I have a .bmp image with a map. What i know:
Height an Width of bmp image
dpi
Map Scale
Image Center's coordinates in meters.
What i want:
How can i calculate some points of image (for example corners) in meters.
Or how can i change a pixel distanse to meters?
What i do before:
For sure i know image center coordinates in pixels:
CenterXpix = Widht/2;
CenterYpix = Height/2;
But what i gonna do to find another corners coordinates. Don't think that:
metersDistance = pixelDistance*Scale;
is a correct equation.
Any advises?
If you know the height or width in both meters and pixels, you can calculate the scale in meters/pixel. You equation:
metersDistance = pixelDistance*Scale;
is correct, but only if your points are on the same axis. If your two points are diagonal from each other, you have to use good old pythagoras (in pseudocode):
X = XdistancePix*scale;
Y = YdistancePix*scale;
Distance_in_m = sqrt(X*X+Y*Y);
Related
I'm trying to resize and reposition a ROI (region of interest) correctly from a low resolution image (256x256) to a higher resolution image (512x512). It should also be mentioned that the two images cover different field of view - the low and high resolution image have 330mm x 330mm and 180mm x 180mm FoV, respectively.
What I've got at my disposal are:
Physical reference point (in mm) in the 256x256 and 512x512 image, which are refpoint_lowres=(-164.424,-194.462) and refpoint_highres=(-94.3052,-110.923). The reference points are located in the top left pixel (1,1) in their respective images.
Pixel coordinates of the ROI in the 256x256 image (named pxX and pxY). These coordinates are positioned relative to the reference point of the lower resolution image, refpoint_lowres=(-164.424,-194.462).
Pixel spacing for the 256x256 and 512x512 image, which are 0.7757 pixel/mm and 2.8444 pixel/mm respectively.
How can I rescale and reposition the ROI (the binary mask) to correct pixel location in the 512x512 image? Many thanks in advance!!
Attempt
% This gives correctly placed and scaled binary array in the 256x256 image
mask_lowres = double(poly2mask(pxX, pxY, 256., 256.));
% Compute translational shift in pixel
mmShift = refpoint_lowres - refpoint_highres;
pxShift = abs(mmShift./pixspacing_highres)
% This produces a binary array that is only positioned correctly in the
% 512x512 image, but it is not upscaled correctly...(?)
mask_highres = double(poly2mask(pxX + pxShift(1), pxY + pxShift(2), 512.,
512.));
So you have coordinates pxX, and pxY in pixels with respect to the low-resolution image. You can transform these coordinates to real-world coordinates:
pxX_rw = pxX / 0.7757 - 164.424;
pxY_rw = pxY / 0.7757 - 194.462;
Next you can transform these coordinates to high-res coordinates:
pxX_hr = (pxX_rw - 94.3052) * 2.8444;
pxY_hr = (pxY_rw - 110.923) * 2.8444;
Since the original coordinates fit in the low-res image, but the high-res image is smaller (in physical coordinates) than the low-res one, it is possible that these new coordinates do not fit in the high-res image. If this is the case, cropping the polygon is a non-trivial exercise, it cannot be done by simply moving the vertices to be inside the field of view. MATLAB R2017b introduces the polyshape object type, which you can intersect:
bbox = polyshape([0 0 180 180] - 94.3052, [180 0 0 180] - 110.923);
poly = polyshape(pxX_rw, pxY_rw);
poly = intersect([poly bbox]);
pxX_rw = poly.Vertices(:,1);
pxY_rw = poly.Vertices(:,2);
If you have an earlier version of MATLAB, maybe the easiest solution is to make the field of view larger to draw the polygon, then crop the resulting image to the right size. But this does require some proper calculation to get it right.
Having this coordinate system:
And this dominant vertical vanishing point:
I would like to rotate the image around x axis so the vanishing point is at infinity. That means that all vertical lines are parallel.
I am using matlab. I find the line segmentes using LSD and the vanishing point using homogeneous coordinates. I would like to use angle-axis representation, then convert it to a rotation matrix and pass this to imwarp and get the rotated image. Also would be good to know how to rotate the segments. The segments are as (x1,y1,x2,y2).
Image above example:
Vanishin point in homogenous coordinates:
(x,y,z) = 1.0e+05 * [0.4992 -2.2012 0.0026]
Vanishin point in cartesian coordinates (what you see in the image):
(x,y) = [190.1335 -838.3577]
Question: With this vanishing point how do I compute the rotation matrix in the world x axis as explained above?
If all you're doing is rotating the image so that the vector from the origin to the vanishing point, is instead pointing directly vertical, here's an example.
I = imread('cameraman.tif');
figure;imagesc(I);set(gcf,'colormap',gray);
vp=-[190.1335 -838.3577,0]; %3d version,just for cross-product use,-ve ?
y=[0,1,0]; %The vertical axis on the plot
u = cross(vp,y); %you know it's going to be the z-axis
theta = -acos(dot(vp/norm(vp),y)); %-ve ?
rotMat = vrrotvec2mat([u, theta]);
J=imwarp(I,affine2d (rotMat));
figure;imagesc(J);set(gcf,'colormap',gray); %tilted image
You can play with the negatives, and plotting, since I'm not sure about those parts applying to your situation. The negatives may come from plotting upside down, or from rotation of the world vs. camera coordinate system, but I don't have time to think about it right now.
EDIT
If you want to rotation about the X-axis, this might work (adapted from https://www.mathworks.com/matlabcentral/answers/113074-how-to-rotate-an-image-along-y-axis), or check out: Rotate image over X, Y and Z axis in Matlab
[rows, columns, numberOfColorChannels] = size(I);
newRows = rows * cos(theta);
rotatedImage = imresize(I, [newRows, columns]);
I have an image and the spatial reference object of that image.
Now i want to crop the image by coordinates according to the spatial reference object.
The function Imcrop can only crop according the pixel coordinates. Is there a way to crop based on the world coordinates?
I tried to use Imcrop and compute for the new reference object but I get lost in the coordinate transformation.
An example of the reference object after warping an Image.
imref2d with properties:
XWorldLimits: [-775.4357 555.5643]
YWorldLimits: [-488.3694 523.6306]
ImageSize: [1012 1331]
PixelExtentInWorldX: 1
PixelExtentInWorldY: 1
ImageExtentInWorldX: 1331
ImageExtentInWorldY: 1012
XIntrinsicLimits: [0.5000 1.3315e+03]
YIntrinsicLimits: [0.5000 1.0125e+03]
What I actually want to do is to crop the image such that the point (0,0) is the center of the cropped image.
According to you spatial reference each pixel has a dimensions of 1 x 1 in world coordinates. Therefore if you want to convert between world coordinate (Xw,Yw) and image coordinate (Xi,Yi) do the following:
Xi = round(abs(-775.4357 - Xw))
Yi = round(abs(-488.3694 - Yw))
So if you want to crop the image such that the real world coordinate (0,0) will be the center of the new cropped image and the size of the new image will be width on height than the rectangle for imcrop will be
[(755 - width) (488 - height) width height]
I have a list of coordinates, which are generated from another program, and I have an image.
I'd like to load those coordinates (making circular regions of interest (ROIs) with a diameter of 3 pixels) onto my image, and extract the intensity of those pixels.
I can load/impose the coordinates on to the image by using;
imshow(file);
hold on
scatter(xCoords, yCoords, 'g')
But can not extract the intensity.
Can you guys point me in the right direction?
I am not sure what you mean by a circle with 3 pixels diameter since you are in a square grid (as mentioned by Ander Biguri). But you could use fspecial to create a disk filter and then normalize. Something like this:
r = 1.5; % for diameter = 3
h = fspecial('disk', r);
h = h/h(ceil(r),ceil(r));
You can use it as a mask to get the intensities at the given region of the image.
im = imread(file);
ROI = im(xCoord-1:xCoord+1; yCoord-1:yCoord+1);
I = ROI.*h;
I have to display 5x5 degrees pie chart on a leaflet map. I can display pie chart using the great leaflet-dvf library, but I have to provide the radius in pixel, and it is static so far.
I would like to have it dynamic so that at any zoom level, the pie chart fills the 5x5 square (aka radius = 5x5 degree length).
How can I know the length in pixel of the side of a 5x5 degrees square, depending on the zoom level?
Thanks
This will work out metres per pixel
metresPerPixel = 40075016.686 * Math.abs(Math.cos(map.getCenter().lat / 180 * Math.PI)) / Math.pow(2, map.getZoom()+8);
I used the following page that provides the pixel per meter on a leaflet map depending on the zoom level:
http://wiki.openstreetmap.org/wiki/Zoom_levels
Then I computed the length of a 5x5 square at the equator: 556000 meters.
Then I store the length ratio for zoom level = 0:
$scope.lengthRatio = 556000 / 156412 // in meter / pixel
Finally, I get the radius of a pie chart depending on the zoom level ($scope.mapZoom):
var radius = $scope.lengthRatio * Math.pow(2,$scope.mapZoom)) / 2
The /2 is because I want the radius and not the diameter
Simple!