I've got my 2D/3D conversion working perfectly, how to do perspective
- by user346992
Although the context of this question is about making a 2d/3d game, the problem i have boils down to some math.
Although its a 2.5D world, lets pretend its just 2d for this question.
// xa: x-accent, the x coordinate of the projection
// mapP: a coordinate on a map which need to be projected
// _Dist_ values are constants for the projection, choosing them correctly will result in i.e. an isometric projection
xa = mapP.x * xDistX + mapP.y * xDistY;
ya = mapP.x * yDistX + mapP.y * yDistY;
xDistX and yDistX determine the angle of the x-axis, and xDistY and yDistY determine the angle of the y-axis on the projection (and also the size of the grid, but lets assume this is 1-pixel for simplicity).
x-axis-angle = atan(yDistX/xDistX)
y-axis-angle = atan(yDistY/yDistY)
a "normal" coordinate system like this
--------------- x
|
|
|
|
|
y
has values like this:
xDistX = 1;
yDistX = 0;
xDistY = 0;
YDistY = 1;
So every step in x direction will result on the projection to 1 pixel to the right end 0 pixels down. Every step in the y direction of the projection will result in 0 steps to the right and 1 pixel down.
When choosing the correct xDistX, yDistX, xDistY, yDistY, you can project any trimetric or dimetric system (which is why i chose this).
So far so good, when this is drawn everything turns out okay. If "my system" and mindset are clear, lets move on to perspective.
I wanted to add some perspective to this grid so i added some extra's like this:
camera = new MapPoint(60, 60);
dx = mapP.x - camera.x; // delta x
dy = mapP.y - camera.y; // delta y
dist = Math.sqrt(dx * dx + dy * dy); // dist is the distance to the camera, Pythagoras etc.. all objects must be in front of the camera
fac = 1 - dist / 100; // this formula determines the amount of perspective
xa = fac * (mapP.x * xDistX + mapP.y * xDistY) ;
ya = fac * (mapP.x * yDistX + mapP.y * yDistY );
Now the real hard part... what if you got a (xa,ya) point on the projection and want to calculate the original point (x,y).
For the first case (without perspective) i did find the inverse function, but how can this be done for the formula with the perspective. May math skills are not quite up to the challenge to solve this.
( I vaguely remember from a long time ago mathematica could create inverse function for some special cases... could it solve this problem? Could someone maybe try?)
