Actual screen size of iMacs, ACD's, etc.

sequitur

I'm trying to find the actual width and height of screens-not including bezel and chin. How does one figure the dimensions of a (diagonal) 21.5", 27", 30", or whatever?



I know the square of the hypotenuse is equal to the sum of the square of the other two sides, but that doesn't give me the answer in reverse - not for screens anyway. Well, maybe, using 16:10 or 16:9, but that's a lot of trouble.



I've tried googling with no luck.



Thanx

Marvin

21.5^2 = w^2 + h^2

w = 16/9 x h



21.5^2 = (16/9 x h)^2 + h^2

h^2 (1+(16/9)^2) = 21.5^2

h = sqrt(21.5^2/(1+(16/9)^2)) = 10.5"

w = 18.7"



So the 21.5" iMac panel with a 16:9 aspect should be 18.7" wide and 10.5" tall. Give or take some rounding. That's usually the viewable area of the screen, not the dimensions of the front of the iMac.

sequitur

Quote:

Originally Posted by Marvin


21.5^2 = w^2 + h^2

w = 16/9 x h



21.5^2 = (16/9 x h)^2 + h^2

h^2 (1+(16/9)^2) = 21.5^2

h = sqrt(21.5^2/(1+(16/9)^2)) = 10.5"

w = 18.7"



So the 21.5" iMac panel with a 16:9 aspect should be 18.7" wide and 10.5" tall. Give or take some rounding. That's usually the viewable area of the screen, not the dimensions of the front of the iMac.

Neat. Now I can just plug in the numbers and get an answer.

Marvin, is there anything you don't know? Your title should be global moderator/guru. Mel Gross and a few others, also, should have that title. I've said this many times: "you guys are awesome."

You're not given enough credit by AI forum. Members take you for granted.

Thanks.

sequitur

Using Marvin's formula, the viewing area of the 27" iMac is:



H= 13.24"



W= 23.5"



That is IF i did it correctly. I tried the formula with my 22" ACD, and it worked out.



if anyone else worked it out, what did you arrive at?