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June 27, 2005 In three previous articles, I discussed arithmetic with square matrices, as a prelude to some other topics having to do with tensors, such as the transformation of vectors. If we draw two mutually perpendicular lines, one east-west called the X axis, and one north-south called the Y axis, their intersection being called the origin, we can use this as a two-dimensional coordinate system. Any point on our piece of paper can be located by X and Y ordinates. Thus if we have a point 3 meters (or other units) to the east of the origin, we say X=3; if it's west, we say X=-3. If the point is 4 meters north of the origin, we say Y=4: if it's south, we say Y=-4. We can identify any point in our universe (or piece of paper) with a pair of numbers. We always put X first and Y second, taking care not to omit minus signs. For example, if a point, P1, is 4 meters east and 7 meters north, P1=[4,7]. If it's northwest, we say P2=[-4,7]. If it's southwest, we say P3=[-4,-7]. If it's southeast, we say P4=[4,-7]. In many applications, for instance, navigation, we are concerned only with two dimensions. A ship goes towards various points of the compass, without going up or down. However, in most applications, for instance, mechanical engineering, flight mechanics or robotics, there's movement in three dimensions, so we add a third, vertical axis Z, passing through the origin. In flight mechanics, Z is positive downward, whereas in other applications Z is positive upward, so here I'll just use positive Z upward. So if point P1 is above the paper five meters, we have, P1=[4,7,5], where the Z ordinate is always third. If P1 were below the paper, we would have P1=[4,7,-5]. Our coordinate system need not be oriented east-west, north-south and up-down. We can turn it in any direction we like, and it still suffices as a frame of reference whereby all the points in the three-dimensional world can be unequivocally identified. The triplets of numbers, like P1=[4,7,5], are called vectors, in this case, position vectors or radius vectors, it being supposed that a unit of distance applies to each of the components of the vector. Vectors may also be used in conjunction with velocities, accelerations and forces, all of which are characterized by having both magnitude and direction, in contradistinction to quantities like temperature and electrical charge, which have magnitude without direction. The magnitude of a vector is gotten by extracting the square root of the sum of the squares of the components. /P1/=sqr(4^2+5^2+7^2)=9.48683. /P1/, the magnitude of P1, is the direct distance, or radius of the point from the origin. If the elements of P1 had been velocities instead of distances, the magnitude would have been 9.48683 meters per second (or other units), instead of 9.46863 meters. Now suppose we have two identical coordinate systems superimposed on each other with their axes lying precisely along the same lines. The only difference is that one is going to remain where it is. Let us say we're using compass points. Then we have X, Y and Z as east-west, north-south and up-down, as before. The stationary coordinate system is called earth-fixed. The other coordinate system will change as a mechanism moves. For instance, consider an airplane with a perfectly level fuselage pointing north at high noon. Then the axes of the plane coincide at high noon to the earth-fixed axes. But when the plane starts flying upwards and northwest, the body-fixed axes, i.e., those of the plane, no longer coincide with the earth-fixed axes, which, though moving with the plane, maintain their earth-fixed orientation. Suppose we measure the angle between each axis of the system X', Y', Z', the body-fixed system, and each axis of the earth-fixed system X, Y, Z. We put them in a box as follows. [ 41.3212, 79.0688, 50.7681] [103.2565, 13.3047, 91.1162] [ 51.7437, 82.5084, 140.7461] This in itself is not a tensor: it's just a table. From it, though, a tensor can be formed, which we may call C, displayed in algebraic form below. [c11, c12, c13] [c21, c22, c23] [c31, c32, c33] The values of the elements c11 through c33 are the cosines of the angles listed in the table. The first digit in each suffix in the tensor represents an axis in the earth-fixed system while the second represents an axis in the body-fixed system, with the correspondence that 1 denotes X or X', 2 denotes Y or Y' and 3 denotes Z or Z'. The element itself is the cosine of the angle between the thus-specified axes. [ .75102, .18963, .63246] [-.22931, .97316, -.01948] [ .61918, .13038, -.77435] These cosines are not independent of each other but must gibe mathematically. The sum of the squares of the cosines in any row or column must equal one. But the sum of the products of corresponding elements in any two different rows or columns must equal zero. In other words, we have the following requirements. a11^2 + a12^2 + a13^2 = 1 a21^2 + a22^2 + a23^2 = 1 a31^2 + a32^2 + a33^2 = 1 a11^2 + a21^2 + a31^2 = 1 a12^2 + a22^2 + a32^2 = 1 a13^2 + a23^2 + a33^2 = 1 a11 x a12 + a21 x a22 + a31 x a32 = 0 a11 x a13 + a21 x a23 + a31 x a33 = 0 a12 x a13 + a22 x a23 + a32 x a33 = 0 a11 x a21 + a12 x a22 + a13 x a23 = 0 a11 x a31 + a12 x a32 + a13 x a33 = 0 a21 x a31 + a22 x a32 + a23 x a33 = 0 Translating these algebraic equations into numbers, we have these. .75102^2 + .18963^2 + .63246^2 = 1 -.22931^2 + .97316^2 + -.01948^2 = 1 .61918^2 + .13038^2 + -.77435^2 = 1 .75102^2 + -.22931^2 + .61918^2 = 1 .18963^2 + .97316^2 + .13038^2 = 1 .63246^2 + -.01948^2 + -.77435^2 = 1 .75102 x .18963 + -.22931 x .97316 + .61918 x .13038 = 0 .75102 x .63246 + -.22931 x -.01948 + .61918 x -.77435 = 0 .18963 x .63246 + .97316 x -.01948 + .13038 x -.77435 = 0 .75102 x -.22931 + .18963 x .97316 + .63246 x -.01948 = 0 .75102 x .61918 + .18963 x .13038 + .63246 x -.77435 = 0 -.22931 x .61918 + .97316 x .13038 + -.01948 x -.77435 = 0 Returning to our vector, P= [4,7,5], which is called a row vector, we can, for convenience, show it as a column vector on the left side of our direction cosine tensor C, as follows. [p1] [c11, c12, c13] [p2] [c21, c22, c23] [p3] [c31, c32, c33] In order to multiply to get the product P x C, we add up the three products gotten by multplying the elements of P by the same-row elements of each column of C. In other words, if we designate the vector thereby obtained as Q, we have the following. p1 x c11 + p2 x c21 + p3 x c31 = q1 p1 x c12 + p2 x c22 + p3 x c32 = q2 p1 x c13 + p2 x c23 + p3 x c33 = q3 Translating this into numbers, we have the following. 4 x .75102 + 7 x -.22931 + 5 x .61918 = 4.49481 4 x .18963 + 7 x .97316 + 5 x .13028 = 8.22254 4 x .63246 + 7 x -.01948 + 5 x -.77435 = -1.47827 Thus, we arrive at vector Q = [4.49481, 8.22254, -1.47827]. If a point in our imaginary airplane had position vector P with respect to the body-fixed axes, the axes of the plane itself, we find that it has position vector Q with respect to the earth-fixed axes, when the orientation of the plane is such that it is accurately described by direction cosine tensor C. Fortunately, the inverse of the cosine tensor is merely the tranpose, so if we want to transform vectors in the opposite direction, we merely use this tensor, C*, transpose and inverse of C. [ .75102, -.22931, .61916] [ .18963, .97316, .13038] [ .63246, -.01948, -.77435] Multiplying Q x C* returns us to P. ------------ About the author Thomas Keyes: I have written two books: A SOJOURN IN ASIA (non-fiction) and A TALE OF UNG (fiction), neither published so far. I have studied languages for years and traveled extensively on five continents. Email: udikeyes@yahoo.com Tell a friend about this site! ------------ All articles are EXCLUSIVE to Useless-Knowledge.com and are not allowed to be posted on other websites. ARTICLE THIEVES WILL BE PROSECUTED! |
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