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Rotating a circle around its x-axis

  1. Oct 15, 2008 #1

    Let's say I have a flat circle on the x-y axes, and I am looking at it from along the z-axis.
    now let's say I rotate it along the x-axis by A(alpha) degrees, and take a 2D picture, I would receive an ellipse, right?
    now let's say I would like to have a function from an angle T(theta), which gives me a point on the original flat circle, to an angle G(gama), which gives me a point on the projected ellipse.

  2. jcsd
  3. Oct 15, 2008 #2


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    So you have a circle in a xy-plane tilted an angle T about the x-axis? Yes, its projection onto the xy-plane is an ellipse.

    Draw a picture: a coordinate system with the y-axis horizontal, the z- axis vertical and the x-axis coming out of the page at you Draw a line through the origin of the yz-plane at angle T to the y-axis. Dropping a perpendicular from that line (representing the tilted y-axis) to the original y-axis gives you a right triangle with angle T. The near side is clearly the projected y-coordinate in the xy-plane and the hypotenuse is the old "y" value (call it y') then "cos(T)= (near side)/(hypotenuse)"= y/y' so y= y'cos(T). Because there is no "tilt" along the x-direction, x= x'.

    The circle, in the tilted xy-plane, now called x',y', is x'2+ y'2= R2. Since x= x and y'= y/cos(T), projeced down to the xy-plane, that becomes
    x2+ y'2/(cos2(T)= R2, an ellipse with "half-axis" in the x direction 1/R and "half-axis" length in the y-direction 1/(R cos(T)).
  4. Oct 15, 2008 #3
    Thanks for your swift response, but I think I didn't explain well what I want...
    See attached a circle and a tilted circle (ellipse). in the circle you see 12 equal points on the perimeter. in the ellipse you see the same 12 (drew only 6, lazy me) points. created by the perspective created by the 3D-ness of the rotation around the x-axis.

    Note: You need to look at the ellipse drawing as a 2D projection of the 3D view. the top point, where all lines meet is the horizon in a 3D perspective view. the lines that meet there are parallel lines in the 3D world, and they cut the circle in the same points as the ones in the circle drawing.

    My question is, how do I calculate T' from T?

    Attached Files:

  5. Oct 22, 2008 #4
    Nobody? I drew a diagram! come on!!! ;-)
  6. Oct 22, 2008 #5
    T' the new angle (in the ellipse frame), relates to x' and y' (the x and y coordinates to the ellipse)... and if you know x' and y' as functions of x and y, which are functions of a radial distance and T (the original polar angle), you really just have to hack through some algebra and trig functions to get T' as a function of T -- substitutions if you aren't elegant and notice tricks. And, like HallsofIvy notes, it might be nicest to keep things where you reference the angle to one coordinate-axis, like x (as HallofIvy chose), that is not compressed by the projection process... that might let you eliminate the radial distance. That would probably make what you wish algebraically cleaner to perform. (edited to add: Note, check limits... with high enough tilt, the circle becomes a line, so your angle, depending on how you reference it, probably should become either 90 degrees or 0.)
    Last edited: Oct 22, 2008
  7. Dec 22, 2008 #6
    After exactly two months (!) I returned to this thread, and suddenly realized why I didn't fully understand what the answers were trying to tell me.

    The thing is, the X axis isn't the same before and after the tilt.

    The desired result is a "perspective" diagram, and not an "isometric" diagram. for example, take a look at my drawing in the previous post, you'll see that the left most and right most points of the circle, aren't the left most and right most points of the ellipse...

    maybe an easier way to look at this problem is, that the axis of tilt isn't the X axis, but a horizontal line touching the lowest point of the circle.

    any help?
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