I'm converting a 3d coordinate sequence into a bvh file.
I made the HIERARCHY using the first frame(not T-pose), and I want to use the rest of the frames to make the MOTION part.
I don't know the exact meaning of the euler angle used in the bvh format.
For example, when there are two vectors v1 and v2 as shown in the picture below
v1 = (X1, Y1, Z1) v2 = (X2, Y2, Z2)
The two vectors were the same length and rotated from v1 to v2.
How can I get rotation(Euler angle in bvh format) between 2 vectors?
I don't know if this answers the "meaning" of the Euler angle in BVH format, but in BVH the Euler angles of a bone in a given frame represent the rotation of the bone from the base position, not from the previous frame. The base position is the OFFSET defined in the HIERARCHY. You should check out Motion Capture File Formats Explained by Meredith and Maddock, they offer a more in-depth discussing about the BVH file format.
To find a rotation matrix to rotate vector v1 onto vector v2 you can use the equation provided in Jur van den Berg's answer here. Then, you can recover the Euler Angles using Shoemake's algorithm, note that the order of rotation will change the equations. But if your bone's base position does not lie on v1, you need to compute the rotation from its base position and not from v1.
Related
I only know my camera rotation R, but without the translational component T. Intrinsics are known, and I have a set of 3D points P with their 2D correspondences Q. Is there a method to compute T directly, and at best is differentiable in P.
I have found an approximate solution in https://arxiv.org/pdf/1611.09813v5.pdf
I have a 2D shape (a circle) that I want to extrude along a 3D curve to create a 3D tube mesh.
Currently the way I generate cross-sections along the curve (which form the basis of the resulting mesh) is to take every control point along the curve, create a 3D transform matrix for it, then multiply the 2D points of my circle by those curve-point matrices to determine their location in 3D space along the curve.
To create the matrix (from 3 vectors), I use the tangent on the curve as the up vector, world-up ([0,1,0]) as the forward vector, and the cross product of the up/forward vectors as the right vector. All three vectors are also orthogonalized during the process to create the final matrix.
The problem comes when my curve tangent is identical to the world-up axis. Ie, my tangent vector is [0,1,0] and the world-up is [0,1,0]....since the cross product of two parallel vectors is not explicit....the resulting extruded mesh has artifacts along those areas of the curve (pinching, twisting, etc).
I thought a potential solution would be to use the dot product of the curve tangent and the world-up as an interpolation value to shift my forward vector from world-up to world-right...in other words, as a curve tangent approaches [0,1,0], my forward vector approaches [1,0,0]...but that results in unwanted twisting along the final mesh as well.
How can I extrude my shape along a curve in a consistent manner that has no flipping/artifacts/twisting? I know it's possible since various off-the-shelf 3D applications can do it...I'm just not sure how.
One way I would approach this is to consider my tangent vector to the 3D curve as actually being a normal vector of the plane I am interested into.
Let's say, the tangent vector is
All you need now is two other vectors that are othoghonal to it, so let's.
Let's construct v like so:
(rotating the coordinates). Because v is the result of the cross product of u and something else, you know that v is orthogonal to u.
(This method will not work if u have equal x,y,z coordinates, in that case, construct the other vector by adding random numbers to at least two variables, rince&repeat).
Then you can simply construct w like before:
normalize and go.
In a 2D plane, I have 2 objects (A and B) with 2 coordinates. Their centers are A(xA, yA) and B(xB, yB) (and C(xC, yC)=C(xB, yA) as AC parallel to the OX line and BC is perpendicular on AC). I can manipulate the rotation of an object and I have access to all usual math operations and can use degrees and radians.
I researched but I couldn't find anything explained in detail.
I also tried using the math formula with arccos formula as follows:
I tried to calculate the distance from A to B (AB) using the Pythagoras theorem, then calculate A to C (AC), then calculate cos(angle)=AC/AB, so the final angle to which I would need to rotate object A towards B is arccos(AC/AB).
Problem is this sounds insanely buggy as you can probably get a lot of digits and ruin everything.
So how can I do this? Please explain mathematically. Thanks!
The simplest way to find the angle between two points is to take their arctangent (a.k.a. inverse tangent). You were on the right track with using cosine, but tangent simplifies the process by not requiring the distance between the points to be known.
As such, you'll want to use an atan2 function in your choice of language. The C# Math.Atan2, for example:
double angle = Math.Atan2(B.Y - A.Y, B.X - A.X);
Note: This particular function returns the angle in radians.
Do you want to rotate object A towards B with C like center of the rotation ?
If it's the case you have only to rotate with an angle of 90 degrees because your triangle is special. But if you want to apply a rotation with a specific angle around a specific center you have to use a transformation TRT.
You will find more explanation here.
I have a 3d object which is free to rotate along x,y and z axis and it is then saved as a transform matrix. In a case where the sequence of rotation is not known and the object is rotated for more than 3 times (eg :-if i rotate the object x-60degress, y-30 degrees, z-45 degrees then again x->30 degrees), is it possible to extract the angles rotated from the transform matrix?.I know that it is possible to get angles if the sequence of rotation is known, but if I have only the final transform matrix with me and nothing else, is it possible to get the angles rotated(x,y,and z) from the transform matrix ?
Euler angle conversion is a pretty well known topic. Just normalize the matrix orientation vectors and then use something like this c source code.
The matrix is the current state of things it has no knowledge of what the transformation has been in the past. It does not know how the matrix was built. You can just take the matrix into and decompose it into any pieces you like, as long as:
The data do not overlap. For example:Two X turns after each other is indistinguishable form each other (no way to know if its 1 2 or three different rotations summed).
The sequence order is known
A decomposition can be built out of the data (for example scale can be measured)
I have an image of a 3D rectangle (which due to the projection distortion is not a rectangle in the image). I know the all world and image coordinates of all corners of this rectangle.
What I need is to determine the world coordinate of a point in the image inside this rectangle. To do that I need to compute a transformation to unproject that rectangle to a 2D rectangle.
How can I compute that transform?
Thanks in advance
This is a special case of finding mappings between quadrilaterals that preserve straight lines. These are generally called homographic transforms. Here, one of the quads is a rectangle, so this is a popular special case. You can google these terms ("quad to quad", etc) to find explanations and code, but here are some sites for you.
Perspective Transform Estimation
a gaming forum discussion
extracting a quadrilateral image to a rectangle
Projective Warping & Mapping
ProjectiveMappings for ImageWarping by Paul Heckbert.
The math isn't particularly pleasant, but it isn't that hard either. You can also find some code from one of the above links.
If I understand you correctly, you have a 2D point in the projection of the rectangle, and you know the 3D (world) and 2D (image) coordinates of all four corners of the rectangle. The goal is to find the 3D coordinates of the unique point on the interior of the (3D, world) rectangle which projects to the given point.
(Do steps 1-3 below for both the 3D (world) coordinates, and the 2D (image) coordinates of the rectangle.)
Identify (any) one corner of the rectangle as its "origin", and call it "A", which we will treat as a vector.
Label the other vertices B, C, D, in order, so that C is diagonally opposite A.
Calculate the vectors v=AB and w=AD. These form nice local coordinates for points in the rectangle. Points in the rectangle will be of the form A+rv+sw, where r, s, are real numbers in the range [0,1]. This fact is true in world coordinates and in image coordinates. In world coordinates, v and w are orthogonal, but in image coordinates, they are not. That's ok.
Working in image coordinates, from the point (x,y) in the image of your rectangle, calculate the values of r and s. This can be done by linear algebra on the vector equations (x,y) = A+rv+sw, where only r and s are unknown. It will boil down to a 2x2 matrix equation, which you can solve generally in code using Cramer's rule. (This step will break if the determinant of the required matrix is zero. This corresponds to the case where the rectangle is seen edge-on. The solution isn't unique in that case. If that's possible, make special exception.)
Using the values of r and s from 4, compute A+rv+sw using the vectors A, v, w, for world coordinates. That's the world point on the rectangle.