Subject:
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Re: 3D geometry question
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Newsgroups:
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lugnet.cad
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Date:
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Wed, 13 Oct 1999 21:24:55 GMT
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Viewed:
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566 times
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Fredrik Glöckner skrev i meddelandet ...
> Hi, I have this problem I hope someone may solve.
>
> Say I have a vector v with length normalized to 1. For simplicity,
> picture this vector as starting in the origin. Now, I need a
> rotation matrix M with the following properties:
>
> o If I rotate a part with one edge going from (0,0,0) to (0,-1,0)
> (i.e., pointing upwards), this edge should coincide with the vector v
> after the transformation
>
> o M should be a proper transformation, i.e., no scaling or shearing.
>
> Thanks for any help!
What about:
Assume V = (Vx, Vy, Vz)
B = arctan(Vx/Vz)
(Better with arctan2(Vz, Vx) if available, otherwise you have to adjust to the
right quadrant manually)
Matrix for rotation around Y:
( cos(B) 0 sin(B) )
( 0 1 0 )
( -sin(B) 0 cos(B) )
Now transform V with this matrix, giving V'
A = arctan(V'y/V'z)
Matrix for rotation around X:
( 1 0 0 )
( 0 cos(A) sin(A) )
( 0 -sin(A) cos(A) )
Now multiply the two matrices, to get your final result.
(I checked with Excel, and it seems to work)
--
Anders Isaksson, Sweden
BlockCAD: http://user.tninet.se/~hbh828t/proglego.htm
Gallery: http://user.tninet.se/~hbh828t/gallery.htm
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Message has 1 Reply:
 | | Re: 3D geometry question
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| Sorry, I just found out that I solved for (0,0,-1), not (0,-1,0), but that's just a symmetry change :-)) -- Anders Isaksson, Sweden BlockCAD: (2 URLs) (25 years ago, 13-Oct-99, to lugnet.cad)
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Message is in Reply To:
| | 3D geometry question
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| Hi, I have this problem I hope someone may solve. Say I have a vector v with length normalized to 1. For simplicity, picture this vector as starting in the origin. Now, I need a rotation matrix M with the following properties: o If I rotate a part (...) (25 years ago, 13-Oct-99, to lugnet.cad)
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