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Parameterization of an ellipse?

  1. Aug 4, 2012 #1
    Suppose I have a vector space. There is circle with center at the origin of the vector space. There is also a line L going through the origin at some angle. On the circle is a point moving around the circle at a constant speed. The vector from the center to the point makes some angle with the line L. What is that angle as a function of time? I'm too dumb to figure it out.
  2. jcsd
  3. Aug 4, 2012 #2


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    A vector space does not have an "origin", it has a zero vector. You seem to be confusing "vector space" with coordinate system. Also you have titled this "parameterization of an ellipse" yet there is no ellipse in your post. That makes this difficult to understand!

    If I do understand it correctly, you write the coordinates of the point moving around the circle [itex](x, y)= (r cos(\theta), r sin(\theta))[/itex]. One can show that "moving around the circle at constant speed" (which I interpret as meaning the same distance on the circumference in the same time) is the same as moving at a constant angular velocity (the same angle in the same time). That is, [itex]\theta= \omega t[/itex] where [itex]\omega[/itex] is the constant angular velocity.

    That means we have [itex](x, y)= (r cos(\omega t), r sin(\omega t))[/itex]. If the line, L, makes angle [itex]\phi[/itex] with the positive x-axis (the slope of the line is [itex]tan(\phi)[/itex]) then (x, y) makes angle [itex]\theta- \phi= \omega t- \phi[/itex] with line L.
  4. Aug 5, 2012 #3
    Aha. Suppose you are aligned with line L. That is, the origin of the coordinate system is in the center of your belly and line L is going out of the top of your head. The circle will look to you like an ellipse. So I thought that might be reflected in the solution somehow.
  5. Aug 5, 2012 #4
    I'm not sure I follow your reasoning, but the only way to make a circle appear like an ellipse in 2D is to stretch one of the axes. In 3D, the ellipse can be a projection of a rotated circle on the 2D plane.
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