2D case Just like the dot product is proportional to the cosine of the angle, the determinant is proportional to its sine. So you can compute the angle like this: dot = x1*x2 + y1*y2 # Dot product between [x1, y1] and [x2, y2] det = x1*y2 – y1*x2 # Determinant angle = atan2(det, dot) … Read more
I know I will never get a 100% precise value of these numbers, but AT LEAST I was expecting to get the same “unprecise” value of sin and cos of complementary angles. Why? The are calculated differently, so different floating point errors will occur (and accumulate). What you see is not an error; FP arithmetic … Read more
A context strokeStyle can be a gradient: // create a gradient gradient = ctx.createLinearGradient(xStart, yStart, xEnd, yEnd); gradient.addColorStop(0.0,”blue”); gradient.addColorStop(1.0,”purple”); // stroke using that gradient ctx.strokeStyle = gradient; Example code and a Demo using a gradient strokeStyle: http://jsfiddle.net/m1erickson/w46ps/ <!doctype html> <html> <head> <link rel=”stylesheet” type=”text/css” media=”all” href=”https://stackoverflow.com/questions/22223950/css/reset.css” /> <!– reset css –> <script type=”text/javascript” src=”http://code.jquery.com/jquery.min.js”></script> <style> … Read more
You want the arctangent: dy = ey – cy dx = ex – cx theta = arctan(dy/dx) theta *= 180/pi // rads to degs Erm, note that the above is obviously not compiling Javascript code. You’ll have to look through documentation for the arctangent function. Edit: Using Math.atan2(y,x) will handle all of the special cases … Read more
The Camera.Parameters getHorizontalViewAngle() and getVerticalViewAngle() functions provide you with the base view angles. I say “base”, because these apply only to the Camera itself in an unzoomed state, and the values returned by these functions do not change even when the view angle itself does. Camera.Parameters p = camera.getParameters(); double thetaV = Math.toRadians(p.getVerticalViewAngle()); double thetaH … Read more
What you want to use is often called the “perp dot product”, that is, find the vector perpendicular to one of the vectors, and then find the dot product with the other vector. if(a.x*b.y – a.y*b.x < 0) angle = -angle; You can also do this: angle = atan2( a.x*b.y – a.y*b.x, a.x*b.x + a.y*b.y … Read more
You can either rotate your bitmap when you draw it by using a matrix: Matrix matrix = new Matrix(); matrix.setRotate(angle, imageCenterX, imageCenterY); yourCanvas.drawBitmap(yourBitmap, matrix, null); You can also do it by rotating the canvas before drawing: yourCanvas.save(Canvas.MATRIX_SAVE_FLAG); //Saving the canvas and later restoring it so only this image will be rotated. yourCanvas.rotate(-angle); yourCanvas.drawBitmap(yourBitmap, left, top, … Read more
This gives a signed angle for any angles: a = targetA – sourceA a = (a + 180) % 360 – 180 Beware in many languages the modulo operation returns a value with the same sign as the dividend (like C, C++, C#, JavaScript, full list here). This requires a custom mod function like so: … Read more
You forgot to add the center point: result.Y = (int)Math.Round( centerPoint.Y + distance * Math.Sin( angle ) ); result.X = (int)Math.Round( centerPoint.X + distance * Math.Cos( angle ) ); The rest should be ok… (what strange results were you getting? Can you give an exact input?)
I’m going to assume you’re using C++, but the answer should be the same if you’re using C or Python. The function minAreaRect seems to give angles ranging from -90 to 0 degrees, not including zero, so an interval of [-90, 0). The function gives -90 degrees if the rectangle it outputs isn’t rotated, i.e. … Read more