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- Path: newsfeeds.ans.net!btco!newsadm
- From: Shalom Reich <sqr1874@acf4.nyu.edu>
- Newsgroups: comp.lang.c++
- Subject: Re: Fahrenheight to Celsius?
- Date: Thu, 29 Feb 1996 15:27:22 -0500
- Organization: Bankers Trust Company
- Message-ID: <31360C2A.6183@acf4.nyu.edu>
- References: <4fvthq$2bl@newserv.agcs.com> <4fra3q$ddt@newsbf02.news.aol.com> <1996Feb26.060213.23285@lafn.org> <4h1v08$eks@druid.borland.com> <xmsb.825531341@shadow>
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-
- Maurice S. Barnum wrote:
- >
- > pete@borland.com (Pete Becker) writes:
- >
- > >In article <1996Feb26.060213.23285@lafn.org>, an234@lafn.org says...
- > >>
- > >>F = C*(9/5) +32
- > >>C = (F-32)*(5/9)
- >
- > >He was better off without this answer, which is wrong.
- >
- > my favorite conversion equations, which i learned from
- > reading an article by isaac asimov, [snip]
- >
- > F = (C + 40) / 1.8 - 40
- > C = (F + 40) * 1.8 - 40
- >
-
- You have the Fs and Cs mixed up in your equations. Otherwise,
- they are exactly equivalent to the original formulas.
-
- Given F(32) = C(0) = freezing point of water,
- and F(212) = C(100) - boiling point of water
-
- then the range in F is 180 degrees and the range in C is 100 degrees.
-
- Since 180/100 = 1.8 = 9/5 and 100/180 = 5/9 you get the ratios
- between the two scales. All that is left is the scaling (i.e. F
- "begins" at 32 and C "begins" at 0 - at least as far as water boiling
- and freezing are concerned).
-
- If we take the two equations for C, we can show them equivalent.
-
- [1] C = (F-32)*(5/9)
-
- C = (F-32)/1.8
- C = (F/1.8) - (32/1.8)
-
- Similarly,
-
- [2] C = (F + 40) / 1.8 - 40 (F and C switched as in my comment above)
-
- C = (F/1.8) + (40/1.8) - 40
- C = (F/1.8) + (40/1.8) - (72/1.8) (convert 40 to (40*1.8)/1.8)
- C = (F/1.8) - (32/1.8)
-
- Which is exactly what the first equation was equal to.
-
- Checking the F equations is left as an excercise (sp?) for the reader.
-
- Shalom Reich
-