🚀 go-pugleaf

>so if we denote the wind-chill index by te (for temp. equiv.), we
>merely need to solve
>
>(a+b*sqrt(v)-c*v)*(s-t)=(a+b*sqrt(4)-4*c)*(s-te)
>
>for te.
>
>have fun,  george tintera



my apologies, George.  You are absolutely right.  I did the
algebra wrong.

here's the result:

sk:/home/alpha/sysop/eab/junk 112 % chill -10 10
The wind chill is -33.41 for a temp of -10.00 and a wind of 10.00 mph.




Here's my c program:

#include    <math.h>
#include    <stdlib.h>

main(argc,argv)
int     argc;
char    *argv[];
{
double  temp,wind,chill;
if (argc!=3) {
    printf("Usage: %s temp wind\n",argv[0]);
    exit(-1);
    }
temp=atof(argv[1]);
wind=atof(argv[2]);
chill=48.07 + .4741 * temp + 1.855 * wind - .0203 * temp * wind
      - 27.74 * sqrt(wind) + .3035 * temp * sqrt(wind);
printf("The wind chill is %.2f for a temp of %.2f and a wind of %.2f mph.\n",chi
ll,temp,wind);
}

Edward A. Bertsch (eab@msc.edu) Minnesota Supercomputer Center, Inc.
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