Keith M. Hunter
British Imperial Units of Measure & the Earth Form
As was previously discussed in the essays concerning the proof of a once existent earth year of 360 days, it was shown that there was a strong tie-in between the actual length of the earth tropical year, and the primary British imperial units including the inch, foot, fathom, and the ideal geographical mile. Set against one another, these units compare as follows:
12 inches = 1 foot
6 feet = 1 fathom
1000 fathoms = 1 ideal geographical mile
Therefore: 1 ideal geographical mile = 6000 feet
Under this system then it can be seen that the primary unit that is the foot, is 1/6000th of a single ideal geographical mile, and furthermore, that whilst the earth was (at one time) possessed of a 360 day orbit, as previously established, the equator of the planet was precisely 21600 IGM. Moreover, with the subsequent increase to the equator occurring in direct proportion to an increase in the tropical year, the current measure of the earth equatorial circumference was given as follows:
The earth tropical year = 365.2421897 days
365.2421897 / 360 = 1.01456163805
21600 x 1.01456163805 = 21914.531382 ideal geographical miles
The suggestion here then is that the very frequency of the earth orbit i.e. the number of days per year, actually determines the length of an ideal geographical mile, for thus, by dividing up the circular equator by 21914.531382, one defines the length of a single ideal geographical mile, including also the foot, which is set at 1/6000th of each IGM. In this way one sets the standard for a whole series of units of measure for length; all in accordance with the orbital frequency of the earth.
The question is, set against such a standard, just how accurately do the primary measures of the British Imperial system correspond? Basically, if one were to take 6000 British feet as a unit of length, and divide the circumference of the earth by this measure, would there be an exact numerical answer of 21914.531382? The analysis, as was indeed given in a previous essay, revealed the following:
Current circumference of the Earth at the equator: 24902.4 statute miles.
Converted into ideal geographical miles: (24902.4 x 5280) / 6000 = 21914.112 IGM
Therefore: 21914.531382 / 21914.112 = 1.0000191375…
What this result implies is that the British Imperial units of measure as a whole are slightly greater in length than what can be regarded as the correct ideal standard – as based upon the tropical year frequency. Even so,in percentage terms the discrepancy is very small indeed, a point which strongly supports the contention that the Imperial units of measure are in fact based upon the earth equator as a primary standard, which itself is determined from the real value of the number of days contained within the earth tropical year.
That being said however, should one still have doubts, it so transpires that an even greater level of correspondence is to be had between the British Imperial units and the ideal noted standard. And this is revealed, by none other than a very careful evaluation of the orbit of the moon.
The main problem that one has with the analysis (as detailed above) concerns the fact that one must employ a physical measure for the earth as it currently standards, in order to compare it to the ideal noted standard.
But just what exactly should one take to be the ‘true’ level or
circumference of the physical earth equator? The answer will no doubt
rest upon what one holds to be the actual material radius of the planet.
But what is that? Should one choose for example the average sea level
height covering the earth, or maybe an average of the height of the
land; or perhaps a combination of the two?
One can see then from this that it is very difficult to truly
compare just how accurately the Imperial units of measure relate to the
set ideal standard, without being sure of what constitutes the true
boundary of the physical earth form. Is there still a way forward
though? Most assuredly there is: By evaluating the Imperial units
against the length of the moon’s mean point of approach, i.e. the
distance value of its orbital semi-major axis.
The primary advantage from this is that such an orbital measure is a far more ‘clean’ and exacting measure, in that it avoids the geophysical boundary considerations as with the earth form. Moreover, and even more importantly, the semi-major axis of the Moon orbit is the mathematical equivalent to the equatorial radius of the earth, which itself is the semi-major axis component of the physical earth ellipsoid form. The very fact of this validates its choice in light of the proportional laws, as detailed previously.
In proceeding with the analysis, one requires therefore a most
exacting value for the length of the moon semi-major axis. Fortunately,
as a result of the Apollo moon missions, there is one. For when the
astronauts landed on the moon during the Apollo program, they set up
special mirrors on the surface of the moon so that scientists could beam
light waves direct to the surface of the mirrors and reflect them back
to the earth, so as to track its orbit to a high level of accuracy. This
allowed them to determine a very precise measure for the moon
In kilometres = 384404 +/- 0.5
Converting this figure into British ideal geographical miles one has the following:
384404 x 0.6213711922 x 0.88 = 210194.663154474944 IGM
Where PE (e) & CE (m) are ratios:
PE (e) = Physical Equator, Earth (present) / Physical Equator, Earth (past)
CE (m) = Celestial Equator, moon (present) / Celestial Equator, moon (past)
4 & 3 are powers
By using the previously noted ratio of change for the earth
tropical year (that which transformed it from 360 days to its present
value) with the above law, one can thus determine exactly just how many
true ideal geographical miles would constitute the full length of the
current moon semi-major axis. This is achieved in two stages. First by
determining the value for the ideal moon semi-major axis, and secondly,
by transforming this value to determine the current value:
1) The ideal moon semi-major axis: 1296000 / (2 x (22/7)) = 206181.818181818181…
2) Using the 3 to 4 power law, the current moon-semi-major axis is determined as follows:
365.2421897 / 360 = 1.01456163805
1.01456163805 to the 4th power = 1.05953119567
1.05953119567 to the 3rd root = 1.01946248613
And: 206181.818181818181…x 1.01946248613 = 210194.628959981772
The above derived value then must represent the true and exacting
standard that defines the length of an actual ideal geographical mile
unit. How accurately then does the system of Imperial units of measure
compare to this standard? Evaluating the above value with that given
previously for the moon semi-major axis in terms of British IGM units,
the following is to be had:
210194.663154474944 / 210194.628959981772 = 1.000000162680…
This indeed is a most incredible result, with the level of
accuracy being significantly greater than with the physical earth
evaluation. And with this result there can be only one conclusion:
THE TRUE STANDARD, UPON WHICH THE FULL SET OF BRITISH IMPERIAL
UNITS OF MEASURE IS BASED, IS THE EARTH TROPICAL YEAR FREQUENCY I.E. THE
NUMBER OF SOLAR DAYS CONTAINED WITHIN ONE ORBIT ABOUT THE SUN, AS A
THE FULL FREQUENCY OF THE EARTH YEAR IN DAYS, MULTIPLIED BY 60
AND APPLIED TO THE EQUATOR OF THE EARTH, DEFINES THE IDEAL GEOGRAPHICAL
MILE. AND FROM THIS, ALL OTHER UNITS OF MEASURE OF THE IMPERIAL SYSTEM
ARE DERIVED, AS VARIOUS FRACTIONAL SUB DIVISIONS.
With the level of discrepancy between the British Imperial units
of measure and the ‘ideally generated’ numerical values of the
earth-moon system being practically none existent, no other conclusion
is possible. Whoever thus established the Imperial units was quite
simply a genius, for through this system one is able to naturally – in a
true physical sense – combine units of time (which is a frequency) and
units of apparent spatial extension.
Imperial units of measure as a whole are thus not merely arbitrary units of distance, but units that are tied in to a real natural physical cycle: that of the earth orbit about the sun itself.