Graphing a polar curve with uniform speed

Question

I would like to draw an animation of a polar curve (a spiral) being graphed. I am using javascript and canvas. Currently, I am using setInterval to call a draw function, which graphs an x and y coordinate found from a parametric representation of the polar curve (x and y in terms of theta). I am incrementing theta by 0.01, from 0 to 2*pi, once for every call to draw(). The problem is that I wish for the animation to draw the same amount of the curve for each call to draw, so that the drawing appears to progress with uniform speed. It doesn't matter if the time between each call to draw is different; I just need the speed (in terms of pixels drawn / # of calls to draw) to be constant for the entire awing. In other words, I need the arc length of the segment of the polar graph drawn for each call to draw to be the same. I have no idea how to go about this. Any help/sugestions would be greatly appreciated. Thanks

Answer 1

Let f(z) be the theta variable you are referring to in your question. Here are two parametric equations that should be very similar to what you have:

x(f(z)) = f(z)cos(f(z))
y(f(z)) = f(z)sin(f(z))

We can define the position p(f(z)) at f(z) as

p(f(z)) = [x(f(z)), y(f(z))]

The speed s(f(z)) at f(z) is the length of the derivative of p at f(z).

x'(f(z)) = f'(z)cos(f(z)) - f(z)f'(z)sin(f(z))
y'(f(z)) = f'(z)sin(f(z)) + f(z)f'(z)cos(f(z))

s(f(z)) = length(p'(f(z))) = length([x'(f(z)), y'(f(z))])

= length([f'(z)cos(f(z)) - f(z)f'(z)sin(f(z)), f'(z)sin(f(z)) + f(z)f'(z)cos(f(z))])

= sqrt([f'(z)cos(f(z))]² + [f(z)f'(z)sin(f(z))]² + [f'(z)sin(f(z))]² + [f(z)f'(z)cos(f(z))]²)

= sqrt(f'(z) + [f(z)f'(z)]²)

If you want the speed s(f(z)) to be constant at C as z increases at a constant rate of 1, you need to solve this first-order nonlinear ordinary differential equation:

s(f(z)) = sqrt(f'(z) + [f(z)f'(z)]²) = C

http://www.wolframalpha.com/input/?i=sqrt%28f%27%28z%29+%2B+%5Bf%28z%29f%27%28z%29%5D%5E2%29+%3D+C

Solving this would give you a function theta = f(z) that you could use to compute theta as you keep increasing z. However, this differential equation has no closed form solution.

In other words, you'll have to make guesses at how much you should increase theta at each step, doing binary search on the delta to add to theta and line integrals over p(t) to evaluate how far each guess moves.

Answer 2

Easier method - change the parameter to setInterval proportional to the step arc length. That way you don't have to try to invert the arc length equation. If the interval starts getting too large, you can adjust the step size, but you can do so approximately.