Get closest Vector3 position from a GameObject and two Transforms (and the line inbetween them)

Question

Hi,

I need to get a Vector3 inside the range between two Transforms, having both Transform as the limit position regarding a third GameObject floating aroung the scene.

I drew an example hoping it would help.

alt text

All those Cubes would be the same GameObject and it's possible positions, I mean, it could be anywhere... but the closest V3 would be constrained inside the line between Transform 1 and Transform 2.

I'm working in C#, but if you have the answer in JS I don't care, just need the logic or initial direction in how to solve this problem.

Thanks in advanced!

Answer 1

Best Answer

Answer by dkjunior · Oct 16, 2015 at 01:11 AM

Basically what you need is to draw a perpendicular from a given point to a line. Here's a function to do that (first solution that came to my head, this can probably be done more elegantly).

 private Vector3 ClosestPoint(Vector3 limit1, Vector3 limit2, Vector3 point)
     {
         Vector3 lineVector = limit2 - limit1;
 
         float lineVectorSqrMag = lineVector.sqrMagnitude;
 
         // Trivial case where limit1 == limit2
         if(lineVectorSqrMag < 1e-3f)
             return limit1;
 
         float dotProduct = Vector3.Dot(lineVector, limit1 - point);
 
         float t = - dotProduct / lineVectorSqrMag;
 
         return limit1 + Mathf.Clamp01(t) * lineVector;
     }

The function takes two limiting points (as in your example), and the location of the floating object. It returns a point on the line between two limits that is closest to the floating object.

Add comment · Show 4 · Share

Smireles · Oct 16, 2015 at 02:45 AM 1

Share

Thank you very much. This function is returning exactly what I was looking for. Good thing my drawing didn't confused you!

I need to study your answer to understand what are you doing; but it does exactly what I was asking for. Again, thank you!

PS: What's up with that 1e-3f ?? I understand that is to check if there is no distance between limit1 and limit2, but is weird to see that. Is it the same to check for something like this:

if (Vector3.Distance(limit1, limit2) == 0.0F)

dkjunior Smireles · Oct 16, 2015 at 09:50 AM 0

Share

Glad it worked for you!

The 1e-3f trick is the practical approach for perfor$$anonymous$$g floating point comparison operations. The thing with floating-point operations is that they will lose precision as you work with them. For example, if you take a (0, 0, 0) point and perform a bunch of transforms on it (move/rotate a few times) so that it eventually returns to its original location, it's very unlikely you will get exactly (0, 0, 0). What you will likely get is something like (0.000001, -0.00000012, 0.0000005). For all practical reasons it's still zeroes, but if you perform a vanilla comparison with 0.0, it won't match and give you many headaches. So whenever you need to compare floating point values for equality it is recommended that you test not for an exact match but for them being within very small distance from each other.

Regarding the algorithm itself, it's basically using line equation in a parameterized form:

p = limit1 + (limit2 - limit1) * t

where each point p belonging to this line is represented by certain value of parameter t. It then tries to find such a value t so that the distance between the point on the line p (corresponding to that t) and your floating point is $$anonymous$$imal. Once found it puts t into line equation to get the point on the line you are looking for. Clamping of t is done so that the resulting point always stays between the limits.

Dave-Carlile dkjunior · Oct 16, 2015 at 12:19 PM 0

Share

Unity also has the $$anonymous$$athf.Approximately function that does this same thing in a slightly more readable way.

Show more comments

Answer 2

 // input values
 Transfrom a;
 Transform b;
 Transform[] gameobjects;
 
 // output values
 Transform closest = null;
 Vector3 pos;
 
 float minDist = float.PositiveInfinity;
 Vector3 posA = a.position;
 Vector3 posB = b.position;
 Vector3 d = (posB - posA).normalized;
 for(int i = 0; i < gameobjects.Length; i++)
 {
     Vector3 p = gameobjects[i].position;
     Vector3 v = p - posA;
     Vector3 point = posA + Vector.Project(v, d);
     float dist = (point - p).sqrMagnitude;
     if (dist < minDist)
     {
         minDist = dist;
         closest = gameobjects[i];
         pos = point;
     }
 }

Untested but should work.

edit
It's hard to tell from your drawing if an object that actually is close to the "ray" that your two transforms defines but it's projected point is outside the line segment what distance you want to calculate.

One way would be to define two Planes which specifiy the ends of the line:

 Plane p1 = new Plane(d,posA);
 Plane p2 = new Plane(-d,posB);

Inside the for loop you would do:

     Vector3 v = p - posA;
     Vector3 point = posA + Vector.Project(v, d);
     if (!p1.GetSide(p))
         point = posA;
     if (!p2.GetSide(p))
         point = posB;
     float dist = (point - p).sqrMagnitude;
     // ....

That way you get a capsule shape around your line.

Get closest Vector3 position from a GameObject and two Transforms (and the line inbetween them)

2 Replies

Your answer

Follow this Question

Related Questions