**What We Can Learn From**

**The M****easures of Spherical Earth**

**And It’s **__Drop Distance Constants__

__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 a^{2} + b^{2} = c^{2}. 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) = .**66**5280 foot

### .**66**__5280__ 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(.**66****66**00 *foot*). Again, in this *foot* measure we see a link measure of **7.92** *inches* (.**66**’); plus .0**792** inch (.00**66**’), 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 16^{th} 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(.**66**70400 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)(.**66**69300 foot, or 8.003160 inches).

### 5________________(25)(.**66**67584 foot, or 8.0011008 inches).

### 6________________(36)(.**66**689568 foot, or 8.00274816 inches).

### 99 ______________(9801)(.**66**6730774 foot, or 8.000769293 inches).

### 229 ______________(229)^{2}(.**66**6278187 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 *mile* “*ddc*” 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 / **7,936.5**07937000… 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 = .**66**5280 foot

### .**66**5280 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 = .**66**9 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* 2^{2} X 8”; or 32”. At 3 miles out it drops 3^{2} X 8”; 4 miles 4^{2} X 8”; and at 5 miles it drops *exactly* 5^{2} X 8”. But at 6 miles out the perfection of *the square of the distance times 8”* breaks down; the drop should be 6^{2} 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

*correspond. Otherwise, imagine what the chances are that an*

__quantities__*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)**

again

**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:

**Mile****, **any 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)7920^{2} “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**)**7920**^{2}

Circumference = (**X**)**7920**^{2}(pi)

Surface Area = 4(pi)[(**X**/2)**7920**^{2}]^{2}

Volume = [4(pi)[(**X**/2)**7920**^{2})^{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 32^{nd} 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 32 ^{nd} of an inch* is a

**2.0**unit long line

*minus*.

**4125**unit. This produces two lines in

*the same ratio as the meter measure (*

__exactly__*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 1.0^{2} (one *square* unit) in the form of a square has an edge-length equal to 1.0^{1} (one *line*al 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,

**10.6347**2310… sq. ft. =** 1531.40**01… sq. in. (and)

Earlier we saw that **1.06347**6563… sq. ft. = (**.2475’** + **.4125’** + **.37125’**)^{2}

and now this equation:

** 41.25 **in. X** 37.125 **in. =** 1531.40**62… sq. in.

Look, the results of these equations are *identical* to a **99.999**6001…% 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. **

*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.**

POSTSCRIPT:

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 **52,80**0” 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 .**9999**48… approach to **7920**).

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

I) (**5600**^{2 }/ 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).