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- Path: druid.borland.com!usenet
- From: pete@borland.com (Pete Becker)
- Newsgroups: comp.lang.c++
- Subject: Re: Fahrenheight to Celsius?
- Date: 28 Feb 1996 20:15:32 GMT
- Organization: Borland International
- Message-ID: <4h2d54$m70@druid.borland.com>
- 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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- In article <xmsb.825531341@shadow>, xmsb@shadow.borland.com says...
- >
- >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, is not only simpler,
- > but avoids having to worry as much about the "obvious"
- > problem when written as C++:
- >
- >F = (C + 40) / 1.8 - 40
- >C = (F + 40) * 1.8 - 40
- >
- > for reasons i've never understood, these are not the
- > equations one will find in most books.
-
- I think these are much more elegant and easier to remember. The drawback that
- I see is that they don't fit the physical analog of what's going on. It's easy
- to picture two thermometers side by side and see that you have to adjust the
- values to map between 0-100 and 32-212. It's not at all clear that that's
- what's happening with this form.
- -- Pete
-
-