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To: JRandomFreeper

I follow you up to 0.019685” = 0.5 mm . Not so sure what 2*60*C signifies.

The idea is that the eccentricity of earth’s orbit is 0.0167, but this is the “off centeredness”, which is the distance of the focus of the orbit from the center, in terms of the radius, or more precisely, the semimajor axis, and a circle shifted by this amount gives an excellent approximation to earth’s orbit. Considered as an ellipse, the minor axis is sqrt( 1 - e^2) times the major axis, or approximately ( 1 - 1/2 e^2 ) or (1 - 0.00014)

But 0.00014 X 8” is 0.001” which is to be compared to the pencil line width of 0.02” . So when I say “easily”, I mean that the pencil line width is twenty times as great as the difference between the major and minor diameters of the earth’s orbit, accurately drawn.

35 posted on 10/28/2012 11:09:13 PM PDT by dr_lew

To: dr_lew
2 light minutes per inch, per the scale you specified (orbit of earth drawn as a circle on 8 1/2 x 11 paper).

Multiply by 60 to get light seconds per scale inch.

Multiply by C (speed of light) to get answer in whatever unit of measure you use for C. I use 186000 miles per second. Multiply by the paper line width. That's the width of the line, to scale.

Cook math. Order of magnitude. Close enough. I don't launch Mars orbiters, and if I did, I'd keep the units the same, even if we were using furlongs per fortnight.

/johnny

37 posted on 10/28/2012 11:25:54 PM PDT by JRandomFreeper (Gone Galt)

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