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☖ Forums → Discussion Forum → Math Problem
Math Problem
2006-09-23, 3:31 AM #1
The thread about 4D vision got me thinking about a fun little math problem. What is the perceived one-dimensional length (w) of a square as it is rotated? I know the minimum length will be the length of the square l and the max will be 2[sup]1/2[/sup]*ll. I work it out...

[http://lpritchett.org/shhhh/math.GIF]

Code: 
w = a + b
b = sinθ * l
a = cosθ * l

w = cosθ * l + sinθ * l
w = l (cosθ + sinθ)


Yay. But now I have a problem. This works up to θ = π, but not past it because cosθ and sinθ start to become negative, meaning you get negative lengths. Is there any way I can restrict so I can have a nice function? All I have is:
w = l (|cosθ| + |sinθ|)

Also, try it for other shapes! I haven't been able to figure it out for triangles yet!
Ban Jin!
Nobody really needs work when you have awesome. - xhuxus
2006-09-23, 9:56 AM #2
I'm confused, are you trying to measure the vertical line there? The lines you've got going across and notation really aren't clear at all.
$do || ! $do ; try
try: command not found
Ye Olde Galactic Empire Mission Editor (X-wing, TIE, XvT/BoP, XWA)
2006-09-23, 10:17 AM #3
Reminds me of circular motion. Seeing as it's the weekend and I'm not getting paid for this maths I can't be arsed to sit down and do it properly - but there's nothing wrong with modulus signs like you have in as is.
2006-09-23, 1:04 PM #4
Originally posted by SMOCK!:
Also, try it for other shapes! I haven't been able to figure it out for triangles yet!

You should come up with a general formula that will work for any perfect polygon. That would be cool. :D
"it is time to get a credit card to complete my financial independance" — Tibby, Aug. 2009
2006-09-23, 1:29 PM #5
It can be done algorithmically much more simply. Algebra and calculus aren't always the answer you know.
Wikissassi sucks.
2006-09-23, 7:04 PM #6
47
MastaX

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