Skip to main content

Rectangle window to viewport mapping

Short answer

The viewing transformation matrix has sx=sy=1/6, translation (0.1667,-0.3333). The rectangle points map to A'(0.333,-0.167), B'(0.833,-0.167), C'(0.833,0.333), D'(0.333,0.333); only the portion inside the window maps to viewport region (0.5,0)-(0.833,0.333).

Suppose there is a rectangle whose coordinates are A(1,1), B(4,1), C(4,4), D(1,4) and the window coordinates are (2,2), (5,2), (5,5), (2,5) and the given viewport location is (0.5,0), (1,0), (1,0.5), (0.5,0.5). Calculate the viewing transformation.

Given: window: xwmin = 2, ywmin = 2, xwmax = 5, ywmax = 5. Viewport: xvmin = 0.5, yvmin = 0, xvmax = 1, yvmax = 0.5. The rectangle is A(1,1), B(4,1), C(4,4), D(1,4).

Scale factors

sx = (1 − 0.5)/(5 − 2) = 0.5/3 = 1/6 = 0.1667
sy = (0.5 − 0)/(5 − 2) = 0.5/3 = 1/6 = 0.1667

Viewing transformation matrix

M = T(xvmin, yvmin) · S(sx, sy) · T(−xwmin, −ywmin)

    | 1 0 0.5 |   | 1/6  0   0 |   | 1 0 −2 |   | 1/6   0   0.1667 |
  = | 0 1  0  | · |  0  1/6  0 | · | 0 1 −2 | = |  0   1/6  −0.3333 |
    | 0 0  1  |   |  0   0   1 |   | 0 0  1 |   |  0    0     1     |

(0.5 − 2/6 = 0.1667 and 0 − 2/6 = −0.3333.)

Mapped points

xv = 0.5 + (x − 2)/6, yv = (y − 2)/6

Point(xv, yv)
A (1,1)(0.333, −0.167)
B (4,1)(0.833, −0.167)
C (4,4)(0.833, 0.333)
D (1,4)(0.333, 0.333)

Answer: the transformed points are A'(0.333, −0.167), B'(0.833, −0.167), C'(0.833, 0.333), D'(0.333, 0.333). The parts of the rectangle with x < 2 or y < 2 lie outside the window, so after clipping only the portion x from 2 to 4, y from 2 to 4 is displayed, mapped to the viewport region (0.5, 0) to (0.833, 0.333).

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Computer Graphics chapter 3 questions →