Thread: Do CRT's have 12 percent more area than widescreens (given the same diagonal)?

I calculated that given a rectangle with 16:9 dimensions and one with 4:3 that the 4:3 one will have 12 percent more area given the same diagonal length. If anyone's interested, please check my result.

Here's my work:
(16x)^2+(9x)^2=d^2
(4y)^2+(3y)^2=d^2

A(d)=16x*9x=144/337*d^2 and the other: A(d)=4y*3y=12/25*d^2

thus the ratio = 12/25*337/144 ~ 1.12

Hey lamp23.

If you are looking at setting the smaller sides equal for both sets then the closest monitor to a square will have the highest area and its got to do with the square being the optimal choice for area given a set of two sides (i.e max area with minimal perimeter has a square as a solution).

This is in line with your working out.

Originally Posted by lamp23

I calculated that given a rectangle with 16:9 dimensions and one with 4:3 that the 4:3 one will have 12 percent more area given the same diagonal length. If anyone's interested, please check my result.

Here's my work:
(16x)^2+(9x)^2=d^2
(4y)^2+(3y)^2=d^2

A(d)=16x*9x=144/337*d^2 and the other: A(d)=4y*3y=12/25*d^2

thus the ratio = 12/25*337/144 ~ 1.12

That is nothing to do with CRT technology, you can/could get a 16:9 CRT TV as well as 4:3. But that for a given advertised screen size it would be smaller that the same advertised screen size LCD, plasma, ... TV because of about a 25mm strip around the edge of the screen was covered by the bezel.

.

Originally Posted by chiro

Hey lamp23.

If you are looking at setting the smaller sides equal for both sets then the closest monitor to a square will have the highest area and its got to do with the square being the optimal choice for area given a set of two sides (i.e max area with minimal perimeter has a square as a solution).

This is in line with your working out.

Thanks! I forgot about that fact from calculus.