Z pass using expressions

Dan Ebberts
August 30, 2005 at 7:42 pm

Try this:

start = 0;
vanish = 1000;
linear(position[2],start,vanish,100,0)

Adjust start and vanish to suit.

Dan
Cory06
August 30, 2005 at 7:52 pm

Wow!! You are good. That
Cory06
August 30, 2005 at 8:05 pm

While I have your attention, I was wondering if i could ask one more question. if i want one object A to to derive it’s z position from object B’s z postion, can i do that while leaving Object A’s X&Y independant with an expression??

only if you have time

thanks again
Dan Ebberts
August 30, 2005 at 8:14 pm

Like this:

p = thisComp.layer(“object B”).position;
[value[0],value[1],p[2]]

Dan
Cory06
August 30, 2005 at 9:18 pm

Thanks again!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

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