> Fahrenheit (since this is a non-linear transform)
huh?
huh?
1. F(c1 + c2) = F(c1) + F(c2)
2. F(k*c1) = k*F(c1)
F(c) = 9/5 * c + 32
F(c1 + c2) = 9/5 * (c1 + c2) + 32
F(c1) + F(c2) = (9/5 * c1 + 32) + (9/5 * c2 + 32) = 9/5 * (c1 + c2) + 64
F(k*c1) = 9/5 * (k*c1) + 32
k*F(c1) = k * (9/5 * c1 + 32) = 9/5 * (k*c1) + k*32
Because it's harder to calculate mentally
One can approximate 2lb to 1kg roughly, 1mi ~ 1.6km, but for Fahrenheit this is harder
One solution is to establish a set point ( 70F = 21.1C, approximate to 20C ) then work in differences, since 1.8deltaF = 1deltaC so, 2dF ~ 1dC (or 10dF ~ 5dC)
Hence, 90F is approximately 30C (exact value, 32.2C)
I would think this would be called linear, but that mean something more specific?
For example, yards to meters is a linear transform. 3 yards converted to meters + 2 yards converted to meters = 5 yards converted to meters. This is not true of temperature. Adding 32F to 32F gives 64F, which is not the same as adding 0C to 0C.