Ok, I am a math rock. Or more specifically, a geometry rock. I need to know two things: 1. On a 16:9 TV how can I figure the diagnal screen size of a 4:3 image. 2. Conversely, how big will 16:9 image be on a 4:3 TV. I need to know this so I can compare and contrast TV ratios. Thanks.

For the 16:9 set, the best way to figure out what size the 4:3 picture would be is to calculate the height. to do that, take the diag size of the 16:9 set and multiply it by 0.49. Now to get the 4:3 TV that has that height, divide that by 0.6. Sooo.. for a 43" WS it would be 43 * 0.49 / 0.6 = 35.12. So approx a 35" 4:3 picture. To figure out what diag size a 16:9 movie would have on a 4:3 set, you want to figure out the width of the 4:3 set. To do that, multiply the diag size by 0.8. Then to get a 16:9 TV with that width, divide by 0.872. Soo for a 43" 4:3 set, a 16:9 image would be.. 43 * 0.8 / 0.872 = 39.45. Hoe this helps..

Its simple trig, or maybe geometry, no I think its trig. Lets see if I remember this right: You calculate the height and width of the 16:9 screen based on the diagnol. That would be 16x^2 + 9x^2 = diagnol^2, solve for x and multiple by 16 and 9 to get your height and width. Well once you have the height, you know the height of your 4:3 image, and multiplying by 1.3333333 gives you the width of the 4:3 image and thus once again using pythagoras you can calculate the diagnol with a^2 + b^2 = c^2 where a is the width, b is the height, solve for c and that will be the diagnol. I hope that first part is right, I really need to write it out to be sure. And if I wrote it all out I could probably come up with a single formula to solve for it based on the diagnol of the 16:9 TV. Oh, and for the second, you just do more or less the same, except that you have a fixed width rather than height. Funny story, before I got my 50" Mits I decided to calculate the height and width so I could determine seating distance. Well obviously its 40"x30". Ok, it should've been obvious to me since I've got a degree in CS & Math and the pythagorian theorem is sometimes referred to as the 3, 4, 5 rule. For those who aren't aware of it, a=3, b=4 and c=5 is the simplest solution to a^2 + b^2 = c^2.

Since I never throw anything away at work, I found my notes when I figured out the dimensions for my 55" 16:9 set. Overall width is 48" and the height is 27". In 4:3 mode the width is 36" and the height is of course 27" so the 4:3 diagnol is 45". Going off that, it's safe to say that the 4:3 diagnol is 82% of the 16:9 diagnol. For my 50" 4:3 TV, the 16:9 height is 22.5". The diagnol is 46". So on a 4:3 set the 16:9 diagnol is 92% of the 4:3 diagnol. Finally, if you want a 16:9 TV that has a certain 4:3 diagnol, take that 4:3 diagnol you desire and multiply by 1.22. Thus if you want a 60" 4:3 image, you'll need a 73" 16:9 TV. I hope that helps.