Problem :
I rotate the landscape image anti clockwise 1 times at 90degree. After rebuilding the image, it does not look it has been rotated 1 times in anti clockwise position. It provides clockwise 90 degrees image.
I used below code :
var encoder = await BitmapEncoder.CreateAsync(BitmapEncoder.PngEncoderId, fileStream);
encoder.SetPixelData(
BitmapPixelFormat.Bgra8,
BitmapAlphaMode.Ignore,
(uint)renderTargetBitmap.PixelWidth,
(uint)renderTargetBitmap.PixelHeight,
DisplayInformation.GetForCurrentView().LogicalDpi,
DisplayInformation.GetForCurrentView().LogicalDpi,
pixelBuffer.ToArray());
encoder.BitmapTransform.Rotation = BitmapRotation.Clockwise90Degrees;
await encoder.FlushAsync();
There is no AntiClockwise90Degrees in BitmapRotation
How to solve this problem?
BitmapRotation.Clockwise90Degrees will rotate 90° clockwise.
BitmapRotation.Clockwise270Degrees will rotate 90° anticlockwise (360° - 90° = 270°)
Related
The text positioning (y) is determined by the TextPlusYExtractionStrategy perfectly when the rotation angle is 0 (portrait). In case of rotation angle is 90 degree (landscape) the position y provided from left to right instead of top to bottom. How to consider the rotation to get the text position y in when the page is 90 degree.
The below code helped me to resolve my issue.
if(rotation == 90){
chunkY = chunk.getLocation().getStartLocation().get(Vector.I1);
}
Constructor of TextPlusYExtractionStrategy can take the rotation angle, and use the Vector.I1 which is x co-ordinate of the chunk.
The issue will solve only the landscape (rotation=90) issue to find the Y coordinate of the search text.
I have the following animation:
I want the circle containing the ETH / USD animation to rotate around the centerpoint of all these circles, but the text itself I want to stay perfectly horizontal. How do I do this?
If it helps, I'm using the svg.js library (https://github.com/svgdotjs/svg.js)
Rotate the circle counterclockwise around its own center whiile it is rotating clockwise around the main center.
Let group be a group containing the small circle with radius r and the text. Let them be centered on (0,0). Let (cx, cy) be the central rotation point. Loop endlessly with even pace in a full circle taking 15s:
var group = draw.group();
group.circle(r, 0, 0);
group.text('ETH / USD');
group.animate(15000, '-').rotate(-360).loop()
.animate(15000, '-').rotate(360, cx, cy).loop();
I've got a binary image with an object and a rotated rectangle over it, found with cv2.findContours and cv2.minAreaRect. The image is normalized to [0;1]
What is the most efficient way to count non-zero area within the bounding rectangle?
Create new zero values Mat that has the same size of your original image.
Draw your rotated rectangle on it in (fillConvexPoly using the RotatedRect vertices).
Bitwise_and this image with your original mask
apply findnonzero function on the result image
You may also apply the previous steps on ROI of the image since you have the bounding box of your rotated rectangle.
According to Humam Helfawi's answer I've tuned a bit suggested steps, so the following code seems doing what i need:
rectangles = [(cv2.minAreaRect(cnt)) for cnt in contours]
for rect in rectangles:
rect = cv2.boxPoints(rect)
rect = np.int0(rect)
coords = cv2.boundingRect(rect)
rect[:,0] = rect[:,0] - coords[0]
rect[:,1] = rect[:,1] - coords[1]
area = cv2.contourArea(rect)
zeros = np.zeros((coords[3], coords[2]), np.uint8)
cv2.fillConvexPoly(zeros, rect, 255)
im = greyscale[coords[1]:coords[1]+coords[3],
coords[0]:coords[0]+coords[2]]
print(np.sum(cv2.bitwise_and(zeros,im))/255)
contours is a list of points. You can fill this shape on an empty binary image with the same size using cv2.fillConvexPoly and then use cv2.countNonZero or numpy.count_nonzero to get the number of occupied pixels.
Here some examples of twisted triangle prisms.
I want to know if a moving triangle will hit a certain point. That's why I need to solve this problem.
The idea is that a triangle with random coordinates becomes the other random triangle whose vertices all move between then
related: How to determine point/time of intersection for ray hitting a moving triangle?
One of my students made this little animation in Mathematica.
It shows the twisting of a prism to the Schönhardt polyhedron.
See the Wikipedia page for its significance.
It would be easy to determine if a particular point is inside the polyhedron.
But whether it is inside a particular smooth twisting, as in your image, depends on the details (the rate) of the twisting.
Let's bottom triangle lies in plane z=0, it has rotation angle 0, top triangle has rotation angle Fi. Height of twisted prism is Hgt.
Rotation angle linearly depends on height, so layer at height h has rotation angle
a(h) = Fi * h / Hgt
If point coordinates are (x,y,z), then shift point to z=0 and rotate (x,y) coordinates about rotation axis (rx, ry) by -a(z) angle
t = -a(z) = - Fi * z / Hgt
xn = rx + (x-rx) * Cos(t) - (y-ry) * Sin(t)
yn = ry + (x-rx) * Sin(t) - (y-ry) * Cos(t)
Then check whether (xn, yn) lies inside bottom triangle
I have a question regarding the projection of an image over a set of 3D points. The image is given to me as a JPG, together with position and attitude information of the camera relative to a cartesian coordinate system (Xc,Yc,Zc and yaw, pitch, roll), as well as the horizontal and vertical field of view (in degrees).
Points are given using solely their 3d position in the same coordinate system (Xp,Yp,Zp).
In my coordinate system, Z is up. To project the image onto the points, I
compute the vector from camera to each point
Vector3 c2p = (Xp,Yp,Zp)-(Xc,Yc,Zc);
rotate c2p according to my camera's attitude (quaternion):
Vector3 c2pCamFrame = getCamQuaternion().conjugate().rotate(c2p);
compute azimuth and elevation from the camera's "center ray" to the point:
float azimuth = atan2(c2pCamFrame.x(),c2pCamFrame.y()));
float elevation = atan2(c2pCamFrame.z(),sqrt(pow(c2pCamFrame.x(),2)+pow(c2pCamFrame.y(),2)));
if azimuth and elevation are within the field of view, I assign the color of the corresponding pixel to the point.
This works almost perfectly, and the "almost" motivates my question. Let me show you:
I cannot figure out why the elevation of the projection is distorted. In the bottom right of the image, you can see that points outside the frustum (exceeding the elevation) actually become colored - and this distortion is null at an azimuth of 0 degrees and peaks at the left and right edges of the image, creating the pillow distortion.
Why does this distortion appear? I'd love to understand this problem both in geometrical as well as mathematical terms. Thank you!
The field of view angles are only valid on the principal axes. But you can do it the other way around. I.e. calculate the x/y bounds from the angles:
maxX = tan(horizontal_fov / 2)
maxY = tan(vertical_fov / 2)
And check
if(abs(c2pCamFrame.x() / c2pCamFrame.z()) <= maxX
&& abs(c2pCamFrame.y() / c2pCamFrame.z()) <= maxY)
Additionally, you might want to check if the points are in front of the camera:
... && c2pCamFrame.z() > 0
This assumes a left-handed coordinate system.