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 20-06-2004, 07:46 AM #1 Gun-Kata Registered User     Join Date: Jan 2004 Location: Los Angeles Posts: 68 Thanks: 0 Thanked 0 Times in 0 Posts Position Attribute in Expression Question (from Learning Maya Dynamics 5) float \$nz = 3* noise (nurbsPlane1ParticleShape.position*.1 + time); nurbsPlane1ParticleShape.position = nurbsPlane1ParticleShape.position0 + <<0,\$nz, 0>>; Hi, that expression is used to control the y value of each particle in a grid of particles that has been parented to a nurbs plane. The result looks kind of like cloth blowing in the wind. What I don't understand is how the part that is in red can possibly be used to assign each particle its own y coordinate. Isn't position a vector attribute? How then can it yield a scalar value for \$nz? And furthermore, how is it that it yields a different y value for each particle in the field? Last edited by Gun-Kata : 20-06-2004 at 07:49 AM.
 20-06-2004, 02:12 PM #2 kbrown Moderator     Join Date: Sep 2002 Location: London, UK Posts: 3,198 Thanks: 0 Thanked 8 Times in 8 Posts The answer lies in the second line. The scalar value is used in a vector <<0, \$nz, 0>>. Why does it yield to a different y value? Well, because the vector is added to the initial position of a given particle. __________________ Kari - My Website - My IMDB Do a lot, Fail a lot and Learn a lot!
 20-06-2004, 07:14 PM #3 Gun-Kata Registered User     Join Date: Jan 2004 Location: Los Angeles Posts: 68 Thanks: 0 Thanked 0 Times in 0 Posts Thanks! What I still don't understand though is how an equation that uses a vector quantity with three distinct numerical components (nurbsPlane1ParticleShape.position) can yield a single decimal numerical value (float \$nz). If, for example, nurbsPlane1ParticleShape.position is equal to <<2,3,4>> and that value is then inserted into the equation, then shouldn't the equation also yield a vector quantity? ps. Nice Quake III model btw.
 21-06-2004, 02:57 AM #4 kbrown Moderator     Join Date: Sep 2002 Location: London, UK Posts: 3,198 Thanks: 0 Thanked 8 Times in 8 Posts Ah... Now I see what you mean. It is because of automatic type conversion. In this case the .position vector is automatically converted to a float by calculating a magnitude (length) of it. These two will yield to the same result: float \$nz = <<1, 2, 3>>; float \$nz = mag(<<1, 2, 3>>); __________________ Kari - My Website - My IMDB Do a lot, Fail a lot and Learn a lot!
21-06-2004, 03:35 AM   #5
Gun-Kata
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 Originally posted by kbrown Ah... Now I see what you mean. It is because of automatic type conversion. In this case the .position vector is automatically converted to a float by calculating a magnitude (length) of it. These two will yield to the same result: float \$nz = <<1, 2, 3>>; float \$nz = mag(<<1, 2, 3>>);
Now I get it. Thanks!

 21-06-2004, 08:06 AM #6 mind_raper Registered User   Join Date: Nov 2002 Location: Pakistan Posts: 57 Thanks: 0 Thanked 3 Times in 3 Posts kool explaination Kbrown... ...... __________________ Learn the Rules|Then| Burn the Rules Book

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