Curve fitting points in 3D space

Question

Trying to find functions that will assist us to draw a 3D line through a series of points.

For each point we know: Date&Time, Latitude, Longitude, Altitude, Spee

Answer 1

You can find an approximation of a line that intersects points in 3d and 2d space using a Hough Transformation algorithm. I am only familiar with it's uses in 2d however but it will still work for 3d spaces given that you know what kind of line you are looking for. There is a basic implementation description linked. You can Google for pre-mades and here is a link to a 2d C implementation CImg.

The algorithm process (roughly)... First you find equation of a line that you think will best approximate the shape of the line (in 2d parabolic, logarithmic, exponential, etc). You take that formula and solve for one of the parameters.

y = ax + b

becomes

b = y - ax

Next, for each point you are attempting to match, you plugin the points to the y and x values. With 3 points, you would have 3 separate functions of b with respect to a.

(2, 3)  : b =  3 -  2a
(4, 1)  : b =  1 -  4a
(10, -5): b = -5 - 10a

Next, the theory is that you find all possible lines which pass through each of the points, which is infinitely many for each individual point however when combined in an accumulator space only a few possible parameters best fit. In practice this is done by choosing a range space for the parameters (I chose -2 <= a <= 1, 1 <= b <= 6) and begin plugging in values for the variant parameter(s) and solving for the other. You tally up the number of intersections from each function in an accumulator. The points with the highest values give you your parameters.

Accumulator after processing b = 3 - 2a

    a: -2   -1    0    1
b: 1                   
   2                  
   3              1   
   4                  
   5         1        
   6                  

Accumulator after processing b = 1 - 4a

    a: -2   -1    0    1
b: 1              1     
   2                  
   3              1   
   4                  
   4                  
   5         2        
   6                  

Accumulator after processing b = -5 - 10a

    a: -2   -1    0    1
b: 1              1     
   2                  
   3              1   
   4                  
   5         3        
   6                  

The parameter set with the highest accumulated value is (b a) = (5 -1) and the function best fit to the points given is y = 5 - x.

Best of luck.