CODEWITHSUNDEEP
javamathlibgdx
Aeglasin
Aeglasin

Reputation: 165

Calculation position of a Vector between two others

I think my current problem is related to intercept theorems that everybody learned at school, but that was quite some time ago and I am kind of getting stuck halfway.

My Problem:

I have two Vectors in a 2 dimensional space called v1 and v2.
What I want to know are the coordinates of a vector v3 which is on the line segement between v1 and v2.
What I know about v3 is the length of the line segement between v1 and v3 and v2 and v3. I my thoughs are always circling about the using the slope between v1 and v2 to determine the coordinates of v3 but at this point I'm getting stuck.

This the actual code I used for the issue:

private Vector2 getPointOnLine(Vector2 v1, Vector2 v2, float distance) throws Exception
{
    double lengthOfSegment=Math.floor(Math.pow(v2.x - v1.x, 2)+Math.pow(v2.y-v1.y,2));

    Vector2 slopeXY=v2.sub(v1);
    float slope=1/slopeXY.x;
    slopeXY.x=slopeXY.x*slope;
    slopeXY.y=slopeXY.y*slope;
}

The actuell problem is game related, I want to let an allied unit (v2) follow the player(v1), but always keep a specific distance (distance).

I'm using LibGDX, are there by chance any library functions that can do this work for me?

Upvotes: 2

Views: 1178

Answers (2)

Edward Doolittle
Edward Doolittle

Reputation: 4100

All you need is the ratio of the distances d(v1,v3) and d(v1,v2) = d(v1,v3) + d(v3,v2). Let

r = d(v1,v3)/(d(v1,v3) + d(v3,v2))

Then 

v3 = v1 + r(v2-v1)

is the answer to your question.

Proof: If d(v1,v3)=0, r=0 and v3=v1. If d(v3,v2)=0 then r=1 and v3 = v1+v2-v1 = v2. The formula for v3 is linear, and it passes through both v1 and v2, so v3 is always on the line between those vectors. Finally, norm(v3-v1) = r * norm(v2-v1) = d(v1,v3) and a similar calculation for norm(v2-v3).

Upvotes: 0

Straw1239
Straw1239

Reputation: 689

I think you are trying to interpolate and get a vector which is a 'mix' of v1 and v2. This will be on the line between them. (Assuming v1 and v2 are treated as position vectors) You can do this by taking a weighted average of the vectors, but if the distance is given in absolute terms rather than relative, then you will need to convert:

Actually, it turns out libGDX vectors can do this for you!:

private Vector2 getPointOnLine(Vector2 v1, Vector2 v2, float distance) 
{
    double length = Math.hypot(v1.x - v2.x, v1.y - v2.y);
    distance /= length;
    return v1.cpy().lerp(v2, distance);

}

Vector2.lerp is a linear interpolation method provided by libGDX

Upvotes: 1

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