Integer-only math

arnold113

Is there a work around for this equation? (x = diameter / square root of 2). I've got a on going project
that I need to be able to obtain the largest square contained inside a circle. I'm using the BS2P40 with v2.5
editor. I need to be able to start with a dia. of say 15.3" and use 1.414 as the divisor (square root of 2).
I have built a portable band saw sawmill and I'm using the BS2p40 to control operation of the sawmill. I need of find the largest square (cant) that can be sawed from a log. Any help will be appreciated.
Thanks, Arnold

Bean

The basic idea is to scale everything up by some constant.
You want the constant to be as large as possible without overflowing (+/- 32767).
So you might use a constant of 1,000. That would allow a log with a 32" diameter.

Then instead of dividing by root 2, multiply by 1/root 2 and use the ** operator.
The ** operator multiplies two values and returns the upper 16 bits of the result.
So it is like A**B = ((A*B) / 65536) it is okay if A*B is larger than 32767 because it is held as 32 bits internally.

So 1/root 2 = 0.70710678 we multiply that by 65536 for the ** operator to get 46341 (rounded to closest int).

So the equation becomes x = 15300 ** 46341 which would make x equal to 10818 (or 10.818") and is calculated as ((15300 * 46341) / 65536).

I hope this helps,

Bean

arnold113

Thanks, Bean. That was just what I needed. Now I can get on to the next problem, what ever it turns out to be.
Thanks again.
Arnold