Since they say this is not what is there. I am making claim on this teaching tool for our solar system.

Step 1) From a starting point A measure a horizontal distance of sq rt of 2 divide by 4 then times 1000 and mark a point 353.5534 units and label this point G2a. It is not important what units are used or how large or small they are. It is totally about ratios.

Step 2) From same starting point A measure a distance this time of sq rt of 5 divide by 4 then times it by 1000 and mark a second point along same line at distance of 559.017 of the same units. Label point G2b

Step 3) Take difference between G2a and G2b and mark a third point this distance in the same horizontal distance and label point G2c. Thus distance from G2a to G2c is (sq rt of 5 / .004) divided by (sq rt of 2 / .004) times 2 and equals 205.4636 x 2 or 410.9272 units. We will call this unit we have used "The Barone Cubit" This Barone Cubit contains 20.62 Barone Inches thus this first distance is 410.9272 x 20.62 or 8473.32 Barone Inches which from now on we will simply call inches.

Step 4 From point G2c measure in the same "easterly " direction a distance of sq rt of 3 / 0.004 or 433.013 units and label point G1b

Step 5 Design the pyramid that will have this point, G1b as its center. Height of this pyramid is to be is to be sq rt of 3 / 3 then times 10,000 to obtain height of 5773.5027 "inches". Angle to be used in the pyramid, G1, is to be The Pi angle of 4 / Pi or 1.27324 this will yield a half base of 5773.5027 / 1.27324 or 4534.4984. Measure this distance along our horizontal line from point G2b, still heading "east" and label point G1c. From same point G1b this time measure distance going "west " an exact same distance of 4534.4984 and label this point G1a. Thus "Pyramid G1" is distance from G1a to G1c and is equal to 2 x 4534.4984 and is 9068.996821 "inches" . This is exactly the same as saying base is Pi / 2 then divided by sq rt of 3 then times 10,000. This divided by 20.62 inches equals 439.8137 "cubits"

Step 6 Make G3, a pyramid drawn from our original point A eqaul to 1 when G1 is eqaul to 1 / Pi x 10 then minus 1 or 0.31831 x 10 = 3.1831 - 1 = 2.1831. Thus G3 + G1 = 3.1831 and dividing we find that G3 = 1 which is 9068.996821 / 2.8131 or 4154.18513 inches. This divided by 20.62 yields a pyramid base of 201.46388 cubits

First half is complete:

From starting point A to SE corner of G3, labelled G3c, is 201.46388 "cubits" . From this point to G2a is 353.5534 - 201.4639 = 152.0895 cubits

From G2a to G2c is 410.9272. From G2c to center of G1 or G1b is 433.013 G1a is 433.013 - (1/2 of 439.8137 or 219.90685) = 213.10615 cubits Thus the spaces between pyramids eqaul 152.0895 + 213.1062 = 365.1957 and within 0.05434 of a day to the correct length of the year as we know it today.

Total horizontal distance from Point A or G3a to G1c is:

201.4639 + 152.0895 + 410.9272 + 213.10615 + 439.8137 = 1417.40045 cubits or 29226.7973 inches.

[imagizer.imageshack.com]

[imagizer.imageshack.com]

Vertical plan to follow after coffee break.

db

"There is nothing as impenetrable as a closed mind"
and ..." if everything is a coincidence what is the point of studying or measuring or analyzing anything ?" db