How do 3D rotational matricies work?
471695
They are a set of Matricies that map one set of {x,y,z} to another set of {x',y',z'} through a rotation angle.  In oder to use them you need to know your starting set of coordiantes and the three angles through which you want to rotate.\n\nOne such set of angles that forms a perfectly good rotation basis are the Euler angles.\n\nIf you know the rotations, you can actually write the whole transformation as one matrix (usefull if you have to apply it over and over again.
627220
I don't grasp this sphere scenario but \n\n-----\n\nLet n be this normal vector. Choose any vector v with\n\nv ≠ 0 and v, n not collinear\n\ncompute\n\na = v x n and b = a x n\n\nthen {n,a,b} is a orthogonal basis. If necessary normalize a, b to get an orthonormal basis.\n\n-----\n\nAnother way: Take the standard basis {e1,e2,e3}, replace one base vector with n (avoid coplanarity) and apply the Gram–Schmidt process:\n\nhttp://en.wikipedia.org/wiki/Gram%E2%80%93Schmidt_process\n\n-----\n\nremarks: Both methods leave one degree of freedom (rotation around n). In each case you have to choose. That's quite annoying. If you find an algorithm without such a decision please tell me.\n\n-----\n\ncross product: a x b
897839
The angular change ratio from the planar field
914296
Use properties of dot product to get angles.\n\nv1*v2 = |v1|*|v2|*cos(theta)\n\nIf you first normalize your vectors then |v| = 1 it makes the calculations easier.
890959
I think I know what is confusing you. Total angular momentum of your system M=M1+M2=I1*w1+I2*w2, while I1=I2, w2=-w1, Thus M=I*0=0. Angular momentum is a vector, because w (angular speed) is a vector. How could we describe the difference of the system with w=0 and w not equal to zero? If w=0, then kinetic energy of the system E=0. When they are rotating, kinetic energy of the system E=E1+E2=I1*w1^2/2+I2*w2^2/2, which is not 0, as both w’s are squared. That means you have to do some work to obtain a certain angular speed w no matter what direction disks would spin.
1130182
Um... I think maybe pull... if something is rotating/turning it is creating it's own little gravity field (I think!) therefore making it easier to pull.  I'm no scientist but this is an educated guess.
471467
Yes. A square has an order of rotational symmetry of 4. \nThis means that if you were to rotate a square 360 degrees (1 complete revolution) about the intersection of its diagonals, you will get 4 as the number of times it takes on the initial form in terms of position and alignment.
228730
You can describe rotations and the effect they have on each point in the figure mathematically by means of simple trigonometric equations which can be condensed into a matrix. Imagine (or better, draw) a point on the plane in coordinate system X0,Y0, located at X0=1, Y0=0. Draw a line to the point from the origin. Now draw another coordinate system X1,Y1 with the same origin as X0,Y0 but rotated CW by an angle Theta. (Make Theta small, say 30 degrees.) What are the coordinates of our point in the new system? The answer is a function of Theta. We find that the point is above the X1 axis and nearer the Y1 axis. Specifically, X1 = X0*cosine(Theta) and Y1 = X0*sine(Theta). \nNow if you repeat this with a new point at X0=0, Y0=1, you find that X1 = -Y0*sine(Theta) and Y1 = Y0*cosine(Theta).\nNow the good part. You can put these four equations together to find X1,Y1 for any value of X0,Y0 (not just points located on one axis or the other). The result is\nX1=X0*cosine(Theta) - Y0*sine(Theta)\nY1=X0*sine(Theta) + Y0*cosine(Theta)\nNext you form a rotation matrix R from the Theta-dependent parts of these equations:\nR=[cosine(Theta)___ -sine(Theta)]\n__[sine(Theta) ___ cosine(Theta)] \nNow when you multiply any vector P0, representing any point in X0,Y0 coordinates, by this matrix R using the rules of matrix multiplication, you are actually using the above two equations, and the result is a vector P1 which is the same point represented in X1,Y1 coordinates. Expressed compactly, P1 = R*P0. P0 is said to be transformed by R to P1.\nOf course, this is a 2-dimensional transformation, and R is a 2-by-2 matrix. Sophisticated 3-D computer graphics use a 3-by-3 matrix implementing very similar equations to do 3-D transformations. This is how you can see convincing 3-D models of objects in rotation.
953148
For found an answer watch here:\nhttp://en.wikipedia.org/wiki/Transition_matrix
730213
Let u, v be the rotated x, y, rotated counter-clockwise by an angle t.\n\nThen x = u*cos(t) + v*sin(t), and y = -u*sin(t) + v*cos(t).\n\nPlug these into the equation (abbreviating c=cos(t), s=sin(t))\n\n(uc+vs)^2 + 4(uc+vs)(vc-us) + 4(vc-us)^2 + (5*sqrt(5))(vc-us) + 5\n= u^2*c^2 + 2*uvcs + v^2*s^2 + 4*uvc^2 + 4*v^2*cs - 4*u^2*cs - 4*uv*s^2 + 4*v^2*c^2  - 8*uvcs + 4*u^2*s^2 + 5*sqrt(5)(vc-us) + 5\n= u^2(c^2-4cs+4s^2) + uv(2cs + 4c^2 - 4s^2 - 8cs) + v^2(s^2 + 4cs + 4c^2) + 5*sqrt(5)(vc-us) + 5\n\nso to eliminate the xy term, we must have 2cs + 4c^2 - 4s^2 - 8cs = 0, but 2cs = sin(2t) and c^2 - s^2 = cos(2t), so we have\n4cos(2t) = 3sin(2t), or tan(2t) = 4/3, so t = arctan(4/3)/2 = 0.4636 radians.\n\nThe resulting equation is 5v^2 - 5u + 10v + 5 = 0.
86452
I don't know for specific .. but here is a link for a 3d max tutorials ..they are helpful\nhttp://3dlinks.com/links.cfm?categoryid=13&subcategoryid=105
911487
The equations of motion for the angular displacement of each disk are given by:\n\nDisk A: R1(t) = (1/2)*alpha1*t² + w1*t +Ri1\nDisk B: R2(t) = (1/2)*alpha2*t² + w2*t +Ri2\n\nFor part (a), we're looking for the time, T, for which R1(T) = R2(T)\n\n(1/2)*alpha1*T² + w1*T +Ri1 = (1/2)*alpha2*T² + w2*T +Ri2\n\nFrom the problem statement, we obtain the following information:\n\nRi1 = Ri2\nalpha1 = 0\nw1 = 9.5 rad/s\nalpha2 = 2.2 rad/s²\nw2 = 0\n\nSubstituting these values and simplifying, we obtain:\n\nw1*T= (1/2)*alpha2*T²\n\nw1= (1/2)*alpha2*T\n\nT = 2*w1 / (alpha2)\n\nT = 2(9.5 rad/s) / (2.2 rad/s²)\n\nT = 8.64 s\n\nTechnically, part (b) is impossible to say without knowing how the reference line is defined, but I'll give it a whirl. If the reference line is measured from the center of each disk to its outer edge, then yes.\n\nHowever, if the reference line is placed across the diameter of the disk, then no. In this case, the reference lines will be aligned when the angular displacement of Disk A is 180° or pi radians ahead of Disk B. To solve for this time, simply add pi to Disk B's equation of motion:\n\n(1/2)*alpha1*T² + w1*T +Ri1 = (1/2)*alpha2*T² + w2*T +Ri2 + pi\n\nSubstituting out known values again leads us to:\n\nw1*T= (1/2)*alpha2*T² + pi\n\nor:\n\n(1/2)*alpha2*T² - w1*T + pi = 0\n\n1.1*T² - 9.5*T + pi = 0\n\nUsing the quadratic formula, we find that \n\nT = .344 seconds or 8.292 seconds\n\nThus, if the reference line is directed across the entire disk, then they are momentarily aligned twice before having the same angular displacement.\n\n(This added after a while)\n{BTW, monohan47130 - no I didn't. Please read what I wrote. If the reference lines are defined along the radius only, then you are correct - there is only one point at which the disks have the same angular displacement. But, reference lines are typically made along the entire circle (diameter). This means that they will be aligned 3 times total in 8.64 seconds :) }
1058187
for a rigid rotor, the transition lines are equally spaced in the wavenumber space. However, this is not always the case, except for the rigid rotor model. \n\nfor non-rigid rotor model, we need to consider changes in the moment of inertia of the molecule due to centrifugal distortion and the effects of vibration on the rotation.\n\nthe formulas are hard to write here, better to look at the references in \nhttp://en.wikipedia.org/wiki/Rotational_spectroscopy\nor elsewhere.
1035212
I know of two:\n\nTranslation and rotation: \n\n1) translation :\n\n Take a rectangle and cut a diagonal and extended the lines  from all three side of the triangle at right angle to it  from the side that is made from cutting the rectangle until those lines equal the lenght of the side of the triangle drawn from  and it  form three boxes  around all three sides of the triangle. Now let take the big side an calulate the square  of it ,then calulate the squares of the other two sides and add themup,  it should equal the square of the big side and that proves the pythagorean theorem. \n\nRotation: \n\nSee the show : Project Mathematics by Cal tech : Entitled "The Pythagorean Theorem" it explain better  than I can, It worth watching. Very good show on Math & Geometry
1164324
To rotate a two dimension image, you need to put you camera in third dimension. \n\nPut a perpendicular on both of rectangles (centers of rectangles) and project them to intersect. Intersection point would be Camera Point A.
792270
Hi I assume you mean an unsymmetrical item. I also assume you want a best fit. I have done similar things on a computer but not by hand. The computer finds three point most distant from the centroid and uses these points as pivots for three orthogonal planes. Not sure of the rest.
655905
It's all "Vector Algebra"\n\nAny OpenGL book would go over the basics of the manipulation math involved.