I found a copy of Berriman and got it yesterday. I could only find a 1953 first edition and I notice that it is dedicated 'to David' - hmm. Excellent book. I see where he says most of the things I was saying in the Greaves thread and some of the things I said in my article. Of course I agree with his statement that 6000 Greek feet = 6250 Roman feet = 5000 remen = one minute of latitude and I agree that this goes a ways towards answering the question. I think I followed along about the square radius and the acre and cubit A and the k proportion, but maybe you can help me with this:
"The area ratio of the English acre to the Roman jugerum is exactly (8/5) if the jugerum is reckoned as equivalent to the square on 96 royal cubits rated 20 + (5/8) inches. If the English acre is identified with the geodetic acre, then the radius of the spherical earth can be rated 100,000 acre-sides in terms of the side of this acre regarded as as square. This acre-side measures 96 sq rt(8/5) royal cubits, and if the royal cubit is now rated 20 sq rt 2 digits in terms of a digit defined as (1/54) meter, then the radius and the circumference of the spherical earth are linked by this geometric metrology. In terms of the digit defined as (1/54) meter, the circumference measures 2160 million digits: in terms of the acre-side, it measures 1536pi/(sq rt 5) million digits. If 1.4062 is written for pi/(sq rt 5) these expressions are equal, but a more accurate value for this ratio would be 1.405." pp. 172-173
I get the 2160 million digit circumference, and I get that 1536pi/(sq rt 5) million is approximately the same, but I am having trouble seeing where 1536 comes from and what the digit/acre-side relation is, if that is what he is saying.