New TV, Different Aspect Ratio ... Same Height?
Date: 01/04/2010 at 11:54:50 From: Anthony Subject: What size HD TV equals my old picture size Hi, Dr Math. How do I figure out what size HD TV to buy so that its screen is at least as tall as my old TV's was? My old TV measures 29 inches diagonally. I know that right triangles can be used to compare the two screens, but I don't know how to compute the vertical dimension knowing only the diagonal of the old TV; and I don't know the ratio of an old set to a new HD set.
Date: 01/04/2010 at 20:45:58 From: Doctor Rick Subject: Re: What size HD TV equals my old picture size Hi, Anthony. The old analog standard (NTSC in the US) had an aspect ratio of 4:3 for the actual display area of a picture tube. The new HD standard has an aspect ratio of 16:9. Thus, in the old set, if the height is H, then the width is (4/3)H, and the diagonal is sqrt(H^2 + ((4/3)H)^2) = H*sqrt(1 + 16/9) = (5/3)H So we can set up this relation between your old set's picture height and its diagonal: (5/3)H = 29" Isolating H and simplifying gives us: H = (3/5)*29" = 17.4" Now, in an HD set with a picture height of 17.4 inches, the width is 17.4" * 16/9 = 30.93" The diagonal is thus sqrt(17.4^2 + 30.93^2) = 35.5" To get a picture height at least as tall as your old set, you'll need a new set with a diagonal of 35.5 inches or greater. In general, if the old diagonal is D, the old height is (3/5)D, and the new diagonal is D*sqrt(337)/15 = 1.2238 - Doctor Rick, The Math Forum http://mathforum.org/dr.math/
Date: 01/05/2010 at 10:44:44 From: Anthony Subject: Thank you (What size HD TV equals my old picture size) Thanks, Dr. Rick. The information you gave me certainly pointed me in the right direction. Because I haven't done this type of math in a long time, and there were a lot of details omitted from your response, a friend and I spent another hour trying to fill in those gaps! It was an interesting exercise, as I am very much out of practice. I need a LOT of help with the mathematical processes, but I think I get it now. My brain is now sore, and I think I need a nap, so in the future I will just multiply the old diagonal by 1.2238, as you said. Thank you for your time and guidance!
Date: 01/05/2010 at 11:32:40 From: Doctor Rick Subject: Re: Thank you (What size HD TV equals my old picture size) Hi, Anthony. You're welcome. I gave the final answer so you wouldn't *need* to fill in the gaps (you could have just trusted me!), but I'm gratified that you were interested enough in the math to put in the effort to understand it. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/
Date: 01/05/2010 at 13:52:08 From: Anthony Subject: Thank you (What size HD TV equals my old picture size) Oh, I trusted you, but the root of my question was *how* do I figure it out. This question has started a small firestorm of lost productivity at work as the self-proclaimed math wizards in my office are now trying to arrive at the solution from different directions. Want to join in? The camps are split along algebra and trig lines. As for me ... I'm just going to watch something on my new HD TV. Thanks again :)
Date: 01/05/2010 at 14:04:11 From: Doctor Rick Subject: Re: Thank you (What size HD TV equals my old picture size) Hi, Anthony. Go ahead and relax in front of the TV, but you might have one of the wizards in the trig camp write in and tell me how they can possibly bring trigonometry into this when no angles are involved. I think that the most they can do is to "re-invent" Pythagoras, coming back to the same solution via a roundabout path. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/
Date: 01/06/2010 at 16:13:07 From: Doctor Rick Subject: Re: Thank you (What size HD TV equals my old picture size) Hi, Anthony. I got the attachment, and it's as I expected: the Pythagorean theorem is hidden in the trigonometry. Your colleague did the following: *-----------------* | / | | 5 / | | / |3 (showing ratio of height to width, 3:4) | / | | / theta | *-----------------* 4 tan theta = 3/4 theta = 36.87 deg *-----------------* | / | | 29 / | | / |x (similar figure with known diagonal) | / | | / theta | *-----------------* sin theta = x/29 0.6 = x/29 x = 17.4 The Pythagorean theorem applies to trig functions as sin^2(theta) + cos^2(theta) = 1 I can use this relation to find the sine of an angle whose tangent is known: dividing both sides by sin^2(theta), I get 1 + 1/tan^2(theta) = 1/sin^2(theta) This also required the fact that sin(theta)/cos(theta) = tan(theta). Then solve for sin(theta): sin(theta) = 1/sqrt(1 + 1/tan^2(theta)) When tan(theta) = 3/4, we find sin(theta) = 1/sqrt(1 + 16/9) = 1/sqrt(25/9) = 3/5 That's why we get such a simple value for the sine of theta; this is a 3-4-5 right triangle. I labeled the diagonal of the first figure with length 5 accordingly. The result for x then follows by similar triangles: x/29 = 3/5 x = 29*3/5 = 17.4 The remainder of the work is a similar application of trig to the 16:9 rectangle: tan(omega) = 9/16 omega = 29.36 deg sin(omega) = 17.4/d_new d_new = 35.49... My work: d_new^2 = 17.4^2 + [17.4*(16/9)]^2 = 17.4^2 (1 + 256/81) d_new = 17.4*sqrt(1 + 256/81) = 35.49... Both methods work; I can't say one method is superior to the other. I prefer mine aesthetically because it requires only a square root, and I feel that it shows connections that are hidden by the trigonometry. The other point of view is that, given that you have a scientific calculator, the work is simpler when you bring the angles into the picture. To each his own! - Doctor Rick, The Math Forum http://mathforum.org/dr.math/
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