rotation changes vector – driving along

[Dan Ebberts]

I think this would be a complicated expression. I gets back to the issue that expressions have no memory. Assuming that the rotation and speed have been changing, the expression has no easy way of knowing what position it calculated on the previous frame. The solution is that at each frame, the expression has to calculate everything that has happend at each previous frame (starting at th4e first frame) and keep a running tally of where it was at each frame. Certainly possible, but not simple.

Dan

[Filip Vandueren]

I’m not sure if I understand that “never go negative in Y”, this would only occur if the rotation would be more than +/- 90°, so you have that under control.

What would happen if the rotation would yield a negative Y, would Y be 0 ?

I think this does what you need, If I understand correctly:

position:

fd=thisComp.frameDuration;
speedSlider=thisComp.layer("Speed").effect("Speed")("Slider");

p=position.valueAtTime(0);

for (t=0;time>t; t+=fd) {
sp=speedSlider.valueAtTime(t)*fd;
vector=toCompVec([0,-sp],t);

if (vector[1]>0) {
vector=[vector[0],0];
}
p+=vector;
}

p;


as you see in Line 2, I reference a Slider called “Speed” on a layer “Speed”.
You can keyframe or wiggle that slider, and also the rotation of this layer can be keyframed or wiggled

[Filip Vandueren]

BTW, if we forget the -Y stuff, this is very easily translated into 3D:


fd=thisComp.frameDuration;
speedSlider=thisComp.layer("Speed").effect("Speed")("Slider");

p=position.valueAtTime(0);

for (t=0;time>t; t+=fd) {
sp=speedSlider.valueAtTime(t)*fd;
vector=toWorldVec([0,-sp,0],t);

if (vector[1]>0) {
vector-=[0,vector[1],0];
}

p+=vector;
}

p;


Thanks to the ToWorldVec(), this takes into account orientation and x-y-z-rotation

[Christopher R. green]

[Filip Vandueren] “I’m not sure if I understand that “never go negative in Y”, this would only occur if the rotation would be more than +/- 90°, so you have that under control.”

Never go positive in Y, that is, ‘down’.

But now I’m going to see what you came up with …

[Christopher R. green]

Um…
Perfect?

That is really cool, Flip.

Dan, you must’ve over-thought the problem!

This is going into my “favorite tricks to steal from others” folder!

I’m giddy.

DANG!

nice.

Thank you.

[Christopher R. green]

You can pretty much disregard the +y thing. I just put that in in case it made any difference in the way the solution evolved. It is, of course, dependent on how the rotation is done.

Thanks again!

cg