# Calculate arc length of a logarithmic spiral between two points.

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)]

• I think that there is not enough information to properly answer this question. There can be uncountably many functions that over a certain distance $[0;d]$ have either these given properties: intersections, first derivative (the angles cited) and second derivative, i.e. it's up-concavity on $(-\delta;d+\delta)$ – alandella Sep 24 '14 at 20:42
• Could think of it as part of a continuous spiral? Would that simplify the problem? – Omar Khan Sep 24 '14 at 20:59
• Could we generate a range of curvatures for which the above angles hold true? Sorry, I'm quite poor with my math. Thank you for your forbearance :) – Omar Khan Sep 24 '14 at 21:16
• No problem ;) By intuition, I'd say that if we assume that the curve belongs to this family:$$\vec{r}(t)=(af(t)\cdot\cos t)\mathbf{i}+(af(t)\cdot\sin t)\mathbf{j}$$ where $a>0,a\in\mathbb{R}$ then the problem could be solved just knowing $af(x)$, i.e. once the curve's complete equation is known. – alandella Sep 24 '14 at 21:23

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.

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 $$s = \int\frac{dr}{\sin\phi} = R \frac{(e^{\theta \tan{\phi} }-1)} {\sin{\phi}}$$