I try to do a thing that should be nothing more than a two-dimensional, linear interpolation but currently I fail finding the correct approach. To describe the problem a bit simplified: there is a drawing area with a size of 3000x3000 pixels where I have to draw e.g. a horizontal line. To do that I'm drawing dots or short lines from every pixel position to the next pixel position which then forms a line.
Now a correction has to be applied to the whole thing where correction information can be found in a (for this example simplified) 4 by 4 array, where every element contains a pair of coordinates describing the values after correction. So a neutral array (with no correction) would look like this:
0,0 1000,0 2000,0 3000,0
0,1000 1000,1000 2000,1000 3000,1000
0,2000 1000,2000 2000,2000 3000,2000
0,3000 1000,3000 2000,3000 3000,3000
A real correction table would contain other coordinates describing the correction to be done:
So as input data I have the coordinates of points on the line without correction, the fields values without correction and the correction data. But how can I calculate the lines points now applying the correction values to it so that a distorted line is drawn like shown in right side if the image? My current approach with two separate linear interpolations for X and Y does not work, there the Y-position jumps on a cells border but does not change smoothly within a cell.
So...any ideas how this could be done?
You have to agree on an interpolation method first. I would suggest either bilinear or barycentric interpolation. In one of my previous posts I visualized the difference between both methods.
I'll concentrate on the bilinear interpolation. We want to transform any point within a cell to its corrected point. Therefore, all points could be transformed separately.
We need the interpolation parameters u and v for the point (x, y). Because we have an axis-aligned grid, this is pretty simple:
u = (x - leftCellEdge) / (rightCellEdge - leftCellEdge)
v = (y - bottomCellEdge) / (topCellEdge - bottomCellEdge)
We could reconstruct the point by bilinear interpolation:
p2 p4
x----x
| o |
x----x
p1 p3
o = (1 - u) * ((1 - v) * p1 + v * p2) + u * ((1 - v) * p3 + v * p4)
Now, the same formula can be used for the corrected points. If you use the original points p1 through p4, you'll get the uncorrected line point. If you use the corrected cell points for p1 through p4, you'll get the corrected line point.
Related
I am trying to make a graph of the brightness of a pixel vs the distance from center of that pixel. To do so I used for loops to check each pixel for these values. But when adding them to my array I find that I can't. One of the issues is I have to define the array size first so no values get placed in the right spot. I believe everything else to be working except adding values to the arrays.
I've tried various methods of concatenation to add the values of each pixel to the array. I didn't have any more solutions to try.
folder3 = 'C:\Users\slenka\Desktop\Image_Analysis\Subtracted';
cd('C:\Users\slenka\Desktop\Image_Analysis\Subtracted');
subtractedFiles = [dir(fullfile(folder3,'*.TIF')); dir(fullfile(folder3,'*.PNG')); dir(fullfile(folder3,'*.BMP')); dir(fullfile(folder3,'*.jpg'))];
numberOfSubImages= length(subtractedFiles);
for b = 1 : numberOfSubImages
subFileName=fullfile(folder3, subtractedFiles(b).name);
chartImage=imread(subFileName);
[chartY, chartX, chartNumberOfColorChannels] = size(chartImage);
ccY= chartY/2;
ccX= chartX/2;
c=[ccX,ccY];
distanceArray=zeros(1,chartX);
intensityArray=zeros(1,chartY);
f=1;
g=1;
for y=1:chartY
for x=1:chartX
D = sqrt((y - c(1)) .^ 2 + (x - c(2)) .^ 2);
grayScale= impixel(chartImage, x, y);
distanceArray(f)=[D];
intensityArray(g)=[grayScale];
f=f+1;
g=g+1;
end
end
xAxis=distanceArray;
yAxis=intensityArray;
plot(xAxis,yAxis);
end
I'm expecting 2 arrays one full of the data values for the light intensity of each pixel in the image, and another for that pixels distance from the center of the image. I am wanting to plot these two arrays as the y and x axis respectively. At the moment the actual results is an entirely empty array full of zeros.
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.
I have two lines, one straight and one curvy. Both have an arbitrary number of x and y values defining the lines - the number of x and y values are not the same for either line. I am attempting to get separate distances of points between the curved line coordinates and the straight line coordinates. You can think of discrete integration to get a better picture of what I'm talking about, something along the lines of this: http://www.scientific-solutions.ch/tech/origin/products/images/calculus_integral.gif
By adding the different distances, I would get the area. The part on which I am stuck is the actual synchronization of the points. I can simply compare the x and y values of the straight and curve coordinates every ten indices for example because the curved coordinates are time dependent (as in the points do not change at a general rate). I need a way to synchronize the actual coordinates of the two sets of points. I thought about interpolating both sets of points to a specific number of points, but again, the time dependence of the curved set of points makes that solution void.
Could someone please suggest a good way of doing this, outlining the basics? I really appreciate the help.
Code to be tried (pseudo):
xLine = [value1 value2 ...]
yLine = [value1 value2 ...]
xCurve = [value1 value2 ...]
yCurve = [value1 value2 ...]
xLineInterpolate = %interpolate of every 10 points of x until a certain value. same with yLineInterpolate, xCurveInterpolate and yCurveInterpolate.
Then, I could just take the same index from each array and do some algebra to get the distance. My worry is that my line values increase at a constant rate whereas my curve values sometimes do not change (x and y values have different rates of change) and sometimes do. Would such an interpolation method be wrong then?
If I understand correctly, you want to know the distance between a straight line and a curve. The easiest way is to perform a coordinate transformation such that the straight line is the new x-axis. In that frame, the y-values of the curved line are the distances you seek.
This coordinate transformation is equal to a rotation and a translation, as in the following:
% estimate coefficients for straight line
sol = [x1 ones(size(x1))] \ y1;
m = sol(1); %# slope
b = sol(2); %# y-offset at x=0
% shift curved line down by b
% (makes the straight line go through the origin)
y2 = y2 - b;
% rotate the curved line by -atan(m)
% (makes the straight line run parallel to the x-axis)
a = -atan(m);
R = [+cos(a) -sin(a)
+sin(a) +cos(a)];
XY = R*[x2; y2];
% the distances are then the values of y3.
x3 = XY(1,:);
y3 = XY(2,:);
You need to use interpolation. I don't see how the time-dependence is relevant here - perhaps you are thinking of fitting a straight line to both curves? That's a bad idea.
You can do a simple interpolation for any curve just by assuming that every two adjacent points are connected by a straight line. This can be shown to be a reasonable approximation for the curve.
So, let's say you are looking at (x1,y1) and (x2,y2) which are adjacent to each other and now you choose an x3 that is between x1 and x2 (x1 < x2 < x3), and want to find the y3 value.
A simple way to find y3 is the following:
p=(x3-x1)/(x2-x1)
y3=y1+p*(y2-y1)
The idea is that p shows the relative position between x1 and x2 (0.5 would be the middle, for example), and then you use p as the relative position between y1 and y2.
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 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