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Physik: Calculation of rim deflection and spoke tension on a bicycle wheel
Freigegeben von matroid am Mo. 02. März 2015 13:43:14
Verfasst von MontyPythagoras -   2721 x gelesen [Gliederung] [Statistik] Druckbare Version Druckerfreundliche Version


Calculation of rim deflection and spoke tension on a bicycle wheel


The purpose of this article is to calculate the deflection of a bicycle rim and the spoke tension depending on the relative position of the spoke in the bicycle wheel. The object of this investigation is a classical bicycle wheel with common steel spokes.
As opposed to other investigations that have been made in the past using FEA calculations or analytic calculations applying a discrete number of concentrated loads representing the spokes, the objective of this article is to calculate the deflection of the rim by replacing the discrete number of spokes by a distributed load, thus facilitating the development and solution of a differential equation.
In addition, all other articles known to the author apply the reaction force from the road as one concentrated load, unlike in reality where the tire distributes the load across a length of a few centimeters. The effect of this simplification is being investigated in this article.
The results show that contrary to a common misconception of the upper spokes carrying the load, it is in fact the relief of the pretension in just a very narrow section of the wheel directly underneath the hub, while the rest of the spokes barely notice. As was to be expected, the distributed load from the tire leads to a more widely spread deflection of the rim, but the effect is rather small and it does not alter the characteristic of the tension distribution in the spokes significantly.
All in all, the article confirms the results found in the other papers, by using an entirely different mathematical approach.

Table of contents

A. Introduction
B. Basic mechanical equations
  1. Distributed spoke stiffness
  2. Bending of the rim
  3. Equilibrium of forces
C. Application of a concentrated load
  1. Boundary conditions and symmetry of the differential equation
  2. Exact differential equation
  3. Linearized ordinary differential equation (ODE)
  4. Solution of the linearized differential equation
  5. Approximation of the solution for real rims
  6. Illustration of the results
  7. Transformation of the non-periodic linear solution into a Fourier series
D. Generalization towards a realistic load distribution
  1. Derivation of a realistic load
  2. Convolution of the linear ODE solution and a parabolic distributed load
E. Summary of results
F. Conclusion
G. Literature
H. Appendix

A. Introduction

Bicycle wheels with steel spokes have been around almost ever since the introduction of air inflated tires. Little changes had been made technologically in the decades before computer aided design and finite element analysis became available and new materials were developed and used in bicycle wheels, foremost in the area of professional bicycle racing, mainly high-tensile aluminum alloys and carbon. Although disc wheels or full carbon wheels with just 3 to 5 massive carbon spokes exist since a couple of years as a result of this development, the classical bicycle wheels with steel spokes are still the equipment of choice even with the professionals in most types of races. They are robust, single spokes are readily available as low-cost spare parts, and the arrangement of the spokes allows easy centering and truing.
Although this type of wheel exists since many decades, there are still as many misconceptions among the general laymen about how the wheels carry the load resting on them. Does the hub "dangle" on the upper spokes or rest on the lower spokes, or both? Is the rim stiff by comparison and the spokes are soft or vice versa?
The ways the mechanical properties of the wheels have been investigated have either been experimental or mostly via finite element analysis ([1], [4]). Finite element analysis could be described as a "mathematical experiment", a hybrid between experimenting and calculating, yielding a result only for that particular 3D-model, virtually crafted of the chosen materials and so forth, but usually not allowing conclusions beyond the obvious. Just one mathematical paper about an analytic calculation with discrete numbers of spokes [2] is known to the author.
It is obvious that a calculation incorporating the true number of spokes is very close to reality, but it comes with a multitude of complications. The rim must be divided into a discrete number of segments, each of which described by an own set of equations. It is already difficult enough to calculate the effect of a concentrated load on the rim in this manner, which is like pinching it with a razor edge, let alone incorporating the true load distribution transmitted by an inflated tire.
But also in the FEA analyses ([1], [4]), the authors only examine the effect of a concentrated load on just one node of the FE net.
Therefore, all the literature cited here made the same simplification, namely reducing the load on the rim to a single point of contact. Thus, the objective of this article is twofold:
a) to develop a set of equations describing the deflection of the entire rim and the variation of spoke tension under a concentrated load, in order to confirm the results found in the literature,
b) use this set of equations to evaluate the effect of a realistic load distribution as exerted by the tire.

General note: for better readability, the author has intentionally abbreviated or left out some of the derivations in the main article, unless they were either absolutely necessary for the deductive reasoning at that point or they could be presented within just a few lines. In the other cases, only the problem is presented, followed by a brief description of the approach, and the result. The full derivation is cross-referenced and given in the appendix. Example: a notation like (→H.1) refers to appendix H, point 1.

B. Basic mechanical equations

1. Distributed spoke stiffness

Mechanically, the spokes in a bicycle wheel need to be regarded as steel springs that have an inherent spring rate. Therefore, we have a discrete number of rim segments that ranges somewhere between 12 and 40, and the discrete loads make it impossible to set up one "universal" differential equation that is valid throughout the entire rim circle. Hence, it is necessary to replace the discrete number of spokes by a generalized "distributed stiffness", like some sort of disc that only transmits radial tension like an almost infinite number of evenly distributed spider threads.
Every spoke has a given spring rate <math>K</math>, such that the force resulting from straining the spoke is

<math>\displaystyle F=K\Delta l</math>

where <math>\Delta l</math> is the spoke's length increment. With an evenly distributed stiffness, we define in a very similar way:

(1)   <math>\displaystyle q=k\Delta l</math>

The local infinitesimal force acting on a short segment of the rim then is:

(2)   <math>\displaystyle \text{d}F=q\cdot R\text{d}\varphi</math>

where <math>R</math> is the unstressed, "nominal" radius of the rim's neutral axis.
We now want to define <math>k</math> in such a way that the force it exerts when increasing the rim radius evenly is the same as a single spoke would when strained by the same amount. Let <math>n</math> be the number of spokes, then the angular range of one spoke is <math>\frac{2\pi}{n}</math>. See the illustration below:

If we bear in mind that the direction of the distributed load is always radial, we obtain:

<math>\displaystyle F=kR\Delta l\intop_{-\frac{\pi}{n}}^{\frac{\pi}{n}}\cos\varphi\text{d}\varphi=K\Delta r</math>

<math>\displaystyle 2kR\sin\frac{\pi}{n}=K</math>

So the radial distributed stiffness of the spokes is

(3)   <math>\displaystyle k=\frac{K}{2R\sin\frac{\pi}{n}}</math>

Since <math>n</math> generally is at least 12 or higher, it is a good approximation to state

<math>\displaystyle \sin\frac{\pi}{n}\approx\frac{\pi}{n}</math>

(4)   <math>\displaystyle k\approx\frac{nK}{2\pi R}</math>

Let <math>r</math> be the generalized polar coordinate radius of the deflected rim, then (1) can more precisely be written as:

(5)   <math>\displaystyle q=k(r-R)</math>

2. Bending of the rim

In a mechanical sense, the rim can be regarded as a curved beam. The bending moment will cause the rim to increase or decrease its nominal radius, such that the difference between local curvature <math>\kappa</math> (which is the reciprocal value of the local bending radius) and the nominal curvature is proportional to the bending moment:

<math>\displaystyle M_b=EI\left(\kappa-\frac{1}{R}\right)</math>

  <math>E</math> = Young's module of elasticity of the rim material (i.e. aluminum)
  <math>I</math> = moment of inertia of the rim cross section
  <math>R</math> = nominal radius of the rim's neutral axis
  <math>\kappa</math> = local curvature
In polar coordinates, <math>\kappa</math> can be noted as:

<math>\displaystyle \kappa=\frac{r^2+2r'^2-rr''}{(r^2+r'^2)^{\frac{3}{2}}}</math>

so that we have

(6)   <math>\displaystyle M_b=EI\left(\frac{r^2+2r'^2-rr''}{(r^2+r'^2)^{\frac{3}{2}}}-\frac{1}{R}\right)</math>

3. Equilibrium of forces

We have a look at the free body diagram of a short section of the rim:

There is a normal force <math>N</math>, a shearing force <math>S</math> and a bending moment <math>M_b</math>. Moreover, there is the distributed load exerted by the spokes. In reality, the spokes are pretensioned, but since the pretension is constant all around the rim, all it would result in is a negligible diminution of the whole rim to a smaller circle. Therefore, the distributed load is assumed to be zero on an unstressed wheel, and any stress resulting from bending of the rim can be positive or negative. A positive value means that the true pretension of the spokes is even being increased, because the rim at that point has been bent further away from the hub, straining the spokes in the process, while a negative value means in fact just a reduction in pretension.
We set up the equilibrium of forces:

<math>\displaystyle \sum F_{\text{vert}}=-N_{\varphi+\text{d}\varphi}\sin(\varphi+\text{d}\varphi)+N_\varphi\sin(\varphi)+S_{\varphi+\text{d}\varphi}\cos(\varphi+\text{d}\varphi)-S_\varphi\cos(\varphi)-qr\cos(\varphi)\text{d}\varphi=0</math>

<math>\displaystyle \text{d}\left(S\cos\varphi\right)-\text{d}\left(N\sin\varphi\right)-qr\cos\varphi\text{d}\varphi=0</math>

<math>\displaystyle \frac{\text{d}}{\text{d}\varphi}\left(S\cos\varphi\right)-\frac{\text{d}}{\text{d}\varphi}\left(N\sin\varphi\right)-qr\cos\varphi=0</math>

<math>\displaystyle S'\cos\varphi-S\sin\varphi-N'\sin\varphi-N\cos\varphi-qr\cos\varphi=0</math>

There are both sine and cosine terms, but the equations must be valid independently from the respective angle, which is why we can separate the equation into two parts that we individually set to zero:

(7)   <math>\displaystyle S+N'=0</math>


(8)   <math>\displaystyle S'-N-qr=0</math>

We differentiate the latter equation:

<math>\displaystyle S''-N'-(qr)'=0</math>

and add the result to (7), which yields:

(9)   <math>\displaystyle S''+S=(qr)'</math>

For the equilibrium of moments we get:

<math>\displaystyle \sum M=M_{b, \varphi+\text{d}\varphi}-M_{b, \varphi}-S_\varphi r\text{d}\varphi=0</math>

(10)   <math>\displaystyle M_b'=Sr</math>

(<math>q</math> does not have an impact on the equilibrium of forces because its effect is proportional to <math>(\text{d}\varphi)^2</math>, which renders it negligible for <math>\text{d}\varphi\rightarrow 0</math>).

C. Application of a concentrated load

1. Boundary conditions and symmetry of the differential equation

A concentrated load generally represents a discontinuity. While we have replaced the numerous singular forces exerted by the spokes by a distributed load, we still have one concentrated load left and that is the force acting on the bottom of the wheel. Any differential equation will have a discontinuity at this point as well, unless the point of force is moved to the boundaries of the differential equation, where the discontinuity becomes irrelevant. It is particularly helpful to divide the force into two halves as well, because in this way we keep everything symmetrical.

Because of the symmetry we can deduce a number of important constraints.
a) The deflection of the rim must be an even function of <math>\varphi</math>:

(11)   <math>\displaystyle r(\varphi)=r(-\varphi)</math>

b) Because of (11), any derivative of odd order of the deflection must be zero at <math>\varphi=0</math>:

(12)   <math>\displaystyle r^{(2m+1)}(0)=0</math>

c) Since the rim must not be kinked, the first derivative of the deflection must be zero at the point of contact as well:

(13)   <math>\displaystyle r'(\pi)=0</math>

d) The force at the end of the rim halves is

(14)   <math>\displaystyle S(\pi)=-\frac12F</math> and <math>\displaystyle S(-\pi)=\frac12F</math>

2. Exact differential equation

In order to get a better overview of the system of differential equations, we collect a few of the several equations from above.

(15)   <math>\displaystyle q=k(r-R)</math>

(16)   <math>\displaystyle S''+S=(qr)'</math>

(17)   <math>\displaystyle M_b'=Sr</math>

In order to eliminate variables, we solve (17) for <math>S</math> and differentiate twice:

<math>\displaystyle S=\frac{M_b'}{r}</math>

<math>\displaystyle S'=\frac{M_b''}{r}-\frac{M_b'r'}{r^2}</math>

<math>\displaystyle S''=\frac{M_b'''}{r}-2\frac{M_b''r'}{r^2}-\frac{M_b'r''}{r^2}+2\frac{M_b'r'^2}{r^3}</math>

Together with (15), we now insert these equations into (16) and get:

(18)   <math>\displaystyle \frac{M_b'''}{r}-2\frac{M_b''r'}{r^2}-\frac{M_b'r''}{r^2}+2\frac{M_b'r'^2}{r^3}+\frac{M_b'}{r}=k(r^2-Rr)'=k(2r-R)r'</math>

In addition to this, we have equation (6):

<math>\displaystyle M_b=EI\left(\frac{r^2+2r'^2-rr''}{(r^2+r'^2)^{\frac{3}{2}}}-\frac{1}{R}\right)</math>

We could now differentiate this equation three times and insert the obtained derivatives into (18). Thus, it is in fact possible to reduce the system to just one non-linear ordinary differential equation <math>f(r,r',...,r^{(5)})=0</math>, but it does not take much fantasy to see how complicated the ODE will be. Therefore, it is necessary to reduce the complexity of this system through linearizing. The exact equations can however be used to verify (by means of a numerical integration) how well the approximate linearized solution matches "reality".

3. Linearized ordinary differential equation (ODE)

Real life experience tells us that the deflection of the rim is not visible to the naked eye. If it does become visible, the spoke tension is definitely the rider's least concern. Therefore, it can be rightfully stated that the polar radius <math>r</math> is approximately equal to the nominal rim radius R:

(19)   <math>\displaystyle r\approx R</math>

We can conclude that

(20)   <math>\displaystyle r(r-R)\approx R(r-R)</math>

holds true as well. Moreover, the first derivative of <math>r</math>, which describes the local rate of change of the polar radius, must be very small by comparison, too:

(21)   <math>\displaystyle r'\ll r</math>

While completely aware that the following equations are just approximations, we keep using the "equal" sign nonetheless for simplicity reasons from here on.
With the above simplifications we can now linearize (16) and (17):

(22)   <math>\displaystyle S''+S=k\left(R(r-R)\right)'=kR\cdot r'</math>

(23)   <math>\displaystyle M_b'=R\cdot S</math>

Equation (6) can be simplified by eliminating <math>r'</math> because of (21):

<math>\displaystyle M_b=EI\left(\frac{r^2-rr''}{r^3}-\frac{1}{R}\right)</math>

<math>\displaystyle M_b=EI\left(\frac{r-r''}{r^2}-\frac{1}{R}\right)=EI\frac{R(r-r'')-r^2}{R r^2}=EI\frac{-r(r-R)-Rr''}{R r^2}</math>

We now utilize (20) in the numerator and (19) in the denominator and find:

(24)   <math>\displaystyle M_b= EI\frac{-R(r-R)-Rr''}{R^3}=EI\left(-\frac{r+r''}{R^2}+\frac1R\right)</math>

We differentiate once and insert into (23), so we note:

<math>\displaystyle M_b'=-\frac{EI}{R^2}(r'+r''')</math>

(25)   <math>\displaystyle S=-\frac{EI}{R^3}(r'+r''')</math>

All that is left to be done now is to combine (22) and (25), by differentiating the latter equation twice and insert into the first:

<math>\displaystyle S''=-\frac{EI}{R^3}(r'''+r^{(5)})</math>

<math>\displaystyle -\frac{EI}{R^3}(r'+2r'''+r^{(5)})=kR\cdot r'</math>

<math>\displaystyle r^{(5)}+2r'''+\left(1+\frac{kR^4}{EI}\right)r'=0</math>

(26)   <math>\displaystyle r^{(5)}+2r'''+(1+\psi)r'=0</math>

Hence, we have reduced the original set of equations to a homogeneous linear ODE of order 5. As a matter of fact, since <math>r</math> does not appear in the above ODE, it is really only a linear ODE of order 4 of the function <math>r'</math>.
In this respect, it is important to note that the equation depends only on the parameter

(27)   <math>\displaystyle \psi=\frac{kR^4}{EI}=\frac{KR^{3}}{2EI\sin\frac{\pi}{n}}</math>

which is the ratio between spoke stiffness <math>kR</math> and rim stiffness <math>\frac{EI}{R^3}</math>. If <math>\psi</math> were approximately 1, both rim and spokes would contribute to the total stiffness of the wheel in the same magnitude, and if <math>\psi</math> were significantly less than 1, the rim would be dominant. But in reality it turns out that <math>\psi</math> is extremely large. Real values appear to range between 500 and 2000.
Therefore, it is safe to say that with real bicycle wheels, even the modern ones with high profiles, it is mostly the spokes that define the stiffness of the wheel as a whole.

4. Solution of the linearized differential equation

If we regard <math>r'</math> as the sought function, we are dealing with a linear ODE of order 4, the characteristic equation of which has four complex roots:

<math>\displaystyle \lambda^4+2\lambda^2+1+\psi=0</math>

<math>\displaystyle (\lambda^2+1)^2+\psi=0</math>

The complex roots of this polynomial are: (→H.1)

<math>\displaystyle \lambda_{1..4}=\pm a\pm b\boldsymbol i</math>


(28)   <math>\displaystyle a=\sqrt{\frac12\left(\sqrt{1+\psi}-1\right)}</math>

(29)   <math>\displaystyle b=\sqrt{\frac12\left(\sqrt{1+\psi}+1\right)}</math>

These quantities have the following special properties:

(30)   <math>\displaystyle a^2+b^2=\sqrt{1+\psi}</math>

(31)   <math>\displaystyle 2ab=\sqrt{\psi}</math>

(32)   <math>\displaystyle a^2-b^2=-1</math>

The general solution of (26) is any linear combination of <math>e^{\pm a\varphi}\sin b\varphi</math> and <math>e^{\pm a\varphi}\cos b\varphi</math>. However, in this particular case we know that <math>r</math> is an even function, which in turn means that <math>r'</math> is an odd function. Thus, it is more convenient to use the hyperbolic sine and cosine instead of the exponential function, because they are odd and even functions as well, while the exponential function is neither. Since <math>r'</math> is odd, we can rule out any even combination (cosh × cos or sinh × sin), and the solution can only be:

(33)   <math>\displaystyle r'=r_1\sinh a\varphi\cos b\varphi+r_2\cosh a\varphi\sin b\varphi</math>

<math>r_{1,2}</math> need to be retrieved from the boundary conditions. Because (14) involves the shearing force <math>S(\pi)</math> at the point of contact, we need to calculate <math>S</math> by using (25), which in turn requires double differentiation of (33). Without showing in detail, we get

(34)   <math>\displaystyle r''=(r_{1}a+r_{2}b)\cosh a\varphi\cos b\varphi+(r_{2}a-r_{1}b)\sinh a\varphi\sin b\varphi</math>


<math>\displaystyle r'''=\left(r_1(a^2-b^2)+2r_2ab\right)\sinh a\varphi\cos b\varphi+\left(r_2(a^2-b^2)-2r_1ab\right)\cosh a\varphi\sin b\varphi</math>

Here we use (31) and (32) to find

(35)   <math>\displaystyle r'''=\left(-r_1+r_2\sqrt{\psi}\right)\sinh a\varphi\cos b\varphi+\left(-r_2-r_1\sqrt{\psi}\right)\cosh a\varphi\sin b\varphi</math>

Inserting (33) and (35) into (25) yields

(36)   <math>\displaystyle S=\frac{EI\sqrt{\psi}}{R^3}\left(-r_2\sinh a\varphi\cos b\varphi+r_1\cosh a\varphi\sin b\varphi\right)</math>

Finally, we use the boundary conditions (13) and (14) to find <math>r_{1,2}</math>:

(37)   <math>\displaystyle r_1\sinh a\pi\cos b\pi+r_2\cosh a\pi\sin b\pi=0</math>


(38)   <math>\displaystyle \frac{EI\sqrt{\psi}}{R^3}\left(-r_2\sinh a\pi\cos b\pi+r_1\cosh a\pi\sin b\pi\right)=-\frac12F</math>

This set of linear equations leads to: (→H.2)

(39)   <math>\displaystyle r_{1}=-\frac{FR^{3}}{2EI\sqrt{\psi}}\cdot\frac{\cosh a\pi\sin b\pi}{\sinh^{2}a\pi+\sin^{2}b\pi}</math>

(40)   <math>\displaystyle r_{2}=\frac{FR^{3}}{2EI\sqrt{\psi}}\cdot\frac{\sinh a\pi\cos b\pi}{\sinh^{2}a\pi+\sin^{2}b\pi}</math>

(41)   <math>\displaystyle r_1^2+r_2^2=\frac{F^2R^6}{4E^2I^2\psi}\cdot\frac{1}{\sinh^{2}a\pi+\sin^{2}b\pi}</math>

As the last step, we integrate (33) to get <math>r</math>:

<math>\displaystyle r=\frac{1}{a^{2}+b^{2}}\left[(r_{1}a-r_{2}b)\cosh a\varphi\cos b\varphi+(r_{2}a+r_{1}b)\sinh a\varphi\sin b\varphi\right]+C</math>

and with (30):

<math>\displaystyle r=\frac{1}{\sqrt{1+\psi}}\left[(r_{1}a-r_{2}b)\cosh a\varphi\cos b\varphi+(r_{2}a+r_{1}b)\sinh a\varphi\sin b\varphi\right]+C</math>

There is no simple way to determine the constant of integration <math>C</math>, because neither <math>r(0)</math> nor <math>r(\pi)</math> nor the radius at any other particular angle <math>\varphi</math> are known. However, it can be shown that <math>C=R</math> by calculating the elastic energy stored in the deflection of the rim and the elongation of the spokes and equating it with the energy spent when exerting the force <math>F</math> along the local deflection at <math>\varphi=\pi</math> (→H.3).

(42)   <math>\displaystyle r-R=\frac{1}{\sqrt{1+\psi}}\left[(r_{1}a-r_{2}b)\cosh a\varphi\cos b\varphi+(r_{2}a+r_{1}b)\sinh a\varphi\sin b\varphi\right]</math>

The compression of the rim at the point of contact then is: (→H.3)

(43)   <math>\displaystyle r(\pi)=R-\frac{FR^{3}}{4EI\sqrt{\psi}\sqrt{1+\psi}}\cdot\frac{b\sinh2a\pi+a\sin2b\pi}{\sinh^{2}a\pi+\sin^{2}b\pi}</math>

which can also be written as

(44)   <math>\displaystyle r(\pi)=R-\frac{FR^{3}}{2EI\sqrt{\psi}\sqrt{1+\psi}}\cdot\frac{b\sinh2a\pi+a\sin2b\pi}{\cosh2a\pi-\cos2b\pi}</math>

According to (42), the deflection at the top of the rim (<math>\varphi=0</math>) is

<math>\displaystyle r(0)=R+\frac{r_1a-r_2b}{\sqrt{1+\psi}}</math>

and with (39) and (40)

(45)   <math>\displaystyle r(0)=R-\frac{FR^{3}}{2EI\sqrt{\psi}\sqrt{1+\psi}}\cdot\frac{a\cosh a\pi\sin b\pi+b\sinh a\pi\cos b\pi}{\sinh^{2}a\pi+\sin^{2}b\pi}</math>

5. Approximation of the solution for real rims

As shown above, for real rims the parameter <math>\psi</math> exceeds values of 500. In (44) and (45), the hyperbolic sine and cosine quickly grow with increasing <math>\psi</math>, because they are basically exponential functions, while the trigonometric sine and cosine are bounded between -1 and 1:

<math>\displaystyle \lim_{\psi\rightarrow\infty}\frac{b\sinh2a\pi+a\sin2b\pi}{\cosh2a\pi-\cos2b\pi}=\lim_{\psi\rightarrow\infty}\frac{b\sinh2a\pi}{\cosh2a\pi}=\lim_{\psi\rightarrow\infty}b\tanh a\pi</math>

<math>\displaystyle \tanh a\pi\approx1</math>

<math>\displaystyle r(\pi)\approx R-\frac{FR^{3}b}{2EI\sqrt{\psi}\sqrt{1+\psi}}</math>

<math>\displaystyle r(\pi)\approx R-\frac{\sqrt{2}FR^{3}}{4EI}\cdot\frac{\sqrt{\sqrt{1+\psi}+1}}{\sqrt{\psi}\sqrt{1+\psi}}=R-\frac{\sqrt{2}FR^{3}}{4EI}\cdot\sqrt{\frac{\sqrt{1+\psi}+1}{\psi+\psi^{2}}}</math>

(relative deviation less than <math>10^{-9}</math> for <math>\psi\ge 500</math>), which can be drastically simplified to

(46)   <math>\displaystyle r(\pi)\approx R-\frac{\sqrt{2}FR^{3}}{4EI}\psi^{-\frac{3}{4}}=R-\frac{\sqrt{2}}{4\:\sqrt[4]{k^{3}EI}}F</math>

(relative deviation less than 2,1% for <math>\psi\ge 500</math>). If we introduce a wheel spring rate based on Hooke's law as

<math>\displaystyle C_{s}=\frac{\Delta F}{\Delta r}\qquad \frac{1}{C_s}\approx\frac{\sqrt{2}}{4\:\sqrt[4]{k^{3}EI}}</math>

then we can note this spring rate as

<math>\displaystyle C_s\approx\sqrt[4]{64k^3EI}</math>

Taking (3) and (27) into account, we conclude

<math>\displaystyle C_s\approx\sqrt[4]{64\frac{K^3EI}{8R^3\sin^3\frac{\pi}n}}</math>

<math>\displaystyle C_s\approx\sqrt[4]{4\frac{K^4}{\sin^4\frac{\pi}n}}\sqrt[4]{\frac{2EI\sin\frac{\pi}n}{KR^3}}</math>

(47)   <math>\displaystyle C_s\approx\frac{\sqrt2\psi^{-\frac14}}{\sin\frac{\pi}n}K</math>

The deflection at the top of the rim generally tends towards zero for large <math>\psi</math>. For instance, at <math>\psi=500</math>, the deflection is just one 12000th of the deflection at the contact point.

6. Illustration of the results

As sort of a plausibility check, we look at the deformation of a wheel with extremely soft spokes. In this case, the rim would have to stay in perfect circular shape and move as a whole. In the following diagrams, the blue line always represents the linearized ODE, while the thin red line represents the numerical calculation of the exact ODE. The latter has been overlaid over the linearized solution in just one half of the diagram in order to get a good notion of how large or rather how small the deviation between the exact and the linearized solution is. The values have been magnified and automatically scaled such that the radius of the rim equals 1 and the deflection at the point of contact equals 0.2. The double black line represents the original unstressed rim.

Wheel deformation, <math>\psi=0.01</math>, exaggerated
A difference between the exact and the approximate solution can hardly be seen, even at this magnification. There is just a very slight deviation at x<-0.8 in xy-coordinates.

Rim deflection, <math>\psi=0.01</math>, scaled to -1 at the point of contact
The slight deviation showing in the previous diagram becomes more obvious here.
Let us now have a look at a realistic wheel:

Wheel deformation, <math>\psi=500</math>, exaggerated
No visible difference between the exact ODE and the linearized ODE whatsoever. It is evident that the wheel is basically only deflected in the direct vicinity of where the singular force is exerted. One has to keep in mind that what looks like a heavy dent is really only a very slight deflection that is not visible to the naked eye. The magnitude of exaggeration is immense.

Rim deflection, <math>\psi=500</math>, scaled to -1 at the point of contact
The red line had to be dashed in order to render the blue line visible at all.
As can be seen in this diagram, the load rests on just a small section of the wheel. At around 141° from the top, which is around 39° from the contact point, the spoke tension becomes zero and inverts from relative compression (which is relief of pretension) to even increased tension until the angle of 88° (92° from point of contact), and so forth:

Rim deflection, <math>\psi=500</math>, zoomed in
The maximum increase in tension is approximately 5% of the maximum compression.
This effect can be seen in the previous diagram, too. The wheel is compressed at the point of contact and it bulges outwards in the neighboring segments.

7. Transformation of the non-periodic linear solution into a Fourier series

Because of the nature of this problem, the polar radius function ought to be periodic with the period <math>2\pi</math>. However, equation (42) is nonperiodic, it is only valid for <math>-\pi\le\varphi\le\pi</math>. Therefore, it is helpful for later calculations to turn (42) into a Fourier series, which by default is periodic. Moreover, from here on we substitute <math>r-R</math> by the measure of deflection <math>f</math>:

<math>\displaystyle f(\varphi)=r(\varphi)-R</math>

Considering that (42) is an even function, we can write <math>f</math> in terms of a Fourier series as

(48)   <math>\displaystyle f_N(\varphi)=\frac12A_0+\sum_{m=1}^NA_m\cos(m\varphi)</math>


(49)   <math>\displaystyle A_m=\frac1{\pi}\intop_{-\pi}^{\pi}f(\varphi)\cos(m\varphi)\text d\varphi=\frac2{\pi}\intop_0^{\pi}f(\varphi)\cos(m\varphi)\text d\varphi</math>

are the Fourier coefficients. With (42) we obtain: (→H.4)

(50)   <math>\displaystyle A_m=\frac{(-1)^{m+1}FR^3}{\pi EI\left( \psi+(m^2-1)^2\right)}</math>

and in particular

<math>\displaystyle A_0=\frac{-FR^3}{\pi (1+\psi)EI}</math>

Finally, we can write <math>f_N(\varphi)</math> as a periodic Fourier series:

(51)   <math>\displaystyle f_N(\varphi)=-\frac{FR^3}{\pi EI}\left(\frac1{2(1+\psi)}-\sum_{m=1}^N\frac{(-1)^{m+1}}{\psi+(m^2-1)^2}\cos(m\varphi)\right)</math>

or, alternatively:

(52)   <math>\displaystyle f_N(\varphi)=-\frac{FR^3}{\pi EI}\left(\frac1{2(1+\psi)}+\sum_{m=1}^N\frac{\cos\left(m(\pi-\varphi) \right)}{\psi+(m^2-1)^2}\right)</math>

In the diagram below one can see how well the Fourier series matches the solution of the linearized ODE even after only 20 summands.
Rim deflection, <math>\psi=500</math>, centered around the point of contact and scaled to -1

D. Generalization towards a realistic load distribution

1. Derivation of a realistic load

In reality, the tire transmits the load reaction force from the road to the rim. If we ignore the tube whose only purpose is to keep the air in place, the tire touches the rim only at the rim flange.
Let us have a look at the wheel cross section in "free air" and in contact with the road:

The free tire tries to take up as much room as possible under the inner pressure, it is being pressed into a circular cross section (a torus). At the rim flange, it exerts a tension (blue arrow) that is tangential to the tire cross section border line. This tension is called the poloidal tension. At this point, the tension is pulling the rim downwards, because the horizontal components of the tensions on both sides of the rim cancel each other out (the tire can only pull at the rim flange, it cannot and does not push). One must bear in mind that the total force acting on the rim is zero, because the tension integrated around the full circumference of the rim yields zero, as long as the entire wheel is lifted up in the air and does not touch the ground.
In the area where the wheel stands on the road the tire is compressed. It still tries to take up as much room as possible, but in the contact zone it cannot help but evade sideways. The part of the tire that is in free air is still toroidal, but the direction of the tension has changed. Compared to the original direction (green arrow), it now pulls more horizontally (blue arrow). The change in tension, symbolized by the red arrows, causes the rim effectively to be pushed upwards against the load on the hub.
The tension in the tire is not a constant, it depends on its location in the tire cross section and its direction. We have the poloidal tension acting on the rim, and we have the toroidal tension, which is acting longitudinally:

On the left we look at the tire cross section straight-on, while on the right we have a perspective view. The toroidal tension has been left out for clarity in the left drawing. Note: the dimensions <math>r</math> and <math>R</math> must not be confused with those used in chapters B and C.
We set up the equilibrium of forces in the lateral direction on the spherical trapezoid. It is not a rectangle, because the "lower" base edge is longer than the "upper" base edge, due to the fact that the lower edge is further away from the torus center line than the upper one. Let the thickness of the tire be <math>t</math>, then we get:

<math>\displaystyle \sigma_{p}(\alpha+\text{d}\alpha)\cdot\left(R_c+r\cos(\alpha+\text{d}\alpha)\right)\text{d}\varphi\cdot t\cdot\cos(\alpha+\text{d}\alpha)-\sigma_{p}(\alpha)\cdot\left(R_c+r\cos(\alpha)\right)\text{d}\varphi\cdot t\cdot\cos(\alpha)\\\qquad\qquad+p\cdot r\text{d}\alpha\cdot\left(R_c+r\cos(\alpha+\frac{1}{2}\text{d}\alpha)\right)\text{d}\varphi\cdot\sin(\alpha+\frac{1}{2}\text{d}\alpha)=0</math>

<math>\displaystyle \frac{\text{\text{d}}}{\text{d}\alpha}\left(\sigma_{p}\left(R_c+r\cos\alpha\right)\cos\alpha\right)+p\frac{r}{t}\left(R_c+r\cos\alpha\right)\sin\alpha=0</math>

We integrate with respect to <math>\alpha</math>:

<math>\displaystyle \sigma_{p}\left(R_c+r\cos\alpha\right)\cos\alpha-p\frac{r}{t}\left(R_c\cos\alpha+\frac{1}{2}r\cos^{2}\alpha\right)=c</math>

For <math>\alpha=\frac{\pi}{2}</math> we immediately find <math>c=0</math>, and we can divide by <math>\cos\alpha</math>:

<math>\displaystyle \sigma_{p}\left(R_c+r\cos\alpha\right)-p\frac{r}{t}\left(R_c+\frac{1}{2}r\cos\alpha\right)=0</math>

(53)   <math>\displaystyle \sigma_{p}=\frac{r}{t}\cdot\frac{R_c+\frac{1}{2}r\cos\alpha}{R_c+r\cos\alpha}p</math>

The toroidal tension does not play a role in the lateral equilibrium of forces, because it only consists of a radial and a longitudinal component.
Since we are specifically looking for the tension at the rim flange, we introduce the rim flange radius <math>R_{rim}</math> and note the following geometric relation:

<math>\displaystyle R_{rim}=R_c+r\cos\alpha</math>

Hence, from (53) we get

<math>\displaystyle \sigma_p=\frac rt\cdot\frac{R_{rim}-\frac12r\cos\alpha}{R_{rim}}p</math>

(54)   <math>\displaystyle \sigma_p=\frac rt\left(1-\frac r{2R_{rim}}\cos\alpha\right)p</math>

If we want to determine the distributed load resulting from this tension, we must calculate the difference between free state of the tire and compressed state. Only the radial component of the tension vector is of interest, because the lateral components cancel each other out. With the index "0" denoting the uncompressed state of the tire, we find the distributed load <math>q_l</math> to be

<math>\displaystyle q_l=2\left(\sigma_{p,0}\sin\alpha_0-\sigma_p\sin\alpha\right)t</math>

(55)   <math>\displaystyle q_l=\fracp{R_{rim}}\left(r_{t}\left(2R_{rim}-r_{t}\cos\alpha_{0}\right)\sin\alpha_{0}-r\left(2R_{rim}-r\cos\alpha\right)\sin\alpha\right)</math>

with <math>r_t</math> being half the uncompressed tire width, see the following image:

In the contact zone between the tire and the road, <math>r</math> and <math>\alpha</math> are dimensions that vary with the longitudinal angle. From the cross section of the wheel we obtain the following relations:

<math>\displaystyle w_{r}=r_{t}\sin(\pi-\alpha_{0})</math>

where <math>w_r</math> is half the rim width, so that

(56)   <math>\displaystyle \alpha_{0}=\pi-\arcsin\frac{w_{r}}{r_{t}}</math>

In addition, we have

<math>\displaystyle w_{r}-w_{c}=r\sin\alpha</math>

As described above, we assume that the tire evades sideways and the arc length of the tire in the cross section is constant:

<math>\displaystyle w_{c}+r\alpha=r_{t}\alpha_{0}</math>

Summation of the last two equations leads to

<math>\displaystyle w_{r}+r\alpha=r_{t}\alpha_{0}+r\sin\alpha</math>

(57)   <math>\displaystyle r=\frac{r_{t}\alpha_{0}-w_{r}}{\alpha-\sin\alpha}</math>

Moreover, we have

<math>\displaystyle h=r+r\cos(\pi-\alpha)=r(1-\cos\alpha)</math>

<math>\displaystyle h=(r_{t}\alpha_{0}-w_{r})\frac{1-\cos\alpha}{\alpha-\sin\alpha}</math>


<math>\displaystyle h_{0}=r_{t}+\sqrt{r_{t}^{2}-w_{r}^{2}}</math>

Comparing the compressed state at any given longitudinal angle <math>\varphi</math> and the uncompressed state at <math>\varphi_c</math> shows that

<math>\displaystyle (R_{r}+h_{0})\cos\varphi_{c}=(R_{r}+h)\cos\varphi</math>

(58)   <math>\displaystyle \varphi=\arccos\left(\frac{R_{r}+r_{t}+\sqrt{r_{t}^{2}-w_{r}^{2}}}{R_{r}+(r_{t}\alpha_{0}-w_{r})\frac{1-\cos\alpha}{\alpha-\sin\alpha}}\cos\varphi_{c}\right)</math>

We can now insert (57) into (55) and have

(59)   <math>\displaystyle q_{l}=\frac{p}{R_{rim}}\left(r_{t}\left(2R_{rim}-r_{t}\cos\alpha_{0}\right)\sin\alpha_{0}-\frac{r_{t}\alpha_{0}-w_{r}}{\alpha-\sin\alpha}\left(2R_{rim}-\frac{r_{t}\alpha_{0}-w_{r}}{\alpha-\sin\alpha}\cos\alpha\right)\sin\alpha\right)</math>

Equations (58) and (59) describe the distributed load dependent on the parameter <math>\alpha</math>. <math>q_l</math> cannot explicitely be written as a function of <math>\varphi</math> because (58) cannot be solved for <math>\alpha</math>.
Directly underneath the hub at <math>\varphi=0</math>, the compression of the tire hits its peak, decreasing in both directions along the circumference of the wheel until the tire is in free air. A parabola and the cosine function have been fitted as elementary approximations:

Both the parabola and the cosine function appear to be very good approximations to the real load distribution, at least for a realistic compression of around 4mm, with the parabola being the slightly better one.

For heavy but unrealistic compression, that had better be avoided on real bicycles, there is an extended "center zone".

2. Convolution of the linear ODE solution and a parabolic distributed load

In a linear system like this, the superposition principle allows to calculate the effect of a distributed load by convolving the effect of the dirac delta function (which in mechanical terms is the concentrated load) and the distributed load function. The formula of the rim deflection (52) then needs to be convolved with a weighting function over the period <math>2\pi</math>:

(60)   <math>\displaystyle (f*w)(\varphi)=\frac1{2\pi}\intop_{-\pi}^{\pi}f(\varphi-\tau)w(\tau)\text d\tau</math>

For reasons of simplicity, we move the point of contact from the angle <math>\pi</math> to zero, such that from here on we redefine

<math>\varphi</math> as <math>\pi-\varphi</math>

and (52) turns into

(61)   <math>\displaystyle \displaystyle f(\varphi)=\lim_{N\rightarrow\infty}f_N(\varphi)
 =-\frac{FR^3}{\pi EI}\left(\frac1{2(1+\psi)}+\sum_{m=1}^{\infty}\frac{\cos m\varphi}{\psi+(m^2-1)^2}\right)</math>

Since the parabola turned out to be the better approximation of a realistic load distribution, we define

(62)   <math>\displaystyle q_l=\begin{cases}q_0\left(1-\frac{\varphi^2}{\varphi_c^2}\right) & |\varphi|\le\varphi_c\\0 & |\varphi|>\varphi_c\end{cases}</math>

with <math>\varphi_c</math> being the "contact" angle, meaning half the angular span at which the tire touches the ground. We want to define the load such that the vertical total force resulting from it equals the concentrated load <math>F</math>:

<math>\displaystyle \intop_{-\varphi_c}^{\varphi_c}q_0\left(1-\frac{\varphi^2}{\varphi_c^2}\right)R\cos\varphi\text d\varphi=F</math>

By means of partial integration we find

<math>\displaystyle q_0R\frac{4(\sin\varphi_c-\varphi_c\cos\varphi_c)}{\varphi_c^2}=F</math>

<math>\displaystyle q_0=\frac{F\varphi_c^2}{4R(\sin\varphi_c-\varphi_c\cos\varphi_c)}</math>

The weighting function then is:

<math>\displaystyle w(\varphi)=\frac{q_{load}(\varphi)2\pi R}F</math>

(63)   <math>\displaystyle w(\varphi)=\begin{cases}\frac{\pi(\varphi_c^2-\varphi^2)}{2(\sin\varphi_c-\varphi_c\cos\varphi_c)} & |\varphi|\le\varphi_c\\0 & |\varphi|>\varphi_c\end{cases}</math>

We now convolve (61) and (63) as described in equation (60) and obtain the rim deflection under distributed load:

<math>\displaystyle f_l(\varphi)=(f*w)(\varphi)=\frac1{2\pi}\intop_{-\varphi_c}^{\varphi_c}-\frac{FR^3}{\pi EI}\left(\frac1{2(1+\psi)}+\sum_{m=1}^{\infty}\frac{\cos m(\varphi-\tau)}{\psi+(m^2-1)^2}\right)\frac{\pi(\varphi_c^2-\tau^2)}{2(\sin\varphi_c-\varphi_c\cos\varphi_c)}\text d\tau</math>

which eventually leads to: (→H.5)

(64)   <math>\displaystyle f_l(\varphi)=-\frac{FR^3}{\pi (\sin\varphi_c-\varphi_c\cos\varphi_c)EI}\left(\frac{\varphi_c^3}{6(1+\psi)}+\sum_{m=1}^{\infty}\frac{\sin m\varphi_c-m\varphi_c\cos m\varphi_c}{m^3\left(\psi+(m^2-1)^2\right)}\cos m\varphi\right)</math>

Finally, we want to evaluate the change of spoke pretension. Due to the spoke's inherent spring rate <math>K</math>, the spoke force <math>F_s</math> is

<math>\displaystyle F_s(\varphi)=Kf(\varphi)</math>

With (64) we obtain

<math>\displaystyle F_{s}(\varphi)=-F\cdot\frac{KR^{3}}{\pi EI}\cdot\frac{1}{\sin\varphi_{c}-\varphi_{c}\cos\varphi_{c}}\left(\frac{\varphi_{c}^{3}}{6(1+\psi)}+\sum_{m=1}^{\infty}\frac{\sin m\varphi_{c}-m\varphi_{c}\cos m\varphi_{c}}{m^{3}\left(\psi+(m^2-1)^2\right)}\cos m\varphi\right)</math>

Using (3) and then (27), we eventually find the following equation:

<math>\displaystyle F_{s}(\varphi)=-F\cdot\frac{2kR^{4}\sin\frac{\pi}{n}}{\pi EI}\cdot\frac{1}{\sin\varphi_{c}-\varphi_{c}\cos\varphi_{c}}\left(\frac{\varphi_{c}^{3}}{6(1+\psi)}+\sum_{m=1}^{\infty}\frac{\sin m\varphi_{c}-m\varphi_{c}\cos m\varphi_{c}}{m^{3}\left(\psi+(m^2-1)^2\right)}\cos m\varphi\right)</math>

(65)   <math>\displaystyle F_{s}(\varphi)=-F\left(\frac{1}{\pi}\sin\frac{\pi}{n}\right)\cdot\frac{\psi}{\sin\varphi_{c}-\varphi_{c}\cos\varphi_{c}}\left(\frac{\varphi_{c}^{3}}{3(1+\psi)}+2\sum_{m=1}^{\infty}\frac{\sin m\varphi_{c}-m\varphi_{c}\cos m\varphi_{c}}{m^{3}\left(\psi+(m^2-1)^2\right)}\cos m\varphi\right)</math>

Just like in chapter B., we utilize the approximation <math>\sin\frac{\pi}{n}\approx\frac{\pi}{n}</math>, such that we finally have

(66)   <math>\displaystyle F_{s}(\varphi)\approx-\frac{F}{n}\cdot\frac{\psi}{\sin\varphi_{c}-\varphi_{c}\cos\varphi_{c}}\left(\frac{\varphi_{c}^{3}}{3(1+\psi)}+2\sum_{m=1}^{\infty}\frac{\sin m\varphi_{c}-m\varphi_{c}\cos m\varphi_{c}}{m^{3}\left(\psi+(m^2-1)^2\right)}\cos m\varphi\right)</math>

With (47), we also get a rough approximation of the change in spoke tension at the point of contact:

<math>\displaystyle F_s(0)\approx -K\frac F{C_s}</math>

<math>\displaystyle F_s(0)\approx -F\frac{K\sin\frac{\pi}n}{\sqrt2K\psi^{-\frac14}}=-F\frac{\sin\frac{\pi}n}{\sqrt2}\psi^{\frac14}</math>

(67)   <math>\displaystyle F_s(0)\approx -\frac Fn \cdot \frac{\pi}{\sqrt2}\psi^{\frac14}</math>

In the following diagram we see the impact of the realistic load distribution on the rim deflection:

Realistic rim deflection, <math>\psi=500</math>, <math>\varphi_c=9°</math>
Evidently, the realistic load distribution does not affect the rim deflection by much, compared to the concentrated load. It broadens the compression span by a little (from around 39° to 40°), and it cushions away the compression peak at the center of the load.
The above diagram however represents a rather modern rim with very high rim stiffness (<math>\psi=500</math>) and a short contact area (<math>\varphi_c=9°</math>). In the following diagram, the same effect is shown for a rather shallow and soft rim (<math>\psi=5000</math>) and a longer contact area (<math>\varphi_c=15°</math>):

Realistic rim deflection, <math>\psi=5000</math>, <math>\varphi_c=15°</math>

The effect is much stronger in this case, the peak is reduced by approximately 25%, and the compression span increases from 22° to 27°. Generally, it can be said that this effect increases with growing values of both <math>\psi</math> and <math>\varphi_c</math>:

Lastly, we have a look at the magnitude of the reduction in spoke tension, relative to the load <math>F</math>, according to equation (66):

Reduction in spoke tension at <math>\varphi=0°</math>

Practical example: on a wheel with 24 spokes, <math>\psi=1000</math>, contact angle <math>\varphi_c=9°</math> under a load of 50kg, we read a value of 12, which means that the bottom spoke is relieved by approximately 12/24×50kg = 25kg.
The less the relief of pretension, the better for the longevity of the spokes and the smaller the risk of spoke failure. The above diagram shows clearly that the modern high profile rims are advantageous in this respect, as was to be expected.

E. Summary of results

Following is a collection of the most important equations.

Distributed spoke stiffness:

<math>\displaystyle k=\frac{K}{2R\sin\frac{\pi}{n}}\approx\frac{nK}{2\pi R}</math>

Stiffness parameter (ratio between spoke and rim stiffness):

<math>\displaystyle \psi=\frac{kR^4}{EI}=\frac{KR^{3}}{2EI\sin\frac{\pi}{n}}</math>

Deflection of the rim at the point of contact under concentrated load:

<math>\displaystyle f(0)=-\frac{FR^{3}}{2EI\sqrt{\psi}\sqrt{1+\psi}}\cdot\frac{b\sinh2a\pi+a\sin2b\pi}{\cosh2a\pi-\cos2b\pi}</math>


<math>\displaystyle a=\sqrt{\frac12\left(\sqrt{1+\psi}-1\right)}</math>

<math>\displaystyle b=\sqrt{\frac12\left(\sqrt{1+\psi}+1\right)}</math>

Approximate "spring rate" of the wheel (without tire):

<math>\displaystyle C_s\approx\frac{\sqrt2\psi^{-\frac14}}{\sin\frac{\pi}n}K</math>

Deflection of the wheel under concentrated load, written as Fourier series:

<math>\displaystyle \displaystyle f(\varphi)=-\frac{FR^3}{\pi EI}\left(\frac1{2(1+\psi)}+\sum_{m=1}^{\infty}\frac{\cos m\varphi}{\psi+(m^2-1)^2}\right)</math>

Deflection of the wheel under realistic load, written as Fourier series:

<math>\displaystyle f_l(\varphi)=-\frac{FR^3}{\pi (\sin\varphi_c-\varphi_c\cos\varphi_c)EI}\left(\frac{\varphi_c^3}{6(1+\psi)}+\sum_{m=1}^{\infty}\frac{\sin m\varphi_c-m\varphi_c\cos m\varphi_c}{m^3\left(\psi+(m^2-1)^2\right)}\cos m\varphi\right)</math>

Change of spoke tension:

<math>\displaystyle F_{s}(\varphi)\approx-\frac{F}{n}\cdot\frac{\psi}{\sin\varphi_{c}-\varphi_{c}\cos\varphi_{c}}\left(\frac{\varphi_{c}^{3}}{3(1+\psi)}+2\sum_{m=1}^{\infty}\frac{\sin m\varphi_{c}-m\varphi_{c}\cos m\varphi_{c}}{m^{3}\left(\psi+(m^2-1)^2\right)}\cos m\varphi\right)</math>

Rough approximation of the relief of spoke tension at the point of contact:

<math>\displaystyle F_s(0)\approx -\frac Fn \cdot \frac{\pi}{\sqrt2}\psi^{\frac14}</math>

F. Conclusion

The rim deflection of a classical bicycle wheel under both a concentrated and a more realistic distributed load can be depicted as Fourier series, based on the solution of a linear ODE of forth order.
With real world bicycle wheels, the spokes can be regarded as the defining elements pertaining to the overall stiffness of the wheel. The distributed load can be approximated by a parabolic curve for realistic tire pressures. Surprisingly, the difference between concentrated and distributed load is rather small. The most significant difference lies in the reduction of the deflection peak, but among the modern high profile rims, this effect accounts for just a single digit percentage, while on old-fashioned low profile rims the difference may be as high as 20% and more.
With the chosen hypothetical examples, the "bottom" spokes in the range of ±30° to ±40° experience a reduction in pretension, while beyond this span, the spokes up to the ones that are horizontal even suffer an increased tension. The softer the rim by comparison, the narrower are the alternating spans of relieved and increased pretension.
One effect described several times ([1], [3], [4]) does not become apparent in the solutions found here, and that is a slight, constant increase in tension of the spokes that are not compressed. That would have meant that the deflection of the rim would need to stay positive in the upper half of the wheel, literally enlarging its diameter as a whole. With this mathematical model, the rim deflection oszillates around zero in these spokes. This might be due to real world physical effects that were not accounted for in the equations in chapter B.
All in all, the results found here are in very good accordance with the ones given in the literature.

G. Literature

[1] Ng, Jinny: "Finite Element Analysis of a Bicycle Wheel: The Effects of the Number of Spokes on the Radial Stiffness", Rensselaer Polytechnic Institute, 2012
[2] José María Mínguez, Jeffrey Vogwell: "Radial stiffness of a bicycle wheel - An analytical study", Universidad del País Vasco / University of Bath, 2008
[3] Brandt, Jobst: "The bicycle wheel" 3rd edition, Avocet, Inc., 1993, ISBN-13: 978-0960723669
[4] Unknown author: "Bicycle Wheel Analysis", private web page: www.astounding.org.uk/ian/wheel/index.html , 2013

H. Appendix

1. (Biquadratic function)

<math>\displaystyle \lambda^{4}+2\lambda^{2}+1+\psi=0</math>

<math>\displaystyle \lambda=a+b\boldsymbol i</math>

<math>\displaystyle \lambda\mathbb{\in C},\quad a,b\in\mathbb{R}</math>

<math>\displaystyle \left(a^{4}+4a^{3}b\boldsymbol{i}-6a^{2}b^{2}-4ab^{3}\boldsymbol{i}+b^{4}\right)+2\left(a^{2}+2ab\boldsymbol{i}-b^{2}\right)+1+\psi=0</math>

<math>\displaystyle \left(a^{4}-6a^{2}b^{2}+b^{4}+2a^{2}-2b^{2}+1+\psi\right)+\left(4a^{3}b-4ab^{3}+4ab\right)\boldsymbol{i}=0</math>

Both brackets set to zero individually. Second bracket:

<math>\displaystyle 4a^{3}b-4ab^{3}+4ab=0</math>

<math>\displaystyle a^{2}-b^{2}+1=0</math>

<math>\displaystyle b^{2}=a^{2}+1</math>

First bracket:

<math>\displaystyle a^{4}-6a^{2}(a^{2}+1)+(a^{2}+1)^{2}+2a^{2}-2(a^{2}+1)+1+\psi=0</math>

<math>\displaystyle a^{4}-6a^{4}-6a^{2}+a^{4}+2a^{2}+1+2a^{2}-2a^{2}-2+1+\psi=0</math>

<math>\displaystyle -4a^{4}-4a^{2}+\psi=0</math>

<math>\displaystyle a^{4}+a^{2}-\frac{1}{4}\psi=0</math>

<math>\displaystyle a^{2}=-\frac{1}{2}+\sqrt{\frac{1}{4}+\frac{1}{4}\psi}=\frac{1}{2}\left(\sqrt{1+\psi}-1\right)</math>

(only positive root, because <math>a</math> is real and <math>a^2</math> must be positive).

<math>\displaystyle a=\sqrt{\frac{1}{2}\left(\sqrt{1+\psi}-1\right)}</math>

<math>\displaystyle b=\sqrt{\frac{1}{2}\left(\sqrt{1+\psi}+1\right)}</math>

<math>a</math> and <math>b</math> defined positive. Complete solution:

<math>\displaystyle \lambda_{1..4}=\pm a\pm b\boldsymbol i</math>

2. (System of two linear equations)

<math>\displaystyle r_1\sinh a\pi\cos b\pi+r_2\cosh a\pi\sin b\pi=0</math>

<math>\displaystyle r_{2}=-r_{1}\frac{\sinh a\pi\cos b\pi}{\cosh a\pi\sin b\pi}</math>

<math>\displaystyle \frac{EI\sqrt{\psi}}{R^3}\left(-r_2\sinh a\pi\cos b\pi+r_1\cosh a\pi\sin b\pi\right)=-\frac12F</math>

<math>\displaystyle \frac{EI\sqrt{\psi}}{R^{3}}\left(r_{1}\frac{\sinh a\pi\cos b\pi}{\cosh a\pi\sin b\pi}\sinh a\pi\cos b\pi+r_{1}\cosh a\pi\sin b\pi\right)=-\frac{1}{2}F</math>

<math>\displaystyle r_{1}\left(\sinh^{2}a\pi\cos^{2}b\pi+\cosh^{2}a\pi\sin^{2}b\pi\right)=-\frac{FR^{3}}{2EI\sqrt{\psi}}\cosh a\pi\sin b\pi</math>

with <math>\cosh^{2}x=1+\sinh^{2}x</math>

<math>\displaystyle r_{1}\left(\sinh^{2}a\pi\cos^{2}b\pi+\sinh^{2}a\pi\sin^{2}b\pi+\sin^{2}b\pi\right)=-\frac{FR^{3}}{2EI\sqrt{\psi}}\cosh a\pi\sin b\pi</math>

and with <math>\cos^{2}x+\sin^{2}x=1</math>

<math>\displaystyle r_{1}\left(\sinh^{2}a\pi+\sin^{2}b\pi\right)=-\frac{FR^{3}}{2EI\sqrt{\psi}}\cosh a\pi\sin b\pi</math>

<math>\displaystyle r_{1}=-\frac{FR^{3}}{2EI\sqrt{\psi}}\cdot\frac{\cosh a\pi\sin b\pi}{\sinh^{2}a\pi+\sin^{2}b\pi}</math>

<math>\displaystyle r_{2}=\frac{FR^{3}}{2EI\sqrt{\psi}}\cdot\frac{\sinh a\pi\cos b\pi}{\sinh^{2}a\pi+\sin^{2}b\pi}</math>

<math>\displaystyle r_1^2+r_2^2=\frac{F^2R^6}{4E^2I^2\psi}\cdot\frac{\cosh^2 a\pi\sin^2 b\pi+\sinh^2 a\pi\cos^2 b\pi}{(\sinh^{2}a\pi+\sin^{2}b\pi)^2}</math>

<math>\displaystyle r_1^2+r_2^2=\frac{F^2R^6}{4E^2I^2\psi}\cdot\frac{\sinh^{2}a\pi+\sin^{2}b\pi}{(\sinh^{2}a\pi+\sin^{2}b\pi)^2}</math>

<math>\displaystyle r_1^2+r_2^2=\frac{F^2R^6}{4E^2I^2\psi}\cdot\frac{1}{\sinh^{2}a\pi+\sin^{2}b\pi}</math>

3. (Calculation and comparison of elastic energy)

In the linearized form, the elastic energy is

<math>\displaystyle W=2\intop_0^\pi\frac{M_b^2}{2EI}R\text d\varphi+2\intop_0^\pi \frac k2(r-R)^2R\text d\varphi</math>

with the first integral being the energy stored in the rim and the second that in the spokes. With (24):

<math>\displaystyle W=\intop_0^\pi\frac{E^2I^2}{EI}\left(\frac{r+r''}{R^2}-\frac{1}{R}\right)^2R\text d\varphi+\intop_0^\pi k(r-R)^2R\text d\varphi</math>

<math>\displaystyle W=\frac{EI}{R^3}\intop_0^\pi\left(r+r''-R\right)^2\text d\varphi+kR\intop_0^\pi (r-R)^2\text d\varphi</math>

<math>\displaystyle k=\psi\frac{EI}{R^4}</math> (from equation (27))

<math>\displaystyle W=\frac{EI}{R^3}\intop_0^\pi\left(r+r''-R\right)^2\text d\varphi+\frac{EI}{R^3}\psi\intop_0^\pi (r-R)^2\text d\varphi</math>

<math>\displaystyle W=\frac{EI}{R^3}\intop_0^\pi\left(r''+r-R\right)^2+\psi(r-R)^2\text d\varphi</math>

If the elastic energy in the rim and the spokes matches the energy spent by the force <math>F</math>, no additional constant of integration is required in equation (42). With a linear spring, the energy spent is

<math>\displaystyle W=\frac12Fs</math>

In this particular case, we need to show that the following equation valid:

<math>\displaystyle \frac{EI}{R^3}\intop_0^\pi\left(r''+r-R\right)^2+\psi(r-R)^2\text d\varphi=-\frac12F\cdot\left(r(\pi)-R\right)</math>

("minus" on the right of the equal sign because <math>r(\pi)-R<0</math>).

<math>\displaystyle \intop_0^\pi\left(r''+r-R\right)^2+\psi(r-R)^2\text d\varphi=-\frac{FR^3}{2EI}\left(r(\pi)-R\right)</math>

It is particularly useful to calculate the first bracket independently:

<math>\displaystyle r''+r-R=\frac{1}{a^{2}+b^{2}}\left(\begin{array}{l}
\left((a^{2}+b^{2})(r_{1}a+r_{2}b)+(r_{1}a-r_{2}b)\right)\cosh a\varphi\cos b\varphi\\
\qquad\qquad+\left((a^{2}+b^{2})(r_{2}a-r_{1}b)+(r_{2}a+r_{1}b)\right)\sinh a\varphi\sin b\varphi

<math>\displaystyle r''+r-R=\frac{1}{a^{2}+b^{2}}\left(\begin{array}{l}
\left((a^{2}+1+b^{2})r_{1}a+(a^{2}+b^{2}-1)r_{2}b\right)\cosh a\varphi\cos b\varphi\\
\qquad\qquad+\left((a^{2}+1+b^{2})r_{2}a-(a^{2}+b^{2}-1)r_{1}b\right)\sinh a\varphi\sin b\varphi

With <math>a^2+1=b^2</math> we get

<math>\displaystyle r''+r-R=\frac{1}{a^{2}+b^{2}}\left(\begin{array}{l}
\left(2ab^{2}r_{1}+2a^{2}br_{2}\right)\cosh a\varphi\cos b\varphi\\
\qquad\qquad+\left(2ab^{2}r_{2}-2a^{2}br_{1}\right)\sinh a\varphi\sin b\varphi

<math>\displaystyle r''+r-R=\frac{2ab}{a^2+b^2}\left((r_1b+r_2a)\cosh a\varphi\cos b\varphi-(r_1a-r_2b)\sinh a\varphi\sin b\varphi\right)</math>

and with (30) and (31):

<math>\displaystyle r''+r-R=\frac{\sqrt{\psi}}{\sqrt{1+\psi}}\left((r_1b+r_2a)\cosh a\varphi\cos b\varphi-(r_1a-r_2b)\sinh a\varphi\sin b\varphi\right)</math>

We now calculate the integrand:

<math>\displaystyle (r''+r-R)^{2}+\psi(r-R)^{2}=\frac{\psi}{1+\psi}\left(\begin{array}{l}
\left[(r_{1}b+r_{2}a)\cosh a\varphi\cos b\varphi-(r_1a-r_2b)\sinh a\varphi\sin b\varphi\right]^{2}\\
\qquad\qquad+\left[(r_{1}a-r_{2}b)\cosh a\varphi\cos b\varphi+(r_{2}a+r_{1}b)\sinh a\varphi\sin b\varphi\right]^{2}

<math>\displaystyle (r''+r-R)^{2}+\psi(r-R)^{2}=\frac{\psi}{1+\psi}\left((r_{1}b+r_{2}a)^{2}+(r_{1}a-r_{2}b)^{2}\right)\left(\cosh^{2}a\varphi\cos^{2}b\varphi+\sinh^{2}a\varphi\sin^{2}b\varphi\right)</math>

<math>\displaystyle (r''+r-R)^{2}+\psi(r-R)^{2}=\frac{\psi}{1+\psi}\left(r_{1}^{2}(a^{2}+b^{2})+r_{2}^{2}(a^{2}+b^{2})\right)\left(\cosh^{2}a\varphi-\cosh^{2}a\varphi\sin^{2}b\varphi+\sinh^{2}a\varphi\sin^{2}b\varphi\right)</math>

<math>\displaystyle (r''+r-R)^{2}+\psi(r-R)^{2}=\frac{\psi}{1+\psi}(a^{2}+b^{2})(r_{1}^{2}+r_{2}^{2})\left(\cosh^{2}a\varphi-\sin^{2}b\varphi\right)</math>

<math>\displaystyle (r''+r-R)^{2}+\psi(r-R)^{2}=\frac{\psi}{\sqrt{1+\psi}}\cdot\frac{F^{2}R^{6}}{4E^{2}I^{2}\psi}\cdot\frac{1}{\sinh^{2}a\pi+\sin^{2}b\pi}\left(\frac{\cosh2a\varphi+1}{2}+\frac{\cos2b\varphi-1}{2}\right)</math>

<math>\displaystyle (r''+r-R)^{2}+\psi(r-R)^{2}=\frac{F^{2}R^{6}}{8E^{2}I^{2}\sqrt{1+\psi}}\cdot\frac{1}{\sinh^{2}a\pi+\sin^{2}b\pi}\left(\cosh2a\varphi+\cos2b\varphi\right)</math>

We can now easily integrate and find

<math>\displaystyle \frac{F^{2}R^{6}}{8E^{2}I^{2}\sqrt{1+\psi}}\cdot\frac{1}{\sinh^{2}a\pi+\sin^{2}b\pi}\left[\frac{1}{2a}\sinh2a\varphi+\frac{1}{2b}\sin2b\varphi\right]_{0}^{\pi}=-\frac{FR^{3}}{2EI}\left(r(\pi)-R\right)</math>

<math>\displaystyle -\frac{FR^{3}}{8EIab\sqrt{1+\psi}}\cdot\frac{b\sinh2a\pi+a\sin2b\pi}{\sinh^{2}a\pi+\sin^{2}b\pi}=r(\pi)-R</math>

<math>\displaystyle -\frac{FR^{3}}{4EI\sqrt{\psi}\sqrt{1+\psi}}\cdot\frac{b\sinh2a\pi+a\sin2b\pi}{\sinh^{2}a\pi+\sin^{2}b\pi}=r(\pi)-R</math>

According to (42), we also have

<math>\displaystyle r(\pi)-R=\frac{1}{\sqrt{1+\psi}}\left[(r_{1}a-r_{2}b)\cosh a\pi\cos b\pi+(r_{2}a+r_{1}b)\sinh a\pi\sin b\pi\right]</math>

Therefore, we have to show that

<math>\displaystyle -\frac{FR^{3}}{4EI\sqrt{\psi}\sqrt{1+\psi}}\cdot\frac{b\sinh2a\pi+a\sin2b\pi}{\sinh^{2}a\pi+\sin^{2}b\pi}=\frac{1}{\sqrt{1+\psi}}\left[(r_{1}a-r_{2}b)\cosh a\pi\cos b\pi+(r_{2}a+r_{1}b)\sinh a\pi\sin b\pi\right]</math>

With the ratio

<math>\displaystyle r_{2}=-\frac{\sinh a\pi\cos b\pi}{\cosh a\pi\sin b\pi}r_{1}</math>

that can be obtained from (39) and (40), we conclude

<math>\displaystyle -\frac{FR^{3}}{4EI\sqrt{\psi}}\cdot\frac{b\sinh2a\pi+a\sin2b\pi}{\sinh^{2}a\pi+\sin^{2}b\pi}=r_{1}\left[\left(a+\frac{\sinh a\pi\cos b\pi}{\cosh a\pi\sin b\pi}b\right)\cosh a\pi\cos b\pi+\left(-\frac{\sinh a\pi\cos b\pi}{\cosh a\pi\sin b\pi}a+b\right)\sinh a\pi\sin b\pi\right]</math>

Using (39), many variables are eliminated:

<math>\displaystyle \frac{b\sinh2a\pi+a\sin2b\pi}{2\cosh a\pi\sin b\pi}=\left(a+\frac{\sinh a\pi\cos b\pi}{\cosh a\pi\sin b\pi}b\right)\cosh a\pi\cos b\pi+\left(-\frac{\sinh a\pi\cos b\pi}{\cosh a\pi\sin b\pi}a+b\right)\sinh a\pi\sin b\pi</math>

<math>\displaystyle \frac12(b\sinh2a\pi+a\sin2b\pi)=\left(\begin{array}{l}
(a\cosh a\pi\sin b\pi+b\sinh a\pi\cos b\pi)\cosh a\pi\cos b\pi\\
\qquad\qquad+(b\cosh a\pi\sin b\pi-a\sinh a\pi\cos b\pi)\sinh a\pi\sin b\pi

<math>\displaystyle \frac12(b\sinh2a\pi+a\sin2b\pi)=a(\cosh^2a\pi-\sinh^2a\pi)\sin b\pi\cos b\pi+b(\cos^2b\pi+\sin^2b\pi)\sinh a\pi\cosh a\pi</math>

<math>\displaystyle b\sinh2a\pi+a\sin2b\pi=2a\sin b\pi\cos b\pi+2b\sinh a\pi\cosh a\pi</math>

Q.E.D., the energies are equal, no additional constant of integration for <math>r</math> is required.
This extensive derivation has brought forth a useful formula for the compression of the rim at the point of contact, as sort of a side effect:

<math>\displaystyle r(\pi)=R-\frac{FR^{3}}{4EI\sqrt{\psi}\sqrt{1+\psi}}\cdot\frac{b\sinh2a\pi+a\sin2b\pi}{\sinh^{2}a\pi+\sin^{2}b\pi}</math>

4. (Calculation of Fourier coefficients)

Product-to-sum identity, based on angle addition theorems:

<math>\displaystyle \cos(x)\cos(y)=\frac12\left(\cos(x-y)+\cos(x+y)\right)</math>

gives us

<math>\displaystyle \intop\cosh(a\varphi)\cos(b\varphi)\cos(m\varphi)\text d\varphi=\frac12\intop\cosh(a\varphi)\left(\cos\left((b-m)\varphi\right)+\cos\left((b+m)\varphi\right)\right)\text d\varphi</math>

Furthermore we find by partial integration:

<math>\displaystyle \intop\cosh(a\varphi)\cos(b\varphi)\text d\varphi=\frac a{a^2+b^2}\sinh(a\varphi)\cos(b\varphi)+\frac b{a^2+b^2}\cosh(a\varphi)\sin(b\varphi)+c</math>


<math>\displaystyle \intop_0^{\pi}\cosh(a\varphi)\cos\left((b-m)\varphi\right)\text d\varphi=\frac a{a^2+(b-m)^2}\sinh(a\pi)\cos\left((b-m)\pi\right)+\frac {b-m}{a^2+(b-m)^2}\cosh(a\pi)\sin\left((b-m)\pi\right)</math>

We use the angle addition theorems again:

<math>\displaystyle \cos\left((b-m)\pi\right)=\cos(b\pi)\cos(m\pi)+\sin(b\pi)\sin(m\pi)=(-1)^m\cos(b\pi)</math>

<math>\displaystyle \sin\left((b-m)\pi\right)=\sin(b\pi)\cos(m\pi)-\cos(b\pi)\sin(m\pi)=(-1)^m\sin(b\pi)</math>

so that

<math>\displaystyle \intop_0^{\pi}\cosh(a\varphi)\cos\left((b-m)\varphi\right)\text d\varphi=\frac {a(-1)^m}{a^2+(b-m)^2}\sinh(a\pi)\cos(b\pi)+\frac {(b-m)(-1)^m}{a^2+(b-m)^2}\cosh(a\pi)\sin(b\pi)</math>

and in the same way

<math>\displaystyle \intop_0^{\pi}\cosh(a\varphi)\cos\left((b+m)\varphi\right)\text d\varphi=\frac {a(-1)^m}{a^2+(b+m)^2}\sinh(a\pi)\cos(b\pi)+\frac {(b+m)(-1)^m}{a^2+(b+m)^2}\cosh(a\pi)\sin(b\pi)</math>

Adding these two equations we note

<math>\displaystyle \frac{a}{a^2+(b-m)^2}+\frac{a}{a^2+(b+m)^2}=a\frac{(a^2+b^2+m^2+2bm)+(a^2+b^2+m^2-2bm)}{(a^2+b^2+m^2+2bm)(a^2+b^2+m^2-2bm)}</math>

<math>\displaystyle \frac{a}{a^2+(b-m)^2}+\frac{a}{a^2+(b+m)^2}=\frac{2a(a^2+b^2+m^2)}{(a^2+b^2+m^2)^2-4b^2m^2}</math>


<math>\displaystyle \frac{b-m}{a^2+(b-m)^2}+\frac{b+m}{a^2+(b+m)^2}=\frac{(b-m)(a^2+b^2+m^2+2bm)+(b+m)(a^2+b^2+m^2-2bm)}{(a^2+b^2+m^2+2bm)(a^2+b^2+m^2-2bm)}</math>

<math>\displaystyle \frac{b-m}{a^2+(b-m)^2}+\frac{b+m}{a^2+(b+m)^2}=\frac{2b(a^2+b^2-m^2)}{(a^2+b^2+m^2)^2-4b^2m^2}</math>

which ultimately leads to

<math>\displaystyle \intop_{0}^{\pi}\cosh(a\varphi)\cos(b\varphi)\cos(m\varphi)\text{d}\varphi=(-1)^{m}\left(\begin{array}{l}

By using

<math>\displaystyle \sin(x)\cos(y)=\frac12\left(\sin(x-y)+\sin(x+y)\right)</math>

and an analogous derivation we also get

<math>\displaystyle \intop_{0}^{\pi}\sinh(a\varphi)\sin(b\varphi)\cos(m\varphi)\text{d}\varphi=(-1)^{m}\left(\begin{array}{l}

When we combine the above equations with (42) and (53), we obtain

<math>\displaystyle A_m=\frac{2(-1)^m}{\pi\sqrt{1+\psi}}\left(\begin{array}{l}

<math>\displaystyle A_m=\frac{2(-1)^m}{\pi\sqrt{1+\psi}}\left(\begin{array}{l}

<math>\displaystyle A_m=\frac{2(-1)^m}{\pi\sqrt{1+\psi}}\left(\begin{array}{l}
\frac{(a^2+b^2)\left ( (a^2-b^2)r_1-2abr_2 \right )+m^2(a^2+b^2)r_1}{(a^{2}+b^{2}+m^{2})^{2}-4b^{2}m^{2}}\sinh(a\pi)\cos(b\pi)\\
\qquad\qquad+\frac{(a^2+b^2)\left (2abr_1+(a^2-b^2)r_2 \right )+m^2(a^2+b^2)r_2}{(a^{2}+b^{2}+m^{2})^{2}-4b^{2}m^{2}}\cosh(a\pi)\sin(b\pi)

<math>\displaystyle A_m=\frac{2(-1)^m}{\pi}\left(\begin{array}{l}

The boundary condition (37) offers a significant simplification:

<math>\displaystyle A_m=\frac{2(-1)^m}{\pi}\left(\begin{array}{l}

A_m=\frac{4ab(-1)^m}{\pi\left ( (a^{2}+b^{2}+m^{2})^{2}-4b^{2}m^{2} \right )}\left(-r_2\sinh(a\pi)\cos(b\pi) +r_1\cosh(a\pi)\sin(b\pi)\right)</math>

The other boundary condition (38) can be used to reduce the entire equation to

<math>\displaystyle A_m=\frac{(-1)^{m+1}FR^3}{\pi EI\left ( (a^{2}+b^{2}+m^{2})^{2}-4b^{2}m^{2} \right )}</math>

Lastly, we simplify the denominator:

<math>\displaystyle (a^2+b^2+m^2)^2-4b^2m^2=(a^2+b^2)^2+2(a^2+b^2)m^2+m^4-4b^2m^2=(a^2+b^2)^2+2(a^2-b^2)m^2+m^4</math>

The equations (30) and (32) come in handy again:

<math>\displaystyle (a^2+b^2+m^2)^2-4b^2m^2=1+\psi-2m^2+m^4</math>

<math>\displaystyle (a^2+b^2+m^2)^2-4b^2m^2=\psi+(m^2-1)^2</math>

<math>\displaystyle A_m=\frac{(-1)^{m+1}FR^3}{\pi EI\left ( \psi+(m^2-1)^2 \right )}</math>

5. (Convolution of rim deflection and parabolic load)

<math>\displaystyle f_l(\varphi)=\frac1{2\pi}\intop_{-\varphi_c}^{\varphi_c}-\frac{FR^3}{\pi EI}\left(\frac1{2(1+\psi)}+\sum_{m=1}^{\infty}\frac{\cos m(\varphi-\tau)}{\psi+(m^2-1)^2}\right)\frac{\pi(\varphi_c^2-\tau^2)}{2(\sin\varphi_c-\varphi_c\cos\varphi_c)}\text d\tau</math>

<math>\displaystyle f_l(\varphi)=-\frac{FR^3}{4\pi (\sin\varphi_c-\varphi_c\cos\varphi_c)EI}\left(\frac1{2(1+\psi)}\intop_{-\varphi_c}^{\varphi_c}(\varphi_c^2-\tau^2)\text d\tau+\sum_{m=1}^{\infty}\frac{\intop_{-\varphi_c}^{\varphi_c}\cos m(\varphi-\tau)(\varphi_c^2-\tau^2)\text d\tau}{\psi+(m^2-1)^2}\right)</math>

<math>\displaystyle f_l(\varphi)=-\frac{FR^3}{4\pi (\sin\varphi_c-\varphi_c\cos\varphi_c)EI}\left(\begin{array}{1}\frac{2\varphi_c^3}{3(1+\psi)}+\sum_{m=1}^{\infty}\frac{\intop_{-\varphi_c}^{\varphi_c}\cos m\varphi\cos m\tau(\varphi_c^2-\tau^2)\text d\tau}{\psi+(m^2-1)^2}\\ \qquad\qquad+\sum_{m=1}^{\infty}\frac{\intop_{-\varphi_c}^{\varphi_c}\sin m\varphi\sin m\tau(\varphi_c^2-\tau^2)\text d\tau}{\psi+(m^2-1)^2}\end{array}\right)</math>

The integral in the second sum is zero, because it is an odd function:

<math>\displaystyle f_l(\varphi)=-\frac{FR^3}{4\pi (\sin\varphi_c-\varphi_c\cos\varphi_c)EI}\left(\frac{2\varphi_c^3}{3(1+\psi)}+\sum_{m=1}^{\infty}\frac{\cos m\varphi\intop_{-\varphi_c}^{\varphi_c}\cos m\tau(\varphi_c^2-\tau^2)\text d\tau}{\psi+(m^2-1)^2}\right)</math>

Through partial integration:

<math>\displaystyle \intop_{-\varphi_c}^{\varphi_c}\cos m\tau(\varphi_c^2-\tau^2)\text d\tau=4\frac{\sin m\varphi_c-m\varphi_c\cos m\varphi_c}{m^3}</math>

<math>\displaystyle f_l(\varphi)=-\frac{FR^3}{4\pi (\sin\varphi_c-\varphi_c\cos\varphi_c)EI}\left(\frac{2\varphi_c^3}{3(1+\psi)}+4\sum_{m=1}^{\infty}\frac{\sin m\varphi_c-m\varphi_c\cos m\varphi_c}{m^3\left(\psi+(m^2-1)^2\right)}\cos m\varphi\right)</math>

<math>\displaystyle f_l(\varphi)=-\frac{FR^3}{\pi (\sin\varphi_c-\varphi_c\cos\varphi_c)EI}\left(\frac{\varphi_c^3}{6(1+\psi)}+\sum_{m=1}^{\infty}\frac{\sin m\varphi_c-m\varphi_c\cos m\varphi_c}{m^3\left(\psi+(m^2-1)^2\right)}\cos m\varphi\right)</math>

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Calculation of rim deflection and spoke tension on a bicycle wheel [von MontyPythagoras]  
Abstract The purpose of this article is to calculate the deflection of a bicycle rim and the spoke tension depending on the relative position of the spoke in the bicycle wheel. The object of this investigation is a classical bicycl
[Die Arbeitsgruppe Alexandria katalogisiert die Artikel auf dem Matheplaneten]

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" Physik: Calculation of rim deflection and spoke tension on a bicycle wheel" | 2 Kommentare
Für den Inhalt der Kommentare sind die Verfasser verantwortlich.

Re: Calculation of rim deflection and spoke tension on a bicycle wheel
von Slash am Mo. 02. März 2015 14:33:48


kann man den Artikel als Weiterführung deines ersten Fahrradreifen-Artikels betrachten? Wegen der professionellen Darstellung und der Sprache - Hast du das für die Uni oder speziell für den MP gemacht?

Gruß, Slash\(\endgroup\)


Re: Calculation of rim deflection and spoke tension on a bicycle wheel
von MontyPythagoras am Mo. 02. März 2015 15:38:08

Hi Slash,
es ist nicht direkt eine Weiterführung, aber der Ursprung liegt tatsächlich in jenem Artikel. Da ich selbst Triathlon betreibe, habe ich ein starkes Interesse an der gesamten Fahrradtechnik, und wenn man etwas berechnen kann, um so besser.
Ich hatte im Kommentar-Bereich des früheren Artikels eine Diskussion mit zwei Usern geführt, und quasi nebenbei entstand ein Gespräch über den Verlauf der Speichenspannung während einer Radumdrehung. Ich selbst war bis dahin der Meinung, dass die Speichenspannung sinusförmig verläuft, was sie aber nun einmal nicht tut. Darauf aufmerksam gemacht wurde ich von einem niederländischen Leser, der jenen Artikel gelesen hatte und anregte, zur Speichenspannung eine mathematische Untersuchung durchzuführen. Da er selbst nach eigenem Bekunden besser Englisch als Deutsch spricht, schlug er vor, den Artikel auf Englisch zu schreiben, da viele Untersuchungen, die in der Vergangenheit auf Deutsch verfasst wurden, aufgrund dieser sprachlichen Barriere im Ausland kaum wahrgenommen wurden.
Und so kam es zu diesem Machwerk. Die Ergebnisse waren vorher eigentlich schon hinlänglich bekannt, sei es durch FEM-Analysen oder einfach experimentelle Untersuchungen, aber ich habe tatsächlich auch in der internationalen Literatur keine rein mathematischen Berechnungen gefunden.
Ich habe einige Jahre in USA gelebt und insofern tut mein Alltags-Englisch nicht weh im Ohr (oder Auge), aber ich bin auch kein Muttersprachler und es war das erste Mal, dass ich es in einem solchen rein wissenschaftlichen Kontext verwendet habe. Ich hoffe, es ist ganz gut geworden.




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