Is there an Algorithm for calculating the drawn distance between equally spaced lines?

OK, well this is possible. Whether this is the best way or not, I would not know, but it will work. First let's establish a base line (representing the bottom edge of your diagram). Draw a sketch representing a side view of this.

Suppose your eye-level is at point A, at a height of, say 5 feet, and your feet are point B (directly below A, of course) So AB = 5 feet.

And suppose that the first line of tiles is at point C, 5 feet in front of you. So BC = 5 feet.

Imagine a vertical sheet of glass standing at point C. The distances which you require to find are the heights, h, at which your line of sight to the lines between tile-edges intersect the sheet of glass ; let this point on the glass be K.

Consider a particular line, L, and let it's distance from B be x.

There is a triangle ABL, whose height = 5, and base = x.

Another triangle, similar to ABL, is KCL, whose height = h, and base = (x - 5)

So 5/x = h/(x - 5), from which

h = 5(x - 5) / x

Since we are imagining that the bottom edge of your diagram is at a distance of 5 feet, then the distances BL to the first, second, third, etc horizontal lines are 6, 7, 8 feet, and so on. So you can prepare a table :

. x . . . . .5(x - 5) / x . . . . . differences

---- . . . --------------- . . . . --------------

. 6 . . . . . . 0.833

. 7 . . . . . . 1.429 . .. . . . . . . 0.596

. 8. . . . . . .1.875 . .. . . . . . . 0.446

. 9 . . . . . . 2.222 . .. . . . . . . 0.347

. etc

So whatever measurement you select to represent the width of a tile is multiplied by the number in the "difference" column to give the distance to the second, third, fourth lines of tiles. Which is what I think that you are seeking.