Drawing a perspective checkered floor, is there a formula to calculate the drawn distance (for example 1 ft.) as it descends down the perspective line. Say we assume viewing the first tile the 1 ft. actual spans 8 inches, as projected on tho the 2 dimensional paper, how "tall" should the 2nd, 3rd, 4th, etc. tiles be respectively?
ignoramus
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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.