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

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:

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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。

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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。

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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：

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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 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：

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

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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

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）

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Problem 3-6
例题 3-6

Information:
信息：

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。

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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：

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) + ⎢

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

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。

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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

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

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：

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：

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

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：

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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

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。

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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。

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

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：

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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.

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

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。

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