11/3/17 The Measures of Spherical Earth

What We Can Learn From

The Measures of Spherical Earth

And It’s Drop Distance Constants

          In the diagram above we see the Earth. For purposes here we will regard it simply as a geometrical sphere the size of the Earth. Without the image of the Earth superimposed on it, what then is depicted is simply the cross-section of a sphere. We see there is a horizontal line tangent to the circle’s circumference and perpendicular to its radius. The other radii extending from the sphere’s center up through the circumference  to the line above the sphere connect to points at increasing distances out from the point tangent on its surface. As the distance of any point on the line extending out from the fixed point on the surface of the sphere grows larger, the point directly below on the surface curvature of the sphere grows more distant from the point above on the straight line.

         On an Earth size sphere, at a distance of one mile from the point of tangency the surface of the earth drops about eight inches. I say “about” eight inches because (as we will see later) the exact amount varies depending on both the radius chosen for the sphere as well as on the choice of the “unit” used for calculating that same measurement (miles, yards, feet, inches, or millimeters). Wikipedia states that distances from points on the surface of Earth to the center range from 6,353 km to 6,384 km (3,947–3,968 mi). According to most sources, the mean radius of Earth is about 3,959 miles. We can use this radius measurement and the Pythagorean Theorem to easily calculate the drop distance due to curvature on an idealized Earth size sphere and get a more precise measure.

          The theorem states that in any right-triangle, after labeling the hypotenuse as side “c”, then sides a2 + b2 = c2. If the radius perpendicular to the horizontal line in the above diagram is side “a”, then the distance out from the surface point is side “b”. Thus the radius “squared”, plus the distance out from the point “squared”, equals the “square” of the distance from the center of the sphere to the point directly above the surface along the line or plane. The square-root of this quantity (side “c”) minus the radius is the drop distance. For the drop distance at one mile out from the point (on an Earth size sphere) we do the following calculations:

(3,959 miles)2 + (1 mile)2 = 15,673,681.00 sq. miles + 1 sq. mile = 15,673,682 sq. mile

(15,673,682 sq. miles)1/2 = 3959.000126 miles

drop distance = .000126 mile = 5280 feet(.000126) = .665280 foot

.665280 foot = 7.98336 inches

It’s important to note that this drop distance can be read as .66 foot, plus exactly 5,280 one-millionths foot. This .66 foot is 7.92 inches; which is 1.0 link on a surveyor’s chain (of 66 feet and comprised of 100 links). A distance of 80 chains is 1.0 mile, which also is divided into 5280 feet. Notice that this quantity of feet in one mile is mirrored in the “5,280” one-millionths of a foot over .66 foot in the drop distance measure.

We can also see this drop distance at one mile equaling “one link” of a survey chain (7.92”) plus .063360 inch. And, since 63,360 inches comprise one mile, the drop distance is simply “one-link” plus “a one-millionth mile”.

Two miles from the tangent point and the drop distance is 2.666400 feet; this is also 4(.666600 foot). Again, in this foot measure we see a link measure of 7.92 inches (.66’); plus .0792 inch (.0066’), which together equal 7.99920 inches. Once again, this is an uncanny correspondence to the surveyor’s chain, which we should remember dates back to the 16th century. When we consider that this 3959 radius measure is the result of today’s measurements of planet Earth it becomes a bit mysterious.

         At a distance of three miles the drop is 9(.6670400 foot, or 8.004800 inches). When we subtract one 7.92” link measure (.66’) we’re left with (.00704’). Together, nine of these measures is .063360’ (remember, 63,360 inches/mile). As we continue mile after mile out from the reference point on the sphere’s surface we can see the various changes in the drop distance constants (ddc).

4 miles____________(16)(.6669300 foot, or 8.003160 inches).

5________________(25)(.6667584 foot, or 8.0011008 inches).

    6________________(36)(.66689568 foot, or 8.00274816 inches).

          99 ______________(9801)(.666730774 foot, or 8.000769293 inches).

         229 ______________(229)2(.666278187 foot, or 7.77533824 inches).

         What can we learn from the above calculations? Certainly it confirms the statement that there is a constant of about an eight inch drop in curvature for every mile away from the point of tangency on a sphere the size of earth. But this eight inch constant must be multiplied by the square of the distance. At 6 miles away, for example, there are 36 of these about 8” units in the drop distance to the surface below. But again, what is most remarkable up to this point is the commensuration of the drop distance constant with the surveyor’s chain of 66 feet with 100 links, each link measuring 7.92”. As we will soon see this is no coincidence; this is by intentional design. What I mean is that when man’s measures were designed to be secretly based upon the simple geometry of form the Earth itself became fundamental to the measuring rod. Let’s see the evidence leading me to this conclusion.

         For starters, if the 3,959 mile mean radius of Earth was just 1.0 mile greater it would become 3,960 miles making its diameter 7,920 miles. Certainly this IS the measure at many places on Earth since we are told 3959 is a “mean”. And we know for certain that this can’t be an exact measurement because there is always a (+/-) margin of error in any act of measuring. To assume at least a (+/-) one mile margin for this mean measurement is probably reasonable especially in light of Wikipedia’s above stated range of radial measures.

         Of course, this “quantity” 7,920 (as miles) for the diameter of the Earth shows it to be a power of the same 792.0 “quantity” (as inches) for the length of a survey chain. Again, can all this be just a coincidence? I didn’t think so, and continued to follow the trail of numbers.

         Let’s look at some of the data I found regarding the drop distance constants for similar, but different size spheres.


         As we’ve seen above, a sphere having a 3,959 “mile” radius (such as Earth) has a drop distance constant (at a distance of 1.0 of these mile “units”) equal to .000126 mile. But so does a sphere with a 3,960 mile radius. In fact, when calculated in mile-units spheres ranging in radius from 3,953 miles to 3,984 miles have exactly the same .000126 mileddc” at an initial distance of one mile. A sphere smaller by just 1.0 mile (3,952) has a .000127 mile drop distance constant. Likewise, 1.0 mile larger than 3,984 miles (3,985) and this measure becomes .000125 mile and applies equally to spheres with radii up to and including 4,016 miles. So an Earth size sphere with a 3,959 mile radius has what is essentially a curvature constant defined by a .000126 quantity, which is also equally expressed as

1 / 7,936.507937000…

         This is a particularly significant quantity for its degree of congruence to the quantity .79375, which relates the size of 1/32 inch compared to 1.0 millimeter: .03125 / .03937007874 = .79375 / 1.0. What this equation shows us is that the two systems of measurement, the inch-measure (im) and meter-measure (mm), reconcile with one another through a single length: the inch. One system divides this measure into 32 parts; the other into 25.4 parts.

25.4 / 32 = .79375

         With the drop distance constant (.000126) written as 1 / 7,936.507937000… the metric system’s quantifier (79375) clearly stands out. How closely do these two quantities conform?

7,936.507937000… / 7937.5 = .9998750157

In fact, this metric property of Earth’s ddc is repeated in the small quantity .007937 added to the initial 7,936.5 portion:

.007937 / .0079375 = .99993700

         Here’s a special case example, again showing how “metric” is literally built into Earth’s measures: if Earth’s radius is 3958.59012… miles then its volume is 259,842,520,000  cubic miles.    This volume quantity is a power of .259842520…    and when 1.0 is added to this quantity it’s reciprocal (1/1.259842520…) equals (.79375). Note that this is the exact relationship between the inch measure and meter measure.


         As mentioned above, this .000126 drop distance constant at one mile applies equally to spheres ranging in radius from 3953 to 3984. With a 3952 radius the constant changes to .000127. This is equally expressed as 1 / 7,874.01574800… and here again the metric system’s unique markers clearly shout out to us since

7,874.01574800 / 2 = 3,937.007874

exposing a “quantity” that is this time an exact power of 100 meters in inches (i.e. 39.37007874 inches equals exactly 1 meter).

         There are radius measures that are extremely close to the size of Earth’s mean radius (3959 miles) where the drop distance constant is .000127 rather than, as just demonstrated, .000126. At 3959.222 miles the drop distance constant is .000127. But if the radius has another “2” (3959.2222) it reverts back to .000126. Radii of 3959.xxxxxxx, such as .66; .9999; .888888; .77777; 555; .4444; .33333, and others all have a .000127 drop distance constant. So there is inherent to Earth, based on its chosen system of measurements, a blend of these two specific drop distance constants, both of which show that the “meter measure” is actually built into the structure of geometric quantitative relationships.

         Let’s visualize what these numbers are describing. In one way, we see that one mile out from the point of origin there is a .000127 mile drop to the sphere’s surface. With 5280 feet in one mile we see that .000127(5280’) = .67056 foot, the drop distance, which is also 8.04672 inches.

         But another way of looking at the same numbers tells us how many drop distance units out from the point of origin defines the extent of one mile. Since .000127 also equals 1 / 7,874.01574800… there are 7,874.01574800… 8.04672 inch units in one mile. Remember, just above we saw that 7,874.01574800 / 2 = 3,937.007874 which means there are two lengths, each containing 3,937.007874 8.04672 inch drop unit measures in one mile:

63,360 (in./mi) / 8.04672 inches = 7,874.01574800… = 2(3,937.007874)

         CLEARLY, when the measures of Earth are established in miles, feet, and inches the measure known today as a meter (39.37007874 inches) occurs itself as a natural unit.

*         This is further proof that the metric system and inch system are both inherently built into the geometry of form and that history has been purposely led astray by omitting this fact. Our global systems of measures have been purposely designed to mimic this mathematics.

Taking a Closer Look at Similar Size Spheres

          Earlier it was stated that the exact amount of drop distance calculated varied slightly depending on the choice of the “unit” used for calculating the same measurement. What follows here is about as good a sample of this as any. This is typical of all near-Earth sized spheres. The example below starts first with the mile as the unit.

(3959 miles)2 + (1 mile)2 = (3959 miles + drop distance)2

15,673,681 square miles + 1 square mile = 15,673,682 square miles

(15,673,682 square miles)1/2 = 3959.000126 miles

This shows a drop distance of .000126 mile

.000126 X 5280 feet = .665280 foot

.665280 X 12 inches = 7.98336 inches

Now, using the exact same Pythagorean Theorem, but with the 3959 mile radius converted to yards (1760 yards/mile) as the unit:

(6,967,840 yards)2 + (1760 yards)2 = (6,967,840 yards + drop distance)2

48,550,794,270,000 sq. yards + 3,097,600 sq. yards = 48,550,797,367,600 sq. yards

(48,550,797,367,600 sq. yards)1/2 = 6,967,840.223

This shows a drop distance of .223 yard, which is .000126704546… mile

.223 X 3.0 feet = .669 foot

.669 X 12 inches = 8.028 inches

And again, but this time using feet as the unit:

(20,903,520 feet)2 + (5280 feet)2 = (20,903,520 feet + drop distance)2

436,957,148,400,000. sq. feet + 27,878,400 sq. feet = 436,957,176,300,000. sq. feet

(436,957,176,300,000. sq. feet sq. feet)1/2 = 20,903,520.67 feet

This shows a drop distance of .67 foot, which is .00012689393… mile

.67 X 12 inches = 8.04 inches

And again, but this time using inches as the unit:

(250,842,240 inches)2 + (63,360 inches)2 = (250,842,240 inches + drop distance)2

Solving this equation will show a drop distance of exactly 8 inches, which is

Also .6666666… foot, or .000126262626… mile

.6666666… X 12 inches = 8.0 inches (exact)

         All of the resulting drop distances calculated above for the sphere with a 3959 mile radius are the same as the results calculated for one having a 3960 radius. But when calculated in “inches”, the range of spheres having exactly an 8 inch drop distance, at a point one mile out (in inches), extends from radii of 3928 miles through 3980 miles. Spheres with a radius of 3927 miles and below have an 8.1 inch (ddc); and those 3981 and above have a 7.9 inch (ddc).

         This initial drop is the drop distance constant in its purest form. How it manifests at greater and greater distances out from the reference point seems to be indicative of the mystery associated with the origins of our measures of length. For example, if the sphere’s radius is 3959 miles (calculated in inches) its initial (ddc) of 8” at mile 1 is maintained in its pure form (of exactly 8”) out through the first 5 miles. This means that at 2 miles out it drops exactly 22 X 8”; or 32”. At 3 miles out it drops 32 X 8”; 4 miles 42 X 8”; and at 5 miles it drops exactly 52 X 8”. But at 6 miles out the perfection of the square of the distance times 8” breaks down; the drop should be 62 X 8”, or 288”. It is 288.1”. The pattern continues to deteriorate as we move farther away from the point of origin. The next smaller sphere by one mile in radius, 3958 miles, maintains the pure (ddc) only to 3 miles out; a radius of 3957, or 3956 miles just through 2 miles. And with a 3955 mile radius, not even the second mile produces with the square of the distance times 8” formula.

 One Very Unique “Earth-size” Sphere

          We just saw the tendency for sphere’s a bit smaller than Earth to loose the purity or exactness of their initial (ddc) the farther they deviate from (what we are told is) Earth’s mean radius of 3959 miles. But when just one mile is added to the radius, making it 3960 miles instead, something quite remarkable happens with the calculated quantities.

         The first characteristic to take note of is that the drop distance calculated at the point of each successive whole mile unit results in whole units of inches out through the first 25 miles! The formula multiplying the square of the distance in miles times 8” holds true this entire distance. With respect to all the others, this sphere is truly unique.

         Compare this to just the first 5 miles out for a radius of 3959 miles. When we do the calculations for a radius of 3961 miles, it results in whole inches for only the first 4 miles. And a 3962 mile radius stays whole for just the first 3 miles; a 3963 radius works only to the second mile. Are we getting the picture yet?

Radius = 3955 miles —–   (ddc)wholeunits out to —– 1 mile

Radius = 3956 miles —– (ddc) whole units out to —– 2 miles

Radius = 3957 miles —– (ddc) whole units out to —– 2 miles

Radius = 3958 miles —– (ddc) whole units out to —– 3 miles

Radius = 3959 miles —– (ddc) whole units out to —– 5 miles

Radius = 3960 miles —– (ddc) whole units out to —– 25 miles

Radius = 3961 miles —– (ddc) whole units out to —– 4 miles

Radius = 3962 miles —– (ddc) whole units out to —– 3 miles

Radius = 3963 miles —– (ddc) whole units out to —– 2 miles

Radius = 3967 miles —– (ddc) whole units out to —– 1 miles

         It doesn’t take a mathematician to tell from the above data which size sphere is truly unique. They are all about the same size and fall into the range of measured radii cited above in Wikipedia.

         But the sphere with a 3960 mile radius is an obvious mathematical anomaly to which this measurement system based on miles/feet/inches most comfortably conforms!

         For example, when we look at the (dd) data for this radius calculated in feet out through 25 miles we find whole units of feet occurring at 3, 6, 9, 12, 15, 18, 21 and 24 miles out. These drop distances beginning at 3 miles out are respectively 6, 24, 54, 96, 150, 216, 294 and 384 “feet” (6 times the squares of the first eight numbers). The miles in between, with drops that are not whole foot quantities, all have exactly 8” remainders which are equal to one drop distance constant. Whole numbers of 100 inch units occur at 5, 10, 15, 20, and 25 miles out. These drop distances are respectively 200, 800, 1800, 3200, and 5000 inches, which seem to coincidentally mimic the 2, 8, 18, and 32 quantities of electrons in an atom’s first four shells.

        None of the other radii have consistent remainders or exhibit any of these other unique characteristics.

*         This is further evidence that the measurement system chosen for this Earth a very long time ago was actually pre-designed from simple mathematics. It was not a hit-and-miss, subjective or arbitrary process as history would have us believe. The initial measurement of 7,920 “units” was assigned to its diameter and the rest naturally follows. The names (miles, feet, and inches) are irrelevant as long as their quantities correspond. Otherwise, imagine what the chances are that an exact number of whole units of miles (out from the point, delineated in inches) would be directly above points on the sphere also at exact whole units of feet or inches?

         Those illuminated ones who long ago designed Earth’s measures also were well aware of the simple “natural” mathematical relationships that exist among the quantities 7920, 5280, and 63360, to the application of scale in the geometry governing spheres.

(0.666…) X 7,920 = 5,280     (number of feet in 1.0 mile)

(0.666… X 12” = 8”) X 7,920 = 63,360     (number of inches in 1.0 mile)


1.0 drop distance unit = 0.666…(1.0 foot)

# of ft. 1.0 mile from point = 0.666…(# of dd units in 1.0 mile from point)

 # of feet in 1.0 mile = 0.666…(# of miles in diameter)

         To see a geometric relationship just imagine a single cube with a volume of 63,360 units. It naturally subdivides into eight smaller cubes; each one has a volume of 7,920 units.This is why it was necessary to add an additional 280 feet to the already existing “mile” under Queen Elizabeth I in the late 1500s. Here’s a brief quote of what the online Encyclopedia Britannica said about this change:

Mileany of various units of distance, such as the statute mile of 5,280 feet (1.609 km). It originated from the Roman mille passus, or “thousand paces,” which measured 5,000 Roman feet. About the year 1500 the “old London” mile was defined as eight furlongs. At that time the furlong, measured by a larger northern (German) foot, was 625 feet, and thus the mile equaled 5,000 feet. During the reign of Queen Elizabeth I, the mile gained an additional 280 feet—to 5,280—under a statute of 1593 that confirmed the use of a shorter foot that made the length of the furlong 660 feet.

         Regardless of any validity to the Britannica version, the Parliamentary record shows that an additional 280 feet were added at that time. This is because they needed to bring the furlong and mile into conformance with their coalescing systems of secretly designed measures. Now the furlong is exactly 10 chains, each 66.0 feet, or 792.0 inches making the mile 80 chains (5,280 feet or 63,360 inches). But the “chain” hadn’t been invented yet. It wasn’t until a few years later, in 1620, that Gunter’s chain of 792.0 inches was also adopted by Parliament.

Additional Evidence Showing “Why” These Measures Were Chosen

Here is another formula to calculate the drop distance. It works best for the 3,960 mile radius, but is also accurate for spheres very close to this radius:

_______Distance Out From Point in Feet_______

[1000 miles / Dist. From Point In Miles]7.92

         Divide the distance out from the point in feet, by (1000 miles divided by the distance from point in miles; times 7.92 inches). Note that even though the (ddc) is 8 inches for this range of spheres, it is this 7.92” length, or one link of a survey chain that is the operative constant for the formula above.

         Earlier it was stated that (when calculated in inches) spheres with radii of 3928 miles up through a 3980 mile radius ALL have an initial 8” drop distance at a point one mile out. Now, we know a mile contains 5,280 feet, or 63,360 inches; so how many of these 8” drop distance units constitute one mile?

63,360” / 8” = 7,920

There are 7,920 drop distance units (8”) in every mile measure.

         This applies to the same broad range of radii (3928-3980 miles) and a bit beyond on either side. And if you go out from the reference point exactly 10,000(5)1/2 inches; or (500,000,000)1/2 inches, the drop distance at that point is exactly 1 inch.

*Again, this is further proof that the measurement system chosen for this Earth was a very special geometric design based in inches, feet, and miles and was applied to this sphere by some group long ago. Otherwise, the chances would be astronomical that the distance out to a point, using an exact number of whole units, would find that point situated above the sphere’s surface that is also at a distance that measures an exact number of whole units.

Now when it comes to this next property (once again) it applies only to this sphere with a 3,960 mile radius. To see this, let’s designate the 8” drop distance at one mile out as (X). And we just saw that one mile out also measures 7920(X). This means that in terms of (X) this sphere’s 7920 “mile” diameter is (X)79202 “inches”.

         Measures of Earth Based on its Drop Distance at 1 Mile Out From Surface Point

drop distance = (X)

1 mile distance out = (X)7920

7920 mile diameter = (X)79202

            Circumference = (X)79202(pi)

 Surface Area = 4(pi)[(X/2)79202]2

                     Volume = [4(pi)[(X/2)79202)3]]/3

How perfect is that? Check it for yourself by substituting 8” for (X).

         Instead of the 8 inch drop distance being the “unit” (as in the example immediately above) let’s use the Earth’s 7920 mile diameter. Now, measure the mile out from the point and the drop distance in terms Earth’s diameter.

Earth’s diameter = 7920 miles

                            One mile out = 1/ 7920 (the diameter in miles)

                            Drop distance = 1/(7920)2 (the diameter in miles)

          If the diameters of the spheres with a 3959 or 3961 mile radius are substituted in the formula above the drop distance results are incorrect as they should be 8 inches and are not. They are respectively 8.00202… and 7.99798… inches. And just as with the many previous examples, the farther away from the sphere with a 3960 mile radius the farther the formula deviates from the correct result. This proves once more that this is truly a unique radial measure when the “quantities” for miles, feet, and inches are its subdivisions.

          The formula 800(792”)2 also produces the correct measurement in inches for Earth’s diameter, but again only for this special sphere. And the ratio between the 8” drop distance at one mile, and a mile, . . . is the exact same ratio as that between the one mile measurement and the Earth’s diameter. . . if it is 7920 miles exclusively.

The drop distance is to 1 mile as 1 mile is to the diameter of Earth


          For anyone who has read this far, it must be abundantly clear that it is impossible for these measures to have evolved in the haphazard, chaotic, subjective and arbitrary manner portrayed by historians; both past and present.

          The simple math presented in the previous sections has proved that (when calibrated in inches) spheres with radii ranging from 3928 miles to 3980 miles all drop exactly 8.0 inches at a distance of 1.0 mile. So, our sphere with its 3960 mile radius and 7920 mile diameter, not only falls right in there, but as we have seen, is far superior to any other radii in its consistency in relationships among significant mathematical quantities.

Once this “ruler” had been established, over time it was implemented into society. Here’s how it was applied to our measurements for the surface of Earth.

 Surface Measures on A Sphere

With A 7920 Mile Diameter

Gunter’s Chain

          Earlier it was stated that in 1620 England’s Parliament codified into law a unit of measure ever since known as Gunter’s chain in honor of its “inventor” Edmund Gunter. This is the surveyor’s chain, which was used like a tape measure by surveyors well into the 1960’s in the United States. Gunter divided his chain’s length of 66 feet into 100 links. Since it is also 792.0 inches, each link is 7.920 inch. One mile is a length of 80 chains. Survey measurements were recorded in numbers of chains and links.

          These were the “overt” measures introduced to the public and codified into law. But behind these numbers is a parallel hidden system derived directly from the geometry of form. It is a system that long ago was applied not only to land measures but to weights, volumes, and both the Fahrenheit and Celsius temperature scales along with the weightsand sizes of our coins and paper money. . . and more. But for now, we will explore how this system of secreted geometry was applied to measures of land on the surface of this spherical earth.


         The data previously presented regarding the drop distance behaviors of various near-Earth-sized spheres shows unequivocally that the quantities 7920 and 3937.007874 to be literally built into the drop distance geometry of all spheres in this size range. These quantities were transferred by the illuminati to measures of terrestrial lengths and areas. The first became the 792.0 inch long survey chain for measurements made in miles, feet, and inches. And the second, much later, became the 39.37007874 inch long meter. In this system the mile measure is the primary unit and the meter measure arises secondarily.

Note: There are 63,360 inches in one mile and 633,600 chains in Earth’s diameter measurement if in miles it is 7920.

         Gunter was probably aware of, and party to, the parallel measures hidden in his survey chain. He may even have been one of the illuminati of his day. Regardless, here is the occulted geometry and associated measures secreted in the now obsolete unit called the chain.

         To begin with, the chain of 792.0 inches can also be regarded as (a round) 800.0 “units” each .99 inch. It is also 8.0 units each 99 inches, or 16 units each 49.5 inches which is 4.125 feet. Therefore, 4.125 feet is a built-in natural subdivision of Gunter’s survey chain. (In my book The Geometry of Money bold red fonts indicate quantities in US gold and silver coinage in terms of their weights in grains; 412.5 grains is the gross weight of the US silver dollar).

         80 chains constitute 1.0 mile, which was shown to also be equal to 7,920, or (3,937.007874 X 2) drop distance units (depending on the base-unit of measure chosen for the calculations; again as was shown above in the previous sections).

         The modern measure for the meter is 39.37007874 inches. The nations of the world agreed on this exact quantity in the mid 1950’s decreeing that the inch measure (im) would contain exactly 25.4 millimeters (mm). The closest subdivision to a millimeter (1/25.4 inch) in the inch system is 1/32 of an inch. Therefore, if the length of one millimeter is considered “1.0 unit”, then the length of one 32nd of an inch is .79375 of that unit.

         But the meter measure was already built into geometry, as we now know from having studied in the previous section the behavior of drop distance constants (ddc) for various near Earth-size spheres. But geometry has another much simpler way of comparing these two measures:

         If a 2.0 unit long line is represented by one millimeter, then one 32nd of an inch is a 2.0 unit long line minus .4125 unit. This produces two lines in exactly the same ratio as the meter measure (mm) and the inch measure (im). . . i.e.; (1.0 / .79735).

         It’s not a coincidence that a power of this same 4125 quantity was just shown above to be an integral subdivision of Gunter’s survey chain. So here is a fact that cannot be disputed:

         At the heart of the base units of length for both the metric system (based on the meter and its subdivisions and derivatives) and our customary system (based on the mile, chain, and their subdivisions and derivatives) we find a power of the 4125 quantity playing an indispensable role in forming the two respective units.

         As we can see documented in my book The Geometry of Money, 4125 is a prime quantity at the heart of the geometry of form that will appear time and again throughout all of the illuminati’s systems of weights and measures.

The Square Chain

          When the surveyor’s chain became a recognized unit of length it also became a unit of area: one square chain. If its edges are subdivided into 16 units it subdivides the square chain into 256 internal squares with each edge measuring 4.125 feet. Since there are10 square chains in one acre an entire acre can be perfectly assembled out of nothing but 4.125 foot squares.

         The 792.0 inch length of the chain can also be divided into 8.0 lengths, each measuring 99 inches (2 times 4.125 feet); or into 800 lengths, each .99 inch. Subdividing the chain in this way makes each of the chain’s 100 links equal to 8.0 lengths each .99 inch. We also know that this link is 7.920 inches; or .66 foot. Five of the link’s .99 inch units equal 4.95 inches; and the remaining three equal 2.97 inches; 4.95 inches is .4125 foot, and 2.97 inches is .2475 foot (and 247.5 grains is the weight of the pure gold content in one original US ten dollar gold coin; one “eagle”).

         Now we’ve revealed that the quantities describing the weights of the US foundational gold and silver coins, expressed as portions of a foot, combine to create the length of each link of the survey chain. And with an additional length of .37125 foot (which quantity as 371.25 grains is the pure silver content in one US silver dollar) a very special unit of land measurement has been created.

.2475 + .4125 + .37125 = 1.03125’

         This length of 1.03125 foot, or 33/32 foot, is also 12.375 inches (37.125/ 3; or, 24.75/ 2). And the 66 foot long survey chain contains exactly 64 of these lengths. The square chain can now be viewed as being comprised of (4096) smaller squares, each a bit larger than one square foot. A square with a 12.375 inch edge contains 153.140625 square inches; this is 1.063476563… square feet. It is this square that is the fundamental unit of land subdivision, and as we will see, it comes directly from the geometry of form.


         In the customary system the area measurement of objects, constructs, commercial goods, etc. begins with the square inch as the base. Agglomerations of square inches form the next successive powers; the square foot, the square yard, and square mile.

         Even though the system dividing land area utilizes the square inch as its measurement base unit, the first power of agglomerations after the square inch, in this system, is the square with a 33/32 foot edge-length (12.375”) and an area of 1.063476563… square feet. Two of these edge-lengths equal 24.75”; three equal 37.125”; and four measures 49.5” which we now know as 4.125 feet. Again, all three quantities are powers of the fundamental foundational US gold and silver coinage measures dating back to 1792.

         The square formed with this 4.125 foot edge is the next power of the land-specific area units of measure. There are 256 of these units in 1.0 square chain. Four of these combine into the next land specific unit making the edge of this new square 8.25 feet; or, 99 inches. 64 of these 99 inch squares combined form 1.0 square chain containing 4,356 square feet (66’)2. There are 6,400 square chains in 1.0 square mile; and since 10 square chains equal one acre, a square mile contains 640 acres.

         Now, I just said a few lines above that it is this square with an edge-length measuring 12.375 inches that is the fundamental unit of land subdivision; and, that we will see that it comes directly from the geometry of form. Just what is meant by this statement? After all, haven’t we’ve just seen from where it was derived?

         Yes. But this which I am about to disclose is the “geometry” on which it is all patterned.


         Generally, “land measures” are units of length and area, whereas “coinage measures” describe weight and volume. But in the light of occulted geometry, we find both land and coinage measures derive from the same geometric properties of form.

         Euclid showed us that the fundamental units of both length and area are inherent to a simple “square”. An area of 1.02 (one square unit) in the form of a square has an edge-length equal to 1.01 (one lineal unit). All of “geometry” ultimately is scaled to this basic form, and from it, so too do our measures of land and money.

         Geometry shows us that there is a maximum amount of volume that 1.0 unit of surface is capable of completely enclosing within the confines of its surface. This amount is 0.094031597… of 1.0 volume unit; and this volume will be in the form of a perfect sphere (these quantities are depicted in the image to the left). The mathematical relationship between one square unit of surface and this ideal volume quantity is expressed by this ratio: 1.0 ∕ 0.094031… i.e., 1.0 surface “unit” to 0.094031… “unit” of volume. But there is another equal “reciprocal” mathematical description of this same relationship since:

1.0 ∕ 0.094031… = 10.63472310… ∕ 1.0

          By assigning the name “one foot” to geometry’s originating square of one surface unit, a human scale is imparted to these forms. Now, 1.0 ∕ 0.094031597…means 1.0 “square foot”, and a 0.094031597… portion of 1.0 “cubic foot”. This also means that 10.63472310… ∕ 1.0, its reciprocal and equivalent mathematical description, is also scaled in “feet”. Therefore,

10.63472310… sq. ft. = 1531.4001… sq. in. (and)

Earlier we saw that 1.063476563… sq. ft. =   (.2475 + .4125 + .37125)2

and now this equation:

           41.25 in. X 37.125 in. = 1531.4062… sq. in.

         Look, the results of these equations are identical to a 99.9996001…% approach to perfection! This is clear proof that our systems of land measures, and coinage or monetary measures have been derived from the same geometry of form. We’ve see this now having arrived at these same quantities after following two completely different investigative lines of reasoning. First, after studying the drop distance characteristics of near earth sized spheres and then following the trail from the obviously unique diameter among the bunch. And secondly, we followed the simple geometric relationships between a 1.0 square unit of surface in the form of a sphere and the volume encompassed by it.

          This entire preceding expose has been about a measurement system that was assigned to this sphere on which we live. The data presented here proves its evolution was not random but controlled to fit a mathematical scheme already well known to an elite group of technocrats. And lastly, the data presented here unequivocally proves the existence of the “illuminati” simply by exposing their occulted works.  


         Look at it this way. The French allege they measured the earth’s quadrant from pole to equator running through Paris. Even though I think that’s “fake news”, let’s say they had. This measure of physical earth, divided into 10 million units, became the fundamental yardstick or ruler to which all things metric trace back.

          In exactly the same sense is our older system based on the mile derived from a physical measure of earth. It is its diameter, to which long ago they “decreed” (not measured like the French allege) to have exactly 7920 initial subdivisions we call “miles”. This was their original “yardstick” and the standard to which all of our measures conform.

          There are many examples presented in my previous research showing this “commensuration” of our measures to this exact proportioned sphere. Here’s a few more I left out:

a) 7920 as sum of cube’s edges makes each face plane measure 4356 “sq. units” (if “sq. feet”, then = 1 sq. chain; or, 1/10 acre).

b) A cube has a volume of 7920 units. A tetrahedron with the sum of its edges equal to the sum of this cube’s edges has a 5280(sq. root of 2) volume.

c) If you calculate in inches the drop distance at a point 52,800” out on our sphere having a diameter of 7920 miles (250,905,600”), then the drop distance  is 5.600”.

d) 5600 is the volume of a cube. A tetrahedron with the sum of its edges equal to this cube’s has a volume equal to 5280.

e) 5600 / 5280 = 437.5 / 412.5 = 1 AV Oz. / wt. silver $.

f) 5600 Silver dollars weigh 5280 AV Oz.s.

g) 5600(sq. root of 2) = 7919.5959 (a .999948… approach to 7920).

h) 5600 / (sq. root of 2) = 3959.7979… (.999948… approach to 3960; earth’s radius).

I) (56002 / 2)1/2 = 3959.7979… (.999948… approach to 3960; earth’s radius).

j) (5600(5600 / 2))1/2 = 3959.7979… (.999948… approach to 3960; earth’s radius).



FORBIDDEN KNOWLEDGE? WELL NOT ANY LONGER. Even Tucker Carlson can’t say enough about this “new” news. Either the media is distracting us from some far more important topics, or they’re setting the public up using UFOs for some reason. Regardless, this isn’t “news” since millions of people have been seeing them and recording their sightings going back hundreds, if not thousands of years; myself included.

Late in the afternoon on September 30, 1979, my wife Norma and our nearly 1 year old son Atom, along with myself and our two cats were at anchor aboard “godot” in Hanalei Bay on the island of Kauai. We were situated approximately where you see us in the photo, but on that day the clouds shrouding the mountain extended farther down as I’ve indicated in the photo.

We were below deck as it all began. People on some of the other boats at anchor started screaming and shouting and we immediately rushed topside. Looking shoreward, right at about the tip of the left hand arrow (and about the size of the arrowhead itself) was a bright orange hemisphere, flat on the bottom with a perfect dome above. It was slowly coming directly at us while gradually gaining altitude. By the time it passed over our masthead at about 500 ft, and Norma and I had turned now looking seaward, another deafening crescendo arose from behind us as people screamed and pointed again at a second craft emerging from beneath the cloud.

We saw four of these craft that day, in broad daylight, along with hundreds of other people. They cruised over our boats one after another traveling no more than 25 mph in complete silence at an altitude of no more than four to five hundred feet. They were in formation and separated by about a minute’s time. I got a very good look at them. They were about 150 feet in diameter and half that in height. There was a definite metallic-like structure that could be seen despite what appeared to be a surface layer of brilliant orange plasma.

The next day hundreds of people were still in shock, some in denial, some crying. I listened to the radio all thru that night and the next day; checked all the newspapers; not a word was to be found about what happened in Hanalei. You’d think with Barking Sands and the Pacific Missile Range and all their sophisticated military radar and tracking equipment (RIGHT ON THE OTHER SIDE OF THAT MOUNTAIN) that they would have seen, tracked, and intercepted these strange visitors. The military had plenty of time to do this since we watched the four of them for another 15 minutes as they slowly disappeared over the horizon never gaining any appreciable altitude or speed.

This was a day one never forgets. My life’s view of our reality, and my life itself was forever changed from that moment forth. Perhaps one of you too was there that day, or you know someone who was. In any event, I’m not alone in recognizing “old news” when I see it. Long ago I learned that ‘the Truth is out there”. . . and I do believe as they say that “the Truth will set you free”.

Look to the Sky, and ALOHA