Mars and Venus Assay
by Daniel Browning
In the brief item entitled How Far to Sirius? the opposition time of Earth and Jupiter was
found based on Jupiter's "mass circumference," which is Jupiter's circumference times
Einstein-mass M. This paper further explains the Jupiter mass-circumference, by showing
in detail how the so-called "modern" speed-of-light and the "Louis" speed-of-light
complement each other. It took the French to give us Jupiter, and modern science has
given us the Earth. We will see that the Jupiter mass is an immense and intricate light
processor, as is the Earth, and that there is a wonderful synthesis between the two
entities, like a cosmic clockwork.
Getting Started
Though the Earth and Jupiter opposition time can only be calculated via the wide
reference-frame of Sirius, and this probably because Jupiter is itself a Sun, and so needs a
more distant sun by which to be measured, the oppositions between Earth and Venus, and
Earth and Mars, can be derived from the Jupiter mass without making the long trip to
Sirius.
The first thing to do is to make a clear and unmistakable entry into grid space. Grid space
is a mathematical space where "grid-seconds" and "grid-minutes" correspond to seconds
and minutes in clocked space. Just as there are "miles per second" in clocked space, so
too there are "miles per grid-second" in grid space. Grid space is "tuned" to the number
144,000, which is the modern "grid" speed of light in "arc-minutes (miles) per gridsecond." This base value has many common factors and so it harmonizes with the entire
universe and is not tied to any one planet.
144,000 is the modern grid value for the speed of light. The French value, based on the
work of the astronomers during the era of Louis XIV, is, in miles per grid-second:
L = 140,109.8578 (miles per clock-sec) / 1.293627749 (clock to grid) =
108,307.7090 miles per grid-sec (SOL, Louis)
This is the Louis grid measure in nautical miles (or equivalently arc-minutes) per gridsecond. The number of orbits of this light around the Jupiter-mass is:
K = L (miles per grid-sec) / 1,082,295.369 (mass circumference, Jupiter) =
0.1000722281 orbits, for 1 grid-sec, Jupiter mass
A full orbit of the Jupiter mass, in grid-seconds, is:
α = 1 (orbit) / K (orbits per grid-sec) = 9.992782403 grid-secs, full orbit
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This number will turn out to be very important as we go along. If the grid speed-of-light
is multiplied by 9.992 grid-seconds we get the Jupiter mass-circumference:
β = L (miles per grid-sec) x α (grid-secs) = 1,082,295.369 miles (mass
circumference, Jup)
Lightspeeds in each reference frame, Earth and Jupiter, are separated by factor C:
C = 144,000 (mps, grid SOL, Earth) / 108,307.709 (mps, grid SOL, Jup) =
1.329545250
The factor C boosts the Louis lightspeed (encoded in the Earth orbits) to the modern
lightspeed. Soon we shall discover that ratio C can be used to derive the opposition time
for either Venus or Mars.
The next step is to express the Jupiter lightspeed value in a raw fashion -- in Earth
degrees, irrespective of Earth lightspeed, by converting miles or arc-minutes to degrees
as follows:
λ = L (miles per grid-sec) / 46.41469457 (arc-mins per Earth deg) =
2333.478869 degs, Earth (for 1 grid-sec, Earth)
λ' = λ / C = 1755.095488 degs, Earth (for 1 grid-sec, Jup)
The number of arc-minutes in an Earth degree is derived directly from the French
measure of one-degree of meridian arc, in Louis grid-space.
A "grid" orbit is 360 degrees, because there are 360 degrees in a circle (of course), so the
number of Earth orbits for 1 grid-second, i.e. for the distance that light travels in a gridsecond on Jupiter, is:
J1 = λ' (degs, Earth) / 360 (degs, circle) = 4.875265244 orbits, Earth (for 1 gridsec, Jup-mass)
The measure of "Earth orbits" is just a raw translation of distance, transferring Jupiter
miles to Earth orbits, and does not take into account Earth's own lightspeed values. But it
does automatically include the Louis lightspeed value for Jupiter, which is built into the
Jupiter circumference as the distance travelled in 1 grid-second.
From Sun to Sun
The grid-distance for Jupiter (L = 108,307.7090) and the grid-distance for Earth
(144,000) are not two different descriptions of the same thing, separated by a span of
history lasting three centuries, but indeed are really two different physical lightspeeds.
They might be attributed to the spin rates of the two spinning tops as seen from the
Moon. But in another conception, Jupiter and Earth may be two aspects of the unity that
is our Sun.
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What we are doing is setting up Earth relative to Jupiter, which is to say, we are creating
the "reference frame" for Earth relative to Jupiter, where Earth rotations are referenced to
Jupiter rotations.
For J1 orbits, we have:
J1 (orbits, Earth, for 1 grid-sec, Jup) x 360 (degs per orbit) = 1755.095488 degs,
Earth, for 1 grid-sec, Jup
Next, already knowing the number of Earth orbits for the distance that light travels
around the Jupiter mass, we find the number of orbits in an arbitrary Earth "reference
frame" of 60 grid-seconds, i.e. a grid-minute:
L = J1 (orbits, Earth, for 1 grid-sec, Jup-mass) x 60 (grid-secs) = 292.5159146
orbits, Earth (for 1 grid-min, Jup-mass)
Remember that J1 is referenced to the Jupiter mass at 108,307 lightspeed. When we
multiply by 60 as above, we're expanding the Earth's reference frame from a single gridsecond to a grid-minute. This has the effect of making some number of Jupiter rotations
take place in a wider space of time. In other words, before we had some portion of the
Jupiter circumference fitting into 1 Earth grid-second, but now it fits into 60 Earth gridseconds. The frame has been "expanded." What is the utility of working in grid-minutes?
Later, when we compute the convergences for Venus and Mars relative to Earth and the
Sun, those convergences will be in the grid-minute reference-frame.
If the number of Earth orbits (in a grid-minute) for a rotation of light around the Jupiter
mass (in 1 grid-second) is scaled by the lightspeed ratio we get:
4
J = L (orbits, Earth) x 3 (lightspeed ratio) = 390.0212193 orbits, Earth, for 1 Jup
grid-min
Scaling by
4
3
gives the number of Earth orbits in an "expanded" grid-minute,
corresponding to either Mars relative to Venus, or Venus relative to Mars.
Clocked Time
For the Louis speed-of-light:
o = 140,109.8578 (mps, SOL) / 21615.60127 (miles, Earth circumference) =
6.481885748 orbits per clock-sec, Earth, Louis SOL
Notice that for orbital parameters L and J1:
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J3 = L (orbits, Earth, for 1 grid-min, Jup) / J1 (orbits, Earth, for 1 grid-sec, Jup) =
60 grid-seconds
Eq. B
The superimposition of the two speeds-of-light associated with either tier is:
140,109.8578 (miles per clock-sec) x 186,282.3959 (miles per clock-sec) = 2.61 x
1010 miles per clock-sec
This result may seem rather astounding, but in fact if you look at the derivations of these
numbers in the Sirius paper, you'll see that it's just:
2
3
x 174
... the latter number being the harmonic of the width of Earth's orbit around the Sun in
miles. The number 2.61 is used as a constant for the Planck-Wheeler area, used in
various physics applications related to surface area, but the magnitudes involved are at
the atomic scale.
Analysis
The opposition times for Mars and Venus can be derived in the most simple way from
the Jupiter mass-circumference:
1,082,295.369 x 180 = m x 3 = 584.4394993
1,082,295.369 x 180 = m x 4 = 779.2526656
These results compare favorably to publsihed values. The published value for Venus is
583.9 days. The published value for Mars is 779.9 days. The opposition times of Mars
and Venus do not both fall on the same slope: Mars is pulled to a higher value relative to
Venus, which is pulled towards a lower value relative to the linear mean. Could it be that
Mars and Venus are inverses of each other?
The number of Earth-orbits for Venus, translated to the modern speed of light, is:
α = 40 (orbits, Venus) x C = 53.18180996 orbits (at 144,000 modern grid SOL)
β = α (orbits, grid) x 1.293627749 (grid-to-clock) = 68.79746511 orbits (at
186,282 SOL)
The number of orbits β is a harmonic, in miles, of the Jupiter circumference, and also of
the Moon's circumference.
The number of miles for those Earth orbits is:
V = β x 21615.60127 (miles, Earth circumference) = 1,487,098.574 miles (for 40
orbits, Venus, at modern SOL, clocked)
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The same operations for the Mars orbits are:
α = 30 (orbits, Mars) x C = 39.88635750 orbits (at 144,000 modern grid SOL)
β = α (orbits, grid) x 1.293627749 (grid-to-clock) = 51.59809887 orbits (at
186,282 SOL)
M = β x 21615.60127 (miles, Earth circumference) = 1,115,323.931 miles (for 30
orbits, Mars, at modern SOL, clocked)
At Louis speed of light we get:
V (miles, modern SOL) / C = 1,118,501.664 miles (at Louis SOL, clocked)
M (miles, modern SOL) / C = 838,876.2481 miles (at Louis SOL, clocked)
Now note for Louis grid speed of light:
l = 108,307.7090 (miles per grid-sec Jup) x 60 (grid-secs Earth) = 6,498,462.540
miles
j = l x C = 8,640,000 miles
[INSERT PURE HARMONIC ANALYSIS]
Mass
Mass is a both a distance and a time. First we "demass" the Jupiter circumference. The
raw Jupiter circumference is:
α = 88,736.21766 (miles, diameter) x π = 278,773.0495 miles (circumference)
If this circular distance is multiplied by Jupiter's mass orbit, it's the same as dividing the
clocked speed of light by mass M, which demasses the circumference and leaves raw
miles:
β = α x 0.1294562111 (mass orbits per sec) =
140109.8578 (miles)
M
= 36,088.90275 miles at SOL (demassed circumference)
Converting from clock-seconds to grid-seconds:
Q = β / 1.293627749 (clock-to-grid) = 27,897.44017 miles at SOL (demassed
circum)
This is about one-tenth of the raw Jupiter circumference:
Q x 9.992782403 (grid-secs per orbit) = 278,773.0495 miles, raw circumference,
Jup
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Another derivation for Q is:
Q =
108307.709 (miles per grid-sec)
= 27,897.44017 miles per grid-sec (demassed)
M
Since this is the unit upon which mass M acts, we can call it a "mass unit." Dividing l, the
mileage for Mars for a full grid-minute, in Earth miles, by this massless Jupiter mileage,
1
gives a massed value which then is "un-massed" (again!) by multiplying by M :
8.5
l (miles) / Q (miles per grid-sec, unmassed) = 232.9411767 x 33 = 60 Mars
seconds in an Earth minute
This unmassed result is tuned to the Jupiter mass in terms of massless miles for Earth. So
Mars and mass M are in sync because 60 is a clean factor. Let's check out Venus:
8.5
j (miles) / Q (miles per grid-sec, unmassed) = 309.7058349 x 33 =
79.77271506 Venus seconds in an Earth minute
This is not a clean factor relative to Earth. The ratio of those 79.77 factor-seconds to a
clean 80 factor-seconds is fudge-factor ff:
ff = 80 (factor-seconds) / 79.77271506 (factor-seconds) = 1.002849156 fudge
factor, Venus
Note that:
C = 186,282.3959 / 140,109.8578 = 1.329545250
4
C x ff = 1. 33 = 3
More Fudge Factors
The idea is to apply, at a preliminary stage, a fudge factor to Venus to get the correct
opposition time for Venus, or to Mars. In fact, this entire paper may center around these
fudge factors, because their development and use leads to the simplified opposition-time
calculations.
Fudge factor ff can be derived as a ratio of two other fudge factors -- one higher and one
lower than ff. The first involves the number of miles for the Venus orbit: V. When V,
which is an un-massed value of the Earth's circumference, is multiplied by mass M, and
then further multiplied by 1.5, we get:
V x M x 1.5 = 8,660,162.283
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This is close to the neutral-point (or gravitational null-boundary) of the Sun, which is
one-tenth of the distance from the Earth to the Sun: 8,700,000 miles. When we take the
ratio of the result above to the full distance, we get a new fudge-factor:
fh = 8,700,000 (miles) / 8,660,162.283 (miles) = 1.004600112
This is a high value. Interestingly, it pairs Venus with the Earth-Sun distance. For Mars,
distance M can be multiplied by mass M, and then further multiplied by .75, to get:
M x M x .75 = 3,256,813.669
Since there is solar neutral point at one-tenth the distance from the Earth to the Sun,
measured from the Earth (the Sun is the more massive body and its gravitation extends
past the midway point between Earth and the Sun), and since from the Moon there is a
neutral point which is again one-tenth the distance from the Earth to the Moon (but is
measured from the Moon's surface because the Earth is the more massive body and its
gravitation extends past the halfway point of the two planets), then there must be a
neutral-point for Mars at one-tenth the distance to Mars, which the French give as
32,625,000 miles.
When we take the ratio of the result above to the full distance, we get a third fudge
factor:
fl = 3,262,500 (miles) / 3,256,813.669 (miles) = 1.001745980
This is a low value, and it pairs Mars with the Earth-Mars distance.
Finally, we see that the ratios of the high and low fudge factors is our canonical MarsVenus fudge factor, ff:
1.004600112
ff = fh / fl = 1.001745980 = 1.002849157
The Eightfold Path
Let's look at the many ways that the Mars and Venus opposition times can be derived in
terms of mass and lightspeed values. Borrowing from Buddhism (apologies!) I call this
the "eightfold path."
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Mars
= 779.5
Venus
= 584.6
Mars
= 783.2
Venus
= 587.4
Mars
= 779.5
Venus
= 584.6
Mars
= 783.2
Venus
= 587.4
1.5
x
108,307
x
186,282
x
8.5
33
1.125
x
"
x
"
x
8.5
33
1
x
"
x
"
x
33
8.5
.75
x
"
x
"
x
33
8.5
1.5
x
140,109
x
144,000
x
8.5
33
1.125
x
"
x
"
x
8.5
33
1
x
"
x
"
x
33
8.5
.75
x
"
x
"
x
33
8.5
The products of the lightspeed values in both groups have the same value. The Louis
lightspeeds in the first column are very much tied to mass M through the Jupiter
circumference, in which light travels 108,307.709 miles in 1 grid-second.
By contrast, 186,282 and 144,000 are "un-massed" values related to the Earth's
circumference. So what this table shows is that Mars and Venus oppositions can be
expressed either as two massed values (when the lightspeed pairs are multiplied by mass
M, thereby adding mass to the Earth values); or as two Earth values (when the lightspeed
pairs are multiplied by the inverse of M, thereby cancelling out mass M in the Jupiter
values).
In either case there is a superposition of two lightspeed values. All the important ratios
seem to be present. These very simple ratios are simple only on first inspection; they
blossom and flower into a multitude of relationships that require careful consideration.
Exploring Superpositions
An opposition is the convergence of two orbiting bodies relative to some start point; the
start point rotates around with one of the bodies and converges again at some later time
with the opposing body. The convergence program in the appendix takes two parameters:
a short orbit parameter, and a long one. The idea is as follows: a) mass M plays a large
role in these orbits, and in fact, numerically, is one of the orbiting bodies; and b) the
other orbiting body is in fact the superposition of two or more orbiting bodies.
In other words, the oppositions of Mars and Venus are superpositions of mass-related
wavefronts of light. Let's see how three simple values shed light on the Earth and the
Moon.
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The first value is V, for Venus orbits in miles, at Earth clock-speed. We will only use V
and not M, which gives the same results, the critical ratio being factor C.
The second value is 10 orbits at the Louis clockspeed for light: 1,401,098.578 miles. The
third value is 7.5 orbits, also at the Louis clockspeed: 1,050,823.934 miles. We'll call
these values U and T, for unity and three-quarters, respectively. These are both mass
distances.
Before we go on we must explore the mass orbital value Q.
What Q Is
Recall that Q is:
Q =
108307.709
= 27,897.44017 massless miles per grid-sec
M
These are base units in Earth mileage for mass M. The Jupiter circumference, from which
all of the Earth orbits are derived, is:
β = 9.992782403 (grid-secs, full orbit, Jup mass) x 108,307.709 (miles per gridsec, SOL) = 1,082,295.369 miles, mass circumference, Jup
The quotient Q / β amounts to:
108,307.709 (miles per grid-sec) x 8.5 (M) / [108,307.70933 (miles per grid-sec) x
9.992782403 (grid-secs, full orbit, Jup-mass) x 33 (M)]
Cancelling terms and inverting 9.992:
1
s = M x .1000722280 (orbits per grid-sec, Jup circumference) = 0.02577617997
un-massed orbits per grid-sec, Jup circumference
This is the "short" orbital period used by the convergence program. It's the number of
grid-seconds around Jupiter without the Einstein mass factored in. Next we will break
down the superposition terms given in the previous section in terms of s.
Break It Down
In this section we'll follow a numerical recipe for all of these terms. For Mars:
α = V (miles, at 140) x T (miles, 7.5 orbits at 140) / V (miles, at 186) =
790,363.4228 miles, Mars (roughly 5.625 orbits at 140)
Eq. A
Next we apply s to U:
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λ = s (unmassed orbits per grid-sec, Jup circumference) x U (miles, 10 orbits at
140) = 36,114.96906 (unmassed orbits per clock-sec x 10, at 140)
Eq. B
The result of Equation B, unmassed orbits of the Jupiter circumference (per 10 clockseconds), is a time value. This number times the result of Equation A gives a harmonic of
the Moon's diameter:
α x λ = 2188.464904 miles, Moon diameter (harmonic)
Now let's invert the Venus pair:
∆ = V (miles, at 186) x U (miles, 10 orbits at 140) / V (miles, at 140) =
1,862,823.959 miles, Venus (roughly 13.33 orbits at 140)
Eq. C
Applying s to T:
κ = s (unmassed orbits per grid-sec, Jupiter circumference) x T (miles, 7.5 orbits
at 140) = 27,086.22681 (unmassed orbits per clock-sec x 7.5, at 140)
Eq. D
The result of Equation D is a "low Moon" orbital value. This result multiplied by the
mileage of Equation C gives a harmonic of the Earth's diameter:
∆ x κ = 6877.384482 miles, Earth diameter (harmonic)
For the Earth and Moon diameters, more values can be approximated by computing with
10 orbits instead of 9.992 orbits in s. This doubles the accounting above, and we see that
the averages are:
Earth Diameter
Moon Diameter
6877.384482 +
6882.351879
/ 2 =
6879.868180
2188.464904 +
2186.885360
/ 2 =
2187.675132
The Apollo missions gave the nautical diameter of the Moon via laser altimetry as 2176
miles. From the French survey of Louis XIV, the Earth's diameter is 6880.45 miles. The
Moon values are less accurate than the Earth values, probably because the orbital values
for Venus are specified as Earth orbits, not Moon orbits.
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Two In One
The average Moon diameter, found in the previous section, produces the following
circumference:
2187.675132 (miles, Moon diameter) x π = 6872.784122 miles, Moon
circumference
Venus value V is a superposition of two entities. If one of those entities is the Earth, then
we find the other to be:
1,487,098.574 (miles per Venus orbit, at 186) / 2161.560127 (miles, 1/10 Earth
circumference) = 687.9746510 miles, 1/10 Moon circumference
V is an expression of the Earth and the Moon, both at one-tenth their normal values. The
superposition of the circumferences (orbits) of the Earth and the Moon gives the
opposition time of the Earth relative to the apparent image of Venus.
For Mars, we first take the average of: a) the Moon value just given; with b) the average
found in the previous section:
(687.9746510 + 687.2784122) / 2 = 687.6265315 miles, 1/10 Moon
circumference
Then, if one of the superposition entities is the Moon, the other is:
838,876.2480 (miles per Mars orbit, at 186) / 687.6265315 (miles, 1/10 Moon
circumference) = 1219.959105 miles, circumference
The identity becomes plainer when we take the diameter:
33
1219.959105 / π = 388.3250439 x 10 = 3883.250439 = 1000 x 8.5 = mass
M
In other words, the second body is mass M. Mars is the expression of the Moon and mass
M. But what is mass M really?
The Mass Perspective
In this part we come to an elegant "mass-centric" solution to the opposition problem.
Two things about lightspeeds. First, there are two lightspeeds involved: 144,000 for
Earth, and 108,307.709 for Jupiter. This Earth and Jupiter duality is crucial.
Second, the use of those lightspeeds seems valid in some way, first of all because Jupiter
grid-space is a "grid-second" space in the 105 range. But they are also valid because the
superposition of their values -- the Planck-Wheeler area -- is indeed an area used in
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black-hole research. But grid values for Earth and Jupiter lightspeeds shouldn't be used
as clock-values, which is what we want to measure opposition times with. We do not live
in a grid. Not usually.
So I resolved to use 140,109.8578 for Jupiter and 186,282.3959 for Earth: i.e., the
clocked values. But I also wanted to work with them as a vector surface area. Now, the
square root of some surface area turns out to be the side of the square that encloses that
area -- logically enough. So I would use the square root of the surface area and end up
with a geodesic. This would be used as the scale value for both short and long values.
There was another consideration. As hinted at in the eightfold path, there are many
simple, fundamental ratios that are always cropping up. But one ratio in particular is
crucial:
1. 33
4
= 3
This is a lightspeed ratio:
4
3
. This number is a key factor between speeds-of-light in the
pure ratio world: i.e., 144 to 108. So this number really has a role in the relationship
between Earth and Jupiter. In essence, the entire convergence fraction is scaled
4
universally by 3 .
New Convergence Values
Finally, I really wanted a convergence. The short and long parameters should be
normalized and have the same magnitudes. Logically the Earth and the Moon are two
entities on either side of mass M. Venus is a superposition of the Earth and mass M; Mars
of mass M and the Moon.
The square root of the lightspeed surface-area of the hypercylinder gives a "lightspeed
geodesic":
α =
186282.3959 x 140109.8578 lightspeed geodesic
This is part of a scale factor that will be used after computing epicycles, using the
program from Appendix A.
For Mars, the short and long values are:
3
s = 6876.265315 (miles, Moon orbit) x 4 = 5157.198988 miles
l = 12199.59105 (miles, mass M orbit)
Eq. L (Mars)
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Notice that mass M has been "moon corrected" as in the previous section. You can see
that we're still using the superposition values found in the previous sections for Mars.
For Venus, the short and long values are:
3
s = 12199.59105 (miles, mass M orbit) x 4 = 9149.693288
l = 21615.60127 (miles, Earth orbit)
Eq. M (Venus)
There's a nice symmetry here, because as the eightfold path shows, Mars and Venus are
1
generated by both M and M, and that's what we have here with these l / s ratios. Mass M
reflects on itself through the Earth and the Moon, which seem to be on either side of it. It
turns out that the mass M mileage can be derived by superimposing the Earth and the
Moon:
21615 x 6876 = 12191
Convergence Results and Post Transformation
The epicycle result and synodic period for Mars are:
ct = .732307849316 epicycle result
Lightspeed values are applied along with mass M as a post transformation:
S =
1
= 1.594353914 x 10-9
α (lightspeed geodesic) x M (Einstein mass)
l x ct x S = 1.42437287168 x 10-5 synod, Mars
For Venus:
ct = .733977284313 epicycle result
l x ct x S = 2.52949992908 x 10-5 synod, Venus
Notice that the epicycle results for Mars and Venus are almost the same. This indicates
that the relative motions of the two systems are nearly identical.
The post-transformation and opposition result for Mars is:
142,437.2871 (Mars synod, harmonic) / [ 365.2563604 (days, Earth year) / 2 ] =
779.9304956 days, Mars opposition
The post-transformation is simply an Earth year in days. The Mars opposition is a
superposition of the Moon and mass M, and these things are wheeling around the
amphitheatre of the Earth. The division by two for the Earth year is a mystery.
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The post-transformation and eastern elongation result for Venus is:
2,529,499.929 (Venus synod, harmonic) / 4331.865 (days, Jupiter year) =
583.9286148 days, Venus eastern elongation
Here, the Venus eastern elongation is a superposition of the Earth and mass M, and these
things are careening around the grand promenade of Jupiter, visible from Earth. There
are many things to think about here and caution is the order of the day. Units and
magnitudes need further development -- possibly the subject of a future paper.
Reference
Jones, Barrie William. The Solar System, Pergamon Press, 1984
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APPENDIX A: BASIC PROGRAM
Here is a very simple BASIC program for computing a convergence using what some
call a "continued fraction." I call it a synod, or an epicycle. The epicycle gets two seed
values, l and s, for long and short orbiting bodies. You can use the sidereal periods in
days, for example, of two planets. The program computes an "epicycle result," and then
multiplies the long value by this result to get a convergence time for the two bodies. For
example, the Earth and Moon sidereal periods (long and short respectively) will yield the
lunisolar synodic period of 29.53 days. This program is invaluable for double-checking
opposition times, but I also use it upon occasion with experimental values, as in this
paper.
DIM ct AS DOUBLE, d AS DOUBLE
DIM r AS DOUBLE, t AS DOUBLE
DIM l AS DOUBLE, s AS DOUBLE
s = 1# / 144000#
l = 9.992782403# / 108307.709#
REM: cycle time is ratio of orbits
ct = l / s
REM: take the series:
d = 1#
r = 0#
FOR i = 1 TO 100
d = d / ct
r = r + d
NEXT i
PRINT r
REM: conjunction = result * long orb
t = r * l
PRINT t
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APPENDIX B: Pure Harmonic Analysis
The results of this paper can be checked against pure harmonic values as follows. For
Venus:
1
α = 256 x M x 3 = 331.2941176
ρ = α x 252.9499929 = 838,008.4469 (harmonic)
4
ρ x 3 = 1,117,344.596
Venus shows up on the Jupiter horizon at grid lightspeed 108, embedded as:
( 256 / 3 ) x 252.9499 / 2 ~= 108 Jupiter SOL (harmonic)
A slightly different (but symmetric) set of pure harmonics for Mars gives:
1
β = 81 x M x 4 = 78.61764705
λ = β x 142.4372871 = 1,119,808.436 (harmonic)
3
λ x 4 = 839,856.3270
Mars shows up on the Earth horizon at grid lightspeed 144, embedded as:
( 81 / 4 ) x 142.4372 / 2 ~= 144 Earth SOL (harmonic)
This beautifully shows how the two systems pull to either side of the centrum. Averaging
values we get the centrum, i.e. the clocked Jupiter mass circumference in Louis space,
tuned to the Earth's circumference:
Ah = (1117.344596 + 1119.808436) / 2 = 1118.576516 (vs. 1118.501664)
3
Al = Ah x 4 = 838.9323870 (vs. 838.8762481)
The centrum values are shown at the far right above.
Tripoint Path Publications