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

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);
}
```

**Answer** by AliAzin
·
Sep 25, 2010 at 02:50 PM

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

function $$anonymous$$rOffset(origin : Vector2, distance : float, angle : float) : Vector2{ //o = angle var x : float = origin.x + distance * $$anonymous$$athf.Cos(angle * $$anonymous$$athf.Deg2Rad); var y : float = origin.y + distance * $$anonymous$$athf.Sin(angle * $$anonymous$$athf.Deg2Rad); return Vector2(x,y); } Angle : 0-360

**Answer** by windexglow
·
Sep 25, 2010 at 09:57 PM

```
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);
}
```

**Answer** by SirGive
·
Sep 27, 2010 at 09:32 AM

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.

**Answer** by Proclyon
·
Nov 16, 2010 at 01:41 PM

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

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