Re: [Algorithms] 3d Lines Intersection

[<ro...@do...>]

Jim Offerman  wrote:

>> Is there a good algo for finding a point of intersection between two *3D
>lines*?
>> The lines is specified by their point (x,y,z) and their direction
>(dx,dy,dz)
>
>When I wrote my first renderer, I used a simple 3D intersection algorithm in
>my (very alternative) projection system:
>
>- calculate (x, y) intersection (result <x1, y>)
>- calculate (x, z) intersection (result <x2, z>)
>- if (x1 == x2) the lines intersect at <x1, y, z>
>
>Of the top of my head I can't say whether this algorithm is 100% correct

It is my sad duty to have to inform you (and I dearly hope that the
news causes no undue emotional stress), that this putative algorithm
is wrong.  

Here is the clearest, simplest to understand counterexample that I
could construct (and there are clearly a bazillion others).

Let line 1 be the x axis, so the line with parametric representation
(s, 0, 0)   

Let line 2 be the line with parametric representation (0, t, t+1).
Thus line 2 lies in the yz plane where it has the implicit equation 
z = y+1.

The projection of line 1 on the xy plane is just the x axis.
The projection of line 2 on the xy plane is the y axis,
The intersection of these projections is (0,0,0).

The projection of line1 on the xz plane is the x axis
The projection of line 2 on the xz plane is the z axis
The intersection of these projections is (0,0,0)

So in your notation  x1 = 0 = x2,  and by your algorithm you claim
that the lines intersect at (0,0,0).  In fact they do not.  Line 2
does not even pass through the point (0,0,0),  In fact these two lines
do not intersect at all, ever, nowhere, no way.  In fact, it is a
problem in freshman analytic geometry to show that the distance of
closest approach of these two lines is sqrt(1/2), but I could as
easily have made it a billion units. 

> (I
>am not a math guy...),

Which is no license to post wrong stuff..

> but, if properly implemented, it will certainly be
>fast and provide at least a reasonable estimation.
>

Bah.

I'm growing weary of playing math cop.  Soon gonna turn in my badge.