[Xavier Gomez]

You can use the linear function, this way:

ypos=thisComp.layer("A").transform.position[1];
x = linear(ypos, 600, 300, 20, 100);
[x,x];

Xavier

[Xavier Gomez]

I tried your code and it works.
I get rotations values 0,1.5, 3, 4.5, 6, etc

Another possibility is to defined rotation based on first in the stack:

leader = thisComp.layer(“LEADER”);
leader.rotation + (index-leader.index)*1.5;

Xavier.

[Xavier Gomez]

What is wrong with the third example (‘A’, ‘B’, ‘C’) ?
Can you post the error, if any, or tell which misbehaviour you observe ?

[Xavier Gomez]

If you are on a Text layer or Shape layer, you can use the sourceRectAtTime method, which is way easier to manipulate and also faster.

For a layer with source (like a solid with a mask etc), to find the left most pixel, you should scan the layer with a verical band that is as high as the layer (ie radius[1] = layer.height). Otherwise with a radius of 0.5 you will need ages to finish.

Method 1 is to roll that a thin vertical band by increments of 1 until you hit something.

Method 2 is to start with a sample as big as the layer (so it is not really vertical).
If you hit, then:

step0 : you divide the width (radius x) by 2,
step1 : sample the left part of the image.
step2 : if you hit, go to step 0 again, with left part of the image,
else go to step 2, but with the right part.
And so forth until the radius x becomes smaller than 0.5 (or more if you dont need pixel accuracy);

[Xavier Gomez]

Here is a possibility, not optimal since it calculates the whole every frame.
If your keys are not far apart and the param averageChangesPerSec isnt too big, it should be fine

startVal = 0;
endVal = 100; // max copies
t1 = numKeys>0 ? key(1).time : inPoint;
t2 = numKeys>1 ? key(2).time : outPoint;
averageChangesPerSec = 1.5;
seed1 = index+1000;
seed2 = index+2000;
a = 3; // controls time variations (close to 1 = small, bigger=more disparity)
b = 10; // controls value variations (same)

if (time<=t1){startVal;}
else if (t2<=time) {endVal;}
else{
numSteps =Math.ceil((t2-t1)*averageChangesPerSec);
t=0; times = [0];
v=0; values = [0];
for (n=1; n<=numSteps; n++){seedRandom(t+seed1, true);t+=random(1,a); times[n]=t; seedRandom(t+seed2, true);v+=random(1,b); values[n]=v;};

t = (time-t1)/(t2-t1) * t;
x = (endVal-startVal) / v;

a=0; b=numSteps; while(b-a>1){j=Math.floor((a+b)*0.5); if (times[j]<=t){a=j;} else{b=j;};};

startVal + x*linear(t, times[a], times[b], values[a], values[b]);
};

Xavier

[Xavier Gomez]

This should work:

startVal = 0;
endVal = 100; // max copies
DTmin = 0.2; // DTmin/max : time interval min/max between changes
DTmax = 0.8;
DNmin = 2; // DNmin/max : change amount min/max
DNmax = 5;

s=t=inPoint;
m=n=startVal;
while (n

Xavier

[Xavier Gomez]

Yes, if the A,B,C points are somewhere deep inside a shape layer and the point M is sommewhere else, you are stuck as i think it is not possible to convert A, B, C in that case. At least i dont know how to do it.
It should be fine otherwise.

Side note: i just noticed that the variable called BC is not even used, so you can erase the corresponding line even in the “slimmed” version.

Xavier

[Xavier Gomez]

I can’t tell.
The expression i gave assumes that all point coordinates A, B, C and M are in the same system. Depending on your set up you might have to use one of the space transform functions (toComp/fromComp etc).

Also, that expression is unnecessarily heavy. If you have lots of points you can change it to this one, which does the same thing without extra stuff:

A = thisComp.layer("A").transform.position.value;
B = thisComp.layer("B").transform.position.value;
C = thisComp.layer("C").transform.position.value;
M = thisComp.layer("Point").transform.position.value;
AB = sub(B,A);
BC = sub(C,B);
CA = sub(A,C);
AM = sub(M,A);
u = AB[0]*CA[1]-AB[1]*CA[0];

c = 100 * (AM[0]*AB[1]-AM[1]*AB[0]) / u;
b = 100 * (AM[0]*CA[1]-AM[1]*CA[0]) / u;
a = 100-(b+c);

letters= ["a", "b", "c"];
values = [a,b,c];
for (k=0; k<3; k++) letters[k] += " : " + (values[k]<10 ? " " : "") + values[k].toFixed(1) + "%";
letters.join("\n");

Xavier

[Xavier Gomez]

Have you renamed something after that ?

It is kind of weird that your expression uses single quotes, as normally AE uses double ones.

One thing to know is that AE recognises single quotes, but, when renaming layers/properties, comps etc, the expression engine only converts expressions that use double quotes.

[Xavier Gomez]

The expression below is for a text layer (display all 3 coordinates in 1 single text) :

A = thisComp.layer("A").transform.position.value;
B = thisComp.layer("B").transform.position.value;
C = thisComp.layer("C").transform.position.value;
M = thisComp.layer("Point").transform.position.value;
AB = sub(B,A); if (AB.length<3) AB.push(0);
BC = sub(C,B); if (BC.length<3) BC.push(0);
CA = sub(A,C); if (CA.length<3) CA.push(0);
AM = sub(M,A); if (AM.length<3) AM.push(0);
u = cross(AB, CA);
U = dot(u,u);

c = dot(u, cross(AM, AB))/U*100;
b = dot(u, cross(AM, CA))/U*100;
a = 100-(b+c);

letters= ["a", "b", "c"];
values = [a,b,c];
for (k=0; k<3; k++) letters[k] += " : " + (values[k]<10 ? " " : "") + values[k].toFixed(1) + "%";
letters.join("\n");

It’s not optimal as it does unnecessary calculations, but trying to optimize would make it less compact.

Xavier