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Hi Don. I revisited Atalanta Fugiens. I found that - if you accept the Maier "constraint" (that one end of the dividers rests on the very centre of the larger circle) then the angle is just over 22.5 degrees and it ain't Mercury (by 100 or so miles). If you don't accept the Maier constraint it gets worse: using Gary's figures - and letting the scale be 1 : 720 miles - Mercury will be 4.21111r (using yours 4.211). Half of this + I.5 (1/2 moon) gives us a triangle we can quantify. Divide by our (assumed) 23.5 tan and we get 8.29221 (8.292082). So we now have a side for a larger similar triangle in 8.29221 (8.292082) + 4.211111 (4.211) = 12.50332 (12.50308). We can now work out our triangle base to a bottom corner of the Maier triangle by X 23.5 tan = 5.4366* (5.4364939).

Next, we turn to the larger circle centre. It's 8.5 down from the Moon top, 5.5 from the Earth top. Our 12.50332 (12.50308) minus this = 4.00332 (4.00308). So we now have opp and adj of a triangle based on the model centre and a base triangle corner : 4.00332 (4.00308) sq + 3.9366* (3.9364939) - we took off the lunar 1.5 to find this triangle side length - gives us 5.6145695 (5.61432397) hypotenuse from the Earth centre. And there's the problem: this box does not fit in that circle. How do I know? The Earth is 11. It's radius is 5.5.

geoffss
the square roots of isis
[geoffss.weebly.com[]]

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Maier's Atalanta Fugiens XX1 544 geoffss 18-Jul-19 11:02
Re: Maier's Atalanta Fugiens XX1 - message to Don Barone 89 geoffss 23-Jul-19 09:54
Re: Maier's Atalanta Fugiens XX1 - message to gary osborn 98 geoffss 30-Jul-19 20:37


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