Gravity isn't constant

How did they get the camera back if there is no gravity at the dome?

Quote from: inquisitive on February 04, 2019, 02:27:40 AM

Quote from: wise on February 04, 2019, 02:22:34 AM

Quote from: georgeashetha1 on January 31, 2019, 01:53:15 AM

If gravity, as we know it, doesn't exist, but rather is a constant acceleration upwards at ~9.8m/s^2: how can gravity differ around the earth. It is known and proven that the gravitational acceleration is around 9.82m/s^2 in the north, while it's closer to 9.78m/s^2 around the equator. How can this be explained by the flat earth theory?

constant:

f:9,80m/s^2 +-0,10 difference caused by distance to dome. So what?

We can measure the f constant and can create a dome distance however we want freely. How re explains it distance to center and creates a distance, we can do it for dome too. No problem.

Please provide details of a dome, where are the edges?

We are not talking about edges. It may be finitive or infinitive. The important thing is distance, as altitude to dome. It has estimated as an about value here as:



73 miles highness.

Somewhere in the earth dome is 72 miles, somewhere it as 74 miles. Then formula changes.

9,81 m/s^2 for 73 miles.

If you see a 9,80 m/s^2, so the dome distance is:

72,93 kms.

As you see that, dome explains everything and

f (flat earth constant accelerate)= 9,81 m/s^2 average.

So rocket moving at a high rate of speed just hit the dome and stopped without being compacted/crushed/destroyed from the impact. 

How did they get the camera footage?

As you see that, dome explains everything ...

Except for everything the dome doesn't explain.

Quote from: John Davis on January 31, 2019, 09:27:33 AM

This is a map of gravity anomalies - where the round earth model fails to predict the observed gravitational pull. Thank you for providing us with yet another proof that the earth is not a globe.

Is a tennis ball a perfect sphere? Of course it isn't. Is it close enough that a reasonable person would describe it as a sphere? Yes.

Is the earth a perfect oblate ellipsoid? Of course it isn't. Is it close enough that we can define an ellipsoid accurate enough to produce maps and use GPS - yes.

If you believe the earth is flat, then is it perfectly flat? Of course not - the same argument holds.

The round earth model is, as you point out in your response, not based on a perfect sphere. Thus, I'm failing to see why the above map is not a map of where the round earth fails to match empirical measurements given your response. I further fail to see why it is not assumption then that the earth is in fact a globe, in spite of empirical measurements saying otherwise, to then assume further that this is due to local mass distribution and not some other factor.

Thank you to all whom have provided further evidence that the round earth model does match up with empirical readings.

Thank you to all whom have provided further evidence that the round earth model does match up with empirical readings.

Fortunately measured distances prove a round earth.

Quote from: John Davis on February 04, 2019, 08:34:48 AM

Thank you to all whom have provided further evidence that the round earth model does match up with empirical readings.

Fortunately measured distances prove a round earth.

To say measured distances prove a round earth, you must show no other solution exists and that the empirical data matches. I have provided one - the surface of the earth is non-euclidean (edit: it is elementary to provide an infinite amount of such disproof by examples.) You also have not shown it matches empirical data, and there are several counter examples to this. One such is that the transatlantic cables do not match the predicted length given the path they lay upon. Many more are available both within our library and from other sources.

I think what you mean is that you have a strong feeling that the earth is round - one that is against verifiable and reproducible evidence.

Quote from: inquisitive on February 04, 2019, 09:14:00 AM

Quote from: John Davis on February 04, 2019, 08:34:48 AM

Thank you to all whom have provided further evidence that the round earth model does match up with empirical readings.

Fortunately measured distances prove a round earth.

To say measured distances prove a round earth, you must show no other solution exists and that the empirical data matches. I have provided one - the surface of the earth is non-euclidean. You also have not shown it matches empirical data, and there are several counter examples to this. One such is that the transatlantic cables do not match the predicted length given the path they lay upon. Many more are available both within our library and from other sources.

I think what you mean is that you have a strong feeling that the earth is round - one that is against verifiable and reproducible evidence.

In other words, you say that the Earth is “a shape” which is non defined... Much like all of FEH.  Your infinite flat Earth or anything near infinite flat is impossible.  Explain star trails with 2 rotational axi both above the surface of the Earth as observed.  Draw a geometric shape which has the star trails we see with their rotations.    If your “model” doesn’t fit the most basic of observation then it is rubbish and your energies would be better spent supporting models which are possible.

Quote from: inquisitive on February 04, 2019, 09:14:00 AM

Quote from: John Davis on February 04, 2019, 08:34:48 AM

Thank you to all whom have provided further evidence that the round earth model does match up with empirical readings.

Fortunately measured distances prove a round earth.

To say measured distances prove a round earth, you must show no other solution exists and that the empirical data matches. I have provided one - the surface of the earth is non-euclidean (edit: it is elementary to provide an infinite amount of such disproof by examples.) You also have not shown it matches empirical data, and there are several counter examples to this. One such is that the transatlantic cables do not match the predicted length given the path they lay upon. Many more are available both within our library and from other sources.

I think what you mean is that you have a strong feeling that the earth is round - one that is against verifiable and reproducible evidence.

We have proof.  Path of the sun, satellite operation.

Maybe you could explain how dish angles point to a satellite over the equator.

Where is the cable length information?  Details from installers and operators please.

constant:

f:9,80m/s^2 +-0,10 difference caused by distance to dome. So what?

More like +-0.3. But more importantly, just why would the distance to the dome cause any difference and why would it be that shape?

We can measure the f constant and can create a dome distance however we want freely. How re explains it distance to center and creates a distance, we can do it for dome too. No problem.

No. Your explanation is entirely post-hoc. You have no justification for the distance to the dome causing any difference nor for why the dome should vary based upon that.

For RE, the vast majority of the variation comes from Earth being an oblate spheroid, which comes from gravity and Earth's rotation. So no, RE doesn't just force the distance to fit. It predicts it based upon simple gravity and rotation.

It has estimated as an about value here as:
73 miles highness.

That rocket doesn't hit a dome. We have gone over this many times. It uses a yo-yo despin mechanism to stop its rotation after the engine burns out. It then continues moving upwards.

Thus, I'm failing to see why the above map is not a map of where the round earth fails to match empirical measurements given your response.

Then perhaps you should try reading and understanding what people say?
It is showing error between the RE model and empirical measurements.
This in no way shows the RE model isn't correct. All it does is show the RE model is not perfectly accurate without any simplifications nor approximations as no one in their right mind would claim it to be.

So no, it doesn't show that Earth isn't a globe or that that would be an assumption.
It shows Earth is not a perfect oblate spheroid without any variations such as due to mountains or non-uniformity in the density of rock/whatever is below the surface.

I have provided one - the surface of the earth is non-euclidean

Which is to say it is the surface of a sphere or a shape closely matching a sphere.
That is effectively an admission that it is round.
If Earth is flat, it must be Euclidean.

One such is that the transatlantic cables do not match the predicted length given the path they lay upon.

And this claim is based upon what?

I think what you mean is that you have a strong feeling that the earth is round - one that is against verifiable and reproducible evidence.

No, one which is consistent with all verifiable and reproducible evidence, which is yet to be contradicted by any actual evidence.

Quote from: inquisitive on February 04, 2019, 02:27:40 AM

Quote from: wise on February 04, 2019, 02:22:34 AM

Quote from: georgeashetha1 on January 31, 2019, 01:53:15 AM

If gravity, as we know it, doesn't exist, but rather is a constant acceleration upwards at ~9.8m/s^2: how can gravity differ around the earth. It is known and proven that the gravitational acceleration is around 9.82m/s^2 in the north, while it's closer to 9.78m/s^2 around the equator. How can this be explained by the flat earth theory?

constant:

f:9,80m/s^2 +-0,10 difference caused by distance to dome. So what?

We can measure the f constant and can create a dome distance however we want freely. How re explains it distance to center and creates a distance, we can do it for dome too. No problem.

Please provide details of a dome, where are the edges?

We are not talking about edges. It may be finitive or infinitive. The important thing is distance, as altitude to dome. It has estimated as an about value here as:



73 miles highness.

Somewhere in the earth dome is 72 miles, somewhere it as 74 miles. Then formula changes.

9,81 m/s^2 for 73 miles.

If you see a 9,80 m/s^2, so the dome distance is:

72,93 kms.

As you see that, dome explains everything and

f (flat earth constant accelerate)= 9,81 m/s^2 average.

Wise, the guy who posted the video above, posted a follow up video explaining what you see in the first video. His closing comment: "So it didn't hit the dome..."

And as you can see, when they found the rocket it seemed to be in pretty good shape considering it would have slammed into the dome at over 3000 mph.

Watch:

The original question keeps being avoided. It is not really interesting whether the gravitational deviations are predictable or not. The thing that is crucial for believers in the flat earth is to explain how there can be deviations in the first place.

Imagine a big plane being accelerated upwards at 1m/s². Now imagine placing a few objects spread around the base of the plane, on the edges and in the center. The objects will have a constant distance to eachother, all having an acceleration upwards equaling the acceleration of the plane. Even if you bend the plane, they all will still have a constant acceleration. If you instead imagine the plane being a creased and irregular sheet of paper. No matter how much you fold the paper, all points on the paper will have the same acceleration as the paper, 1m/s².

In order for the acceleration to be irregular, the distance between points or objects on the plane would have to increase. This would lead to stretching and tearing of the plane or paper.

The same logic applies to the earth. A constant accelerating force upwards is not possible, since the acceleration is observed to have deviations that can be explained by gravity on a globe, taking the centrifugal force in account. Of couse the gravity is impossible to predict perfectly, since the interior of the earth is not fully mapped.

I do not really understand what you guys mean with distance to the dome. Is "the dome" pushing the earth downwards with a constant force, which is different at different places? That would mean that the gravitational pull would increase as you move closer to the force, right? The only way to explain where the deviations come from, would be that they are faked by the scientist and the instruments observing them. This would further disprove flat earth theory through the reasoning of Occam's razor, since the deviations have been observed long before the Cold War, which if I understand correctly would be the reason "Round Earth" was "imprinted" to the public in the first place.

Finally: Instead of picking on minor mistakes in my text, please answer the major question first:

How can the flat earth have deviations in its upwards acceleration without "breaking"?

Well, you have a common, but incorrect (mis-)understanding of how the acceleration values would be measured. That's the first problem.

Well, you have a common, but incorrect (mis-)understanding of how the acceleration values would be measured. That's the first problem.

so the same weigh scale taken to different locations on earth.
calibrated.
then taken that the measured scale weight and dividing it by mass weight is incorrect.
it's three numbers.
F = MxA.
F = scaled weight
M = mass weight
A = accel.

where's the error/
or does F= MA not work?

M = mass weight

The mass doesn't have a fixed weight.

Quote

M = mass weight

The mass doesn't have a fixed weight.

Coreect

You not realizin/ understanding this is the issue

I'm not the one that just said mass=mass weight 

Because in laymans terms pwiple dropped the term force in weight vs mass.

In all itents and purposes we generalize 9.8m/s^2 as kg or lb.   
Not newtons or lbf.
So mass weight is mass.
Thanks for pointing that pointless tidbit out.

I mean, we are discussing differences of acceleration, and you are clearly equating M with "mass weight".  I think it's pretty relevant to mention that you do not seem to know what you are talking about.

I mean, we are discussing differences of acceleration, and you are clearly equating M with "mass weight".  I think it's pretty relevant to mention that you do not seem to know what you are talking about.

Only if you eant to play semantics

Jackb or curious is free to correct me.

Still doesnt ref georges point or the same point made with jane in the ua discussion.
But keep in distracting.
Serves tour ouropose

I guess we can weight (haha) for someone else to point out your error then, because I cannot make any headway against gibberish.

I already addressed George's point. like ypu, he has not thought out his "problem" thoroughly.

F7wisithwhurjhtvrdufur

Yes
My thumbs do need to learnto type bettrr

Well, you have a common, but incorrect (mis-)understanding of how the acceleration values would be measured. That's the first problem.

There are 2 methods, one easier than the other.
The first is to measure the weight of an object and how it varies around the globe.
The second is to measure acceleration by measuring the motion, with it accurately timed (note: for this purpose, a pendulum would be a type of motion you can measure, not an accurate timing source, as pendula are effected by gravity/acceleration in the UA model which influences their period).

so the same weigh scale taken to different locations on earth.
calibrated.
then taken that the measured scale weight and dividing it by mass weight is incorrect.
it's three numbers.
F = MxA.
F = scaled weight
M = mass weight
A = accel.

I would leave it as mass, not mass weight. I haven't seen the term mass weight used anywhere. The closest I have seen is discussions of inertial mass vs gravitational mass, and laypeople using mass and weight interchangeably as if they are the same thing.
So in this context we have the force, which is its weight, and we have its mass. We find the acceleration by a=F/m.

that's why i said mass weight was layman.
ah well
ok stand corrected.
still doesn't dispute geof's point.

W=w*a isn't "layman". It betrays a gross misunderstanding of the entire subject which you happened to Google to try to seem brilliant. You may, for all I know, in fact be brilliant, but you clearly do not know enough about the topic we are discussing here well enough to have a meaningful discussion.

Quote from: Ski on February 07, 2019, 01:30:16 PM

Well, you have a common, but incorrect (mis-)understanding of how the acceleration values would be measured. That's the first problem.

so the same weigh scale taken to different locations on earth.
calibrated.
then taken that the measured scale weight and dividing it by mass weight is incorrect.
it's three numbers.
F = MxA.
F = scaled weight
M = mass weight
A = accel.

where's the error/
or does F= MA not work?

Should I bring up gravity is used to find oil and minerals?

It comes as no surprise to learn you're fully onboard the "earth will rip itself apart because of variations in measured g"-train.  Birds of a sad feather and what not, I suppose.  Do you measure weight by w=w*a, too?

W=w*a isn't "layman". It betrays a gross misunderstanding of the entire subject which you happened to Google to try to seem brilliant. You may, for all I know, in fact be brilliant, but you clearly do not know enough about the topic we are discussing here well enough to have a meaningful discussion.

Quote from: Themightykabool on February 07, 2019, 02:14:43 PM

Quote from: Ski on February 07, 2019, 01:30:16 PM

Well, you have a common, but incorrect (mis-)understanding of how the acceleration values would be measured. That's the first problem.

so the same weigh scale taken to different locations on earth.
calibrated.
then taken that the measured scale weight and dividing it by mass weight is incorrect.
it's three numbers.
F = MxA.
F = scaled weight
M = mass weight
A = accel.

where's the error/
or does F= MA not work?

F = ma.
I clearly labelled it.
Youre playing semantics.