MATLAB - get angle between arrays of points - arrays

For the sake of illumination analysis, based on this document, I am trying to determine three things for an array of lights and a series of points on a solid surface:
(Image key: big blue points are lights with illumination direction shown, small points are the points on my surface)
1) The distances between each of the lights and each of the points,
2) the angles between the direction each light is facing and the normal vectors of all of the points:
Note in this image I have replicated the normal vector and moved it to more clearly show the angle.
3) the angles between the direction each light is facing, and the vector from that light to all of the points on the solid:
Originally I had nested for loops iterating through all of the lights and points on the solid, but am now doing my best to do it in true MATLAB style with matrices:
I have found the distances between all the points with the pdist2 function, but have not managed to find a similar method to find the angles between the lights and all the points, nor the lights and the normal vectors of the points. I would prefer to do this with matrix methods rather than with iteration as I have been using.
Considering I have data set out, where each column of Lmat has my x,y,z position vectors of my lights; Dmat gives x,y,z directions of each light, thus the combination of each row from both of these matrices fully define the light and the direction it is facing. Similarly, Omega and nmat do the same for the points on the surface.
I am fairly sure that to get angles I want to do something along the lines of:
distMatrix = pdist2(Omega, Lmat);
LmatNew = zeros(numPoints, numLights, 3);
DmatNew = zeros(numPoints, numLights, 3);
OmegaNew = zeros(numPoints, numLights, 3);
nmatNew = zeros(numPoints, numLights, 3);
for i = 1:numLights
LmatNew(:,i,1) = Lmat(i,1);
LmatNew(:,i,2) = Lmat(i,2);
LmatNew(:,i,3) = Lmat(i,3);
DmatNew(:,i,1) = Dmat(i,1);
DmatNew(:,i,2) = Dmat(i,2);
DmatNew(:,i,3) = Dmat(i,3);
end
for j = 1:numPoints
OmegaNew(j,:,1) = Omega(j,1);
OmegaNew(j,:,2) = Omega(j,2);
OmegaNew(j,:,3) = Omega(j,3);
DmatNew(:,i,1) = Dmat(i,1);
DmatNew(:,i,2) = Dmat(i,2);
DmatNew(:,i,3) = Dmat(i,3);
end
angleMatrix = -dot(LmatNew-OmegaNew, DmatNew, 3);
angleMatrix = atand(angleMatrix);
angleMatrix = angleMatrix.*(angleMatrix > 0);
But I am getting conceptually stuck trying to get my head around what to do after my dot product.
Am I on the right track? Is there an inbuilt angle equivalent of pdist2 that I am overlooking?
Thanks all for your help, and sorry for the paint images!
Context: This image shows my lights (big blue points), the directions the lights are facing (little black traces), and my model.

According to MathWorks, there is no built-in function to calculate the angle between vectors. However, you can use trigonometry to calculate the angles.
Inputs
Since you unfortunately didn't explain your input data in great detail, I'm going to assume that you have a matrix Lmat containing a location vector of a light source in each row and a matrix Dmat containing the directional vectors for the light sources, both of size n×3, where n is the number of light sources in your scene.
The matrices Omega and Nmat supposedly are of size m×3 and contain the location vectors and normal vectors of all m surface points. The desired result are the angles between all light direction vectors and surface normal vectors, of which there are n⋅m, and the angles between the light direction vectors and the vectors connecting the light to each point on the surface, of which there are n⋅m as well.
To get results for all combinations of light sources and surface points, the input matrices have to be repeated vertically:
Lmat = repmat(Lmat, size(Omega,1), 1);
Dmat = repmat(Dmat, size(Omega,1), 1);
Omega = repmat(Omega, size(Lmat,1), 1);
Nmat = repmat(Nmat, size(Lmat,1), 1);
Using the inner product / dot product
The definition of the inner product of two vectors is
where θ is the angle between the two vectors. Reordering the equation yields
You can therefore calculate the angles between your directional vectors Dmat and your normal vectors Nmat like this:
normProd = sqrt(sum(Dmat.^2,2)).*sqrt(sum(Nmat.^2,2));
anglesInDegrees = acos(dot(Dmat.',Nmat.')' ./ normProd) * 180 / pi;
To calculate the angles between the light-to-point vectors and the directional vectors, just replace Nmat with Omega - Lmat.
Using the vector product / cross product
It has been mentioned that the above method will have problems with accuracy for very small (θ ≈ 0°) or very large (θ ≈ 180°) angles. The suggested solution is calculating the angles using the cross product and the inner product.
The norm of the vector product of two vectors is
You can combine this with the above definition of the inner product to get
which can obviously be reordered to this:
The corresponding MATLAB code looks like this:
normCross = sqrt(sum(cross(Dmat,Nmat,2).^2,2));
anglesInDegrees = atan2(normCross,dot(Dmat.',Nmat.')') * 180/pi;

Related

Using each element of a vector in a series of calculations

I am trying to write MATLAB code that uses a set of variables in a vector in a calculation. I am trying to run the same formula with each value in the vector and then store each result in a new vector.
The goal is to calculate and plot the cost of constructing a water tank based on various radius sizes. In the calculations I have a cylindrical tank and a hemispherical top. The exact value for the volume of the tank is 500m^3. The cost of the tank is $400/m^2 surface area for the hemispherical top and $300/m^2 surface area for the cylindrical body. I know I need to use element wise operators, however I am getting strange, unrealistic results which leads me to believe I am using these incorrectly.
rTank = 2:0.5:10;
h = ((250./(pi.*rTank(:)))-((rTank(:).^2)./3));
cost = ((2*pi*400.*(rTank(:).^2))+(2*pi*h(:).*300.*rTank(:)));
plot(rTank,cost)
I am expecting a curve of all positive values between radii 2m and 10m, with positive values for cost. For some reason I am getting negative values for results, and according to the resulting plot, the cost of the water tank is free when the radius is 8m, which makes no sense.
Filter out h<0
rTank = 2:0.5:10;
h = 250./(pi*rTank)-1/3*rTank.^2;
good_h=h(h>0);
good_rTank=rTank(h>0);
cost = 2*pi*400*good_rTank.^2 + 2*pi*good_h*300.*good_rTank;
plot(good_rTank, cost)

What exactly is the output quaternion of slerp?

I'm trying to implement SLERP (described by Ken Shoemake in "Animating Rotation with Quaternion Curves)
I've read up on the topic on wikipedia (topic: quaternions, 1 and 2) and other sites and also searched stackoverflow about this problem. It seems like I understand the theory behind it, but oversee one small detail. I will use w for the scalar value of the quaternion
So initially I have two 3D vectors. Each vector has a representation in two coordinate systems (C and C'). My goal is to find a third representation of these vectors in the system "halfway" the initial two.
So what I do is I find the rotation matrix, which transform the vectors from C to C', which seems to work out quite fine.
My next step is to transform this matrix into a quaternion, which also works.
Now my issue is with the formula of slerp, which is:
slerp(q1, q2; u) = ((sin(1-u) * t)/ (sin t)) * q1 + (sin(ut)/sin t) * q2
(sorry can't upload images yet for a better representation: see source 1)
so I guess here u = 0.5, q1 is the vector I would like to rotate (with w=0) and q2 equals the quaternion I calculated previously. Theta is calculated from the dotproduct of the normalized vector and the (already) normalized quaternion.
So what I expect is that I get back a vector, rotated either from C to the third coordinate system or from C' to the third coordinate system.
My issue now is, that I don't see, how I will get a vector and not a quaternion. Meaning, how is it possible, that I will get a quaternion with (w=0), as by simply multiplying q2 with this factor won't set w to 0. Or is it something else I will get from this function?
What am I overseeing here?
Thanks for your help!
Seems like I figured it out. For someone with the same understanding problem:
slerp simply interpolates between two orientations, meaning between two actual rotations. So in my case, q1 is the quaternion corresponding to the identity matrix (so [1, 0, 0, 0]). q2 is the rotation. theta is still 0.5.
With the quaternion I get from this, I have to calculate the rotation with q^-1 v q. Where v is my vector I want to rotate. This can be calculated using the Hamilton product.

Uniformly sampling on hyperplanes

Given the vector size N, I want to generate a vector <s1,s2, ..., sn> that s1+s2+...+sn = S.
Known 0<S<1 and si < S. Also such vectors generated should be uniformly distributed.
Any code in C that helps explain would be great!
The code here seems to do the trick, though it's rather complex.
I would probably settle for a simpler rejection-based algorithm, namely: pick an orthonormal basis in n-dimensional space starting with the hyperplane's normal vector. Transform each of the points (S,0,0,0..0), (0,S,0,0..0) into that basis and store the minimum and maximum along each of the basis vectors. Sample uniformly each component in the new basis, except for the first one (the normal vector), which is always S, then transform back to the original space and check if the constraints are satisfied. If they are not, sample again.
P.S. I think this is more of a maths question, actually, could be a good idea to ask at http://maths.stackexchange.com or http://stats.stackexchange.com
[I'll skip "hyper-" prefix for simplicity]
One of possible ideas: generate many uniformly distributed points in some enclosing volume and project them on the target part of plane.
To get uniform distribution the volume must be shaped like the part of plane but with added margins along plane normal.
To uniformly generate points in such volumewe can enclose it in a cube and reject everything outside of the volume.
select margin, let's take margin=S for simplicity (once margin is positive it affects only performance)
generate a point in cube [-M,S+M]x[-M,S+M]x[-M,S+M]
if distance to the plane is more than M, reject the point and go to #2
project the point on the plane
check that projection falls into [0,S]x[0,S]x[0,S], if not - reject and go to #2
add this point to the resulting set and go to #2 is you need more points
The problem can be mapped to that of sampling on linear polytopes for which the common approaches are Monte Carlo methods, Random Walks, and hit-and-run methods (see https://www.jmlr.org/papers/volume19/18-158/18-158.pdf for examples a short comparison). It is related to linear programming, and can be extended to manifolds.
There is also the analysis of polytopes in compositional data analysis, e.g. https://link.springer.com/content/pdf/10.1023/A:1023818214614.pdf, which provide an invertible transformation between the plane and the polytope that can be used for sampling.
If you are working on low dimensions, you can use also rejection sampling. This means you first sample on the plane containing the polytope (defined by your inequalities). This later method is easy to implement (and wasteful, of course), the GNU Octave (I let the author of the question re-implement in C) code below is an example.
The first requirement is to get vector orthogonal to the hyperplane. For a sum of N variables this is n = (1,...,1). The second requirement is a point on the plane. For your example that could be p = (S,...,S)/N.
Now any point on the plane satisfies n^T * (x - p) = 0
we assume also that x_i >= 0
With these given you compute an orthonormal basis on the plane (the nullity of the vector n) and then create random combination on that bases. Finally you map back to the original space and apply your constraints on the generated samples.
# Example in 3D
dim = 3;
S = 1;
n = ones(dim, 1); # perpendicular vector
p = S * ones(dim, 1) / dim;
# null-space of the perpendicular vector (transposed, i.e. row vector)
# this generates a basis in the plane
V = null (n.');
# These steps are just to reduce the amount of samples that are rejected
# we build a tight bounding box
bb = S * eye(dim); # each column is a corner of the constrained region
# project on the null-space
w_bb = V \ (bb - repmat(p, 1, dim));
wmin = min (w_bb(:));
wmax = max (w_bb(:));
# random combinations and map back
nsamples = 1e3;
w = wmin + (wmax - wmin) * rand(dim - 1, nsamples);
x = V * w + p;
# mask the points inside the polytope
msk = true(1, nsamples);
for i = 1:dim
msk &= (x(i,:) >= 0);
endfor
x_in = x(:, msk); # inside the polytope (your samples)
x_out = x(:, !msk); # outside the polytope
# plot the results
scatter3 (x(1,:), x(2,:), x(3,:), 8, double(msk), 'filled');
hold on
plot3(bb(1,:), bb(2,:), bb(3,:), 'xr')
axis image

I need to translate 3d points relative to a triangle as if the triangle was somewhere else

I posted this on twitter a while ago but seeing how none of my followers appears to be a math/programming genius, I'll try my luck here as well. I got here because I found this which might contain part of my solution.
I described my problem in the following pdf document, containing a picture of what I'm trying to achieve.
To give some more details, I divided the pentagon's of a dodecahedron (12 pentagons) into triangles (5/pentagon, 60 triangles in total), then collected a set of data points relative to each of these triangles.
The idea is to generate terrain meshes for each individual triangle.
To do so, the data must be represented flat, in a 32K x 32K square (idTech4 Megatexture)
I have vaguely heard of transformation matrices, which when set up properly, could do the trick of passing all the data points trough them to have them show up in the right place.
I looked at this source code here but I don't understand how I'm supposed to get the points in and/or out of there, not to mention how to do the setup so I can present each point in turn and get the result point back.
I got as fas as identifying the point that belongs in the back right corner. All my 3D points are originally stored in latitude / longitude pairs. I retrieve the 3D vectors this way:
coord getcoord(point* p)
{
coord c;
c.x=cos(p->lat*pi/180.l) * cos(p->lon*pi/180.l);
c.y=cos(p->lat*pi/180.l) * sin(p->lon*pi/180.l);
c.z=sin(p->lat*pi/180.l);
return c;
};
My thought is that if I can find the center of my triangle, and discover how to offset my angles so the vector from the center of my sphere to the middle of the triangle moves to 90N then my points would already be in the right plane if I rotated them all along the same angles. If I then convert them all to 3d and subtracti the radius from y, they'll be at the correct y position as well.
Then all I'd need to do is the rotation, the scaling, and the moving to the final position.
There are several kinds of 'centers' for a triangle, I think the one I need is the one that is equidistant to the corners of the triangle (Circumcenter?)
But then there might be an easier approach to the whole problem so while I continue my own research, perhaps some of you can help pointing me in the right direction.
It appears as if some sample data is in order, here are a few of these triangles in obj file format:
v 0.000000 0.000000 3396.000000
v 2061.582356 0.000000 2698.646733
v 637.063983 1960.681333 2698.646733
f 1 2 3
And another:
v -938.631230 2888.810129 1518.737455
v 637.063983 1960.681333 2698.646733
v 1030.791271 3172.449325 637.064076
f 1 2 3
You will notice that each point is at a distance of 3396 from 0,0,0
I mentioned 'on the sphere' meaning that the face away from the center of the sphere is the face that needs to become the 'top' when translated into the square.
Theoretically all these triangles should in fact have identical sizes, but due to rounding errors in the math that generated them, this might not be entirely true.
If I'm not mistaken I already took measures to ensure that the first point you see here is always the one opposite the longest border, so it's the one that should go in the far left corner (testing the above 2 samples confirms this, but I'm measuring anyway just to be sure)
Both legs leading away from this point should theoretically have the same length as well, but again rounding errors might slightly offset that.
If I've done it correctly then the longer side is 1,113587 times longer than the 2 shorter sides. Assuming those are identical, then doing some goal seeking in excel, I can deduct that the final points, assuming I was just translating this triangle, should look like:
v 16384.000000 0.000000 16384.000000
v -16384.000000 0.000000 9916.165306
v 9916.165306 0.000000 -16384.000000
f 1 2 3
So I need to setup the matrix to do this transformation, preferably using the 4x4 matrix as explained below.
I would recommend using transform matrices. The 3d transform matrix is a 4x4 data structure which describes a translation and rotation (and possibly a scale). Once you have a matrix you can transform a point like so
result.x = (tmp->pt.x * m->element[0][0]) +
(tmp->pt.y * m->element[1][0]) +
(tmp->pt.z * m->element[2][0]) +
m->element[3][0];
result.y = (tmp->pt.x * m->element[0][1]) +
(tmp->pt.y * m->element[1][1]) +
(tmp->pt.z * m->element[2][1]) +
m->element[3][1];
result.z = (tmp->pt.x * m->element[0][2]) +
(tmp->pt.y * m->element[1][2]) +
(tmp->pt.z * m->element[2][2]) +
m->element[3][2];
int w = (tmp->pt.x * m->element[0][3]) + (tmp->pt.y * m->element[1][3])
+ (tmp->pt.z * m->element[2][3]) + m->element[3][3];
if (w!=0 || w!=1)
result.x/=w; result.y/=w; result.z/=w;
This will transform the 3D point pt by the matrix m. If you now a little matrix math you'll see i'm just multiplying my origin point as a vector against the matrix (and doing a little normalization if it is a skew matrix.) Matrices can be multiplied together to form complicated transformations so they are very useful.
For details on making matrices suggest reading this link.
http://en.wikipedia.org/wiki/Transformation_matrix

WPF: Getting new coordinates after a Rotation

With reference to this programming game I am currently building.
alt text http://img12.imageshack.us/img12/2089/shapetransformationf.jpg
To translate a Canvas in WPF, I am using two Forms: TranslateTransform (to move it), and RotateTransform (to rotate it) [children of the same TransformationGroup]
I can easily get the top left x,y coordinates of a canvas when its not rotated (or rotated at 90deg, since it will be the same), but the problem I am facing is getting the top left (and the other 3 points) coordinates.
This is because when a RotateTransform is applied, the TranslateTransform's X and Y properties are not changed (and thus still indicate that the top-left of the square is like the dotted-square (from the image)
The Canvas is being rotated from its center, so that is its origin.
So how can I get the "new" x and y coordinates of the 4 points after a rotation?
[UPDATE]
alt text http://img25.imageshack.us/img25/8676/shaperotationaltransfor.jpg
I have found a way to find the top-left coordinates after a rotation (as you can see from the new image) by adding the OffsetX and OffsetY from the rotation to the starting X and Y coordinates.
But I'm now having trouble figuring out the rest of the coordinates (the other 3).
With this rotated shape, how can I figure out the x and y coordinates of the remaining 3 corners?
[EDIT]
The points in the 2nd image ARE NOT ACCURATE AND EXACT POINTS. I made the points up with estimates in my head.
[UPDATE] Solution:
First of all, I would like to thank Jason S for that lengthy and Very informative post in which he describes the mathematics behind the whole process; I certainly learned a lot by reading your post and trying out the values.
But I have now found a code snippet (thanks to EugeneZ's mention of TransformBounds) that does exactly what I want:
public Rect GetBounds(FrameworkElement of, FrameworkElement from)
{
// Might throw an exception if of and from are not in the same visual tree
GeneralTransform transform = of.TransformToVisual(from);
return transform.TransformBounds(new Rect(0, 0, of.ActualWidth, of.ActualHeight));
}
Reference: http://social.msdn.microsoft.com/Forums/en-US/wpf/thread/86350f19-6457-470e-bde9-66e8970f7059/
If I understand your question right:
given:
shape has corner (x1,y1), center (xc,yc)
rotated shape has corner (x1',y1') after being rotated about center
desired:
how to map any point of the shape (x,y) -> (x',y') by that same rotation
Here's the relevant equations:
(x'-xc) = Kc*(x-xc) - Ks*(y-yc)
(y'-yc) = Ks*(x-xc) + Kc*(y-yc)
where Kc=cos(theta) and Ks=sin(theta) and theta is the angle of counterclockwise rotation. (to verify: if theta=0 this leaves the coordinates unchanged, otherwise if xc=yc=0, it maps (1,0) to (cos(theta),sin(theta)) and (0,1) to (-sin(theta), cos(theta)) . Caveat: this is for coordinate systems where (x,y)=(1,1) is in the upper right quadrant. For yours where it's in the lower right quadrant, theta would be the angle of clockwise rotation rather than counterclockwise rotation.)
If you know the coordinates of your rectangle aligned with the x-y axes, xc would just be the average of the two x-coordinates and yc would just be the average of the two y-coordinates. (in your situation, it's xc=75,yc=85.)
If you know theta, you now have enough information to calculate the new coordinates.
If you don't know theta, you can solve for Kc, Ks. Here's the relevant calculations for your example:
(62-75) = Kc*(50-75) - Ks*(50-85)
(40-85) = Ks*(50-75) + Kc*(50-85)
-13 = -25*Kc + 35*Ks = -25*Kc + 35*Ks
-45 = -25*Ks - 35*Kc = -35*Kc - 25*Ks
which is a system of linear equations that can be solved (exercise for the reader: in MATLAB it's:
[-25 35;-35 -25]\[-13;-45]
to yield, in this case, Kc=1.027, Ks=0.3622 which does NOT make sense (K2 = Kc2 + Ks2 is supposed to equal 1 for a pure rotation; in this case it's K = 1.089) so it's not a pure rotation about the rectangle center, which is what your drawing indicates. Nor does it seem to be a pure rotation about the origin. To check, compare distances from the center of rotation before and after the rotation using the Pythagorean theorem, d2 = deltax2 + deltay2. (for rotation about xc=75,yc=85, distance before is 43.01, distance after is 46.84, the ratio is K=1.089; for rotation about the origin, distance before is 70.71, distance after is 73.78, ratio is 1.043. I could believe ratios of 1.01 or less would arise from coordinate rounding to integers, but this is clearly larger than a roundoff error)
So there's some missing information here. How did you get the numbers (62,40)?
That's the basic gist of the math behind rotations, however.
edit: aha, I didn't realize they were estimates. (pretty close to being realistic, though!)
I use this method:
Point newPoint = rotateTransform.Transform(new Point(oldX, oldY));
where rotateTransform is the instance on which I work and set Angle...etc.
Look at GeneralTransform.TransformBounds() method.
I'm not sure, but is this what you're looking for - rotation of a point in Cartesian coordinate system:
link
You can use Transform.Transform() method on your Point with the same transformations to get a new point to which these transformations were applied.

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