[Tiago Cav]

I understand, Good observation, HSL can be better for this.

Now that You taught me the base to work with colors, I can develop over that.

You’ve helped enough, thanks again.

[Tiago Cav]

I got some interesting results with an expression that I created rapidly using the color red. I could go on, but it would be a very complex work, and results might be satisfactory, however, I would find many problems because of having to combine the three color channels extensively for simple colors.

Just one more thing:

When you asked me about the color be exactly the same; this suggests that After Effects might make an mode recognize colors within its range where a non-accurate color could be considered the same (of the same family)?

Thanks.

// TEST RED COLOR

c = effect("Color Control 2")("Color")
if ((clamp(c[0]*2, 0,2) >1 && clamp((value[0]*2),0,2) >1) || (clamp(c[0]*2, 0,2) <1 && clamp((value[0]*2),0,2) <1))
[0,0,0,1]
else
value

[Tiago Cav]

(Updated)
One more thing;
I need to propose something more daring.
Could determine half range of the same color (such as all Red variations, for example), for all #seven colors (Red; Violet; Blue; Cyan; Green; Yellow; Orange)?

In other words,
a division between Light and Dark in each color separately.

If “Color Control 1” is a color, (Red for example), Dark Red; #cc0000 rgb(204, 0, 0)
and
“Color Control 2” is a variation of Dark Red; #990000 rgb(153, 0, 0)

“Color Control 1” = black; “Color Control 2” keep color.

Another condition which result is the same:
If “Color Control 1” is Light Red; #ff8080 rgb(255, 128, 128)
and
“Color Control 2” is a variation of Light Red; #ff4d4d rgb(255, 77, 77)

See example (Updated).jpg

And:
If “Color Control 1” is dark and the “Color Control 2” is light (or inverted) variation of the same color (which can be #seven colors), keep color in “Color Control 1” and in “Color Control 2”.

Thanks. I want to know, at least, if all this is possible.

[Tiago Cav]

Yes Dan, exactly equal.

YES! Work Perfect; Thank you!!! 🙂

[Tiago Cav]

Yes! Very good! I understand. I make some tests.

I realized that actually reading is what matters most in rendering. After all, countless paintings will be read, among other data, for only then be summarized in a few MB recording, could be higher if you render in lossless then we’d GB of data exported, but still, I think the storage device does more reading than recording.

I did not see much difference if the footage is that Disk (SSD) or another (HD), as you said, the SSD has a technology that can read randomly, that’s enough. And, is RAM memory responsible for loading the Footage, from what I saw the use of discs: in use with After Effects (SSD) is 0%, and the disk where was Footage (HD) is also 0%, since when importing this Footage, and when I preview, just see the Ram memory being full.

Thanks!

[Tiago Cav]

I have a SSD with system Windows 10 installed, and After Effects CC 2015.

[Tiago Cav]

Hey Dan, you have a good intuition.
Same not knowing exactly what I wanted, by me not knowing express myself right, managed to give me the road that would lead to the right.
So I was able.

Thank You Very Much! Always very efficient.

// This is the main part of where were the problem in expression.
var a="Scale";
var b="Position";
var c="Side";
var names=[a, b, c];
for (var i=0; i<names.length; i++){
try{
var e = t.effect(names[i])("Layer").name;
}catch(err){
var e = 0;
}
}

[Tiago Cav]

Genial.
Thank you, Dan!

[Tiago Cav]

Example:
var myComp = app.project.activeItem;
var myLayer = myComp.layer("Footage").source.layer("Video");
myComp.duration = (myLayer.outPoint - myLayer.inPoint);

[Tiago Cav]

Just a more thing, please.

I want to to access a layer (Footage) which is inside of Comp (A) and Comp (A) is within Main Comp (app.project.activeItem;)

That is, use a layer which is inside of a Comp, that is into other Comp, in Script.

I only got it, through Project Panel with: app.project.item;
And cannot by Comp Selected: app.project.activeItem;

Thanks.