[Dan Ebberts]

Try replacing the last line with:

clamp(d[0] + d[1],-20,20);

Dan

[Dan Ebberts]

Something like this should work as long as the markers are farther apart than the keyframes:


m = thisComp.marker;
t = 0;
if (numKeys > 1 && m.numKeys > 0){
n = m.nearestKey(time).index;
if (time < m.key(n).time) n--;
if (n == 1)
t = key(1).time + time - m.key(1).time
else if (n == 2)
t = key(2).time - (time - m.key(2).time);
}
valueAtTime(t)


[Dan Ebberts]

If I understand how you have this set up, you need to adjust the time by the offset of the precomp’s start time within the main comp, like this (not tested):

C = comp(“Slide3”);
L = C.layer(thisComp.name);
p = C.layer(“SW_Robot_1”).transform.position;
d = p.valueAtTime(time + L.startTime) – p.valueAtTime(time + L.startTime – .1);

(d[0] + d[1]);

Dan

[Dan Ebberts]

If you want the whole animation triggered at each marker (rather than centered at each marker), it would be more like this:


m = thisComp.marker;
t = 0;
if (numKeys > 1 && m.numKeys > 0){
d = key(2).time - key(1).time;
n = m.nearestKey(time).index;
if (time < m.key(n).time) n--;
if (n > 0){
tm = time - m.key(n).time;
if (tm < d){
t = linear(tm,0,d,key(1).time,key(2).time);
}else{
t = linear(tm,d,d*d,key(2).time,key(1).time);
}

}
}
valueAtTime(t)


Dan

[Dan Ebberts]

I’m not sure if this is what you’re after, but give it a try:


m = thisComp.marker;
t = 0;
if (numKeys > 1 && m.numKeys > 0){
d = key(2).time - key(1).time;
tm = m.nearestKey(time).time;
if (time < tm){
t = linear(time,tm-d,tm,key(1).time,key(2).time)
}else{
t = linear(time,tm,tm+d,key(2).time,key(1).time)
}
}
valueAtTime(t)


Dan

[Dan Ebberts]

I haven’t test this, but I think it would be:


if (marker.numKeys > 1){
t1 = marker.key(1).time;
t2 = marker.key(2).time
linear(time,t1,t2,0,t2).toFixed(0);
}else
0


Dan

[Dan Ebberts]

>reverse this so it is the other way round

What do you mean by that, exactly?

[Dan Ebberts]

No, like this:


C = comp("Scene 01");
ctrl = C.layer("controls").effect("Slider Control")("Slider");
L = C.layer(thisComp.name);
val = Math.round(ctrl.valueAtTime(time+L.startTime));
val == index ? 100 : 0


Dan

[Dan Ebberts]

These days, that last line should be modified:

val == index ? 100 : 0

Dan

[Dan Ebberts]

It’s tricky, but if your graphs are true functions of x (no instances of multiple y values for the same x) you can probably do it with an iterative solution like this simple binary search:


iterations = 12;
L = thisComp.layer("Path 1");
p = L.content("Shape 1").content("Path 1").path;
xc = thisComp.layer("Cursor").transform.position[0];
p0 = L.toComp(p.pointOnPath(0));
p1 = L.toComp(p.pointOnPath(1));
if (xc < p0[0]){
p0;
}else if (xc > p1[0]){
p1;
}else{
lower = 0;
mid = .5;
upper = 1.0;
for ( i = 0; i < iterations; i++){
pCur = L.toComp(p.pointOnPath(mid));
if (pCur[0] == xc) break;
if (pCur[0] > xc){
upper = mid;
}else{
lower = mid
}
mid = (upper+lower)/2;
}
[xc,pCur[1]]
}


Bumping the iterations value should make it more accurate at the expense of rendering performance.

Dan