I hope you can help me. My problem is with the collada's skinning equation:
v += {[(v * BSM) * IBMi * JMi] * JW}
n: The number of joints that influence vertex v
BSM: Bind-shape matrix
IBMi: Inverse bind-pose matrix of joint i
JMi: Transformation matrix of joint i
JW: Weight of the influence of joint i on vertex v
Each vertex "v" must be calculated (i.e. through a "for" bucle). But, it`s not very, very slow if I have a mesh of 10,000 vertices or more? This must be calculated ever in real time? No other way to calculate "v"?
Thank you very much. :-)
You can probably use a threshold JW - for each vertex v, you may skip further computation for any joint i on v where JWi is below some threshold.
Also you could precompute IBMi * JMi for each joint once, right?
Related
I have a vertex (x, y, z) and I want to calculate the screen location where this point would be rendered on my viewport. Something like Ray Picking, just more or less the other way around. I don't think I can use gluProject because at the time I need the projected point my matrices are restored to identities.
I would like to stay independent from OpenGL, so no extra render pass. This way I'm sure it would only be some math like the ray picking thing. I've implemented that one and it works well, so I want to project a vertex the same way.
Of course I have camera pos, up and lookAt vectors and fovy. Is there any source of information about this? Or does anyone know how to work this out?
If your know your matrices (or at least know how to construct them), you can compute screen location for a vertex by multiplying its position with the matrices and then performing viewport transformation:
vProjected = modelViewPojectionMatrix * v;
if (
// check that vertex shouldn't be clipped.
-vProjected.w <= vProjected.x && vProjected.x <= vProjected.w &&
-vProjected.w <= vProjected.y && vProjected.y <= vProjected.w &&
-vProjected.w <= vProjected.z && vProjected.z <= vProjected.w
) {
vProjected /= vProjected.w;
vScreen.x = VIEWPORT_W * vProjected.x / 2 + VIEWPORT_CENTER_X;
vScreen.y = VIEWPORT_H * vProjected.y / 2 + VIEWPORT_CENTER_Y;
}
Note that, as per OpenGL convention, (0, 0) is lower left corner, not upper left one.
Any math library with verctor and matrix operations can help you with that. For example, mathfu or glm.
UPD. How you can construct modelViewProjectionMatrix given camera position and orientation and projection params? We need two matrices (let's assume that model matrix is just an identity, i.e. vertex positions a given already in world coordinate system). First one would be the view matrix, which takes into account camera position and orientation. Here I'll be using mathfu since I'm more familiar with it, but almost every math library design with 3D graphics in mind has the same functions:
viewMatrix = mathfu::mat4::LookAt(
cameraLookAtPosition,
cameraPosition,
cameraUpVector
);
The second one would be projection matrix:
projectionMatrix = mathfu::mat4::Perspective(fovy, aspect, zNear, zFar);
Now modelViewProjectionMatrix is just a product of those two:
modelViewProjectionMatrix = projectionMatrix * viewMatrix;
Note that matrix multiplication is not commutative, in other words A * B != B * A. So order in which matrices are multiplied is important.
A university assignment requires me to use the Vertex Coordinates I have to calculate the Normals and the Tangent from the Normal values so that I can create a Object Space to Texture Space Matrix.
I have the code needed to make the Matrix, and the binormal but I don't have the code for calculating the Tangent. I tried to look online, but the answers usually confuse me. Can you explain to me clearly how it works?
EDIT: I have corrected what I wrote previously as clearly I misunderstood the assignment. Thank you everyone for helping me see that.
A tangent in the mathematical sense is a property of a geometric object, not of the normalmap. In case of normalmapping, we are in addition searching for a very specific tangent (there are infinitely many in each point, basically every vector in the plane defined by the normal is a tangent).
But let's go one step back: We want a space where the u-direction of the texture is mapped on the tangent direction, the v-direction on the bitangent/binormal and the up-vector of the normalmap to the normal of the object. Thus the tangent for a triangle (v0, v1, v2) with uv-coordinates (uv1, uv2, uv3) can be calculated as:
dv1 = v1-v0
dv2 = v2-v0
duv1 = uv1-uv0
duv2 = uv2-uv0
r = 1.0f / (duv1.x * duv2.y - duv1.y * duv2.x);
tangent = (dv1 * duv2.y - dv2 * duv1.y) * r;
bitangent = (dv2 * duv1.x - dv1 * duv2.x) * r;
When having this done for all triangles, we have to smooth the tangents at shared vertices (quite similar to what happens with the normal). There are several algorithms for doing this, depending on what you need. One can, for example, weight the tangents by the surface area of the adjacent triangles or by the incident angle of them.
An implementation of this whole calculation can be found [here] along a more detailed explaination: (http://www.opengl-tutorial.org/intermediate-tutorials/tutorial-13-normal-mapping/)
I have a plane in my 3d space and I want to move it somewhere else, so I use glTranslate to do so.
The planes vertex data is: (0,0,0), (1,0,0), (1,1,0) and (0,1,0).
I translate the object to the position of (2,0,0) through the use of glTranslatef(2.0, 0.0, 0.0).
After the translation the point data is unchanged so if I was to want to collide with my plane the visual position is not its actual position.
Is there a way to get the point data from the MODELVIEW_MATRIX or at least a way to find out what the new values are after the glTranslate?
Don't respond with just add 2.0 to the actual values to move it because what if I want to the use glRotate etc. I still want the points locations.
If you really don't want to maintain your own transformation matrix, you can get the current modelview matrix with:
GLfloat mat[16];
glGetFloatv(GL_MODELVIEW_MATRIX, mat);
You can then apply this matrix to your vertices with a standard matrix multiplication. Keep in mind that the matrix is arranged in column-major order. With an input vector xIn, the transformed vector xOut is:
xOut[0] = mat[0] * xIn[0] + mat[4] * xIn[1] + mat[8] * xIn[2] + mat[12];
xOut[1] = mat[1] * xIn[0] + mat[5] * xIn[1] + mat[9] * xIn[2] + mat[13];
xOut[2] = mat[2] * xIn[0] + mat[6] * xIn[1] + mat[10] * xIn[2] + mat[14];
Keeping track of the current transformation matrix in your own code is really a better approach, IMHO. Aside from eliminating glGet() calls, which can be harmful to performance, it gets you on a path to using modern OpenGL (Core Profile), where the matrix stack and all related calls do not exist anymore.
You can create a matrix from your translation and rotation, so that you can use the matrix to transform the coordinates.
There're many libraries to help you create such matrix and transform coordinates.
I'm wondering how a precise algorithm can be written to compute the frontier of the surface of intersection between a parametric surface f : R^2 --> R^3 and a triangulated mesh.
I've thought to a first approach:
nStepsU = 100
nStepsV = 100
tolerance=0.01 // pick some sensical value
intersectionVertices={}
for u from minU to maxU in nStepsU:
for v from minV to maxV in nStepsV:
for v in verticesInMesh:
if euclidean distance( f(u,v), v ) < tolerance:
add vertex v in a set
connect the vertices in intersectionVertices with a line strip
draw the vertices in intersectionVertices
This algorithm, is very simple but slow (n^3) and does not keep in account that the topography of the mesh is based on triangles so the output points are points of the mesh and not points computed exploiting the intersection of surface with the triangles and is heavily dependent of the tolerance one has to set.
Has someone some better idea or can one drive me to a suitable library for this purpose?
I would iterate over each triangle, and compute the intersection of the triangle with the surface. I would use a geometry shader which takes the triangles as input, and outputs line strips. For each vertex in the triangle, compute the signed distance to the surface. Then iterate over the edges: If there are two vertices where h has different signs, the edge between these vertices intersects with the surface. While I'm sure the exact intersection can be computed, the easiest solution would be to interpolate linearly, i.e.
vec3 intersection = (h0 * v1 + h1 * v0) / (h0 + h1);
Then output each intersection as a vertex of your line segment.
The code I posted here can get you started. If you want to just draw the result, you will probably run into the same problem that I described in that question. If you need the vertices on the client, you can use transform feedback.
Edit: I just did a little test. As the distance function I used
float distToHelicoid(in vec3 p)
{
float theta = p.y / 5 + offset.x / 50;
float a = mod(theta - atan(p.z, p.x), 2*PI) - PI; // [-PI, PI[
if (abs(a) > PI/2)
a = mod(theta - atan(-p.z, -p.x), 2*PI) - PI;
return a;
}
Since there is no inside/outside, and this distance function goes from -90° to 90°, you can only emit vertices if the sign goes from small negative to small positive or vice versa, not when it flips from 90° to -90°. Here I simply filtered out distances where abs(dist) > 45°:
The clean way would be to determine the index of the closest revolution. E.g. [-pi, pi] would be revolution 0, [pi, 3pi] = revolution 1, etc. You would then only emit if two distances refer to the same revolution.
If your surface is always helicoid, you can try to project everything on a cylinder around axis Y.
The surface of helicoid consists of lines orthogonal to the surface of that cylinder and after projection you will get a spiral. After projection of 3D triangle mesh onto that cylinder you will get 2D triangle mesh (note that some areas may be covered with several layers of triangles).
So the task becomes finding triangles in 2D triangle mesh intersecting the spiral which is simpler. If you are OK with approximations, you can segment that spiral and use some kind of tree to find triangles intersecting the spiral.
When you have a triangle intersecting part of spiral, its intersection will be a segment, you can just recalculate 3D coordinates of the segment and set of these segments is your intersection line.
I have a parametric curve, say two vectors of doubles where the parameter is the index, and I have to calculate the angle of the tangent to this curve at any given point (index).
Any suggestion or link about how to do that?
Thanks.
Here's a short formula, equivalent (I think) to pau.estalella's answer:
m[i] = (y[i+1] - y[i-1]) / (x[i+1] - x[i-1])
this approximates, reasonably well, the slope at the point (x[i], y[i]).
Your question mentions the "angle of the tangent". The tangent line, having slope m[i], makes angle arctangent(m[i]) with the positive x axis. If this is what you're after, you might use the two-argument arctangent, if it's available:
angle[i] = atan2(y[i+1] - y[i-1], x[i+1] - x[i-1])
this will work correctly, even when x[i+1] == x[i-1].
I suggest you check out the Wikipedia article on numerical differentiation for a start. Before you go much further than that, decide what purposes you want the tangent for and decide whether or not you need to try more complex schemes than the simple ones in the article.
The first problem you run into is to even define the tangent in one of the vertexes of the curve. Consider e.g. that you have the two arrays:
x = { 1.0, 2.0, 2.0 };
y = { 1.0, 1.0, 2.0 };
Then at the second vertex you have a 90-degree change of direction of the line. In that place the tangent isn't even defined mathematically.
Answer to gregseth's comment below
I guess in your example the "tangent" at the second point would be the line parallel to (P0,P2) passing through P1... which kind of give me the answer : for any point of index N the parallel to (N-1, N+1) passing through N. Would that be a not-too-bad approximation?
It depends on what you are using it for. Consider for example:
x = { 1.0, 2.0, 2.0 };
y = { 1.0, 1000000, 1000000 };
That is basically an L shape with a very high vertical line. In your suggestion it would give you an almost vertical tangent. Is that what you want, or do you rather want a 45-degree tangent in that case? It also depends on your input data how you sould define it.
One solution is get the two vectors connection to the vertex, normalize them and then use your algorithm. That way you would get a 45-degree tangent in the above example.
Compute the first derivative: dy/dx. That gives you the tangent.
The tangent to a smooth curve at a point P is the parametric straight line P + tV, where V is the derivative of the curve with respect to "the parameter". But here the parameter is just the index of an array, and numerical differentiation is a difficult problem, hence to approximate the tangent I would use (weighted) least squares approximation.
In other words, choose three or five points of the curve around your point of interest P (i.e. P[i-2], P[i-1], P[i], P[i+1], and P[i+2], if P==P[i]), and approximate them with a straight line, in the least squares sense. The more weight you assign to the middle point P, the more close the line will be to P; on the other hand, the more weight you assign to the extremal points, the more "tangent" the straight line will be, that is, the more nicely it will approximate you curve in the neighborhood of P.
For example, with respect to the following points:
x = [-1, 0, 1]
y = [ 0, 1, 0]
for which the tangent is not defined (as in Anders Abel's answer),
this approach should yield a horizontal straight line close to the point (0,1).
You can try to compute the tangent of an interpolating curve that passes through the given points (I'm thinking of a cubic spline, which is pretty easy to derive) or compute the tangent directly from the data points.
You can find a rough approximation of the derivative in the following manner
Let a curve C pass through points p1,p2 and p3. At point p2 you have two possible tangents: t1=p2-p1 and t2=p3-p2. You can combine them by simply computing their average: 0.5*(t1+t2)
or you can combine them according to their lengths (or their reciprocal 1/length)
Remember to normalize the resulting tangent.
In order to compute the angle between the tangent and the curve, remember that the dot product of two unit vectors gives the cosine of the angle between them. Take the resulting tangent t and the unit vector v2=|p3-p2|, and acos(dot(t,v2)) gives the angle you need.