Cubit with cubes

Okay, last one for tonight… since I upset √2 with the previous formula, I thought I’d have a go using square roots, but didn’t find anything yet. Maybe another day.

But… I did find something using cube roots… in fact, the cube roots of 2, 3, 5, and something derived from 7 …. and the accuracy is the best yet, even if the formula is horribly complex. Beauty and the beast, I suppose.

Cubit in cubes…

I suppose we could write 10 as cube root of 1000 as well.

The comparison table now looks as follows:

MethodValueDifference from π/6Abs difference
π/60.523598775598300.000000000000000.00000000000000
φ²/50.523606797749980.000008022151680.00000802215168
Ave Year0.523607797087850.000009021489560.00000902148956
π – φ²0.52355866483991-0.000040110758390.00004011075839
10²/φ²π²ⅇ²0.523764440990030.000165665391720.00016566539172
cube roots0.523600350171940.000001574573650.00000157457365

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