Cumulus Support

@colinpb neat trick,
which is essentially what I do except I use
(9/5 * C) + 32 directly
which of course is equivalent to your
(10/5 * C) +32 - (1/5 * C)

Actually, yours can be improved by, rather than asking how many 5's in the C, just dividing your first result by 10 (i.e. move the decimal point) and subtracting that.

so for example 37C ..... 37*2 = 74....... + 32 = 106...... - 7.4 = 98.6F
or .............. 21.6C.... 21.6*2 = 43.2..... + 32 = 75.2.... - 4.32 = 70.88 F

gluepack wrote:Actually, yours can be improved by, rather than asking how many 5's in the C, just dividing your first result by 10 (i.e. move the decimal point) and subtracting that.

so for example 37C ..... 37*2 = 74....... + 32 = 106...... - 7.4 = 98.6F
or .............. 21.6C.... 21.6*2 = 43.2..... + 32 = 75.2.... - 4.32 = 70.88 F

Neat, and the other way round it works approximately (as it only uses a factor of 0.55 whereas 5/9 = 0.5555555...)

98.6F - 32 = 66.6... 66.6 / 2 = 33.3 + 3.3 = 36.7 ~ 37C

When NZ was first discussing the change to Metric, our Senior Math teacher(tutor) made some interesting observations in support of number systems with higher bases.

With metric base 10, there are only 2 divisors which result in whole numbers - 2 and 5.
with base 12 there are 4 divisors which result in whole numbers - 2, 3, 4, 6
with base 16 there are 3 divisors which result in whole numbers - 2, 4, 8

This was used to propose that we should be using a base 12 'decimal' system as the only magic about base 10 metric is that that 10 has the '0'. So if there were 2 extra symbols then all the advantages of the metric number system could be enhanced.
For example you could count 0,1,2,3,4,5,6,7,8,9,y,z,10,11,12,13,14,15,16,17,18,19,1y,1z,20,21, and so on, and the 'duodeceimal' system (as it was called) would would be an improved metric system.

Of course the suggestion was not practical as the primary reason for the change is standardis(z)ation, not the math (apart from not having the extra digits)!