Offset animation, then loopOut.

[Yoan Boisjoli]

Hi Griffin! Is the property only one value or an array of 2 or more?

[Dan Ebberts]

I’m not real clear on exactly what you’re asking, but it might end up being something like this:

tOffset = effect("Slider Control")("Slider");
t1 = key(1).time;
t2 = key(2).time;
dur = t2 - t1;
t = (time - (t1 + tOffset))%dur;
valueAtTime(t1+t)

[Griffin Englander]

Thanks guys.

Hi Yoan! It’s a value of 1 in this case mate.

Dan, I’ll try and explain a little better.

I have a tall strip of numbers pre-comped (think an odometer), and this precomp has 3 keyframes on its Y position. Say two keyframes to go from 0 – 100, then a third keyframe to dictate the pause before loopOut(“offset”) comes in and exponentially cycles the animation.

What I’m hoping for, is to use another slider to offset the point at which that whole animation starts.

Sorry if that’s even less clear, but thank you for helping.

[Yoan Boisjoli]

Alright I got something like that to work. I’m animating the slider with hold keyframes but feel free to try something else. I’m joining a screencast of how it’s behaving and my ae file also.

// Time Offset from Slider
var offset = effect("Slider Control")("Slider")/10; //dividing it by 10 becasue it's sensitive
// Total duration of the loop (from first to last keyframe)
var firstKeyTime = key(1).time;
var lastKeyTime = key(numKeys).time;
var loopDuration = lastKeyTime - firstKeyTime;
// Current Time adjusted by the offset
var adjustedTime = time - offset;
// Calculate the looped time
var loopedTime = (adjustedTime - firstKeyTime) % loopDuration + firstKeyTime;
// Ensure loopedTime doesn't go below the first keyframe time
loopedTime = (loopedTime < firstKeyTime) ? firstKeyTime : loopedTime;
// value at the looped time
valueAtTime(loopedTime);

loopOdometer.mp4 638 KB MPEG-4 - Click to view

• Copy Download Link

Loopest.aep 108 KB After Effects Project Files - Click to view

• Copy Download Link

[Griffin Englander]

Thank you, Yoan. Legend

I can make that work.

[Brie Clayton]

Thank you for solving this, Yoan!