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:

  1. Magnitude
  2. Vector (or Direction)
  3. 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 Uel(x) stored in a spring is given by

{\displaystyle 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 Uel can be visualized as a parabola on the Ux-plane such that Uel(x) = 1/2kx2. 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:

{\displaystyle {\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.

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

or

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:

{\displaystyle 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:

  1. Vector
  2. Magnitude
  3. 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

{\displaystyle 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 = GME/r2 (where ME is the mass of the Earth and rE is the radius of the Earth).

The supposed local value for gravity on the earth is g = GME/r2 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 (Fd) and the buoyancy is equal to the downward force of gravity (FG) 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 {\displaystyle 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.

area = Δv

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.

Screen Shot 2017-06-19 at 9.17.25 AM

If we multiply both sides of the definition of acceleration, a = Δv / Δt , by the change in time, Δt, we get Δv = aΔt.

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

Δv = aΔt = (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.

Screen Shot 2017-06-19 at 9.26.42 AM

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:

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

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:

tΔp = tmΔv

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:

If “Δvm = maΔt”

Then “ΔtW = mΔv” or “ΔtW = maΔt”

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  ma = Δv / Δt · m.  This results in:

Δvm = ΔtF

ΔtF = maΔt = Δvm

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.

 

Distance to Horizon Equation Video

I wanted to create a short video that presents the ideas behind an equation that I came up with that properly presents what I call “Distance to Horizon”.  I supposed the point of the equation is to have an accurate way to measure distances on ball since certain inaccuracies in other equations have been used to dismiss the evidence presented by flat earth folk. Once you have an accurate way to measure, it makes even the most entrenched skeptic open to the evidence.

Another reason for this equation is to present what our experience should be like if we lived on a ball.  Of course, one equation can only go so far.  Empirical evidence needs to be collected and real scientific work need to be done to show the truth of our situation.  With certain work already accomplished by individuals in the flat earth community, any contribution towards greater truth is a good thing.

Hopefully this video can help explain this equation a bit better.  Please let me know if any other work needs to be done on this equation.

Flat Earth – The Horizon, Curvature and Angles

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.

BallEarthMath

 

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 ((1/(COS(θ)/R))-R). 

flatplanetrig_newv9
Click me

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.”

 

Polaris Paradox – Flat Earth Trigonometry

The problem of Polaris has been a stubborn one for those investigating the Flat Earth. The crux of this problem has to do with viewing angles, distances and elevation.

Using trigonometry, one should be able to measure the height of any object from any particular distance. Unfortunately, the math just doesn’t seem to add up. Using the supposed radius of the Earth – which is 3959.16 miles – we should be able to figure out the height of Polaris based upon the viewing angle and distance from the North Pole. The two assumptions – the radius of Earth and the distance from the North Pole – are generally agreed to values from both FE (Flat Earth) and GE (Globe Earth) people.

The viewing angle (“VA”) is where the problem starts – and ultimately will be resolved.   GE theory states that the viewing angle of Polaris is equal to the particular latitude the observer views Polaris (ie. 49th parallel has a viewing angle of 49°). The distance from the 49th parallel to the North Pole is 2,597.55 miles or the radius of the Earth at that parallel.   In the GE theory, the viewing angle is dependant upon the curvature of the Earth.

In the traditional FE view, Polaris is approximately 3600 miles above the North Pole.   However, the viewing angle from the Equator is supposed to be 1° but according to traditional trigonometry, the viewing angle should be around 42° – Hence the paradox (or in GE theory, proof of a globe).

In examining this problem, I began by using a classic trigonometry set and drew, in 10° increments, the viewing angles from an object at 3600 miles above the North Pole. Several interesting anomalies appeared that, in the end, helped me resolve this problem.

20160320_105030

You can see from this image that the distances between the viewing angles are not equal. In GE theory, the distances between viewing angles are equal since the curvature is doing the work. I summed up these observations as follows:

  • distance between degrees on a sphere/circle are equal (degrees of parallel)
  • distance between degrees on a flat plane are not equal (this is important)
  • As the height of an object from a flat plane decreases, the angle of view decreases and tends towards infinity (law of perspective using geometry). As the observer increases distance from the object, the angle of view decreases.
  • The viewing angle is inversely proportional to the distance from the object. As the viewing angle doubles, the distance to the object is reduced by half.
  • An object of 3959 miles from a flat plane would have a viewing angle of 10° and would be at a distance of 22,962.2 miles.

Observations 1-4 are all perfectly logical and fit well with the FE model. However, the 5 observation does not fit with known distances whether FE or GE. There is the possibility the FE model is incorrect but direct observations have shown that there is no curvature. We are right back in the middle of the paradox.

In an effort to resolve this confounding riddle, I began to model distances, heights and viewing angles in Excel and look for patterns or answers of some kind. After a few weeks of tinkering I developed this model:

https://docs.google.com/spreadsheets/d/11rVPrOIpXaej5OZxiUo13NdKwXUuHrI3Ul8IwBmOqaE/edit?usp=sharing

There are 2 assumptions in the model:

  • The radius of the Earth (3.959.16 miles). All other numbers are generated using standard trigonometry and are without opinion or conjecture.
  • There are 90° between the North Pole and the Equator

The model is defined by 1° increments (1-89) and uses TAN, COS and ATAN functions to obtain either an angle or a distance. There are two main sections separated by a blue line. The left hand section takes each viewing angle (starting at 1°) use a TAN function (H/TAN(VA)) to derive the distance. For example, and object that is 69.101 miles above the observer, would have a VA of 1° and a distance of 3958.79 miles. This equation is applied to each VA up to 89°.

I noticed that the VA and the distance are related to each other (see observations 3 & 4) up to and including the 32°. After that, the relationship doubles and the distance an observer is required to travel to double the VA is 4x the distance. I added a column that calculates the distance whenever the distance doubles starting at 1°. The distances correlate well but not perfectly. Plus, any differences increase as the distance decreases up to the 32° and then returns to normal after that. The distances are variable and change as the height of the object changes. By doing this, the actual VA is maintained and the distance alters the equation mentioned above. Another column (Apparent VA) was finally added but I will return to that one later as it is directly related to the solving of the paradox.

The right hand column uses the radius of the Earth as the fix value (rather then the VA in the left side). To obtain the actual viewing angle based upon distance from object, I used an ATAN function – ATAN(H/R).   The common value between both sides is the object height. I derived the radius of the Earth at each degree by the following equation [Radius of Earth*COS((Degree)/180*3.14159)]. This essentially flattens out the Earth into a series of concentric circles.

Once all these relationships were in place, all I had to do was change one single value – the height – to see how the entire model behaved. The biggest pattern that I observed was the “compression” of VA as the distance and height increased. For example, at a relatively low height of 2 miles, all of the distances and VA matched the degree relative to the equator. However, as the height increased the VA began to “compress” at the lower VA values. As I increased the height, the VA differential (difference between the degrees from the equator and the actual VA) increased. You can observe the graph “Angle Differential” begin to form a SIN wave as the height increases. I haven’t taken the VA into decimal increments at the top and the bottom but my guess is that the pattern repeats.

So how important is this VA “compression”? As it turns out, it makes all the difference in the world. The model suggests that VA on a flat plane do not operate the same as VA on curved surfaces. As the distance from the object increases, the change in the actual viewing angle per degree increases at a slower rate.   Take the above example of 69.101 miles – the rate of change from 1° to 16° is only 1° of VA (33° to 34°). As we can see, the VA does not increase at the same rate as the degrees from the equator. Therefore, on a flat plane we would expect a variable VA per degree from the equator whereas on a curved surface it would not be variable.

At 69.101 the VA from the equator is equal to 1°. However, as the height increases from this point, the VA becomes “compressed”. What do I mean by “compresses”? If you examine the “actual viewing angle” column on the right side of the model, you will notice that for the first 70° there is only a 3° change in VA. Within those first 3° the observer will notice very little change in the height of the object even over a great distance. It is only in the last few miles (from 76° to 89°) that any real movement in the object would be noticeable.

The phenomenon becomes even more exaggerated as the object increases in height. As I continually increased the height, I found that the height of Polaris would be 2,513.5 miles above the North Pole. At this height, the VA from the equator would actually be between 32° and 33°; all the remaining degrees are hidden from view since they are “compressed” into a small area below that degree. Of course no actual compression is happening but it is a phenomenon of perspective on a flat plane.

How can this be possible?

I was contemplating the problem of “compressed” VA but I could find anything that worked with the trigonometry – until I saw this awesome video by p-brane:

This video provides the mechanism with which the VA becomes “compressed” for objects near the horizon.

The human eye and perspective

An important piece to this puzzle is within the nature of the human eye. I have included two major references that the reader can take the time to read. The first is from Ian P. Howard (Perceiving in Depth, Volume 3: Other Mechanisms of Depth Perception, Volume 3 Chapter 26.4 .1(Effect of Height in the field of view)) and the second is Zetetic Astronomy, by ‘Parallax’ (pseud. Samuel Birley Rowbotham), [1881] chapter 14.

When looking at the horizon with the naked eye, (as opposed to using a telescope or binoculars) there are various laws of perspective that need to be considered.

Let’s take another example with an object at a height of 2.571 miles above the observer. To achieve a VA of 1° the observer would need to be 147.29 miles from the object. The observer would then have to move half that distance – 73.65 miles – to achieve a VA of 2°. However, if the observer traveled half the distance again – 36.82 miles – the VA would become 4°. This continues at the same rate until the VA is 32° at which the observer is merely 4.60 miles from the object. To achieve a 64° VA the observer would have to travel 4 times the distance – 1.15 miles from the object.

As we can see, the non-linear changes would have a direct impact on the VA based upon distance. We can put this model into practice through the observation and measurement of distant objects in relation to their height. For example, from Vancouver, BC, the distance to Mount Baker is approximately 68.39 miles. This would mean the VA would be approximately 1.71°. As an aside, if we assume a curvature of the planet, 1/3rd of Mt. Baker should be below the horizon when observed from Vancouver. In fact, 3,184 feet of 10,781 feet of the mountain would be below the horizon. As anyone from Vancouver has seen with their own eyes, the entire height of Mt. Baker can been seen (from base to peak) from 68.39 miles.

Take into consideration that 97% of the distance between Vancouver and Mt. Baker is traveled in the first 3rd of the VA (up to the 32°). The remaining 68° of VA occur during the last 3% of the journey. This will have an impact on how the human eye perceives objects at a distance. The VA is not constant. Because of this objects in the sky will appear higher or lower than they really are.

For example, an airplane flying overhead at 500 miles/hr will approach the observer slowly at first and then begin to accelerate as the distance decreases.   The airplane will reach maximum velocity (from the point of view of the observer) when it is directly overhead. The plane will then begin to decrease velocity as it moves away. We all know that the speed of the plane has not changed but the VA is changing. The more distance the plane gets the slower it appears to go. If we take this example and apply it to static objects (like a mountain) the same rules apply. However, the change in VA is due to the observer. As the observer approaches the mountain, the VA changes but at an inconsistent rate. At a great distance the mountain will appear to “rise” up from the horizon at a very slow rate until the first 32° of VA are completed. After that the mountain will begin to “rise” at an accelerated rate.

If we now apply these observations to Polaris, we can see that 97% of the VA is far behind the Equator. According to the model, the star will rise at a much faster and consistent rate after the 32° (which the actual degree of parallel of the Equator). You have to imagine an observer that is 147,298 miles from the North Pole. At that distance the actual VA is 1°. It would take 97% of the journey before the star will begin to “rise” up from the horizon. After that point, the star “rises” and a relatively consistent rate (albeit not at 1° per degree of parallel – but close).

Another important observation that needs to be taken into consideration is the human eye itself. It is documented that the total VA that the eye is able to perceive if the observer is looking directly at the horizon is 60°. In other words, the total field of view is only 60°. Also, the field of view also contains the ground beneath our feet. So within that 60° we have 100% of the field of view.

It is important to note that the model being presented is scalable for any object at any height (assuming the radius of earth being valid). This means we can use this model to accurately map the surface of the Earth using an object at a constant height (ie. Polaris).

Some thoughts on Gravity and Newton’s Laws

Over the past few weeks I’ve been thinking about gravity and the various laws that Newton proposed.  One of the main questions I have pertains to the equation for F (force) which is F=ma and the so-called law of universal gravitation which is:

F1 = F2 = G⋅m1 x m2 / r(squared)

And this is put into normal language as:

The force is proportional to the product of the two masses and inversely proportional to the square of the distance between them

meaning, if the mass increases then the product increases.  For example, as a mass accelerates the value of F will increase.  This can be seen in the extreme example of a mass that approaches the speed of light;  as it approaches, it’s mass approaches an infinite value.  The difficulty with this concept (as it is only a concept without any empirical evidence) is Newton’s 3nd law which states that “for every force there is an equal and opposite force”.

If you take the 2nd law and apply it to any accelerating mass, you increase the total amount of force which also an increase in energy (e=mc2).

Now if you take the law of universal gravitation (as shown above), and apply the 2nd law to m2, the total force between the two objects will increase.  you can ignore the increase of r(radius) since the increase in radius is insignificant relative to the size of m1.  In other words, an object like a rocket could not possibly launch if gravity acts according to the prevailing theory since the acceleration adds to the objects total energy (as shown above).  By increasing energy you increase the force between m1 and m2 with a “…a force that is proportional to the product of the two masses” and with equal force (as per the 2nd law).  Therefore for every pound of thrust an equal amount of force is brought to bear between the rocket and the earth.

This would also apply to anything accelerating away from the center of the earth (ie. rapidly raising my hand above my head).  In other words, gravity should be acting like a brake against the accelerating body.  To “break free” (think of an inverted pendulum flywheel) of the gravitational force, an object would have to accelerate with a force greater than m1 * acceleration.   Therefore, if we take two equal masses (m3 = m4) and accelerate one of them (m4) it would require the second mass (m4) to accelerate at a greater value than m3 * acceleration.  Essentially, it is an application of e=mc2.  This a far better explanation as to why the mass of an object increases as it approaches the speed of light – the mass itself is not increasing but the affect of gravity increases proportionally to the accelerated mass  therefore the effective mass increases.

The great irony here is that this makes gravity all but impossible since any object on the surface of a globe (ie. Earth) would be held fast against the surface.  Anything pushing against gravity would encounter massive (no pun intended) resistance (like blood flow, plant growth, etc) to the point where no life could form.   Nor could objects be buoyant.  An object floating on the ocean surface is essentially accelerating away from the center of the earth (until it finds equilibrium at the ocean surface).  The gravity of the Earth is far greater then the total outward thrust of the buoyant object (ie. air inflated beach ball).  In a nutshell, buoyancy would be completely overwhelmed by gravity.  As well, the lift experience by commercial airplanes would also be insufficient to overcome gravity.

However, no matter how reasonable this line of thinking is, many gravity apologists will simply drag out their favorite solution:

Einstein – The grand-daddy of excuses

If we push aside Newton for the moment a look at what Einstein proposed, we are actually in a less favorable possible (if you believe in gravity).  Firstly, Space-Time needs to bend or be distorted to create this magical affect.  Looking at the area around an object (like a planet), we see it is spherical.  The so-called gravity “well” needs to encompass the entire planet not just a portion underneath.  I talked about this in a previous post.  The point being, the distortion of space time is not like this:

It needs to be an gravity sphere.  So what about the distortion around objects on the earths surface?  Is this not what causes gravity?  Does space-time wrap around a cube or a oddly shaped stone?

The explanation is somewhat specious:

Bodies with spatial extent

If the bodies in question have spatial extent (rather than being theoretical point masses), then the gravitational force between them is calculated by summing the contributions of the notional point masses which constitute the bodies. In the limit, as the component point masses become “infinitely small”, this entails integrating the force (in vector form, see below) over the extents of the two bodies.  

In this way it can be shown that an object with a spherically-symmetric distribution of mass exerts the same gravitational attraction on external bodies as if all the object’s mass were concentrated at a point at its centre.[2] (This is not generally true for non-spherically-symmetrical bodies.)

https://en.wikipedia.org/wiki/Newton’s_law_of_universal_gravitation

Right.  This explanation effectively reduces all objects to single point masses and runs gravitational vectors between two bodies.  It’s really a rather grotesque idea.  Then the gravity within a body is nullified since all the internal objects of a “single mass” are counted as one:

  • The portion of the mass that is located at radii r < r0 causes the same force at r0 as if all of the mass enclosed within a sphere of radius r0 was concentrated at the center of the mass distribution (as noted above).

  • The portion of the mass that is located at radii r > r0 exerts no net gravitational force at the distance r0 from the center. That is, the individual gravitational forces exerted by the elements of the sphere out there, on the point at r0, cancel each other out.

As a consequence, for example, within a shell of uniform thickness and density there is no net gravitational acceleration anywhere within the hollow sphere.

So now we have a hollow uniform body and only the surface itself has gravity.  So if all the gravitational forces are cancelled out as per the explanation above, then all the gravitational forces most somehow come from the surface.  How is that even possible? Geometrically, the math flattens out the sphere by pushing everything to the surface so the concept of a larger mass having greater density and therefore greater gravity is expunged and we are left with flat plane – in essence.

So does the math coincide with reality?  If the math says all the gravity is on the surface, then is it really on the surface?  I mean really only on the surface.  If yes, then how could a surface, no matter how big, generate a sufficiently potent gravity field as to warp space-time?

Watching Planes Approaching an Airport – never seen on the horizon

I was thinking about how airplanes would approach an airport and how they should look if on a globe.  I was looking at various sites that talk about when a commercial airline (flying at 30,000 feet) starts it’s decent.  At about 300 miles out the plane will start descending at about 500 feet per minute.   If you could look 200 miles out with a powerful enough telescope, the horizon would equal about 5.6 miles.  In other words no airplane could be seen since at that distance since 5.6 miles is the highest a plane normally flies.
With that descent time, a plane would reach ground in just under 1 hour.  It would be assumed that the cruising speed would also decrease since the average speed is about 500 miles per hour.  Assuming that the average speed during decent is about 300 miles per hour and 500 feet is dropped per minute then at what distance should the plane be visible?
After 30 minutes the plane will have dropped 15,000 feet (2.84 miles) and be 30 minutes out and 150 miles from the airport.  Again, if you could look out that far with a telescope the plane would be at or below the horizon.  Over the next 15 minutes the plan would drop another 7500 feet and be 75 miles out.   At that point the plan should still be at or below the horizon.  Over the next 7.5 minutes the plane would drop 3750 feet and be 37.5 miles out and 937.5 feet above the ground.  Again, the plan would not be visible.  Over the proceeding 3.75 minutes, the plane would be 18.75 miles out and 234 feet above the ground and still not visible.  However, it is mostly likely that the decent and speed would be throttle to match the runway so the final few miles one would finally allow the plane to appear at the horizon.  So what we would see would be the plane shoot upward from the horizon and then rapidly drop down.
Of course this is an extreme example to demonstrate the idea.  But the point it that if you watch airplanes approach airports they are always very high in the sky and slowly descend over time.  They never shoot up from the horizon.  What we do see is planes ascending or descending but never appear at the horizon.
If someone was to film airplanes as they approach the airport,  I can pretty much guarantee that not a single plan will shoot up from the horizon and then curve downwards toward the airport.
On another note, whenever you see jet streams they are always straight.  At plane at 30,000 feet will be visible for a shot period of time before it starts curving towards the horizon.  They should have a curve in them.  No ones sees that either.

Simple experiment to show Earth is either Spinning or Stationary

How do we know if the world is spinning or stationary?  Do current images from space or the material in text books prove the Earth is spinning?  Only if you believe the source to be valid.  If, however, you are a natural scientist you would want to verify the veracity of the statements by repeating the experiments.

Using these two examples, let’s look at what is experienced when a person walks upon the earth.  If the earth is spinning it is comparable to the conveyor belt example except that the earth’s axis acts as the central wheel that “belt” of the earth’s surface or ground moves around.  It cannot be the train example since the floor of the train is not in motion.  We can see that additional energy is provided by the moving ground whereas no additional energy is provided to the person who is beside the belt.  We could then create an experiment where a person would leap against the direction of the belt (both on the belt and off).   We could place an object 5 feet from the person both on and off the belt and ask them to leap towards the object.  Since the additional energy of the belt is continually compelling the object forward, we would theorize that the total distance between the person and the object would be less than then person leaping beside the moving belt.

Why is leaping important in this experiment?  If the person simply walks on the belt, the forward momentum will be continually added to the individual since there is always contact with the belt.  By disengaging with the belt in a direction against the motion, we are subtracting (for a brief moment) the forward momentum of the belt and allowing the belt to move underneath.  It would be possible to land nearer or even surpass the object whereas the person leaping beside the belt would gain no advantage like that.   The question is how quickly does the forward momentum of the belt dissipate once the person leaps?  If the person can leap (towards the object like a long jump) with an acceleration of 1 m/s/s and can stay airborne for 2 seconds then a total distance of 2m can be achieved (with a final velocity of 2m/s).  Since the belt is moving at 1 m/s then in the first second the momentum has been overcome and the object has moved 1m closer.  In the 2nd second the object moves 2m closer if we add the motion of the belt and the acceleration of the person.  The total distance covered would be 3 meters.  In the case of the person beside the belt, they would only be able to cover 2m if they leaped with the same acceleration.  (you can work out the acceleration equation if you want)

We could increase the acceleration and distance by using something like a canon.  If we use the same arc and acceleration the ball should land at a greater distance from the canon if on a moving belt than if stationary.  If we assume that the acceleration of the canon is 10/m/s/s, the tennis ball can reach 80 m/s velocity and has a total airtime of 10 seconds, we can calculate the total distance travelled (640m).  If the velocity of the belt is 1 m/s then within 1 second the momentum of the belt has been overcome and we can add an additional 9 seconds of belt movement to the distance for a total of 649m.  If we then turn around and fire the canon in the opposite direction then the distance travelled by the canon is subtracted from the total distance of the ball (1m/s x 10 seconds = 10m).  This would mean that the total distance between the canon and the ball would be 630m.

If we take the above example and apply it to a spinning earth then a similar result must take place.  Taking the spin as the same as the conveyor belt one need only launch an object (a tennis ball thrower would suffice) to the west and then to the east.  If the objects land at similar distances from the thrower then we are not spinning.  However, if they land at different distances then we must be spinning.

Response to Spherical Geometry and Empirical Evidence

In response to my latest post, I received a comment that due to refraction, objects at certain distance would be visible even if they are below the horizon.  He also talked about adjusting curvature every 8-10 miles.  I’ve started reading his blog but the reason for the adjustment is not clear.  However, I think he is talking about this. In the end, the height of the person and the object being observed are taken into account.  Here is the original comment:


“You are using an improper formula for curvature that has to be constantly corrected for every 8-10 miles. The correct forumula for curvature is found here. https://chizzlewit.wordpress.com/2015/05/13/working-with-the-curvaure-of-a-spherical-earth/ Also you are over water which tends to make refraction a bit of a problem. You will find however that with the height of you the observer on the deck as you pointed out and the height of the lands you put in that it does indeed fit in with the sphere earth model especially when you account for refraction. If you want to use a crude formula for distances between lighthouses and boats that includes for refraction you can use this one though its only an average correction for refraction. http://www.pajack.com/stories/pitts/viewdistance.html I suggest you try this with surveying equipment though if you are at sea the choppiness of the water will make it tough. Also being over water with refraction conditions makes it tough. You will find though that if you use surveying equipment you can clearly measure landmarks and see the curvature of the earth. https://www.youtube.com/watch?v=IOFmLHHwtic


To be fair to the individual, I wanted to address the objections empirically and see if he had valid points.  You can read my response here:


“I appreciate your comment. The main issue with the video is that it was taken with a crappy Samsung smart phone. It is possible to take a stable view from land (from either side) which is what I plan to do so any shakiness could be taken out. This will require a telescope with a camera to bring a sharper focus at a greater length. There are distances of 70 miles or more that can be observed from Vancouver. Specifically from Kits beach to Denman Island.

I would like to address your blog and the concept within them. If we wish to proceed using a scientific process (which must involve empirical evidence not just math), we must apply your math to observations.

If I misrepresent your ideas or math in anyway please forgive as I’m simply trying to apply them the best I can. If there are errors, please point them out and we can move forward from that. If we disagree, then the empirical evidence must be the final judge, not our opinion. Do you agree with that?

If you can apply your formula to the distances given in my post (29 and 33 miles) that would provide me with an actual representation of how your formula represents curvature. From there we can compare the two formulas and see the differences. Are you ok with doing that?

Refraction is defined as “the bending of a wave when it enters a medium where its speed is different. The refraction of light when it passes from a fast medium to a slow medium bends the light ray toward the normal to the boundary between the two media.”

As I’m looking towards Vancouver from the deck of the ferry, I would not be looking into water which would be a slower medium. As well the refraction requires the object be within the slower medium (or at least in between my eye and the object in question). When looking at this wiki article, the superior mirage is the only explanation that you are referring to that could bend light (if you have a different one, please explain). It would require that the object in question is within an area of higher temperature than the observer. As well, if this object (like a building) occupies various temperature strata then the object would appear quite distorted since different temperature regions would overlap and the image would not make much sense to the eye (https://en.wikipedia.org/wiki/Mirage#/media/File:Superior_and_inferior_mirage.svg). The video that I shot shows objects that are consistent in shape and size without any distortion. With both superior and inferior mirages the objects are inverted and/or distorted. The other issue is that the temperature differential required for a superior mirage is significant:

“The “resting” state of the Earth’s atmosphere has a vertical gradient of about -1° Celsius per 100 metres of altitude. (The value is negative because it gets colder as altitude increases.) For a mirage to happen, the temperature gradient has to be much greater than that. According to Minnaert,[1] the magnitude of the gradient needs to be at least 2°C per metre, and the mirage does not get strong until the magnitude reaches 4° or 5°C per metre. These conditions do occur with strong heating at ground level, for example when the sun has been shining on sand or asphalt, commonly generating an inferior image”

The temperature differential from the bottom of UBC to the peak would require a consistent temperature change over 81 meters. According to the above formula, the temperature from the bottom of UBC (let’s say 12 degrees C) to the peak would be impossible. It makes sense at small distances (like asphalt) since the temperature difference can be significant. However, the temperature from sea level to 81 meters will not be 4-5 degrees per meter. So unless you see an error in that formula, we can safely rule out a superior mirage. So once I can see your math using your equations then we can move forward with the next discussion. Hopefully I will be able to get a sharper image next time that will show more detail.

Thanks again for your comment.”


It occurred to me that the image of Chicago across lake Michigan

was described as a superior mirage.  As I wrote in my post above, the temperature differential between the lake surface and the highest object would require at least 2 degrees per meter.  If you think about it, an object 30 meters high (~100 feet) would require 60 degrees difference for the affect to occur.  And that is not a very strong affect.  According to the above article, you would need 4-5 degrees for it really be visible.  There simply is not that kind of temperature differential in such short distances.  As well, the object would need to be inverted or distorted which is obviously not the case in this video.  Near the end of the video, they show a “duct” between the hot air and the cold air.  For the affect to work, the temperature difference must be so great as to bend the light.  The distance between the hot air and cold air is greater than 1 meter.  The important thing to remember is that the “temperature gradient” must change not just one portion but the entire length of the object.  In the case of a small object on the horizon at a great distance (70 miles), atmospheric distortion would occur and you would see inverted images.  But if the object(s) in question are visible at 70 miles at a greater distance above the horizon (like a tall building) then temperature must be that much greater.  Finally, if the image in question was taken with a camera with a zoom lens it completely removes the possibility of a mirage since the focal point of the lens would bypass the affect as seen in this video:

As the lens sharpens the image, any distortion would be removed and any mirage would disappear.

You can read about Fata Morgana to see how these distance object are explained within the commonly accept ideas about the curvature of the earth.  However, mirages are distinctive by their distorted image and changing conditions.  In the above video of Chicago, there is no distortion.