Converting Celsius to Fahrenheit

[GRobLewis]

"At what temperature are the Fahrenheit and Celsius readings the same? People will look at you with newfound respect when you reveal the astonishing answer: minus 40."

Which explains why the easiest way to convert between the two scales (which for some inexplicable reason is not taught in schools) is this:

• Start with a temperature in either C or F
• Add 40
• Multiply by either 9/5 (for C to F) or 5/9 (for F to C)
• Subtract 40

Done.

[BrotherCadfael]

And you didn't learn this in seventh grade?

[Leaffan]

Um, the actual conversion from C to F is multiply by 9/5 and add 32.

To go the other way, subtract 32 and multiply by 5/9.

The easy way for most everyday temperatures is:

C - F - double and add 30
F - C - subtract 30 and divide my 2.

[johnpost]

Originally Posted by GRobLewis

"At what temperature are the Fahrenheit and Celsius readings the same? People will look at you with newfound respect when you reveal the astonishing answer: minus 40."

Which explains why the easiest way to convert between the two scales (which for some inexplicable reason is not taught in schools) is this:

• Start with a temperature in either C or F
• Add 40
• Multiply by either 9/5 (for C to F) or 5/9 (for F to C)
• Subtract 40

Done.

it is taught in the less confusing schools.

[GRobLewis]

Originally Posted by Leaffan

Um, the actual conversion from C to F is multiply by 9/5 and add 32.

To go the other way, subtract 32 and multiply by 5/9.

The easy way for most everyday temperatures is:

C - F - double and add 30
F - C - subtract 30 and divide my 2.

My formula is equivalent to yours and is correct (try it and see). I find it's just easier to remember and to compute than the one that's usually taught (yours).

[Leaffan]

Originally Posted by GRobLewis

My formula is equivalent to yours and is correct (try it and see). I find it's just easier to remember and to compute than the one that's usually taught (yours).

I'm sure I could figure out algebraically why they're equivalent; however you are correct. I was never taught this method.

[tim-n-va]

They are equivalent because both are doing the same thing at a generic level.

1) Get a reference point where you know what the temperature is on both scales.

2) Find out how far the temperature you want to covert is above/below that point.

3) Multiple by the appropriate ratio for the direction of conversion

4) Add that to the reference point on the target scale.

Using 32 and 0 is commonly taught because in one step you are just adding or subtracting zero. You can construct a formula with -40 (as shown upthread), with 100 and 212 or any other point where you know the tempature on both scales.

[Kenm]

I use this, which is easier now with smart phones and tablets.

It's tough hauling a desktop machine everywhere, what with its separate keyboard, heavy tower and 200 miles/321.8694 kilometers of extension cord. (I still do, though the extension cord causes problems at the U.S. border, but I'm old school.)

[tim-n-va]

For really old school, for anything where the units have a common "zero" point, just muliply by 1. The trick is that the "one" has to be a fraction where the numerator is in the target unit and the denominator is in the original unit.

Feet to inches use 12 inches / 1 foot or 36 inches / 3 feet, etc. Any place you know the equal values works.

[GRobLewis]

Originally Posted by Kenm

I use this, which is easier now with smart phones and tablets.

It's tough hauling a desktop machine everywhere, what with its separate keyboard, heavy tower and 200 miles/321.8694 kilometers of extension cord. (I still do, though the extension cord causes problems at the U.S. border, but I'm old school.)

There's a very nice iOS app called PCalc that does this and a lot more.

[iamnotbatman]

I'm better at adding and subtracting than dividing, so I've always found that the easiest way for me to do a rough conversion in my head is to remember that 10 degrees C is 50 degrees F, and that for every 10 degrees C you add/substract 18 degrees F.

So... what is 25 degrees C in F? 50 + 18 + 9 = 77.

Going the other way is nearly as easy. For every 20 degrees F you add/subtract 11 degrees C.

So... what is 25 degrees F in C? 10 - 11 - 3 = -4.

Maybe it's silly, but to each his own.

[TriPolar]

When I see a Celsius temperature I throw the thermometer away and get a real one.

[Learjeff]

Yup, 40' is the same for F and C. But, gee, whenever I tell people that, they just look at me like I'm a nerd.

It's a simple algebra problem, suitable for a first course.

Mentally multiplying by 9/5 is easy: double, then subtract 1/10 of the doubled value. E.g.:

9/5 * 53 = 106 - 10.6 = 95.4

No doubt there's a good mental way to go the other way but I don't know it. Closest I can come is to halve the number and then add 1/9 of the halved number (but cheat, and add 1/10 as close enough).

Anyway, I think it's easier to add or subtract 32, than to add 40 and then subtract it. Still, the add-40, subtract-40 trick is a nice one and no doubt some folks find it easier.

[Ignatz]

I remember and use the formula

C = 5/9 (F-32)

[Kenm]

I run out to the car and punch the buttons to switch the dashboard readout/outside temperature display from metric to imperial.

Unless it's November to April.

[Irishman]

You are correct, that works. Thanks tim-n-va for the explanation.

[John W. Kennedy]

Originally Posted by Ignatz

I remember and use the formula

C = 5/9 (F-32)

Nothing wrong with it, but remember that there have to be two formulae:

C = 5/9 (F - 32)
F = 9/5 C + 32

Some people find it easier to memorize:

C = 5/9 (F + 40) - 40
F = 9/5 (C + 40) - 40

As Zathras would say, “At least is symmetry.”

[Irishman]

Originally Posted by John W. Kennedy

Nothing wrong with it, but remember that there have to be two formulae:

Technically, it is only one equation, just whether it is arranged to solve for C of F. You can plug the numbers in the alternate arrangement and still solve for the correct answer. Just takes algebra.

But the trick with the standard formula is remembering which you add 32 and which you subtract 32, and which you 5/9 and which you 9/5. This alternate removes one of those complications, because you add then subtract 40 in both cases. Then you just have to keep straight the scaling factor.

[johnpost]

scaling is easy because you know F has 180 degrees and C 100; you then know if you want more degrees or fewer and then to multiply or divide by 1.8.