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Rickey
03-23-2003, 04:18 PM
Hi everyone,

I'd really appreciate it if somebody could help me with this.

I need to turn a lot of gears in a machine with enormous precision. As reverence I have the diameter of the gears, of course ( gear1 = 12,65cm, Gear2 = 6,75 cm, etc. ). How can I use the diameter of a gear to drive the expression?

Hope somebody understands that!

thanks in advance.

Rick

ericsmith
03-25-2003, 11:38 AM
All you need to do is add a multiplier value to the rotation expression for the gears. If I am not mistaken, the diameters are not the issue, but rather the number of teeth. So if your large gear has 30 teeth, an your small gear has 18 teeth, the multiplication factor would be 18 / 30 or .6.

So the expression for gear two (the one with 18 teeth) would be:

gear1.rotation.X * .6

(X would be the rotational channel you are animating, H, P, or B)

Mylenium
03-25-2003, 12:37 PM
Yepp, the number of teeth is the most important thing. I also would suggest to use dynamic division just in case you have strange ratios (8/7) or something like that. Even though this does not obviously affect calculations it can help you 'cos internally LW will work mor precise than manually writing 1.11111112 or something like that. You could also combine the number of teeth and rotational angles to be on a very, very save side ;) .

Mylenium

Heimhenge
03-25-2003, 12:44 PM
If you have gears all with common divisors you'll be fine using the simple multiplier method. If you have gears with like 7 and 13 teeth, even if LW calculates internally to double precision (and I'm pretty sure it does), after a finite number of rotations that rounded-off repeating decimal will eventually cause a loss of synch between the gears.

Sielan
03-26-2003, 01:49 AM
Off course you can use follower instead of expressions. I think it's simpler.