Finding offset position relative to actor angle

[Cherno]

My math skills are limited, so I have come here for help

A want to find a Vector3 that is offset from an actor's position relative to the actor's angle. So, maybe 16 units to the front and 32 units to the right.

I figured out how to find the x offset (forwards/back), but I don't know about the right-left offset maths.

Code: Select all

int offsetX = 16;//forward-backward
int offsetY = 32;//right-left
Vector3 offsetPos = (pos.x + cos(angle) * offsetX, pos.y + sin(angle) * offsetX, pos.z);
			

[KeksDose]

You can easily do the same thing you did forwards a second time to go sideways by adding or subtracting 90° from your angle:

Code: Select all

vector3 forwards  = (cos(angle),      sin(angle),      0);
vector3 sideways  = (cos(angle - 90), sin(angle - 90), 0);
vector3 offsetPos = pos + offsetX * forwards + offsetY * sideways;

When you get used to that, you can also use that cos(x + 90) = -sin(x) and sin(x + 90) = cos(x) to find out that you can write sideways in terms of forwards like this:

Code: Select all

vector3 sideways = (forwards.y, -forwards.x, 0);

I like to think of forwards and sideways as local x and y axes, with offsetX and offsetY being the coordinates there.

[Cherno]

Thank you very much.