据的最简单方法是首先计算每个排队位置的平均车头时距。然后将平均车头时距与排队位置作图，并通过数据绘制最佳曲线（目测）。在计算每个位置的平均车头时距时，请注意并非所有排队位置的观测值数量都相同。

Queue Position
排队位置

Cycle Number
周期编号

Total
总计

Headway
车头时距

No. of
车头时距数量

Headways
车头时距

Ave
平均.

Headway
车头时距

1

2

3

4

5

6

7

8

9

10

1

3.6

3.4

3.2

3.5

3.5

3.3

3.6

3.5

2.4

3.5

33.5

10

3.4

2

2.8

2.7

2.6

2.7

2.5

2.6

2.9

2.6

2.7

2.8

26.9

10

2.7

3

2.2

2.4

2.3

2.1

2.5

2.4

2.4

2.4

2.6

2.4

23.7

10

2.4

4

2.0

2.2

2.1

2.1

2.3

2.1

2.0

2.2

2.2

2.2

21.4

10

2.1

5

2.1

1.9

2.0

2.2

2.1

2.0

2.1

1.8

1.9

1.8

19.9

10

2.0

6

1.9

2.0

2.1

2.0

1.8

2.1

2.0

1.8

2.0

1.7

19.4

10

1.9

7

1.9

2.0

1.8

2.1

1.9

1.9

2.1

1.9

2.0

2.0

19.6

10

2.0

8

x

2.1

1.8

1.9

2.0

2.0

2.0

1.8

x

1.9

15.5

8

1.9

9

x

1.8

x

2.0

x

2.0

1.9

x

x

1.8

9.5

5

1.9

10

x

1.9

x

1.8

x

x

2.0

x

x

1.8

7.5

4

1.9

AVG HEADWAY vs QUEUE POSTION
平均间距与队列位置

3.5

3.0

2.5

2.0

1.5

0 2 4 6 8 10

Queue Position
排队位置

From the figure, the saturation headway is the extension of the flat portion of the curve, which, in this case, starts with the 5th queue position. From the figure, the saturation headway is: 1.95 s.
从图中可以看出，饱和车头时距是曲线平坦部分的延伸，在本例中，从第 5 个队列位置开始。从图中可以看出，饱和车头时距为：1.95 秒。

1

2

The first four headway involve some component of start-up lost time. The start- up lost time is the difference between the actual headway and the saturation headway (1.95 s) in each case. Reading the actual headways from the figure, these values are:
前四个车头时距包含启动损失时间的部分。启动损失时间是每种情况下实际车头时距与饱和车头时距 (1.95 秒) 之间的差值。从图中读取实际车头时距，这些值为：

Position 1: 3.40 – 1.95 = 1.45 s
位置 1：3.40 – 1.95 = 1.45 秒

Position 2: 2.70 – 1.95 = 0.75 s
位置 2：2.70 – 1.95 = 0.75 秒

Position 3: 2.35 – 1.95 = 0.40 s
位置 3：2.35 – 1.95 = 0.40 秒

Position 4: 2.10 – 1.95 = 0.15 s
位置 4：2.10 – 1.95 = 0.15 秒

SUM 2.75 s
总计 2.75 秒

The start-up lost time is, then, 2.75 s.
因此，启动损失时间为 2.75 秒。

The saturation flow rate is computed from the saturation headway as:
饱和流量由饱和车头时距计算得出：

Problem 20-2
习题 20-2

In general, the capacity of an intersection approach is given by:
通常情况下，交叉口的方法容量由以下公式给出：

In this case, C = 75 s (given), s = 3600/h = 3600/2.48 = 1,452 veh/hg, g = G+Y-tL = 40 + 4 – (2.3+1.1) = 40.6 s. Then:
在这种情况下，C = 75 秒（已知），s = 3600/小时 = 3600/2.48 = 1,452 辆/小时，g = G+Y-tL = 40 + 4 – (2.3+1.1) = 40.6 秒。那么：

c = 1452 (40.675)= 786 veh / h / ln
c = 1452 (40.675) = 786 辆 / 小时 / 车道

As there are three lanes on this approach, the total capacity of the approach is 3*786 = 2,358 veh/h.
由于该路口有三个车道，该路口的总容量为 3*786 = 2,358 辆/小时。

Problem 20-3
问题 20-3

The equation suggests a start-up time of 2.04 s/veh and a saturation headway of
该公式表明起步时间为 2.04 秒/辆，饱和间隔为

2.35 s/veh, which translates to a saturation headway of 3600/2.35 = 1,532 veh/hg.
2.35 秒/辆，这相当于饱和间隔为 3600/2.35 = 1,532 辆/小时/车道。

2

3

Problem 20-4
问题 20-4

The maximum sum of critical lane volumes is computed as follows:
最大关键车道流量总和计算如下：

Vc =

= 1,334 veh/ h
= 1334 车/小时

Problem 20-5
例题 20-5

For the intersection of Problem 20-5, the maximum sum of critical lane volumes is:
对于例题 20-5 的交叉路口，临界车道流量之和最大为：

If a 1 x 1 intersection is used, the sum of critical lane volumes is 800 + 1400 = 2200 veh/h > 1504 veh/h (NG). If the 1400 veh/h is divided into 2 lanes, then the sum of critical lane volumes becomes 800 + 700 = 1500 veh/h < 1504 veh/h. Therefore, this design (2-lane each direction E and W, 1 lane each direction N and S) would be adopted
如果使用 1 x 1 交叉路口，临界车道流量之和为 800 + 1400 = 2200 车/小时 > 1504 车/小时（不合格）。如果将 1400 车/小时分成 2 个车道，则临界车道流量之和变为 800 + 700 = 1500 车/小时 < 1504 车/小时。因此，将采用此设计方案（东西向各 2 车道，南北向各 1 车道）。.

Problem 20-6
例题 20-6

For the intersection of Problem 20-6, the maximum sum of critical lane volumes is:
对于例题 20-6 的交叉路口，临界车道流量之和最大为：

If we assume that both left-turn phases use a single exclusive lane, then 200 + 300 = 500 veh/h of this total are already utilized, leaving 1252-500 = 752 veh/h for the through movements. If each of the LT bays have two lanes, they would utilize 500/2 = 250 veh/h of the total, leaving 1252-250 = 1002 veh/h for the through vehicles.
如果我们假设左转相位都使用一条专用车道，那么总共 200 + 300 = 500 车/小时已经被利用，剩下 1252-500 = 752 车/小时用于直行交通。如果每个左转车道有两条车道，它们将利用总车流量的 500/2 = 250 车/小时，剩下 1252-250 = 1002 车/小时用于直行车辆。

3

4

The table that follows shows the total critical lane volumes for the through movements, given various lane configurations:
下面表格显示了不同车道配置下，直行车辆的总关键车道流量：

Volume
数量

No. of Lanes
车道数

Volume
数量

No. of Lanes
车道数

Sum
总和

1000

1

700

1

1,700 (NG)

1000

2

700

1

1,200 (NG)

1000

2

700

2

850

1000

3

700

2

684

If double LT lanes are used, then the 2 x 2 option would work. If single LT lanes are provided, the 3 x 2 option for through vehicles is the minimum required.
如果使用双左转车道，那么 2 x 2 的选项将适用。如果提供单左转车道，则直行车辆的 3 x 2 选项是最低要求。

Problem 20-7
问题 20-7

As seen in the previous problem, a 2 x 1 design produces a sum of critical lane volumes of 1,500 veh/h. Them minimum and desirable cycle lengths are computed as follows:
正如上一个问题所示，2 x 1 的设计产生 1,500 辆/小时的关键车道流量总和。然后，计算最小和理想的周期长度如下：

The minimum cycle length is clearly 60 s (the actual cycle length provided). A “desirable” cycle length cannot be provided, as there is not enough effective green time in an hour. The cycle length should probably be increased to the maximum practical limit (considering nearby signals and other issues).
最小周期长度显然是 60 秒（提供的实际周期长度）。由于每小时有效的绿灯时间不足，因此无法提供“理想”的周期长度。周期长度应该增加到最大的实际限制（考虑到附近的信号灯和其他问题）。

Problem 20-8
问题 20-8

The equivalent is computed by assuming that the 20 through vehicles in one lane are equivalent to the 10 through and 5 left-turning vehicles in the other, as both utilize the same amount of green time:
等效值是通过假设一条车道上的 20 辆直行车辆相当于另一条车道上的 10 辆直行车辆和 5 辆左转车辆计算出来的，因为两者都使用了相同的绿灯时间：

20 = 10 + 5 E
20 = 10 + 5E

4

5

There were 35 vehicles observed, of which 5 were left-turning vehicles. The proportion of left turns is, therefore, 5/35 = 0.143. The left-turn adjustment factor is then computed as:
观察到 35 辆车，其中 5 辆为左转车辆。因此，左转比例为 5/35 = 0.143。然后计算左转调整系数为：

Variables that affect the left-turn equivalent include the opposing through flow, and the number of lanes over which it is distributed.
影响左转等效值的变量包括对向直行车流量以及其分布的车道数。

Problem 20-9
例题 20-9

There are two ways to compute the value. The first is to use the through-vehicle equivalent directly:
计算该值的方法有两种。第一种是直接使用直行车辆当量：

Type of Vehicle
车辆类型

No. of Vehicles
车辆数量

Equivalent Equivalent No.
等效车辆数

Through Vehs
直行车辆

1350*0.92=1242

1.00 1,242

Left-Turn Vehs
左转车辆

1350*0.08= 108

2.70 292

Total
总计

1,534 tvu/h
1,534 辆/小时

A second approach would be to compute the left-turn adjustment factor as:
第二种方法是计算左转调整系数：

Then, the equivalent through volume is found as:
然后，等效直行交通量为：

Problem 20-10
例题 20-10

The left-turn adjustment factor is:
左转调整系数为：

The saturation flow rate for all vehicles is: s = 1,700 * 0.806 = 1370 veh / hg / ln
所有车辆的饱和流量为：s = 1,700 * 0.806 = 1370 辆/小时/车道

5

6

The saturation headway is:
饱和车头时距为：

Effective green time is used at a rate of one vehicle every “s” seconds. The capacity of the lane is, therefore:
有效绿灯时间以每“s”秒一辆车的速度使用。因此，车道的通行能力为：

c = s * ⎜⎝(gC
 = 1370 * (4575)= 822 veh / h
c = s * ⎜⎝(gC= 1370 * (4575)= 822 车/小时

Problem 20-11
问题 20-11

The capacity of the approach is:
该方法的容量为：

The v/c ratio is, therefore, 500/906 = 0.552. This value is quite low. For a v/c ratio this low, Webster’s Delay Equation is appropriate for use in predicting delay:
因此，v/c 比为 500/906 = 0.552。这个值相当低。对于这么低的 v/c 比，韦伯斯特延迟方程适用于预测延迟：

Operating with a v/c ratio of 1.05, this intersection experiences both uniform and overflow delay. In this v/c range, the v/c ratio is best estimated using Akcelik’s equation.
以 1.05 的 v/c 比运行，该交叉路口同时存在均匀延误和溢流延误。在这个 v/c 范围内，v/c 比最好使用 Akcelik 方程估算。

In order to make these computations, the capacity and saturation flow rate for the intersection must be computed. This is done using the given demand flow rate, and the known v/c ratio:
为了进行这些计算，必须计算交叉路口的容量和饱和流量。这是使用给定的需求流量和已知的 v/c 比来完成的：

6

7

c = 800 /1.05 = 762 veh / h
c = 800 / 1.05 = 762 车/小时

s = 762 / 0.60 = 1,270 veh/ hg
s = 762 / 0.60 = 1270 车/小时

Uniform delay is estimated using Webster’s Delay Equation for the simplified
均匀延误是使用韦伯斯特延误方程估算的，用于简化的

case in which v/c = 1.00. Remember that the maximum v/c ratio that can be used in Webster’s Equation is 1.00:
v/c = 1.00 的情况。请记住，韦伯斯特方程中可使用的最大 v/c 比是 1.00：

UD = 0.50 C[1 − (g / C)] = 0.50 * 75 * (1 − 0.60) = 15.0 s / veh
UD = 0.50 C[1 − (g / C)] = 0.50 * 75 * (1 − 0.60) = 15.0 秒/车

This delay applies to any time period for which the demand situation is as stated. Akcelik’s equation can be used to estimate the overflow delay that occurs over the ½ hour, or over the first 5 minutes of the hour. Note that the saturation flow rate is expressed as veh/s in Akcelik’s equation:
此延误适用于需求情况如所述的任何时间段。Akcelik 方程可用于估算在半个小时内或在小时的前 5 分钟内发生的溢流延误。请注意，在 Akcelik 方程中，饱和流量以车/秒表示：

Xo = 0.67 + (⎜ sg ⎞⎟ = 0.67 + (⎜ (1270 / 3600) * (75 * 0.60) ⎞⎟ = 0.6965

The total delay is the sum of the uniform delay and the overflow delay:
总延误是均匀延误和溢流延误之和：

d30 = 15.0 + 15.8 = 30.8 s / veh d5 = 15.0 + 5.0 = 20.0 s / veh
d30 = 15.0 + 15.8 = 30.8 s/车 d5 = 15.0 + 5.0 = 20.0 s/车

Even though there is an overflow situation, the delays are not very high because
即使存在溢出情况，延时也不高，因为

of two reasons: (a) the demand is only 5% higher than the capacity, and (b) the condition exists for only ½ hour. Delay in the first five minutes of overflow is obviously less than the average over ½ hour, as the queue has not yet fully developed.
两个原因：（a）需求仅比容量高 5%，（b）这种情况仅存在 30 分钟。溢流前五分钟的延误显然小于 30 分钟的平均延误，因为队列尚未完全形成。

7

8

To determine the most appropriate equation for use in predicting delay, the v/c ratio for the hour should be considered:
为了确定用于预测延误的最合适的方程，应考虑每小时的 v/c 比率：

c = s * ⎜⎝(gC
 = 3250 * (55100 = 1,788 veh/ hg )
c = s * ⎜⎝(gC= 3250 * (55100 = 1,788 车/小时)

v/ c = 2,0001,788 = 1.12
v/c = 2,000/1,788 = 1.12

This value is significantly higher than 1.00. In this case, the simple theoretical equations for overflow delay maybe employed. As in the previous example (20- 12), the simplified equation for Uniform Delay (Webster’s Equation) can be used. This value applies to all time periods during which the stated conditions exist:
此值明显高于 1.00。在这种情况下，可以使用简单的溢流延误理论方程，如前面的示例（20-12）所示，可以使用均匀延误的简化方程（韦伯斯特方程）。此值适用于存在上述条件的所有时间段：

UD = 0.50C⎢1 − ⎜⎝(gC

 = 0.50 *100 * (1 − 0.55) = 22.5 s/ veh
= 0.50 * 100 * (1 − 0.55) = 22.5 秒/车

Overflow delay for the full hour is computed as:
整小时的溢流延误计算如下：

where X = v/c = 1.12, and 3600 is the number of seconds in an hour. The same formula is used to estimate the overflow delay during the first 15 minutes of the hour. In this case, T = 60*15 = 900 s:
其中 X = v/c = 1.12，3600 是小时内秒数。相同的公式用于估计小时前 15 分钟内的溢流延误。在这种情况下，T = 60 * 15 = 900 秒：

The overflow delay in the last 15 minutes of the hour is estimated as:
小时最后 15 分钟的溢流延误估计为：

where T1 = 45*60 = 2,700 s, and T2 = 60*60 = 3,600 s:
其中 T1 = 45*60 = 2700 秒，T2 = 60*60 = 3600 秒：

= 378 s / veh
= 378 秒/车

8

9

Total delays must add the UD to OD to obtain:
总延误必须将 UD 和 OD 相加才能得到：

dhour = 22.5 + 216.0 = 238.5 s / veh dfirst15 = 22.5 + 54.0 = 76.5 s / veh dlast15 = 22.5 + 378.0 = 400.4 s / veh
dhour = 22.5 + 216.0 = 238.5 秒/车 dfirst15 = 22.5 + 54.0 = 76.5 秒/车 dlast15 = 22.5 + 378.0 = 400.4 秒/车

Obviously, delay during the first 15 minutes is less severe than delay in the last 15 minutes, as there is no residual queue at T = 0, and there is a substantial residual queue at T = 45 minutes.
显然，前 15 分钟的延误比最后 15 分钟的延误轻微，因为在 T = 0 时没有剩余排队，而在 T = 45 分钟时存在大量剩余排队。

9

1

Traffic Engineering, 4th Edition
《交通工程》，第四版

Roess, R.P., Prassas, E.P., and McShane, W.R.
Roess，R.P.，Prassas，E.P.和 McShane，W.R.

Solution to Problems in Chapter 21
第 21 章习题解答

Problem 21-1
习题 21-1

The change (or yellow) interval is timed to allow a driver that is too close to the signal to stop safely to proceed through the intersection safely. It is computed as follows:
变换（或黄灯）时间间隔的设置是为了允许距离信号灯过近而无法安全停车的驾驶员安全通过交叉路口。其计算方法如下：

The clearance (or all red) interval can be found using one of several equations. In this case, given a significant pedestrian crossing movement, vehicles should be cleared beyond the far crosswalk during the all red interval. Therefore:
可以使用多种公式之一来计算清除（或全红）时间间隔。在本例中，鉴于存在大量的行人通行，车辆应在全红时间间隔内驶过远侧人行横道。因此：

Problem 21-2
习题 21-2

The solution depends upon whether the local jurisdiction allows pedestrians in the crosswalk during the yellow and all-red intervals. Assuming that pedestrians are permitted in the crosswalk during the yellow and all –red intervals, Gp ≤ G+Y for each phas
解答取决于当地管辖区是否允许行人在黄灯和全红时间间隔内通行人行横道。假设允许行人在黄灯和全红时间间隔内通行人行横道，则对于每个阶段，Gp ≤ G+Y。

Phase A: 30.0 ≤ 18.0+4.5 = 22.5 ? NO
A 阶段：30.0 ≤ 18.0+4.5 = 22.5？否

Phase B: 15.0 ≤ 60.0+4.0 = 64.0 ? YES
B 阶段：15.0 ≤ 60.0+4.0 = 64.0？是

This is a classic situation. During the longer vehicular phase, pedestrians cross a narrower side street (Phase B). During the shorter vehicular phase, pedestrians generally cross a wider main street (Phase A). Pedestrians, therefore, need more green time during the shorter phase than they do during the longer phase.
这是一个典型的案例。在较长的车辆通行阶段，行人穿过较窄的侧街（B 阶段）。在较短的车辆通行阶段，行人通常穿过较宽的主街（A 阶段）。因此，行人在较短的阶段需要比在较长的阶段更多的绿灯时间。

The solution to this problem is to lengthen the cycle length so that the proportioning of green time remains the same, while satisfying the minimum pedestrian needs in both phases. To satisfy the Phase A pedestrian requirement,
解决这个问题的方法是延长周期长度，使绿灯比例保持不变，同时满足两阶段的最小行人需求。为了满足阶段 A 行人的要求，

1

2

the green time must be increased to at least 30.0 – 4.5 = 25.5 s. However, to maintain the current balance of green time, the green time in Phase B would also have to be increased:
绿灯时间至少必须增加到 30.0 – 4.5 = 25.5 秒。但是，为了保持当前绿灯时间的平衡，阶段 B 的绿灯时间也必须增加：

This leads to a cycle length of (25.5+4.5)+(85.0+4.0) = 119 s. If this is a pre-timed signal, a cycle length of 120 s would be used. Keeping the split of green times constant, the green times become:
这会导致一个循环长度为 (25.5+4.5)+(85.0+4.0) = 119 秒。如果这是一个预定信号，则会使用 120 秒的循环长度。保持绿灯时间的比例不变，则绿灯时间变为：

All pedestrian requirements are met. Yellow and all-red timings are unchanged. If the jurisdiction decided to allow pedestrians in the crosswalk only during the green, then the same computations would be repeated, but gA would have to be at least 30.0 s.
所有行人需求都已满足。黄灯和全红灯时间保持不变。如果管辖区决定只允许行人在绿灯期间过人行横道，那么将重复进行相同的计算，但 gA 至少必须为 30.0 秒。

Problem 21-3
问题 21-3

Step 1 – Phasing
步骤 1 – 配相

This is a simple intersection of two one-way streets. A two-phase signal will suffice, as there are no opposed left turns. Note that the WB right turns and NB left turns are in exclusive lanes, which will have to be treated as a separate lane group for timing.
这是一个两条单向街道的简单交叉口。由于没有对向左转，使用两相信号就足够了。请注意，西向右转和北向左转都在专用车道上，这将必须作为单独的车道组进行配时。

The ring diagram for the proposed signalization follows.
拟议的信号控制方案的环图如下。

2

3

Step 2 – Convert Volumes to Through Vehicle Equivalents
步骤 2 – 将流量转换为车辆当量

NB left turns can be treated in several ways: (a) as unopposed left turns (ELT = 1.05 – Table 21.1), (b) as opposed left turns with “0” opposing volume (ELT = 1.1- Table 21.1), or (c) as right turns against a low pedestrian flow (ELT = 1.21 – Table 21.2). As the left turns from this one-way street will immediately encounter pedestrians in a manner similar to right turns, option “c” will be used.
北向左转可以用几种方式处理：(a) 视为非对向左转 (ELT = 1.05 – 表 21.1)，(b) 视为对向左转，且“0”为对向流量 (ELT = 1.1 – 表 21.1)，或 (c) 视为低行人流量的右转 (ELT = 1.21 – 表 21.2)。由于从这条单向街道驶出的左转车辆将立即遇到行人，其方式类似于右转，因此将使用选项“c”。

Step 3 – Determine the Sum of Critical Lane Volumes
Step 3 – Determine the Sum of Critical Lane Volumes 步骤 3 – 确定关键车道流量之和

This is done by inserting the lane volumes from Step 2 into the ring diagram of Step 1:
这是通过将步骤 2 中的车道流量插入步骤 1 的环图中来完成的：

3

4

363

526

454

400

Vc = 526+454 = 980 tvu/h
Vc = 526 + 454 = 980 tvu/h

Step 4 – Determine “Yellow” and “All Red” Intervals
步骤 4 – 确定“黄色”和“全红”间隔

The following speeds are used in these determinations:
下列速度用于这些确定：

S85A = 50 + 5 = 55 mi/h S15A = 50 – 5 = 45 mi/h
S85A = 50 + 5 = 55 英里/小时 S15A = 50 – 5 = 45 英里/小时

S85B = 40 + 5 = 45 mi/h
S85B = 40 + 5 = 45 英里/小时

S15B = 49 – 5 = 35 mi/h Then:
S15B = 49 – 5 = 35 英里/小时 然后：

4

5

Step 4 – Determine Lost Times
步骤 4 – 确定损失时间

If the standard values for l 1 = 2.0 s and e = 2.0 s, then the lost times are equal to the sum of the yellow and all-red intervals in the signal:
如果 l1 的标准值为 2.0 秒，e 为 2.0 秒，则损失时间等于信号中黄色和全红间隔之和：

L = (4.7 + 1.2) + (4.4 + 1.1) = 11.4 s / cycle
L = (4.7 + 1.2) + (4.4 + 1.1) = 11.4 秒/循环

Step 5 – Determine the Appropriate Cycle Length
步骤 5 – 确定合适的周期长度

Step 6 – Splitting the Green
步骤 6 – 分割绿色

The total effective green time in the cycle is 40.0 – 11.4 = 28.6 s. This is split in proportion to the critical lane volumes for Phase A (526) and Phase B (454):
周期内总有效绿灯时间为 40.0 – 11.4 = 28.6 秒。此时间将按 A 相位 (526) 和 B 相位 (454) 的临界车道流量比例分配：

Check : 15.4 + 13.2 + 11.4 = 40.0 OK
检查：15.4 + 13.2 + 11.4 = 40.0 正确

As the standard values for l 1 and e were used, actual greens (G) and effective greens (g) are the same.
由于使用了 l 1 和 e 的标准值，实际绿灯时间 (G) 和有效绿灯时间 (g) 相同。

Step 6 – Check Pedestrians
第 6 步 – 检查行人

The default for “low” pedestrian activity is 50 peds/h. Thus:
“低”行人活动量的默认值为 50 peds/h。因此：

5

6

The minimum pedestrian crossing time required is:
所需的最小行人过街时间为：

GpA (Xing EW Street) = 3.2 +
A 相位行人绿灯时间 (东西向街道) = 3.2 +

= 3.2 + 0.2 + 15.0 = 18.0 s
= 3.2 + 0.2 + 15.0 = 18.0 秒

The adequacy of the vehicular signal timing depends upon which pedestrian policy is in effect. Using the most lenient policy, pedestrians may be in the crosswalk during yellow and all red intervals. Then:
车辆信号配时的充分性取决于有效的行人策略。使用最宽松的策略，行人可以在黄灯和所有红灯期间穿过人行横道。那么：

G = 18.0 s ≤ 15.4 + 4.7 +1.2 = 21.3 s OK
G = 18.0 秒 ≤ 15.4 + 4.7 + 1.2 = 21.3 秒 正确

G = 17.0 s ≤ 13.2 + 4.4 + 1.1 = 18.7 s OK
G = 17.0 s ≤ 13.2 + 4.4 + 1.1 = 18.7 s，符合要求

pB

If pedestrians are only allowed in the crosswalk during yellow intervals, but not all red intervals:
如果行人只允许在黄灯时间段内通行，而不是所有红灯时间段内通行：

G = 18.0 s ≤ 15.4 + 4.7 = 20.1 s OK
G = 18.0 s ≤ 15.4 + 4.7 = 20.1 s，符合要求

G = 17.0 s ≤ 13.2 + 4.4 = 17.6 s OK
G = 17.0 s ≤ 13.2 + 4.4 = 17.6 s，符合要求

pB

If pedestrians are only allowed in the crosswalk during green periods:
如果行人只允许在绿灯时间段内通行：

G = 18.0 s ≤ 15.4 s NG
G = 18.0 s ≤ 15.4 s，不符合要求

G = 17.0 s ≤ 13.2 s NG
G = 17.0 s ≤ 13.2 s，不符合要求

pB

In the latter case, the signal cycle would have to be increased to accommodate the same relative split in green times, while providing for safe pedestrian crossing times.
在后一种情况下，信号周期将不得不增加，以适应绿灯时间的相同相对分配，同时确保行人安全通行时间。

Problem 21-4
问题 21-4

Step 1 – Phasing
步骤 1 – 配相

Left-turn volumes should be checked to determine whether protected LT phasing will be necessary. These are checked using three criteria: a) minimum LT flow rate, b) cross-product rule, and c) ITE criteria. This is done below:
应检查左转车流量，以确定是否需要设置左转保护相位。这通过三个标准进行检查：a) 最小左转车流量，b) 交叉乘积规则，以及 c) ITE 标准。如下所示：

6

7

EB LT: vLT = 210 veh/h > 200 veh/h Protected phase needed.
EB 左转：vLT = 210 车/小时 > 200 车/小时 需要保护相位。

WB LT: vLT = 350 veh/h > 200 veh/h Protected phase needed.
WB 左转车道：vLT = 350 车/小时 > 200 车/小时 需要设置保护相位。

NB LT:
NB 左转车道：

vLT = 30 veh/h < 200 veh/h
vLT = 30 车/小时 < 200 车/小时

x-prod = 30*(265/1) = 7,950 < 50,000 ITE (Figure 18.1)
x-prod = 30*(265/1) = 7,950 < 50,000 ITE（图 18.1）

Criterion not met. Criterion not met. Criterion not met.
未满足标准。未满足标准。未满足标准。

SB LT:
SB 左转车道：

vLT = 15 veh/h < 200 veh/h
vLT = 15 车/小时 < 200 车/小时

x-prod = 15*(250/1) = 3,500 < 50,000 ITE (Figure 18.1)
x-prod = 15*(250/1) = 3,500 < 50,000 ITE（图 18.1）

Criterion not met. Criterion not met. Criterion not met.
未满足标准。未满足标准。未满足标准。

From these results, a protected LT is needed for the E-W arterial, while permitted phasing will be ok for the N-S arterial. The latter is important, as the one-lane approaches do not provide for exclusive LT lanes, which are required to implement many protected LT solutions.
根据这些结果，东西向主干道需要设置左转保护相位，而南北向主干道允许设置允许相位即可。后者很重要，因为单车道进近不允许设置专用左转车道，而许多左转保护相位解决方案都需要专用左转车道。

As there is significant imbalance between the two LT movements needing protection, a split LT phasing will be used. The NEMA phasing approach will be used:
由于需要保护的两个 LT 运动之间存在显著的不平衡，因此将采用分阶段的 LT 方案。将采用 NEMA 分相方法：

A1

A2

A3

B

Phase/Ring Diagram
相位/环图

Step 2: Convert to Through Vehicle Units
步骤 2：转换为车辆通行单位

Demand volumes must now be converted to “through vehicle units” using the equivalents of Tables 21.1 and 21.2. This is done below:
现在必须使用表 21.1 和 21.2 的当量将需求量转换为“车辆通行单位”。以下进行此转换：

7

8

Mvt
运动

Vol
交通量

(veh/h)
(辆/小时)

Equiv
当量

(T 21.1/21.2)
(表 21.1/21.2)

Vol. (tvu/h)
车辆通行量 (tvu/小时)

No. of Lanes
车道数

Vol/Lane (tvu/h)
每车道通行量 (tvu/小时)

EB L
东向

210

1.05

221

1

221

EB TH
东经高

775

1.00

775

2

417

EB R
东经右

45

1.32*

59

WB L
西经左

350

1.05

368

1

368

WB TH
WB TH 西向车道

750

1.00

750

2

408

WB R
WB R 西向车道右转

50

1.32*

66

NB L
NB L 北向车道左转

30

3.31**

99

1

382

NB TH
NB TH 北向车道直行

250

1.00

250

NB R
NB R 北向车道右转

25

1.32*

33

SB L
SB L 南向车道左转

15

3.13**

47

1

325

SB TH
SB TH 南向车道直行

265

1.00

265

SB R
SB R 南向车道右转

10

1.32*

13

* for modest pedestrian activity ** interpolated
* for modest pedestrian activity ** interpolated * 用于适度行人活动 ** 插值

Step 3 – Determine the Sum of Critical Lane Volumes
Step 3 – Determine the Sum of Critical Lane Volumes 步骤 3 – 确定关键车道流量之和

The volume per lane (in tvu/h) for each portion of the phase diagram is entered. The critical path through the signal is the one that produces the highest sum of critical lane volumes.
为相位图的每个部分输入每条车道的通行量（以 tvu/h 为单位）。信号中的关键路径是产生最高关键车道通行量总和的路径。

221

368

407

417

382

325

A1

A2

A3

B

Critical Lanes ?
关键车道？

221 + 407 = 628 tvu/h or
221 + 407 = 628 tvu/h 或

368 + 417 = 785 tvu/h√

382 tvu/h√ or 325 tvu/h
382 tvu/h√ 或 325 tvu/h

Σ = 785+382 = 1,167 tvu/h

8

9

The sum of the critical lane volumes is 1,167 tvu/h.
关键车道通行量之和为 1,167 tvu/h。

Step 4: Determine Yellow and All-Red Times
第 4 步：确定黄灯和全红灯时间

Yellow times are computed using Equation 21-2. The ring diagram shows that this is a 3-phase signal phase plan. Thus, there are 3 yellow times, two for the E- W arterial, one for the N-S arterial. The two E-W yellow times are based upon an 85th percentile approach speed of 35+5 = 40 mi/h. The N-S yellow is based upon an 85th percentile approach speed of 30+5 = 35 mi/h:
黄灯时间使用公式 21-2 计算。环状图显示这是一个三相信号相位方案。因此，共有 3 个黄灯时间，东西向主干道两个，南北向主干道一个。东西向的两个黄灯时间基于 85 百分位数的接近速度 35+5 = 40 英里/小时。南北向的黄灯时间基于 85 百分位数的接近速度 30+5 = 35 英里/小时：

Because there is moderate pedestrian presence, the all-red phases should be timed to clear vehicles beyond the far pedestrian crosswalk. Again, there are 3 phases and 3 all-red times, two for the E-W arterial, and one for the N-S arterial. The E-W all-reds are based upon clearing a pedestrian distance of 24 ft (2 lanes) plus 10 ft = 34 ft. The N-S all-red is based upon clearing a pedestrian distance of 24+16+24 plus 10 ft = 74 ft. The E-W street has a 15th percentile speed of 35 – 5 = 30 mi/h; the N-S street has a 15th percentile speed of 30-5 = 25 mi/h. Then:
由于人行道上的人流量中等，因此应将全红相位时间调整为清除远端人行横道外的车辆。再次强调，共有 3 个相位和 3 个全红时间，用于东西方向的主干道，一个用于南北方向的主干道。东西方向的全红时间是根据清除 24 英尺（2 车道）的人行道距离加上 10 英尺 = 34 英尺计算的。南北方向的全红时间是根据清除 24 + 16 + 24 英尺的人行道距离加上 10 英尺 = 74 英尺计算的。东西方向街道的 15% 百分位速度为 35 – 5 = 30 英里/小时；南北方向街道的 15% 百分位速度为 30-5 = 25 英里/小时。然后：

A yellow and all-red transition occurs twice in Phase A (for the E-W street). Both rings have a transition at the end of Phase A3. One ring has a transition at the end of Phase A1, the other ring at the end of Phase A2. Combined these two count as one full transition.
在 A 相位（对于东西方向街道）中，黄灯和全红灯转换两次。两个环在 A3 相位结束时都有转换。一个环在 A1 相位结束时转换，另一个环在 A2 相位结束时转换。 这两个转换算作一个完整的转换。

9

10

Step 5: Determine Lost Time Per Cycle
第 5 步：确定每循环的延误时间

Assuming that standard values of start-up lost time (2 s) and encroachment time (2.0 s) are used, the lost time is the sum of the yellow and all-red phases, or (3.9+1.0) + (3.9+1.0) + (3.6+2.6) = 16.0 s
假设使用标准的启动延误时间（2 秒）和入侵时间（2.0 秒），延误时间是黄灯和全红灯相位的总和，或者 (3.9+1.0) + (3.9+1.0) + (3.6+2.6) = 16.0 秒

Step 6: Determine the Appropriate Cycle Length
第 6 步：确定合适的循环长度

An appropriate cycle length is estimated using Equation 21-11:
使用方程式 21-11 来估算合适的循环长度：

Step 7: Spitting the Green
第 7 步：分配绿灯时间

The 75-s cycle length includes 16 s of lost time, leaving 75 – 16 = 59 s of effective green time to allocate in proportion to the critical lane volumes in the critical path of the signal. The critical path and resulting critical lane volumes are as follows:
75 秒的循环长度包括 16 秒的延误时间，剩余 75 – 16 = 59 秒的有效绿灯时间可按信号关键路径中的关键车道流量进行分配。关键路径和由此产生的关键车道流量如下：

Phases A1+A2 Phase A3
A1+A2 相位 A3 相位

Phase B
B 相

Total
总计

368 tvu/h
368 tvu/小时

417 tvu/h
417 tvu/小时

382 tvu/h
382 tvu/小时

1,167 tvu/h
1,167 tvu/小时

Then:
然后：

gA1+ A2 = 59 * (368 /1167) = 18.6 s gA3 = 59 * (417 /1167) = 21.1 s gB = 59 * (382 /1167) = 19.3 s
gA1 + A2 = 59 * (368 / 1167) = 18.6 秒 gA3 = 59 * (417 / 1167) = 21.1 秒 gB = 59 * (382 / 1167) = 19.3 秒

The totals should add up to 59 s. 18.6+21.1+19.3 = 59.0 s. Adding the 16 s of yellow and all red brings us to the 75-s cycle length.
总和应加起来为 59 秒。 18.6 + 21.1 + 19.3 = 59.0 秒。 将黄色和所有红色的 16 秒相加，得出 75 秒的周期长度。

10

11

Because of the phase plan used, timing the critical path does not fully specify the signal timing. Phases A1 and A2 must be specified to complete the process. These phases, however, exist only on the non-critical path through the signal. The sum of Phases A1, A2, and A3 is known: 18.6+21.1 = 39.7 s. On the non-critical path, Phase 1A has a lane volume of 221 tvu/h, while the sum of Phases A2 and A3 has a lane volume of 407 tvu/h. Therefore:
由于所使用的相位方案，关键路径的时间并不完全指定信号时间。必须指定 A1 和 A2 相位才能完成该过程。然而，这些相位仅存在于信号的非关键路径上。已知 A1、A2 和 A3 相位的总和：18.6 + 21.1 = 39.7 秒。在非关键路径上，1A 相位的车道流量为 221 tvu/小时，而 A2 和 A3 相位的车道流量总和为 407 tvu/小时。因此：

The timing of Phase A2 cannot be computed directly, as no one movement uses this phase exclusively. This is, in effect, the overlap phase. However, the sum of Phases A1, A2, and A3 is known, and the length of Phases A1 and A3 is known. By inference:
由于没有一个动作只使用此相位，因此无法直接计算 A2 相位的时间。这实际上是重叠相位。但是，已知 A1、A2 和 A3 相位的总和，以及 A1 和 A3 相位的时间。通过推断：

gA2 = 39.7 −14.0 − 21.1 = 4.6 s
gA2 = 39.7 − 14.0 − 21.1 = 4.6 秒

Step 8: Pedestrian Requirements
第 8 步：行人需求

Pedestrian requirements are estimated using Equation 21-15:
行人需求使用等式 21-15 估算：

The default value for “moderate” pedestrian activity (Table 21.2) is 200 peds/h in each crosswalk. With a cycle length of 75 s, the number of pedestrian per cycle in each crosswalk is:
“中等”行人活动（表 21.2）的默认值为每个斑马线 200 人/小时。如果周期长度为 75 秒，则每个斑马线每周期行人数为：

Pedestrians cross the 24-ft N-S street during Phase A3 only. The minimum
行人仅在 A3 阶段穿越 24 英尺南北街。最小

pedestrian green time is:
行人绿灯时间为：

The green for Phase A3 is 21.1 s, which is more than the required 10.3 s. This is safe for pedestrians.
A3 阶段的绿灯时间为 21.1 秒，超过了所需的 10.3 秒。这对行人来说是安全的。

11

12

Pedestrians cross the 64-ft E-W street during Phase B. The minimum pedestrian green time is:
行人穿越 64 英尺东西街是在 B 阶段。最小行人绿灯时间为：

GpB = 3.2 +

= 3.2 + 1.1 + 16.0 = 20.3 s
= 3.2 + 1.1 + 16.0 = 20.3 秒

Phase B is 19.3 s, which is less than that needed. If, however, pedestrians are permitted to be in the crosswalk during yellow and/or all-red phases, the situation maybe deemed safe. G+y = 19.2+3.6 = 22.8 s; G+y+ar = 22.8+2.6 = 25.4 s, both of which are more than the required 20.3 s.
B 相位是 19.3 秒，这小于所需时间。然而，如果允许行人在黄色和/或全红相位期间在人行道上通行，则情况可能被认为是安全的。G+y = 19.2 + 3.6 = 22.8 秒；G+y+ar = 22.8 + 2.6 = 25.4 秒，两者都超过了所需的 20.3 秒。

Discussion
讨论

Technically, the signal timing as proposed is acceptable and safe for pedestrians. It should be noted, however, that the overlap Phase A2 is quite short – 4.6 s. Given the greater complexity that this type of phasing presents to drivers, it is wise to question whether or not it is justified, given the short overlap provided.
从技术上讲，建议的信号定时对行人来说是可接受且安全的。然而，需要注意的是，重叠相位 A2 非常短——4.6 秒。鉴于这种类型的相位对驾驶员来说更为复杂，考虑到提供的短重叠，有必要质疑其是否合理。

If the proposed ring diagram is examined, a simple 3-phase signal with a simultaneous left-turn phase for the E-W street results in the same sum of critical lane volumes as the original timing. Thus, the cycle length and splits would be unaffected (except that both LT phases would be 18.6 s, and there would be no Phase A2.
如果检查建议的环形图，则对于东西街来说，具有同时左转相位的简单三相信号会产生与原始定时相同的关键车道体积总和。因此，周期长度和分割不会受到影响（除了两个 LT 相位都将为 18.6 秒，并且没有 A2 相位。

In this case, there is no advantage to the overlapping phase plan. This is because the higher EW LT movement and the higher EW TH movement on are opposite approaches, which places them on the same ring of the signal. The advantage of using a split LT phasing is significant only when the higher LT and higher TH volume are on the same approach.
在这种情况下，重叠相位方案没有任何优势。这是因为较高的东西方向左转运动和较高的东西方向直行运动位于信号的相反方向，这使它们位于信号的同一环上。仅当较高的左转和较高的直行量位于同一方向时，使用分割式左转相位才具有显著优势。

In the end, a simple 3-phase signal would be recommended.
最后，建议采用简单的三相信号。

Problem 21-5
问题 21-5

Check Left Turns
检查左转

This problem features an intersection between a major artery and a significant one-way street or artery. Because of the one-way street, there is only one potentially-opposed left turn in the EB direction. This left turn should be protected because:
本问题包含主要干道与重要单行道或干道的交叉点。由于单行道的原因，EB 方向只有一个潜在的相反左转。该左转应受保护，因为：

12

13

vLT = 275veh/ h > 200 veh / h
vLT = 275 辆/小时 > 200 辆/小时

There is not WB left turn, and there is no flow of any kind in the SB direction. The left-turn from the one-way street is not opposed by a vehicular flow, but is opposed by a pedestrian flow, much as a right turn is. The value of ELT for this movement can be treated in any one of three ways, as a (a) protected LT (ELT = 1.05), (b) a permitted LT with no opposing flow (ELT = 1.1), or (c) a right turn against moderate pedestrian flow (ELT = 1.32). While these will yield slightly different signal timings, any of the three are logically acceptable. For this solution, a value of 1.1 will be used.
没有向西的左转车道，并且南向没有任何车辆通行。单行道的左转不受车辆阻碍，但受到行人阻碍，就像右转一样。此车道的左转延误时间（ELT）可以有三种处理方式：(a) 受保护的左转 (ELT = 1.05)，(b) 允许左转，但无阻碍车辆通行 (ELT = 1.1)，或 (c) 右转，遇到中等强度行人阻碍 (ELT = 1.32)。尽管这些方法会略微影响信号计时，但所有三种方法在逻辑上都是可接受的。本方案将采用 1.1 的值。

Conversion of Demand Flows to tvu’s
需求流转换为 tvu

The table below illustrates the computation of through-vehicle-equivalents and their assignment to lane groups and lane volumes.
下表说明了通过车辆当量计算及其分配给车道组和车道流量。

Approach
方法

Mvt
运动

Vol
交通量

(veh/h)
(辆/小时)

Equiv. (T21.1,2)
当量 (T21.1,2)

Vol
交通量

(tvu/h)

Lane
车道

Group
组

Vol
交通量

(tvu/h)

Vol/Ln (tvu/h/ln)
车道流量 (tvu/h/车道)

EB

LT TH
左转车道 左转绿灯

275

1,100

1.05 1.00

289

1,100

289

289

1,100

550

WB

TH RT
直行车道 右转绿灯

730

100

1.00 1.32

730

132

862

431

NB

LT TH RT
左转车道 左转绿灯 直行车道 右转绿灯

45

900

25

1.10 1.00 1.32

50

900

33

983

328

Ring Diagram and Sum of Critical Lane Volumes
环状图和关键车道流量总和

With only one protected left turn in the EB direction, a leading EB green will be used with no corresponding WB lag. The ring diagram is shown below:
由于东向只有一个保护左转，因此将使用东向领先绿灯，而没有相应的西向滞后。环状图如下所示：

13

14

289

550

\

431

328

Vc = 720+328 = 1,048 tvu/h
Vc = 720 + 328 = 1,048 pcu/h

Yellow and All Red Times and Lost Times
黄灯和全红时间以及损失时间

With an average speed of 40 mi/h on all approaches, the 85th percentile speed is estimated to be 40+5 = 45 mi/h, while the 15th percentile speed is estimated to be 40-5 = 35 mi/h. Then:
所有进路的平均速度为 40 英里/小时，则第 85 百分位速度估计为 40 + 5 = 45 英里/小时，而第 15 百分位速度估计为 40 - 5 = 35 英里/小时。然后：

Because there are three phases on the critical path, there are three sets of lost times in this cycle. As the standard default values for start-up lost time and encroachment time (2 s ea) are being used, lost time is equal to the sum of the yellow and all red intervals, and effective green is equal to actual green. Then:
由于关键路径上有三个阶段，因此该周期内有三组损失时间。由于使用启动损失时间和侵占时间（各 2 秒）的标准默认值，因此损失时间等于黄灯和全红时间间隔的总和，有效绿灯时间等于实际绿灯时间。然后：

14

15

L = (4.3 + 1.2) + (4.3 + 1.2) + (4.3 + 2.0) = 17.3 s

Cycle Length
周期长度

The cycle length is estimated as:
周期长度估计为：

Splitting the Green
绿色分割

In a 75-s cycle, with 17.3 s of lost time, the amount of effective green time to be allocated is:
在 75 秒的周期内，损失时间为 17.3 秒，则应分配的有效绿灯时间为：

gtot = 75.0 −17.3 = 57.7 s

Then:
然后：

Checking Pedestrian Safety
检查行人安全

The minimum green needed for pedestrians to safely cross the street is given by:
行人安全过街所需的最小绿灯时间由下式给出：

The default volume of pedestrians for “moderate” activity is 200 peds/h per crosswalk. Therefore:
“中等”活动的行人流量默认为每个交叉路口每小时 200 人次。因此：

15

16

Npeds =

= 3.9 peds/ cycle ⇒ 4 peds/ cycle
= 3.9 人/循环 ⇒ 4 人/循环

To safely cross the E-W artery, pedestrians need:
为了安全地穿过东西方向的道路，行人需要：

These pedestrians cross during Phase B, which has a green time of 18.1 s. The yellow is 4.3 s, and all red is 2.0 s. If pedestrians are permitted to cross during the yellow and all red in addition to the green, the total available crossing time is 18.1+4.3+2.0 = 24.4 s, which is sufficient. If pedestrians are permitted to cross only during green, or during green and yellow, a deficiency would exist, and some adjustment to the signal timing would be needed.
这些行人会在 B 相位通行，该相位的绿灯时间为 18.1 秒。黄灯时间为 4.3 秒，全红灯时间为 2.0 秒。如果允许行人在黄灯和全红灯期间以及绿灯期间通行，总可通行时间为 18.1 + 4.3 + 2.0 = 24.4 秒，这足够了。如果仅允许行人在绿灯期间或绿灯和黄灯期间通行，则会出现不足，需要对信号计时进行一些调整。

To safely cross the NS artery, pedestrians need:
为了安全地穿过南北方向的道路，行人需要：

These pedestrians cross the street during Phase A2, which has a green time of
这些行人会在 A2 相位过马路，该相位的绿灯时间为

22.1 s, which is sufficient.
22.1 秒，这足够了。

Problem 21-7
问题 21-7

The unique feature of this intersection is the offset. The offset will cause two specific recommendations:
该十字路口的独特特征是偏移量。偏移量将导致两个具体的建议：

1. Despite the results of any left-turn checks, the N-S street must get a fully protected LT phase. This will avoid potential accidents between left-turn vehicles and opposing through vehicles, which are in much more dangerous positions than in a standard 90o intersection.
1. 尽管任何左转检查的结果，南北街必须获得完全保护的左转相位。这将避免左转车辆和对向直行车辆之间的潜在事故，而这些车辆在标准 90 度交叉路口中的位置更为危险。

2. The clearance time for the E-W street will be quite long due to the position of the crosswalks. Vehicles will have to clear 120 ft.
2. 由于人行横道的方位，东西街的通行时间会相当长。车辆必须通过 120 英尺的路段。

Left Turns
左转

16

17

As the geometry already requires a protected LT phase for the N-S street, this need not be further compared to criteria. The E-W LTs, however, must be checked to see whether they will also require protection:
由于几何形状已经要求南北街拥有一个受保护的左转相位，因此无需进一步与标准进行比较。然而，必须检查东西方向的左转，看看它们是否也需要保护：

WB: 180 veh/h < 200 veh/h No
西往：180 辆/小时 < 200 辆/小时 否

Xprod = 180*(700/2) = 63,000 > 50,000 Yes
Xprod = 180*(700/2) = 63,000 > 50,000 是

EB: 80 veh/h < 200 veh/h No
东往：80 辆/小时 < 200 辆/小时 否

Xprod = 80*(825/2) = 33,000 No ITE Criteria: Yes (barely)
Xprod = 80*(825/2) = 33,000 否 ITE 标准：是（勉强）

The N-S street has reasonable balance between the two left turns (90, 120), so we will use an exclusive LT lane. Also, with a confusing geometry, and simple signalization plan would be preferred. The E-W left turns are quite different, so a NEMA phasing plan with an exclusive LT phase followed by a leading green will be used.
南北街在两个左转（90 度，120 度）之间具有合理的平衡，因此我们将使用专用的左转车道。此外，由于几何形状复杂，简单的信号控制方案将更受欢迎。东西方向的左转则大相径庭，因此我们将使用带有专用左转相位，接着是领先绿灯的 NEMA 相位方案。

Converting Volumes to tvu’s
将流量转换为 tvu

Approach
方法

Mvt
运动

Volume (veh/h)
车流量 (辆/小时)

Equiv (T21.1,2)
当量 (T21.1,2)

Vol
交通量

(tvu/h)

Lane Grp
车道组

Vol
交通量

(tvu/h)

Vol/Ln (tvu/h/ln)
车道流量 (tvu/h/车道)

EB

L T R
左转右转

80

700

65

1.05 1.00 1.21

84

700

79

84

779

84

390

WB

L T R
左转右转

180

825

95

1.05 1.00 1.21

189

825

115

189

940

189

470

NB

L T R
左转右转

120

750

40

1.05 1.00 1.21

126

750

48

126

798

126

399

SB

L T R
左转右转

90

680

75

1.05 1.00 1.21

95

680

91

95

771

95

386

Ring Diagram and Sum of Critical Lane Volumes
环状图和关键车道流量总和

The ring diagram is shown on the next page, with appropriate demand volumes shown for each portion of each ring.
环状图显示在下一页，其中显示了每个环的每一部分的相应需求量。

17

A

B

C1

C2

C3

95

126

399

386

84

189

470

390

18

95 or 126*
95 或 126*

399* or 386
399 乘以或 386

84 + 470 = 554 Or
84 加 470 等于 554 或

189 + 390 = 579*

Vc = 126+399+579 = 1,104 tvu/h
Vc = 126 + 399 + 579 = 1,104 tvu/h

Yellow and All Red Intervals, Lost Times
黄灯和全红时间段，损失时间

The average speed on both streets is 35 mi/h. All yellow times will, therefore, be the same:
两条街道的平均车速为 35 英里/小时。因此，所有黄灯时间都将相同：

Even though there are only a few pedestrians, the awkward geometry gives adequate reason to time the all red intervals to allow vehicles to clear the far crosswalks, i.e., using P instead of w. N-S vehicles would then have to clear 50 + 10 = 60 ft. E-W vehicles would have to clear 120 – 10 = 110 ft (note that the 120 ft on the diagram goes from the far edges of both crosswalks). Then:
即使行人很少，但由于几何形状不规则，有充分理由将全红时间设置为允许车辆驶离远端人行横道，即使用 P 代替 w。那么，南北向车辆需要驶离 50 + 10 = 60 英尺。东西向车辆需要驶离 120 – 10 = 110 英尺（请注意，图中 120 英尺是从两个远端人行横道的边缘测量的）。然后：

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This is a four-phase signal, and there will be four sets of yellow and all red intervals, two for each street (as each street has protected LT phases). The total of all yellow and all-red phases is, therefore:
这是一个四相信号，将有两组黄灯和全红时间段，每条街道各一组（因为每条街道都有受保护的左转相位）。所有黄灯和全红相位的总和为：

(3.9+2.9)+(3.9+2.9)+(3.9+1.8)+(3.9+1.8) = 25.0 s
(3.9+2.9)+(3.9+2.9)+(3.9+1.8)+(3.9+1.8) = 25.0 秒

As we are using the standard default values for start-up lost time and encroachment (each 2.0 s), this is also equal to the total lost time in the cycle.
由于我们使用的是启动损失时间和占用时间的标准默认值（各 2.0 秒），这也等于循环的总损失时间。

Cycle Length
周期长度

The cycle length is computed as:
周期长度计算如下：

C = L

( V ⎞
（V⎞

⎝ 1615 * PHF * v/ c
⎝1615 * PHF * v/c,

Splitting the Green
绿色分割

The 100 s cycle length has 25 s of lost time, leaving 110 – 25 = 85 s of effective green time to assign to the four phases on the critical path, as follows:
100 秒的周期长度有 25 秒的损失时间，留下 110 – 25 = 85 秒的有效绿灯时间分配给关键路径上的四个阶段，如下所示：

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We must also establish the boundary between Phases C1 and C2, which occurs only on the non-critical path. The total length of Phases C1, C2, and C3 is 14.6+30.0 = 44.6 s. Then:
我们还必须确定 C1 和 C2 阶段之间的界限，这仅发生在非关键路径上。C1、C2 和 C3 阶段的总长度为 14.6 + 30.0 = 44.6 秒。然后：

gC2 = 44.6 − 30.0 − 6.7 = 7.9 s
gC2 = 44.6 − 30.0 − 6.7 = 7.9 秒

Checking Pedestrian Safety
检查行人安全

Pedestrians crossing the E-W street do so during Phase B. The crosswalk length is 50 ft. Then:
行人横穿东西向街道是在 B 阶段进行的。人行横道长度为 50 英尺。然后：

Npeds = 50 /(3600 /110) = 1.5 peds/ cycle
Npeds = 50 /（3600 / 110）= 1.5 个行人/周期

Pedestrians crossing the N-S street do so during Phase C3. The crosswalk length is 60 ft. Then:
穿越南北方向街道的行人，在 C3 阶段通行。人行横道长度为 60 英尺。然后：

GpC3 = 3.2 + (0.27 *1.5) + ⎜⎝(

 = 18.6 s < 30.0 s OK
= 18.6 秒 < 30.0 秒 OK

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