Re: [Plib-users] <<Code review>> ssgEntity::isectTest ( sgSphere *s, sgMat4 m, int testNeeded )

[Steve B. <sjb...@ai...>]

Rahul Choudhury wrote:

> sgVec3 center_vec ;
> float sum_radii = s->getRadius() + tmp.getRadius() ;
> sgSubVec3 ( center_vec, s->getCenter(),
> tmp.getCenter() ) ;
> 
> if ( center_vec[0] > sum_radii ||
>    center_vec[1] > sum_radii ||
>    center_vec[2] > sum_radii ) {
>         return SSG_OUTSIDE ;
>    }
> 
> I think the last line should be modified to
> take absolute values of center_vec[0],
> center_vec[1] and center_vec[2]

Yep - I think you are right about that one.
 
> Next thing ssg code does is taking squares
> of each of the radius, adding them and
> compares them with the square of the
> distances between the centers
> 
> Or in other words, it says
> if d^2 >= radiousOne^2 + radiousTwo^2
> then the given sphere lies outside the
> entity. (here d is the distance
> between the two centers of spheres,
> radiousOne is the radius of the
> entity sphere and radiousTwo is the
> radius of second sphere).
> 
> In stead I think we should say
> if d^2 >= (radiousOne + radiousTwo)^2
> or d >= (radiousOne + radiousTwo)
> then the given sphere lies outside the
> entity.

I think I could save time by saying:

  d^2 >= (r1+r2)^2

...but in order to use:

  d >= r1+r2

...I'd have to compute d using a sqrt(). Since
a sqrt is MUCH slower than a multiply (or even the
TWO multiplies I actually did) - it's more efficient
to compare the squares of the ranges than the ranges
themselves.

However, if I compute (r1+r2)^2, and that test fails
then I'd have to compute r1^2 and r2^2 later.  It's
a fine line as to which would be more efficient - and
it depends mainly on the statistics of sphere-to-sphere
contacts.

Good catch on the first one though - the code would
never have broken - but it will be faster with your
change.

-- 
Steve Baker                  http://web2.airmail.net/sjbaker1
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