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• pairwise crossproduct in Python?

  - by user248237

  how can I get the list of cross product pairs from a list of arbitrarily long lists in python? e.g. a = [1,2,3] b = [4,5,6] crossproduct(a,b) should yield [[1,4], [1, 5], [1,6], ...] thanks.

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• MediaElement seems to disable lock screen when playing a video

  - by CrossProduct

  I am working on an application that plays a video using the MediaElement component. Now, I would like that if the user is idle, the lock screen appears, as configured by the user in the Settings of the device. If I don't play a video, the lock screen indeed appears. But, when a video is playing no lock screen appears. I cannot find any information on this. Currently I set the idle detection modes like this: PhoneApplicationService.Current.ApplicationIdleDetectionMode = IdleDetectionMode.Enabled; PhoneApplicationService.Current.UserIdleDetectionMode = IdleDetectionMode.Enabled; I am a bit lost now. The only solution I can think of is running a timer myself and stop the video playback after a certain time. (but there seems to be no API calls to receive the configured lock timeout.) Any suggestions are welcome, thanks.

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• Arcball 3D camera - how to convert from camera to object coordinates

  - by user38873

  I have checked multiple threads before posting, but i havent been able to figure this one out. Ok so i have been following this tutorial, but im not using glm, ive been implementing everything up until now, like lookat etc. http://en.wikibooks.org/wiki/OpenGL_Programming/Modern_OpenGL_Tutorial_Arcball So i can rotate with the click and drag of the mouse, but when i rotate 90º degrees around Y and then move the mouse upwards or donwwards, it rotates on the wrong axis, this problem is demonstrated on this part of the tutorial An extra trick is converting the rotation axis from camera coordinates to object coordinates. It's useful when the camera and object are placed differently. For instace, if you rotate the object by 90° on the Y axis ("turn its head" to the right), then perform a vertical move with your mouse, you make a rotation on the camera X axis, but it should become a rotation on the Z axis (plane barrel roll) for the object. By converting the axis in object coordinates, the rotation will respect that the user work in camera coordinates (WYSIWYG). To transform from camera to object coordinates, we take the inverse of the MV matrix (from the MVP matrix triplet). What i have to do acording to the tutorial is convert my axis_in_camera_coordinates to object coordinates, and the rotation is done well, but im confused on what matrix i use to do just that. The tutorial talks about converting the axis from camera to object coordinates by using the inverse of the MV. Then it shows these 3 lines of code witch i havent been able to understand. glm::mat3 camera2object = glm::inverse(glm::mat3(transforms[MODE_CAMERA]) * glm::mat3(mesh.object2world)); glm::vec3 axis_in_object_coord = camera2object * axis_in_camera_coord; So what do i aply to my calculated axis?, the inverse of what, i supose the inverse of the model view? So my question is how do you transform camera axis to object axis. Do i apply the inverse of the lookat matrix? My code: if (cur_mx != last_mx || cur_my != last_my) { va = get_arcball_vector(last_mx, last_my); vb = get_arcball_vector( cur_mx, cur_my); angle = acos(min(1.0f, dotProduct(va, vb)))*20; axis_in_camera_coord = crossProduct(va, vb); axis.x = axis_in_camera_coord[0]; axis.y = axis_in_camera_coord[1]; axis.z = axis_in_camera_coord[2]; axis.w = 1.0f; last_mx = cur_mx; last_my = cur_my; } Quaternion q = qFromAngleAxis(angle, axis); Matrix m; qGLMatrix(q,m); vi = mMultiply(m, vi); up = mMultiply(m, up); ViewMatrix = ogLookAt(vi.x, vi.y, vi.z,0,0,0,up.x,up.y,up.z);

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• C# Vector maths questions

  - by Mark

  Im working in a screen coordinate space that is different to that of the classical X/Y coordinate space, where my Y direction goes down in the positive instead of up. Im also trying to figure out how to make a Circle on my screen always face away from the center point of the screen. If the center point of my screen is at x(200) y(300) and the point of my circle's center is at x(150) and y(380) then I would like to calculate the angle that the circle should be facing. At the moment I have this: Point centerPoint = new Point(200, 300); Point middleBottom = new Point(200, 400); Vector middleVector = new Vector(centerPoint.X - middleBottom.X, centerPoint.Y - middleBottom.Y); Vector vectorOfCircle = new Vector(centerPoint.X - 150, centerPoint.Y - 400); middleVector.Normalize(); vectorOfCircle.Normalize(); var angle = Math.Acos(Vector.CrossProduct(vectorOfCircle, middleVector)); Console.WriteLine("Angle: {0}", angle * (180/Math.PI)); Im not getting what I would expect. I would say that when I enter in x(150) and y(300) of my circle, I would expect to see the rotation of 90 deg, but Im not getting that... Im getting 180!! Any help here would be greatly appreciated. Cheers, Mark

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• How to rotate a set of points on z = 0 plane in 3-D, preserving pairwise distances?

  - by cagirici

  I have a set of points double n[] on the plane z = 0. And I have another set of points double[] m on the plane ax + by + cz + d = 0. Length of n is equal to length of m. Also, euclidean distance between n[i] and n[j] is equal to euclidean distance between m[i] and m[j]. I want to rotate n[] in 3-D, such that for all i, n[i] = m[i] would be true. In other words, I want to turn a plane into another plane, preserving the pairwise distances. Here's my code in java. But it does not help so much: double[] rotate(double[] point, double[] currentEquation, double[] targetEquation) { double[] currentNormal = new double[]{currentEquation[0], currentEquation[1], currentEquation[2]}; double[] targetNormal = new double[]{targetEquation[0], targetEquation[1], targetEquation[2]}; targetNormal = normalize(targetNormal); double angle = angleBetween(currentNormal, targetNormal); double[] axis = cross(targetNormal, currentNormal); double[][] R = getRotationMatrix(axis, angle); return rotated; } double[][] getRotationMatrix(double[] axis, double angle) { axis = normalize(axis); double cA = (float)Math.cos(angle); double sA = (float)Math.sin(angle); Matrix I = Matrix.identity(3, 3); Matrix a = new Matrix(axis, 3); Matrix aT = a.transpose(); Matrix a2 = a.times(aT); double[][] B = { {0, axis[2], -1*axis[1]}, {-1*axis[2], 0, axis[0]}, {axis[1], -1*axis[0], 0} }; Matrix A = new Matrix(B); Matrix R = I.minus(a2); R = R.times(cA); R = R.plus(a2); R = R.plus(A.times(sA)); return R.getArray(); } This is what I get. The point set on the right side is actually part of a point set on the left side. But they are on another plane. Here's a 2-D representation of what I try to do: There are two lines. The line on the bottom is the line I have. The line on the top is the target line. The distances are preserved (a, b and c). Edit: I have tried both methods written in answers. They both fail (I guess). Method of Martijn Courteaux public static double[][] getRotationMatrix(double[] v0, double[] v1, double[] v2, double[] u0, double[] u1, double[] u2) { RealMatrix M1 = new Array2DRowRealMatrix(new double[][]{ {1,0,0,-1*v0[0]}, {0,1,0,-1*v0[1]}, {0,0,1,0}, {0,0,0,1} }); RealMatrix M2 = new Array2DRowRealMatrix(new double[][]{ {1,0,0,-1*u0[0]}, {0,1,0,-1*u0[1]}, {0,0,1,-1*u0[2]}, {0,0,0,1} }); Vector3D imX = new Vector3D((v0[1] - v1[1])*(u2[0] - u0[0]) - (v0[1] - v2[1])*(u1[0] - u0[0]), (v0[1] - v1[1])*(u2[1] - u0[1]) - (v0[1] - v2[1])*(u1[1] - u0[1]), (v0[1] - v1[1])*(u2[2] - u0[2]) - (v0[1] - v2[1])*(u1[2] - u0[2]) ).scalarMultiply(1/((v0[0]*v1[1])-(v0[0]*v2[1])-(v1[0]*v0[1])+(v1[0]*v2[1])+(v2[0]*v0[1])-(v2[0]*v1[1]))); Vector3D imZ = new Vector3D(findEquation(u0, u1, u2)); Vector3D imY = Vector3D.crossProduct(imZ, imX); double[] imXn = imX.normalize().toArray(); double[] imYn = imY.normalize().toArray(); double[] imZn = imZ.normalize().toArray(); RealMatrix M = new Array2DRowRealMatrix(new double[][]{ {imXn[0], imXn[1], imXn[2], 0}, {imYn[0], imYn[1], imYn[2], 0}, {imZn[0], imZn[1], imZn[2], 0}, {0, 0, 0, 1} }); RealMatrix rotationMatrix = MatrixUtils.inverse(M2).multiply(M).multiply(M1); return rotationMatrix.getData(); } Method of Sam Hocevar static double[][] makeMatrix(double[] p1, double[] p2, double[] p3) { double[] v1 = normalize(difference(p2,p1)); double[] v2 = normalize(cross(difference(p3,p1), difference(p2,p1))); double[] v3 = cross(v1, v2); double[][] M = { { v1[0], v2[0], v3[0], p1[0] }, { v1[1], v2[1], v3[1], p1[1] }, { v1[2], v2[2], v3[2], p1[2] }, { 0.0, 0.0, 0.0, 1.0 } }; return M; } static double[][] createTransform(double[] A, double[] B, double[] C, double[] P, double[] Q, double[] R) { RealMatrix c = new Array2DRowRealMatrix(makeMatrix(A,B,C)); RealMatrix t = new Array2DRowRealMatrix(makeMatrix(P,Q,R)); return MatrixUtils.inverse(c).multiply(t).getData(); } The blue points are the calculated points. The black lines indicate the offset from the real position.

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