1

Flat Earth Theory / Re: Center of gravity of objects on a flat earth

« on: August 28, 2020, 07:03:34 AM »

You need to ask yourself how 100kg-kid is "... supported above the ground".  He isn't suspended by levitation, he's only there because he's an integral part of a rigid ASSEMBLY comprising him, the 50kg-kid, and a beam, all attached to the planet by a pivot.  We assume that the beam itself is symmetrical and its mass is evenly distributed, so the COM of the ASSEMBLY is to the right of the pivot, because that's the end the 100kg-kid is sitting.   

If the planet accelerated up at 9.81 m/s/s, that force is going to be felt at the pivot of the ASSEMBLY.  The inertia of the ASSEMBLY acts at its COM which to the right of the upward accelerating force.  The 2 forces form a couple which rotates the ASSEMBLY clockwise until the planet hits 100kg-kid's butt, at which time the accelerating force is now distributed to both sides of the COM, via the pivot and his butt. 

Stop thinking of him in isolation.  He's just a part of something bigger. 

And have you noticed I've stopped calling him fat-kid.

I didn't fully explain myself, my bad.

Assume we have a seesaw with the fulcrum in the middle with 100N downward force on the left and 50N downward force on the right (IOW, gravity).  It will rotate counter clockwise.

Now reverse the direction of the force (universal acceleration), with 100N on the left upward force and 50N on the right upward force.  It will rotate clockwise.  The heavier side will be suspended in the air, instead of falling to the ground.

My point was that inertia will slow down the rotation, maybe even stop it, but it won't reverse the rotation from clockwise to counterclockwise.

Even in a vacuum, the  acceleration is going to produce an upward force on the fulcrum through a normal force exerted by the ground  Since the force does not occur at the center of mass of the seesaw, a torque is exerted .This will cause the seesaw to rotate exactly as it would in a gravitational field,

It’s been 100 years since The Equivalence Principle was proposed.  If there was a hypothetical experiment that would have violated it, I think someone would have figured it out by now.

Edit:  Reading through the above threads, with the seesaw/teetertotter example, we need to consider torque.  Simply looking at the forces doesn't tell the whole story, because the seesaw system as a whole its no longer a rigid body.  As mentioned above, the degree to which the seesaw rotates, is going to depend not just on the mass balance at either end, but where the fulcrum is placed.  This is a separate issue from the CoM, as we can make a simplification that the mass of the board connecting the children and the  mass and that of the fulcrum are small in comparison to the weight of the two children.  But if we shift the position of the fulcrum, it changes the behavior without affecting the CoM or the forces involved. 

2

Flat Earth Theory / Re: Erathosnes on Diameter (from the Wiki)

« on: August 12, 2020, 11:41:19 PM »

So it is Imagination Land then. That just creates an assumption that there is a place where there are no forces.

Lets go into Imagination Land and try to get bodies to travel from Point A and Point B in a straight line.

Think of a rocket ship in space that blasts its engines for five seconds:



What makes you think it is possible to create an engine that fires perfectly evenly from all points? If the engine starts in the slightest manner from one side or gives off the slightest bit of more energy on one side then the rocket builds too much momentum on one side and goes spiraling off to one side, either quickly or slowly. It doesn't reach the Point B destination.

The only way to go straight is for the rocket to be constantly controlled and navigated. AKA, an artificial result.

Even in something as simple as the game of Pool, it is rather difficult to hit a ball to go straightly to the Point B hole:



You have to hit it in just the right spot to get it to go the way you want it to go. Luckily in pool you are only shooting a few feet.

What if you are shooting a ball at a Point B hole which is miles away? Ignoring the fact that you don't have the arm strength to do that, and ignoring everything about surface friction, to hit a pool ball perfectly for a distance ranging in miles and get into a hole is still very improbable. The ball is more likely to go off in some random direction.

So again, straight line trajectories between two points do not naturally occur in nature. They are highly unnatural.

Ok, so you don't believe in straight line trajectories.  So this leads to the question, how can we measure anything? 

How do we know that the light reflected from the ruler we just measured the length of something as 20 centimeters or whatever, isn't nature just playing tricks on us?

3

Flat Earth Theory / Re: Erathosnes on Diameter (from the Wiki)

« on: August 04, 2020, 06:35:40 PM »

Herein lies the problem.  It will be impossible for an observer on a flat disc to ever observe the Sun at/below the horizon, at the same time another observer at another point of the disc observes it directly overhead, (or really, any point substantially above the horizon). Let me illustrate.

This is an entirely different argument related to the nature of light and perspective and how they operate at large scales, which you are proposing to be indisputably true.

We observe that the sun sets, therefore there must be a mechanism which makes this happen.

See: https://wiki.tfes.org/Sunrise_and_Sunset

All of this is rather unconnected to this discussion of the model, unless you want to prove your assumed axioms.

My assume axioms are basic Euclidean space (the 3d dimensional extension of 2d Euclidean plane geometry), and that light, at least this particular circumstance of the Sun to Earth observer, is (mostly) following a straight line path.

If FET is going to advance alternative axioms, such as non-Euclidean geometries, or that the light from the Sun to Earth does not travel in a straight line path in the circumstances we have here, this needs to be stated.  It also should be quantified, when making quantifiable predictions such as measuring the diameter of the Earth based on measurements like Erathosnes.   Otherwise it is impossible to work out the logical conclusions of the theory.

So I would ask the following questoins

1) Is Euclidean space an accurate representation of reality, at least in most circumstances? 

2) If the answer to (1) is no, what geometry model is FET using?  (hyperbolic and elliptic are two alternative geometries)

3) If the answer to question (1) is yes, does light follow straight line path, at least in some circumstances, such that your re able to determine the actual position of objects (X,Y,Z) coordinates in three dimensional space given multiple simultaneous observations from observers located some distance part?  Or alternatively, multiple observations over time from different perspectives assuming the object is stationary?

4

Flat Earth Theory / Re: Erathosnes on Diameter (from the Wiki)

« on: August 04, 2020, 05:55:34 PM »

First off, this measurement was not made at the north pole, but much further south (in Egypt.  Also degrees, as a unit of measure of an angles, have no definition along a straight line. They only have a definition as fraction of the circumference of a circular. (The full circumference represents an angle of 360 degrees, half circle 180 degrees, etc.).  They are also defined by the intersection of two lines, because you can construct a circular arc of radius 1 originating at the intersection point which begins at one line at ends at the other. You can do this physically with a protractor.   So I'm not sure how you can justify labelling the north pole as 90 degrees (or any other degree measurement, really), assuming a flat earth model. 

I also do not see the intersection of any lines, or anything that represents an angle, which correspond to the red line that is supposed to represent 1/25 of the radius, so I'm not how the measurement of 7°12' is coming into play here.

7°12' is 1/25th of 180 degrees.

If you move North to South from one side to the Sun's area of light to the other side the Sun will rise from the horizon and make 180 degrees arc over your head.

When you say "move North to South", I presume you mean from a given point towards the closest point on the outer edge of the disc (which would be away from the North Pole).   

Herein lies the problem.  It will be impossible for an observer on a flat disc to ever observe the Sun at/below the horizon, at the same time another observer at another point of the disc observes it directly overhead, (or really, any point substantially above the horizon). Let me illustrate.   Keep in mind this is a side view.



Observer O observe the sun directly overhead (90 degrees).  Observer A will observe at some angle < 90 degrees. Observer B will observe the sun at a some angle < 90 degrees.  As we imagine observer B getting farther and farther away from A and O, the angle keeps diminishing, this is true.  But you cannot imagine even getting close to the horizon (an angle of 0 degrees), until observer B is very far away. 

But we can better than just intuitive observations - we can actually do the math to determine how far B must be from A or O, in order to observe the sun at some given angle. 

I'll use the following convention.  ang(X) is the angle formed at point X, between the line from X to the sun, and line of the ground.  In terms of Erathosnes, it would be 90 - the angle of the shadow formed in his stick experiment.   For example, ang(O) = 90 degrees.  dist(X, Y) is the distance from point X to Y.  Also, though I didn't label it, I'll use S for the sun.

This following requires trigonmetry.  Since we have a right triangle (one where one of the angles is 90 degrees), we can easily solve for lengths if we know one of them and angle between one of the sides. 
In a right triangle, there will two line segments which are short, and a third line segment which is longer than the other two.  This is called the hypotenuse.   

There is a cool function, called "sine", which is defined for a given angle, assuming the hypotenuse is 1.  It is writen as sin(X), reading as "the sine of X"  sin(X) would be the length of the side opposite that angle.  Since the hypotenuse is 1, sine can never assume a value > 1.  In a real right triangle, where the hypotenuse > 1, we can simply multiply sin(X) by the hypotenuse to get the length of the opposite side. 

There is also a similar function called "cosine", similar to sin.  This gives you the length of the adjacent side to the angle.   It is written as cos(X).   cos(X) * length of hypotenuse  gives  the length of the side

Also I shoudl note, when using angles in the sine and cosine functions, you need to plug them in as "radians".  Radians are defined as the length of the circular arc formed by the angle assuming a radius of 1.   The circumfrence of a half circle is defined to be PI, which is roughly 3.141527.   This correspondes to 180 degrees.  So the conversion from degrees to radians is radians = PI * degrees / 180.

Looking at the diagram again, we can see both triangles share a common side, which is dist(O, S).  We can use trigonometry to solve for the dist(B, O), if we have enough information.

In this case, we know ang(A), we know ang(O), we know dist(A, O), and we are presuming ang(B)  The issue is to determine the dist(B, O) which satisfies the observations. 

Now lets write down some equations, just going by the trigonometry

1) dist(B, S)*sin(B) = dist(O, S)
2) dist(B, S)*cos(B) = dist(B, O)
3) dist(A, S)*sin(A) = dist(O, S)
4) dist(A, S)*cos(A) = dist(A, O)

Equation (2) is the one we want to solve for, dist(B, O).  However, we do not know dist(B, S).  But we can use equation (1) to solve for dist(B,S).  Dividing both sides by sin(B), we get

5) dist(B,S) = dist(O,S)/sin(B).

Now the problem is we don't know dist(O,S).  But we can use equations 3) and 4) to solve for it.  Equation 3) gives us the exac trelation - the only problem is that we don't know dist(A,S).  But we can use equation 4, dividing both sides by cos(A), we get

6) dist(A, S) = dist(A,O)/cos(A)

Plugging this back into (3), we get

3) dist(A, S) * sin(A) = dist(O, S)
7) (dist(A,O)/cos(A)) * sin(A) = dist(O, S)

Now plugging this back into equation (5)

5) dist(B,S) = dist(O,S)/sin(B).
9) dist(B,S) = [ dist(A,O)/cos(A)) * sin(A) ] /sin(B).

Now we plug this back into (2) to solve for dist(B, O)

2) dist(B, S)*cos(B) = dist(B, O)
10) [[ dist(A,O)/cos(A)) * sin(A) ] /sin(B) ] * cos(B) = dist(B.O)

And we ended up with a mess, however a mess that can be calculated.

Now let's plug in Erathosnes numbers, and a reasonable number for ang(B), like lets say 5 degrees.  He observe an angle of 7.2 degrees, which means that ang(A) = 90 - 7.2 = 82.8 degrees angle of the sun above the horizon.  The distance, dist(A, O) was 500 miles.

So dist(B,O) = [[ (dist(A,O)/cos(A)) * sin(A) ] /sin(B) ] * cos(B) 
                   = [[ (500 / cos(A * PI/180)) * sin(A * PI/ 180) ] / sin (B * PI / 180) ] * cos ( B*PI/180)
                   = 45239.09 miles.

The claimed diameter of the Earth in the wiki was 25,000 miles.   And we weren't actually measuring the diameter, but something less than the radius.  What that means the diameter woud have to be more than twice our number, so > 90,000 miles.

And this is assuming an angle of 5 degrees, as a I felt that was reasonable.  But lets see what happens to our numbers as they get closer to 0 degrees.

For 2 degrees, we have dist(B, O) = 113339 miles ( to the nearest mile)
For 1 degrees, we have dist(B, O) = 226748 miles (to the nearest mile )
For 0.1 degrees we have dist(B, O  = 2267711 miles (to the nearest mile)

So we can see whats going on, as we get closer and closer to 0, the distance is increasing very quickly.  In fact looking back out our equation, since we are dividing by sin(B), it cannot be zero otherwise the value is undefined.  sin(B) is only zero when B itself is 0, i.e, the sun is at the horizon.  The model predicts that we can never get far enough away, to actually observe this. 

Now, I suppose you could introduce that the idea that the light from the sun is doing something funny which causes it not to move in a straight line (refracting, accelerating, etc.) causing the observation you get.   But, when do that, it trashes the basic geometry of the problem, such that he observations themselves worthless without knowing what the light is doing/ why it is doing that.       

Therefore we can take the known distance and multiply by 25 to get that distance across the sunlight area. If we think that the Sun is moving in a circle like the Monopole model we can double our figure for the total diameter.

It's really just measuring the sunlight area on one side and then multiplying by two to get dimensions for a model.

I determined that the sun goes around the Monopole model because I read the FAQ.

I determined that the sun would rise and eventually set when you move towards it from one side of its area of light to the other based the sun's rising and setting over the course of a day.

I determined that the Sun would move consistently or near-consistently in the sky by observation.

The sun moving consistently in the sky at a given point, over the course of the year, is another problem (it doesn't, it varies over the course of a year at any given point), but I'll put that issue aside for now.  Right now I want to focus on the basic geometrical reasoning.

So, in conclusion, it is my position that, assuming a flat Earth, Eratosthenes experiment can tell you nothing about the diameter of the Earth. 

As a addendum, I'll note that I am not trying to necessarily argue for an alternate mode such as RET.  I am simply working out the claims of FET, as claimed in that section of the Wiki, to its rational conclusions given basic geometrical reasoning.

5

Flat Earth Theory / Re: Erathosnes on Diameter (from the Wiki)

« on: August 04, 2020, 07:05:33 AM »

First off, this measurement was not made at the north pole, but much further south (in Egypt.  Also degrees, as a unit of measure of an angles, have no definition along a straight line. They only have a definition as fraction of the circumference of a circular. (The full circumference represents an angle of 360 degrees, half circle 180 degrees, etc.).  They are also defined by the intersection of two lines, because you can construct a circular arc of radius 1 originating at the intersection point which begins at one line at ends at the other. You can do this physically with a protractor.   So I'm not sure how you can justify labelling the north pole as 90 degrees (or any other degree measurement, really), assuming a flat earth model. 

I also do not see the intersection of any lines, or anything that represents an angle, which correspond to the red line that is supposed to represent 1/25 of the radius, so I'm not how the measurement of 7°12' is coming into play here.

6

Flat Earth Theory / Re: Erathosnes on Diameter (from the Wiki)

« on: August 04, 2020, 12:57:25 AM »

If the sun moves across the sky consistently or near-consistently (by observation), and if the 500 mile distance between those two points is 1/25th of the total longitude radius of the Monopole model (90°S - 90°N), then the Monopole model can be computed to have a radius of 12,250 miles. Double for the diameter to get 25,000 miles.

I'm still not quite following you - could you provide a diagram for what you mean?  I also don't see anything in the wiki about the consistency of the motion of the sun coming into play here.

7

Flat Earth Theory / Erathosnes on Diameter (from the Wiki)

« on: August 03, 2020, 09:40:42 PM »

In the Wiki it is stated

https://wiki.tfes.org/Eratosthenes_on_Diameter

We can use Eratosthenes' shadow experiment to determine the diameter of the flat earth.

Syene and Alexandria are two North-South points with a distance of 500 miles. Eratosthenes discovered through the shadow experiment that while the sun was exactly overhead of one city, it was 7°12' south of zenith at the other city.

7°12' makes a sweep of 1/25th of the FE's total longitude from 90°N to 90°S (radius).

Therefore we can take the distance of 500 miles, multiply by 25, and find that the radius of the flat earth is about 12,250 miles. Doubling that figure for the diameter we get a figure of 25,000 miles.

However, I am confused about how this claim can be made.  Here is an simple diagram of the situation (no tto scale)



It is claimed, that on a flat earth, that if we know D, the distance between Syene and Alexandria, we know the angle of A, that we can compute De (diameter of the earth in a 'flat disc' model) 

However, I do not see how this can be done.  Without going into the geometry. Imagine extending De to any aribtrary length.  It will not change the angle A, nor the distance D.  Therefore, for any given A or D, maps to any De > D at least (which would be obvious)

8

Flat Earth Community / Re: ... and so easy to prove to yourselves.

« on: July 15, 2020, 05:03:34 AM »

Well, Jeb- it works for me, but apparently, any video evidence is unacceptable here on TFES, unless
posted by a FE member. (Am I wrong here ? Can't seem to be able to get a straight answer).

From my reading of the posts, I don't get that impression, no.   But there is some truth in that pictures, especially still frames without context, can be deceiving.  And the truth is not always obvious or intuitive - for example special relativity is highly counter-intuitive. Another example, light refraction has been used to explain why Rowbathan observed what he did at the Beford Level.   It is true that light bends in some circumstances.   So,  examples like those show why it is difficult for someone not formally trained especially, to understand when/why/how these phenomena occur.

9

Flat Earth Community / Re: ... and so easy to prove to yourselves.

« on: July 14, 2020, 07:36:32 PM »

I'm not sure if it would help much.

Here's a hobbyist weather balloon that got to 110000 feet.



There would be a number of objections raised

1) Lens or image distortions

2) Bending of light

3) The curve is what they would expect from a "flat disc" model of the earth.  Since clouds are going to be obscuring a good portion of the view, you will not be able to show that only one side of the Earth is visible at once. 

Edit: I double checked that video - if you pause it at the right time, you can actually see the curve of the earth invert, when the camera is rolling around.  I think this is probably due to exposure, but it would be an example of something they would probably bring up.

10

Flat Earth Theory / Re: Doubt in Universal Acceleration

« on: July 13, 2020, 11:14:07 PM »

If the Pound-Rebka experiment was carried out in an accelerating frame (without gravity) such as a rocket or in this case the Earth accelerating upwards (UA), then the results would be somewhat different. This is due to the 'blueshift drift' effect for accelerating frames. The formula for the expected z values is given by:    which is time dependent.

The plot below shows blueshift (z) against time, if the same experiment was carried out in gravity (red line) and an accelerating frame such as UA (blue line). For the accelerating frame, we have blueshift drift, and for gravity no blueshift drift would be detected. The time axis is in seconds and goes to six months.

Using the Equivalence Principle as evidence to support UA is fine for things like how objects fall, projectile motion etc... but when it comes to the nature of light and how we observe its Doppler shifts over periods of time then it no longer holds. This flaw within the Equivalence Principle is a way to distinguish between an accelerating frame and gravity.

Hmm, but I thought the idea of the EP, was that there was no local experiment you can do which differentiated between free fall and an accelerating frame relative to some observer who is not accelerating or experiencing free fall.

Now the catch here is "local", because as gravity falls off with the square of the distance, the acceleration is not constant, so your experiment/measurement has to take this into account.  Was that the point of the experiment ?   

This leads to another hypothetical experiment, which might be a bit easier to understand.  I was reading the Wiki on UA.  Apparently the "gravitational anomalies" as determined by things like scale measurements are dismissed out of hand because the scales aren't re calibrated (which is kind of the point, but lets ignore that for now) 

I don't see a claim whehter UA is predicting an acceleration which is invariant with altitude, or makes some attempt to account for it, if it can be observed.   

Edit:  Sorry, there does appear to be a reference to "Celestial gravitation" which causes the acceleration to fall off with altitude, so apparently some celestial bodies exhibit gravitational pull but apparently it does not apply to Earth, at least not the same way).   However, I can't find how this would be quantified.   Gravitational theory predicts that g should fall off with altitude, with the square of the distance between the two bodies.

I'm wondering if an experiment whereby free fall was measured, with say with a ball bearing in a vacuum tube, perhaps suspended by an electromagnet, released with electronics which was tied to a precise timer which is demagnetizes and stops when a sensor at the bottom is hit.

In constant acceleration, the velocity v =  a * t, assuming initial velocity 0.  Which means the position, we'll call height h, is the antideriviative, so is h =  0.5 * a * t^2, again assuming initial height of 0.  Solving for t, we get t = sqrt(h / (0.5 * a))

If the tube were 1 m, then at the surface of the Earth (g = 9.806 m/s^2), the free fall time should be the sqrt (1 m /(0.5 * 9.806 m/s^2)) = 0.4516 s, or 451.6 ms.   At an altitude of 10 km, which is typical of airlines flight, g is predicted to be closer to 9.776 m/s^2, which yields a drop time of 1/(0.5 * 9.776 m/s^2) =   452.3 ms, so a difference of 0.7 ms.  Not exactly easy to measure, but with decent electronics, shouldn't be hard at all, as its only a frequency of ~1500 Hz. 

Of course we would have to be sure the  plane is not accelerating, and that the tube was perfectly level.