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I'm doing this :
box {<-10,0,-7>,<10,7,7>
matrix < 1, -.5, 0,
0, 1, 0,
0, 0, 1,
0, 0, 0 >
}
..which shears the box so the top is no more parallel to the ground
plane. However, I have a problem determining at which angle it is now !
I would have tought that this gives a 30 degrees angle (arcsinus 0.5),
but placing a 30-degrees rotated object on top of the box proves it
wrong ! How do I calculate the exact angle ?
Fabien.
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Fabien Mosen wrote:
> I'm doing this :
>
> box {<-10,0,-7>,<10,7,7>
> matrix < 1, -.5, 0,
> 0, 1, 0,
> 0, 0, 1,
> 0, 0, 0 >
> }
>
> ..which shears the box so the top is no more parallel to the ground
> plane. However, I have a problem determining at which angle it is now !
> I would have tought that this gives a 30 degrees angle (arcsinus 0.5),
> but placing a 30-degrees rotated object on top of the box proves it
> wrong ! How do I calculate the exact angle ?
>
> Fabien.
It's not arcsin, but arctangent. If you rotate another object by
degrees(atan2(-0.5,1))*z you will get the right angle.
Josh English
eng### [at] spiritone com
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Josh English wrote:
> It's not arcsin, but arctangent. If you rotate another object by
> degrees(atan2(-0.5,1))*z you will get the right angle.
Ooops ! Right ! Why did I use sinus ? Now that I think about
it, I know that I pictured (in my mind) the trigonometric circle
with the tangent drawn in dots.
Fabien.
..slapping myself..
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