Wide Screen Math

This post was published 10 years, 6 months ago. Due to the rapidly evolving world of technology, some concepts may no longer be applicable.

I was recently asked an interesting question about how the area of a widescreen compares with the area of a traditional display. Now, usually, one hears that you get to see additional parts of the picture on a widescreen, so that was my conditioned response. However, I decided to delve into the math a bit, to see if that proved to be true. As it turned out, you don’t get more area on a widescreen display.

A traditional display has a ratio (width:length) of 4:3, screen size is typically given as the diagonal (or if we are thinking triangles, the hypotenuse).

d represent the diagonal
w represent the width
h represent the height
A represent the area

We get two equations:

Using the Pythagorean theorem:
1: d=\sqrt{w^2+h^2}

Using the ratio of length to width:
2: w=\frac{4\cdot h}{3}

Equation 3, comes from substituting 2 into 1, we get:
d=\sqrt{(\frac{4\cdot h}{3})^2+h^2}
d=\sqrt{\frac{25\cdot h^2}{9}}
d=\frac{5\cdot h}{3}
h=\frac{3\cdot d}{5}

Equation 4 comes from substituting 3 into 2 we get w, which we will use to find area:
w=\frac{4\cdot \frac{3\cdot d}{5}}{3}
w=\frac{4\cdot d}{5}

Therefore, the area of the 4:3 display (equation 5) is:
A=w\cdot h
A=\frac{4\cdot d}{5}\cdot \frac{3\cdot d}{5}

Following the same procedure for the 16:9 display, we get:
2: w=\frac{16\cdot h}{9}
3: h=\frac{9\cdot d}{\sqrt{337}}
4: w=\frac{16\cdot d}{\sqrt{337}}
5: A=\frac{144}{337}d^2

The ratio of the area of the 4:3 display to the 16:9 display, therefore, is:

In other words, you get about 12% extra screen area on a traditional, 4:3 display, compared to a widescreen 16:9 display. Not exactly a revelation perhaps, not something that immediately springs to mind. The advantage (of the widescreen), perhaps comes in the form of usable area for an image displayed in widescreen, where the 4:3 display will likely have less usable area.

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