Apologies if my example is a bit contrived, I've been trying to figure out how to apply matrix transformations to this coordinate problem, and I'm not getting the right output. I need to understand the math for this (some pseudo code might be helpful). See graphic on bottom.
Assume a game world has four properties World => ( ORIENTATION = angle, MAP_SIZE = [mw,ml], CELL_SIZE = [cw, cl])
Now assume we have a World instance called World_A with the following properties:
World_A = ( -90, [4096,4096], [64, 64])
In other words, World_A has a 2-dimensional, 4096 x 4096 pixel map with an initial orientation of -90 degrees (inverse Cartesian plane), which can be divided by 64 x 64 named cells (denoted by their calculated subcoordinates [p,q] e.g. Cell_p_q), and an active cursor (denoted by its rendered position [x,y])
Using the coordinates of an active cursor [x,y] we can calculate which cell the cursor is over (some Cell_p_q) and designate that cell the Active_Cell which is then highlighted. Example: if the cursor is at [8,16] then we determine that it's over Cell_1_1 and Active_Cell = Cell_1_1.
Now-- let's add a special mirror world called World_B where World_B = Isometric(World_A)
Both worlds share the same Cursor with position [x,y].
How do I transform the World_A( Orientation, Map, Cells) to the isometric equivalent of World_B( Orientation, Map, Cells), such that when the Cursor is over either map, the right Active_Cell is highlighted?
Here's a quick screenshot of what I'm experimenting with; you can see in this graph that World_A is overlaying World_B coordinates. My math is brute-force contrived, but it works, what I want is to understand the matrix transformation way of doing this.
matrix.identity();,matrix.scale(sqrt(2) / 2, sqrt(2) / 4, 1);,matrix.rotate(0, 0, 1, -45); //degrees,matrix.inverse();in that order. Then you just multiply your position vector with that matrix(x, y).multiply(matrix);. The result of the final calculation should be your coordinates in iso space. \$\endgroup\$