Parametric Sphere FormulaDate: 7/7/96 at 20:31:40 From: Anonymous Subject: Parametric Sphere Formula How are circles generally described in 3D space instead of on the plane? In 2D a parametric form is x=rsin(theta), y=rcos(theta) What is the equivalent in 3D? I know a sphere centered at the origin can be written as: x^2 + y^2 + z^2 = r^2 Ideally I want a parametric form of a circle in 3 space that passes through a particular 3 points. Date: 7/7/96 at 23:25:30 From: Doctor Pete Subject: Re: Parametric Sphere Formula As you may already know, the parameterization of a surface in 3-space requires two parameters. We can derive such a parameterization for the form of a circle in 3 space that passes here from the one for the circle. Consider x = v cos(u) y = v sin(u) u = [0, 2*Pi) , z = 0 which describes a circle of radius v in the xy-plane. Now, let's vary v from 0 to some constant radius r. We obtain a series of concentric circles in the xy-plane. These circles correspond to what would happen if we sliced our desired sphere with radius r perpendicular to the z-axis. So visualize what this looks like in the xz-plane. The hemisphere above the xy-plane now looks like a semicircle or radius r in the xz-plane. For some v, we see that the sphere has z-coordinate sqrt(r^2-v^2), where we take the positive square root. It follows that x = v cos(u) y = v sin(u) u = [0, 2*Pi) z = sqrt(r^2-v^2) v = [0, r] parameterizes a hemisphere above and including the xy-plane. To get the whole sphere, we could mirror it in the xy-plane and let the square root take on both positive and negative values, but this is not very elegant. Rather, consider replacing v with r cos(v). As this new v goes from 0 to Pi/2, r cos(v) goes from r to 0. So let this v go from -Pi/2 to Pi/2, and z = sqrt(r^2-r^2 cos^2(v)) = r sin(v). Hence x = r cos(v) cos(u) y = r cos(v) sin(u) u = [0, 2*Pi) z = r sin(v) v = [-Pi/2, Pi/2] parameterizes the whole sphere of radius r in 3-space (verify this by substituting this into the Cartesian formula and simplifying). The point here is that there generally exists more than one parameterization for a surface, just as in the 1-parameter case. As for your second question, this is not so simple. If I wanted to find a parameterization of a circle that passes through three arbitrary (non-collinear) points in 3-space, I would first translate the coordinate system so that one of the points is at the origin, then apply a few rotations in 3-space to get all 3 points in the xy-plane, and finally, calculate the parameterization that gives the circle passing through these three points. Then one would apply the inverse transformations on this parameterization to obtain the desired result. To obtain these transformations, it might help to find the equation of the plane that passes through the three points. This is done by taking the general equation ax+by+cz = 1, plugging in the coordinates, obtaining a system of three equations in the unknowns a,b,c, and solving for these. Needless to say, this is a task much better suited for a computer. -Doctor Pete, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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