offset on local y-axis of auto positioned and rotated object between two points

[Steve Sierra]

Hi,

Here’s a way to do it :

– parent C to A.
– seperate position dimensions for all three objects.
– add a slider control to C.
– in C’s X Position, put this expression :

var a = (thisComp.layer(“B”).transform.xPosition-thisComp.layer(“A”).transform.xPosition);
var s = effect(“Slider Control”)(“Slider”);
a*s/100;

-in C’s Y Position, put this one :

var a = (thisComp.layer(“B”).transform.yPosition-thisComp.layer(“A”).transform.yPosition);
var s = effect(“Slider Control”)(“Slider”);
a*s/100

Does that help ?
Cheers !

[Lars Jandel]

Hi Steve,

thanks for your reply!
Your set up works great for the “x”-position between A and B. But I was aiming for the “y”-offset. ????
Maybe you have an idea for that as well. Plus I would prefer to have C not parented, but everything set via expressions. ATM for centering, I use in the position of C:

p1 = thisComp.layer(“A”).transform.position;
p2 = thisComp.layer(“B”).transform.position;
(p1+p2)/2

…yes, I will add a slider to that later on as well. ????
In the rotation of C, I use:

p1 = thisComp.layer(“A”).transform.position;
p2 = thisComp.layer(“B”).transform.position;
vec = p2 – p1;
radiansToDegrees(Math.atan2(vec[1], vec[0]))

Now, it would be fantastic to add a “y”-offset to the position via a slider, which works in a rotated mode as well.

Thanks, Lars.

[Steve Sierra]

Hi again,

With the same setup as my first post, you can do this :

-add a second slider to C.
– replace X expression :

var a = (thisComp.layer(“B”).transform.xPosition-thisComp.layer(“A”).transform.xPosition);
var s = effect(“Slider Control”)(“Slider”);
var s2 = effect(“Slider Control 2”)(“Slider”);
(a*s/100) + s2;

– replace Y expression :

var a = (thisComp.layer(“Null 2”).transform.yPosition-thisComp.layer(“Null 1”).transform.yPosition);
var s = effect(“Slider Control”)(“Slider”);
var s2 = effect(“Slider Control 2”)(“Slider”);
(a*s/100) + s2;

Does that help ?
😉

[Steve Sierra]

I just saw my last post doesn’t work… The offset needs to be rotated….

[Lars Jandel]

????
Was just about to answer…Sliding funny 😉

[Steve Sierra]

I get very funny slides mixing x and y….

here’s one I think works :

Same setup, but unliked C from A and added A pos to my expressions.

in X:
var a = (thisComp.layer(“B”).transform.xPosition-thisComp.layer(“A”).transform.xPosition);
var aY = (thisComp.layer(“B”).transform.yPosition-thisComp.layer(“A”).transform.yPosition);
var s = effect(“Slider Control”)(“Slider”);
var s2 = effect(“Slider Control 2”)(“Slider”);
(a*s/100) + thisComp.layer(“A”).transform.xPosition – aY*s2/100

in Y:
var a = (thisComp.layer(“B”).transform.yPosition-thisComp.layer(“A”).transform.yPosition);
var aX = (thisComp.layer(“B”).transform.xPosition-thisComp.layer(“A”).transform.xPosition);
var s = effect(“Slider Control”)(“Slider”);
var s2 = effect(“Slider Control 2”)(“Slider”);

a*s/100 + thisComp.layer(“A”).transform.yPosition + aX*s2/100

Any better ?

[Lars Jandel]

YES – it works!!!
Thanks so much Steve ☺
I was looking through the forums and trying stuff myself, but couldn’t figure it out…
And your solution is just as I like it, without limitations (using a linear expression) – perfect!

Thanks again, Lars. 🙂

[Lars Jandel]

As a follow up:
I will definitely use the solution – it does, what I was aiming for and is very dynamic.
Now I was just wondering, if it would be possible to achieve a fixed offset on the y-axis. Right now, the offset works in proportion to the distance between A and B: As they get closer the offset shrinks and vice versa. As said, it is nice and cool to bring all objects to the same point when compressing the distance.
As an alternative, it would be great to set a fixed offset in pixels via the slider. I played a bit with the expression, but as it is quite hard for me to bend my head around these vector(?) calculations, I did not succeed… ????
Don’t know, if this is possible..?

[Steve Sierra]

Hi Lars,

I can’t seem to get what you need…. not a vector pro either ????
I’ll continue playing with the expression when I have a bit of time.

Meanwhile, maybe one of the real expressioneers can help you out !

Good luck !

[Lars Jandel]

Thanks Steve!
The first solution is already great.
Yeah, maybe there is a completely different approach – let’s see, if so speaks up. 😉
Thanks again for your time and one working set up – will be used for sure!