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Rotated Spheroid

  1. Jul 15, 2005 #1
    If I take a spheroid (slightly flattened sphere) with the following equation:

    x2/a2 + y2/b2 + z2/a2 - 1 = 0

    and rotate it by arbitrary angles about the x-, y-, and z-axes, the new equation for the rotated spheroid is:

    Ax2 + By2 + Cz2 + Dxy + Exz + Fyz + Gx + Hy + Iz + J = 0

    How do I find the outline of its projection on the xy-plane?
    Last edited: Jul 15, 2005
  2. jcsd
  3. Jul 15, 2005 #2


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    use mathematica? macaulay?
  4. Jul 15, 2005 #3
    I derive all equations by hand. I'm looking for a step-by-step solution to this problem.
  5. Jul 15, 2005 #4


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    ok. but sometimes equations for the image of a projection are a little hard to come by. thats why the mechanism of grobner bases and the program macaulay were written.
  6. Jul 16, 2005 #5
    It seems to me that you're simply looking for all of the points on this surface such that the vector e_3=(0,0,1) is tangent to it at that point. The vector space at (x,y,z) of the surface is given by the kernel of this 1-form:

    2Ax dx + 2By dy + 2Cz dz + Dx dy + Dy dx + Ex dz + Ez dx
    +Fy dz + Fz dy + G dx + H dy + I dz

    So, we need to find all of the points where, if we plug e_3 into the above 1-form, we get zero. This boils down to:

    2Cz + Ex + Fy + I = 0.

    So, you need to find the intersection of the surface with the above plane in 3-space. This will be the curve (in 3-space) that projects to the boundary of the projection on the xy-plane -- clearly some sort of ellipse, since it will be a conic section (its equation is quadratic in x and y) and it is bounded.
    Last edited: Jul 16, 2005
  7. Jul 18, 2005 #6
    Ok. But in your solution, the tangent plane has just as many unknowns as the original equation. Not a complete solution I'm afraid.
  8. Jul 18, 2005 #7
    I'm not quite sure what you mean here. If you fix a point (x,y,z) on the surface, the tangent space at that point is given by all vectors [tex](v_1, v_2, v_3)\in R^3[/tex] such that

    [tex]2Ax v_1 + 2By v_2 + 2Cz v_3 + Dx v_2 + Dy v_1 + Ex v_3 + Ez v_1
    +Fy v_3 + Fz v_2 + G v_1 + H v_2 + I v_3 = 0 [/tex]

    which is in fact a 2-diml. subspace of [tex]R^3[/tex] given the definitions of the constants A, B, ... , I (keep in mind that you have to regard (x,y,z) as fixed). The point we're looking for are those points on the surface which will have a tangent vector of (0,0,1). Plugging that into the above eqn. gives the plane equation that I got before.

    So, in other words, to find all of the points on your surface that project to the boundary of the domain in the xy-plane, you need to solve this set of equations:

    Ax2 + By2 + Cz2 + Dxy + Exz + Fyz + Gx + Hy + Iz + J = 0
    2Cz + Ex + Fy + I = 0

    This is the intersection of two transverse surfaces, so the result will be a curve in 3-space. Once you find those, the curve in the xy-plane is just the same points, ignoring the z-coordinates of each.

    Not a very easy task. But I would like to point out that your original equation defines a much broader surface than just the spheroid. Any quadric surface (e.g. hyperboloids of one or two sheets, general ellipsoids, cones, etc.) satisfies this equation: Ax2 + By2 + Cz2 + Dxy + Exz + Fyz + Gx + Hy + Iz + J = 0. So, you might benefit by finding out some constraints on those constants.
  9. Jul 19, 2005 #8
    Ok. The surface normal is always perpendicular to the z-axis, and rotates about it, such that the partial derivative of the rotated spheroid equation F with respect to z is zero.

    Fz(x,y,z) = Ex + Fy + 2Cz + I = 0

    What you said. Solving for z:

    z = -(I + Ex + Fy) / (2C)

    and substituting into F yields an ellipse of the form:

    Kx2 + Ly2 + Mxy + Nx + Oy + P = 0

    So I obviously found the outline of the projection of the rotated spheroid. Converting to parametric form should allow me to find the extrema and plot the curve. Not so hard after all!
  10. Jul 20, 2005 #9
    Well, it was unclear what specifity you wanted. Every ellipse has the form of Kx2 + Ly2 + Mxy + Nx + Oy + P = 0, as well as every other conic section (parabolas, hyperbolas, pairs of intersecting lines).

    You can easily argue that the result will be an ellipse without using any calculus whatsoever. However, you have not answered what type of ellipse you end up with. Clearly, the major and minor axes are limited by the values of a and b. But, is there a restriction on the eccentricity of the resulting ellipse?
  11. Jul 20, 2005 #10
    The constants would be known exactly by the original spheroid equation prior to rotation, i.e. semi-major and semi-minor axes. These would be changed in a known and exact way by the angles of rotation, resulting in exact values for the constants of the final equations. Unless there is a problem with floating point errors (which is a distinct possibility) then constraints on the range of allowed values for the constants can be ignored. When converting to parametric form, the only constraint would be 0 <= t <= 2*pi. But that has nothing to do with eccentricity, more with the range of the independent variable t. Otherwise it wouldn't make sense to plot a curve outside of the range in which it exists. For example, you can't plot a unit circle for a range of x-values greater than 1.

    What were you suggesting?
    Last edited: Jul 20, 2005
  12. Jul 20, 2005 #11
    I'm just suggesting that in order to solve the problem you need to put in the constants K, L, M, ..., P in terms of a, b, and the rotation angles. Now, you may have already done that and just not typed it up here. But, that's what I meant when I said that it's not that easy of a problem.

    However, I haven't done it myself. So, it might indeed be easy.
  13. Jul 20, 2005 #12
    Looks like I made a few not-so-obvious typos, and that was the cause of my problem. My methods were correct. Now that the typos are fixed, everything checks out, the results plot nicely. Cool. This problem had me stumped for quite a while.
    Last edited: Jul 21, 2005
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