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Hooke’s Law and the Missing Energy of Gravity.

Recently I’ve been going back and forth with various folks in the twitter realm about the fundamentals of physics. The biggest stumbling block is the issue of forces and how to define them. I stated in a previous post that you need 3 specific attributes to calculate any force equation:

Magnitude
Vector (or Direction)
Time

I then present a simple example of a 4000kg car accelerating at 10 m/s² for 10s. The Impulse of Force of the car at each second is:

Mass x acceleration x time¹ = Impulse of Force¹ (¹ is all reference frames from 1 – 10 seconds). Multiplying Force · Time is known as impulse and is a valid way of expressing a force.

4000kg · 10m/s² · 1s = 40,000 kg·m/s (velocity = 10 m/s) = 200,000 Joules
4000kg · 10m/s² · 2s = 80,000 kg·m/s (velocity = 20 m/s) = 400,000 Joules
4000kg · 10m/s² · 3s = 120,000 kg·m/s (velocity = 30 m/s) = 600,000 Joules
4000kg · 10m/s² · 4s = 160,000 kg·m/s (velocity = 40 m/s) = 800,000 Joules
4000kg · 10m/s² · 5s = 200,000 kg·m/s (velocity = 50 m/s) = 1,000,000 Joules
etc…
4000kg · 10m/s² · 10s = 400,000 kg·m/s (velocity = 100 m/s) = 2,000,000 Joules

However, where I caught the consternation of others in the twitter realm was due to confounding of terms that are very specific in the physics world. We all experience the affects of forces all day, everyday in our lives and we generally don’t break them down into units of measurement, types of units, work vs. force, etc. So my way of describing what I see as a fatal flaw in gravity was met with irate rantings.

For one, I was pointing out a potential problem with something they hold in an almost sacred sense (gravity). And secondly, I was not using terminology correctly for their taste. I don’t necessarily blame them for wanting exact language, but I would argue that they most likely understood the point I was trying to make but did not want point me in the right direction. But, for the most part however, they were decent. In a way, by them resisting what I was presenting forced me to fix and clarify my position.

Force, rate of change, work and energy

What I saw was that over time the acceleration due to gravity should equate to an increase in force on the mass involved. However, the concept of force is important since in the case of gravity it is, indeed, constant. I wasn’t arguing that gravity is increasing but the net forces are increasing. I kept arguing that time must be included in the equation or it doesn’t represent reality (though it was valid see impulse of the force). But again, my use of the words net force with time was incorrect with respect to physics as we are taught today and net forces has a specific meaning. It wasn’t until I recalled kinetic and potential energy (work) and how it relates back to force that I was able to present my case in a language that would be acceptable.

From the car example above, if F = ma, then the rate of change is 40,000 N which means the force is constant. But this is counter-intuitive to most people since they know the car is traveling faster with each second. However, for acceleration to exist at all, energy must be constantly applied. If we accept the conservation of energy law, then that energy must be going somewhere. In this case it is being expressed as kinetic energy.

In the case of gravity acting on an object or a person on the earth’s surface, gravity is that constant force or energy that is supplying the acceleration. Either the object or person needs to move or the energy needs to be converted into something else like heat or sound or whatever. It is the amount of energy that is increasing not the force itself. We can conclude then, that time multiplies how much energy is in a system with respect to the force being applied.

We can all bear witness to the fact that we are not heating up or emitting noise or expressing some other form of energy release under the stress if gravity. The counter argument is that since there is no movement there can be no acceleration. But this is a fallacy since the height of the person is equal to the “x” value in Hooke’s law and the motion is expressed as the compression of the human frame.

There is also a distinction between kinetic energy and potential energy. A car driving at 100 km/h has a specific amount of kinetic energy but a spring under increasing compression has potential energy that is being stored in the spring itself. The spring has a maximum compression it can reach before it will begin to become crushed under an increasing load.

Hooke’s law is only a first-order linear approximation to the real response of springs and other elastic bodies to applied forces. It must eventually fail once the forces exceed some limit, since no material can be compressed beyond a certain minimum size, or stretched beyond a maximum size, without some permanent deformation or change of state. Many materials will noticeably deviate from Hooke’s law well before those elastic limits are reached.

Wikipedia

Even though the spring no longer has motion the amount energy will increase as the load capacity is transferred to less elastic structures like the metal of the spring or the surface the spring is sitting on.

In other words, the increase of potential energy will propagate to the surrounding environment and begin to compress the weakest structures first. In the case of a human being standing on the surface of the earth, the acceleration due to gravity would continue to propagate through the human frame as per Hooke’s law. I have had many discussions with folks who insist that the forces between the earth and the human being cancel each other out. But they are confusing force with work.

Any equal and opposite reaction is within the frame of the human being (like a spring) and the ground they are standing on. The opposing forces balance out, but this would not stop the acceleration; the balanced forces would only resist the downward pull of gravity and subsequently increase the potential energy. Since gravity is supposed to be an acceleration and is supposed to be pulling at a force proportional to the mass of the earth, the human being wouldn’t stand a chance. It is precisely because of the equal and opposite reaction and the balancing of forces that potential energy is possible. So they are confusing the forces involved with the energy (work) being applied to the mass.

Potential Energy

assuming conservation of energy: if F=ma and a = 9.8 m/s² and m > 0 then ΔPE > 0. where’s all that energy going? We should all be crushed by now.

Spring energy

The potential energy $U el (x)$ stored in a spring is given by

$U_{\mathrm {el} }(x)={\tfrac {1}{2}}kx^{2}$

which comes from adding up the energy it takes to incrementally compress the spring. That is, the integral of force over displacement. Since the external force has the same general direction as the displacement, the potential energy of a spring is always non-negative.

This potential $U el$ can be visualized as a parabola on the $Ux$ -plane such that $U el (x) = 1 / 2 kx 2$ . As the spring is stretched in the positive $x$ -direction, the potential energy increases parabolically (the same thing happens as the spring is compressed). Since the change in potential energy changes at a constant rate:

${\frac {d^{2}U_{\mathrm {el} }}{dx^{2}}}=k\,.$

Note that the change in the change in $U$ is constant even when the displacement and acceleration are zero.

What does this mean? It means, a constant and catastrophic amount of potential energy should be building up in every object on the earth’s surface. As mentioned above, this does not violate F=ma since we are not talking about an increase in the force of gravity but an increase in potential energy due to that constant force.

If we calculate the “elastic” capacity of the human frame and equate it to “k” and multiplied that by “x” which would be the height of the person (since they should be getting compressed by gravity) and then multiply by 1/2 we should get the potential energy stored or expressed within the human frame. Also, since the value for “k” of the human being is less than “k” for ground, the human being would be crushed into the ground. However, as we all know, we aren’t springs, and won’t bounce back from such an event. Most of the energy will be released in the form of noise and heat. Not a pretty picture.

The modern theory of elasticity generalizes Hooke’s law to say that the strain (deformation) of an elastic object or material is proportional to the stress applied to it. However, since general stresses and strains may have multiple independent components, the “proportionality factor” may no longer be just a single real number, but rather a linear map (a tensor) that can be represented by a matrix of real numbers.

Wikipedia

So we can all say with a high level of certainty that we are not being crushed by gravity, therefore, gravity must be falsified. We exist in a non-gravitational realm.

The Motionless Earth – The Horizon, Curvature and Angles

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Over the past few months I’ve been working on the curvature equation for a circle. It doesn’t seem like very exciting stuff but it has enormous implications. During this process I thought I had found this equation but it turned out to be incorrect. I had fixed what I thought was the error but that turned out to be incorrect as well.

The nagging problem stems from the way one would experience curvature if they truly lived on a ball. All the current methods of finding curvature don’t really have a good explanation and upon further study are shown to be calculating something other than curvature. So what do we mean by curvature? I will answer this question and provide a new and rational equation for curvature and show why the other equations do not work.

Ball Earth Math

This famous document that has made the rounds in the flat earth community is not so much incorrect as it calculates an irrelevant number. Though it does properly calculate the value for X (or the “drop” along the axis), X is not the value we are looking for.

As you can see, the line X is tilted so it is parallel to the axis. This is where the calculation becomes misleading. What we really need to do is extend the line in order to intersect the line of site projected from point 0 (in the Ball Earth Map document). Please see my curvature image to see an example of this.

Distance² x 8 / 12

The equation “distance² x 8 inches / 12” does not solve for curvature but only for the hypotenuse of the distance: “Height = R-R*(cos(θ))” and “Length = sin(θ)*R”. Since these two values only represent the length and height of the distance travelled along the hypotenuse; and, since you cannot have a circle without height and length of equal value, you cannot have an arc without height and length of unequal value; therefore, curvature cannot simply be height (or the amount of “drop” along the axis) nor the value of the hypotenuse.

As an example, 1° of circumference is equal to 24901/360°=69.1 miles. If we plug in that value to the equation we get “√((3959-3959*(cos(1°))² + (sin(1°)*3959))² = 69.09 miles. This is not the value for curvature. However, this does resolve the hypotenuse. The equation was being used to measure curvature by plugging in the arc distance (the distance travelled along the ball) not the hypotenuse; therefore, the resultant value will only solve the value of the hypotenuse. All we are calculating with this equation is the point at which two different lines of site intersect on a ball not the height required for an object to be visible from point A.

Two points would intersect if we “forced the line” and built along the ground from each point (A and G). The two lines would have to be 3959 miles long. This is equivalent to a building at point D being 1639 miles in height. Since we build either along the surface or perpendicular to the Earth’s surface (i.e. buildings), we need to calculate something else.

Line of Site, the Hypotenuse and θ

So what are we actually trying to calculate? If the hypotenuse is not the distance nor the length along the axis, then what is? The confounding problem is related to the line of site of the observer. Once we establish the position of the observer, all the other pieces fall into place. If you examine the image below, you will notice that a line of site moves away from point A towards infinity. So we need to take the observer as being at point A and does not move.

Next, several dashed lines at various degrees have been drawn until they intersect with the line of site originating at point A. The equation ((1/(COS(θ)/R))-R) calculates the hypotenuse from θ to the point at which it intersects with the line of site from point A minus the radius. What we are calculating is the height necessary for an object to be above the surface, at a particular arc distance away or at angle θ, in order for it to be visible to the observer at point A. This is the key.

As an example (if the ball theory is to work), as a ship goes over the horizon, the mast (and the rest of the boat) will begin to tilt as per angle θ. Though this is a tiny angle at first, it nonetheless must follow that angle. The mast does not start tilting back to stay parallel with the axis of the ball, it stays fixed to the boat. As the boat continues along the circumference, the mast would need to extend in height to remain visible to the observer at point A. This is an effect of curvature.

In the diagram below, I put the original curvature calculation beside the equation I have proposed. You can see that at smaller distances the two values are very similar. However, as you move past 2° the values begin to diverge more rapidly.

A simple way to calculate the “distance to horizon” is to divide the distance travelled by 69.17 miles which equals θ and plug that into ((R/(COS(θ))-R)=Cur.

Curvature

So what do we mean by curvature? For example, if I travelled from point A to G, I would have experienced 6225 miles of the total circumference but there is not 6225 miles of curvature between point A and G. If you look at line D₁ – D₀, you will see that there is only 1159 miles of arc height (or the maximum height of the arc between two points). It also happens to be the value of X at 45° or 90°/2. This works for any arc length.

If you know the distance travelled, you can calculate the equivalent θ travelled. Using a variation of the Ball Earth Math equation, R-R(COS(θ/2)), we can solve for X which gives us the maximum height of the arc.

At a distance of 6225 miles (or ¼ of the globe) the maximum height of the arc has been shown to be 1159 miles. In a similar fashion, the ship and the observer would have to lift off the surface at 45° and travel for 1159 miles to see each other. However, we can see that all this is doing is altering the line of site for both the boat and the observer.

Conclusion

The definition of curvature is the degree to which a curve deviates from a straight line, or a curved surface deviates from a plane. The curvature of a circle is defined mathematically as the reciprocal of the radius (ie. κ = 1/r). All this tells us is that as the circle becomes larger κ becomes smaller. This does not help us figure out what the effect of curvature is to a person living on a ball.

However, by showing rational examples I have demonstrated the effect of curvature is the height necessary for an object to be above the surface, at a particular arc distance away or at angle θ, in order for it to be visible to the observer at point X. Without altering the line of site of the observer and keeping the object (in this case a boat) on the surface of the earth, we can measure the effect of curvature.

I would propose then, that what we are looking for is effective curvature and it is defined as “The height necessary for an object to be above the surface, at a particular arc distance away or at angle θ, in order for it to be visible to the observer at point X and X being a stationary position.”

A preponderance of evidence

First, let me perfectly clear about a few things: There is no Outer Space; There are no planets; stars are not burning thermonuclear giants; There are no nebulas; comets and asteroids are not chucks of rock flying through the void of space. In fact all of modern astronomy is a farce. It’s all Science Fiction. You must be saying I’m totally daft. How can I possibly say or even think such absurd thoughts? You must also be asking why in the world would the venerated space agencies present space as a reality if it is, in fact, a fiction? What would possibly motivate them to sustain such a farce and how could they fool seemingly intelligent people in the reverent halls of modern academia. This is impossible you say.

Another thing that needs to be clarified: Intelligence or the ability to recite information in a logical manner does not equate to the truth. If I can describe in minute detail the people, places and events that occur in the Lord of the Rings or Harry Potter, it does not mean they exist, have ever exited or will exist. I could take a course in “The Physics of Wands in Harry Potter”, read hundreds of esteemed books, write papers, submit thesis, get peer reviewed, and ultimately receive a doctorate, but this will tell me nothing about the real world regardless how much I believed it to be true. Knowledge is not knowing.

The preponderance of evidence (a concept used by Bob from Globebusters) is so overwhelming that any rational and independent human being would have to take notice:

Total lack of curvature
Flight paths on a globe
Eclipses on a globe
Absence of spin
Gravity
Corpuscular rays
Real pictures of Earth from space
Bubbles in space
Antarctic Treaty
Etc, etc, etc, etc

If you are completely unwilling to examine the evidence presented and just accept the word of so-called experts, then I would suggest that you should be a possession of the managing powers of the world. By willing handing over your will, thoughts and feelings to an outside agency, you have demonstrated your inability to manage your own person and require intervention by such powers. Discernment is a requirement to understand the truth. In fact, you should not continue to read this post or any other piece of evidence if that is the case. You are not ready for the truth. If that is the case then you must believe what you are being told and continue in your blind ignorance. Stop reading now.

do-not-read

However, if you are capable of examining ideas with an open and independent mind, then please continue reading. I would expect that any ideas put forward here are examined and properly verified using your own means. I’m simply putting forward ideas.

Weightlessness, Free Fall and Outer Space: more proof of a Stationary Earth

I wanted to revisit the whole concept of weightlessness as it pertains to so-called “outer space”. Let’s examine the key points that define how weightlessness is supposed to arise when an object or person is in “space”.

Assumption #1: All Objects in Orbit are in Free Fall

It all starts with a hypothesis by Newton himself (or at least attributed to him) that describes how a cannon ball fired with sufficient velocity in “outer space” would continue to move as per his 1st law of motion and would be kept indefinitely in orbit due to the gravitational pull of the earth and would subsequently “free fall” around the planet.

If we can get past this untested hypothesis and allow for the suspension of disbelief, we can try to drill down into the misconceptions related to it.

First we need to examine what the cannon ball is doing while revolving around the earth. If fired at a sufficient speed and unimpeded by any other object or force, the cannon ball will move tangentially to the surface of the earth while being “turned” around the globe.

In typical free-fall, the acceleration of gravity acts along the direction of an object’s velocity, linearly increasing its speed as it falls toward the Earth, or slowing it down if it is moving away from the Earth. In the case of an orbiting spacecraft, which has a velocity vector largely perpendicular to the force of gravity, gravitational acceleration does not produce a net change in the object’s speed, but instead acts centripetally, to constantly “turn” the spacecraft’s velocity as it moves around the Earth. Because the acceleration vector turns along with the velocity vector, they remain perpendicular to each other. Without this change in the direction of its velocity vector, the spacecraft would move in a straight line, leaving the Earth altogether

Wiki

If we apply the same logic to the cannon ball, we can see that it should “turn” around the earth but not “fall” to the earth. So here is a major misconception: the cannon ball is “turning” around the earth not “falling” around the earth. In other words, there is no free fall of the cannon ball. Why is this important? Simply because floating “astronauts” are impossible if there is no free fall.

Let’s break this down a little:

The important word to focus on is “perpendicular”. Since the velocity of the orbiting object (i.e. cannon ball, space craft, etc) is perpendicular to the force of gravity and is hypothetically kept in orbit due to a sufficient velocity along with the centripetal force of gravity, no free fall is possible.

Compare this to an object in free fall where the direction of velocity is in the direction of gravity or parallel to the force of gravity. For something to free fall it has to actually fall. When other explanations use elevators or thrust, they are assuming the objects in question are parallel to gravity or to the thrust of the rocket.

Assumption #2: Gravity keeps an airplane from flying off into space

The same action takes place closer to home when a commercial jetliner “turns” around the earth. The jetliner has constant thrust which overcomes the drag coefficient of air and the “downward” force of gravity. But don’t take my word for it, even NASA agrees:

Weight
Weight is a force that is always directed toward the center of the earth. The magnitude of the weight depends on the mass of all the airplane parts, plus the amount of fuel, plus any payload on board (people, baggage, freight, etc.). The weight is distributed throughout the airplane. But we can often think of it as collected and acting through a single point called the center of gravity. In flight, the airplane rotates about the center of gravity.

Flying encompasses two major problems; overcoming the weight of an object by some opposing force, and controlling the object in flight. Both of these problems are related to the object’s weight and the location of the center of gravity. During a flight, an airplane’s weight constantly changes as the aircraft consumes fuel. The distribution of the weight and the center of gravity also changes. So the pilot must constantly adjust the controls to keep the airplane balanced, or trimmed.

https://www.grc.nasa.gov/www/k-12/airplane/forces.html

Lift
To overcome the weight force, airplanes generate an opposing force called lift. Lift is generated by the motion of the airplane through the air and is an aerodynamic force. “Aero” stands for the air, and “dynamic” denotes motion. Lift is directed perpendicular to the flight direction. The magnitude of the lift depends on several factors including the shape, size, and velocity of the aircraft. As with weight, each part of the aircraft contributes to the aircraft lift force. Most of the lift is generated by the wings. Aircraft lift acts through a single point called the center of pressure. The center of pressure is defined just like the center of gravity, but using the pressure distribution around the body instead of the weight distribution.

The distribution of lift around the aircraft is important for solving the control problem. Aerodynamic surfaces are used to control the aircraft in roll, pitch, and yaw.

Weight force and gravity force are the same thing. Therefore, if the weight has been overcome then so has gravity. So gravity cannot be used as an excuse for keeping the airplane from flying off into space. Again, we can consult the folks at NASA:

As the plane’s engines move the wings forward, the air has to flow both over and under them. The wings are designed so that net effect is that the air pushes them upward, countering the downward pull of gravity.

https://spaceplace.nasa.gov/review/dr-marc-technology/rockets.html

So the upward forces of lift have cancelled out the downward pull of gravity. People don’t experience weightlessness while flying, so why would the astronauts? We are in a bit of a conundrum. The aeroplane has overcome gravity yet the passengers aren’t floating around like astronauts nor has the plane flew of into “space”. How is this possible?

Assumption #3: Cancelling of forces does not mean weightlessness

The force of gravity at 400 km high is about the same as on the surface of the earth, therefore, there is no major difference between an jetliner “turning” around the earth and an object “turning” around in “outer space”. Let’s assume a 75kg object or person:

La te xi t 1 5

And in orbit:

La te xi t 1 6

The mass of an average jetliner is about 351,534 kg while the mass of the ISS is about 419,600 kg. To calculate the centrifugal force you need to take the linear speed squared and divide by the radius. In both case, the radius is equal to radius of the earth + the cruising altitude. Since the airliner is going at a slower speed (~ 500 mph) it will require a constant thrust to stay aloft since the the buoyancy force of the atmosphere is constantly pushing the airplane downward (the buoyancy force operates in both directions depending on the object and the fluid it is contained in – but this will be discussed later in this post).

Some will argue that the centrifugal and centripetal forces must balance out. I completely agree with that statement. Earth’s gravity is a constant force (I keep hearing). It can’t pull any harder than 9.8 m/s². It’s kind of like having a weak string attached to a sphere you are swinging around your head. If the string breaks, the sphere will fly off. But until that point the sphere is fixed in place (i.e. orbit). If you could place a marble inside the sphere and begin to swing the sphere around, what would happen to the marble? It all depends upon the orbital velocity. If the sphere orbits at the exact velocity to ensure the string doesn’t break but it also doesn’t allow it to go slack, the marble would not float in the middle of the sphere, it would push against the outer edge of the sphere. In fact, any amount of angular velocity would force the marble to the outside edge. In the same way, the astronauts should remain fixed to the ceiling (outer edge) of the ISS.

Let’s look at an example:

For a satellite to stay in orbit, this centrifugal force must balance gravity. Thus,

.

Simplifying, we find that

,

and hence that

.

Thus for a satellite 1000 km above the Earth (and hence 7,400 km from the center of the Earth) traveling in a circular orbit,

(the mass of the Earth is 5.967´ 10²⁴ kg).

http://www.mso.anu.edu.au/pfrancis/roleplay/MysteryPlanet/Orbits/

No complaints about the logic or the math with that example. If we take the exact same equation and apply it to the ISS, here’s what we get:

Supposedly, the ISS is traveling at a velocity of 27,600 km/h (7.6 km/s). The centrifugal force and acceleration acting on the ISS at that velocity at an altitude of 400 km is:

F_c = m a_c

= m v² / r

Mass of ISS * Velocity of ISS ² / (Radius of Earth + Altitude)

Centrifugal Force = 419,000 kg * 7657 m/s ² / 6,771,000 m = 3472 kN

Centrifugal acceleration = 7657 m/s ² / 6,771,000 m = 8.6 m/s²

We can see that the centrifugal and centripetal forces are balanced but the effect would not be a free fall but cancelling of forces – just like a person who is flying in an airplane. Do we float around? No. Since there is no change in distance in the direction of gravity, then a = 0 and if that is the case then F = 0. The balancing of force does not mean weightlessness.

During orbit, the astronauts should be standing on the ceiling of the ISS and have a weight that is 89% of their weight on the surface of the Earth (being forced to the outer edge by angular velocity). However, if the ISS could maintain its altitude without orbiting, then the astronauts would have to be standing on the floor of the ISS. THEY WHOULD NOT BE FLOATING. As I mentioned earlier, the concept of floating is incorrect since the proposed reason for floating is because they should be falling. Anyone who has been on a plane during bad turbulence will attest that during the sudden loss of altitude, objects will crash into the ceiling of the plane. This is not because they are weightless but because the ground beneath their feet has suddenly dropped away. The “vomit comet” plane shows people falling because the plane itself is falling (it is dropping parallel to the so-called force of gravity) not because the people have suddenly become weightless.

In contrast to this, the ISS is not falling since it is NOT losing altitude. In fact, the only way for the astronauts to be floating (the way we are shown in countless videos) is if the ISS was in a constant free fall (which it is not nor could it be). For the ISS to maintain the altitude it does while maintaining it’s supposed path around the Earth, it must maintain a constant speed. Any variation from this would result in either the ISS crashing to Earth or flying off into “space”. Hypothetically, the ISS would need to be creating an artificial Lagrange Point.

The Lagrange points mark positions where the combined gravitational pull of the two large masses provides precisely the centripetal force required to orbit at the same angular velocity (essentially, the speed of the orbit) and thus remain in the same relative position (or altitude).

This artificial Lagrange Point is reverse engineered from a mathematical model that requires the ISS to orbit at an absurd velocity. The speed is required to make the math work as opposed to the math describing reality.

The Angular Velocity Problem

Since the ISS is not in a free fall but rather orbiting at a constant altitude, the astronauts should experience the effects of angular velocity. Hypothetically, the force of gravity is always perpendicular to the floor of the ISS which means the centripetal force is always acting upon the astronauts in the opposite direction – meaning towards the ceiling of he ISS. In this hypothetical situation, the forces are balanced but what does this mean in reality versus a mathematical model?

If we compare this to a car travelling with a constant velocity in a straight line along a road, no effects of that motion would be experienced by the driver. If, however, the driver turns a corner, the effects will be felt as a pull in the opposite direction of the turn. Assuming a constant turn rate and velocity, an object sitting on the dashboard will only begin to move once the forces of friction and inertia are overcome by the angular velocity forces – the object breaks free of the forces of friction and inertia and moves away from it’s static location or in the case of the ISS, it flies of into “Space”. In this example, the forces of friction and inertia are equivalent to “gravity” and the angular velocity is the motion of the ISS. Too much velocity and ISS will fly off into “Space”; too little and it will fall to the ground. But until then, the object stays put on the dashboard surface and the ISS stays in “orbit” even if the opposing forces are equal.

In reality, the ISS would function much like a fast moving aeroplane. In the hypothetical environment of “Outer Space”, there is no atmosphere to slow it down but there also isn’t an atmosphere to create lift. So how does it stay aloft? From the artificial Lagrange point I wrote about earlier. If an aeroplane was able to travel at the same velocity and constant altitude (it doesn’t matter what altitude since we have artificially removed the atmosphere) as the ISS, it would not cause the passengers to become weightless. It would allow the plane to stay aloft without additional thrust just like the ISS is supposedly doing – Another artificial Lagrange point. The only difference would be a slight change in speed due to the increased “gravity” at a lower altitude.

The G-Force problem

There is a serious problem with the description of g-force as written in wikipedia:

The gravitational force, or more commonly, g-force, is a measurement of the type of acceleration that causes a perception of weight.

In the case of an object on the earth’s surface, weight is produced by the supposed force of gravity acting towards the centre and the opposing force of the earth itself pushing back on the object. (how the earth can pull an object towards the centre and simultaneously push back with the same amount of force is equally baffling)

When the g-force acceleration is produced by the surface of one object being pushed by the surface of another object, the reaction force to this push produces an equal and opposite weight for every unit of an object’s mass. The types of forces involved are transmitted through objects by interior mechanical stresses. The g-force acceleration (except certain electromagnetic force influences) is the cause of an object’s acceleration in relation to free fall.^[1]^[2]

However, the g-force is not apparent on the surface of the earth:

Gravitation acting alone does not produce a g-force, even though g-forces are expressed in multiples of the acceleration of a standard gravity. Thus, the standard gravitational acceleration at the Earth’s surface produces g-force only indirectly, as a result of resistance to it by mechanical forces. These mechanical forces actually produce the g-force acceleration on a mass. For example, the 1 g force on an object sitting on the Earth’s surface is caused by mechanical force exerted in the upward direction by the ground, keeping the object from going into free fall. The upward contact force from the ground ensures that an object at rest on the Earth’s surface is accelerating relative to the free-fall condition. (Free fall is the path that the object would follow when falling freely toward the Earth’s center). Stress inside the object is ensured from the fact that the ground contact forces are transmitted only from the point of contact with the ground.

How can an object at rest still be accelerating? This paragraph is full of contradictory concepts. For the ground to be pushing up, there must be a continuous force pulling down. You can’t be accelerating if you are at rest. And this has NOTHING to do with the supposed spin of the earth.

Some would argue that if they throw an object up in the air, the point where the object comes at rest in mid-air (i.e. maximum height before falling back down), the acceleration due to gravity is acting upon the object even though it is at rest. This example is making liberal use of the word “rest”. If an object was thrown with a slight arc (even a tiny one), the object would never be “at rest” since the object would follow an arc pattern and be in continuous motion. That the arc of an object throw directly upward is zero is simply a mathematical description of one possible scenario where the object transitions from upward to downward in a single moment. For any rational person, an object thrown into the air is always in motion until it is actually at rest on the surface of the earth. To say it is “at rest” while in motion is sophistry exemplified.

I have written about this absurdity in a previous post so I won’t go too much detail. Suffice to say that if a person is to stand on a scale, the acceleration due to gravity MUST continue to act until the object is at rest. Meaning the spring should continue to compress regardless of the mass of the object. The only object, according to the current gravitational hypothesis that can be “at rest” is the atom/particle at the very center of the earth (gravity is zero). It is this magical object that is able to pull everything into the center while pushing back with an equal amount of force. All objects on the earth’s surface should be in a continual downward pull towards the center. What stops the downward compression at the very center of the earth? The explanation is that the gravitational forces cancel each other out. This doesn’t even work mathematically. If we take the formula to calculate the volume of a sphere (V = (4/3)πr) and simply subtract the total volume of the earth from the volume of the earth at half the radius, we can see that the total volume on the periphery is many times greater than the center:

V = (4/3)π(3959 mi) = 2.6×10¹¹ = Volume of earth

V = (4/3)π(1975.9 mi) = 3.23×10¹⁰ = Volume of earth at half distance

Volume of earth at periphery = 2.27×10¹¹

There is an order of magnitude more volume at the periphery than the center. In other words, the center would be pulled towards the periphery due to the greater mass. This would make planetary formation through gravitational attraction impossible. For as more mass accumulated on the periphery, the less attractive the force in the center.

The Atmospheric Pressure Problem

The amount of air above a persons head pushes down with a force of approximately 7 pounds / square inch. The average human head is about 21 inches in diameter or π x r² = 3.14 x 10.5² = 346 square inches. At 7 pounds per inch, there should be 2,424 lbs / square inch of additional weight on the head of each human being.

If I stand on a scale and weigh 72kg and then pick up 2 weights of 10kg each, the scale would show an increase of 20 kg to 92kg. For the same reasons the weight of every human being on earth should be their weight + the weight of the atmosphere above their head. This is what the theory of gravity must predict to be true. However, no one experiences an additional 2,424 lbs of weight on their head. This is another falsification of gravity.

We also need to differentiate between air pressure and air mass (or weight). Pressure is caused by air being forced into an enclosed space which causes more air molecules of air to be present in the same space. Air weight is the downward force caused by gravity. What causes air pressure to be perpendicular to the force of gravity? According to the theory of gravity, once an object is in contact with the surface (in this case the human head) it will experience resistance from the skull which pushes back with an equal amount of force thereby stopping the downward motion. The earth is not a pressurized container since the only thing holding the atmosphere in is gravity and gravity is in one direction only – downward. Therefore, there is no known force perpendicular to the force of gravity which can explain air pressure.

Since air cannot be lighter or denser than itself, density and buoyancy cannot be be used to explain this problem away.

The Net-Force Argument

That the net force between an object a rest on the surface of the earth is zero, does not mean there is no force between them. It’s mathematical slight of hand to make it seem like no force is present. If that was the case then all objects would weigh zero.

The key word is NET. As I wrote earlier, if I pickup an additional 20kg of weight the NET force between myself and the ground is still zero even though the scale has registered an increase in weight.

If I place my hand on the ground and place 2000kg on it, my hand would be crushed even though the NET force is equal to zero.

Another way to think about this is to examine how a balancing scale works. If I place a 1kg weight on both scales the forces are balanced – or equal to zero. However, there is a force of 1kg still pushing down on each scale. That they are balanced does not negate the forces in question here which is implied with the word ZERO. We can test this by removing one of the weights. The scale will become imbalanced and tip in the direction of the remaining weight.

Buoyancy – The separating force

Many of the globe folks comment that if buoyancy is responsible for the downward force on objects, what is it that makes them “fall” downward rather than “upward”. The answer is that objects do “fall” upward – Helium balloons for example.

Objects of different densities separate at a constant rate plus or minus the addition of external forces. What we experience as “down” is really the separation of lighter material from more dense material. Objects go in opposite directions which creates “up” and “down”.

So why is “up” above our heads and “down” everything below our feet? Because of the density of the average human is greater then air. We, as physical humans, will tend towards objects of equal density – the ground beneath our feet vs. atmosphere above our heads. If we became lighter then the density of the atmosphere, “down” would become what we experience as “up”.

Just as “down” for a helium balloon is “up” for humans. Similarly, if a beach ball filled with air is forced under water to any depth, it will rise at a specific rate. But it is “falling” “down” towards a location of equal density or equilibrium.

Just as it takes energy to submerge the beach ball, it also takes energy to place a human above the surface of the ground. The atmosphere is constantly pushing the human towards the ground and simultaneously pushes a helium balloon towards the sky.

That’s it for now.

The Sun never sets on the Motionless Earth

The title of this blog might seem, at first, to be debunking the motionless earth; it is however, a vital hypothesis towards a proper theory of the earth’s shape. My previous posts have presented various concepts that try to show many inconsistencies with the heliocentric model. The problem with such an effort is that a proper model for the flat earth that fits the empirical data is never properly developed. My goal is to present a model that does, indeed, do this.

What do I mean by the ‘Sun never sets’? Well, in the heliocentric model the sun is literally “setting” or disappearing below the horizon. Due to the hypothesized ball shape of the earth and its associated spin, the sun will eventually become obscured by the curved surface of the ball (i.e. you can’t see through solid ground).

In the flat earth, the sun is always at the same height (with some minor changes as will be explained later), regardless of the date or time, at all locations on the plane; hence, it never sets. So how does the sun function on a flat plane to create seasons, timezones, night, day, etc? Many current models have been presented and many of these questions have been answered to varying degrees of satisfaction. The most confounding problem that faces the flat earth model is the trigonometric disconnect between the latitudes, angle of incidence and height of the sun – the sun is just not where it’s supposed to be!

This problem provides heliocentric proponents with the most devastating argument against the flat earth because even with a obviously lack of curvature, the trigonometry matches the ball model better than the flat model. So we are in a kind of stalemate – neither model can claim victory.

However, I have developed a hypothesis for the sun that not only fits the flat earth model but also a methodology for mapping the earth we live on.

I wanted to add a video by Jeranism which perfectly visualizes what I’m trying to present. Please watch this video before reading the rest of this post since it will help you understand how such a phenomenon is possible.

Hypothesis of the Sun

The sun is electromagnetic
The maximum radius of the sun’s radiation is approximately 6225 miles in all directions (gases in atmosphere are excited by the presence of the sun and then go into quiescence as the sun moves away) hence you get a glow in the distance even after sun is no longer in view. Please watch this video by Dan Dimension:

At a distance of 6225 miles from the observer, due to laws of perspective and refraction, the sun will appear to be at the horizon (~ 1°). At a distance greater than 6225 miles the sun will begin to “set” or “rise” as it approaches.
Since the sun is at the same height (with specific increases and decreases with respect to seasons), regardless of the position of the observer, we must apply the laws of perspective to obtain an accurate altitude of the sun.
The sun is between 4200 – 4600 miles above the plane depending on the date. It is this fluctuation that produces the analemma.
The curvature calculation (Ball Earth Math) is really a perspective calculation (The Effect of Perspective). It calculates the reduction in size of an object as it moves away from the observer. This value must be subtracted from the trigonometric calculation. This is due to the fact that standard trigonometry defines 2D objects while we exist in a 3D environment; hence, perspective and angular resolution must be taken into account. Please watch this video by Curious Life:

For example, at 49.2827° N, 123.1207° W (Vancouver, Canada), the sun appears to be approximately 40.6° above the horizon. Since there are 69.15 miles per degree of latitude, this places Vancouver ~ 3388.35 miles from the equator. On the spring equinox at solar noon, standard trigonometry would calculate the altitude as 53.4°. However, we must also subtract the effect of perspective:

[tan^-1(Height of Sun / distance to equator)] = Altitude (in degrees)

tan^-1(4400 / 3388) = 53.4°

(Height of Sun – (Height of Sun(cos(θ))) = Effect of Perspective

(4400 – (4400(cos(49.2°))) = 1,525 miles

Height of Sun – Effect of Perspective = Perceived height

4400 – 1525 = 2875 miles

tan^-1(2875 / 3388) = 40.3°

Scaled Perspective

I’m certain that many will object to this model by arguing that objects at a shorter distance do not present this kind of affect. In other words, an object that is 100 feet away at 10 feet in height will not display the effect of perspective. That the effect is so infinitesimal does not mean it’s not present. For example, an observer 67.15 miles from an object 2 miles in height would only measure an effect of perspective of 1.6 feet – That’s a 1.6 foot difference over 10,560 feet. In other words, if it’s almost imperceptible at 67.15 miles, it’s definitely imperceptible at 100 feet.

So what am I basing this scale on? Well…the sun, moon and stars. They are the only objects that are at a sufficient distance and height to show the effect. Ultimately, the position, height and motion of the celestial objects in the sky are either due to perspective or curvature. Since no curvature has been show to exist (at least empirically), then we are left with perspective.

This video by Wide Wake perfectly shows how the sun becomes distorted as it moves away from the observer (it becomes egg shaped). The video by Jeranism shows the mechanism by which this affect occurs and why it appears to drop.

I’ve uploaded the Excel workbook for anyone to use. I was able to use both Date & Time and Stellarium values for the Sun’s height at various location on the earth. The values match almost perfectly with the hypothesis. There are 4 tabs in the workbook as I had to calculate the equinox and both solstices. I added a short-distance tab to calculate objects closer than 1°. The result is that the effect of perspective is accumulative with both distance AND height. In other words, as the object increases in height and distance, the effect is more pronounced. In the example above, a mountain that is 67.15 miles away and at a height of 2 miles will have an insignificant amount of this affect since the accumulated affect is limited to relatively small values.

The workbook is a little too complex to put directly into the post so I’ve provided it for people to download. Just click on image below to download:

With an accurate way to measure the distance and heights of the celestial objects, we should be able to map the surface of the earth using the formula above.

How Airplane Trim Debunks Curvature

My previous post “How ascending planes debunk the globe – part 2” describes, using simple trigonometry, how the ascent rate of an airplane is impossible on a globe. The rebuttal to this was that the airplane somehow still “follows the curve” while ascending. However, an airplane cannot simultaneously “follow the curve” which requires the trim to be constantly in the upward position (Nose-Down), while at the same time ascending which, requires the trim to be in the completely opposite position (Nose-Up).

Therefore, during take-off and descent, the total amount of curvature that SHOULD be accounted for by the trim is non-existent. Just take some time and think about that.

elevator-trim

What, exactly, is trim or the trimming of an airplane. This is from Wikipedia:

The use of trim tabs significantly reduces pilots’ workload during continuous maneuvers (e.g. sustained climb to altitude after takeoff or descent prior to landing), allowing them to focus their attention on other tasks such as traffic avoidance or communication with air traffic control.

Both elevator trim and pitch trim affect the small trimming part of the elevator on jet airliners. The former is supposed to be set in a certain position for a longer time, while the pitch trim (controlled with the landing pilot’s thumb on the yoke or joystick, and is thereby easy to maneuver) is used all the time after the flying pilot has disabled the autopilot, especially after each time the flaps are lowered or at every change in the airspeed, at the descent, approach and final. Elevator trim is most used for controlling the attitude at cruising by the autopilot.

As you can see, once the elevator trim is set, it is rarely adjusted by either the autopilot or the pilot. Even if we allow for the trim to be constantly adjusted, it is not designed to nor does it ever “constantly adjust for curvature”. However, if we assume that curvature does exist, to satisfy those who insist on curvature, then the trim would have to account for it. You can’t have one without the other folks.

Here is the curvature math and geometry:

This is because a curve is a non-linear shape that requires a constant change in pitch to maintain a constant altitude and thereby “follow the curve”. For each change in pitch, the trim must add each subsequent change to maintain the non-linear shape. I’ve used the curvature math above but also included the speed of the aircraft to calculate “Instantaneous curvature” – or how much curvature must be accounted for over a certain period of time and velocity. Assuming an aircraft with a velocity of 500 mph, the amount of curvature increases by 0.309 inches / second. This might seem like an inconsequential value at first however, this is an accumulative value that increases over time. For example, after the first 30 seconds, the amount of curvature that the trim would need to adjust for is 10 feet. After 760 seconds (12 minutes), the trim needs to adjust for 1.41 miles of curvature. If the trim is constantly adjusting for curvature, then no drastic motions would be felt by passengers and crew. However, this would cause the trim to be in an impossible state after just a few minutes of flight. But this is somewhat of an academic exercise since neither pilots nor the autopilot adjust the trim in this fashion.

This question was posed to a pilot in a forum and part of his answer was:

You said that, if the plane was trimmed for a straight and level flight, it would ‘gain altitude’ while flying as the earth surface ‘fell away’ due to the curvature of the earth. Well, that would probably happen in a perfectly motionless atmosphere where the plane would fly dead ahead, and over time gain altitude (provided it has sufficient thrust) as the earth curves away from under the airplane.

In reality, a constant altitude must be kept using the standard pressure and that means a fixed distance to the earth center of gravity is maintained, making the path of the plane a curved one.

http://www.askcaptainlim.com/flying-the-plane-flying-90/1301-does-a-jet-aircraft-need-to-constantly-adjust-nose-down-to-follow-the-curvature-of-the-earth.html

Even the pilot acknowledges that the aircraft would “…gain altitude over time” if the trim was set. He follows up saying that “In reality…” this does not happen. Instead of saying to himself, “Humm…that contradicts the spherical hypothesis of the globe, maybe it’s wrong”, he evokes “gravity” and “standard pressure”. But neither the “standard pressure” nor “gravity” can fix this conundrum since the trim must continually be altered to “follow the curve” and we shall see in a moment, “gravity” is not the answer either. He talks about the trim by saying:

As such, the flight controls are constantly moving very subtly to maintain the correct attitude.

I would agree but he’s not applying a spherical model to the response, only a flat surface model. If he was flying over a sphere, the trim would have to adjust every second.

Here is the formula used:

((R / cos(((v*t)/5280)/69.1))-R) * 5280)

R = Radius of Spherical Earth

How should an aircraft maintain altitude on a sphere?

For every 69.1 miles of flight, a 1° change in pitch must take place. If the pilot makes only a single 1° change in pitch over a 500 mile trip (a total of 7.2° of pitch change) then the airplane will be 6.2° off of level. A single 1° pitch change from a particular starting position produces a slope not a curve. To maintain a curve, the pitch change MUST increase with time. So to travel 6,219 miles on a sphere, the trim of the aircraft would have to account for 90° from the starting position. This is regardless of gyroscopes or accelerometers. Unless the aircraft “follows the curve”, it will increase in altitude and to “follow the curve”, the trim MUST be in a constantly increasing, upward position. No pilot would ever adjust the trim to such a degree since it would cause a major imbalance in the aircraft. In other words, this could not happen nor does it happen because we don’t live on a sphere, rather, it is indicative of a flat surface.

if a pilot or autopilot had to change pitch to maintain “level” flight (by pushing forward on the column or modifying the trim) the pilot or autopilot could not nor would not pull back on the column nor put the trim back to the original position once the adjustment was made otherwise, they would continue to fly in a tangent to the surface of the sphere and subsequently start gaining altitude again.

Again, any pilot would tell you, the trim is not continuously adjusted for curvature nor are they continuously pushing forward on the column. The point being that once trim is set to a particular level (especially the elevator trim), it is left alone. This totally contradicts what should happen on a sphere but is complete what we would expect on a flat surface.

Gravity to the Rescue!

Now some of the globe folks (including pilots) will evoke gravity to fix this rather intractable problem. However, to evoke gravity at this point and say that, “IT keeps the plane level”, is to ignore the obvious: the aircraft has already overcome the supposed effects of gravity otherwise it could not be in flight – Even NASA agrees.

Weight is a force caused by the supposed acceleration due to gravity and the lift forces have already overcome the weight of the aircraft and thereby, gravity. Level flight is, either by line-of-site (i.e. pilot flying level to horizon) or via the gyro, not due to gravity. However (and for the sake of argument), if gravity was responsible for level flight on a sphere, by constantly forcing the aircraft level (or tangent to the sphere), then the ability of the aircraft to ascend or descend would be impossible since it could never increase it’s angle of attack from level. This, of course, does not happen in reality.

Starting Position

The problem of arguing whether the earth is spherical or flat using everyday experiences (i.e. flight) is that IF the earth is FLAT, those who do subscribe to a spherical earth will be using experiences of a FLAT earth to try and describe a spherical one. In other words, they use actual events and then describe those events through the lens of a spherical model (i.e. since the earth is a sphere, and planes fly level, it’s due to gravity). Instead, they should be describing events AS THEY SHOULD HAPPEN on a sphere not on a FLAT plane. The model should be predictive not reactive. In other words, does the spherical model describe our experiences?

For example, the gyroscope is known for its ability to maintain rigidity in space and provides empirical evidence for a flat surface rather than a sphere. However, the subscribers to the spherical model say that the gyroscope is modified during flight to account for the change in pitch. There are several problems with this.

1) The aircraft MUST “follow the curve” regardless of what the instruments say or bad things will happen.

2) To “follow the curve” a real change in pitch MUST happen. It cannot be ignored by modifying the gyro.

3) Even if they gyro was modified to continually adjust for pitch changes, it does not alter the fact that the trim MUST be continually changing position to adjust for real pitch changes.

4) A small aircraft, like a Cessna, does not modify or have the ability to modify their gyros during flight (i.e. Schuler Tuning) which debunks the notion that this happens with larger, commercial aircraft. But even if we allow for that, it doesn’t fix the problems with the spherical model as has been explained.

How ascending airplanes debunk the globe – Part 2

In my last post about how an ascending plane debunks the globe, I wrote that the GE folks will use the argument that the plane is tracking with the curve therefore, curvature is of no consequence.

Even though I have the mathematical proof of this impossibility, I was challenged anyway. It seems the only way to provide definitive proof, is to show that the rate of climb, the climb angle and the speed of the aircraft make it impossible for it to track with the curvature.

As the aircraft ascends, it does so at a specific angle and speed. Knowing these values, we can map the actual path of the aircraft as it ascends. For the aircraft to track with curvature, it would need a much smaller climb angle since the curvature alone provides 4.37 miles of altitude.

It must also be noted that commercial aircraft like 747s gain altitude by aiming the nose upward and gaining lift through forward motion; they do not act like balloons and float upward. When the aircraft starts to take-off it angles the nose upward and ascends away from the ground. In order for the aircraft to track with the curvature it would have to stay perpendicular to the axis of the earth. As the distance from the starting point increases, the smaller the climb angle would need to be.

The final altitude is 6.62 miles which leaves 2.25 miles that the aircraft would need to climb. To calculate the necessary angle we would take the inverse TAN of 2.25 miles / 185.97 miles.

2.25 miles / 185.97 miles = .012 miles

TAN ^-1 (.012 miles) = 0.69°

This means that the average angle over the entire ascent would have to be 0.69°. Unfortunately for GE folks, this is not what happens in the real world. During take off most 747 aircraft use a climb rate of about 12° but reduce that by about 0.5° every minute or so. There is obviously variability based upon certain conditions but this is sufficient for this case. By recording the level on a flight to France I can confirm this reality with going into speculative territory.

I can say for certain that the pitch of the aircraft for the first 12 minutes was between 4-12° and the final 13 minutes ranged between 1-3° making an average of 0.69° impossible.

At the end of the day for an aircraft to ascend on a curved surface while tracking with that curve, it must be perpendicular to the axis of the earth and ascend like a balloon.

How the climb rate of an airplane debunks the globe

I have watched and read FlatEarth (“FE”) content that talks about the issue with flight paths, times and other anomalies that are difficult or impossible on a GlobeEarth (“GE”). However, many GE folks tend to write-off these anomalies by evoking gravity. Though the gravity excuse can be rebuked, it gives the GE folks a free pass. This pass is about to be taken away.

My previous work on curvature calculations has given me the idea to track the flight path of a 747-200 class aircraft during the ascent phase of the journey. On a recent trip to France I recorded the angle of ascent using a bubble level and screen capture software. This allowed me to compare (in real-time) the ascent angle over time with the flight data from Flight Aware. The majority of the ascent was between 1-3°. It lost about 1° of angle every minute. During the initial take off the angle was much higher at around 12° but it quickly reduced to 2° in about 12 minutes. The final 13 minutes was at a nominal angle of 1°. A quick calculation shows that TAN(2°) x 185 miles (6.46 miles) is close to the 6.62 miles achieved after 25 minutes. In other words, the bubble level and the recorded values from flight aware seem to correlate.

If you have an account at flight aware, you can take a look at the actual flight taken. I will add the data tables here for those who don’t wish to create an account.

The purpose of this post is to show that if there is curvature to the earth, the climb rate over time and the curvature of the earth must place any average international flight at more than double the recorded cruising altitude. This is because the amount of curvature during the ascent phase of the example flight below would be approximately 4.37 miles even without adding any climb at all.

All of the data is publicly available and anyone can do the same calculations I’m going to do. In fact, any international flight can be used for this experiment.

If you take a look at this graph:

you can see the ascent is in the first 25 minutes and 25 seconds of the flight (01:59:02 – 02:24:27). You can compare this to the table data below:

An examination of this data eliminates a potential objection by GE folks that plane is flying with the curvature of the earth since the plane cannot be tracking with the curvature and ascending at the same time. By it’s very definition, the plane is moving away from the ground (ascending), not tracking with it as it should during cruising altitude. This might seem obvious to most logically minded folks but we are dealing with hardened attitudes that will use any argument, no matter how illogical, to avoid the conclusions reached here.

Since the plane is ascending, and if we are living on a ball, we must take into account the ascending and curvature rates at the same time. There is no getting away from this. This also destroys the idea that gravity keeps the plane moving along the curve, since if this was the case, a plane could never ascend at all. The ascent angle over time is the key.

For example, if a plane took off and ascended at a nominal climbing rate of 1 meter / hour at an average speed of 450 miles / hour, it would reach a cruising altitude of 4.7 miles in 26 minutes. If it continued at this rate for 52 minutes (or an additional 26 minutes), it would be at a cruising altitude of 19.8 miles. The plane can ascend because it’s under it’s own power. It has already overcome the force of gravity or it would fall to the ground. But I digress….

Using a real example, if we include curvature into the calculation with the recorded climb rate, the plane should reach a cruising altitude of 35,000 feet (or 6.62 miles) in approximately 13 minutes. This was calculated using 51 specific data points. Each point contains time, latitude, longitude, course, direction, kts (knots), km/h, meters, climb rate and reporting facility. By calculating the speed by the time interval, I can calculate the total distance travelled during each time interval.

The first 12 data points are ignored for distance travelled since the plane took off in a westerly direction and then turned around to a northeast direction. We still need to include the altitude reach during this time frame which is approximately 1.26 miles. This altitude was reached at 2:02:17 pm.

To accurately calculate the distance travelled, I had to measure the time difference between each data point (between 15-60 seconds) and multiply by the speed of the plane at that interval. This eliminated the use of an average speed which would most likely be used as an additional rebuttal against this proposal.

If you examine the table below, you can see each time interval along with the height and curvature reached.

The 7 columns of the far right show the curvature that would occur over the distance travelled. The distance travelled was calculated by converting the speed to nautical miles/min (i.e 281 kts / 60 min = 4.68 nm/min).

I then divided the time frame for each data point by 60 seconds to get the fraction of a minute that the plane travelled during that time frame (ie. 25 seconds / 60 seconds = .41 min)

Once I had that value, I was able to multiply that by the plane’s current speed to get the distance travelled (i.e. 4.68 nm/min × .41 min = 1.95 nm). I then converted the value from nautical miles to miles (i.e. 1.95 nm × 1.15 = 2.24 miles).

Once I had the distance at each interval I could calculate the amount of curvature in feet. Whew !!!

Since curvature is accumulative, the total arc distance from the starting point to the specific time frame must be included not just the distance travelled in that particular time frame.

I finally converted the last 21 data points into miles of curvature. Since the plane was already 1.26 miles in altitude when it started heading northeast, I had to add that value to the curvature accumulated from 02:02:17 pm onward. I calculated both the altitude of the plane without curvature and one with curvature.

As you can see from the table, the cruising altitude should have been reached at approximately the 13 minute mark. Since the plane does not stop ascending at the 13 minute mark but continues for an additional 12 minutes, we are left with the inescapable conclusion that there is no curvature.

I’ve included a link to a post that calculates the distance covered during take off and the time taken to reach cruising altitude for comparison purposes.

https://aviation.stackexchange.com/questions/14357/how-long-after-takeoff-for-a-boeing-747-400-to-reach-cruise-speed

Falsification of the Universal Law of Gravity

Note: Just as my previous post, I had to fix some errors in my formulas. However, the conclusions are the same.

As a follow up to my previous post, I wanted to take a closer look at the gravitational constant of “Big G”. Upon examination, I found that though Newton used Kepler’s laws of planetary motion, the relationship between them is rather ambiguous. For example, Kepler’s laws deal with the velocity of a planet around the sun with respect to time – meaning: The time it takes to complete an orbit and the area covered is proportional.

There is no force involved since he was not concerned with the mass of the objects. The time it takes for a planet to revolve around the sun increases with distance, hence his proportional law of 1/r². But, again, this has nothing to do with a force acting on the object, rather just the proportions of the ellipse and the time taken to traverse it.

Newton, however, was completely concerned about the mass of an object. He took the proportions with respect to time and converted it into a force: Gravity. By replacing the proportions with mass he removed time from the equation.

The gravitational constant specifically is:

6.67 ×10⁻¹¹ m³⋅kg⁻¹⋅s⁻²

6.67 x 10⁻¹¹ m³/kg/s²

It is a constant that is applied over time as an acceleration but time is not factored in. An acceleration over time, by it’s very definition, is an increase in velocity over time. You can’t have a change in velocity without a change in time.

The equation for the “law of gravity” is:

$F=G{\frac {m_{1}\times m_{2}}{r^{2}}}\,.$

However, all force equations require 3 attributes to comply with reality and no variable can be isolated:

Vector
Magnitude
Time

The classic force equation F=ma violates this requirement since it excludes the time attribute. Math is a descriptor language so it must describe reality not abstractions. For example, a 4,000kg car accelerating at 10 m/s² for 10 seconds cannot be described by the equations F=ma since it excludes the change in time while it was accelerating. All it can describe is an abstracted force not a rational one.

Velocity, acceleration, force and momentum all violate these basic requirements since they only include partial attributes. For example:

a = Δv / Δt – must include a magnitude to describe reality – ma = Δv / Δt · m

v = d/t – must include a magnitude to describe reality – mv = d/t · m

F = ma – must include time to describe reality – ΔtF = maΔt

p = mv – must include time to describe reality – tp = mvt

So we must write the “law of gravity” as:

ΔtF = maΔt = GmM/r² · Δt = Δvm

This translates into a significant problem for gravity since the velocity or force involved must increase with respect to time. Neither the mass nor the acceleration due to gravity changes, just the velocity and the time the acceleration is applied. A secondary issue is that the acceleration will increase as the distance between them decreases. In the case of gravity, the acceleration never stops since it is supposedly inherent to matter.

No object could ever separate from any other object of any mass if time is applied. We neither observer nor experience what should be happening which is a self-evident falsification of the universal law of gravity. Simply stated: Gravity is Dead.

The Falsification of Terminal Velocity, Gravity and the Source of Mass

Note on update July 2nd: I had to make an update to this post as I had erred in some of the equations. The conclusions are the same but now the formulas are correct. Mainly, I had neglected to balance the force equations on both side which resulted in much to-do in the twitter realm.

Before I get into the rational explanation for the falsification of terminal velocity, I need to make it clear what I’m not saying. My experience with posting controversial ideas tends to make one focused on the negative response rather than the positive.

I’m not saying that their isn’t a maximum velocity that objects in free fall achieve. This is experimentally demonstrable and not the point of this post. What I am saying is that the explanation for this phenomenon is incorrect and the mathematical description is invalid.

Some definitions:

Big G

$F=G{\frac {m_{1}\times m_{2}}{r^{2}}}\,.$

The constant of proportionality, $G$ , is the gravitational constant. Colloquially, the gravitational constant is also called “Big G”, for disambiguation with “small g” ( $g$ ), which is the local gravitational field of Earth (equivalent to the free-fall acceleration).^[2]^[3] The two quantities are related by $g = GM E / r 2$ (where $M E$ is the mass of the Earth and $r E$ is the radius of the Earth).

The supposed local value for gravity on the earth is $g = GM E / r 2$ which translates to 9.8 m/s². For the purposes of this proposal, small ‘g’ is sufficient.

Terminal Velocity

The current explanation for terminal velocity as per wiki:

Terminal velocity is the highest velocity attainable by an object as it falls through a fluid (air is the most common example). It occurs when the sum of the drag force (F_d) and the buoyancy is equal to the downward force of gravity (F_G) acting on the object. Since the net force on the object is zero, the object has zero acceleration.^[1]

There is a significant issue with this explanation which ties into the invalid structure of the mathematical derivation.

According to the standard model, the only downward force that is acting on the object is the force of gravity and any opposite force will result in the reduction of the downward force of gravity itself. Objects can only accelerate in free fall; they can never achieve a constant velocity. Again, this is a rebuttal to the current explanation of terminal velocity not a dismissal of terminal velocity itself.

If the net force acting on the object is zero, then the object will simply stop accelerating and since the only downward force is an acceleration, the object should stop moving (akin to neutral buoyancy). Any objection to this conclusion requires either the addition or subtraction of forces. For example, some will argue that once the net forces are in balance the final velocity achieved at that point, is terminal velocity. They will also pull out Newton’s 1st law:

An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force

We need to take apart all the assumptions about a free falling object in order to see why terminal velocity as it is currently understood doesn’t work. As mentioned before, a free falling object only has the force of gravity acting on it therefore the speed is an acceleration not a constant velocity so once the acceleration is zero due to drag, it should simply slow down to zero. If there was no unbalanced force of drag, the object would continue to fall at an increasing rate. A simple analogy: two cars tied together and accelerating in opposite directions would go nowhere but the forces are still active. There are no additional forces present. The cars wouldn’t just start drifting at a constant velocity towards one car or the other.

So where does the velocity value even come from? To figure that out we need to see how it is derived (from Wiki):

Derivation for terminal velocity

Using mathematical terms, defining down to be positive, the net force acting on an object falling near the surface of Earth is (according to the drag equation):

$F_{{net}}=ma=mg-{1 \over 2}\rho v^{2}AC_{{\mathrm {d}}}$

At equilibrium, the net force is zero (F = 0);

$mg-{1 \over 2}\rho v^{2}AC_{{\mathrm {d}}}=0$

Solving for v yields

$v={\sqrt {\frac {2mg}{\rho AC_{{\mathrm {d}}}}}}$

If you examine this formula closely you will see where the error creeps in. It’s not that the math itself is incorrect, but the conclusions are incorrect relative to the math.

There are two distinctive forces present in this formula – gravity and drag. The portion of the equation that describes the force of gravity is “mg”. The second portion of the formula is “1/2 pv²AC” which describes the drag forces. This is the upward force due to drag which counter-balances the supposed force of gravity.

To solve for “v” we need to isolate it like this:

mg = 1/2 pv²AC (multiply both side by 2)

2mg = pv²AC (divide both sides by pAC)

2mg / pAC = v² (get the square root of each side)

v = √ 2mg / pAC

The variable “v” is a function of the upward force of drag so it cannot be isolated out in real life. It’s part of the accelerating upward force. From a mathematical point of view, it simply describes the upward velocity of the air at a specific point in time. But for the equation to balance out we need to also take the square root of the force of gravity and the mass of the object. That doesn’t even make sense. What does the square root of mass even describe in real life?

So even though the math works (meaning it balances out) it has nothing to do with the actual events we would observe. The fact that the resultant force is called a velocity rather than an acceleration is also interesting since the downward force on any object at anytime is gravity (which is an accelerating force) according to the standard model.

Secondly, the article goes on to state that buoyancy effects can be subtracted from the mass of object.

Buoyancy effects, due to the upward force on the object by the surrounding fluid, can be taken into account using Archimedes’ principle: the mass $m$ has to be reduced by the displaced fluid mass $\rho V$ , with $V$ the volume of the object. So instead of $m$ use the reduced mass $m_{r}=m-\rho V$ in this and subsequent formulas.

This is in direct conflict with the concept of mass since the mass of an object is always constant; it is the supposed force of gravity that gives the object weight so it must be the weight that is displaced NOT the mass of the object. From Wiki:

Assuming Archimedes’ principle to be reformulated as follows,

$\text{apparent immersed weight} = \text{weight} - \text{weight of displaced fluid}\,$

then inserted into the quotient of weights, which has been expanded by the mutual volume

$\frac { \text{density}} { \text{density of fluid} } = \frac { \text{weight}} { \text{weight of displaced fluid} }, \,$

yields the formula below. The density of the immersed object relative to the density of the fluid can easily be calculated without measuring any volumes:

$\frac { \text {density of object}} { \text{density of fluid} } = \frac { \text{weight}} { \text{weight} - \text{apparent immersed weight}}\,$

(This formula is used for example in describing the measuring principle of a dasymeter and of hydrostatic weighing.)

According to the standard model, the weight of an object is equal to the mass times the force of gravity:

weightf

There are two important factors that need to be considered when looking at drag and buoyancy on a falling object.

In fluid dynamics, the drag equation is a formula used to calculate the force of drag experienced by an object due to movement through a fully enclosing fluid. The formula is accurate only under certain conditions: the objects must have a blunt form factor and the system must have a large enough Reynolds number to produce turbulence behind the object.

So the key to drag is turbulence. This is one of the effects that slows the object down as it falls. Turbulence also increases over time. There is zero turbulence when object is suspended in air and motionless and an increasing turbulence over time once the object is released.

Of particular importance is the $u^{2}$ dependence on flow velocity, meaning that fluid drag increases with the square of flow velocity. When flow velocity is doubled, for example, not only does the fluid strike with twice the flow velocity, but twice the mass of fluid strikes per second. Therefore the change of momentum per second is multiplied by four. Force is equivalent to the change of momentum divided by time.

As the object falls, the flow velocity (meaning the total amount of air pushing back on the object) and subsequently the total mass of air will increase. The result is an increasing increase in turbulence behind the object. The total turbulence is directly proportional to the maximum velocity an object can sustain in free fall minus any buoyancy effects.

So the proper formula for buoyancy effects would be mg – pV or W – pV. Therefore, the proper formula for terminal velocity would be the weight of an object minus buoyancy effects minus drag or v = [W – (pV)] – (1/2 u²pAC). We can see that this is simply a momentum calculation that subtracts two opposing effects – buoyancy and drag.

The reason for an increasing increase of velocity (acceleration) during the initial release of the object is due to the turbulence requiring time to increase. But as the turbulence increasingly increases to a maximum value, the object begins to slow down from it’s initial acceleration in direct proportion to the value of turbulence.

We could postulate that the acceleration of an object in free fall would be unlimited if turbulence was zero. Therefore, the acceleration of an object due to the force of gravity does NOT have to be evoked to explain an object in free fall but only the difference between the density of air and the falling object. It is the turbulence that is slowing down the object. Once the density of the falling object meets the density of the ground, the falling stops. Thus, there is no downward force acting on the object just a difference in density.

If there was indeed an accelerating force due to gravity, the object on the ground should continue to move towards the center of the earth. Why should the acceleration due to gravity stop just because air gave way to ground? The fact that we do not continue to accelerate towards the center of a ball when, by all theoretical and mathematical requirements, we should be, is a clear falsification of gravity.

Gravity – an imaginary force

I further postulate that the mass and the weight of an object are the same thing and gravity is an imaginary force that is only applied to imaginary celestial objects (planets and other imaginary celestial objects must all be in free fall around some other object). By direct observation, an object at rest on the surface of the earth is no longer subject to the imaginary force of gravity since the weight of an object must continually increase due to the constant acceleration.

If Newton’s third law is to be followed, the object “at rest” (but not really) on the surface of the Earth must be continually pushing back with an “equal and opposite reaction” over time since an acceleration is the change of velocity over time. So with each second the net force being applied to the object must increase. Let’s take an example from Khan Academy:

(https://www.khanacademy.org/science/physics/one-dimensional-motion/acceleration-tutorial/a/what-are-acceleration-vs-time-graphs)

What does the area represent on an acceleration graph?

The area under an acceleration graph represents the change in velocity. In other words, the area under the acceleration graph for a certain time interval is equal to the change in velocity during that time interval.

$It might be easiest to see why this is the case by considering the example graph below which shows a constant acceleration of 4 m/s² for 9s.$

If we multiply both sides of the definition of acceleration,

Plugging in the acceleration 4 m/s² and the time interval 9s we can find the change in velocity:

= = (4m/s²)(9s) = 36 m/s

Multiplying the acceleration by the time interval is equivalent to finding the area under the curve. The area under the curve is a rectangle, as seen in the diagram below.

The area can be found by multiplying height times width. The height of this rectangle is 4 m/s², and the width is 9s. So, to find the area also gives you the change in velocity.

area = (4m/s²)(9s) = 36 m/s

The area under any acceleration graph for a certain time interval gives the change in velocity for that time interval.

We can now take the acceleration due to gravity (9.8 m/s²) and apply the same model. A person standing on the surface of the Earth must experience a constant change in velocity:

= = (9.8m/s²)(1s) = 9.8 m/s

= = (9.8m/s²)(2s) = 19.6 m/s

= = (9.8m/s²)(3s) = 29.4 m/s

= = (9.8m/s²)(∞s) = ∞ m/s

According to the standard model of gravity, a person in free fall is weightless, therefore their mass is not subject to measurement. Mass as a function of weight is only applicable when there is resistance to counteract free fall. As the mass of the person begins to impact the surface of the Earth, they begin to weigh something and that mass is subject to an increasing velocity which translates into an increasing weight.

How do we measure weight? This can be measured on a standard scale. A scale has springs that can compress and translate the downward pressure into a weight. As long as the springs can handle the increasing weight, so will the person increase velocity and we can measure the increase. But the force of the mass of a person is dependent upon the velocity at a particular time since:

tp = tmv

Of course this is not a common way to express momentum since it is a change in momentum due to a change in velocity. However, in this case, the change in momentum is being translated into a change in weight:

= m

Then “W = mv” or “W = m

So what is a kilogram or a pound? It is an arbitrary unit of measurement based upon various assumptions. The primary assumption is that weight is a function of gravity and mass. However, we can clearly see that if velocity is constantly changing due to the force of gravity then the resultant weight must increase. It is required by the standard model of gravity that the mass and acceleration of the object are fixed and do not change. It’s the velocity that must be changing over time.

The objection to this proposal is that time cannot be a function of Weight or of Force. The standard way to express the equation is F = ma. However, this equation contains within it a time function as m. This results in:

Δvm = ΔtF

ΔtF = maΔt =

Therefore, if there is no change in velocity, as anyone can observe, then we are left with:

F = W = m

So what is mass? How does something weigh more than something else? That is our next topic.

The Foundation of Mass

What we are left with are two attributes of mass – Density and Buoyancy. Gravity, in this instance, is no longer an accelerating force inherent to the mass of an object but a synonym for mass itself. There is no intrinsic force attributable to mass.

Let’s start with a quote from Newton himself:

“It is inconceivable that inanimate Matter should, without the mediation of something else, which is not material, operate upon, and affect other matter without mutual contact. Gravity should be innate, inherent and essential to matter, so that one body may act upon another at a distance thru a vacuum, without the mediation of any thing else, by and through which their action and course may be conveyed from one to another, is to me so great an absurdity that I believe no man who has in philosophical matters a competent faculty of thinking can ever fall into (for) it. Gravity [mass] must be caused by an agent acting constantly according to certain laws; but whether this agent be material or immaterial, I have left to the consideration of my readers.” – Sir Isaac Newton, Letters to Bentley, 1692

We are indeed considering his words. What we can extract from this is that gravity or mass is the result of something not the cause of something and a “mediating” entity must be involved.

Here is a quote from Ken Wheeler:

Before embarking on explaining magnetism, first the all important word field must be rigorously and scientifically defined with ambiguity and misconceptions fully removed. A field (Greek: χώρα) is a conjugate nonspatial attribute between the subject, Ether, and the object causing the perturbation (or EM induction), either as gravity [weight/mass], electricity, magnetism, or electromagnetism. A field has relevance only as a relational perturbation between the subject and the object causing its appearance; a field is specified only as regards to: electricity, magnetism, gravity, matter, and dielectricity (and another unmentioned for another article); however as it must be necessitated electricity, dielectricity, gravity [weight/mass], and magnetism are by their very principle not different from the Ether itself. The very term ‘magnet’ merely denotes the electrified mass formerly ‘not-a-magnet’; as such magnetism and magnet are abstractions or distinctions without a difference except as relates to the coherent field charges of the before and after mass. Conceptual disambiguation between a magnet and magnetism itself cannot be enjoined, and is merely a fallacious reification of connotative abstractions.

Ken Wheeler: Uncovering the Missing Secrets of Magnetism p.27

Everything seems to come down to the historically dismissed Ether. It must be noted that even the high priest of gravity, Newton, proposed an Ether. The persistent objection to the dismissal of gravity is always, “what is the alternative”? That an empirically based science that includes the Ether is a valid alternative and has always been a valid alternative seems to elude those so certain of gravity. It’s the implications of an Ether that frightens them so.

I will not put words into Mr. Wheeler’s mouth so it is important to note that he does not dismiss Gravity per say but the origins of it. Based upon Mr. Wheelers writings, I can safely say that he is NOT a proponent of the Flat Earth so I can only assume that he believes in planets, comets, gravity, etc. So in no way am I linking him to the Flat Earth theory. But I digress…

Current scientific thought requires gravity to be an inherent property of mass itself rather than an affect of a mass within the Ether (like the waves on an ocean is the ocean but not inherent to the ocean; something else causes waves). If there is NOT an intrinsic force to mass then much of modern astronomy falls apart.

Density and Matter

So what causes mass? In mathematical terms it would be p * V = m (density * Volume = mass). In this sense, mass is a function of density and volume. We know that volume is a function of geometry so what is density?

Magnetism is purely radiative, is the termination of electrification and the end-of-road byproduct of dielectricity. Dielectricity comes before everything else in the four-part schema of Force Unification. Dielectricity and magnetism are the two co-principles of the universe. So how do you get magnetism out of dielectricity, since magnetism requires a subject to emanate from or itself is the termination point of either mass in movement or electricity as it terminates? The answer is that dielectricity terminates into the creation of matter, which itself then has in this conjugate relationship, magnetism as its radiative principle (the proton as found in hydrogen, the most abundant element is magnetically dominant, is the polarized charging dynamo for its discharge plane of interatomic magneto-dielectric volume). Creation (dielectricity) and radiation (magnetism), and their two byproducts, electricity and mass, or gravity [density] as centripetal attributes choate to mass / matter.

Ken Wheeler: Uncovering the Missing Secrets of Magnetism p.38

What Ken Wheeler is calling “gravity” in this instance would simply be the density attribute of mass. A dielectric object (not-a-magnet mass) has a specific density due to the “centripetal attributes choate” to it. It is the centripetal/centrifugal attributes of the object that ultimately give objects their density and subsequently makes objects heavier or lighter than others.

A complete thesis developed and proven by Ken Wheeler is available for anyone to read. It is incumbent upon the read to disprove his empirically based science before dismissing the concept of an Ether. This is true science and if you fancy yourself a true scientist or a rational human being, you will need to apply the proper scientific rigor rather than falling into Scientism consensus.

Gravity: The Baron Munchausen of Forces

The Missing force in gravity

Two bodies that are interacting require “something” to push against. For example, in a classic tug-o-war there are two teams pulling against each other using a rope. However, for it to work, they must push against the ground not air or space in order to pull. The classic F=ma equation usually cited does not give the fulcrum from which each body is pushing against. They can’t push against air or space while pulling at the rope. Hence, there is no “equal and opposite” reaction since there is nothing to react against. There is no push to the pull.

Essentially we have an equation that has the earmarks of Baron Munchausen

Tug-O-War2

Centripetal force is not gravity. This is a common misconception. Gravity is an independent force due to the mass of the object. People usually link things together which are not related. For example, angular momentum gives rise to the centripetal and centrifugal forces. The centripetal force is “pulling” at the the centrifugal force and the centrifugal force is “pulling” against the centripetal force and gravity at a rate of ω²r, where ω is the angular speed (=linear speed/r) but it does not cause gravity.

In other words, they have to subtract the emergent centripetal and centrifugal forces from the force of gravity to get the resultant force. For example, a 72kg person will exert 2.36N of centrifugal force at the equator while gravity exerts (pulls) 705N of force giving a resultant force of 702.64N. But for gravity to pull it must push against something at that’s what’s missing. The earth must be fixed in space to pull.

The only thing that gravity can push against is space and space must have zero mass. For space would need sufficient mass to resist the motion of any and all celestial bodies (galaxies, suns, planets, etc). Therefore, space would no longer be space but mass. So the conundrum for an astrophysicist is the necessity of space and the simultaneous necessity for space to have mass. They are mutually exclusive.

For this reason, gravity violates Newton’s 3rd law.

Spacetime, curvature and Relativity nonsense

Since gravity is a function of mass, any supposed curvature of space is the result of the mass of the object not the other way around. Also, curvature of space around an object is impossible. This would require the supposed “fabric” of space is pulled inward towards the object. What exactly is it pulling on? Space by it’s very definition is empty. How can a force pull on emptiness? Even if gravity was granted such magical powers, it would then, by necessity, require space to have mass and if it has mass then it is not space. As well, time is an abstraction and a force can’t pull on an abstraction.

Even if we grant gravity the magical powers to pull on emptiness it still needs to push against something to pull on something. We get into a circular argument where the space around the earth is being pulled on while simultaneously it is resisting the motion of the earth itself. Hence the allusion to Baron Munchausen pulling himself up by his own pigtails.

The Inertial mass required to fix the earth in space would negate space itself. So regardless of which theory of gravity is evoked, it violates Newton’s 3rd law.

Density/Buoyancy and the falsification of Gravity

Density and buoyancy are a function of the differences between the mass of objects. For example, an object more dense than air will drop to the ground due to the displacement of the surrounding air. Air under compression can push against a heavier object since the air is not being displaced. The mass of an object is also distinct from the force of gravity.

Here we get to the crux of the problem? If a 72kg person is standing on the earth’s surface, they should experience a constant acceleration of 9.8 m/s/s. But this acceleration does not stop just because they are standing on the earth’s surface. Their “weight” will only be 72kg (702.64N) for one second. What happens at the 2nd second? Their “weight” would double since in the 1st second it is 702.64N + 702.64N in the 2nd second (the “weight” of a person in free fall is zero). In other words, we should all be compressed piles of mush. All mass should be pulled relentlessly into the centre of the earth. Therefore, since the acceleration due to gravity is not what we observe, it must be shown as having been falsified.

Other celestial observations, like the gravity bending light, are non-sensical since the bending of light would require either space (which is empty and massless) or light to have mass. Neither of these have mass. The motions of celestial bodies (planets, suns, comet, galaxies, etc) are equally non-sensical since they require gravity and since gravity has been shown to be falsified, the motions of the lights in the sky are due to other phenomenon.

And finally, outer space, planets, comets, etc are all constructs of the theory of gravity. Without gravity they fall apart. Welcome to the flat earth.

There is an alternative to gravity that has been elegantly presented by Ken Wheeler: https://archive.org/details/magnetism1small

Refractivity, Polaris and the Motionless Earth

Refractivity seems to be the last refuge of the GE theory. Since objects at a distance that should be well below the horizon can be brought into focus using a zoom leans on a camera, there is little an advocate of the GE can do other than claim it is all due to refraction.

It would not be a wild assumption that most people are ignorant of the mechanics of refraction. I’m no expert either, but I will put the concept to a test to see to what degree refraction can impact objects at a distance. Let’s start with a definition (from Wiki):

re·frac·tion

rəˈfrakSH(ə)n/

noun

PHYSICS

the fact or phenomenon of light, radio waves, etc., being deflected in passing obliquely through the interface between one medium and another or through a medium of varying density.

change in direction of propagation of any wave as a result of its traveling at different speeds at different points along the wave front.

measurement of the focusing characteristics of an eye or eyes.

The density of air at sea level is very similar regardless of temperature. It would require extreme heat or cold to vary the density to an significant degree. Here is a chart that shows the changes in refractivity with respect to range and height of the target (radar).

$refractivity$

Here is the original document for those interested.

Pressure and water-vapor content decrease rapidly with altitude, while temperature decreases slowly, refractivity and decreases with altitude. Thus, velocity increases with altitude, and rays bend downward. Dominant change in refractivity occurs with along vertical. Key point: not the actual refractivity, but changes (gradient) in refractivity cause rays to bend.

As anyone can see from the above graph, refraction has little impact at both lower heights and the greater the range (which will become important later).

Most curvature experiments performed use an object at a distance that is brought back into focus using an optical zoom. Even if we calculate for curvature error due to refraction (typically 14%), we are left with 86% of the object that is visible without refraction. How can we account for that?

As an approximate compensation for refraction, surveyors measuring longer distances than 300 feet subtract 14% from the calculated curvature error and ensure lines of sight are at least 5 feet from the ground, to reduce random errors created by refraction. (Wiki)

In other words, refraction cannot be used by GE theory to account for objects being visible at a distance that should be below the horizon. So when surveyors are correcting for error, what exactly are they correcting for if they are not on a ball? The predictive model that I created for Polaris could possibly hold the answer. Instead of curvature that is being corrected for, the surveyors are actually correcting for perspective on a flat plane.

I would bet that the error corrections in both radar and other visible light spectrums will be satisfied by the predictive model that was mentioned above. For example, an object at a great height (say 100,000 feet) but at a close range (<1 mile & > 45°) will become obscured at the “top” due to perspective whereas the same object will become obscured at the “bottom” as the range increases and the viewing angle is < 45°.

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Force, rate of change, work and energy

Potential Energy

Spring energy

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Ball Earth Math

Distance² x 8 / 12

Line of Site, the Hypotenuse and θ

Curvature

Conclusion

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Weightlessness, Free Fall and Outer Space: more proof of a Stationary Earth

Assumption #1: All Objects in Orbit are in Free Fall

Assumption #2: Gravity keeps an airplane from flying off into space

Assumption #3: Cancelling of forces does not mean weightlessness

The G-Force problem

The Atmospheric Pressure Problem

The Net-Force Argument

Buoyancy – The separating force

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Hypothesis of the Sun

Scaled Perspective

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How should an aircraft maintain altitude on a sphere?

Gravity to the Rescue!

Starting Position

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ΔtF = maΔt = GmM/r² · Δt = Δvm

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Big G

Terminal Velocity

Derivation for terminal velocity

Gravity – an imaginary force

(https://www.khanacademy.org/science/physics/one-dimensional-motion/acceleration-tutorial/a/what-are-acceleration-vs-time-graphs)

What does the area represent on an acceleration graph?

area = Δv

Δv = aΔt = (4m/s²)(9s) = 36 m/s

area = (4m/s²)(9s) = 36 m/s

The area under any acceleration graph for a certain time interval gives the change in velocity for that time interval.

Δv = aΔt = (9.8m/s²)(1s) = 9.8 m/s

Δv = aΔt = (9.8m/s²)(2s) = 19.6 m/s

Δv = aΔt = (9.8m/s²)(3s) = 29.4 m/s

Δv = aΔt = (9.8m/s²)(∞s) = ∞ m/s

tΔp = tmΔv

If “Δvm = maΔt”

Then “ΔtW = mΔv” or “ΔtW = maΔt”

Δvm = ΔtF

ΔtF = maΔt = Δvm

F = W = m

The Foundation of Mass

Density and Matter

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The Missing force in gravity

Spacetime, curvature and Relativity nonsense

Density/Buoyancy and the falsification of Gravity

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Here is the original document for those interested.

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= = (4m/s²)(9s) = 36 m/s

= = (9.8m/s²)(1s) = 9.8 m/s

= = (9.8m/s²)(2s) = 19.6 m/s

= = (9.8m/s²)(3s) = 29.4 m/s

= = (9.8m/s²)(∞s) = ∞ m/s

tp = tmv

= m

Then “W = mv” or “W = m

ΔtF = maΔt =