[mlyle]

12.34567N 89.01234W is 14 digits, 2 points, and 2 signs or letters.

Earth is about 2^49 square meters. If you want precision of ~8 square meters, you're going to need to convey 46 bits of information (or use an optimization that gives less resolution over oceans, etc, but that's only going to be a small win-- a bit or two).

46/3.3 = 14 digits, but the latitude/longitude coding is a little less efficient because it's not uniform. A plus code does it in 11 characters (not including the plus) worth each 4.32 bits =~ 47.5 bits.

[permo-w]

if Earth sqm = 2^49: to address a single metre, we'd need 49 bits. log_10(2^49) = 14.75 so we'd need 15 spoken decimal digits for 1 sqm. for 8m do log_10(2^47)

[mlyle]

> if Earth sqm = 2^49: to address a single metre, we'd need 49 bits.

Yes-- that's why I said it was 46 bits for an 8 square meter area (because this is about the size of area you care about for this problem-- W3W is 9 square meters).

> for 8m do log_10(2^47)

No, because 2^3 = 8, so one does log_10(2^(49-3)) and gets 13.8. 46 / 3.3 =~ 13.9 works because 1/log_10(2) =~ 3.3.