Thursday, 13 September 2012

Rob does math (is never pretty)

A while ago, a friend asked on Facebook how big a new widescreen TV should be if he wanted the height of the picture to be about the same as his old 32-inch TV, or something like that. I didn’t answer at the time, because I thought it was ridiculous. You want the width to be the same? Or maybe it was the area. It struck me as odd, because at the very least, you want your new TV to have the same picture area, or ideally, the same height.

The old standard TV picture is proportioned at about 4x3.  The newer widescreen TVs are typically about 16x9.  So if you maintained the same width, the height of the picture would shrink proportionally. So I declined to answer the question.
On the other hand, this has been simmering at the back of my head for weeks. Maybe months.  If I wanted to preserve the height of my picture (and add width to make up the new proportion), how would I do that?

So to distract myself this evening, I think I figured this out.

To do it my way (keep the height the same, add the width, starting diagonal dimension about 32 inches):  The 32 inch diagonal is the hypotenuse of a right triangle.  The two other legs of the triangle have the proportion of 4x and 3x.  Pythagoras, I think, teaches us that they length of the hypotenuse (32) squared (=1024) equals the square of the lengths of the other two sides (4x)^2 + (3x)^2.  At this point, I was stymied, since for the life of me I couldn’t remember how to take the square of a complex quantity. 

After some Googling, I finally figured out that (4x)^2 (4x-squared) is 16(x^2).  That is, both terms in the quantity square.  If this is wrong, please let me know and I’ll start over.
So now, armed with this, I know that 16(x^2) plus 9(x^2) = 25(x^2) = 1024 (parentheses here just to remind me I’m now just squaring my variable.  Divide both sides by 25 and we get x^2 = (approximately) 41.  The square root of 41 is about 6.4. 

So the proportions of the sides of my television are about 4 x 6.4 = 25.6 inches and 3 x 6.4 = 19.2 inches. I don’t mind doing a little approximating, since I assume the 32 inch diagonal I started with is also an approximation. 

So, to buy a widescreen with the same vertical height, I need one that is about 19.2 inches high, and it will be a little over 34 inches wide because (19.2/9)*16.  The diagonal dimension will be something like 39.something inches.

So having worked all this out, I built myself a little spreadsheet.  You put in the diagonal dimension of your old 4x3 TV and it tells you the diagonal dimension of a widescreen TV if the vertical dimension is the same, the horizontal dimension is the same, and if you want to maintain the picture area.

Me so smrt sometimes. Unless I got the original quantity wrong, in which case never mind.

I wonder if I could  somehow build an app out of this… Hmm….

1 comment:

Rob Hagiwara said...

Update: It turns out I have 21" tv, on a stand that is at most 26 inches wide, in a corner of the room which won't accommodate much more than that. So it looks like we're back to the 32" tv idea.