Date: Mon, 20 Apr 1998 12:09:07 -0500 (CDT) From: Paul Burkhardt <burkhard@aries.scs.uiuc.edu> To: cp@opus.hpl.hp.com Message-Id: <aabcdefg1325$foo@default> Subject: Re: Celcius to Farenheight and vice versa
Hi Jose,
Please allow me to provide perhaps a more intuitive method, which actually
derives the temperature conversion formula and doesn't require any
experimentation. :)
All that you must remember, is the freezing and boiling point range in
both Celsius and Fahrenheit.
32-212F vs 0-100C
>From this range, you can see that F changes linearly with C, because each
increment in F corresponds to a constant multiple increment in C. In other
words, one degree increase in C, corresponds to a 9/5 degree increment in
F. This is easily seen by taking ratios of different temperature ranges.
Full: 180/100 = 9/5
Half: 90/50 = 9/5
Quarter: 45/25 = 9/5
Right away, you see that there is a linear relationship, as denoted by the
constant multiple, 9/5. We then recall our formula for a straight line:
y = mx + b, where m is the slope of the line and b is a constant.
Changing some labels:
F = mC + b
The slope is that constant increment multiple that was noted earlier. We
can then get the constant, b, by using the boundary conditions. At C=0,
the constant must be 32 to correspond with the freezing point in F. This
constant holds for all other values of C. You can then manipulate the
formula to get the C from F conversion.
F = 9/5C + 32 or C = 5/9F - 32
Paul Burkhardt
> I've got lots of feedback on my question and now I feel it's my turn.
> Several people who wrote to me didn't know the formula of conversion
> from Farenheight to Celcius and vice versa. It's not a very "nice"
<snip>
> 3.- Feel free to distribute this message. I would feel very honored
> if it is incorporated into any of the CP FAQ or WEB pages.
>
> I sincerely hope this helps.
>
> Best regards,
>
> ()()()()()
> ()* || () Jose Gengler
> ()=3D []=3D()
> () || () jose.gengler@usa.net
> ()()()()()
>
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