One-point perspective and point with negative depth - math

Question

I am trying to create a game using one-point perspective. Everything works fine for points within the view but goes wrong with the negative depth. I understand the perspective as shown on the following picture (source).
In general, I took a point at some distance from the left of the right vertical edge of the frame along the lower horizontal line (5 points in this case), join it with the O' point (line H'O') and where the line intersects the vertical line (at point H') is the depth line (of 5 in this case). This works well even for negative depth (as the line H'O' intersect the vertical line below the viewpoint). However, if the depth is more then is the distance of O' (that mean the point would be on the right from the O') the line flip and the H' end on top of the viewpoint (although it should end up below).
How should I correct it, so the point with negative depth is transformed correctly (mean from 3D space to 2D space)?
EDIT
This image is probably better.
My question is how to handle points with negative depth (should end up below the screen) higher then is a distance of transversal.

Answer 1

The points to the right of the point O', along the line determined by the lower edge of the frame, correspond to points that are behind the observer, so technically, the observer cannot see them. To see the points behind you, means that you have to turn around, so you need to change the position of the screen. Draw a copy of the black square frame to the right of the point O', so that the new square is the mirror symmetric image of the original frame square with respect to the line orthogonal to the horizon line and passing trough the point O'.
Edit: The points with negative depth to the right of point O' (i.e. a point behind the observer) is supposed to be mapped above the horizontal blue line. This is the right way to go.
I assume your coordinate system in three dimensions has its origin at the lower right corner of the square frame on your picture. The x axis (I think how you measure width) runs along the lower horizontl edge of the frame, while the y axis (what you call height) is along the right vertical edge of the frame. The depth axis is in three dimensions and it's perpendicular to the plane of the square frame (so it is parallel to the ground). It starts from the lower right corner of the frame square. Assume that the distance of point O' from the right vertical edge of the square is S and the coordinates of the point C are {C1, C2} (C1 is the distance of point C from the right vertical edge and C2 is the distance of C from the lower horizontal edge of the square).
Given the coordinates {w, h, d} (w - width, h - height, d - depth) of a point in three dimensions, its representation on the two dimesnional square screen is gievn by the formulas:
x = (S*w + C1*d)/(S+d)
y = (S*h + C2*d)/(S+d)
So the points you gave as an example in the comments are
P1 = {h = 5, w = 5, d = 5} and P2 = {h = 5, w = 5, d = -10}
Their representation on the screen is
P1_screen = {(S*5 + C1*5)/(S+5), (S*5 + C2*5)/(S+5)}
P2_screen = {(S*5 - C1*10)/(S-10), (S*5 - C2*10)/(S-10)}
whatever your parameters S, C1 and C2 are. The representation of the (infinte) line connecting points P1 and P2 is represented on the screen as the (infinite) line connecting the points P1_screen and P2_screen. However, if you want the 2D representation of the visible part of the segment that connects P1 and P2, then you have to draw the (infinite) line between P1_screen and P2_screen and exclude the following two segment: segment [P1_screen, P2_screen] and the segment from P2_screen along the line up towards the upper top edge. You have to draw on the screen only the segment from the infinite line connecting P1_screen and P2_screen which starts from P1_Screen and goes down towards the lower horizontal edge of the screen.