Its hard for me to put into words so please bear with me.
Given a line of a certain length, how could I calculate the the arc length of a logarithmic spiral given that it intersects the line at two different angles.
I hope that I haven't confused anyone. Please refer to the attached diagram for clarification.
UPDATE:
Curve as part of spiral:
[Edited to specify logarithmic spiral (@MartinG)]
On
Outline solution:
Assume a logarithmic spiral. In polar coordinates: $$ r = Re^{\theta\tan{\phi}} $$ where $R$ is the radius at the first point and $\phi$ is the pitch angle of the spiral.
Consider the triangle formed by the origin $O$ and the two points $A, B$ at the ends of the arc. The angle at $O$ is $\theta$. Call the other two angles $A$ and $B$.
Relate the two angles specified in the problem ($\alpha$ and $\beta$) to $A$, $B$ and $\phi$.
Apply the sine rule to the triangle, and note that $\theta = \alpha + \beta$.
Derive an equation in $\tan{\phi}$ and known quantities.
Solve for $\tan{\phi}$ (using numerical techniques) then calculate R.
Finally calculate the arc length $$ s = \int\frac{dr}{\sin\phi} = R \frac{(e^{\theta \tan{\phi} }-1)} {\sin{\phi}} $$
If you know the number of rounds it has made before it subtends $n\pi+\frac{18\pi}{180}$ to $(n+1)\pi -\frac{12*\pi}{180}$, then you can use the parametric form of a spiral to get the curve length. The parametric form of a spiral is as below:
$$ x = tcost$$ and $$ y = tsint$$
Put the point 1 as (x,y) and the other point ( x+ 200,y)
$$x = (n\pi+18\pi/180)cos(n\pi+18\pi/180)$$
the other point is $$x+200 = ((n+1)\pi - 12\pi/180)cos(((n+1)\pi - 12\pi/180)$$
You can find n and then solve the below integral to get the length of the curve
The length of the curve is given by the equation $$L = \int_{t_1}^{t_2} \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2}dt$$
Hope that helps.