Projecting a point onto a path

Question

Suppose I have an ordered array containing points (lat, lon) describing a path, and I also have a point (lat, lon) describing my current location.

How can I project the point onto the path (and place the point in the appropriate place in the array)?

What I tried is just simply by searching for the nearest two points and assume it's in the middle of them. It's a good guess, but sometimes fails.

What would be a good way of doing this?

Answer 1

I see it like this:

• p0,p1 are path line segment endpoints
• p is your position
• q' closest point on line in 3D cartessian
• q is q' corrected by spherical projection

So:

1. convert points to 3D cartessian coordinates

2. compute perpendicular distance from point and line

   • q'=p0+(dot(p-p0,p1-p0)*(p1-p0)/(|p-p0|*|p1-p0|))
   • perpendicular_distance = |p-q'|

3. find segment with smallest perpendicular_distance

   and use only it for the rest of bullets

4. compute q

   If you use sphere instead of ellipsoid then you already know the radius if not then either compute the radius algebraically or use average:

   r=0.5*(|p0-(0,0,0)|+|p1-(0,0,0)|)
   

   assuming (0,0,0) is Earth's center. You can also be more precise if you weight by position:

   w=|q'-p0|/|p1-p0|
   r=(1-w)*|p0-(0,0,0)|+w*|p1-(0,0,0)|
   

   now just correct the position of q'

   q=q'*r/|q'|
   

   set vector q' as q with size r if it is not obvious enough. Also |p0-(0,0,0)|=|p0| obviously but I wanted to be sure you get how I got it ...

5. convert q from Cartesian to spherical coordinates

[Notes]

• |a| is size of vector a done like: |a|=sqrt(ax*ax+ay*ay+az*az)

• dot(a,b) is dot product of vectors a,b done like: dot(a,b)=(a.b)=ax*bx+ay*by+az*bz

  if your path is not too complex shaped then you can use binary search to find the closest segment. For distance comparison you do not need the sqrt ...

Answer 2

Find the path segment which is closest to the current position.

Distance from point to a line