The Sun never sets on the Flat Earth

The title of this blog might seem, at first, to be debunking the flat 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.solar-analemma-070000-UTC
  • 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:

Screen Shot 2018-04-29 at 10.29.24 AM

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.

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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:

Screen Shot 2017-09-01 at 11.12.22 AM

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:

Screen Shot 2017-08-27 at 10.49.42 PM

Screen Shot 2017-08-27 at 10.50.03 PM

Screen Shot 2017-08-27 at 10.50.22 PM

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.

flightawarev2

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

 

 

 

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.

 

Gravity: The Baron Munchausen of Forces

 BaronMunch

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 Flat 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
  1. 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°.

 

Polaris Paradox – Update

I’ve been working on a new model based upon our conversation.  I found a bunch of errors in my original model that I missed the first time.

 

I was mingling FE with GE values which made the model not work properly.  Looking at it again, I built the model with the assumption of an FE.  There are only 2 assumptions:

 

1) Radius of FE a is 12,482 miles
2) The height of Polaris is 3959 miles (or radius of GE)

 

I fixed the graphs as well.  One of them didn’t provide any meaningful information so I removed it.

 

From this model I’ve developed FE latitudes which have nothing to do with GE values.  The latitude values are a function of radius and the height of polaris – [FE Radius]*COS((FE Latitude)/180*3.14159) = FE Radius @ latitude.

 

 

The viewing angle and perspective are an important concept in this model.  Here are some axioms that we can test:

 

– Viewing angle and latitude are not equal
– Objects at a distance increase in height at an unequal rate as the object approaches assuming a constant velocity.
– A viewing angle of 0-45 degrees is obscured from the bottom-up.
– A viewing angle of 45-90 degrees is obscured from the top-down.
– The obscured area subsequently obscures a proportional amount of viewing angle.
– At 45 degrees the obscured values are equal at the top and bottom

 

I’ve highlighted sections that have interesting or important values.  I have a hypothesis that the GE radius (3959 mi) was used since the distance to Polaris is 3959 miles.  They just rotated the model 90 degrees.

 

I’m added in some actual locations (Toronto, Vancouver, Quito, Monterey) and compared the model against real-world distances.  It fits quite well.  I think the Viewing Angle issue is resolved by the model since it shows how perspective obscures the top and bottom.

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.”

 

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.