Calculating Logarithmic Spiral with Constant Arc Length

What is a logarithmic spiral? How is it different from Archimedean spiral?

In an Archimedean spiral, the distances of intersection points along a line that goes through the origin of the spiral are constant. An illustration of an Archimedean spiral follows:

Archimedean Spiral
An archimedean spiral. Source: http://faculty.nps.edu/ncrowe/spie10.htm

Whereas in a logarithmic spiral, the distances of intersection points along a line that goes through the origin of the spiral are growing exponentially. An illustration of a logarithmic follows:

Logarithmic Spiral
A logarithmic spiral. Source: https://sites.google.com/site/conchshellphysicalmodel/home

Here is another illustration of both spirals:

An archimedean spiral. Source: http://mathworld.wolfram.com/ArchimedesSpiral.html
An archimedean spiral. Source: http://mathworld.wolfram.com/ArchimedesSpiral.html
A logarithmic spiral. Source: http://mathworld.wolfram.com/LogarithmicSpiral.html
A logarithmic spiral. Source: http://mathworld.wolfram.com/LogarithmicSpiral.html

The polar equation of a logarithmic spiral is:

r=ae^{b \theta}

where r is the radius from the origin, \theta is the angle from x-axis, and a & b are arbitrary constants.

The common method of creating a logarithmic spiral is by keeping the \theta constant for each step (a step means a single point). However, there is another method, which I recently worked out for a computer programming project of mine.

The method requires that the length of the arc between two points on two subsequent steps to be constant, instead of the angle. It gave me a problem when I needed to calculate the radius for an arbitrary step. To use the above equation, I must first know the \theta of that particular step. And this proved to be a problem. Why? Because I could not find any reference for this method anywhere on the internet (and I tried so many different search terms with Google, but somehow I could not find what I was searching for).

Then I decided to try to come up with a formula by myself. Before continuing, I should admit that I am not good in Math. And what I did maybe was not an ideal approach to the problem, but anyway, I worked for me. So, how to calculate the radius of an arbitrary step if the arc length between two subsequent steps must be constant?

First, how to calculate the arc length between two points?

Let’s say that the two points are A and B, in polar coordinates:

A=(r,\theta_1)

B=(r{}

Then the arc length between A and B is:

l = \int_{\theta_1}^{\theta_2} \sqrt {r^2 + \left({\frac{dr}{d\theta}}\right)^2}\,d\theta

If:

r{}

then:

l = \int_{\theta_1}^{\theta_2} \sqrt {r^2 + \left(r{}

 

Using the equation for a logarithmic spiral in polar coordinate, we calculate r^2 as:

r=(ae^{b \theta})^2

r=a^2e^{2b \theta}

And calculating r{} as:

r{}

r{}

r{}

We can find the differential of exponent as:

\frac{d(e^u)}{d\theta}=e^u\frac{du}{d\theta}

So, we continue calculating r{} as:

r{}

r{}

And calculating (r{} as:

(r{}

(r{}

Then we substitute r^2 and (r{} into the integral, so we calculate the arc length between A and B as:

l = \int_{\theta_1}^{\theta_2} \sqrt {a^2e^{2b \theta} + a^2b^2e^{2b \theta}}\,d\theta

 

To solve the integral, I used Wolfram Aplha, as follows:

http://www.wolframalpha.com/input/?i=integrate+sqrt%28%28%28a%5E2%29e%5E%282bx%29%29+%2B+%28%28a%5E2%29%28b%5E2%29%28e%5E%282bx%29%29%29%29

l = \int_{\theta_1}^{\theta_2} \sqrt {a^2e^{2b \theta} + a^2b^2e^{2b \theta}}\,d\theta=\frac{\sqrt{a^2(b^2+1)e^{2b\theta}}}{b}+c

l=\frac{\sqrt{a^2(b^2+1)e^{2b\theta_2}}}{b}-\frac{\sqrt{a^2(b^2+1)e^{2b\theta_1}}}{b}

l=\frac{\sqrt{a^2(b^2+1)e^{2b\theta_2}}-{\sqrt{a^2(b^2+1)e^{2b\theta_1}}}}{b}

So, we calculate the arc length between two points as:

l=\frac{\sqrt{a^2(b^2+1)}\left({\sqrt{e^{2b\theta_2}}}-{\sqrt{e^{2b\theta_1}}} \right )}{b}

Okay. Now, let’s find out how to calculate the angle of an arbitrary step (or point).

Let’s say that the first step from origin is at an angle of \theta_1, second step from origin is at an angle of \theta_2, and the n^{th} step from origin is at an angle of \theta_n.

Arc length between \theta_1 and \theta_2 is:

l_2 =\frac{\sqrt{a^2(b^2+1)}\left({\sqrt{e^{2b\theta_2}}}-{\sqrt{e^{2b\theta_1}}} \right )}{b}

and arc length between \theta_1 and \theta_n is:

l_n =\frac{\sqrt{a^2(b^2+1)}\left({\sqrt{e^{2b\theta_n}}}-{\sqrt{e^{2b\theta_1}}} \right )}{b}

 

The arc length for every step is constant, so the arc length between \theta_1 and \theta_n (n steps) is:

l_n=nl_2

\frac{\sqrt{a^2(b^2+1)}\left({\sqrt{e^{2b\theta_n}}}-{\sqrt{e^{2b\theta_1}}} \right )}{b}=n\left( \frac{\sqrt{a^2(b^2+1)}\left({\sqrt{e^{2b\theta_2}}}-{\sqrt{e^{2b\theta_1}}} \right )}{b}\right)

\sqrt{a^2(b^2+1)}\left({\sqrt{e^{2b\theta_n}}}-{\sqrt{e^{2b\theta_1}}} \right )=n\left( \sqrt{a^2(b^2+1)}\left({\sqrt{e^{2b\theta_2}}}-{\sqrt{e^{2b\theta_1}}} \right )\right)

\sqrt{a^2(b^2+1)}\left({\sqrt{e^{2b\theta_n}}}-{\sqrt{e^{2b\theta_1}}} \right )=\sqrt{a^2(b^2+1)}\left(n \left({\sqrt{e^{2b\theta_2}}}-{\sqrt{e^{2b\theta_1}}} \right )\right)

\sqrt{a^2(b^2+1)} can be eliminated from both sides of the equation, so:

{\sqrt{e^{2b\theta_n}}}-{\sqrt{e^{2b\theta_1}}}=n \left({\sqrt{e^{2b\theta_2}}}-{\sqrt{e^{2b\theta_1}}} \right )

{\sqrt{e^{2b\theta_n}}}=n \left({\sqrt{e^{2b\theta_2}}}-{\sqrt{e^{2b\theta_1}}} \right )+{\sqrt{e^{2b\theta_1}}}

e^{2b\theta_n}=\left(n \left({\sqrt{e^{2b\theta_2}}}-{\sqrt{e^{2b\theta_1}}} \right )+{\sqrt{e^{2b\theta_1}}} \right )^2

2b\theta_n=ln\left(\left(n \left({\sqrt{e^{2b\theta_2}}}-{\sqrt{e^{2b\theta_1}}} \right )+{\sqrt{e^{2b\theta_1}}} \right )^2 \right )

\theta_n=\frac{{}ln\left(\left(n \left({\sqrt{e^{2b\theta_2}}}-{\sqrt{e^{2b\theta_1}}} \right )+{\sqrt{e^{2b\theta_1}}} \right )^2 \right )}{2b}

 

Conclusion:

For a logarithmic spiral with constant arc length between steps, given \theta_1\theta_2 and b, we can calculate the angle for any arbitrary steps (\theta_n) as follows:

\theta_n=\frac{{}ln\left(\left(n \left({\sqrt{e^{2b\theta_2}}}-{\sqrt{e^{2b\theta_1}}} \right )+{\sqrt{e^{2b\theta_1}}} \right )^2 \right )}{2b}

 

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