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From: woodhams@phoenix.Princeton.EDU (Michael Woodhams)
Newsgroups: comp.sys.handhelds
Subject: Re: HP 48 Yards.FeetInches
Message-ID: <10737@idunno.Princeton.EDU>
Date: 13 Jun 91 23:56:15 GMT
References: <2855ad7d:3432comp.sys.handhelds@hpcvbbs.UUCP>
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In article <2855ad7d:3432comp.sys.handhelds@hpcvbbs.UUCP> akcs.joehorn@hpcvbbs.UUCP (Joseph K. Horn) writes:
>Any mathematicians out there?  I need your help.
>
>All the code I've seen written to handle Feet.Inches pulls the number
>apart into its int and frac parts, handles them, and then recombines.
>
>But some years ago, this algorithm for Hours.MinSec was published:
>
>              +-----------------------------------------+
>              | HMS->(x) = x + FP(x)/1.5 + FP(x*100)/90 |
>              +-----------------------------------------+
>
>The fact that it works mystifies me.  My question is: If this kind of
>algorithm works for H.MS, can similar ones be created for any arbitrary
>fractional systems, like Yards.FeetInches?  If so, how?

HMS->(H.MMSS) = H.MMSS + 0.MMSS/1.5 + 0.SS/90
              = H + MM/100 + SS/10000 + MM/150 + SS/15000 + SS/9000
              = H + MM(1/100+1/150) + SS(1/10000+1/15000+1/9000)
              = H + MM/60 + SS/3600 as required.

The solution to the yards, feet, inches problem is obvious: use metrics.

Let YFI->(Y.FII) convert yards.feet_inches to decimal yards, and
assume it is of the form

YFI->(x) = x + FP(x)/a + FP(x*10)/b

and solve for a and b:

YFI->(Y.FII) = Y.FII + 0.FII/a + 0.II/b
             = Y + F/10 + II/1000 + F/(10*a) + II/(1000*a) + II/(100*b)
             = Y + F*(a+1)/(10*a) + II*(a*b+b+10*a)/(1000*a*b)
             = Y + F/3 + II/36
so (a+1)/(10*a) = 1/3     =>   a=3/7
1/36 = (a*b+b+10*a)/(1000*a*b)
     = (3*b/7+b+30/7)/(3000*b/7)
     = (3*b+7*b+30)/(3000*b)
     = (b+3)/(300*b) => b=9/22
so 
YFI->(x) = x + FP(x)*7/3 + FP(x*10)*22/9

The reverse transformation is left as an exercise for the student.