start/stop expression at specific time

[Mike Clasby]

ok, somebody will have a more sophisticated solution, but to keep that expression value (Position) you can always split the layer (Edit>Split Layer) then on that upper layer set a Position keyframe (or whatever property the expression is on), then disable the expression (click the equal sign next to the stopwatch). You should be good to go.

[Dan Ebberts]

What’s the code for your expression?

Dan

[Atomicrabbit]

[Dan Ebberts] “What’s the code for your expression?

Dan”

I found this code from a post. It’s to make a falling leaf effect, but I modified the code a little. I’m also using a checkbox to start the expression.

position
———————————————————————————–
control = effect(“Checkbox Control”)(“Checkbox”) ;
whattime=time-3;
if (control == 1){
sp=2;
amp=150;
gravity = 50;
x = Math.sin(whattime*sp)*amp;
y = Math.sin(whattime*1.5)*50;
value + [x,y];
}else{
value;
}

scale
————
control = effect(“Checkbox Control”)(“Checkbox”) ;
whattime=time-3;
if (control == 1 || value > 40){
gravity = 50;
z= whattime*whattime
value – [z,z];
}else{
value;
}

rotation
————————————————————————————
control = effect(“Checkbox Control”)(“Checkbox”) ;
whattime=time-3;
if (control == 1){
sp=2;
amp=5;
x = Math.sin(whattime*sp)*amp;
value – x;
}else{
value;
}

[Dan Ebberts]

Give this a try (for position):

chkBx = effect(“Checkbox Control”)(“Checkbox”);

n = 0;
if (chkBx.numKeys > 0){
n = chkBx.nearestKey(time).index;
if (chkBx.key(n).time > time){
n–;
}
}

t = 0;
curTime = time;

if (n > 0){
for (i = n; i > 0; i–){
if (chkBx.key(i).value > 0){
t += (curTime – chkBx.key(i).time);
}
curTime = chkBx.key(i).time;
}
if ((curTime > 0) && (chkBx.key(1).value > 0)) t += curTime;
}else{
if (chkBx.value > 0) t = time;
}

sp=2;
amp=150;
gravity = 50;
x = Math.sin(t*sp)*amp;
y = Math.sin(t*1.5)*50;
value + [x,y];

Dan

[George Costakis]

I doubt anyone will see this, but what about for this expression?

veloc = 7;
amplitude = 80;
decay = .7;

amplitude*Math.sin(veloc*time)/Math.exp(decay*time)

right now it starts at the beginning of the comp even if i try the split and duplicate layer trick. any help would be great.
-george

[Brennan Shappell]

I also am trying to get the decaying pendulum expression to start at a time specified(not the beginning of the comp).
Thanks for any help!

veloc = 5;
amplitude = 20;
decay = .9;

amplitude*Math.sin(veloc*time)/Math.exp(decay*time)

[Dan Ebberts]

This is one way:


beginTime = 4; // start at 4 seconds
if (time < beginTime){
0
}else{
t = time - beginTime;
veloc = 5;
amplitude = 20;
decay = .9;
amplitude*Math.sin(veloc*t)/Math.exp(decay*t)
}


Dan

[Brennan Shappell]

Thanks for the absurdly fast turn around time on that, Dan =]
Works like a charm.

[Duca Simone luchini]

Hallo Dan, I didn’t understand in this your code where is the END parameter where the expression stops. I Understood that at 4 second starts expression (beginTime = 4;)
BUT where stops???
For instance, in your expression (), if the comp is 10 seconds and I’d like to activate expression from 2 second to 8 second, what I have to write?

beginTime = 4; // start at 4 seconds
if (time < beginTime){ 0 }
else{ t = time – beginTime;
veloc = 5; amplitude = 20; decay = .9; amplitude*Math.sin(veloc*t)/Math.exp(decay*t) }

THX!

[Dan Ebberts]

It doesn’t actually stop. Eventually it settles out to approximately zero, but the timing depends on the decay variable. If you wanted it to just stop at whatever value corresponds to 8 seconds, you could do this:


beginTime = 4; // start at 4 seconds
endTime = 8;
if (time < beginTime){
0
}else{
t = Math.min(time,endTime) - beginTime;
veloc = 5;
amplitude = 20;
decay = .9;
amplitude*Math.sin(veloc*t)/Math.exp(decay*t)
}


Dan