Originally posted by

**Ristogod**View PostAs an example, imagine a display that is 9 units tall and 16 units wide. Its diagonal will be (9^2+16^2)^(1/2) = 12 units.

If the width and height are scaled by a factor α, then the new width will be 16α, the height will be 9α and the diagonal will be ((9α)^2+(16α)^2)^(1/2)=12α

If the horizontal and vertical resolution is fixed to λ (pixels/unit), the number of horizontal pixels will be 16αλ and the number of vertical pixels will be 9αλ. The total number of pixels will be, 144*(α^2)(λ^2)

Then doubling the diagonal length while keeping λ the same, will require twice as many pixels in the horizontal and vertical directions and 4 times the number of total pixels.

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