Conference Paper $\cdot$ June 2012
会议论文 $\cdot$ 2012 年 6 月

DOI: $10.2514/6.2012-2650$ CITATIONS READS
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AIAA 2012-2650 美国建筑师协会 2012-2650

John C. Vassberg* 约翰·瓦斯伯格*

Long Beach, California 90803
加利福尼亚州长滩 90803

Aerodynamics Lecture 空气动力学讲座

${30}^{\text{th }}$ AIAA Applied Aerodynamics Conference
${30}^{\text{th }}$ AIAA 应用空气动力学会议

New Orleans, Louisiana 路易斯安那州新奥尔良

28 June, 2012 28 六月， 2012

Abstract 抽象

Aerodynamic characteristics and flight dynamics of boomerangs are investigated. A basic aerodynamic model, developed in the 1960's, is expanded upon using Blade Element Theory. The new aerodynamic model is coupled with a gyroscope model for rudimentary analyses. Some significant findings are made regarding the radius of a boomerang's circular flight path, the required inclination angle of its axis-of-rotation, its trim state, as well as its dynamic stability. These discoveries provide a basic understanding of how the interplay between aerodynamic forces and moments, and gyroscopic precession combine to return the boomerang to its rightful owner by way of a circular flight path.
研究了回旋镖的空气动力学特性和飞行动力学。1960 年代开发的基本空气动力学模型在使用叶片元件理论的基础上进行了扩展。新的空气动力学模型与陀螺仪模型相结合，用于基本分析。关于回旋镖圆形飞行路径的半径、旋转轴所需的倾斜角、配平状态以及动态稳定性，取得了一些重要的发现。这些发现为空气动力和力矩以及陀螺进动之间的相互作用如何结合，通过圆形飞行路径将回旋镖送回其合法所有者提供了基本理解。

A traditional V-shaped boomerang design is developed as a case study for further detailed analyses. Unsteady Reynolds-averaged Navier-Stokes solutions provide accurate aerodynamic characteristics of the subject boomerang. The high-fidelity aerodynamic model is coupled with the equations of motion to provide accurate six-degree-of-freedom simulations of boomerang flight dynamics. Boomerang orientation during its flight trajectory is described by the classical Euler angles.
传统的 V 形回旋镖设计被开发为案例研究，用于进一步的详细分析。非稳态 Reynolds 平均 Navier-Stokes 解可提供对象回旋镖的精确空气动力学特性。高保真空气动力学模型与运动方程相结合，以提供回旋镖飞行动力学的精确六自由度仿真。回旋镖在其飞行轨迹中的方向由经典的欧拉角描述。

Copyright (C) 2012 by John C. Vassberg.
版权所有 （C） 2012 John C. Vassberg。

Published by the American Institute of Aeronautics and Astronautics with permission.
经许可由美国航空航天学会出版。

*Boeing Technical Fellow, Fellow AIAA, Boomerang Novice
*波音技术研究员、AIAA 研究员、Boomerang 新手

$a,b,c,d$ Arbitrary Coefficients
$a,b,c,d$ 任意系数

b Coordinate Vectors of Body Frame
b Body Frame 的坐标向量

C Blade Reference Chord C Blade 参考和弦

2D Lift Coefficient $=\frac{lift}{qC}$
2D 提升系数 $=\frac{lift}{qC}$

$C_{l0}$ 2D Lift Coefficient at $\alpha =0$
$C_{l0}$ 2D 提升系数 $\alpha =0$

$C_{l\alpha }$ 2D Lift-Curve Slope
$C_{l\alpha }$ 2D 提升曲线坡度

3D Lift Coefficient $=\frac{Lift}{qS}$
3D 提升系数 $=\frac{Lift}{qS}$

3D Rolling-Moment Coefficient
3D 滚动力矩系数

$C_P$ Pressure Coefficient $=\frac{P-P_{\mathrm{\infty }}}{q_{\mathrm{\infty }}}$
$C_P$ 压力系数 $=\frac{P-P_{\mathrm{\infty }}}{q_{\mathrm{\infty }}}$

$g$ Gravitational Acceleration $\simeq 32\frac{ft}{se{c}^2}$
$g$ 引力加速度 $\simeq 32\frac{ft}{se{c}^2}$

i Coordinate Vectors of Inertial Frame
i 惯性系的坐标矢量

$I$ Moment of Inertia
$I$ 转动惯量

$L$ Lift  $L$ 电梯

$L_0$ Basic Lift  $L_0$ 基本提升

$L_{\alpha }$ Lift due to Angle-of-Attack $\eta$ Fraction of Blade Radius $=\frac{r}{R}$ .
$L_{\alpha }$ 由于叶片半径 $=\frac{r}{R}$ 的攻角 $\eta$ 分数 .

$LE$ Airfoil Leading Edge
$LE$ 翼型前沿

$m$ Mass  $m$ 质量

$M$ Rolling-Moment
$M$ Rolling-Moment （滚动力矩）

$M_0$ Basic Rolling-Moment
$M_0$ 基本滚动力矩

$M_{\alpha }$ Rolling-Moment due to Angle-of-Attack $\chi$ ND Rotational Velocity $=\frac{\omega R}{V}$
$M_{\alpha }$ 由攻角 $\chi$ ND 旋转速度 $=\frac{\omega R}{V}$ 引起的滚动力矩

n Node Vector of Body Frame
n Body Frame 的节点向量

$q$ Dynamic Pressure $=\frac{1}{2}\rho V^2$
$q$ 动态压力 $=\frac{1}{2}\rho V^2$

$r$ Radial Coordinate
$r$ 径向坐标

$R$ Reference Radius of Blade
$R$ 叶片的参考半径

${\mathcal{R}}_p$ Radius of Circular Flight Path
${\mathcal{R}}_p$ 圆形飞行路径半径

s Vector in Rotation Plane $\mathrm{\perp }$ n
s 旋转平面 $\mathrm{\perp }$ n 中的向量

$S$ Reference Area $=RC$
$S$ 参考区域 $=RC$

Transformation Matrix 变换矩阵

$TE$ Airfoil Trailing Edge
$TE$ 翼型后缘

$V$ Velocity  $V$ 速度

$x$ Streamwise Cartesian Coordinate
$x$ Streamwise Cartesian Coordinate （流式笛卡尔坐标）

Lateral Cartesian Coordinate
横向笛卡尔坐标

$z$ Vertical Cartesian Coordinate
$z$ 垂直笛卡尔坐标

$ND$ Non-Dimensionalized
$ND$ 无量纲

2D Two Dimensional 2D 二维

2 Three Dimensional 2 三维

$\alpha$ Angle-of-Attack
$\alpha$ 攻角

Euler Angle: Orbital-Plane Inclination
Euler Angle： 轨道-平面倾斜

$\varphi$ Euler Angle: Right Ascension
$\varphi$ Euler Angle： Right Ascension （欧拉角）：赤经

$\hat{\chi }$ Lessor of $\chi$ or $1;=min\left(\chi ,1\right)$ ’
$\hat{\chi }$ $\chi$ 或 $1;=min\left(\chi ,1\right)$ ' 的出租人

$\psi$ Euler Angle: Argument of Perigee
$\psi$ Euler Angle： Argument of Perigee （欧拉角）： Argument of Perigee （近地点参数）

${\psi }_c$ Cut-Off Angle $={\cos }^{-1}\left(-\chi \right)$
${\psi }_c$ 截止角 $={\cos }^{-1}\left(-\chi \right)$

$\omega$ Rotational Velocity
$\omega$ 旋转速度

Boomerangs are deceptively simple-looking devices. But make no mistake, boomerang aerodynamics are rather complex. Not only is the flow unsteady and very three dimensional; it also reverses itself across the wing such that the leading and trailing edges swap roles twice every cycle. Furthermore, the Reynolds number is very low, $\mathcal{O}\left(50,000\right)$ .
回旋镖是看似简单的设备。但不要误会，回旋镖空气动力学相当复杂。不仅流动不稳定且非常三维;它还会在伞体上反转自己，这样前缘和后缘在每个周期中交换角色两次。此外，雷诺数非常低， $\mathcal{O}\left(50,000\right)$ 。

In spite of these challenges and by trial and error, Australian Aborigines managed to develop boomerang technology some 10,000 years ago; see Lorenz, ${}^1{\text{Mauro}}^2$ and Hanson. ${}^3$ In fact,several civilizations developed straight-flying "boomerangs" for hunting and war many millenia ago. A gold-tip throwing-stick was found in King Tutankhamun's tomb. A boomerang was found in the sand dunes of The Netherlands that dates back 2,400 years. An exquisite mammoth-tusk throwing stick found in Poland is about 23,000 years old. This unique artifact has a mass of about $800$ grams,a span of almost $70\mathrm{cm}$ ,a chord of about $6\mathrm{cm}$ and is $1.5\mathrm{cm}$ thick. The ivory of this relic has a density of about 1,900 kg-per-cubic-meter. These ancient throwing sticks have aerodynamic and gyroscopic properties that stabilize their flights and extend their range. However, it was only the Aborigines who created the returning boomerang which they used primarily for recreation and sport, just as we do today. Indirectly, they also used the returning boomerang to aid in hunting fowl. Here, they would throw a boomerang over a flock of ducks floating on a pond. The waterfowl would mistake the boomerang for a hawk and would take flight, yet remain low to avoid the "predator". The low-flying game would then be well within range of hunters waiting nearby. Another unusual, but very utilitarian, use of the boomerang is when the Aborigines rubbed the leading edge against another piece of wood in a sawing motion to ignite a fire.
尽管面临这些挑战，并通过反复试验，澳大利亚原住民还是在大约 10,000 年前成功开发了回旋镖技术;见 Lorenz 和 ${}^1{\text{Mauro}}^2$ Hanson。 ${}^3$ 事实上，许多文明在几千年前就发展出了用于狩猎和战争的直飞“回旋镖”。在图坦卡蒙国王的坟墓中发现了一根金尖投掷棍。在荷兰的沙丘中发现了一个回旋镖，其历史可以追溯到 2,400 年前。在波兰发现的一根精美的猛犸象牙投掷棒大约有 23,000 年的历史。这种独特的文物质量约为 $800$ 克，跨度几乎 $70\mathrm{cm}$ 为，弦约为 $6\mathrm{cm}$ ，而且很 $1.5\mathrm{cm}$ 厚。这件文物的象牙密度约为每立方米 1,900 公斤。这些古老的投掷棒具有空气动力学和陀螺仪特性，可以稳定它们的飞行并扩大它们的射程。然而，只有原住民创造了回归的回旋镖，他们主要用于娱乐和运动，就像我们今天所做的那样。间接地，他们还使用返回的回旋镖来帮助狩猎家禽。在这里，他们会把回旋镖扔到漂浮在池塘上的一群鸭子上。水禽会把回旋镖误认为是鹰，然后飞起来，但为了躲避“捕食者”而保持低位。然后，低空飞行的猎物将完全在附近等待的猎人的射程内。回旋镖的另一个不寻常但非常实用的用途是，原住民以锯切动作将前缘与另一块木头摩擦以点燃火。

Figure 1 provides a small sampling of early ${20}^{\text{th }}$ -century Aboriginal boomerangs. The boomerangs in this figure are probably straight-flying configurations. However, it is hard to know without throwing them - it is possible they are of the returning type. Figure 2 illustrates a wide variety of modern-day boomerangs; all of these examples are returning designs. Of particular note are the Tri-Fly and LED boomerangs, as these are utilized by the author for flight tests and further analysis in this body of work. The middle V-type boomerang of the modern Aboriginal group is similar to a design developed herein for the purpose of high-fidelity analysis using computational fluid dynamics.
图 1 提供了本世纪早期 ${20}^{\text{th }}$ 土著回旋镖的一小部分样本。这个图中的回旋镖可能是直线飞行的配置。但是，如果不抛出它们就很难知道 - 它们可能是 return 类型。图 2 说明了各种各样的现代回旋镖;所有这些示例都是返回的设计。特别值得注意的是 Tri-Fly 和 LED 回旋镖，因为作者将这些用于飞行测试和本工作中的进一步分析。现代原住民群体的中间 V 型回旋镖类似于本文开发的设计用于使用计算流体动力学进行高保真分析的设计。

A number of boomerang world records as of June 2010 are included; see Wikipedia. ${}^4$ The most number of catches made within 5 minutes is 81 by Manuel Schutz, Switzerland, 2005. The quickest five-throw-catch sequence is 14.6 seconds by Adam Ruhf, USA, 1996. The most number of consecutive catches is 2,251 by Haruki Taketomi, Japan, 2009. The maximum time aloft in the unlimited class is 380.6 seconds by Billy Brazelton, USA, 2010. The farthest distance out for a returning flight is 238 meters by Manuel Schutz, Switzerland, 1999. The longest hand-held throw of any object was set in 2005 by David Schummy when he threw a boomerang 427.2 meters (over a quarter-mile). All of these are incredible statistics; little did the Aborigines know just how far their returning sticks could go.
包括截至 2010 年 6 月的许多 boomerang 世界纪录;参见 维基百科. ${}^4$ 5 分钟内捕获的最多渔获数是瑞士 Manuel Schutz 于 2005 年完成的 81 次。最快的五次投球序列是 14.6 秒，由美国 Adam Ruhf 于 1996 年创作。连续捕捞次数最多的是 Haruki Taketomi，日本，2009 年，2009 年。无限级的最高高空飞行时间为 380.6 秒，由美国 Billy Brazelton 于 2010 年提供。返程航班最远的距离是 238 米，由 Manuel Schutz， Switzerland， 1999 提供。大卫·舒米 （David Schummy） 在 2005 年创造了任何物体中最长的手持投掷，当时他投掷了 427.2 米（超过四分之一英里）的回旋镖。所有这些都是令人难以置信的统计数据;原住民几乎不知道他们返回的棍子能走多远。

This paper is organized in the following manner. Section II introduces Euler's classical orientation angles. Section III derives the equations of angular motion associated with boomerang dynamics. Section IV provides a discussion regarding the kinematics of an over-hand boomerang throw. Section V revisits work done in the 1960's regarding the aerodynamics of a boomerang. Section VI develops a new aerodynamic model for boomerangs based on blade element theory. Section VII develops general formulas for the radius of a boomerang's circular flight path, as well as the rotation-plane inclination angle. Section VIII develops a general condition for static trim and proves that dynamic stability is an intrinsic property of the boomerang. Section IX develops a traditional V-type boomerang design for further analysis. Section X provides the two-dimensional steady-state CFD results for the boomerang airfoil. Section XI discusses the three-dimensional unsteady CFD results for the boomerang subjected to a fixed cyclic motion. Section XII provides a full 6- DOF numerical simulation of a boomerang in flight. Section XIII is a summary of conclusions. Throughout the text, references are given as positive superscripts while footnotes are indicated by negative superscripts. Tables of data are embedded within the text while all figures are appended to the end of the paper.
本文的组织方式如下。第 II 节介绍了欧拉的经典取向角。第 III 节推导出了与回旋镖动力学相关的角运动方程。第四部分讨论了上手回旋镖投掷的运动学。第五部分回顾了 1960 年代关于回旋镖空气动力学所做的工作。第六部分基于叶片元件理论开发了一种新的回旋镖空气动力学模型。第七节开发了回旋镖圆形飞行路径半径以及旋转平面倾斜角的一般公式。第八节制定了静态配平的一般条件，并证明了动态稳定性是回旋镖的内在特性。第九部分开发了传统的 V 型回旋镖设计以供进一步分析。第 X 部分提供了 Boomerang 翼型的二维稳态 CFD 结果。第十一节讨论了在固定循环运动作用下回旋镖的三维非稳态 CFD 结果。第 XII 部分提供了飞行中回旋镖的完整 6 自由度数值模拟。第十三节是结论的总结。在整个文本中，参考文献以正上标形式给出，而脚注以负上标表示。数据表嵌入在文本中，而所有图表都附加到论文的末尾。

The classical Euler angles $\varphi ,\theta$ and $\psi$ are shown in Figure 3. Also depicted in Figure 3 are the inertial and body-fixed coordinate systems,labeled as $\left({\mathbf{i}}_{\mathbf{1}},{\mathbf{i}}_{\mathbf{2}},{\mathbf{i}}_{\mathbf{3}}\right)$ and $\left({\mathbf{b}}_{\mathbf{1}},{\mathbf{b}}_{\mathbf{2}},{\mathbf{b}}_{\mathbf{3}}\right)$ ,respectively. In celestial mechanics, the angles $\left(\varphi ,\theta ,\psi \right)$ are used to specify the orientation of an orbit. They are known as the right ascension of the ascending node, the inclination of the orbital plane, and the argument of perigee, respectively. For the boomerang problem,as set-up herein,the axis of rotation is about the ${\mathbf{b}}_{\mathbf{3}}$ vector, $\theta$ describes the bank angle that the plane of rotation makes with the horizon,while the precession rate of ${\mathbf{b}}_{\mathbf{3}}$ about ${\mathbf{i}}_{\mathbf{3}}$ is given by $\varphi$ .
经典的欧 $\varphi ,\theta$ 拉角 和 $\psi$ 如图 3 所示。图 3 中还描绘了惯性坐标系和物体固定坐标系，分别标记为 $\left({\mathbf{i}}_{\mathbf{1}},{\mathbf{i}}_{\mathbf{2}},{\mathbf{i}}_{\mathbf{3}}\right)$ 和 $\left({\mathbf{b}}_{\mathbf{1}},{\mathbf{b}}_{\mathbf{2}},{\mathbf{b}}_{\mathbf{3}}\right)$ 。在天体力学中，角度 $\left(\varphi ,\theta ,\psi \right)$ 用于指定轨道的方向。它们分别被称为升交点的赤经、轨道平面的倾角和近地点的参数。对于回旋镖问题，如本文所述，旋转轴是关于矢量的 ${\mathbf{b}}_{\mathbf{3}}$ ， $\theta$ 描述了旋转平面与地平线形成的倾斜角，而大约 ${\mathbf{i}}_{\mathbf{3}}$ 的 ${\mathbf{b}}_{\mathbf{3}}$ 进动率由下 $\varphi$ 式给出。

Transforming from the inertial frame to the body frame can be accomplished with three rotations in series. Pay attention to the order of rotations as this is crucial. First rotate by $\varphi$ about the ${\mathbf{i}}_{\mathbf{3}}$ -axis,then by $\theta$ about the node vector $\mathbf{n}$ and finally with a rotation of $\psi$ about the ${\mathbf{b}}_{\mathbf{3}}$ -axis. The complete transformation from the inertial frame to the body frame using this three-rotation sequence can be described as
从惯性坐标系到体坐标系的转换可以通过串联三次旋转来完成。注意轮换的顺序，因为这很关键。首先 $\varphi$ 绕 ${\mathbf{i}}_{\mathbf{3}}$ -轴旋转，然后绕 $\theta$ 节点向量 $\mathbf{n}$ 旋转，最后绕 ${\mathbf{b}}_{\mathbf{3}}$ -轴旋转 $\psi$ 。使用这个 3 次旋转序列从惯性系到体系的完整变换可以描述为

Here, $R\left(\alpha ;\mathbf{v}\right)$ represents an elementary rotation of angle $\alpha$ about vector $\mathbf{v}$ . This sequence of rotations is graphically illustrated by Figure 4. Completing the multiplication symbolized by Eqn (1), one obtains the transformation matrix
其中， $R\left(\alpha ;\mathbf{v}\right)$ 表示绕向量 $\mathbf{v}$ 的角度 $\alpha$ 的基本旋转。图 4 以图形方式说明了这种旋转顺序。完成由 Eqn （1） 符号化的乘法，得到变换矩阵

Here, $c$ indicates cosine and $s$ indicates sine. Since Eqn (2) is a pure rotation,its inverse is just its transpose. Hence, to transform from the body frame to the inertial frame, one uses
这里， $c$ 表示余弦和 $s$ 正弦。由于 Eqn （2） 是一个纯旋转，它的逆只是它的转置。因此，要从体系变换到惯性系，可以使用

The next section discusses Euler's equations of angular motion.
下一节讨论 Euler 的角运动方程。

In this section,we begin with the equations of angular motion as defined by Wiese ${}^{15}$ and then adapt them for boomerang dynamics. We start the discussion with a very generalized form of Euler's equations of angular motion, then progress through a series of simplifications which retain only essential elements of the equations needed to describe a level circular flight path typical of a returning boomerang. Further simplifications regarding mass properties are introduced which appropriately continue to capture the low-frequency content of dynamic motion for boomerang flight. While this may filter or blur some of the high-frequency motion within a spin cycle, it greatly reduces the complexity of the ensuing derivations presented throughout this body of work.
在本节中，我们从 Wiese ${}^{15}$ 定义的角运动方程开始，然后根据回旋镖动力学对它们进行调整。我们从欧拉角运动方程的非常概括的形式开始讨论，然后通过一系列简化进行讨论，这些简化只保留了描述返回回旋镖典型的水平圆形飞行路径所需的方程的基本元素。引入了有关质量属性的进一步简化，这些属性适当地继续捕获回旋镖飞行的动态运动的低频内容。虽然这可能会过滤或模糊自旋周期中的一些高频运动，但它大大降低了整个工作中呈现的后续推导的复杂性。

For the boomerang problem, the principal-axis frame naturally aligns with an appropriate body-fixed coordinate system. This is fortuitous as the equations of angular motion are simplest when expressed in the principal-axis body frame because the off-diagonal terms of the moment-of-inertia matrix vanish. In the present work, the origin of the principal-axis frame coincides with the center-of-mass of the boomerang. When the Euler angles are zeroed, the body and inertial frames are aligned; see initial state of Figure 4. In this position,the $Z$ -axis is the axis-of-rotation, $Y$ is in the spanwise direction and $X$ is streamwise. Here,the boomerang reference plane coincides with the $X-Y$ plane where $Z=0$ ,the boomerang’s centerline coincides with the $X$ -axis,the right wing resides in the half-space where $Y\ge 0$ and the boomerang upper-surface resides in the $Z\ge 0$ half-space. For this system,Euler’s equations of angular motion are given as:
对于回旋镖问题，主轴坐标系自然会与适当的身体固定坐标系对齐。这是偶然的，因为在主轴体坐标系中表示时，角运动方程是最简单的，因为惯性矩矩阵的非对角线项消失了。在本工作中，主轴框架的原点与回旋镖的质心重合。当 Euler 角度归零时，物体和惯性坐标系对齐;参见图 4 的初始状态。在这个位置， $Z$ -轴是旋转轴， $Y$ 在翼展方向上， $X$ 是流向的。在这里，回旋镖参考平面与 $X-Y$ 平面重合，其中 $Z=0$ 回旋镖的中心线与 $X$ 轴重合，右翼位于半空间中 $Y\ge 0$ ，回旋镖上表面位于 $Z\ge 0$ 半空间中。对于该系统，欧拉角运动方程为：

Here,the $\omega$ -vector describes the angular rates in the principal body-axis frame. These equations provide the time-rate-of-change of the $\omega$ -vector,yet only constitute one half of the rotational equations of motion. Another set of equations is needed which link the angular rates of the body frame to the orientation angles of the body in the inertial frame. The remaining set of equations is given as:
在这里， $\omega$ -vector 描述了主体轴坐标系中的角速率。这些方程提供了 $\omega$ -向量的时间变化率，但只构成了旋转运动方程的一半。需要另一组方程，将体系的角速率与惯性系中体的方向角联系起来。其余的方程组如下：

These equations transform the $\omega$ -vector given by Eqn (4) to provide the time-rate-of-change of the Euler angles $\left(\dot{\varphi },\dot{\theta },\dot{\psi }\right)$ . In practice,these angular velocities are used to update the orientation angles $\left(\varphi ,\theta ,\psi \right)$ for the body as a function of time.
这些方程变换 $\omega$ 方程 （4） 给出的向量，以提供欧 $\left(\dot{\varphi },\dot{\theta },\dot{\psi }\right)$ 拉角 的时间变化率。在实践中，这些角速度用于更新物体的取向角 $\left(\varphi ,\theta ,\psi \right)$ 作为时间的函数。

If the body is axi-symmetric about one of the principal axes, then further simplifications can be made to the equations of angular motion. So how can a boomerang be axi-symmetric? This question is best answered if one views the boomerang when it is spinning fast, rather than when statically displayed. Just as this motion blurs the visual of the boomerang to appear as a spinning disk, the moments of inertia can be time-averaged in the $X-Y$ plane. Although this simplification alters high-frequency content,it retains sufficient mass properties to accurately describe low-frequency boomerang dynamics.
如果物体绕其中一个主轴轴对称，则可以进一步简化角运动方程。那么回旋镖怎么可能是轴对称的呢？如果人们在回旋镖快速旋转时查看它，而不是在静态显示时查看它，则可以最好地回答这个问题。正如这种运动使回旋镖的视觉效果变得模糊，看起来像一个旋转的圆盘一样，惯性矩可以在 $X-Y$ 平面上进行时间平均。尽管这种简化改变了高频内容，但它保留了足够的质量特性来准确描述低频回旋镖动力学。

For the purpose of this discussion,let’s assume that the body is symmetric about the ${\mathbf{b}}_{\mathbf{3}}$ -axis and a rolling-moment exists purely about the $\mathbf{n}$ -axis. Let’s designate this torque about the nodal line as $M_n$ . See Figure 3 to get familiar with this situation. Note that the $\mathbf{n}$ -axis is perpendicular to the ${\mathbf{b}}_{\mathbf{3}}$ -axis and thus can be described as a linear combination of ${\mathbf{b}}_{\mathbf{1}}$ and ${\mathbf{b}}_{\mathbf{2}}$ .
为了进行本讨论，我们假设物体绕 ${\mathbf{b}}_{\mathbf{3}}$ -轴对称，并且滚动力矩纯粹围绕 $\mathbf{n}$ -轴存在。让我们将关于节点线的扭矩指定为 $M_n$ 。请参阅 图 3 以熟悉这种情况。请注意， $\mathbf{n}$ -axis 垂直于 ${\mathbf{b}}_{\mathbf{3}}$ -axis，因此可以描述为 ${\mathbf{b}}_{\mathbf{1}}$ 和 ${\mathbf{b}}_{\mathbf{2}}$ 的线性组合。

Also define an s-axis which is perpendicular to both the $\mathbf{n}$ -axis and the ${\mathbf{b}}_{\mathbf{3}}$ -axis.
还要定义一个垂直于 $\mathbf{n}$ -轴和 -轴的 ${\mathbf{b}}_{\mathbf{3}}$ s 轴。

Note that the s-axis is not shown in Figure 3.
请注意，图 3 中未显示 s 轴。

Because of symmetry,let’s define a moment of inertia $\left(I_{rr}\right)$ that is constant for any radial in the $\left(\mathbf{n},\mathbf{s}\right)$ plane. Hence,
由于对称性，让我们定义一个惯性矩 $\left(I_{rr}\right)$ ，该矩对于平面中的任何径向都是恒定的 $\left(\mathbf{n},\mathbf{s}\right)$ 。因此

Combining Eqns (4-8) gives:
组合方程 （4-8） 可得到：

If an axi-symmetric body is also planar and homogeneous (i.e., a disk), then the principal moments of inertia are simply:
如果轴对称体也是平面和均匀的（即圆盘），那么主惯性矩就是：

where $m$ is the mass of the disk and $R$ its radius. Now,the equations of angular motion reduced to:
其中 $m$ 是圆盘的质量及其 $R$ 半径。现在，角运动方程简化为：

For the moment,let’s assume that the angular motion is such that $\theta$ is constant,then $\dot{\theta }=\ddot{\theta }=0$ . Now, if the rolling-moment is purely about the $\mathbf{n}$ -axis,then $M_s=0$ . Let’s also assume that there is no moment being applied about the spin axis,i.e., $M_3=0$ . Under these conditions,the equations of angular motion reduce further to:
目前，我们假设角运动是 $\theta$ 恒定的，那么 $\dot{\theta }=\ddot{\theta }=0$ 。现在，如果滚动力矩纯粹是关于 -轴 $\mathbf{n}$ 的，那么 $M_s=0$ 。我们还假设没有关于自旋轴的弯矩，即 $M_3=0$ 。在这些条件下，角运动方程进一步简化为：

One may recognize these equations as those that describe the motion of an inclined toy gyroscope, with its tip resting on a table. Here,the high spin rate of the gyroscope disk remains at a constant $\psi$ ,with the gyroscope slowly rotating around its pivot-point-vertical-axis with a constant precession rate of $\varphi$ ,all the while,maintaining a constant inclination angle $\theta$ . This precessional motion is driven by the moment $M_n$ created by the gyroscope's weight, multiplied by the lateral off-set distance between the pivot point and the gyroscope’s center-of-mass. For boomerang dynamics,the moment $\left(M_n\right)$ is supplied aerodynamically.
人们可能会认出这些方程式是描述倾斜玩具陀螺仪的运动的方程式，陀螺仪的尖端放在桌子上。在这里，陀螺仪盘的高自旋速率保持恒定 $\psi$ ，陀螺仪以恒定的进动率绕其枢轴-垂直-轴缓慢旋转 $\varphi$ ，同时保持恒定的倾角 $\theta$ 。这种进动运动由陀螺仪重量产生的力矩 $M_n$ 乘以枢轴点与陀螺仪质心之间的横向偏移距离驱动。对于回旋镖动力学，力矩 $\left(M_n\right)$ 由空气动力学提供。

If $M_n,\theta$ and $\psi$ are known,then $\varphi$ can be determined using the quadratic formula applied to Eqn (12).
如果 $M_n,\theta$ 和 $\psi$ 已知，则可以使用 $\varphi$ 应用于方程 （12） 的二次公式来确定。

From experience, we know that the direction of the precession is the same as that of the disk spin. Hence, the positive branch of Eqn (13) is physically correct. Further,if $\dot{\varphi }\ll \dot{\psi }$ ,then solution of Eqn (13) can be
根据经验，我们知道岁差的方向与圆盘自旋的方向相同。因此，方程 （13） 的正分支在物理上是正确的。此外，如果 ，则 $\dot{\varphi }\ll \dot{\psi }$ 方程 （13） 的解可以是

approximated by: 近似值：

where the second form is included for the more general case of constant $\theta$ motion. Eqn (14) can be directly derived from Eqn (12) by dropping the ${\dot{\varphi }}^2$ term; the intermediate step of Eqn (13) is included as a formality.
其中第二种形式包括在恒定 $\theta$ 运动的更一般情况下。Eqn （14） 可以通过删除 ${\dot{\varphi }}^2$ 项直接从 Eqn （12） 派生出来;Eqn （13） 的中间步骤作为形式包括在内。

Eqns (12-14) will be used to develop a simplified analysis of a boomerang's flight dynamics to determine the radius of its circular flight path as well as other pertinent relationships.
方程 （12-14） 将用于开发对回旋镖飞行动力学的简化分析，以确定其圆形飞行路径的半径以及其他相关关系。

The next section reviews the kinematics of an over-hand boomerang throw.
下一节回顾了上手回旋镖投掷的运动学。

A brief discussion regarding the kinematics of a boomerang throw is included in this section. The primary objective is to understand how the initial conditions of the boomerang's kinetics are somewhat predetermined by the kinematic motion of an over-hand throw.
本节包括有关回旋镖投掷运动学的简要讨论。主要目标是了解回旋镖动力学的初始条件是如何在某种程度上由上手投掷的运动学预先决定的。

Consider holding a basketball in the cup of your hand and throwing it under-handed as if bowling. The natural motion associated with this toss has the basketball rolling off the fingertips just as the ball is being released. This establishes both a forward velocity as well as a rotational velocity for the basketball. When the basketball makes contact with the ground, its spin rate will remain essentially unchanged. Hence, the natural kinematics of bowling a basketball creates an initial relationship between the ball's velocity and its spin rate.
考虑将篮球放在手杯中，然后像打保龄球一样用手下扔。与这次抛球相关的自然运动使篮球在球被释放时从指尖滚落。这确定了篮球的前进速度和旋转速度。当篮球与地面接触时，其旋转速率将基本保持不变。因此，打篮球保龄球的自然运动学在球的速度和旋转速率之间产生了初始关系。

Here, $V$ is the translational velocity of the basketball, $\omega$ is its rotational velocity and $R$ its radius. While the relationship defined in Eqn (15) is only an approximate one and the person bowling the basketball can bias it one way or the other, it is quite difficult to violate it entirely.
这里， $V$ 是篮球的平移速度， $\omega$ 是它的旋转速度和 $R$ 半径。虽然 Eqn （15） 中定义的关系只是一个近似的关系，并且打篮球的人可能会以某种方式偏向它，但完全违反它是相当困难的。

Let's now consider a boomerang throw and draw similarities with the above example. To start, one holds a blade of the boomerang near its tip, grasping it in the palm of his hand. Because of the width of the hand,the center of this grasp is located around $85\mathrm{\%}-90\mathrm{\%}$ of the blade’s radius. Traditionally,a V-type configuration is held by the lifting blade,with the dingle blade- ${}^1$ pointed forward at the far end of the boomerang. The boomerang throw proceeds in a manner analogous to throwing a small axe or a hatchet over-handed. The plane of rotation is nearly vertical, with only a slight lay-over angle. Hence, the axis of rotation is nearly horizontal. In terms of Euler's orientation angles, this corresponds to an initial inclination angle of the rotation plane in the range:
现在让我们考虑一个 boomerang throw and draw 与上述示例的相似之处。首先，一个人将回旋镖的刀片靠近其尖端，将其握在手掌中。由于手的宽度，这种抓握的中心位于刀片半径的周围 $85\mathrm{\%}-90\mathrm{\%}$ 。传统上，V 型配置由提升刀片保持，dingle 刀片 - ${}^1$ 指向回旋镖的远端。回旋镖投掷的方式类似于双手投掷小斧头或斧头。旋转平面几乎是垂直的，只有一个轻微的放置角度。因此，旋转轴几乎是水平的。就 Euler 的方向角而言，这对应于 范围内旋转平面的初始倾角：

Since the boomerang blade must be firmly held, it does not roll out of the hand as efficiently as does the finger-tip roll of the basketball. Hence, the relationship between the initial translational and rotational velocities is slightly different than before. In the case of the boomerang throw, we draw from experience, make an educated guess, and assume this relationship to be:
由于必须牢牢握住回旋镖刀片，因此它不会像篮球的指尖滚动那样有效地从手中滚出。因此，初始平移速度和旋转速度之间的关系与以前略有不同。在回旋镖投掷的情况下，我们从经验中汲取灵感，做出有根据的猜测，并假设这种关系是：

Here, ${\omega }_0$ and $V_0$ are the boomerang’s initial rotational and translational velocities,respectively. Also define the initial rotational velocity $\left({\omega }_0\right)$ of $\mathrm{Eqn}\left(17\right)$ in a nondimensionalized form; this form will be used throughout the body of this work.
这里， ${\omega }_0$ 和 $V_0$ 分别是回旋镖的初始旋转和平移速度。还要定义 的初始旋转速度 $\left({\omega }_0\right)$ $\mathrm{Eqn}\left(17\right)$ 以无量纲形式;此表格将贯穿本作品的整个正文。

The initial condition of Eqns (17-18) has the forward-going tip moving at $1.85V_0$ and the bottom tip also moving forward but only at a speed of $0.15V_0$ . In practice,a typical throw of a boomerang will yield a translational speed of about $60\mathrm{mph}$ ,with a rotational cadence of about $10\mathrm{Hz}$ .
方程 （17-18） 的初始条件是向前的尖端移动 $1.85V_0$ ，底部尖端也向前移动，但速度为 $0.15V_0$ 。在实践中，典型的回旋镖投掷将产生大约 $60\mathrm{mph}$ 的平移速度，旋转节奏约为 $10\mathrm{Hz}$ 。

In order to verify the assumption made by Eqns (17-18), we analyze a strobed image of a boomerang in night flight; ${}^6$ see Figures 5-6. In Figure 5,note the insert at the bottom-right corner. This insert shows a close-up view of a three-bladed boomerang with an LED strobe light located on each blade at $75\mathrm{\%}$ of its radius. Also in Figure 5 is a red rectangular box that traps a portion of the flight path approximately perpendicular to the camera's view. This segment of the flight path is depicted in Figure 6 and is further analyzed. Since there are three strobed paths, focus on the one centered about the solid bold vertical aqua line. To extract the relationship between translational and rotational velocities of the boomerang at this stage of its flight, a hypocycloid is fit through the strobed curve of interest. The result is shown as the bold gray line that fits the strobed-light path. The hypocycloid corresponding to this path is graphically given by the very-large-radius circular contact surface depicted by the orange line at the bottom of the image and the rolling wheel given by the yellow circle. The green circle in this figure represents the swept area of the boomerang while the gray circle coincides with the location of the LED light. Finally, the ratio of the radii of the gray-to-green circles provides a derived relationship of rotational-to-translational velocities for this flight segment. In this particular case, the initial condition of Eqns (17-18) is exactly verified. In the author's experience, the translational velocity decays slightly faster than does the rotational velocity during the course of a boomerang flight. Hence,the nondimensionalized rotational velocity $\left(\chi \right)$ of a boomerang increases during the course of its mission, from start to finish, so we have:
为了验证 Eqns （17-18） 的假设，我们分析了夜间飞行中回旋镖的频闪图像; ${}^6$ 参见图 5-6。在图 5 中，请注意右下角的插入。该插页显示了三叶片回旋镖的特写视图，每个叶片在其半径处 $75\mathrm{\%}$ 都有一个 LED 频闪灯。图 5 中还有一个红色矩形框，它捕获了大致垂直于相机视图的飞行路径的一部分。图 6 描绘了这段飞行路径，并进行了进一步分析。由于有三条叠放路径，因此请专注于以实线粗体垂直浅绿色线为中心的路径。为了提取回旋镖在飞行的这个阶段的平移速度和旋转速度之间的关系，通过感兴趣的选通曲线拟合下摆线。结果显示为适合频闪光路径的粗灰色线。与该路径对应的近摆线由图像底部的橙色线和黄色圆圈表示的滚动轮以图形方式给出。图中的绿色圆圈代表回旋镖的扫掠区域，而灰色圆圈与 LED 灯的位置相吻合。最后，灰色与绿色圆圈的半径比提供了该飞行段的旋转速度与平移速度的推导关系。在这种特殊情况下，方程 （17-18） 的初始条件得到了精确验证。根据作者的经验，在回旋镖飞行过程中，平移速度的衰减速度比旋转速度略快。因此，回旋镖的无量纲旋转速度 $\left(\chi \right)$ 在其任务过程中从开始到结束都会增加，因此我们有：

${}^{-1}$ The dingle blade is the blade most-influenced by the wake of the other for a V-type boomerang.
${}^{-1}$ 对于 V 型回旋镖，dingle 刀片是受另一个尾流影响最大的刀片。

This property will be used in Section VII to explain why the inclination angle of a properly-design boomerang reduces during flight.
这一特性将在第七节中用于解释为什么设计合理的回旋镖在飞行过程中的倾角会减小。

Due to the natural kinematics of a boomerang throw, it is almost impossible to induce an extremely-high rotational velocity at point of launch. However, if one throws a boomerang slightly downward (like throwing a skipping stone) and bounces it off the ground, a high-spin-rate flight can be achieved. In essence, some of the translational kinetic energy gets impulsively converted to rotational kinetic energy. To establish a rough order-of-magnitude estimate, assume that the instantaneous bounce event conserves kinetic energy. Using the moment of inertia of a disk from Eqn (10), the states before and after the bounce are related as:
由于回旋镖投掷的自然运动学，几乎不可能在发射点产生极高的旋转速度。但是，如果将回旋镖稍微向下投掷（就像投掷跳石）并将其从地面上弹起，则可以实现高旋转速率的飞行。从本质上讲，一些平移动能被冲动转化为旋转动能。要建立一个粗略的数量级估计，假设瞬时反弹事件动能守恒。使用 Eqn （10） 中磁盘的惯性矩，反弹前后的状态相关如下：

where $R$ is the radius of the boomerang,and subscripts 0 and 1 indicate the states before and after the bounce event, respectively. With some rearrangement, substitution of Eqn (17) into Eqn (20) gives:
其中 $R$ 是回旋镖的半径，下标 0 和 1 分别表示 bounce 事件前后的状态。经过一些重新排列，将 Eqn （17） 代入 Eqn （20） 得到：

Hence,if the bounce event reduces the translational velocity $\left(V\right)$ by $50\mathrm{\%}$ ,the rotational velocity $\left(\omega \right)$ will increase by about $75\mathrm{\%}$ . However in nondimensionalized terms,the ratio between rotational and translational velocities change by a factor of about 3.5 due to the impulse of the bounce,to give ${\chi }_1\doteq 3.0$ .
因此，如果弹跳事件使平移速度 $\left(V\right)$ 降低 $50\mathrm{\%}$ ，则旋转速度 $\left(\omega \right)$ 将增加约 $75\mathrm{\%}$ 。然而，在无量纲化的术语中，由于反弹的冲量，旋转速度和平移速度之间的比率发生了大约 3.5 倍的变化，得到 ${\chi }_1\doteq 3.0$ 。

The next section provides some background on work done in the field of boomerang flight dynamics nearly 50 years ago.
下一节提供了近 50 年前在回旋镖飞行动力学领域所做的一些工作的背景。

In the 1960’s,the aerodynamics of boomerangs was studied by Felix Hess. ${}^7$ Hess concluded that for sufficiently rapid rotation, the time-averaged lift and rolling-moment of an advancing boomerang are directly proportional to ${\omega }^2$ and $\omega V$ ,respectively. Hence,
1960 年代，Felix Hess 研究了回旋镖的空气动力学。 ${}^7$ 赫斯得出结论，对于足够快的旋转，前进的回旋镖的时间平均升力和滚动力矩分别与 ${\omega }^2$ 成 $\omega V$ 正比。因此

where $a$ and $b$ are constants of proportionality, $\omega$ is the rotational speed of the boomerang,and $V$ is its flight speed. Note that the coefficients $a$ and $b$ lump together both aerodynamic characteristics and mass properties of a given boomerang. The units of $a$ and $b$ are $lbs-se{c}^2$ .
其中 $a$ 和 $b$ 是比例常数， $\omega$ 是回旋镖的转速，是 $V$ 它的飞行速度。请注意，系数 $a$ 和 $b$ 将给定回旋镖的空气动力学特性和质量特性归为一谈。和 $b$ 的 $a$ 单位是 $lbs-se{c}^2$ 。

Hess utilized his aerodynamic model, given by Eqn (23), in conjunction with the gyroscope model given by Eqn (11), to develop representative flight paths for a family of boomerangs, simply by varying the coefficients $a$ and $b$ . A deficiency of Hess’ aerodynamics model is that it cannot self trim. Further,if manually trimmed, it is neutrally stable. However from practice, we know that boomerangs are dynamically stable over a wide variety of shapes and flight conditions; there must exist an inherent self-trimming capability in their designs. Hence, further work is needed to develop an aerodynamics model that can accommodate this characteristic.
Hess 利用 Eqn （23） 给出的空气动力学模型和 Eqn （11） 给出的陀螺仪模型，只需改变系数 $a$ 和 $b$ ，就可以为一系列回旋镖开发具有代表性的飞行路径。赫斯空气动力学模型的一个缺陷是它不能自我修剪。此外，如果手动修剪，它是中性稳定的。然而，从实践中我们知道，回旋镖在各种形状和飞行条件下都是动态稳定的;他们的设计中必须存在固有的 self-trimning 功能。因此，需要进一步的工作来开发能够适应这一特性的空气动力学模型。

For easier comparison later, Hess' model, given by Eqn (23), is reformulated. In the current work, the rotational speed $\left(\omega \right)$ is nondimensionalized as follows.
为了便于以后比较，我们重新制定了 Eqn （23） 给出的 Hess 模型。在当前工作中，转速 $\left(\omega \right)$ 是无量纲的，如下所示。

where $R$ is the radius of boomerang blades,and $V_{tip}$ is the blade tip speed due to rotation. We also refer to $\chi$ as the tip-speed ratio. Substitution of Eqn (24) into Eqn (23) gives Hess’ aerodynamic model in terms of
其中 $R$ 是回旋镖叶片的半径，是 $V_{tip}$ 旋转引起的刀尖速度。我们也称 $\chi$ 之为叶尖速度比。将 Eqn （24） 代入 Eqn （23） 可得出 Hess 的空气动力学模型，即

$\chi$ as:  $\chi$ 如：

where the bracketed terms are now the constants of proportionality.
其中，括号内的项现在是比例的常数。

Time-averaged relationships for lift and rolling-moment will be significantly enhanced in the next section using Blade Element Theory.
在下一节中，将使用叶片元理论显著增强升力矩和滚动力矩的时间平均关系。

An aerodynamic model is developed herein using Blade Element Theory (BET). This model uses the lift-curve of a boomerang's two-dimensional airfoil section and it is assumed that the boomerang operates at angles-of-attack close to zero. ${}^{-2}$ Hence,the relationship between lift and alpha is essentially linear.
本文使用叶片元件理论 （BET） 开发了空气动力学模型。该模型使用回旋镖二维翼型截面的升力曲线，并假设回旋镖以接近零的攻角运行。 ${}^{-2}$ 因此，提升和 alpha 之间的关系基本上是线性的。

Here, $\alpha$ is the angle-of-attack of the airfoil, $C_{l0}$ is its lift coefficient at $\alpha =0$ and $C_{l\alpha }$ is the slope of the lift curve. These values will be determined numerically by running computational fluid dynamics (CFD) on a representative airfoil section of the boomerang blade over a range of angles-of-attack.
其中， $\alpha$ 是翼型的攻角， $C_{l0}$ 是它的升力系数， $\alpha =0$ $C_{l\alpha }$ 是升力曲线的斜率。这些值将通过在一定迎角范围内对回旋镖叶片的代表性翼型截面运行计算流体动力学 （CFD） 来确定。

Figure 7 provides a top-view schematic of a single,non-tapered blade that rotates about the $Z$ -axis in a counter-clockwise manner. In this figure, $R$ is the radius of the blade, $C$ is its chord and the freestream velocity $\left(V\right)$ is aligned with the $X$ -axis. Also in this schematic, $\psi$ is the angular position of the blade,where $\psi =0$ represents the position when the advancing blade is perpendicular to the freestream flow. $V_n$ is the local normal on-set velocity component to a blade section; it is a function of radial and angular positions and is given by:
图 7 提供了一个俯视图示意图，该叶片以逆时针方式绕 $Z$ 轴旋转。在此图中， $R$ 是叶片的半径， $C$ 是其弦，自由流速度 $\left(V\right)$ 与 $X$ -轴对齐。在此示意图中， $\psi$ 还有叶片的角度位置，其中 $\psi =0$ 表示前进叶片垂直于自由流时的位置。 $V_n$ 是叶片截面的局部法线起始速度分量;它是径向和角度位置的函数，由下式给出：

Here, $\omega =\dot{\psi }$ is the rotational speed of the blade spinning about the $Z$ -axis. In the present work,this rotational speed is nondimensionalized as follows.
这里， $\omega =\dot{\psi }$ 是刀片绕 $Z$ -轴旋转的转速。在本工作中，该转速是无量纲的，如下所示。

Figures 8-10 illustrate how $V_n$ varies within the disk area swept by the blade for rotational speeds of $\chi =\left(0.7,1.0,1.5\right)$ ,respectively. In these images,the freestream velocity is from left-to-right. Note that $V_n<0$ is depicted by the blue-shaded regions which only occur when ${90}^{\circ }\le \psi \le {270}^{\circ }$ . As an aside,the $V_n=0$ contour is a circle of diameter $\frac{1}{\chi }$ .
图 8-10 分别说明了当转速为 $\chi =\left(0.7,1.0,1.5\right)$ 时，叶片扫过的圆盘区域内 $V_n$ 的变化情况。在这些图像中，自由流速度是从左到右。请注意， $V_n<0$ 它由蓝色阴影区域表示，该区域仅在 时 ${90}^{\circ }\le \psi \le {270}^{\circ }$ 出现。顺便说一句， $V_n=0$ 轮廓是一个直径 $\frac{1}{\chi }$ 为 的圆。

In general,the rotational plane of the blade can be at an angle-of-attack $\left(\alpha \right)$ to the freestream velocity $\left(V\right)$ . The velocity component normal to the rotational plane $\left(V_z\right)$ can be approximated as:
通常，叶片的旋转平面可以处于自由流速度 $\left(V\right)$ 的攻角 $\left(\alpha \right)$ 。垂直于旋转平面 $\left(V_z\right)$ 的速度分量可以近似为：

As the blade sweeps in $\psi$ ,the airfoil sections along its span are subjected to a cyclic variation of both normal velocity $\left(V_n\right)$ and apparent angle-of-attack. Designate this effective angle-of-attack as ${\alpha }_n$ and approximate
当叶片扫入 $\psi$ 时，沿其跨度的翼型截面会受到法向速度和 $\left(V_n\right)$ 视攻角的循环变化。将此有效攻角指定为 ${\alpha }_n$ 和 近似值

with: 跟：

${}^{-2}$ Actually,very large angles-of-attack exist,but only at very low dynamic pressures,so the error introduced is minimal.
${}^{-2}$ 实际上，存在非常大的迎角，但仅限于非常低的动态压力下，因此引入的误差很小。

The reason the absolute value of $V_n$ is needed above is because,even when $V_n$ is negative,the sign of lift due to angle-of-attack is solely determined by the sign of $V_z$ . As will be seen later,this complicates matters in the derivations for lift and rolling-moment due to angle-of-attack.
上面需要绝对值 的 $V_n$ 原因是，即使为负数 $V_n$ ，攻角引起的升力符号也完全由 $V_z$ 的符号决定。正如稍后将看到的，这使得攻角引起的升力和滚动力矩的推导变得复杂。

While all boomerangs have more than one blade, with blade element theory each are treated identically; there is no distinction between the force and moments of the lifting and dingle blades. As a consequence, deriving the coefficients of lift and rolling-moment for a single blade is equivalent to doing so for an $N$ -bladed configuration. For example, a 3-bladed boomerang will have 3 times the lift of one of its blades but its lift will also be nondimensionalized by 3 times the reference area of a blade.
虽然所有回旋镖都有不止一个刀片，但在刀片元素理论中，每个刀片的处理方式都是相同的;Lifting 和 Dingle 刀片的力和力矩没有区别。因此，推导出单个叶片的升力和滚动力矩系数等同于推导出 $N$ -bladed 配置的升力和滚动力矩系数。例如，一个 3 叶片的回旋镖将具有其其中一个叶片的 3 倍的升力，但其升力也将由叶片参考面积的 3 倍进行无量纲化。

The next step is to develop a relationship for the dimensional lift generated by the blade as a function of its angular position $\left(\psi \right)$ .
下一步是建立叶片产生的尺寸升力与其角位置的 $\left(\psi \right)$ 函数的关系。

Here,the integration is over the span of the blade, $\rho$ is air density, $C_l\left({\alpha }_n\right)$ is the 2D sectional lift coefficient given by Eqn (26),but now as a function of ${\alpha }_n$ and $C$ is chord length. Eqns (26-30) have also been incorporated into Eqn (31). By observation, note that the total lift of Eqn (31) can be split into a basic lift and a lift due to angle-of-attack. To help keep the remaining derivations manageable, we will develop these lifts separately, then combine later.

In the context of this work, basic lift is defined as that which is generated by an airfoil acting at no angle-of-attack to its onset flow. Starting with the first term of the integrand of Eqn (31), and substituting Eqns (27-28), the basic-lift derivation continues as:

where $q=\frac{1}{2}\rho V^2$ is the freestream dynamic pressure and $S=RC$ is the blade planform area.

To determine a time-averaged value for the basic lift, $L_0\left(\psi \right)$ is integrated over $\psi$ . However,since $L_0$ is symmetric about the $Y$ -axis,this integration can be performed from 0 to $\pi$ .

Comparing Eqn (33) with Eqn (25) illustrates that Hess' aerodynamic model is insufficient for small rotational speeds where $\chi <1.5$ ,even for basic lift.

Before proceeding to the derivation of the lift due to angle-of-attack, it will be necessary to briefly touch on a special integration topic. Consider a line segment on the interval $\left[0,1\right]$ of the form $y\left(x\right)=a+bx$ . If this segment is either completely above or below the $X$ -axis,then an integration of its absolute value is trivial. However,let’s consider the situation where the line starts below the $X$ -axis at $x=0$ and ends above it at $x=1$ . For this case,it can be shown that:

Here, $a$ and $b$ are arbitrary constants. Note that the first term in the square brackets of Eqn (34) is an addition to the result one would get if the integrand had not been made an absolute value.

We now proceed with the derivation of lift due to angle-of-attack, starting with the second term of the integrand of Eqn (31).

Here, $\eta =\frac{r}{R}$ is the nondimensional spanwise coordinate along the blade. Because of the absolute-valued term inside the integral of Eqn (35),the following four cases exist for $0\le \psi \le \pi$ .

For $0\le \psi \le \frac{\pi }{2}$ ,and any value of $\chi >0$ ,the term $\left[\cos \psi +\chi \eta \right]\ge 0$ over the interval of integration,so:

For $\frac{\pi }{2}<\psi \le \pi$ ,and $\chi \ge 1$ ,the term $\left[\cos \psi +\chi \eta \right]$ switches sign,so we apply Eqn (34) to get:

For $\frac{\pi }{2}<\psi \le {\cos }^{-1}\left(-\chi \right)$ ,and $\chi <1$ ,the term $\left[\cos \psi +\chi \eta \right]$ also switches sign,and again we have:

For ${\cos }^{-1}\left(-\chi \right)<\psi \le \pi$ ,and $\chi <1$ ,the term $\left[\cos \psi +\chi \eta \right]<0$ over the interval,so:

In order to determine a time-averaged value for lift due to angle-of-attack, $L_{\alpha }\left(\psi \right)$ is integrated over $\psi$ from 0 to $\pi$ . It is possible to do this considering only two cases; one with $\chi \ge 1$ and the other with $\chi <1$ . However, this does require a little manipulation of Eqns (36-39) when folded into the integration over $\psi$ . For $\chi \ge 1$ ,