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Perspective projection of pyramid
Short answer
COP (0,0,20), view plane z=0 so dp=20. Use xp = 20x/(20−z), yp = 20y/(20−z). Base: (15,15,10)→(30,30,0); (20,20,10)→(40,40,0); (25,15,10)→(50,30,0); (20,10,10)→(40,20,0). Apex (20,15,20) has z=zprp → projection at infinity.
Obtain perspective projection co-ordinates for the pyramid with vertices of base (15, 15, 10), (20, 20, 10), (25, 15, 10), (20, 10, 10) and apex (20, 15, 20); given that Zprp = 20 and Zvp = 0.
Centre of projection at (0, 0, zprp) = (0, 0, 20), view plane z = zvp = 0, so dp = zprp − zvp = 20.
Formulas: xp = x·dp/(zprp − z), yp = y·dp/(zprp − z), zp = 0
Homogeneous matrix:
| 1 0 0 0 |
| 0 1 0 0 |
| 0 0 0 0 |
| 0 0 -1/20 1 |
with h = 1 − z/20, so (xp, yp) = (x_h/h, y_h/h).
| Vertex | z | h = (20 − z)/20 | xp = x/h | yp = y/h | Projected |
|---|---|---|---|---|---|
| A(15,15,10) | 10 | 0.5 | 15/0.5 = 30 | 30 | (30, 30, 0) |
| B(20,20,10) | 10 | 0.5 | 40 | 40 | (40, 40, 0) |
| C(25,15,10) | 10 | 0.5 | 50 | 30 | (50, 30, 0) |
| D(20,10,10) | 10 | 0.5 | 40 | 20 | (40, 20, 0) |
| E apex(20,15,20) | 20 | 0 | 20/0 | 15/0 | undefined |
Apex: its z equals zprp, so it lies in the plane through the centre of projection that is parallel to the view plane. Its projector never meets the view plane (h = 0), so the projection is at infinity. With this data the apex cannot be projected; it would be projected normally if it were in front of the COP (z < 20) and clipped when z ≥ 20.
Answer: base projects to (30,30), (40,40), (50,30), (40,20) on the plane z = 0; the apex is at infinity because z = zprp.
✓ Answer checked by Claude Opus 5.5 (high effort)
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