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Geek Culture / Trig Problem

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Jeff032
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Posted: 5th Mar 2008 18:21 Edited at: 5th Mar 2008 18:22
I've been struggling with this trig problem for a while, and I can't seem to figure it out, even though there is probably an easy solution that I keep missing.

So basically, I have a 3D box. This box has a diagonal of length r. From one side of the box, the angle the diagonal appears to make with vertical is a, and from the other side, the angle the diagonal appears to make with vertical is b.

How can I find the dimensions of the box?

Here is a picture:



Thanks

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Osiris
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Posted: 5th Mar 2008 18:46
Use Sine, Cosine, and Tangent (if needed) to find the dimensions of the two 2d diagrams you have up there.

SOHCAHTOA

SOH = Sine for opposite of the hypotenuse

CAH = Cosine for adjacent of the hypotenuse

TOA = Tangent for opposite of the adjacent


Hope that helps a little, I did one of these in physics a while back.

RIP Max-Tuesday, November 2 2007
You will be dearly missed.
Jeff032
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Posted: 5th Mar 2008 18:51 Edited at: 5th Mar 2008 19:31
I know those functions, but the problem is that I know the length of the diagonal in 3 dimensional space, but I do not know the length that it appears to be in the 2 dimensional projections.

[EDIT]
Not sure if this helps:

(Note, whether 1.15 = ..., 0.71 = ..., 1.73 = ... is not 100% certain, I can only get precision to 2 decimal places, and those seemed to match)

and 'a' and 'b' in my case will always be 0, 22.5, or 45.

diagonal = sqrt(width*width + height*height + length*length)

BiggAdd
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Posted: 5th Mar 2008 21:03 Edited at: 5th Mar 2008 21:18
I've made some Code for you:



All the Trig is between the stared lines. Don't worry yourself about the visual code.

The angle a represents the angle between the red lines
the angle b represents the angle between the blue lines
r is the diagonal length (Green Line)
Use Up and Down arrow Keys to adjust a
Use left and right arrow keys to adjust b
use left and right mouse buttons to adjust r
mouse move is rotate


L# is Size of X
H# is Size of Y
D# is size of Z
Q# is the diagonal of the plane with Dimensions L# and D# (Purple Line)


Hope this helps.

[EDIT]Updated the code to add a 3D box representation[/EDIT]

IanM
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Posted: 5th Mar 2008 22:58
@Jeff032,
Yep, Pythagoras in 3 dimensions is the way to solve the problem - it's simply the 3D distance specified slightly differently.

Jeff032
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Posted: 5th Mar 2008 23:17 Edited at: 5th Mar 2008 23:17
Thanks BiggAdd, but it doesn't seem to work quite right. When both the angles are 45 degrees (so it's a cube), and r is 100, it lists the sides of the cube as being 50 units. They should be 57.735, or 100*sqrt(1/3).

The only other thing is that b is the angle on the side to the left of a (or to the right), instead of on the bottom, but I could probably work with that anyway.



@IanM,
Yeah, thats what I figured, I just haven't found the right way to go about using it. I'll have more time this evening to look at it.

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BiggAdd
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Posted: 5th Mar 2008 23:50 Edited at: 6th Mar 2008 14:24
Sorry you are right! Forget my functions, they are wrong - I was making an assumption about the angles. I shall try come up with a better solution to your problem.

aluseus GOD
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Posted: 8th Mar 2008 04:41
you also need some trig to make use of them angles.

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Osiris
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Posted: 8th Mar 2008 04:48
No, Its: Dem' der' angles'.

RIP Max-Tuesday, November 2 2007
You will be dearly missed.
BiggAdd
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Posted: 8th Mar 2008 06:31
Its impossible to do this with only 2 angles. If you work through by substitution trig equations for other trig equations, you get a messy quadratic involving a lot of Tan functions.
A third angle is needed. I'll post the solution in a minute, just let me write it up.

Mr Tank
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Posted: 8th Mar 2008 17:56 Edited at: 8th Mar 2008 18:34
The height of the box is C = r / sqrt( tan^2 a +tan^2 b + 1 )

The other sides are C tan a and C tan b.



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BiggAdd
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Posted: 9th Mar 2008 03:28
Gah! I knew I was doing something wrong. Now I feel stupid.

Jeff032
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Posted: 9th Mar 2008 03:31
Wow...thanks
Now I feel stupid too.

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