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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

My Dad hates math, but he asked what is Kelvin.

So I got into a conversation about all three systems,

and how centigrade (celsius) has degrees the same

size as Kelvin except Kelvin starts way down 273.16 degrees

lower at the coldest possible temperature. And I told him

that Farenheit degrees are smaller than Celsius degrees and

gave him some examples.

He said there are formulas for the conversion, right?

And I said, yes, they involve the numbers 5, 9, and 32,

and told him I just figure it out each time I convert.

Then I got to thinking. You could use the numbers 5, 9

and 17 and 7/9ths just as well since 0°F is -17.7777°C.

And then I thought, why not go from any number at all

and just compute the difference from that point in both

systems. So another good one to figure the distance

from is -40°F, which is also -40°C.

**igloo** **myrtilles** **fourmis**

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**ryos****Member**- Registered: 2005-08-04
- Posts: 394

This is what happens when you invent a temperature scale before you understand what heat is.

As for your question, I'm not entirely sure I understand. The formula is to multiply by 5/9 and add 32, or 9/5 and subtract, something like that (I can never remember either ). I'm not sure why it's so hoaky, but if there were an easier way we'd all be using it. Right? Right? Oh, forget it.

El que pega primero pega dos veces.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

ryos is nearly right.

Celsius --> x9 --> ÷ 5 --> +32 --> Farenheit.

If you don't mind a bit of inaccuracy, an easy way to remember it is just to multiply by 2 and add 30.

Why did the vector cross the road?

It wanted to be normal.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,664

Maybe I could add "mathsy's approximation" to Conversion of Temperature - Celsius to Fahrenheit

0° = 32° (or 30° using "mathsy's approximation")

10° = 50° (or 50° using "mathsy's approximation")

20° = 68° (or 70° using "mathsy's approximation")

30° = 86° (or 90° using "mathsy's approximation")

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

Here is a fun one.

To go from °F to °C or

to go from °C to °F, either way

you Add 40 first, then do the 9 5 thing and then subtract 40 when done.

So you always add first and subtract last, if you use the 40 method,

regardless if you are going from celsius to Farenheit or backwards.

Here's another one.

In this one you subtract before and after from °F to °C

and you add before and after for the °C to °F.

The number you add or subtract is 11 3/7.

The reason is because 11 3/7 °F equals -11 3/7 °C.

Pretty fun! huh?!

**igloo** **myrtilles** **fourmis**

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**irspow****Member**- Registered: 2005-11-24
- Posts: 457

C(f) = (f - 32)(5/9) = (5f - 160)/9;

F(c) = (9c/5) + 32 = (9c + 160)/5;

K(c) = 273.15 + C;

K(f) = (2297.9 + 5f)/9;

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

Nice equations, irspow, but just one typo I think, the 2297.9 should be 2298.4 perhaps.

**igloo** **myrtilles** **fourmis**

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