Following Quote from this source:
http://www.cambridgeincolour.com/tutorials/image-projections.htm
Equirectangular image projections map the latitude and longitude
coordinates of a spherical globe directly onto horizontal and vertical
coordinates of a grid, where this grid is roughly twice as wide as it
is tall.
I have a 13312 px width and 6656 pixel height Panorama picture. It's a equirectangular projection of a room and have a 2:1 ratio.
I use following formular to calculate the xPosition:
var xPosition = ( panorama.width / 360 ) * azimuth
Azimuth = Phi = Heading = Angle to the left or right
How do I project this now on a 1366x768px browser screen?
I think my results are wrong, because it's not on the point where it should be.. it could be because the sphere has a distortion on the left and right:
Is there any formular to calculate the position with attention to the distortion and scale it to fit on the browser screen? I looked many (MANY) sources to find a solution for this, but they always just say that equirectangular are just lat and long.. they don't consider the distortion.
Last question: To find a special solution, I tryed to put a plane on the circle and expanded the line which shows the alpha angle. I though with Phytagoras I could find the position.. but this didn't worked either.. maybe I did something wrong? Is this the way even possible or am I doing it wrong?
edit
THIS is what I'm actually looking for: http://othree.github.io/360-panorama/three-2d/
The black grid in the background. What is the name of this? For what do I have to google or look for? When you start the 2D Panorama, if you want to get the coordinations of the top right corner of the window, what do you have to do?