Get direction angle after bounce?

Atlanten

I have some actors moving around randomly, bouncing into each other.

Every time they bounce into each other, they change direction.

How do I get to that direction angle, so that I can have the actors pointing forward at all times?

I would imagine it could be calculated using the X and Y velocities, but that's beyond me.

Any other clever (and easy) way to do it?

CodeMonkey

Fixed rotation might be easier. It's in the physics attributes of the actor.

sdpardue

I haven't actually tried this, but here's how I'd basically approach the problem:

Create attributes to hold an X,Y location pair, and to hold an angle:
XOrigin, YOrigin, ActorAngle
Create a rule in which: when your actor collides with another specified actor, save its location as the origin point:
Change Attribute XOrigin = self.Position.X
Change Attribute YOrigin = self.Position.Y
Set a Timer such that after .5 seconds (or whatever amount gives the actor time to move enough to its new position -- might need to experiment with this value), you calculate the angular direction of the actor by using the vectorToAngle function:
Change Attribute ActorAngle = vectorToAngle(self.Position.X-XOrigin, self.Position.Y-YOrigin)
Then, apply this angle to rotate your Actor "forward".
-end of the rule-

There may be a more elegant solution, but this is at least how I'd start out trying to solve it.

Atlanten

Thanks !! I haven't used VectorToAngle before.

Using short timers, the result is always a little jerky or flickering though.

I also tried constraining the rotation to the VectorToAngle output, looks really cool because then the actors actually turn into position. But I'm getting a bit of jitter, probably where it crosses from zero to 360 degrees... Must experiment more...

I have this vague idea that there could be a way of using the X and Y linear velocities... You know, say

velocity.x=0 and velocity.y=100 would mean 90°

x=100 and y=100 would mean 45°

x=-100 and y=0 would mean 180° and so on.

But I don't know what such a relationship would actually look like.