Objective C Float value is converted internally - no idea why - iphone

Am writing a game using cocos 2d which involves lot of Float type properties to determine the vertices(x,y) of my shapes (Square , Triangle). I have a Shape class which simply holds the total number of vertices(int) and the vertices (CGFloat*). I have one more class called ShapeHolder which defines the values for different types of shapes , it has the properties called Shape square; and Shape Triangle. In the ShapeHolder init am storing(hard coding) the vertices information(x,y) as follows , am also printing the values immediately after the definition which works perfectly
CGPoint _square[] = {
CGPointMake(100.0f, 100.0f) ,
CGPointMake(200.0f, 100.0f) ,
CGPointMake(200.0f, 200.0f) ,
CGPointMake(100.0f, 200.0f)
};.
square = [[[Shape alloc] initWithVerticesLength: 4] retain];
square.vertices = _square;
for (int i = 0; i< square.count; i++) {
NSLog(#"Init Square Vertices X Y = %f , %f" , square.vertices[i].x ,square.vertices[i].y );
}
triangle = [[[Shape alloc] initWithVerticesLength: 3] retain];
triangle.vertices = _triangle;
The problem am having is on while retrieving the same values in my OpenGL drawing class to render different shapes , here is the code
allShapes = [[ShapeHolder alloc] init];
for (int i = 0; i< allShapes.square.count; i++) {
NSLog(#"Main Square Vertices X Y = %f , %f" , [[allShapes square]vertices][i].x ,[[allShapes square] vertices][i].y );
}
The value is converted somehow , no idea why please clarify. Am not getting any runtime errors or compile time error which is really bothering my knowledge
Init Square Vertices X Y = 100.000000 , 100.000000
Init Square Vertices X Y = 200.000000 , 100.000000
Init Square Vertices X Y = 200.000000 , 200.000000
Init Square Vertices X Y = 100.000000 , 200.000000
Main Square Vertices X Y = 100.000000 , 100.000000
Main Square Vertices X Y = -1.998910 , 0.000000
Main Square Vertices X Y = 0.000000 , 0.000000
Main Square Vertices X Y = -1.998912 , 0.000000

Looks like you are doing a shallow copy of the array pointer for _square. You need to be doing a deep copy of the vertices into a member array in your object.
The reason your first vertex isn't being changed is that your class is keeping the pointer to the first point in the array. Subsequent vertices in the array are being reclaimed by the memory manager and reused. Possibly luck too.

Related

loop over array matrix matlab

I am detecting circles in an image. I return circle radii and X,Y of the axis. I know how to crop 1 circle no problem with formula:
X-radius, Y-radius, width=2*r,height=2*r using imcrop.
My problem is when I get returned more than 1 circle.
I get returned circle radii in an array radiiarray.
I get returned circle centers in centarray.
When i disp(centarray), It looks like this:
146.4930 144.4943
610.0317 142.1734
When I check size(centarray) and disp it i get:
2 2
So I understand first column is X and second is Y axis values. So first circle center would be 146,144.
I made a loop that works for only 1 circle. "-------" is where I'm unsure what to use to get:
note: radius = r
1st circle)
X = centarray(1)-r;
Y = centarray(3)-r;
Width =2*r;
Width =2*r;
2nd circle)
X = centarray(2);
Y = centarray(4);
Width =2*r;
Width =2*r;
How would I modify the "------" parts for my code? I also would like that if there are 3+ circles the loop would work as Im getting sometimes up to 9 circles from an image.
B = imread('p5.tif');
centarray = [];
centarray = [centarray,centers];
radiiarray = [];
radiiarray = [radiiarray,radii];
for j=1:length(radiiarray)
x = centarray((------))-radiiarray(j); %X value to crop
y = centarray((------))-radiiarray(j); %Y value to crop
width = 2*radiiarray(j); %WIDTH
height = 2*radiiarray(j); %HEIGHT
K = imcrop(B, [x y width height]);
end
My full code, which doesnt work, as I realized why when i saw the way values are stored...:
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% DETECT + GET X Y WIDTH HEIGHT OF CIRCLES
I = imread('p5.tif');
subplot(2,2,1);imshow(I);title('Original Image');
%sharpen edges
B = imsharpen(I);
subplot(2,2,2);imshow(B);title('sharpened edges');
%find circles
Img = im2bw(B(:,:,3));
minRad = 20;
maxRad = 90;
[centers, radii] = imfindcircles(Img, [minRad maxRad], ...
'ObjectPolarity','bright','sensitivity',0.84);
imagesc(Img);
viscircles(centers, radii,'Color','green');
%nuber of circles found
%arrays to store values for radii and centers
centarray = [];
centarray = [centarray,centers];
radiiarray = [];
radiiarray = [radiiarray,radii];
sc = size(centarray);
disp(sc)
disp(centarray)
disp(radiiarray)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%CROP USING VALUE FROM ARRAYS NUMBER OF TIMES THERE ARE CENTERS(number of
%circles)
for j=1:length(radiiarray)
x = centarray((2*j)-1)-radiiarray(j); %X value to crop
y = centarray((2*j))-radiiarray(j); %Y value to crop
width = 2*radiiarray(j); %WIDTH
height = 2*radiiarray(j); %HEIGHT
disp(x)
disp(y)
disp(centarray)
%crop using values
K = imcrop(B, [x y width height]);
%togray
gray = rgb2gray(K);
subplot(2,2,3);imshow(K);title('cropped before bw');
Icorrected = imtophat(gray, strel('disk', 15));
%to black and white
black = im2bw(Icorrected);
subplot(2,2,4);imshow(black);title('sharpened edges');
%read
results = ocr(black);
number = results.Text;
%display value
disp(number)
end
Any help on how to create this kind of loop is appreciated as I just have no more ideas or cant find answer to this..
EDIT
SOLUTION
Hi, answer is to treat matrix as 2 dimensional.
for j=1:length(radiiarray)
x=centarray(j,1)
y=centarray(j,2)
width = radiiarray(j)
height = radiiarray(j)
end
as j increases values update correctly now.
answer is to treat matrix as 2 dimensional.
for j=1:length(radiiarray)
x=centarray(j,1)
y=centarray(j,2)
width = radiiarray(j)
height = radiiarray(j)
end
as j increases values update correctly now.
Thanks for #beaker for his comment! Thats why I figured it out

EaselJS shape x,y properties confusion

I generate a 4x4 grid of squares with below code. They all draw in correct position, rows and columns, on canvas on stage.update(). But the x,y coordinates for all sixteen of them on inspection are 0,0. Why? Does each shape has it's own x,y coordinate system? If so, if I get a handle to a shape, how do I determine where it was drawn originally onto the canvas?
The EaselJS documentation is silent on the topic ;-). Maybe you had to know Flash.
var stage = new createjs.Stage("demoCanvas");
for (i = 0; i < 4; i++) {
for (j = 0; j < 4; j++) {
var square = new createjs.Shape();
square.graphics.drawRect(i*100, j*100, 100, 100);
console.log("Created square + square.x + "," + square.y);
stage.addChild(square);
}
}
You are drawing the graphics at the coordinates you want, instead of drawing them at 0,0, and moving them using x/y coordinates. If you don't set the x/y yourself, it will be 0. EaselJS does not infer the x/y or width/height based on the graphics content (more info).
Here is an updated fiddle where the graphics are all drawn at [0,0], and then positioned using x/y instead: http://jsfiddle.net/0o63ty96/
Relevant code:
square.graphics.beginStroke("red").drawRect(0,0,100,100);
square.x = i * 100;
square.y = j * 100;

Projection of circular region of interest onto rectangle [duplicate]

BOUNTY STATUS UPDATE:
I discovered how to map a linear lens, from destination coordinates to source coordinates.
How do you calculate the radial distance from the centre to go from fisheye to rectilinear?
1). I actually struggle to reverse it, and to map source coordinates to destination coordinates. What is the inverse, in code in the style of the converting functions I posted?
2). I also see that my undistortion is imperfect on some lenses - presumably those that are not strictly linear. What is the equivalent to-and-from source-and-destination coordinates for those lenses? Again, more code than just mathematical formulae please...
Question as originally stated:
I have some points that describe positions in a picture taken with a fisheye lens.
I want to convert these points to rectilinear coordinates. I want to undistort the image.
I've found this description of how to generate a fisheye effect, but not how to reverse it.
There's also a blog post that describes how to use tools to do it; these pictures are from that:
(1) : SOURCE Original photo link
Input : Original image with fish-eye distortion to fix.
(2) : DESTINATION Original photo link
Output : Corrected image (technically also with perspective correction, but that's a separate step).
How do you calculate the radial distance from the centre to go from fisheye to rectilinear?
My function stub looks like this:
Point correct_fisheye(const Point& p,const Size& img) {
// to polar
const Point centre = {img.width/2,img.height/2};
const Point rel = {p.x-centre.x,p.y-centre.y};
const double theta = atan2(rel.y,rel.x);
double R = sqrt((rel.x*rel.x)+(rel.y*rel.y));
// fisheye undistortion in here please
//... change R ...
// back to rectangular
const Point ret = Point(centre.x+R*cos(theta),centre.y+R*sin(theta));
fprintf(stderr,"(%d,%d) in (%d,%d) = %f,%f = (%d,%d)\n",p.x,p.y,img.width,img.height,theta,R,ret.x,ret.y);
return ret;
}
Alternatively, I could somehow convert the image from fisheye to rectilinear before finding the points, but I'm completely befuddled by the OpenCV documentation. Is there a straightforward way to do it in OpenCV, and does it perform well enough to do it to a live video feed?
The description you mention states that the projection by a pin-hole camera (one that does not introduce lens distortion) is modeled by
R_u = f*tan(theta)
and the projection by common fisheye lens cameras (that is, distorted) is modeled by
R_d = 2*f*sin(theta/2)
You already know R_d and theta and if you knew the camera's focal length (represented by f) then correcting the image would amount to computing R_u in terms of R_d and theta. In other words,
R_u = f*tan(2*asin(R_d/(2*f)))
is the formula you're looking for. Estimating the focal length f can be solved by calibrating the camera or other means such as letting the user provide feedback on how well the image is corrected or using knowledge from the original scene.
In order to solve the same problem using OpenCV, you would have to obtain the camera's intrinsic parameters and lens distortion coefficients. See, for example, Chapter 11 of Learning OpenCV (don't forget to check the correction). Then you can use a program such as this one (written with the Python bindings for OpenCV) in order to reverse lens distortion:
#!/usr/bin/python
# ./undistort 0_0000.jpg 1367.451167 1367.451167 0 0 -0.246065 0.193617 -0.002004 -0.002056
import sys
import cv
def main(argv):
if len(argv) < 10:
print 'Usage: %s input-file fx fy cx cy k1 k2 p1 p2 output-file' % argv[0]
sys.exit(-1)
src = argv[1]
fx, fy, cx, cy, k1, k2, p1, p2, output = argv[2:]
intrinsics = cv.CreateMat(3, 3, cv.CV_64FC1)
cv.Zero(intrinsics)
intrinsics[0, 0] = float(fx)
intrinsics[1, 1] = float(fy)
intrinsics[2, 2] = 1.0
intrinsics[0, 2] = float(cx)
intrinsics[1, 2] = float(cy)
dist_coeffs = cv.CreateMat(1, 4, cv.CV_64FC1)
cv.Zero(dist_coeffs)
dist_coeffs[0, 0] = float(k1)
dist_coeffs[0, 1] = float(k2)
dist_coeffs[0, 2] = float(p1)
dist_coeffs[0, 3] = float(p2)
src = cv.LoadImage(src)
dst = cv.CreateImage(cv.GetSize(src), src.depth, src.nChannels)
mapx = cv.CreateImage(cv.GetSize(src), cv.IPL_DEPTH_32F, 1)
mapy = cv.CreateImage(cv.GetSize(src), cv.IPL_DEPTH_32F, 1)
cv.InitUndistortMap(intrinsics, dist_coeffs, mapx, mapy)
cv.Remap(src, dst, mapx, mapy, cv.CV_INTER_LINEAR + cv.CV_WARP_FILL_OUTLIERS, cv.ScalarAll(0))
# cv.Undistort2(src, dst, intrinsics, dist_coeffs)
cv.SaveImage(output, dst)
if __name__ == '__main__':
main(sys.argv)
Also note that OpenCV uses a very different lens distortion model to the one in the web page you linked to.
(Original poster, providing an alternative)
The following function maps destination (rectilinear) coordinates to source (fisheye-distorted) coordinates. (I'd appreciate help in reversing it)
I got to this point through trial-and-error: I don't fundamentally grasp why this code is working, explanations and improved accuracy appreciated!
def dist(x,y):
return sqrt(x*x+y*y)
def correct_fisheye(src_size,dest_size,dx,dy,factor):
""" returns a tuple of source coordinates (sx,sy)
(note: values can be out of range)"""
# convert dx,dy to relative coordinates
rx, ry = dx-(dest_size[0]/2), dy-(dest_size[1]/2)
# calc theta
r = dist(rx,ry)/(dist(src_size[0],src_size[1])/factor)
if 0==r:
theta = 1.0
else:
theta = atan(r)/r
# back to absolute coordinates
sx, sy = (src_size[0]/2)+theta*rx, (src_size[1]/2)+theta*ry
# done
return (int(round(sx)),int(round(sy)))
When used with a factor of 3.0, it successfully undistorts the images used as examples (I made no attempt at quality interpolation):
Dead link
(And this is from the blog post, for comparison:)
If you think your formulas are exact, you can comput an exact formula with trig, like so:
Rin = 2 f sin(w/2) -> sin(w/2)= Rin/2f
Rout= f tan(w) -> tan(w)= Rout/f
(Rin/2f)^2 = [sin(w/2)]^2 = (1 - cos(w))/2 -> cos(w) = 1 - 2(Rin/2f)^2
(Rout/f)^2 = [tan(w)]^2 = 1/[cos(w)]^2 - 1
-> (Rout/f)^2 = 1/(1-2[Rin/2f]^2)^2 - 1
However, as #jmbr says, the actual camera distortion will depend on the lens and the zoom. Rather than rely on a fixed formula, you might want to try a polynomial expansion:
Rout = Rin*(1 + A*Rin^2 + B*Rin^4 + ...)
By tweaking first A, then higher-order coefficients, you can compute any reasonable local function (the form of the expansion takes advantage of the symmetry of the problem). In particular, it should be possible to compute initial coefficients to approximate the theoretical function above.
Also, for good results, you will need to use an interpolation filter to generate your corrected image. As long as the distortion is not too great, you can use the kind of filter you would use to rescale the image linearly without much problem.
Edit: as per your request, the equivalent scaling factor for the above formula:
(Rout/f)^2 = 1/(1-2[Rin/2f]^2)^2 - 1
-> Rout/f = [Rin/f] * sqrt(1-[Rin/f]^2/4)/(1-[Rin/f]^2/2)
If you plot the above formula alongside tan(Rin/f), you can see that they are very similar in shape. Basically, distortion from the tangent becomes severe before sin(w) becomes much different from w.
The inverse formula should be something like:
Rin/f = [Rout/f] / sqrt( sqrt(([Rout/f]^2+1) * (sqrt([Rout/f]^2+1) + 1) / 2 )
I blindly implemented the formulas from here, so I cannot guarantee it would do what you need.
Use auto_zoom to get the value for the zoom parameter.
def dist(x,y):
return sqrt(x*x+y*y)
def fisheye_to_rectilinear(src_size,dest_size,sx,sy,crop_factor,zoom):
""" returns a tuple of dest coordinates (dx,dy)
(note: values can be out of range)
crop_factor is ratio of sphere diameter to diagonal of the source image"""
# convert sx,sy to relative coordinates
rx, ry = sx-(src_size[0]/2), sy-(src_size[1]/2)
r = dist(rx,ry)
# focal distance = radius of the sphere
pi = 3.1415926535
f = dist(src_size[0],src_size[1])*factor/pi
# calc theta 1) linear mapping (older Nikon)
theta = r / f
# calc theta 2) nonlinear mapping
# theta = asin ( r / ( 2 * f ) ) * 2
# calc new radius
nr = tan(theta) * zoom
# back to absolute coordinates
dx, dy = (dest_size[0]/2)+rx/r*nr, (dest_size[1]/2)+ry/r*nr
# done
return (int(round(dx)),int(round(dy)))
def fisheye_auto_zoom(src_size,dest_size,crop_factor):
""" calculate zoom such that left edge of source image matches left edge of dest image """
# Try to see what happens with zoom=1
dx, dy = fisheye_to_rectilinear(src_size, dest_size, 0, src_size[1]/2, crop_factor, 1)
# Calculate zoom so the result is what we wanted
obtained_r = dest_size[0]/2 - dx
required_r = dest_size[0]/2
zoom = required_r / obtained_r
return zoom
I took what JMBR did and basically reversed it. He took the radius of the distorted image (Rd, that is, the distance in pixels from the center of the image) and found a formula for Ru, the radius of the undistorted image.
You want to go the other way. For each pixel in the undistorted (processed image), you want to know what the corresponding pixel is in the distorted image.
In other words, given (xu, yu) --> (xd, yd). You then replace each pixel in the undistorted image with its corresponding pixel from the distorted image.
Starting where JMBR did, I do the reverse, finding Rd as a function of Ru. I get:
Rd = f * sqrt(2) * sqrt( 1 - 1/sqrt(r^2 +1))
where f is the focal length in pixels (I'll explain later), and r = Ru/f.
The focal length for my camera was 2.5 mm. The size of each pixel on my CCD was 6 um square. f was therefore 2500/6 = 417 pixels. This can be found by trial and error.
Finding Rd allows you to find the corresponding pixel in the distorted image using polar coordinates.
The angle of each pixel from the center point is the same:
theta = arctan( (yu-yc)/(xu-xc) ) where xc, yc are the center points.
Then,
xd = Rd * cos(theta) + xc
yd = Rd * sin(theta) + yc
Make sure you know which quadrant you are in.
Here is the C# code I used
public class Analyzer
{
private ArrayList mFisheyeCorrect;
private int mFELimit = 1500;
private double mScaleFESize = 0.9;
public Analyzer()
{
//A lookup table so we don't have to calculate Rdistorted over and over
//The values will be multiplied by focal length in pixels to
//get the Rdistorted
mFisheyeCorrect = new ArrayList(mFELimit);
//i corresponds to Rundist/focalLengthInPixels * 1000 (to get integers)
for (int i = 0; i < mFELimit; i++)
{
double result = Math.Sqrt(1 - 1 / Math.Sqrt(1.0 + (double)i * i / 1000000.0)) * 1.4142136;
mFisheyeCorrect.Add(result);
}
}
public Bitmap RemoveFisheye(ref Bitmap aImage, double aFocalLinPixels)
{
Bitmap correctedImage = new Bitmap(aImage.Width, aImage.Height);
//The center points of the image
double xc = aImage.Width / 2.0;
double yc = aImage.Height / 2.0;
Boolean xpos, ypos;
//Move through the pixels in the corrected image;
//set to corresponding pixels in distorted image
for (int i = 0; i < correctedImage.Width; i++)
{
for (int j = 0; j < correctedImage.Height; j++)
{
//which quadrant are we in?
xpos = i > xc;
ypos = j > yc;
//Find the distance from the center
double xdif = i-xc;
double ydif = j-yc;
//The distance squared
double Rusquare = xdif * xdif + ydif * ydif;
//the angle from the center
double theta = Math.Atan2(ydif, xdif);
//find index for lookup table
int index = (int)(Math.Sqrt(Rusquare) / aFocalLinPixels * 1000);
if (index >= mFELimit) index = mFELimit - 1;
//calculated Rdistorted
double Rd = aFocalLinPixels * (double)mFisheyeCorrect[index]
/mScaleFESize;
//calculate x and y distances
double xdelta = Math.Abs(Rd*Math.Cos(theta));
double ydelta = Math.Abs(Rd * Math.Sin(theta));
//convert to pixel coordinates
int xd = (int)(xc + (xpos ? xdelta : -xdelta));
int yd = (int)(yc + (ypos ? ydelta : -ydelta));
xd = Math.Max(0, Math.Min(xd, aImage.Width-1));
yd = Math.Max(0, Math.Min(yd, aImage.Height-1));
//set the corrected pixel value from the distorted image
correctedImage.SetPixel(i, j, aImage.GetPixel(xd, yd));
}
}
return correctedImage;
}
}
I found this pdf file and I have proved that the maths are correct (except for the line vd = *xd**fv+v0 which should say vd = **yd**+fv+v0).
http://perception.inrialpes.fr/CAVA_Dataset/Site/files/Calibration_OpenCV.pdf
It does not use all of the latest co-efficients that OpenCV has available but I am sure that it could be adapted fairly easily.
double k1 = cameraIntrinsic.distortion[0];
double k2 = cameraIntrinsic.distortion[1];
double p1 = cameraIntrinsic.distortion[2];
double p2 = cameraIntrinsic.distortion[3];
double k3 = cameraIntrinsic.distortion[4];
double fu = cameraIntrinsic.focalLength[0];
double fv = cameraIntrinsic.focalLength[1];
double u0 = cameraIntrinsic.principalPoint[0];
double v0 = cameraIntrinsic.principalPoint[1];
double u, v;
u = thisPoint->x; // the undistorted point
v = thisPoint->y;
double x = ( u - u0 )/fu;
double y = ( v - v0 )/fv;
double r2 = (x*x) + (y*y);
double r4 = r2*r2;
double cDist = 1 + (k1*r2) + (k2*r4);
double xr = x*cDist;
double yr = y*cDist;
double a1 = 2*x*y;
double a2 = r2 + (2*(x*x));
double a3 = r2 + (2*(y*y));
double dx = (a1*p1) + (a2*p2);
double dy = (a3*p1) + (a1*p2);
double xd = xr + dx;
double yd = yr + dy;
double ud = (xd*fu) + u0;
double vd = (yd*fv) + v0;
thisPoint->x = ud; // the distorted point
thisPoint->y = vd;
This can be solved as an optimization problem. Simply draw on curves in images that are supposed to be straight lines. Store the contour points for each of those curves. Now we can solve the fish eye matrix as a minimization problem. Minimize the curve in points and that will give us a fisheye matrix. It works.
It can be done manually by adjusting the fish eye matrix using trackbars! Here is a fish eye GUI code using OpenCV for manual calibration.

Matlab: given 4 points, fit the closest rhombus/square

I need a script that "checks" if 4 given points form a square or a rhombus.
I am working in a QR code segmentation script in which I try to locate the vertex by looking for the non-negative values of a binary image traversing it by rows and columns.
There are some cases in which checking is not neccessary, like in this image:
It is a bit hard to see, but the vertex are labelled as 4 points in green, magenta, cyan and yellow. In this case the script should return the same input points, since no modification is needed.
On the other hand, there are cases in which the vertex are labeled as so:
It can be seen that the magenta and cyan labels rely on the top right corner of the image. This is obviously not correct, but it fullfills the specified condition: traverse each row of the image until you find a row satisfying sum(row)>1 (greater than 1 to avoid single, noisy pixels).
How can I locate the misplaced vertex and place it using the remaining vertex coordinates?
EDIT
Solved the problem. I'm posting the code of the function in case someone needs it:
function correctedCorners = square(corners)
correctedCorners = corners;
X = corners(:,1);
Y = corners(:,2);
sortedX = sort(corners(:,1));
sortedY = sort(corners(:,2));
%% DISTANCES BW POINTS
for i=1:4
for j=1:4
distances(i,j) = sqrt((corners(i,1)-corners(j,1))^2+ (corners(i,2)-corners(j,2))^2);
end
end
%% relationship bw distances
% check corner 1
d11 = distances(1,1);%0
d12 = distances(1,2);%x
d13 = distances(1,3);%sqrt(2)*x
d14 = distances(1,4);%x
bool1 = [(d12*0.8<=d14)&(d12*1.2>=d14) (d12*0.8*sqrt(2)<=d13)& (d12*1.2*sqrt(2)>=d13) (d14*0.8<=d12)&(d14*1.2>=d12) (d14*0.8*sqrt(2)<=d13)&(d14*1.2*sqrt(2)>=d13)];
% check corner 2
d21 = distances(2,1);%x
d22 = distances(2,2);%0
d23 = distances(2,3);%x
d24 = distances(2,4);%sqrt(2)*x
bool2 = [(d21*0.8<=d23)&(d21*1.2>=d23) (d21*0.8*sqrt(2)<=d24)&(d21*1.2*sqrt(2)>=d24) (d23*0.8<=d21)&(d23*1.2>=d21) (d23*0.8*sqrt(2)<=d24)&(d23*1.2*sqrt(2)>=d24)];
% check corner 3
d31 = distances(3,1);%sqrt(2)*x
d32 = distances(3,2);%x
d33 = distances(3,3);%0
d34 = distances(3,4);%x
bool3 = [(d32*0.8<=d34)&(d32*1.2>=d34) (d32*0.8*sqrt(2)<=d31)&(d32*1.2*sqrt(2)>=d31) (d34*0.8<=d32)&(d34*1.2>=d32) (d34*0.8*sqrt(2)<=d31)&(d34*1.2*sqrt(2)>=d31)];
% check corner 4
d41 = distances(4,1);%x
d42 = distances(4,2);%sqrt(2)*x
d43 = distances(4,3);%x
d44 = distances(4,4);%0
bool4 = [(d41*0.8<=d43)&(d41*1.2>=d43) (d41*0.8*sqrt(2)<=d42)&(d41*1.2*sqrt(2)>=d42) (d43*0.8<=d41)&(d43*1.2>=d41) (d43*0.8*sqrt(2)<=d42)&(d43*1.2*sqrt(2)>=d42)];
bool = [bool1; bool2;bool3;bool4];
idx = 0;
for i=1:4
if (sum(bool(i,:))==0)
idx = [idx i];
end
end
if (length(idx)>=2)
for i=2:length(idx)
switch idx(i)
case 1
correctedCorners(1,:) = abs(corners(4,:)-(corners(3,:)-corners(2,:)));
case 2
correctedCorners(2,:) = abs(corners(3,:)-(corners(4,:)-corners(1,:)));
case 3
correctedCorners(3,:) = abs(corners(2,:)+(corners(1,:)-corners(1,:)));
case 4
correctedCorners(4,:) = abs(corners(1,:)+(corners(3,:)-corners(2,:)));
end
end
end
From basic geometry about squares:
TopLeft distance to BotLeft = x
TopLeft distance to TopRight= x
TopLeft distance to BotRight= sqrt(2)*x
Use the same logic for BotLeft to other points, etc.
Allow yourself something like 10-20% margin of error to declare an incorrect point. That is, if TopLeft distance to 2 points outside of the range (80%;120%)*x , and its distance to the third point is outside of the range (80%;120%)*sqrt(2)*x, you can declare the point as placed incorrectly.
In your case, the TopLeft point fails on all distance tests:
0 instead of x to TopRight (about 100% error)
sqrt(2)*x vs x to BotLeft (about 44% error)
x vs sqrt(2)*x to BotRight) (about 31% error)
As long as the rhombus is very similar to a square, a 20% margin of error while treating it as a square should still work.

create cylinders in 3D volumetric data

I'm trying to create a dataset of raw volumetric data consisting of geometrical shapes. The point is to use volume ray casting to project them in 2D but first I want to create the volume manually.
The geometry is consisting of one cylinder that is in the middle of the volume, along the Z axis and 2 smaller cylinders that are around the first one, deriving from rotations around the axes.
Here is my function so far:
function cyl= createCylinders(a, b, c, rad1, h1, rad2, h2)
% a : data width
% b : data height
% c : data depth
% rad1: radius of the big center cylinder
% rad2: radius of the smaller cylinders
% h1: height of the big center cylinder
% h2: height of the smaller cylinders
[Y X Z] =meshgrid(1:a,1:b,1:c); %matlab saves in a different order so X must be Y
centerX = a/2;
centerY = b/2;
centerZ = c/2;
theta = 0; %around y
fi = pi/4; %around x
% First cylinder
cyl = zeros(a,b,c);
% create for infinite height
R = sqrt((X-centerX).^2 + (Y-centerY).^2);
startZ = ceil(c/2) - floor(h1/2);
endZ = startZ + h1 - 1;
% then trim it to height = h1
temp = zeros(a,b,h1);
temp( R(:,:,startZ:endZ)<rad1 ) = 255;
cyl(:,:,startZ:endZ) = temp;
% Second cylinder
cyl2 = zeros(a,b,c);
A = (X-centerX)*cos(theta) + (Y-centerY)*sin(theta)*sin(fi) + (Z-centerZ)*cos(fi)*sin(theta);
B = (Y-centerY)*cos(fi) - (Z-centerZ)*sin(fi);
% create again for infinite height
R2 = sqrt(A.^2+B.^2);
cyl2(R2<rad2) = 255;
%then use 2 planes to trim outside of the limits
N = [ cos(fi)*sin(theta) -sin(fi) cos(fi)*cos(theta) ];
P = (rad2).*N + [ centerX centerY centerZ];
T = (X-P(1))*N(1) + (Y-P(2))*N(2) + (Z-P(3))*N(3);
cyl2(T<0) = 0;
P = (rad2+h2).*N + [ centerX centerY centerZ];
T = (X-P(1))*N(1) + (Y-P(2))*N(2) + (Z-P(3))*N(3);
cyl2(T>0) = 0;
% Third cylinder
% ...
cyl = cyl + cyl2;
cyl = uint8(round(cyl));
% ...
The concept is that the first cylinder is created and then "cut" according to the z-axis value, to define its height. The other cylinder is created using the relation A2 + B 2 = R2 where A and B are rotated accordingly using the rotation matrices only around x and y axes, using Ry(θ)Rx(φ) as described here.
Until now everything seems to be working, because I have implemented code (tested that it works well) to display the projection and the cylinders seem to have correct rotation when they are not "trimmed" from infinite height.
I calculate N which is the vector [0 0 1] aka z-axis rotated in the same way as the cylinder. Then I find two points P of the same distances that I want the cylinder's edges to be and calculate the plane equations T according to that points and normal vector. Lastly, I trim according to that equality. Or at least that's what I think I'm doing, because after the trimming I usually don't get anything (every value is zero). Or, the best thing I could get when I was experimenting was cylinders trimmed, but the planes of the top and bottom where not oriented well.
I would appreciate any help or corrections at my code, because I've been looking at the geometry equations and I can't find where the mistake is.
Edit:
This is a quick screenshot of the object I'm trying to create. NOTE that the cylinders are opaque in the volume data, all the inside is considered as homogeneous material.
I think instead of:
T = (X-P(1))*N(1) + (Y-P(2))*N(2) + (Z-P(3))*N(3);
you should try the following at both places:
T = (X-P(1)) + (Y-P(2)) + (Z-P(3));
Multiplying by N is to account for the direction of the axis of the 2nd cylinder which you have already done just above that step.