Question about the "drop" at 6 miles

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Question about the "drop" at 6 miles
« on: December 15, 2008, 03:46:32 PM »
Rowbotham suggests that the "drop" at 6 miles would be over 20ft.  My calculations show that to be way off. 

Suppose that the earth is a sphere of radius 3963 miles. If you are at a point P on the earth's surface and move tangent to the surface a distance of 1 mile then you can form a right angled triangel as in the diagram. Using the theorem of Pythagoras a2 = 39632 + 12 = 15705370 and thus a = 3963.000126 miles. Thus your position is 3963.000126 - 3963 = 0.000126 miles above the surface of the earth. 0.000126 miles = 12*5280*0.000126 = 7.98 inches. Hence the earth's surface curves at approximately 8 inches per mile.

Now, this could explain some of what he is saying about restoring hulls with a telescope.  His distances are all wrong and he concludes that you can see hulls at 6 miles.

What are your thoughts about this?

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Tom Bishop

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Re: Question about the "drop" at 6 miles
« Reply #1 on: December 15, 2008, 04:36:52 PM »
The RE  isn't a horizontal slope which slopes downwards at 8 inches for every mile. That's just the drop in the first mile. The RE is an arc of a circle which continues to bend away from the observer as distance increases.



Suppose that the earth is a sphere with a radius of 3,963 miles. If you are at a point P on the earth's surface and move tangent to the surface a distance of 1 mile then you can form a right angled triangle as in the diagram.

Looking over a distance of 1 mile, we can use the theorem of Pythagoras:

a2 = 3,9632 + 12 = 15,705,370

and when we square root that figure we get a = 3,963.000126 miles

Thus your position is 3,963.000126 - 3,963 = 0.000126 miles above the surface of the earth.

0.000126 miles = 12 in * 5,280 ft * 0.000126 mi = 7.98 inches

Hence after one mile the earth drops approximately 8 inches.

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Looking over a distance of 6 miles, we can use the theorem of Pythagoras:

a2 = 3,9632 + 62 = 15,705,405

and when we square root that figure we get a = 3,963.004542 miles

Thus your position is 3,963.004542 - 3,963 = 0.004542 miles above the surface of the earth.

0.004542 miles = 5,280 ft * 0.004542 mi = 23.98 feet

Hence after six miles the earth drops approximately 24 feet.


« Last Edit: December 15, 2008, 04:51:37 PM by Tom Bishop »

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Perfect Circle

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Re: Question about the "drop" at 6 miles
« Reply #2 on: December 15, 2008, 06:33:08 PM »
Suppose that the earth is a sphere with a radius of 3,963 miles.
That is where you are mistaken, Tom. First of all, following the rules of significance, you cannot have an result more accurate than your starting data, and that radius is only accurate to the nearest mile. Second of all, the radius of the earth has a variation of about 13 miles depending on your location. The results you obtained have a MASSIVE margin of error with that radius value, so you cannot use them in your experiments as conclusive proof of a flat earth.
Like the sun, the stars are also expanding and contracting their diameter as they spin around the hub every six months.

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Johannes

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Re: Question about the "drop" at 6 miles
« Reply #3 on: December 15, 2008, 10:17:23 PM »
13 miles compared to almost 4000 isn't a lot. It is negligible. Funny how RE makes excuses....
« Last Edit: December 15, 2008, 10:27:47 PM by Johannes Kepler »

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Perfect Circle

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Re: Question about the "drop" at 6 miles
« Reply #4 on: December 15, 2008, 10:47:19 PM »
13 miles compared to almost 4000 isn't a lot. It is negligible. Funny how RE makes excuses....
I think that 13 miles compared to 8 inches would not be considered negligible. Stop trying to derail the topic.
Like the sun, the stars are also expanding and contracting their diameter as they spin around the hub every six months.

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Tom Bishop

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Re: Question about the "drop" at 6 miles
« Reply #5 on: December 16, 2008, 02:27:45 AM »
I think that 13 miles compared to 8 inches would not be considered negligible. Stop trying to derail the topic.

It doesn't really matter if there would be a slight difference in the drop. The experimenters in these water convexity experiments are putting their eyesight as close to the water's surface as possible, so that any curvature, if it existed, would be readily seen over a distance of 6 miles.


Re: Question about the "drop" at 6 miles
« Reply #6 on: December 16, 2008, 03:57:01 AM »
I think that 13 miles compared to 8 inches would not be considered negligible. Stop trying to derail the topic.

It doesn't really matter if there would be a slight difference in the drop. The experimenters in these water convexity experiments are putting their eyesight as close to the water's surface as possible, so that any curvature, if it existed, would be readily seen over a distance of 6 miles.


But wait a second:

Shine a laser over a two mile stretch of water. The beam received will appear slightly higher in altitude than the beam sent. This proves that light bends upwards.

So does light bend or not? Bear in mind that if light bends then the surface of water should appear convex.

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Tom Bishop

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Re: Question about the "drop" at 6 miles
« Reply #7 on: December 16, 2008, 04:18:03 AM »
Quote
So does light bend or not?

That would depend on the Flat Earth model we were talking about.
« Last Edit: December 16, 2008, 04:36:19 AM by Tom Bishop »

Re: Question about the "drop" at 6 miles
« Reply #8 on: December 16, 2008, 04:19:30 AM »
From a completely unbiased perspective, I'll give it a shot with error taken into account...

For these equations, I'll assume there was insignificant altitude variance between object and observer height.

Values:
Min Radius: 3948 miles
Max Radius: 3963 miles
For distance between object and observer, I'll use 2.5% which is what I encountered when using a measuring wheel on loose soil.
Min distance: 5.85 miles
Max distance: 6.15 miles
For simplicity I'll use R, r, D, d for max and min radius and distance.

Equations:
sqrt(R2+D2)-R=0.004772
sqrt(R2+d2)-R=0.004318
sqrt(r2+D2)-r=0.004790
sqrt(r2+d2)-r=0.004334

So 0.004554(+/-0.000472)miles
Which is 24.0(+/-2.5)ft without taking into account a change in altitude between observer and object.

Re: Question about the "drop" at 6 miles
« Reply #9 on: December 16, 2008, 08:21:32 AM »
Thank you, my brain wasn't working.  LOL  :o

Re: Question about the "drop" at 6 miles
« Reply #10 on: December 16, 2008, 03:01:53 PM »
Quote
So does light bend or not?

That would depend on the Flat Earth model we were talking about.

So which model does rowbothams experiments support?

Re: Question about the "drop" at 6 miles
« Reply #11 on: December 16, 2008, 03:17:16 PM »
From a completely unbiased perspective, I'll give it a shot with error taken into account...

For these equations, I'll assume there was insignificant altitude variance between object and observer height.

Values:
Min Radius: 3948 miles
Max Radius: 3963 miles
For distance between object and observer, I'll use 2.5% which is what I encountered when using a measuring wheel on loose soil.
Min distance: 5.85 miles
Max distance: 6.15 miles
For simplicity I'll use R, r, D, d for max and min radius and distance.

Equations:
sqrt(R2+D2)-R=0.004772
sqrt(R2+d2)-R=0.004318
sqrt(r2+D2)-r=0.004790
sqrt(r2+d2)-r=0.004334

So 0.004554(+/-0.000472)miles
Which is 24.0(+/-2.5)ft without taking into account a change in altitude between observer and object.

your calculations are still problematic, R and r each carrie an uncertainty of +/- 0,5 miles, d an D +/-0,005 miles. if you do a clean propagation of uncertainty you have to take these uncertainties into account, too.

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Perfect Circle

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Re: Question about the "drop" at 6 miles
« Reply #12 on: December 16, 2008, 03:33:43 PM »
Quote
So does light bend or not?

That would depend on the Flat Earth model we were talking about.
Pick one and stick with it. RE'ers aren't flip-flopping between multiple earth models like you do.
Like the sun, the stars are also expanding and contracting their diameter as they spin around the hub every six months.

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Tom Bishop

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Re: Question about the "drop" at 6 miles
« Reply #13 on: December 16, 2008, 06:21:19 PM »
Pick one and stick with it. RE'ers aren't flip-flopping between multiple earth models like you do.

Yes you do. One minute gravity is a sub-atomic particle, and the next minute it's a distortion of space.

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markjo

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Re: Question about the "drop" at 6 miles
« Reply #14 on: December 16, 2008, 06:29:00 PM »
Pick one and stick with it. RE'ers aren't flip-flopping between multiple earth models like you do.

Yes you do. One minute gravity is a sub-atomic particle, and the next minute it's a distortion of space.

Or it's a subatomic particle that distorts space-time.
Science is what happens when preconception meets verification.
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Besides, perhaps FET is a conspiracy too.
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It is just the way it is, you understanding it doesn't concern me.

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Tom Bishop

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Re: Question about the "drop" at 6 miles
« Reply #15 on: December 16, 2008, 06:30:40 PM »
Or it's a subatomic particle that distorts space-time.

That's not an existing theory. But feel free to throw another one into your mess of models.

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markjo

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Re: Question about the "drop" at 6 miles
« Reply #16 on: December 16, 2008, 06:57:45 PM »
Or it's a subatomic particle that distorts space-time.

That's not an existing theory. But feel free to throw another one into your mess of models.

As opposed to the finely tuned FE models discussed here?
Science is what happens when preconception meets verification.
Quote from: Robosteve
Besides, perhaps FET is a conspiracy too.
Quote from: bullhorn
It is just the way it is, you understanding it doesn't concern me.

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Tom Bishop

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Re: Question about the "drop" at 6 miles
« Reply #17 on: December 16, 2008, 07:09:10 PM »
As opposed to the finely tuned FE models discussed here?

FET is a model in development. Your model has had thousands of years to get its act together, but your greatest minds still labor to discover the basics.

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markjo

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Re: Question about the "drop" at 6 miles
« Reply #18 on: December 16, 2008, 07:29:16 PM »
As opposed to the finely tuned FE models discussed here?

FET is a model in development. Your model has had thousands of years to get its act together, but your greatest minds still labor to discover the basics.

While your model can't even agree if the FE generates a gravitational field (by any means) or not in the first place.
Science is what happens when preconception meets verification.
Quote from: Robosteve
Besides, perhaps FET is a conspiracy too.
Quote from: bullhorn
It is just the way it is, you understanding it doesn't concern me.

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Tom Bishop

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Re: Question about the "drop" at 6 miles
« Reply #19 on: December 16, 2008, 07:55:10 PM »

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Johannes

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Re: Question about the "drop" at 6 miles
« Reply #20 on: December 16, 2008, 08:11:36 PM »
Quote
So does light bend or not?

That would depend on the Flat Earth model we were talking about.
Pick one and stick with it. RE'ers aren't flip-flopping between multiple earth models like you do.
You don't even understand how the RET works. Please don't criticize us.

Re: Question about the "drop" at 6 miles
« Reply #21 on: December 16, 2008, 11:55:35 PM »
Pick one and stick with it. RE'ers aren't flip-flopping between multiple earth models like you do.

Yes you do. One minute gravity is a sub-atomic particle, and the next minute it's a distortion of space.
It can be both at the same time Tom. That is what you don't understand about the RE Model. Just because you don't understand something does not make it wrong.

What occurred is that in RE physics, people approach the problem of the Fundamental Forces form two different sides. One was as discrete packets that affected the motions of other discreet packets. The other was from examining the effects of these forces.

The effects of Gravity is to distort the geometry of Space-Time. But it appears that it comes in discreet packets. Thus both are correct, in that the effects of gravity are to distort space, and that there exist sub-atomic particles.

It is not two models, but the same one described differently. Just as you could describe a work of art by explaining the emotions it invokes, or by describing the physical properties of it (colour, etc). Both are describing the same thing, just in different ways.
Everyday household experimentation.

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Tom Bishop

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Re: Question about the "drop" at 6 miles
« Reply #22 on: December 17, 2008, 12:04:41 AM »
Quote
It can be both at the same time Tom. That is what you don't understand about the RE Model.

Um, no it can't.

None of the gravity models combine QM's Graviton particle model and GR's distorting space model. Those are two separate mechanisms.

Quote
It is not two models, but the same one described differently.

Wrong.

Quote
The effects of Gravity is to distort the geometry of Space-Time. But it appears that it comes in discreet packets.

What the hell are you talking about? No one observed gravity coming in discrete packets.

Educate yourself.
« Last Edit: December 17, 2008, 12:41:49 AM by Tom Bishop »

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The Yellow

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Re: Question about the "drop" at 6 miles
« Reply #23 on: December 17, 2008, 01:04:59 AM »
Ever heard of String Theory?
It states that gravity can be a particle that effects space-time.
Also before paying out on the complexity of RET, have a look at the sheer insanity of FET's varius phyical properties...
Rowbotham was not right in all of his explanations. It doesn't help he wrote in victorian english, either.

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trig

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Re: Question about the "drop" at 6 miles
« Reply #24 on: December 17, 2008, 01:41:34 AM »
Back to the OP:

The six mile distance was either carefully or naively chosen by Rowbotham as the perfect distance for many of his experiments, but what he did not say is:

- It is next to impossible to level a telescope to the precision necessary to accurately measure those 24 feet.You need an experiment designed to get the telescope leveled with that precision, then you can think about how to see the fall of 24 feet.
- It is a good distance to get a lot of refraction due to differing temperatures of air. With a little bit of bad luck and if you do your measurements at one time of the day, not at four or five different times, you will get a very strange result and will be able to confuse the issue for some time.

A good decision would be to use good telescopes and design your experiments using much longer distances, where the effect of temperature differentials cannot be sustained. For example, the many photos taken across the lake towards Toronto: If you really want to see the curvature of large bodies of water, you can go there and see for yourself the city from the beaches of the lake. Or, you can go see the ships on the sea, but take the time to see them until they are much more than 6 miles away, to be sure.

Re: Question about the "drop" at 6 miles
« Reply #25 on: December 17, 2008, 08:20:02 AM »
It's all physics Tom.  In the thousands of years we have gotten our shit together.  You guys are still wondering, if you sneak past the penguin guards, make it to the edge (if you are traveling on the FE model that isn't an infinite plane) and peer far enough over the edge, if you will see the turtle glaring at you from underneath.  We continue to make discoveries everyday to show how the world and it's forces work together.  When you guys get a map and move past step one, then we can talk about who's theory needs work ok.  When you guys can come together and agree how the 32 mile diameter sun knows when to slow down and when to speed up, when to rotate left and when to rotate right, directing it's spot light to different areas of the world, we can talk about who's theory needs work.  When you guys can agree whether or not mysterious bodies called the antimoon, the shadow object, or multiple shadow objects, or the turtle changing directions cause the tides, we can talk about who's theory needs work.  When you guys can explain why the escape velocity of flat earth decreases with altitude but sustained space flight still isn't possible, we can talk about who's theory needs work.  When you guys can give evidence to anything other then an hundred year old book written by someone who isn't a scientist, we can talk about who's theory needs work.  When you can explain the magic behind how a telescope can restore the hull of a ship over a closed body of water but not an open one, we can talk about who's theory needs work.  I could go on and on, but I won't.
The Earth is Round.

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Re: Question about the "drop" at 6 miles
« Reply #26 on: December 23, 2008, 10:08:31 AM »
Quote
So does light bend or not?

That would depend on the Flat Earth model we were talking about.
Well yes and no. Light bends most significantly in FE, but it should probably be noted that light does bend in the RE model around any massive object, before any RE'er suggests that bendy light was invented exclusively for FE.  ;)
If I was asked to imagine a perfect deity, I would never invent one that suffers from a multiple personality disorder. Christians get points for originality there.

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Re: Question about the "drop" at 6 miles
« Reply #27 on: December 23, 2008, 10:20:43 AM »
Quote
So does light bend or not?

That would depend on the Flat Earth model we were talking about.
Well yes and no. Light bends most significantly in FE, but it should probably be noted that light does bend in the RE model around any massive object, before any RE'er suggests that bendy light was invented exclusively for FE.  ;)

Don't forget refraction due to certain atmospheric conditions as well.  Scientists take refraction into account when calculating the distance to the true horizon, because the horizon you are looking at over the ocean may actually be below your line of sight in RE.