Universal Physics Journal
Authors: Ethan Skyler, Ryan E. Skyler, Robert E.
Understanding the role that forces play in producing Earth's ocean tides remains an ongoing challenge. Even though all tidal elements are known, fitting these elements together in a logical manner that is true to Nature has proved to be an elusive goal. In considering Earth's tidal events, we think there is a considerable advantage to using the Universal Physics perspective regarding the behavior of the action and reaction forces involved. With this thought in mind, let us now launch a new effort to understand the types of forces present, their characteristics, and the roles they serve while working together in causing Earth's tides. (If your interest in the cause of Earth's tides does not extend beyond a quick summary, then scroll down to paragraph 62.)
(2) A tide occurring to portions of Object A usually has as its
beginning, gravitational forces that are being generated individually within Object A's
components of matter caused by an imbalance in their operation due to their reception of gravitational energy
previously emitted in Object A's direction by the myriad of individual
components of the matter of distant Object B. These internal gravitational
forces being generated within Object A's matter act as the cause of
acceleration for portions of Object A's matter with said acceleration
directed toward Object B. Of prime importance where the tides are
concerned is the recognition that it is possible for different portions of Object A's matter
at different rates in Object B's direction. Development of this "different rates" recognition is
key to understanding the forceful process that results in causing Earth's high tides.
(3) This investigation begins by focusing upon a space where just two objects
exist, Earth as Object A
and Earth's moon as Object B. Since Earth's diameter is large and its
distance from the Moon is relatively short, the force-generating
effect of the Moon's gravitational energy is reduced by as much as 3%
as this energy expands and thereby becomes less dense while passing through
nearly 8000 miles of Earth's matter, from near side to far side.
Accordingly there exists a small but still significant difference in the magnitude of the action force of Moon
gravitation within Earth's near side facing the Moon as compared to the lesser
magnitude of average Moon gravitation force being experienced both at
Earth's center of matter and also at every point on a curved average
division of Earth's matter into slightly unequal near and far portions where each
such point is the same distance from the Moon's center of matter as is
Earth's center of matter. This above average action force excess, present only
within Earth's near side portion, is herein
termed the "excess gravitation force" and is the root cause of Earth's
(4) Earth and the Moon, as a two body system, share a center of matter
about which they both orbit. This center of matter is known as the
barycenter. In our event, the Moon will present the same side toward
Earth as it does now while orbiting the barycenter. But Earth will have no rotation what-so-ever
as it's center of matter also orbits the barycenter.
This setup will help to isolate the forces and accelerations
that are responsible for Earth's ocean tides.
(5) The barycenter is an Earth/Moon orbital axis that resides about
1063 miles below Earth's near side surface. The Moon's center of
matter resides on average at 238,855 miles distant from the barycenter about which it orbits in a time period of 27.32 normal Earth days.
Earth's center of matter is 2901 miles deeper underground from the barycenter about which Earth orbits also in 27.32 Earth days. It is
known that the
force of Earth gravitation, being generated individually within the Moon's
myriad of components of matter, is the acceleration/Action force
responsible for causing the Moon to orbit the barycenter in 27.32 days.
In a mutual manner, the force of Moon gravitation being generated individually within
Earth's myriad of components of matter is the
acceleration/Action force responsible for causing Earth's center of matter
to orbit the same barycenter in the same elapsed time. Overall these
two gravitational acceleration/Action forces, though widely separated, are equal in magnitude and
opposite in direction in compliance with Newton's LAW III and
Rule 8 of the Universal Physics Rules for
Force & Motion. Also it is
known that the Moon gravitational forces being generated
within Earth's components of matter are directed toward the Moon's
components, and not directed toward the underground barycenter which is
merely the site of the orbital axis of the Earth/Moon system.
(6) It is also known that, on average, Earth is accelerating in a weightless
manner in the vacuum of space with this acceleration directed toward the
orbiting Moon. Considering
our non-rotating Earth's dual high tides, Earth's ocean water is maintaining a
high tide mound on Earth's side nearest the Moon plus a second high tide mound of ocean
water on Earth's far side. While at first one may think that all portions of
our non-rotating Earth
are experiencing the same rate of Moon-directed acceleration, we find that different rates of acceleration are
required of Earth's ocean waters in order for these dual high tide mounds to
be maintained while slowly traveling around the non-rotating Earth in sync with
Earth's and the Moon's
leisurely orbit of
(7) The challenge here is to determine the arrangement of forces that could possibly be at work to cause different rates of acceleration
for Earth's ocean waters in order for the dual high tide mounds to be
maintained. After all, it is well-known and fully accepted that different objects
in a vacuum accelerate at the same rate when this acceleration is powered by
the internal acceleration/Action force of gravitation. By this "same rate
of acceleration" recognition, Earth's water and land
should accelerate through the vacuum of space at the same rate toward the Moon making high
tide mounding impossible. Yet with Earth displaying dual high tide
mounds, one need not abandon logic to think that the acceleration/Action forces being
experienced by Earth's water are somehow different from the
acceleration/Action forces being experienced by Earth's land.
(8) Returning to the subject of Earth's near side excess
consider that, on average, the gravitational acceleration/Action forces
present at Earth's center of matter and at any point on Earth's curved
average gravitational division, as previously described in paragraph (3), are at the correct magnitude to cause that
portion of Earth's land and water to accelerate at the correct rate which is Earth's average rate of acceleration
toward the Moon. Next consider that the land and
water on Earth's far side are as much as 3963 miles farther away from the Moon than
matter at Earth's center. Here on Earth's far side, the force of Moon
gravitation is less than average. Yet if the actual rate of acceleration
for Earth's far side matter were
to truly be less than average then a large portion of Earth's far side
matter would have to separate or break free from the rest of Earth.
Since this does not happen, then some additional action force must be
present and effective in bringing the magnitude of the total
acceleration/Action force up to that required to match with the average rate of acceleration being experienced by
all matter located at the average gravitational division that divides Earth into
slightly unequal near and far side
portions. But if this additional action force,
predicted to be present on Earth's far side, is also a gravitational force then it
will be of no help in explaining why the land and water located there accelerate at
different rates. To be helpful in solving the "different rates"
puzzle, this additional force has to be some form of Type 3 external (contact) tension force that is
toward the Moon.
(9) Let us now consider the Moon gravitation forces present within Earth's
side, which is the side facing the Moon. Here we are as much as 3963 miles closer
to the Moon than matter located at any position on Earth's average
gravitational division. Due solely to the
reduced distance to the Moon's center of matter, the acceleration/Action force of Moon
gravitation is greater than the average force of Moon gravitation
at Earth's average division. If this greater-than-average force is
truly able to
cause greater-than-average acceleration for Earth's near side
then again some portion of Earth will be forced to separate or break free.
Since this does not occur, then the next question is: "What happens to the
excess force of Moon gravitation that is being generated within the
components of matter of Earth's near side portion?"
We are thinking that it accumulates and thereby stacks up in magnitude while traveling back, in
a serial manner, through Earth's matter to become the Type 3 external (contact) force needed to tug on Earth's
portion thereby effecting its rise to the average rate of acceleration
being experienced at Earth's average gravitational division. If it does provide this external
(contact) force then there should be evidence of Earth's shape being stretched out along a line drawn from
the Moon's center through Earth's center.
This stretching-of-Earth, if real, will serve the same role as Isaac
Newton's "distended" rope that is pulling forward with an external
force on the dragging stone that is equal and opposite to the external
(contact) force the rope is pulling backward on the walking horse.
(10) Standing as proof of this prediction, Earth's shape is already known to stretch
along the previously described central line. Since the excess force of Moon
gravitation being generated individually by the myriad of components of
Earth's near side solid matter is not being terminated as it would be if it
were acting as the cause
acceleration for Earth's near side solid matter, this excess force in the Moon's
direction is free to accumulate as it is transported as an external serial transfer
tension force that passes back through Earth's solid matter to
terminate in the action of forcefully raising
the acceleration rate of the solid matter within Earth's far side up to the average
rate of Earth's average gravitational division.
(11) For the above stated reasons, we accept that the excess action force of Moon gravitation
being generated within Earth's near side portion of solid matter does end up tugging on the solid matter on Earth's
portion. But while
this excess acceleration/Action force is helping to cause an
average acceleration rate for Earth's far side solid matter, we wonder if
it is also
doing the same for Earth's far side water? After some consideration
we draw the conclusion that the accelerational effect
of this external force is being only indirectly applied to Earth's ocean water through the
interface of the water's friction with the ocean floor. This somewhat ineffective frictional method
of force transfer serves to prevent full transfer of the acceleration/Action force
required to cause the average rate of acceleration for Earth's far side
water. Hence Earth's far side ocean water, that is approaching the
high tide mound, is left with a
lower-than-average rate of acceleration than is being experienced by the
far side ocean floor. (The transfer of acceleration/Action forces to the ocean
water from the ocean floor will be greater at greater depths and at its
least near the ocean surface.)
(12) Consider the following event. Hold your forearm out
horizontally with your palm turned up. Pour a single drop of water
into your palm and then quickly pull your hand closer to your body in the
horizontal direction. As you will discover, the external
acceleration/Action force from your hand depends upon the same ineffective
frictional method of force transfer to the water. While the water does
accelerate in your direction, its rate will always be
lower than the acceleration rate of your palm. The result of this
lower rate of acceleration is that the water's position on your palm lags
behind as it automatically shifts backward in the direction of your
forward-accelerating fingers without there being present any event-causing
action force in that backward direction.
(13) In the same manner, due to the external nature of this excess
Type 3 serial transfer
acceleration/Action tension force, Earth's ocean water on the far side
that is approaching the high tide mound experiences a lower rate of acceleration
in the Moon's direction than the average rate
now being experienced by the ocean floor underneath. This allows the ocean water to lag
behind the accelerating ocean floor resulting in the automatic formation of
the high tide mound on Earth's far
side. Notice that here too there exists no backward-directed
tidal-mound-building action force in this event for no such action force is known
to exist in
Nature. There does exist a myriad of backward-directed internal acceleration/Reaction support forces
being reactively generated within the myriad of forward-directed
accelerating components of Earth's matter. But as reaction forces,
they are incapable of acting as the cause of any event. Here the role
of these internal acceleration/Reaction forces is to provide equal and
opposite support and termination for both the internal and external
acceleration/Action forces responsible for
causing forward-directed (Moon-directed) acceleration of Earth's far side
(14) Now that the external-force cause of Earth's far side high tide mound is
what then of the high tide mound on Earth's side nearest the Moon? Here, as much
as 3963 miles
closer to the Moon than Earth's average gravitational division, the acceleration/Action force of Moon gravitation
affecting Earth's near side ocean water is
above average. This force is also above average for
side ocean floor. Yet the acceleration rate of the near
side ocean floor is no greater than what is average at Earth's average
since all excess (non-terminated) Moon gravitational forces being generated within Earth's
near side portion solid matter are being transferred back to augment Earth's
far side acceleration. This leaves Earth's near side
ocean waters, being somewhat restrained by the water's friction with the
accelerating ocean floor, otherwise free to respond to this above average internal force of Moon
gravitation being generated within the myriad of components of matter of Earth's
ocean water. Here a higher-than-average rate of Moon-directed acceleration
is being forcefully caused by the near side excess gravitation force resulting in the forceful building of Earth's
high tide mound.
(15) In summary, it is clear that the external serial transfer tension forces
a major role in causing Earth's near side and far side waters to
accelerate at rates different from the average rate of Earth's solid matter below.
On Earth's far side, the Moon-directed external tension force causes the
increase in the solid matter's acceleration rate up to average which
the ocean water to mound up on Earth's far side as the water's acceleration
rate lags behind. Meanwhile, on Earth's near side, the external tension force,
directed away from the Moon, holds the solid matter's Moon-directed rate of
acceleration down to average while the ocean water's Moon-directed rate remains
above average resulting in the forceful building of Earth's near side ocean mound.
This "different rates" of acceleration for portions of Earth's water compared to Earth's
solid matter at the same general location on Earth is directly caused by the
different characteristics of internal gravitational forces compared to
external tension forces. Moon gravitational forces affect Earth's land
and water equally while tension forces applied to Earth's land have
only a limited effect upon Earth's water. (By introducing the
stretching of Earth, we are acknowledging that Earth's solid matter also
experiences tides. Again, this stretching is due to the presence of
the external tension forces described above. Invisibly, Earth's
atmosphere must also mound up over the ocean tidal mounds,
again due to all the same reasons as does Earth's ocean water.)
(16) Now let us perform some calculations of the forces and
accelerations that are either directly causing or indirectly resulting in
the formation of our non-rotating Earth's high tidal mounds. We will employ
three relatively small portions of Earth's matter
with each portion being an object with a mass rating of 40 million
or in SI terms, 18.1 million kg which is close to 1/3 the mass rating of the
carelessly-operated Titanic ocean liner. Earth Object EO1 will be positioned at
Earth's near side facing the Moon. Earth Object EO2 will reside at Earth's
center of matter. Earth Object EO3 will reside at Earth's far side
from the Moon. All three objects will remain on the line previously
drawn from the Moon's center of matter through Earth's center of matter.
In order to stay on this line, like the barycenter embedded deep within
Earth's matter, objects EO1 and EO3 will need to magically slide sideways through
the non-rotating Earth's matter as all three travel in a circular manner
about the Earth/Moon barycenter as they maintain their alignment with the orbiting Moon.
(17) We will begin by embedding object EO1 into Earth's near side
surface facing the Moon. Here we will calculate the above average
force of Moon gravitation being actively generated within the myriad of
components of EO1's matter. During the following series of calculations, we will
specific values plus some general understandings drawn from a number of sources including "Fundamental
Astronomy", Second Enlarged Edition by H. Karttunen, P. Kroger, H. Oja, M.
Poutanen, and K.J.Donner Editors; "In Quest of the Universe"
Second Edition by Karl F. Kuhn; "Classical Mechanics" A Modern Perspective -
Second Edition by Vernon D. Barger & Martin G. Olsson; "University Physics" Eighth Edition by Hugh
D. Young; and "Conceptual Physics" Seventh Edition by Paul G. Hewitt.
Of course, the specific values are subject to improvement in accuracy. Proof of concept is
our goal which, if successful, will continue to remain intact even if minor adjustments of these values
are made. Calculations are performed using
two Microsoft Windows calculators set to
scientific mode and the unique and ever-handy A.P.C. units converter which
includes an acceleration & gravitation calculator.
(18) The values below, in US units, will be used in the calculations
that follow. (Please note that these values may be successfully pasted into
the Windows Calculator just as they are printed below. Thus
1.61,994e+23 is accepted along with the included comma. Also when
using the A.P.C. program, change the "e" for exponent to "d"
Gravitational Distance from Earth Objects to Moon c/m.
Gravitational Constant = 3.321998855540755e-11 using US units ft, lb.m, and lb.f. (Shift the decimal 11 positions left for the true value.
(19) Problem 1. Embed object EO1 into Earth's near side facing
the Moon. Determine the Moon gravitational force being generated
within EO1's matter.
force = Gravitational Constant * Moon Mass * EO2 Mass / Moon Distance2
= (3.32,199,885,554,075,5e-11 lb.f*ft2/lb.m2) * 1.61994e+23 lb.m * 40,000,000 lb.m / (1,240,245,604 ft)2
= (3.32,199,885,554,075,5e-11 lb.f*ft2/lb.m2) * 6.47976e+30 lb.m2 / 1,538,209,158,241,324,816 ft2
= (3.32,199,885,554,075,5e-11 lb.f*ft2/lb.m2) * 4,212,663,344,320.5 lb.m2/ft2
= 139.94 lb.f
(EO1 is 4.6 lb.f above average. See Problem 2 below.)
(20) Problem 2. Determine the Moon gravitational force being
generated within EO2's matter, located at Earth's center. This is the
average force per unit mass of all of non-rotating Earth's accelerating
force = Gravitational Constant * Moon Mass * EO2 Mass / Moon Distance2
(EO2 is average)
(21) Problem 3. Embed object EO3 into Earth's far side, which
is the side opposite to
the Moon. Determine the Moon gravitational force being generated
within EO3's matter.
force = Gravitational Constant * Moon Mass * EO3 Mass / Moon Distance2
= (3.32,199,885,554,075,5e-11 lb.f*ft2/lb.m2) * 1.61994e+23 lb.m * 40,000,000 lb.m / (1,282,096,916 ft)2
= (3.32,199,885,554,075,5e-11 lb.f*ft2/lb.m2) * 6.47976e+30 lb.m2 / 1,643,772,502,016,711,056 ft2
= (3.32,199,885,554,075,5e-11 lb.f*ft2/lb.m2) * 3942005351744.30 lb.m2/ft2
= 130.95 lb.f
(EO3 is 4.39 lb.f below average. After being raised to average, 96.8% of its acceleration/Action force is internal Moon gravitation which directly affects water while the remaining 3.2% is an external tension force which only indirectly affects water.)
(22) Next we need to check if the 135.34 lb.f being
experienced by the EO2 portion embedded at Earth's center is the correct
magnitude of force needed to cause EO2 to orbit the barycenter, with a
2902.25 mile or 15,323,881 ft radius in 27.3217 days or 2,360,595 seconds.
In the process of checking, we will end up knowing Earth's average rate of
Moon-directed acceleration. First we will need
to know EO2's velocity about the barycenter.
(23) Problem 4: Calculate EO2's orbital velocity about the
Velocity = Orbital Circumference / Time
(40.7875 ft/sec is about 28 miles/hour which reveals the leisurely speed at which Earth's axis orbits the Earth/Moon barycenter.)
(24) Problem 5: Now we are ready to calculate the Moon-directed
force required to cause EO2 to orbit the barycenter, 2902.25 miles distant.
We will use the formula developed long ago by Isaac Newton for calculating
the action force in circular events named by Newton as "centripetal force".
We include division by 32 to convert the answer from the absolute force Poundal units to
more commonly used lb.f
force = Mass * Velocity2 / Radius / 32
(25) Problem 6: You can see that the 135.71 lb.f required to
cause EO2 to orbit the barycenter 2902.25 miles distant when its orbital
velocity is 40 ft/sec is quite close to the 135.34 lb.f of Moon gravitation
we calculate as being experienced by EO2 in Problem 2. Are you
wondering what force value we will get if we treat EO2's acceleration as
a linear event? The F=ma formula is also based upon Isaac Newton's work.
We will need EO2's rate of acceleration which in circular events is Acceleration =
Velocity2 / Radius. The answer is already
calculated and underlined in Problem 5 above.
force = Mass
* Acceleration / 32
(26) Now that we have used three different methods to generally verify the
135.34 lb.f average force
being generated within EO2's matter, let's consider how Earth's near side excess acceleration/Action force causes to exist the external tension force that acts equally in
opposite directions as the final step in causing our non-rotating Earth's dual high
tides. Let us take a fresh perspective of this accelerating Earth
event by supposing that our three Earth Objects continue to exist at their
current positions while Earth itself just magically disappears. The
Moon gravitational forces be generated within each object will
continue unchanged. Only now EO1 will begin moving ahead of EO2 in the
Moon's direction since it is free to experience the higher rate of acceleration being
caused by its 4.6 lb.f greater-than-average near side gravitation
Meanwhile, EO3 with its 4.38 lb.f less-than-average Moon gravitation force
will begin lagging behind EO2 with its average acceleration rate of 0.00,010,8564 ft/sec2. How do
you think we could cancel the expansion of the distances between the
three Earth Objects?
(27) At the front of our three object lineup, to reduce EO1's
acceleration rate down to average, one would need to exert a backward pull of 4.6 lb.f.
At the rear, to increase EO3's acceleration rate up to average, one would
need to exert a forward pull of 4.39 lb.f. If we could
establish a mechanical connection between EO1 and EO3 then EO1's excess
gravitation 4.6 lb.f would
be more than enough to provide the required force needed to increase EO3's
rate up to the average being experienced by EO2. As this connection is
being made, EO3's acceleration will switch from being 100% caused by Moon
gravitation to instead being directly caused by a combination of internal
forces of Moon gravitation and external (contact) forces being delivered
through its mechanical connection with EO1.
(28) Suppose we attach one end of an 7926.4 mile long strand of dental
floss to the back side of EO1. The strand will pass freely through an
opening provided in EO2 on its way back to be tied to the front side of EO3.
Since we have the use of magical powers in making Earth disappear, at the
same instant of the disappearance, our dental floss strand/tow cable will be
installed and properly tensioned to pull forward against EO3 with the same
4.6 lb.f magnitude of acceleration/Action force that the strand is pulling
backward against EO1.
(29) The role of the strand as providing a bipole external tension
force experienced by both EO1 and EO3 has already been well-described
by Isaac Newton as follows:
"If a horse draws a stone tied to a rope, the horse (if I may say so) will be equally drawn back towards the stone; for the distended rope, by the same endeavor to relax or unbend itself, will draw the horse as much towards the stone as it does the stone towards the horse, and will obstruct the progress of the one as much as it advances that of the other." (Quote is from PRINCIPIA, Volume I, "The Motion of Bodies" LAW III, pages 13-14.)
(30) Think of our dental strand as Newton's rope while EO1 is the horse and EO3
is the stone. In our event, mixing in Newton's words, the tensioned
strand, through the application of an external (contact) force, is pulling
equally and oppositely on both objects thereby obstructing the
accelerational progress of EO1 as much as it
advances that of EO3. The result is that EO1 and EO3 generally share
the average acceleration rate and acceleration direction being experienced by the central
(31) Let us now restore the Moon and our three Earth Objects, along with the
attached and tensioned dental strand, to the instant Earth
disappeared. (This is required since in Earth's absence, the Moon no longer
follows a orbital path about Earth in space.) In Problem 1, we learned that EO1, in
the lead, is experiencing a Moon gravitational force of 139.94 lb.f.
Without the attached strand, this force will be an internal Type 1
force as EO1 will be accelerating in a weightless manner in the Moon's
direction. But with the 7926.4 mile long strand attached to its back
side, 3% of the Moon gravitational force being actively generated within
each component of EO1's matter is accumulating through the stacking-of
forces effect to reach a maximum forward-directed 4.6 lb.f (pound.force) at
its far side hook. Each individual forward-directed 3% portion from
each component is the result of an internal Type 2 force where it is
generated internally within each component of matter but, unlike a Type 1 force, is
being delivered as an external
pull against components beyond. This Type 2 force is also a monopole force of
origination for the force begins here. (To get the most out
of Event 5, consider Article IV a
(32) The myriad of forward-directed internal Type 2 forces, identified
above, stack up in the backward direction to reach a maximum of 4.6lb.f at EO1's back side hook. The force that collects these individual
Type 2 forces from EO1's components is an external Type 3 bipole force that
is responsible for collecting the full 4.6 lb.f, delivering it to the back
side hook, transferring it along the entire 7926.4 mile dental strand to EO3's
front side hook and finally stacking down as portions of this
acceleration/Action force are experienced by every single component of EO3's matter.
This linear Type 3 bipole force is the same force provided by Newton's rope as the
forward exterior head interfaces with EO1 component's internal Type 2 forces
of gravitation and the equal and opposite rearward exterior head interfaces
with EO3 component's internal Type 2 forces of acceleration/Reaction.
(33) Viewing this event from the action force perspective, we have a
myriad of forward-directed internal monopole Moon gravitational forces of
origination being sourced from within EO1's matter. A total of 97%
of these gravitational forces are Type 1 as they directly cause the action
of acceleration for EO1's components. The remaining 3% of Moon gravitational
forces are Type 2 gravitational forces that also are internally generated but
are externally collected and transferred in the rearward direction by the
forward head of the external Type 3 bipole action force. The departure of this
3% portion totaling 4.6 lb.f leaves EO1 to experience just an average rate of acceleration.
After being transferred back along the dental strand, the rearward head of the external Type 3 bipole force transfers this 3%
force portion to EO3's myriad of components causing internal Type 2
acceleration/Action forces that increase EO3's rate of acceleration
up to average while terminating in a linear manner against internal Type 2 acceleration/Reaction forces being reactively
generated within EO3's forward accelerating components of matter.
(34) Viewing this event from the reaction force perspective,
there are a myriad of rearward-directed internal acceleration/Reaction
forces being generated within each component of EO3's accelerating matter.
Here, 97% are in the form of Type 1 acceleration/Reaction forces that are
caused by and directly interface with EO3's Type 1 a/A forces of Moon
gravitation. These Type 1 acceleration/Reaction forces provide support
and termination for all the Type 1 a/A forces present within EO3's
components of matter. The 3% of the acceleration/Reaction forces
that remain within EO3's matter are the Type 2 a/R forces that are caused by
and provide support and termination for the 4.6lb.f being passed back along
the dental strand. Thus EO3's acceleration/Reaction force total is to
be found by combining the Type 1 and Type 2 a/R forces present.
Forward in EO1's matter, all of the acceleration/Reaction forces being
reactively generated within its own matter are internal Type 1 forces
providing support and termination within each individual component of EO1's
accelerating matter. The Type 2 forces of acceleration/Reaction being
generated within EO3's accelerating components, caused by the bipole
external Type 3 4.6 lb.f being transferred back along the dental strand, are
themselves transferred forward along that same strand all the way to the
site of the original Type 2 Moon gravitational forces located deep within
EO1's matter. Their presence at any point along this Type 3 serial
transfer force path may be observed by severing the path at any location and
inserting a tension scale which will display the magnitude of the equal and
opposite acceleration/Action and supporting acceleration/Reaction forces
(35) Here we have described the entire linear force journey that both
deprives EO1 of 3% of its accelerational force lowering its rate to average
while delivering this excess acceleration/Action force to EO3 thereby increasing EO3's
acceleration rate 3% up to average. The end result is that all three
of our Earth Objects accelerate at the same average rate despite the
inequality of the Moon gravitational forces being experienced
by each object.
(36) Ninety seven percent (97%) of EO3's acceleration/Action force is internal Moon gravitation which, if water were present, would directly cause the water to accelerate in the Moon's direction at the same rate as solid matter. The remaining 3% of the acceleration/Action force EO3 is experiencing is the external tension force provided by the dental floss strand with this contact force accelerating EO3's solid matter while leaving any water only indirectly affected by its friction with the solid matter. Thus in the real world, Earth's far side high tide is formed as the water's moon-directed acceleration is slightly below average and therefore unable to match the slightly higher average rate being experienced by the far side land below.
(37) Our next project will be to calculate the tidal forces being
experienced by ocean waters at various points around Earth's near-side surface.
First we will cause Earth to reappear with the three Earth Objects embedded
in Earth's near side, center and far side as previously described in paragraph (17).
Earth remains non-rotating while its center of matter resumes its leisurely
27.32 day orbit of the Earth/Moon barycenter. Looking back at Earth
along the Earth/Moon centerline, we will first draw a circle around that
centerline on Earth's near-side surface that represents every point on
Earth's surface that is the terminus of a radius drawn at a 10 degree angle
with the Earth/Moon centerline beginning at Earth's center of matter.
Then additional radial circles are drawn every 10 degrees until the last one
reaches the 90 degree angle to the Earth/Moon centerline which includes both
the North and South Poles. In the calculations that
follow we will repeatedly embed EO1 at some point on each radial circle to determine the tidal force experienced at that
position. We will write a Microsoft QuickBasic (R) program to make
these lengthy calculations fun and easy, something not possible back in
Newton's time where all such calculations were laboriously performed long-hand.
|(38) Start the Microsoft QuickBasic Editor (R) and enter the program below exactly as written. Be sure your printer is hooked up and turned on or you may get a "Device Fault Error" when running. Press shift/PrtSc to print the output. Or, better yet, click on the upper left icon on the Quick Basic window, select Edit and then Mark. Then highlight the text you would like to copy to the Windows (c) clipboard. While highlighted, press Enter to complete the Copy. Then you can Paste the screen printout into any Windows application.|
10 CLS : CLEAR
110 REM Earth Radius = 3963.19 mi * 5280ft/mi er = 20,925,656 ft
1200 eo1mg = (gravconst * moonmass * eo1mass) / ((emd - t1sb) ^ 2)
Download a copy of the Near Side Tidal Force Calculator program by clicking Here. Once loaded into your browser, click File and then Save Page As.... Then start the Quick Basic Editor, select File, Open, double click the two dots in the Directory Box, find the folder where you saved the file and then select it to Open.
P.S. You may also copy the program to your clip board and then
paste it into the QuickBasic Editor beginning with the upper-left icon and
then choose Edit... Paste. If all else fails and you have an
hour to spare, open the Quick Basic Editor and type in the program.
(40) Let us put the Near Side Tidal Force Calculator program to work by
calculating a range of tidal position angles on Earth's side nearest
the Moon. Let 10 degrees be our first angle. To arrive at this
angle, first draw the portion of the Earth/Moon centerline that runs from
Earth's near-side surface on down to Earth's center of matter. Then,
forming a 10 degree angle with the Earth/Moon centerline, draw a straight
line back up to Earth's surface. When this 10 degree line returns to
Earth's near-side surface, it matters not whether it is north, south, west
or east of the Earth/Moon centerline. In fact if all possible points
of return at Earth's near-side surface are connected they will form a circle
about the Earth/Moon centerline. In the solution of triangle T1, our Near
Side Tidal Force program determines the length of this return line from Earth's c/m as
T1sc with the radial distance at right angle to the Earth/Moon centerline
represented by side T1sa. When our 10 degree line reaches Earth's
surface, this point marks the location of the 40
million pound-mass Earth Object 1 in our first series of calculations.
The Near Side Tidal Force program determines EO1's distance to the Moon's c/m for Moon
gravitational calculations. It also gives us the vertical and
horizontal component forces that affect EO1 at the 10 degree tidal position angle
on Earth's near-side surface.
(41) The following is the complete printout for the 10 degree
tidal position angle.
Earth Near-Side Tidal Force Calculator
Written by Ethan Skyler 6/12/2009
Enter Tidal Position Angle from Earth/Moon centerline (0.1-89.9 deg)
Angle is formed by following Earth/Moon centerline down to Earth's
c/m and then proceeding directly up to EO1's surface position on
Earth's near side.
Tidal Position Angle ? 10
T1aa = 10.0000 degrees (T1, angle A)
T1ab = 80.0000 degrees (T1, angle B)
T1ac = 90.0000 degrees (T1, angle C)
T1sa = 3,633,702.03 ft (T1, side a)
T1sb = 20,607,748.27 ft (T1, Side b)
T1sc = 20,925,656.00 ft (T1, Side c)
T2aa = 89.8322 degrees (T2, angle A)
T2ab = 0.1678 degrees (T2, angle B)
T2ac = 90.0000 degrees (T2, angle C)
T2sa = 1,240,563,511.73 ft (T2, side a)
T2sb = 3,633,702.03 ft (T2, side b)
T2sc = 1,240,568,833.41 ft (T2, side c)
T3aa = 79.8322 degrees (T3, angle A)
T3ab = 10.1678 degrees (T3, angle B)
T3ac = 90.0000 degrees (T3, angle C)
EO1md = 1,240,568,833.41 ft (Earth Object 1 Moon distance)
EO1mg = 139.867453 lb.f (EO1 Moon gravity force)
EO1gd = 4.527457 lb.f (EO1 excess gravitation force)
T3sa = 4.456353 lb.f (vertical force component.)
T3sb = 0.799241 lb.f (horizontal Tidal Force component.)
Calculate another Near Side Tidal Position Angle, Y / N ?
(42) Where are the locations of the three triangles being calculated
by the Near Side Tidal Force Calculator? First consider the relationship of
Earth to the Moon in the to-scale drawing below. As you can see, the
gravitational lines drawn above and below the Earth/Moon centerline converge
at the Moon and therefore are not parallel to each other. We take this
convergence into account when calculating the vertical and horizontal force
components represented by the adjacent and opposite sides of Triangle T3.
(43) Consult the drawing below that shows the locations of the three
triangles on Earth. Note that Earth Object 1 is tangent to Earth and shares
one corner in each of the three right-angle triangles.
(44) In triangle T1 we set the tidal position angle at 30 degrees so
all three angles are known along with the hypotenuse which is equal to
Earth's radius. Side T1sa is shared with triangle T2. Side T1sb
is used to determine the magnitude of Moon gravitation being experienced by
(45) Shown is Earth's end of triangle T2. We know the length of
T2sb and can determine the length of T2sa along the Earth/Moon centerline.
Then, with Pythagoras's help, we can determine the length of the hypotenuse,
T2sc. Then the tangent of T2aa is calculated.
(46) Our third triangle is one half of a force parallelogram (rectangle in
this case) which reveals the vertical component force and more importantly,
the horizontal component tidal force that together make up EO1's near side
Moon gravitation force. Angle T3ab equals 180 degrees minus the sum of
angles T1ab and T2aa borrowed from our first two solutions which also
reveals angle T3aa. Gravitational calculations reveal the length of Earth’s
near side excess gravitation force vector T3sc. The vertical and horizontal
vector lengths are revealed with the horizontal vector T3sb representing the
direction and magnitude of Earth's Near Side Tidal Force present. Our three
solutions are complete.
(47) While Earth's ocean tides
have as their root cause Earth's near side excess gravitation force which, in the
10 degree tidal position angle calculation listed above, is EO1gd at 4.53
lb.f, notice that only a small percentage of this relatively small force,
the T3sb horizontal component force of 0.8 lb.f, is serving as the
near side tidal force by
helping to force ocean water horizontally in the direction of the Earth/Moon
centerline. Decreasing the Earth weight of the ocean water by a ratio
of 1 over 9 million as does the vertical component T3sa, will, by itself,
have only a small effect on the elevation of Earth's ocean waters.
Therefore, we will assign the term "tidal force" solely to the horizontal
component force, T3sb. This means that only a portion of the
near side excess gravitation force acts as the tidal force and then
only on Earth's side nearest the Moon.
(48) Next let us proceed by recording the T3sb horizontal force
component for each tidal position angle from 10 to 80 degrees.
angle horizontal tidal force
excess gravitation force (near side only)
(49) At the 10 degrees angle, the
vertical component of the excess gravitation force is high in magnitude at
4.53 lb.f yet the horizontal tidal force component is low at 0.80 lb.f.
Then at the 80 degree angle, the horizontal component of the excess
gravitation force is in the right direction yet low in magnitude since it's
location on Earth is so close to the average gravitational division which
again leaves the horizontal tidal force component low at 0.73 lb.f.
(50) Next we will determine the tidal position angle with the greatest tidal force.
horizontal tidal force
(51) Unlike on Earth's near side, there is no tidal force
present to act as the cause of Earth's far side high
ocean tide, which instead results from the ocean waters experiencing
lower-than-average-rates of Moon-directed acceleration compared to the
overall average rate of Moon-directed acceleration being experienced by Earth's solid
matter. Thus there is present no tidal force for us to
calculate as the direct cause of Earth's far side tidal effect. "Yet
the thing is not altogether desperate", to quote an ever-hopeful Isaac Newton.
There is insight to be gained by calculating the magnitude of the "missing force" that if present would supplement the Moon's gravitational force
already present so that jointly they could act to cause Earth's far
side ocean water at that position to accelerate at the average rate already
being experienced by Earth's solid matter. If present, this "missing force" would effectively cancel the far side tidal effect for that position
on Earth's surface. By calculating the magnitude of this "missing
force" at various locations on Earth's far side surface, we will likely be exposed
to the explanation as to why the far side tidal effect varies in a manner
that is comparable to the variations of Earth's near side tidal force.
(52) To calculate the far side "missing force", we will need to
solve a second set of 3 triangles, T4, T5 and T6. During this process, we will embed
Earth Object 3, now composed entirely of sea water, at various tidal
positions on Earth's far side ocean surface. Think of EO3 as a giant
water balloon with just enough air inside to keep it floating on the ocean
surface. Earth is still non-rotating while its center of matter
continues its leisurely 27.32 day orbit of the Earth/Moon barycenter.
To maintain the constant tidal position angle being calculated, EO3 will have to magically slide sideways
as previously described in paragraph (16).
(53) The drawing below displays the far side location and relative
size of the three triangles whose solutions will provide the magnitude of
the horizontally-directed "missing force" for every tidal position angle
from 0.1 to 89.9 degrees. Understand that unlike the tidal force we
calculated using the Near Side Tidal Force program, this "missing force" is
not real. It does not exist. Were it to exist, then EO3 would
accelerate in the Moon's direction at the same average rate being
experienced by EO2 located at Earth's c/m. This
same-rate-of-acceleration for EO3 at every far side position would
effectively cancel Earth's far side tidal mounding. Calculating the
magnitude of the "missing force" will complete our understanding of the
cause and symmetry of Earth's dual tides.
(54) First up is triangle T4, similar to triangle T1 on Earth's near
side. Of interest are the lengths of sides T4sb and T4sa which
together allow us to calculate the magnitude of the Moon gravitation being
generated within EO3's now watery matter.
(55) Next we need to solve the hypotenuse of triangle T5 (T5sc) which
is drawn directly from EO3's c/m to the Moon's c/m. The length of this
line will be used to calculate the magnitude of Moon gravity present within
EO3's matter at its far side position.
(56) We now have enough information to solve the "missing force"
triangle T6. Our goal is to learn the length of T6sa for that fact
will indicate how much extra force in the horizontal direction is required
in order for EO3 to not participate in any tidal bulging or movement
away from its current position on Earth's far side.
(57) As before, start the Microsoft QuickBasic Editor (R) and enter the program below
exactly as written. Be sure your printer is hooked up and turned on or you may
get a "Device Fault Error" when running. Press shift/PrtSc
to print the output. Or, better yet, click on the upper left icon on
the Quick Basic window, select Edit and then Mark. Then highlight the
text you would like to copy to the Windows (c) clipboard. While
highlighted, press Enter to complete the Copy. Then you can Paste the
screen printout into any Windows application.
10 CLS : CLEAR
Download a copy of the Far Side Tidal Effect Calculator program, TIDEFAR1.BAS, by clicking Here. Once loaded into your browser, click File and then Save Page As.... Then start the Quick Basic Editor, select File, Open, double click the two dots in the Directory Box, find the folder where you saved the file and then select it to Open.
P.S. You may also copy the program to your clip board and then paste it into the QuickBasic Editor beginning with the upper-left icon and then choose Edit... Paste. If all else fails and you have an hour to spare, open the Quick Basic Editor and type in the program.
(59) The following is the complete printout for the 10 degree tidal
position angle. First see that EO3's Moon gravity (eo3mg) is less than
EO2's average Moon gravity (eo2mg). The difference is the EO3 "missing
force (eo3mf). Notice how the horizontally-directed component (T6sa)
of the "missing force" is small in magnitude since only a small horizontal
force is needed for EO3 to hold its position, relative to the Earth/Moon
centerline, thereby preventing it from
contributing to tidal mounding.
Earth Far-Side Tidal Effect Calculator
Written by Ethan Skyler 11/01/2009
Enter Tidal Position Angle from Earth/Moon centerline (0.1-89.9 deg)
Angle is formed between the Earth/Moon centerline extended beyond Earth's c/m
and an Earth radius proceeding directly up to EO3's surface position on
Earth's far side.
Tidal Position Angle ? 10
T4aa = 80.0000 degrees (T4, angle A)
T4ab = 10.0000 degrees (T4, angle B)
T4ac = 90.0000 degrees (T4, angle C)
T4sa = 20,607,748.27 ft (T4, side a)
T4sb = 3,633,702.03 ft (T4, Side b)
T4sc = 20,925,656.00 ft (T4, Side c)
T5aa = 89.8376 degrees (T5, angle A)
T5ab = 0.1624 degrees (T5, angle B)
T5ac = 90.0000 degrees (T5, angle C)
T5sa = 1,281,779,008.27 ft (T5, side a)
T5sb = 3,633,702.03 ft (T5, side b)
T5sc = 1,281,784,158.83 ft (T5, side c)
T6aa = 9.8376 degrees (T6, angle A)
T6ab = 80.1624 degrees (T6, angle B)
T6ac = 90.0000 degrees (T6, angle C)
EO3md = 1,281,784,158.83 ft (Earth Object 3 Moon distance)
EO2mg = 135.339996 lb.f (EO2 Average Moon gravity force)
EO3mg = 131.017286 lb.f (EO3 Moon gravity force)
EO3mf = 4.322710 lb.f (EO3 Missing Force)
T6sb = 4.259149 lb.f (vertical Missing Force component.)
T6sa = 0.738560 lb.f (horizontal Missing Force component.)
Calculate Another Far Side Tidal Position Angle, Y / N ?
(60) Now it is time to calculate a range of far side tide-canceling
EO3 horizontal "missing force" components. This should give us an idea
of what allows for Earth's far side tidal mounding to be similar in symmetry
to the tidal mounding on Earth's near side. The near side tidal force
horizontal component values are included for comparison. Notice that
the near side values are slightly greater in magnitude. Perhaps
this is a help in explaining why Earth's near side tidal mounds are
generally greater in magnitude than Earth's far side counterpart.
Far Side Missing Force Near
Side Tidal Force
10 deg = 0.74 lb.f
10 deg = 0.80 lb.f
(61) Finally we will determine the tidal position angle where the
greatest magnitude of the "missing force" horizontal component is required
to prevent tidal mounding. The near side results are included for
Far side horizontal "missing force" component
(62) Moon gravitation is the fundamental cause of Earth's dual ocean tidal mounds. This internal attractive force is above average within Earth's near-side matter, average at Earth's center of matter and below average within Earth's far side matter. The above average tidal force within Earth's near side ocean waters causes an above average accelerated shift in the direction of the Moon effecting the formation of Earth's near side tidal mound. Meanwhile, Earth's solid matter accelerates in the Moon's direction at but one average rate. Excess Moon gravitation present within Earth's near side solid matter is transferred as an external tension force back through Earth's solid matter to effect an increase in the acceleration rate of Earth's far side matter up to average. This external tension force has less effect on Earth's far side ocean waters allowing them to lag behind with a reduced rate of acceleration resulting in the formation of tidal mounding on Earth's far side.
Ryan E. Skyler
Robert E. Skyler
Thanks for your help, encouragement and support,
to download a copy of Event 5: The Physics of Earth's Tides.