Now, 100F to -24F vs 38c to about -30c... how do I ballpark that quickly? There's no easy ratio. Pounds to Kilograms was like what? divide by 2.2? or was it times? That one at least isn't too hard to figure out. If I say 20 pounds or approx... 8-10 kilos... not hard to do. But the Farenheit and Celcius one... I still have no proper ratio to try and ball park it with.
If you're talking about ballparking everyday temperatures, to convert f to c, subtract 30, then divide by 2. So 50F is (50-30)/2 = 10, which is exact.
100F is (100-30)/2 = 35. In reality, 100F is 37.7, so you're 2.7c too low. 0F is (0-30)/2 = -15. In reality, 0F is -17.7, so you're 2.7 too high.
To go the other way - C to F - you do the opposite. Multiply by 2, add 30.
Skip the following unless you're bored:
Reduce the number you're subtracting by by 1 for every 10F (or 5C) you warm up. So if you were gonna ballpark around 100F, you wouldn't subtract by 30, you'd subtract by 25. (100F-25)/2 = 37.5C. You're still off by about 0.2, but whatever. Increase by 5 for ever 50F you cool off - so (0F-35)/2 = -17.5C, which is also off by about 0.2.
Skip the following unless you're
really bored:
More exhaustively, your constant is going to be exactly equal to (320-temperature in F)/9. So for 50F you have (320-50)/9 = 30, which we already know is precise. If you're going to be doing a bunch of calculations around, say, 230F, the previous would say you'd subtract by 13, then divide by 2. But that's a little bit off: (230F-13)/2 = 108.5C, which in reality should be 110C. Oh noes! Instead, subtract by (320-230)/9 = 10. (230-10)/2 = 110C, which once again is precise.
Note that to save time, we can substitute (320-x)/9 directly into our formula:
(x-(320-x)/9)/2
(9x/9+x/9-320/9)/2
(10x/9-320/9)/2
5x/9-160/9
(5x-160)/9
(x-32)*5/9
or from c to f:
x*9/5+32
Which is the slightly more complicated conversion formula you should have been using in the first place.
What? I'm bored.
