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Discussion Starter #1
Here is my question, for any math major out there.


I have a 61" diagonal 4x3 tv, that shows 16x9 HDTV in top and bottom gray bar letterbox.


I also have a 51" diagonal 16x9 HDTV.


Which has the larger 16x9 picture area and by how much?
 

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Even with the sets turned off you can just measure the widths to compare 16:9 picture sizes. Or just measure the heights to compare 4:3 picture sizes.


- Tom
 

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1. A 4:3 display has a diagonal of 5 (this is a good 'ol 3:4:5 right triangle. Note 3^2 + 4^2 = 5^2.) So a 61" diagonal 4:3 display has a width of 61" x 4/5 = 48.8", and a height of 61" x 3/5 = 36.6".


2. A 16:9 display has a diagonal of sqrt(16^2 + 9^2) = 18.36. Assuming your letterboxed 16:9 picture on the 61" 4:3 fills the screen, it is 48.8" wide (from above.) So its diagonal is 48.8" x 18.36/16 = 56".


Answer: Your 61" display has a diagonal of 56" for 16:9 material, which is larger than your 51" HDTV. :)
 

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Discussion Starter #5
Thanks, djs.


Using your calculations, the total 16x9 viewing area of the 51" diagonal is 1103 square inches, or 82% of the 1342 square inches of the 16x9 portion of the 61" 4x3.
 
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