QuarternionsDate: 2/17/96 at 8:10:14 From: Anonymous Subject: Quarternions Hi, Can someone tell me what quarternions are? What are all the various ways of doing 3D transformations? Thanks. Prasad Date: 2/20/96 at 18:30:22 From: Doctor Byron Subject: Re: Quarternions Hi Prasad, A Quaternion is something like a complex number except that it has four components instead of two and looks like: (a+ib+jc+kd) Quaternions further satisfy the rules i^2=j^2=k^2=-1, ij=-ji=k, jk=-kj=i, ki=-ik=j. These quaternions are often used in the generation of four- dimensional fractals. There is an good introduction to these kinds of graphics applications at: http://www.krs.hia.no/~fgill/quatern.html The site even has some sample C programs for generating very impressive fractal graphics. Here is some additional information about quaternion computations. I found it at http://www.dtek.chalmers.se/Datorsys/Project/qjulia/theory/quaternion.html Definition A Quaternion number is an extension to complex numbers invented by Lord William Hamilton. A Quaternion number has two extra imaginary units which are called j and k. We write such a number as q = a + b*i + c*j + d*k, or for short: (a, b, c, d). As long as c and d are equal to zero, q is nothing other then a complex number. These are the multiplication rules for i, j, k: i*i = j*j = k*k = -1 i*j = k, j*i = -k j*k = i, k*j = -i k*i = j, i*k = -j One can see these rules as if i, j, k are three orthogonal unit vectors; the multiplication acts like a right-handed cross product of them. These rules do not cause any real difficulty until we look at multiplication of quaternions. Later on this page we see a very important fact: multiplication/division of quaternions is not commutative. Conjugate, Norm and Absolute value q = (a, b, c, d) conj(q) = (a, -b, -c, -d) norm(q) = a*a + b*b + c*c + d*d abs(q) = sqrt(norm(q)) Addition and subtraction q = (a, b, c, d), p = (x, y, z, w) q + p = (a+x, b+y, c+z, d+w) q - p = (a-x, b-y, c-z, d-w) Scaling q = (a, b, c, d) t = Real value t*q = q*t = (t*a, t*b, t*c, t*d) Multiplication q = (a, b, c, d), p = (x, y, z, w) q * p = (a + b*i + c*j + d*k) * (x + y*i + z*j + w*k) = a * (x + y*i + z*j + w*k) + b*i * (x + y*i + z*j + w*k) + c*j * (x + y*i + z*j + w*k) + d*k * (x + y*i + z*j + w*k) = a*x + a*y*i + a*z*j + a*w*k + b*x*i - b*y + b*z*k - b*w*j + c*x*j - c*y*k - c*z + c*w*i + d*x*k + d*y*j - d*z*i - d*w = (a*x - b*y - c*z - d*w, a*y + b*x + c*w - d*z, a*z - b*w + c*x + d*y, a*w + b*z - c*y + d*x) We can rewrite this if we define two 3-dimensional vectors: qV = (b, c, d) q = (a, qV) pV = (y, z, w) p = (x, pV) q * p = (a*x - dotprod(qV, pV), a*pV + x*qV + crossprod(qV, pV)) Now it is easy to see that multiplication of quaternion is not commutative, since the cross product of the vectors is not commutative. One must take this in consideration while working with quaternion algebra. If we square a quaternion, the cross product is zero: q * q = (a*a - dotprod(qV, qV), 2*a*qV) In fact, one does not have to consider the multiplication order at all when raising a quaternion to any positive integer power. Fast multiplication There are some schemes available that reduce the number of internal multiplications when doing quaternion multiplication. Here is one: q = (a, b, c, d), p = (x, y, z, w) tmp_00 = (d - c) * (z - w) tmp_01 = (a + b) * (x + y) tmp_02 = (a - b) * (z + w) tmp_03 = (c + d) * (x - y) tmp_04 = (d - b) * (y - z) tmp_05 = (d + b) * (y + z) tmp_06 = (a + c) * (x - w) tmp_07 = (a - c) * (x + w) tmp_08 = tmp_05 + tmp_06 + tmp_07 tmp_09 = 0.5 * (tmp_04 + tmp_08) q * p = (tmp_00 + tmp_09 - tmp_05, tmp_01 + tmp_09 - tmp_08, tmp_02 + tmp_09 - tmp_07, tmp_03 + tmp_09 - tmp_06) With this method you get 7 less multiplications, but 15 more additions/subtractions. Generally, this is still an improvement. Inverse and division inv(q) = conj(q)/norm(q) q / p = inv(p)*q or q*inv(p) Here it can be difficult to know which one to choose. I do not use division very often, but when I have to I always use q/p = +inv(p)*q. If I use it all the time, there seems to be no problem. I am not that good at explaining this, if someone knows better please send me an e-mail. As far as your second question about 3D transformations, I hardly know where to begin because there are so many possible ideas in this area. Perhaps you could narrow down your question? For example, if you were interested in learning more about how to use matrices to represent linear transformations in three dimensions, I would be able to send you some much more useful information. Otherwise, my answers will probably end up being too vague. I hope this has been helpful. Good luck with the quaternion material! -Doctor Byron, The Math Forum |
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