Ed Kolis
March 5th, 2007, 01:15 PM
Thought you’d get a kick out of this, guys... but did you know that widescreen monitors have 11% less screen space than regular monitors of the same size? And they charge you more for the widescreen! http://forum.shrapnelgames.com/images/smilies/tongue.gif
See, they measure monitors on the diagonal, which means that the closer you are to a square, the more space you get. 4:3 (a regular monitor) is much closer to square than 16:9 (a “widescreen”).
How much less? Well, the Pythagorean theorem says that X^2 + Y^2 = Z^2, where X and Y are the width and height of the monitor and Z is the diagonal. So since the diagonals are the same, Xr^2 + Yr^2 = Xw^2 + Yw^2, where Xr and Yr are the dimensions of the regular screen and Xw and Yw are the dimensions of the wide screen. Putting in the ratios, we have (4 * R) ^ 2 + (3 * R) ^ 2 = (16 * W) ^ 2 + (9 * W) ^ 2, where R and W are proportionality constants. Multiplying and condensing, 25 * R^2 = 337 * W^2. Remember that the part on the left is the diagonal of the regular screen, while the part on the right is the diagonal of the widescreen. We can solve for W, getting W = 5 * R / sqrt(337).
Now X * Y is the area, of course. So, the area of the regular screen Ar = (4 * R) * (3 * R) = 12 * R^2, while the area of the widescreen Aw = (16 * W) * (9 * W) = 144 * W ^ 2. But W = 5 * R / sqrt(337), so that comes out to Aw = 144 * (5^2 * R^2 / 337), or 3600 * R^2 / 337. The ratio of areas is, thus (drumroll please): (3600 * R^2 / 337) / (12 * R^2) = 300 / 337, or about 89%! The widescreen is wider, but it overcompensates by being shorter, leading to an 11% reduction in surface area!
See, they measure monitors on the diagonal, which means that the closer you are to a square, the more space you get. 4:3 (a regular monitor) is much closer to square than 16:9 (a “widescreen”).
How much less? Well, the Pythagorean theorem says that X^2 + Y^2 = Z^2, where X and Y are the width and height of the monitor and Z is the diagonal. So since the diagonals are the same, Xr^2 + Yr^2 = Xw^2 + Yw^2, where Xr and Yr are the dimensions of the regular screen and Xw and Yw are the dimensions of the wide screen. Putting in the ratios, we have (4 * R) ^ 2 + (3 * R) ^ 2 = (16 * W) ^ 2 + (9 * W) ^ 2, where R and W are proportionality constants. Multiplying and condensing, 25 * R^2 = 337 * W^2. Remember that the part on the left is the diagonal of the regular screen, while the part on the right is the diagonal of the widescreen. We can solve for W, getting W = 5 * R / sqrt(337).
Now X * Y is the area, of course. So, the area of the regular screen Ar = (4 * R) * (3 * R) = 12 * R^2, while the area of the widescreen Aw = (16 * W) * (9 * W) = 144 * W ^ 2. But W = 5 * R / sqrt(337), so that comes out to Aw = 144 * (5^2 * R^2 / 337), or 3600 * R^2 / 337. The ratio of areas is, thus (drumroll please): (3600 * R^2 / 337) / (12 * R^2) = 300 / 337, or about 89%! The widescreen is wider, but it overcompensates by being shorter, leading to an 11% reduction in surface area!