I'm looking for some help in a IT school project. We need to create a programm which can detect roads in a satelite photograph. Our group decided to use a function for detect edges. We search differents solutions and filters on Internet and we decides to use Sobel filter.
We have tried to implement this filter in Scala but it didn't work. We use differents webpages to help us, some of these are on StackOverflow (here). We use this one to help us and try to translate the code : Sobel filter in Ruby.
Start Code --
codeGrey(); // This function transform the RGB in grey level
var sobel_x: Array[Array[Double]] = Array(
Array(-1, 0, 1),
Array(-2, 0, 2),
Array(-1, 0, 1))
var sobel_y: Array[Array[Double]] = Array(
Array(1, 2, 1),
Array(0, 0, 0),
Array(-1, -2, 1))
for (x <- 1 to wrappedImage.height - 2) {
for (y <- 1 to wrappedImage.width - 2) {
var a = (image2D(x - 1)(y - 1) & 0x00FF0000) >> 16
var b = (image2D(x)(y - 1) & 0x00FF0000) >> 16
var c = (image2D(x + 1)(y - 1) & 0x00FF0000) >> 16
var d = (image2D(x - 1)(y) & 0x00FF0000) >> 16
var e = (image2D(x)(y) & 0x00FF0000) >> 16
var f = (image2D(x + 1)(y) & 0x00FF0000) >> 16
var g = (image2D(x - 1)(y + 1) & 0x00FF0000) >> 16
var h = (image2D(x)(y + 1) & 0x00FF0000) >> 16
var i = (image2D(x + 1)(y + 1) & 0x00FF0000) >> 16
var pixel_x =
(sobel_x(0)(0) * a) + (sobel_x(0)(1) * b) + (sobel_x(0)(2) * c) +
(sobel_x(1)(0) * d) + (sobel_x(1)(1) * e) + (sobel_x(1)(2) * f) +
(sobel_x(2)(0) * g) + (sobel_x(2)(1) * h) + (sobel_x(2)(2) * i);
var pixel_y =
(sobel_y(0)(0) * a) + (sobel_x(0)(1) * b) + (sobel_x(0)(2) * c) +
(sobel_y(1)(0) * d) + (sobel_x(1)(1) * e) + (sobel_x(1)(2) * f) +
(sobel_y(2)(0) * g) + (sobel_x(2)(1) * h) + (sobel_x(2)(2) * i);
var res = (Math.sqrt((pixel_x * pixel_x) + (pixel_y * pixel_y)).ceil).toInt
image2D(x)(y) = 0xFF000000 + (res * 65536 + res * 256 + res);
}
}
End Code --
The image returned by this implementation is just an image with black and white pixels and I don't know why. I've got no experience in image processing and we learned Scala 8 weeks ago so that doesn't help.
I'm sorry, my english is not perfect so please forgive me if I didn't write correctly.
I'm not sure I grasp all the details of your solution, anyway here some observation:
consider using vals instead of vars: Scala prefers
immutables and you are not really changing any of those variables.
In scala you can write nested for cycles as a single one over two
variables (check here for details:
Nested iteration in Scala). I think it makes code cleaner.
I presume image2D is the array of arrays in which you are
holding your image. In the last line of your nested for loop you are
changing the current pixel value. This is not good because you will
access that same pixel later when you calculate your a,b,..,h,i
values. The center pixel during current iteration is the side pixel
during next iteration. I think you should write the result in a
different matrix.
Related
I am trying to numerically solve the Klein-Gordon equation that can be found here. To make sure I solved it correctly, I am comparing it with an analytical solution that can be found on the same link. I am using the finite difference method and Matlab. The initial spatial conditions are known, not the initial time conditions.
I start off by initializing the constants and the space-time coordinate system:
close all
clear
clc
%% Constant parameters
A = 2;
B = 3;
lambda = 2;
mu = 3;
a = 4;
b = - (lambda^2 / a^2) + mu^2;
%% Coordinate system
number_of_discrete_time_steps = 300;
t = linspace(0, 2, number_of_discrete_time_steps);
dt = t(2) - t(1);
number_of_discrete_space_steps = 100;
x = transpose( linspace(0, 1, number_of_discrete_space_steps) );
dx = x(2) - x(1);
Next, I define and plot the analitical solution:
%% Analitical solution
Wa = cos(lambda * x) * ( A * cos(mu * t) + B * sin(mu * t) );
figure('Name', 'Analitical solution');
surface(t, x, Wa, 'edgecolor', 'none');
colormap(jet(256));
colorbar;
xlabel('t');
ylabel('x');
title('Wa(x, t) - analitical solution');
The plot of the analytical solution is shown here.
In the end, I define the initial spatial conditions, execute the finite difference method algorithm and plot the solution:
%% Numerical solution
Wn = zeros(number_of_discrete_space_steps, number_of_discrete_time_steps);
Wn(1, :) = Wa(1, :);
Wn(2, :) = Wa(2, :);
for j = 2 : (number_of_discrete_time_steps - 1)
for i = 2 : (number_of_discrete_space_steps - 1)
Wn(i + 1, j) = dx^2 / a^2 ...
* ( ( Wn(i, j + 1) - 2 * Wn(i, j) + Wn(i, j - 1) ) / dt^2 + b * Wn(i - 1, j - 1) ) ...
+ 2 * Wn(i, j) - Wn(i - 1, j);
end
end
figure('Name', 'Numerical solution');
surface(t, x, Wn, 'edgecolor', 'none');
colormap(jet(256));
colorbar;
xlabel('t');
ylabel('x');
title('Wn(x, t) - numerical solution');
The plot of the numerical solution is shown here.
The two plotted graphs are not the same, which is proof that I did something wrong in the algorithm. The problem is, I can't find the errors. Please help me find them.
To summarize, please help me change the code so that the two plotted graphs become approximately the same. Thank you for your time.
The finite difference discretization of w_tt = a^2 * w_xx - b*w is
( w(i,j+1) - 2*w(i,j) + w(i,j-1) ) / dt^2
= a^2 * ( w(i+1,j) - 2*w(i,j) + w(i-1,j) ) / dx^2 - b*w(i,j)
In your order this gives the recursion equation
w(i,j+1) = dt^2 * ( (a/dx)^2 * ( w(i+1,j) - 2*w(i,j) + w(i-1,j) ) - b*w(i,j) )
+2*w(i,j) - w(i,j-1)
The stability condition is that at least a*dt/dx < 1. For the present parameters this is not satisfied, they give this ratio as 2.6. Increasing the time discretization to 1000 points is sufficient.
Next up is the boundary conditions. Besides the two leading columns for times 0 and dt one also needs to set the values at the boundaries for x=0 and x=1. Copy also them from the exact solution.
Wn(:,1:2) = Wa(:,1:2);
Wn(1,:)=Wa(1,:);
Wn(end,:)=Wa(end,:);
Then also correct the definition (and use) of b to that in the source
b = - (lambda^2 * a^2) + mu^2;
and the resulting numerical image looks identical to the analytical image in the color plot. The difference plot confirms the closeness
First of all, sorry for the bad title. I'm not really sure how to title this topic, so feel free to mod it where necessary.
I am drawing X rings inside my stage with given dimensions. To give this some sense of depth, each ring towards the screen boundaries is slightly wider:
The largest ring should be as wide as the largest dimension of the stage (note that in the picture i am drawing 3 extra rings which are drawn outside the stage boundaries). Also it should be twice as wide as the smallest ring. With ring i am refering to the space between 2 red circles.
After calculating an _innerRadius, which is the width of the smallest ring, i am drawing them using
const RINGS:Number = 10; //the amount of rings, note we will draw 3 extra rings to fill the screen
const DEPTH:Number = 7; //the amount of size difference between rings to create depth effect
var radius:Number = 0;
for (var i:uint = 0; i < RINGS + 3; i++) {
radius += _innerRadius + _innerRadius * ((i*DEPTH) / (RINGS - 1));
_graphics.lineStyle(1, 0xFF0000, 1);
_graphics.drawCircle(0, 0, radius * .5);
}
One of the sliders at the bottom goes from 0-100 being a percentage of the radius for the green ring which goes from the smallest to the largest ring.
I tried lerping between the smallest radius and the largest radius, which works fine if the DEPTH value is 1. However I don't want the distance between the rings to be the same for the sake of the illusion of depth.
Now I've been trying to figure this out for hours but it seems I've run into a wall.. it seems like I need some kind of non-linear formula here.. How would I calculate the radius based on the slider percentage value? Effectively for anywhere in between or on the red circles going from the smallest to the largest red circle?
thanks!
[edit]
Here's my example calculation for _innerRadius
//lets calculate _innerRadius for 10 rings
//inner ring width = X + 0/9 * X;
//ring 1 width = X + 1/9 * X;
//ring 2 width = X + 2/9 * X
//ring 3 width = X + 3/9 * X
//ring 4 width = X + 4/9 * X
//ring 5 width = X + 5/9 * X
//ring 6 width = X + 6/9 * X
//ring 8 width = X + 7/9 * X
//ring 9 width = X + 8/9 * X
//ring 10 width = X + 9/9 * X
//extent = Math.max(stage.stageWidth, stage.stageHeight);
//now we should solve extent = X + (X + 0/9 * X) + (X + 1/9 * X) + (X + 2/9 * X) + (X + 3/9 * X) + (X + 4/9 * X) + (X + 5/9 * X) + (X + 6/9 * X) + (X + 7/9 * X) + (X + 8/9 * X) + (X + 9/9 * X);
//lets add all X's
//extent = 10 * X + 45/9 * X
//extent = 15 * X;
//now reverse to solve for _innerRadius
//_innerRadius = extent / 15;
The way your drawing algorithm works, your radii are:
r[i + 1] = r[i] + (1 + i*a)*r0
where a is a constant that is depth / (rings - 1). This results in:
r0
r1 = r0 + (1 + a)*r0 = (2 + a)*r0
r2 = r1 + (1 + 2*a)*r0 = (3 + 3*a)*r0
r3 = r2 + (1 + 3*a)*r0 = (4 + 6*a)*r0
...
rn = (1 + n + a * sum(1 ... n))*r0
= (1 + n + a * n*(n - 1) / 2)*r0
Because you want ring n - 1 to correspond with your outer radius (let's forget about the three extra rings for now), you get:
r[n - 1] = (n + (n - 1)*(n - 2) / 2)*r0
a = 2 * (extent / r0 - n) / (n - 1) / (n - 2)
Then you can draw the rings:
for (var i = 0; i < rings; i++) {
r = (1 + i + 0.5 * a * (i - 1)*i) * r0;
// draw circle with radius r
}
You must have at least three rings in order not to have a division by zero when calculating a. Also note that this does not yield good results for all combinations: if the ratio of outer and inner circles is smaller than the number of rings, you get a negative a and have the depth effect reversed.
Another, maybe simpler, approach to create the depth effect is to make each circle's radius a constant multiple of the previous:
r[i + 1] = r[i] * c
or
r[i] = r0 * Math.pow(c, i)
and draw them like this:
c = Math.pow(extent / r0, 1 / (rings - 1))
r = r0
for (var i = 0; i < rings; i++) {
// draw circle with radius r
r *= c;
}
This will create a "positive" depth effect as long as the ratio of radii is positive. (And as long as there is more than one ring, of course.)
correlation = zeros(length(s1), 1);
sizeNum = 0;
for i = 1 : length(s1) - windowSize - delta
s1Dat = s1(i : i + windowSize);
s2Dat = s2(i + delta : i + delta + windowSize);
if length(find(isnan(s1Dat))) == 0 && length(find(isnan(s2Dat))) == 0
if(var(s1Dat) ~= 0 || var(s2Dat) ~= 0)
sizeNum = sizeNum + 1;
correlation(i) = abs(corr(s1Dat, s2Dat)) ^ 2;
end
end
end
What's happening here:
Run through every values in s1. For every value, get a slice for s1
till s1 + windowSize.
Do the same for s2, only get the slice after an intermediate delta.
If there are no NaN's in any of the two slices and they aren't flat,
then get the correlaton between them and add that to the
correlation matrix.
This is not an answer, I am trying to understand what is being asked.
Take some data:
N = 1e4;
s1 = cumsum(randn(N, 1)); s2 = cumsum(randn(N, 1));
s1(randi(N, 50, 1)) = NaN; s2(randi(N, 50, 1)) = NaN;
windowSize = 200; delta = 100;
Compute correlations:
tic
corr_s = zeros(N - windowSize - delta, 1);
for i = 1:(N - windowSize - delta)
s1Dat = s1(i:(i + windowSize));
s2Dat = s2((i + delta):(i + delta + windowSize));
corr_s(i) = corr(s1Dat, s2Dat);
end
inds = isnan(corr_s);
corr_s(inds) = 0;
corr_s = corr_s .^ 2; % square of correlation coefficient??? Why?
sizeNum = sum(~inds);
toc
This is what you want to do, right? A moving window correlation function? This is a very interesting question indeed …
I am building my first large-scale MATLAB program, and I've managed to write original vectorized code for everything so for until I came to trying to create an image representing vector density in stereographic projection. After a couple failed attempts I went to the Mathworks file exchange site and found an open source program which fits my needs courtesy of Malcolm Mclean. With a test matrix his function produces something like this:
And while this is almost exactly what I wanted, his code relies on a triply nested for-loop. On my workstation a test data matrix of size 25000x2 took 65 seconds in this section of code. This is unacceptable since I will be scaling up to a data matrices of size 500000x2 in my project.
So far I've been able to vectorize the innermost loop (which was the longest/worst loop), but I would like to continue and be rid of the loops entirely if possible. Here is Malcolm's original code that I need to vectorize:
dmap = zeros(height, width); % height, width: scalar with default value = 32
for ii = 0: height - 1 % 32 iterations of this loop
yi = limits(3) + ii * deltay + deltay/2; % limits(3) & deltay: scalars
for jj = 0 : width - 1 % 32 iterations of this loop
xi = limits(1) + jj * deltax + deltax/2; % limits(1) & deltax: scalars
dd = 0;
for kk = 1: length(x) % up to 500,000 iterations in this loop
dist2 = (x(kk) - xi)^2 + (y(kk) - yi)^2;
dd = dd + 1 / ( dist2 + fudge); % fudge is a scalar
end
dmap(ii+1,jj+1) = dd;
end
end
And here it is with the changes I've already made to the innermost loop (which was the biggest drain on efficiency). This cuts the time from 65 seconds down to 12 seconds on my machine for the same test matrix, which is better but still far slower than I would like.
dmap = zeros(height, width);
for ii = 0: height - 1
yi = limits(3) + ii * deltay + deltay/2;
for jj = 0 : width - 1
xi = limits(1) + jj * deltax + deltax/2;
dist2 = (x - xi) .^ 2 + (y - yi) .^ 2;
dmap(ii + 1, jj + 1) = sum(1 ./ (dist2 + fudge));
end
end
So my main question, are there any further changes I can make to optimize this code? Or even an alternative method to approach the problem? I've considered using C++ or F# instead of MATLAB for this section of the program, and I may do so if I cannot get to a reasonable efficiency level with the MATLAB code.
Please also note that at this point I don't have ANY additional toolboxes, if I did then I know this would be trivial (using hist3 from the statistics toolbox for example).
Mem consuming solution
yi = limits(3) + deltay * ( 1:height ) - .5 * deltay;
xi = limits(1) + deltax * ( 1:width ) - .5 * deltax;
dx = bsxfun( #minus, x(:), xi ) .^ 2;
dy = bsxfun( #minus, y(:), yi ) .^ 2;
dist2 = bsxfun( #plus, permute( dy, [2 3 1] ), permute( dx, [3 2 1] ) );
dmap = sum( 1./(dist2 + fudge ) , 3 );
EDIT
handling extremely large x and y by breaking the operation into blocks:
blockSize = 50000; % process up to XX elements at once
dmap = 0;
yi = limits(3) + deltay * ( 1:height ) - .5 * deltay;
xi = limits(1) + deltax * ( 1:width ) - .5 * deltax;
bi = 1;
while bi <= numel(x)
% take a block of x and y
bx = x( bi:min(end, bi + blockSize - 1) );
by = y( bi:min(end, bi + blockSize - 1) );
dx = bsxfun( #minus, bx(:), xi ) .^ 2;
dy = bsxfun( #minus, by(:), yi ) .^ 2;
dist2 = bsxfun( #plus, permute( dy, [2 3 1] ), permute( dx, [3 2 1] ) );
dmap = dmap + sum( 1./(dist2 + fudge ) , 3 );
bi = bi + blockSize;
end
This is a good example of why starting a loop from 1 matters. The only reason that ii and jj are initiated at 0 is to kill the ii * deltay and jj * deltax terms which however introduces sequentiality in the dmap indexing, preventing parallelization.
Now, by rewriting the loops you could use parfor() after opening a matlabpool:
dmap = zeros(height, width);
yi = limits(3) + deltay*(1:height) - .5*deltay;
matlabpool 8
parfor ii = 1: height
for jj = 1: width
xi = limits(1) + (jj-1) * deltax + deltax/2;
dist2 = (x - xi) .^ 2 + (y - yi(ii)) .^ 2;
dmap(ii, jj) = sum(1 ./ (dist2 + fudge));
end
end
matlabpool close
Keep in mind that opening and closing the pool has significant overhead (10 seconds on my Intel Core Duo T9300, vista 32 Matlab 2013a).
PS. I am not sure whether the inner loop instead of the outer one can be meaningfully parallelized. You can try to switch the parfor to the inner one and compare speeds (I would recommend going for the big matrix immediately since you are already running in 12 seconds and the overhead is almost as big).
Alternatively, this problem can be solved in using kernel density estimation techniques. This is part of the Statistics Toolbox, or there's this KDE implementation by Zdravko Botev (no toolboxes required).
For the example code below, I get 0.3 seconds for N = 500000, or 0.7 seconds for N = 1000000.
N = 500000;
data = [randn(N,2); rand(N,1)+3.5, randn(N,1);]; % 2 overlaid distrib
tic; [bandwidth,density,X,Y] = kde2d(data); toc;
imagesc(density);
I want to find the coordinate of an unknown node which lie somewhere in the space which has its reference distance away from 3 or more nodes which all of them have known coordinate.
This problem is exactly like Trilateration as described here Trilateration.
However, I don't understand the part about "Preliminary and final computations" (refer to the wikipedia site). I don't get where I could find P1, P2 and P3 just so I can put to those equation?
Thanks
Trilateration is the process of finding the center of the area of intersection of three spheres. The center point and radius of each of the three spheres must be known.
Let's consider your three example centerpoints P1 [-1,1], P2 [1,1], and P3 [-1,-1]. The first requirement is that P1' be at the origin, so let us adjust the points accordingly by adding an offset vector V [1,-1] to all three:
P1' = P1 + V = [0, 0]
P2' = P2 + V = [2, 0]
P3' = P3 + V = [0,-2]
Note: Adjusted points are denoted by the ' (prime) annotation.
P2' must also lie on the x-axis. In this case it already does, so no adjustment is necessary.
We will assume the radius of each sphere to be 2.
Now we have 3 equations (given) and 3 unknowns (X, Y, Z of center-of-intersection point).
Solve for P4'x:
x = (r1^2 - r2^2 + d^2) / 2d //(d,0) are coords of P2'
x = (2^2 - 2^2 + 2^2) / 2*2
x = 1
Solve for P4'y:
y = (r1^2 - r3^2 + i^2 + j^2) / 2j - (i/j)x //(i,j) are coords of P3'
y = (2^2 - 2^2 + 0 + -2^2) / 2*-2 - 0
y = -1
Ignore z for 2D problems.
P4' = [1,-1]
Now we translate back to original coordinate space by subtracting the offset vector V:
P4 = P4' - V = [0,0]
The solution point, P4, lies at the origin as expected.
The second half of the article is describing a method of representing a set of points where P1 is not at the origin or P2 is not on the x-axis such that they fit those constraints. I prefer to think of it instead as a translation, but both methods will result in the same solution.
Edit: Rotating P2' to the x-axis
If P2' does not lie on the x-axis after translating P1 to the origin, we must perform a rotation on the view.
First, let's create some new vectors to use as an example:
P1 = [2,3]
P2 = [3,4]
P3 = [5,2]
Remember, we must first translate P1 to the origin. As always, the offset vector, V, is -P1. In this case, V = [-2,-3]
P1' = P1 + V = [2,3] + [-2,-3] = [0, 0]
P2' = P2 + V = [3,4] + [-2,-3] = [1, 1]
P3' = P3 + V = [5,2] + [-2,-3] = [3,-1]
To determine the angle of rotation, we must find the angle between P2' and [1,0] (the x-axis).
We can use the dot product equality:
A dot B = ||A|| ||B|| cos(theta)
When B is [1,0], this can be simplified: A dot B is always just the X component of A, and ||B|| (the magnitude of B) is always a multiplication by 1, and can therefore be ignored.
We now have Ax = ||A|| cos(theta), which we can rearrange to our final equation:
theta = acos(Ax / ||A||)
or in our case:
theta = acos(P2'x / ||P2'||)
We calculate the magnitude of P2' using ||A|| = sqrt(Ax + Ay + Az)
||P2'|| = sqrt(1 + 1 + 0) = sqrt(2)
Plugging that in we can solve for theta
theta = acos(1 / sqrt(2)) = 45 degrees
Now let's use the rotation matrix to rotate the scene by -45 degrees.
Since P2'y is positive, and the rotation matrix rotates counter-clockwise, we'll use a negative rotation to align P2 to the x-axis (if P2'y is negative, don't negate theta).
R(theta) = [cos(theta) -sin(theta)]
[sin(theta) cos(theta)]
R(-45) = [cos(-45) -sin(-45)]
[sin(-45) cos(-45)]
We'll use double prime notation, '', to denote vectors which have been both translated and rotated.
P1'' = [0,0] (no need to calculate this one)
P2'' = [1 cos(-45) - 1 sin(-45)] = [sqrt(2)] = [1.414]
[1 sin(-45) + 1 cos(-45)] = [0] = [0]
P3'' = [3 cos(-45) - (-1) sin(-45)] = [sqrt(2)] = [ 1.414]
[3 sin(-45) + (-1) cos(-45)] = [-2*sqrt(2)] = [-2.828]
Now you can use P1'', P2'', and P3'' to solve for P4''. Apply the reverse rotation to P4'' to get P4', then the reverse translation to get P4, your center point.
To undo the rotation, multiply P4'' by R(-theta), in this case R(45). To undo the translation, subtract the offset vector V, which is the same as adding P1 (assuming you used -P1 as your V originally).
This is the algorithm I use in a 3D printer firmware. It avoids rotating the coordinate system, but it may not be the best.
There are 2 solutions to the trilateration problem. To get the second one, replace "- sqrtf" by "+ sqrtf" in the quadratic equation solution.
Obviously you can use doubles instead of floats if you have enough processor power and memory.
// Primary parameters
float anchorA[3], anchorB[3], anchorC[3]; // XYZ coordinates of the anchors
// Derived parameters
float Da2, Db2, Dc2;
float Xab, Xbc, Xca;
float Yab, Ybc, Yca;
float Zab, Zbc, Zca;
float P, Q, R, P2, U, A;
...
inline float fsquare(float f) { return f * f; }
...
// Precompute the derived parameters - they don't change unless the anchor positions change.
Da2 = fsquare(anchorA[0]) + fsquare(anchorA[1]) + fsquare(anchorA[2]);
Db2 = fsquare(anchorB[0]) + fsquare(anchorB[1]) + fsquare(anchorB[2]);
Dc2 = fsquare(anchorC[0]) + fsquare(anchorC[1]) + fsquare(anchorC[2]);
Xab = anchorA[0] - anchorB[0];
Xbc = anchorB[0] - anchorC[0];
Xca = anchorC[0] - anchorA[0];
Yab = anchorA[1] - anchorB[1];
Ybc = anchorB[1] - anchorC[1];
Yca = anchorC[1] - anchorA[1];
Zab = anchorB[2] - anchorC[2];
Zbc = anchorB[2] - anchorC[2];
Zca = anchorC[2] - anchorA[2];
P = ( anchorB[0] * Yca
- anchorA[0] * anchorC[1]
+ anchorA[1] * anchorC[0]
- anchorB[1] * Xca
) * 2;
P2 = fsquare(P);
Q = ( anchorB[1] * Zca
- anchorA[1] * anchorC[2]
+ anchorA[2] * anchorC[1]
- anchorB[2] * Yca
) * 2;
R = - ( anchorB[0] * Zca
+ anchorA[0] * anchorC[2]
+ anchorA[2] * anchorC[0]
- anchorB[2] * Xca
) * 2;
U = (anchorA[2] * P2) + (anchorA[0] * Q * P) + (anchorA[1] * R * P);
A = (P2 + fsquare(Q) + fsquare(R)) * 2;
...
// Calculate Cartesian coordinates given the distances to the anchors (La, Lb and Lc)
// First calculate PQRST such that x = (Qz + S)/P, y = (Rz + T)/P.
// P, Q and R depend only on the anchor positions, so they are pre-computed
const float S = - Yab * (fsquare(Lc) - Dc2)
- Yca * (fsquare(Lb) - Db2)
- Ybc * (fsquare(La) - Da2);
const float T = - Xab * (fsquare(Lc) - Dc2)
+ Xca * (fsquare(Lb) - Db2)
+ Xbc * (fsquare(La) - Da2);
// Calculate quadratic equation coefficients
const float halfB = (S * Q) - (R * T) - U;
const float C = fsquare(S) + fsquare(T) + (anchorA[1] * T - anchorA[0] * S) * P * 2 + (Da2 - fsquare(La)) * P2;
// Solve the quadratic equation for z
float z = (- halfB - sqrtf(fsquare(halfB) - A * C))/A;
// Substitute back for X and Y
float x = (Q * z + S)/P;
float y = (R * z + T)/P;
Here are the Wikipedia calculations, presented in an OpenSCAD script, which I think helps to understand the problem in a visual wayand provides an easy way to check that the results are correct. Example output from the script
// Trilateration example
// from Wikipedia
//
// pA, pB and pC are the centres of the spheres
// If necessary the spheres must be translated
// and rotated so that:
// -- all z values are 0
// -- pA is at the origin
pA = [0,0,0];
// -- pB is on the x axis
pB = [10,0,0];
pC = [9,7,0];
// rA , rB and rC are the radii of the spheres
rA = 9;
rB = 5;
rC = 7;
if ( pA != [0,0,0]){
echo ("ERROR: pA must be at the origin");
assert(false);
}
if ( (pB[2] !=0 ) || pC[2] !=0){
echo("ERROR: all sphere centers must be in z = 0 plane");
assert(false);
}
if (pB[1] != 0){
echo("pB centre must be on the x axis");
assert(false);
}
// show the spheres
module spheres(){
translate (pA){
sphere(r= rA, $fn = rA * 10);
}
translate(pB){
sphere(r = rB, $fn = rB * 10);
}
translate(pC){
sphere (r = rC, $fn = rC * 10);
}
}
function unit_vector( v) = v / norm(v);
ex = unit_vector(pB - pA) ;
echo(ex = ex);
i = ex * ( pC - pA);
echo (i = i);
ey = unit_vector(pC - pA - i * ex);
echo (ey = ey);
d = norm(pB - pA);
echo (d = d);
j = ey * ( pC - pA);
echo (j = j);
x = (pow(rA,2) - pow(rB,2) + pow(d,2)) / (2 * d);
echo( x = x);
// size of the cube to subtract to show
// the intersection of the spheres
cube_size = [10,10,10];
if ( ((d - rA) >= rB) || ( rB >= ( d + rA)) ){
echo ("Error Y not solvable");
}else{
y = (( pow(rA,2) - pow(rC,2) + pow(i,2) + pow(j,2)) / (2 * j))
- ( i / j) * x;
echo(y = y);
zpow2 = pow(rA,2) - pow(x,2) - pow(y,2);
if ( zpow2 < 0){
echo ("z not solvable");
}else{
z = sqrt(zpow2);
echo (z = z);
// subtract a cube with one of its corners
// at the point where the sphers intersect
difference(){
spheres();
translate ([x,y - cube_size[1],z]){
cube(cube_size);
}
}
translate ([x,y - cube_size[1],z]){
%cube(cube_size);
}
}
}