Formula to evenly arrange lines on perspective imitation?

Question

In short: How to calculate h0...h4?

When imitating perspective, which formula is used to calculate h(0) to h(n) distances when known deformation values like either top-width-to-bottom-width ratio, or angle a (whichever parameter is useful for that)?

Answer 1

We can see perpective transformation having axial symmetry around vertical axis.

So transformation of rectangle with coordinates (0,0)-(SrcWdt, SrcHgt) with axial line at SrcWdt/2

gives trapezoid centered vertically with axial line at DstWdt/2 and coordinates of right corners RBX, RBY, RTX, RTY

In this case transformation formulas are a bit simpler (than general case)

X' = DstXCenter + A * (X - XCenter) / (H * Y + 1)
Y' = (RBY +  E * Y) / (H * Y + 1)

And we can calculate coefficients A, E, H without solving of eight linear equation system using only coordinates of two corners of trapezoid.

Delphi code:

 procedure CalcAxialSymPersp(SrcWdt, SrcHgt, DstWdt, RBX, RBY, RTX, RTY: Integer;
                              var A, H, E: Double);
  begin
     A := (2 * RBX - DstWdt) / SrcWdt;
     H := (A * SrcWdt/ (2 * RTX - DstWdt) - 1) / SrcHgt;
     E := (RTY * (H * SrcHgt + 1) - RBY) / SrcHgt;
  end;

Having coefficients, we can apply transformation to any source point PSrc and get coordinates of mapped point. In your case PSrc.X = 0 and PSrc.Y = i * SrcHgt / 5 for i=1..4 will give Y-coordinates od horizontal lines.

procedure PerspMap(SrcWdt, DstWdt, RBY: Integer; A, H, E: Double; PSrc: TPoint): TPoint;
  begin
     Result.X := Round(DstWdt / 2 + A * (PSrc.X - SrcWdt/2) / (H * PSrc.Y + 1));
     Result.Y := Round((RBY +  E * PSrc.Y) / (H * PSrc.Y + 1));
  end;