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# Coordinate Transform

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Hi all, i would like to do the following: I want to plot a graph like a sin or cos, but i want it to be plotted along a path, for example a circle. I thought about transforming the coordinate system to polar, but how could i draw the polar system without transforming back to carthesian? Im a little bit confused here now Thank you! Bye, Gerd

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You have to revert back to cartesian coordinates to draw it; it''s just easier to express it in terms of polar coordinates.

r = a + sin(theta)

where a is the radius of the circle.

Then,

x = r*cos(theta)
y = r*sin(theta)

If you had a general trajectory (other than a circle), the problem would be more complicated, in terms of maths, but we could probably find an algorithm that does it without too much of a problem.

Cédric

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Why is it confusing? Are you asking about plotting it on a computer or by hand? Unless you are using a math package that supports polar plots you are going to have to convert it to rectangular with (x,y)=(r*cos(theta),r*sin(theta)). That is no biggy on a computer. I think most graphing calculators support polar plots or at least provide functions to do the conversion for you. If you are doing it by hand that conversion is a pain in the rump so you use polar graph paper instead. If you can''t find any locally then surely you can find some online.

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cedricl gave a great solution to the wave around a circle problem. I''ll try to summarize an approach for a general path. Wish I could host images to illustrate. For a general path, you need to find the cartesian coordinates of the wave exactly the way cedricl says, x = r*cos(angle), y = r*sin(angle). But these values are represented in a coordinate system that is to be aligned with the path. Thus, a transformation is required.

I''ll attempt to summarize an approach:

let t = a parameter that represents the position along your path (whether that be a circle or anything else). t = 0 at the start of the path and t = 1 at the end of the path. For a circular path, when t = 0, the angle would be 0 and when t = 1, then angle would be 360 degrees.

let l = the wavelength of the sin/cos wave that you''re interested in.
let a = amplitude of the wave
let pi = 3.14159265;

Now, the process, in pseudocode:

float dt = 1/number_of_points;float t = 0.;for (i = 0; i < number_of_points; i++){// calculate the path point  xp = x coordinate of path at t  yp = y coordinate of path at t// calculate the cartesian wave point, in wave space  xw = distance along the path;  yw = a*sin(2*pi*xw/l); // or cos, whatever.// calculate the normal vector to the path as a unit// vector  nx = x coordinate of path normal vector at t  ny = y coordinate of path normal vector at t// now transform (xw, yw) to be aligned with the path.// this could be done using matrices, but I didn''t want// to have to explain my notation. The transformed// version of xw is simply xp + the dot product of// (yw,0) with (nx,ny).  xw'' = xp + yw*nx;// similarly, the transformed version of yw is simply// yp + the dot product of (0,yw) with (nx,ny)  yw'' = yp + yw*ny;// what happened here is that I used the path itself to define// a baseline (xp,yp) coordinate. The wave is a perturbation// of that position normal to the path. In terms of the wave,// given a baseline point, the only adjustment is vertical// relative to the wave, which corresponds to the normal vector// of the path. So I adjusted (xp,yp) along the normal vector// by the amplitude of the wave at that point, the amplitude// of the wave being yw. The horizontal wave position xw is// only used to find the amplitude.// the point to plot is (xw'', yw'')// increment the path position for the next point  t += dt;}

Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.

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