Dan Lewis <[EMAIL PROTECTED]> skrev den Thu, 5 May 2005 21:28:30 -0500:
On Thursday 05 May 2005 07:57 pm, Johnny Andersson wrote: [snip][snip]
I guess you already know that °C=(°F-32)/1.8 => °F=°C×1.8+32.
JohnnyPardon me for jumping back into an OT discussion. Actually there is an easier method of converting Celcius to Fahrenheit and back: three steps. 1) Add 40 2) Multiply by 9/5 to change from Celcius to Fahrenheit or by 5/9 to change from Fahrenheit to Celcius. 3) Subtract 40. Try it: it works. Anyone desiring a mathematical explanation reply directly to me, and I will send you a personal reply. (I learned this from a chemistry professor in the fall of 1960.)
Dan
Dan,
Why is 9/5 easier than 1.8? I thought it was the same thing... ;D
However 5/9 might be easier than 1/1.8 or 0.55555555...
Why is it easier to add 40 and then subtract 40 than it is to just add or subtract 32 once only?
My formula (I know that it's not MINE, but it got to be called something, doesn't it..?):
°F = °C×1.8 + 32
Your method, converted to a simple formula:
°F = (°C + 40)×9/5 - 40 => °F = 9/5×°C + 9/5×40 - 40 => °F = 9/5×°C + 40(9/5 - 1) =>
=> °F = 9/5×°C + 40×4/5 => °F=9/5×°C + 32 => °F = °C×1.8 + 32 = My formula.
Conversion the other way: My formula: °C = (°F - 32)/1.8
Your formula:
°C = (°F + 40)×5/9 - 40 => °C = 5/9×°F + 40×5/9 - 40 => °C = °F×5/9 - (40 - 40×5/9) =>
=> °C = °F×5/9 - 40(1 - 5/9) => °C = °F×5/9 - 40×4/9 => °C = °F×5/9 - 32×5/4×4/9 =>
°C = °F×5/9 - 32×5/9 => °C = °F/(9/5) - 32/(9/5) => °C = °F/1.8 - 32/1.8 =>
=> °C = (°F - 32)/1.8 = My formula.
Yes, you're right, it works, but is it easier? Maybe it's just a matter of personal taste or preference.
Johnny
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