How to get vector X distance from point on Y angle?

Questions & Answers
1

When I modded warcraft3 there was a function called (in jscript syntax) PolarOffset(v : Vector2, distance : float, angle : float) : Vector2

It would generate a point X distance and an Y angle from the input. No matter the angle, it would always be X distance away from the given point.

I've done google searches on it, but it's a wc3-only term so my luck ends there. Does anybody know the proper term for this?

SOLUTION

function PolarOffset(origin : Vector2, distance : float, angle : float) : Vector2{ //o = angle
    var x : float = origin.x + distance * Mathf.Cos(angle * Mathf.Deg2Rad);
    var y : float = origin.y + distance * Mathf.Sin(angle * Mathf.Deg2Rad);
 return Vector2(x,y);
}
2

Take a look at this. You can simply convert between polar and cartesian coordinates.

3 
function PolarOffset(origin : Vector2, distance : float, angle : float) : Vector2{ //o = angle
    var x : float = origin.x + distance * Mathf.Cos(angle * Mathf.Deg2Rad);
    var y : float = origin.y + distance * Mathf.Sin(angle * Mathf.Deg2Rad);
 return Vector2(x,y);
}
4

I do believe that logically, you're speaking about the Distance formula.

Distance between two points = squareroot( (x2-x1)(x2-x1) + (y2-y1)(y2-y1) + (z2-z1)*(z2-z1)). If you set a variable equal to that formula, where x1 is the center and x2 is the position your limiting from. You could then rotate x2 based on x1's coordinate system. I hope I am understanding you question.

5

Isn't what you mean simply the vector length of the radius of a circle? I have no idea what all this polar stuff is everyone seems to be mentioning. But if angle isn't part of the equasion, just ignore it right? Leaving behind one thing, a 2 dimensional line with a length. So why make it more complex than that, since the angle can't change the length