Spirals: Polar, Parametric Equations [RESOLVED]

I am trying to come up with the equation(s) that would describe the curve that would be the result of the following actions.

Say we have a wheel spinning at a constant rotational rate. In this example, let's use 1 revolution per second. Now I overlay a cartesian coordinate system with the origin at the center of the wheel. If I had a marker moving at constant linear velocity from position (0, rMax) along the y axis toward the origin to position (0, rMin). What would be the equation of the resulting curve drawn on the wheel? Let us take the constant linear velocity of my marker as (rMax-rMin) per 0.5 seconds. At this point you can mentally stop the wheel from spinning to examine the resulting curve (i.e. some type of spiral).

I'm guessing that in this case, the curve is like a half circle or half sine wave, but I would also like to know a general form if say the linear velocity or the angular velocity were to be varied. An image is attached. Or at least for a constant linear acceleration of the marker (instead of the constant linear velocity).

I was thinking that this graphical approach could be programmed, so then I thought of VB and the Math forum.

Thanks.

What you have to do is switch to the coordinate system that rotates along with the wheel and express the marker trajectory in that system.

The trajectory equations in the stationary coordinate system are simply:

x(t) = rMax - v0*t (v0 being the constant velocity toward the origin)
y(t) = 0

Let omega be the angular velocity of the wheel (and therefore of the rotating coordinate system) and assume it is rotating CCW.

If you call x' and y' the axes of the rotating coordinate system, then the transformation equations are the well known:

x' = x*cos(omega*t) + y*sin(omega*t)
y' = -x*sin(omega*t) + y*cos(omega*t)

Susbtituting the above x(t) & y(t) in the latter pair of equations, we end up with:

x'(t) = (rMax - v0*t)*cos(omega*t)
y'(t) = -(rMax - v0*t)*sin(omega*t)

so these are the equtions describing the marker trajectory as seen from the wheel or, conversely, the

resulting curve drawn on the wheel.

Thanks krtxmrtz.

That definitely looks like the way I should have approached it. I ended up generating a table of points for discrete rotational steps. I haven't verified if we actually have the same equations, but I wanted to thank you.

Quote:

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Originally posted by Phenix
I haven't verified if we actually have the same equations, but I wanted to thank you.

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You're very welcome! :)