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How does this work within the design proposal as 20.736 inches?

this analysis is something i came across today

[www.secretsinplainsight.com]

The cubit works as follows

20.736 x 1200 = 24883.20 Michell's meridinal circumference

20.736 x 1760 = 36495.36 the GP base

meridinal x 1.4666r(the eclipse unit) = GP base

It is possible to drill into the cubit to understand what is actually in it.

The unit that opens it up is 12 inches / 100 = 0.12 inches or 100th of a foot.

20.736 / 0.12 = 172.8 / 0.12 = 1440 / 0.12 = 12000 / 0.12 = 100000

this is 12/100 4 times as 20736 / 100 to the power 4.

It is possible to find pi in it by using 3 different methods

1 cubit / 6.6

2 meridinal circ / 7920

3 GP base / 11616 being 7920 x 1.4666r (eclipse unit)

The result in all 3 cases is 3.14181818181818r

One thing that seems to work with the cubit is to work out its proximity to 560 and this can be done by multiplying it by 27.

20.736 x 27 = 559.872

So the Earth circumference that aligns with 560 can be calculated.

24883.2 / 559.872 x 560 = 24888.888r and this surely tells us that they were designing the cubit against 560/27 = 20.740740740740

So using a unit of 28/27 = there are 20 of the units in the 20.740740r or 560/27 cubit.

It is also worthy of note that 1.4666r x /28 x 27 = 99/70 as root 2 approximate and that many cubit rods have shown divisions into 27 as well as 28.

perhaps this helps us to understand the thinking of the designers

Edited 3 time(s). Last edit at 26-Mar-19 15:26 by DavidK.

this analysis is something i came across today

[www.secretsinplainsight.com]

**I just checked out the distance between the Shard and Silbury Hill and 123456 metres does fall into it and is is 66.66 nautical miles but you can't just buy a building plot in London against it location to ancient monuments. i wonder what ancient building was there before the Shard?**

Anyway the site is worth a read.Anyway the site is worth a read.

The cubit works as follows

20.736 x 1200 = 24883.20 Michell's meridinal circumference

20.736 x 1760 = 36495.36 the GP base

meridinal x 1.4666r(the eclipse unit) = GP base

It is possible to drill into the cubit to understand what is actually in it.

The unit that opens it up is 12 inches / 100 = 0.12 inches or 100th of a foot.

20.736 / 0.12 = 172.8 / 0.12 = 1440 / 0.12 = 12000 / 0.12 = 100000

this is 12/100 4 times as 20736 / 100 to the power 4.

It is possible to find pi in it by using 3 different methods

1 cubit / 6.6

2 meridinal circ / 7920

3 GP base / 11616 being 7920 x 1.4666r (eclipse unit)

The result in all 3 cases is 3.14181818181818r

One thing that seems to work with the cubit is to work out its proximity to 560 and this can be done by multiplying it by 27.

20.736 x 27 = 559.872

So the Earth circumference that aligns with 560 can be calculated.

24883.2 / 559.872 x 560 = 24888.888r and this surely tells us that they were designing the cubit against 560/27 = 20.740740740740

So using a unit of 28/27 = there are 20 of the units in the 20.740740r or 560/27 cubit.

It is also worthy of note that 1.4666r x /28 x 27 = 99/70 as root 2 approximate and that many cubit rods have shown divisions into 27 as well as 28.

perhaps this helps us to understand the thinking of the designers

Edited 3 time(s). Last edit at 26-Mar-19 15:26 by DavidK.

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