Rotating a Plane about a Point in SpaceDate: 07/03/2004 at 23:26:12 From: Tdon Subject: plane rotation about a point in space What is the equation to rotate a random plane in space around a known point? The plane is defined by three points on a right-triangle. After rotation the first point is to be at the origin of the rotated plane, the second point on the x-positive axis and the third point on the y-positive axis. The axes are defined as X, Y, and Z. The rotary axes are defined as B (rotating around the Y-axis) and C (rotating around the Z-axis). This is for a real life application. It will be used on a computer controlled metal cutting milling machine. I have tried the local college and all of my friends that have any chance of figuring it it out with no luck. I have also tried internet sites that provide math answers but have not yet found the answer I need. Thank you for your help. Date: 07/04/2004 at 09:40:58 From: Doctor Tom Subject: Re: plane rotation about a point in space Hello Tdon, First, let me restate the problem to make certain that I understand it. If I have this wrong, please write back and let me know what I have misunderstood. Originally, you have three points in 3-dimensional space and I assume you have their coordinates in your original coordinate system. To minimize confusion, I will call the three points O, X and Y. Also, you know that the angle XOY is a right angle. Let me label the original coordinates as follows: O = (Ox, Oy, Oz) X = (Xx, Xy, Xz) Y = (Yx, Yy, Yz) After the transformation, you want O to be at the origin, you want X to lie on the x-axis, and you want Y to lie on the y-axis. In addition, I assume that you want a method to figure out where an arbitrary point P = (Px, Py, Pz) will wind up if the three points above are moved rigidly as described. So if the rigid transformation takes P -> Q = (Qx, Qy, Qz), what you're really looking for is a formula to find the values of Qx, Qy and Qz in terms of the twelve variables Ox, Oy, ... Py, Pz, right? First do a rigid translation of the whole system that takes O to the origin: P' = (Px - Ox, Py - Oy, Pz - Oz) X' = (Xx - Ox, Xy - Oy, Xz - Oz) Y' = (Yx - Ox, Yy - Oy, Yz - Oz) and of course: O' = (Ox - Ox, Oy - Oy, Oz - Oz) = (0, 0, 0) Let me name the new translated coordinates as follows: Px - Ox = Px' Py - Oy = Py' and so on. Now calculate the length of the vectors X' amd Y' and let me call them Lx' and Ly' (these are just numbers): Lx' = square_root( (Xx')^2 + (Xy')^2 + (Xz')^2 ) and similarly for Ly'. We can calculate the point Z' as follows: Z' = X' x Y', where the "x" above refers to the vector cross-product. It happens that if you take the cross product of two vectors that lie at 90 degrees from each other, the result will be a vector perpendicular to both and which forms a right-handed coordinate system with them, which is exactly what you want. In other words, this cross product will find a point that should transform to the positive z-axis. Here's the formula for the vector cross-product: (Xx', Xy', Xz') x (Yx', Yy', Yz') = (Xy' Yz' - Xz' Yy', Xz' Yx' - Xx' Yz', Xx' Yy' - Xy' Yx') This cross-product will yield a vector Z' = (Zx', Zy', Zz') As you did before, find Lz', the lengh of Z'. Now you know what you need is a calcuation that takes: X' -> (Lx', 0, 0) Y' -> (0, Ly', 0) Z' -> (0, 0, Lz') in a linear manner. In other words you are searching for a 3x3 matrix M such that: X' M = (Lx', 0, 0) Y' M = (0, Ly', 0) Z' M = (0, 0, Lz') That is equivalent to the matrix equation: [ Xx' Xy' Xz' ] [ Lx' 0 0 ] [ Yx' Yy' Yz' ] M = [ 0 Ly' 0 ] [ Zx' Zy' Zz' ] [ 0 0 Lz'] Let me use the name N to refer to the matrix to the left of M and let me use the name L to refer to the matrix to the right of the "=" sign. We need to solve the following matrix equation for M: N M = L Let N^(-1) be the inverse of matrix N. Then: M = N^(-1)L There you are! So, to transform a point P to where it will be after the translation/ rotation, first subtract off the coordintes of the point O from all your points. Next, calculate the cross-product for a z-like vector and find the lengths of all. Find the entries in the matrices N and L as above. Invert N, and multiply it by L on the right to obtain M. Finally, multiply the vector P' (the one that was translated to the origin) by M and you will have the result you desire: P'M = Q. Good luck! - Doctor Tom, The Math Forum 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/