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Manu says

'Here is one way the Stonehenge builders may have tracked the Saros Cycle:

Each one turn of the Aubrey Circle is 30 months of 27 days.

After each one turn, you back up 13 days/13 IF and count them again: That's 823 days per cycle and it begins anew.

Repeat 7 more times: 8 x 823 = 6584 days

On the first day of the ninth cycle, it's Eclipse Day = 6585 days.

It's almost too easy.'

So 891 works bery easily once the interpretation is made, so the different circles are cross checking each other. A 27 day cycle works 28 day 29.333 day 29.5 day they were all over it and this is what Thom's booka are about, he says they could identifiy the Moon's wobble just prior to an eclipse and the fraction in his calculation is 0.000132.

This is what he says - hold on to your hat.

'... broadly descibed by saying it cosists of 3 cyclical components. There is a large roughly sinusoidal component with a mean period of 27.32 days and an amplitude controlled by the sinusoidal limits shown in fig 2.2. These limits have a period of 18.61 yearsand an amplitude of i = 5 degrees 8 minutes and 43 seconds. Superimposed on this long period component is the small peturbation period of 173.3 days and an amplitude not quite constant, but never getting far from 9 arc minutes.

Near a standstill it is sufficient to consider that these 3 components are additive. Let us write any one of them as y = g cos 2 pi t /P , where t is the time of the maximum of the component considered, P is the period and G the amplitude.'

are you still with the professor? Of course not and he has not got warmed up yet

This guy is a real pro, not an amateur like us which is why I promote his work.

He says

18.61 year component g = i k = 0.000132 and i know this x 10/9 is .000146666r and it is the long term component.

here is another beauty

t max = - k3t0/(k3 + k4) = -0.022 t0.

I recognise 22 as being the imperial.

The he says

' the above is the broad picture; but if we have erected foresights of sufficient accuracy, we shall find that the limits are subject to another oscillation of small amplitude and period 173 days. This small oscillation is associeted with eclipses in such a way that they can only happen when the oscillation is making its greatest contributionto the limits existibg at the time.Eclipse can happen near any maximum of the 173.3 day oscillation'

So i looked for 173.333 and it is the distance in miles from SH to Avebury as 17.333r miles 91520 feet /52 = 1760/ 4 x 3 = 1320.

I have the foot represent 100 units the aubrey at 89760 and 2912 days being represented

2912 x 89760 = 261381120

So does 173.3333r divide into it?

yes leaving 1507968

divide by 816 to get 1848

and by 33 to get 56 the Aubrey.

I got it by understanding snippets of his calculations - he is a real heavyweight.

His version of pi is 5440 / 1733.333r exactly 5280 x 34/33 = 5440 nodal and 173.333r eclipse danger periods.

Pi as 22/7 is 5280/1680 exactly the same reasoning. saros and 28 day moon cycle. 11/7 is the GP design.

he tells us how they did it and gives us the aubrey measurements. Fred hoyle had already published about the Aubrey being an eclipse calculation device. So if the Aubrey does not tell us how to predict eclipses, fortunately it does!! Thom was correct.

SH to avebury 91520 x 4 = 366080 / 1000 = 366.08 solar days x 360000 = 131788800

132000000 - 131788800 = 211200 / 5280 = 40 miles. You can't make this stuff up they really did do it. This is exactly what Michell tells us.

This is test 2 in my second book

[www.amazon.co.uk]

cheers

Dave

Edited 4 time(s). Last edit at 17-Jul-18 10:21 by DavidK.

'Here is one way the Stonehenge builders may have tracked the Saros Cycle:

Each one turn of the Aubrey Circle is 30 months of 27 days.

After each one turn, you back up 13 days/13 IF and count them again: That's 823 days per cycle and it begins anew.

Repeat 7 more times: 8 x 823 = 6584 days

On the first day of the ninth cycle, it's Eclipse Day = 6585 days.

It's almost too easy.'

So 891 works bery easily once the interpretation is made, so the different circles are cross checking each other. A 27 day cycle works 28 day 29.333 day 29.5 day they were all over it and this is what Thom's booka are about, he says they could identifiy the Moon's wobble just prior to an eclipse and the fraction in his calculation is 0.000132.

This is what he says - hold on to your hat.

'... broadly descibed by saying it cosists of 3 cyclical components. There is a large roughly sinusoidal component with a mean period of 27.32 days and an amplitude controlled by the sinusoidal limits shown in fig 2.2. These limits have a period of 18.61 yearsand an amplitude of i = 5 degrees 8 minutes and 43 seconds. Superimposed on this long period component is the small peturbation period of 173.3 days and an amplitude not quite constant, but never getting far from 9 arc minutes.

Near a standstill it is sufficient to consider that these 3 components are additive. Let us write any one of them as y = g cos 2 pi t /P , where t is the time of the maximum of the component considered, P is the period and G the amplitude.'

are you still with the professor? Of course not and he has not got warmed up yet

This guy is a real pro, not an amateur like us which is why I promote his work.

He says

18.61 year component g = i k = 0.000132 and i know this x 10/9 is .000146666r and it is the long term component.

here is another beauty

t max = - k3t0/(k3 + k4) = -0.022 t0.

I recognise 22 as being the imperial.

The he says

' the above is the broad picture; but if we have erected foresights of sufficient accuracy, we shall find that the limits are subject to another oscillation of small amplitude and period 173 days. This small oscillation is associeted with eclipses in such a way that they can only happen when the oscillation is making its greatest contributionto the limits existibg at the time.Eclipse can happen near any maximum of the 173.3 day oscillation'

So i looked for 173.333 and it is the distance in miles from SH to Avebury as 17.333r miles 91520 feet /52 = 1760/ 4 x 3 = 1320.

I have the foot represent 100 units the aubrey at 89760 and 2912 days being represented

2912 x 89760 = 261381120

So does 173.3333r divide into it?

yes leaving 1507968

divide by 816 to get 1848

and by 33 to get 56 the Aubrey.

I got it by understanding snippets of his calculations - he is a real heavyweight.

His version of pi is 5440 / 1733.333r exactly 5280 x 34/33 = 5440 nodal and 173.333r eclipse danger periods.

Pi as 22/7 is 5280/1680 exactly the same reasoning. saros and 28 day moon cycle. 11/7 is the GP design.

he tells us how they did it and gives us the aubrey measurements. Fred hoyle had already published about the Aubrey being an eclipse calculation device. So if the Aubrey does not tell us how to predict eclipses, fortunately it does!! Thom was correct.

SH to avebury 91520 x 4 = 366080 / 1000 = 366.08 solar days x 360000 = 131788800

132000000 - 131788800 = 211200 / 5280 = 40 miles. You can't make this stuff up they really did do it. This is exactly what Michell tells us.

This is test 2 in my second book

[www.amazon.co.uk]

cheers

Dave

Edited 4 time(s). Last edit at 17-Jul-18 10:21 by DavidK.

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