Testing a Point on a SphereDate: 01/14/97 at 11:52:14 From: Kevin Isbister Subject: Testing a point on a sphere Dr. Math, The problem is this: How do you define a curved line segment on the surface of a sphere in a mathematical equation? I know that a sphere is defined as x^2 + y^2 + z^2 = r^2, but how do you describe a segment of the surface of that sphere that runs, say, from A = (3,4,5) to B = (-4,5,3)? ANY point on the surface of the sphere will make the above equation work, but I can't seem to find anything in my calculus textbooks defining what such a segment looks like. More to the point, how does one "test" a point mathematically to see if it falls on this curved line? Obviously, the sample point would have to have the following things true: * must still make x^2 + y^2 + z^2 = r^2 true * -4<= x <= 3 * 4<= y <= 5 * 3<= z <= 5 But even that still describes an area on the surface that has as one of its diagonals the curved segment I'm looking for. How is this limited down? The same process, I assume could be expanded out in some way to test a point to see if it rests in an area on the surface, right? Help! Kevin Isbister Date: 01/14/97 at 12:54:26 From: Doctor Tom Subject: Re: Testing a point on a sphere Hi Kevin, There are a bunch of ways to do this, and I'll give you enough info to get you started. Assuming that you mean the great circle connecting the pair of points, the great circle is the intersection of the sphere with the plane passing through the two points and the center of the sphere. For your particular example, that would be the plane passing through (0,0,0), (3,4,5), and (-4,5,3), or -13x -29y + 31z = 0. You can substitute this into the sphere's equation to get an equation in two variables (any two you choose) that will be on the great circle. Unfortunately, this isn't exactly what you want, since the combination of equations will be satisfied by any point on the great circle. Since you are only interested in the arc on the sphere, there will be inequalities involving x, y, and z to restrict the equation to the arc. Another way is a parametric approach. Your numbers are ugly, so let me illustrate with something simple. Suppose your arc is on the unit sphere (radius 1), and the arc makes up a quarter circle. In the plane, we can write such an arc as follows: A(t) = (cos(t), sin(t), 0), where 0 <= t <= pi/2. I've included the third coordinate since we'll want to move this planar arc into three dimensions. But simple rotations about the origin can put the arc anywhere you want it, and then varying t will give all the points on that arc. See any book on 3D computer graphics to learn how to do rotations like this. -Doctor Tom, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/