3

I made this thing in desmos: https://www.desmos.com/calculator/na9sehjskk

The distance between points changes depending on the speed of the points. Is there a way to keep the distance between them equal at all times?

I tried changing the difference between a and b, b and c etc. so that when their speed increases, the difference decreases, but this only decreased the problem a bit, and the last few points usually went backward when the first points slowed down.

Was my original approach correct and I just need to fine-tune it, or is there a different, better way to do it?

To clarify, I made a parametric equation, then a point that follows that equation as 'a' changes. Then I made another point that does the same thing as 'b' changes. I repeated this a couple of times

a,b,c are variables, and b=a+0.05, c=b+0.05 etc.

Because $\sqrt{(dx/dt)^2+(dy/dt)^2}$ isn't constant, these points spread apart as each point has a different speed from the other at the same time

I want a way to keep the arclength the same between every point.

Edit:

I don't really understand what the linked posts mean, but I did figure out a method of approximation by using the osculating circle at each point. https://www.desmos.com/calculator/n7u0evif9s

This uses the equation for the arc length of a parametric circle, the equation for the radius of the osculating circle to the curve at any point, and a lot of points to keep the arc length equal between points.

all the other changes were for simplicity or to hide the fact that this is an approximation,

aaaaaaaa1234564
  • 41
  • +1 because this animation is fun to watch. But can you please explain the problem here in mathematical terms, for those of us not very familiar with Desmos. – Jair Taylor May 03 '24 at 17:38
  • The usual method is to parameterize with respect to arclength by taking $l(r) = \int_{-3}^r \sqrt{(dx/dt)^2+(dy/dt)^2} , dt$ and then take $x(l^{-1}(s+d)), y(l^{-1}(s+d))$ for $d = 0, 0.05, 0.1$ (or whatever you wish.) But generally speaking arclength is not possible to find analytically, so you will probably need to numerically integrate and numerically invert. I don't know how to do this in Desmos. – Jair Taylor May 03 '24 at 19:48
  • 1
    I have addressed this problem previously here: https://math.stackexchange.com/questions/1371668/equation-to-place-points-equidistantly-on-an-archimedian-spiral-using-arc-length/2216736#2216736, and here: https://math.stackexchange.com/questions/2375940/parametric-curves-with-constant-length-differential/2376794#2376794. – Cye Waldman May 04 '24 at 18:54

0 Answers0