/sci/ - Science & Math

Given the two equations that describe the following figures:

[eqn]x(\theta) = cr\cos(\theta) - qr\cos(c\theta)\\
y(\theta) = cr\sin(\theta) - qr\sin(c\theta)[/eqn]

Where [math]R = 0.5, N = 3, r = \frac{R}{N}, \theta \in [0,2\pi][/math].
The figure on the left has [math]q=0.75[/math] while the figure on the right has [math]q=1.5[/math].

We want to solve the equation [math]O(q) -4.25 =0[/math] by finding the appropriate q.
Given:
[eqn]O(q) = \int_{0}^{2\pi}\sqrt{[x'(\theta)]^2 + [y'(\theta)]^2}\ d\theta\\
x'(\theta) = (R+r)(q\sin(c\theta) - \sin(\theta))\\
[x'(\theta)]^2 = (R+r)^{2}(q^{2}\sin^2(c\theta) -2q\sin(c\theta)\sin(\theta) + \sin^2(\theta))\\
\\
y'(\theta) = (R+r)(q\cos(c\theta) - \cos(\theta))\\
[y'(\theta)]^2 = (R+r)^{2}(q^{2}\cos^2(c\theta) - 2q\cos(c\theta)\cos(\theta) + \cos^2(\theta))\\
[x'(\theta)]^2 + [y'(\theta)]^2 = (R+r)^2(q^2 + 2q\cos(c\theta - \theta) + 1)\\
\Rightarrow O(q) = \frac{2}{3}\cdot\int_{0}^{2\pi}\sqrt{q^2 + 2q\cos(3\theta) + 1}\ d\theta[/eqn]
>>14495510
I had already tried that with my own code to no avail. I'll try it with yours.