RE: Programming question

Thanks for all the replies.
These posts have got some of the wheels turning in my brain.
Just a few comments.

Previously I was think of using a "reverse" lookup table like Zonn
suggested, and I might go that route now after he suggested doing a binary
search on it. (binary search had slipped my mind and I was planning on
starting at one end and just moving through the table, which would have
averaged 128 loops) I'm actually currently doing lots of calculations with
data from lookup tables. I.E.- collision detection.

I had considered using gravity in something like 16 steps, but I still need
the straight line distance to determine which ring I'm in. (as well as to
normalize the vectors)

The problem with doing equations like this
         if (distanceX < 10)
                scaleX = 1;
        if (distanceX < 20)
                scaleX = 2;
        if (distanceX < 30)
                scaleX = 4;
                ...etc...
        speed.x += gravity_constant/scaleX;
is that it doesn't take into consideration the actual distance from the sun.
You could be 500 y units away and 10 x units and it will end up pulling
harder in the x direction. Both vectors must first be normalized to properly
divide the gravitational pull between them.

Rodger, I also couldn't get this to properly work, but if you do find the
proper equation, it might be worth a try.
Take the smaller of xlength or ylength, multiply it time 3, divide
by 2, and add the result to the larger of xlength or ylength. The
result is a close approximation of the square root of the sum of
the squares.

I have also toyed with the idea of doing linear gravity, I'm just afraid
that you won't be able to orbit properly. (I get upset in gravitar because
you can't quite orbit the planet that you can fly in circles around)

Clay, I threw a few numbers into these equations
"real dist" = sqrt(x^2+y^2), "fake dist" = ((X*2)+(Y*2))/4
and in a circular obit, the gravity gets stronger when you have only an x or
y offset and weaker when your at one of the 45 degree angles from the sun.
(making the gravity rings a square turned 45 degrees). But this might be ok
consider how little time it takes to calculate.

thanks
-jeff
Received on Thu Oct 28 16:15:35 1999