FE Experiments

been there, done that, etc.

been there, done that, etc.

Then here, do this problem.  I'm stuck. 
If sin(−5x + 4y + z) = 0, find the first partial derivatives [(∂z)/(∂x)] and [(∂z)/(∂y)] at the point (0, 0, 0).

A. [(∂z)/(∂x)]|(0, 0, 0) =?

B. [(∂z)/(∂y)]|(0, 0, 0) =?

it's been 2 years, man. And my books are in the attic.

wait, hang on. Isn't sin 0 just 0?

wait, hang on. Isn't sin 0 just 0?

Yes but thats not the answer.

when you figure it out, post it here, because I'm interested.

when you figure it out, post it here, because I'm interested.

In about an hour I will be putting it into Mathematica so you have an hour.

This time, get a ruler, or another completely flat and straight object and then line it up with the the horizon line on the last URL I posted. You will see that the far right side curves off your ruler or straight object.

Sorry, but this is good evidence that the earth is indeed round.

That would happen on a Flat earth, too. 

Quote from: MessiahOfFire on February 18, 2008, 07:29:33 PM

This time, get a ruler, or another completely flat and straight object and then line it up with the the horizon line on the last URL I posted. You will see that the far right side curves off your ruler or straight object.

Sorry, but this is good evidence that the earth is indeed round.

That would happen on a Flat earth, too. 

Correct, via mild fish-eye lens effect.

Quote

ok, just what experiments have YOU, Tom Bishop, PERSONALLY done? let's see some of YOUR data, and see if ANYONE else interprets it the same way!

I live along the California Monterey Bay. It is a relatively long bay that sits next to the Pacific Ocean. The exact distance between the extremes of the Monterey Bay, Lovers Point in Pacific Grove and Lighthouse State Beach in Santa Cruz, is 33.4 statute miles. See this map.

On a very clear and chilly day it is possible to see Lighthouse Beach from Lovers Point and vice versa. With a good telescope, laying down on the stomach at the edge of the shore on the Lovers Point beach 20 inches above the sea level it is possible to see people at the waters edge on the adjacent beach 33 miles away near the lighthouse. The entire beach is visible down to the water splashing upon the shore. Upon looking into the telescope I can see children running in and out of the water, splashing and playing. I can see people sun bathing at the shore and teenagers merrily throwing Frisbees to one another. I can see runners jogging along the water's edge with their dogs. From my vantage point the entire beach is visible.

IF the earth is a globe, and is 24,900 English statute miles in circumference, the surface of all standing water must have a certain degree of convexity--every part must be an arc of a circle. From the summit of any such arc there will exist a curvature or declination of 8 inches in the first statute mile. In the second mile the fall will be 32 inches; in the third mile, 72 inches, or 6 feet, as shown in this chart. Ergo; looking at the opposite beach 30 miles away there should be a bulge of water over 600 feet tall blocking my view. There isn't.

Here's the math:



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.

-

Ergo, looking across 30 miles the Pythagorean theorem becomes:

a² = 3963² +30² = 15,706,269

and when we square root that figure we get a = 3,963.113549 miles

Thus your position is 3,963.113549 - 3,963 = 0.113549 miles above the surface of the earth

0.113549 miles = 5,280 ft * 0.113549 mi = 599.53872 feet

Hence after 30 miles the earth drops approximately 600 feet.

-

Whenever I have doubts about the shape of the earth I simply walk outside my home, down to the beach, and perform this simple test. The same result comes up over and over throughout the year under a plethora of different atmospheric conditions.

There are a number of different methods to calculate the drop of the Round Earth. Go ahead and look some up try a few out. You will find that the drop while looking over 30 miles is on the order of 600 feet.

Very thought provoking post, Tom.  But, I think there might be an explanation for your observations in the RE model.  You see, the earth is not a perfect sphere but rather an oblate spheroid.  This means that a cross section of the earth is an ellipse through a north-south plane.  Your math uses a perfectly circular earth.  However, your observations were along a north-south line, so we must take the elliptical curve of the earth into account. 

The radius of the earth increases as one moves further south.  The radius of the earth at the poles is b=3949.901 mi and a=3963.189 at the equator.

The formula for finding the radius of the earth at any latitude is



I looked up the latitude of the two locations you mentioned, and they are 36 deg 38 min and 36 deg and 57 min. I plugged in the numbers.  The radii of the earth at the two locations are 3958.4860 miles and 3958.4157 miles.  So the difference in the radius of the earth is (3958.4860-3958.4157)=.0703 mi=371.1 feet.  As you can see this is quite significant compared to the drop you calculated with a perfect sphere.  This should really be taken into account.  Also, I looked up the straight line distance between your two locations and it is more like 22 miles not 33.  These two factors may very well account for your observations in the RE model.  Get back to me; I'll try to use these numbers and some more math to see how far you could possible see in the RE model.

A few questions:  What kind of telescope did you use?  With my telescope, I have difficulty resolving people at a few miles much less 33 miles.  Do you have any pictures?  Are you sure you were precisely 20 inches above water level?  It seems that the surf would get in the way at that height.  Any more details would be greatly appreciated.

Very thought provoking post, Tom.  But, I think there might be an explanation for your observations in the RE model.  You see, the earth is not a perfect sphere but rather an oblate spheroid.  This means that a cross section of the earth is an ellipse through a north-south plane.  Your math uses a perfectly circular earth.  However, your observations were along a north-south line, so we must take the elliptical curve of the earth into account. 

The radius of the earth increases as one moves further south.  The radius of the earth at the poles is b=3949.901 mi and a=3963.189 at the equator.

The formula for finding the radius of the earth at any latitude is



I looked up the latitude of the two locations you mentioned, and they are 36 deg 38 min and 36 deg and 57 min. I plugged in the numbers.  The radii of the earth at the two locations are 3958.4860 miles and 3958.4157 miles.  So the difference in the radius of the earth is (3958.4860-3958.4157)=.0703 mi=371.1 feet.  As you can see this is quite significant compared to the drop you calculated with a perfect sphere.  This should really be taken into account.  Also, I looked up the straight line distance between your two locations and it is more like 22 miles not 33.  These two factors may very well account for your observations in the RE model.  Get back to me; I'll try to use these numbers and some more math to see how far you could possible see in the RE model.

A few questions:  What kind of telescope did you use?  With my telescope, I have difficulty resolving people at a few miles much less 33 miles.  Do you have any pictures?  Are you sure you were precisely 20 inches above water level?  It seems that the surf would get in the way at that height.  Any more details would be greatly appreciated.

What is word?  Owned? 

Why should we be willing to perform experiments on demand when you're not willing to build a rocket ship and confirm NASA's observation of a Globe Earth?

Firstly, Tom, you are a bot and wouldn't believe us if we did. Secondly, I find it rather amusing that Rowbotham used the principle that standing water is flat to prove the Earth is flat.
Thirdly, as I will be posting soon, it is possible for flat water to exist on a round earth, explaining how the old bedford, as well as all other examples of "flatness" could arise on a globe.

(I am sure that an explanation exists, I have heard someone mention "Local Flatness" or something.

Quote from: Tom Bishop on February 16, 2008, 05:10:05 PM

Why should we be willing to perform experiments on demand when you're not willing to build a rocket ship and confirm NASA's observation of a Globe Earth?

Firstly, Tom, you are a bot and wouldn't believe us if we did. Secondly, I find it rather amusing that Rowbotham used the principle that standing water is flat to prove the Earth is flat.
Thirdly, as I will be posting soon, it is possible for flat water to exist on a round earth, explaining how the old bedford, as well as all other examples of "flatness" could arise on a globe.

(I am sure that an explanation exists, I have heard someone mention "Local Flatness" or something.

Local flatness refers to the way a sphere is a manifold, where at any point it can appear flat locally, to a person, but is actually spherical. It is the same principle with a circle, magnified enough a any point on a circle begins to look like a straight line.

A better explaination for the Bedford "flatness", if it is actually flat, is that it would be a man made structure. Locks, straights, canals and such are dug out of the Earth by humans. There is no reason for a company or building team to try and reproduce the curvature of the Earth when digging these things, so a pretty flat surface could be carved out of the globe of the Earth.

Not wanting to give them any ammo but how exactly does one carve water in to a flat surface as opposed to one that follows the curvature of the Earth?

Not wanting to give them any ammo but how exactly does one carve water in to a flat surface as opposed to one that follows the curvature of the Earth?

I am not sure what you mean, water tends to take whatever shape its container is. If part of the Earth was flattened, and water was to flow into that area, shouldn't it level out depending on the surface under it. Keep in mind the Bedford canal is not very large in itself, so gravitationally speaking the water might tend to do one thing, on such a small scale I don't see any problem.

That's true but how would they dig the canal in such a way that it didn't follow the curvature of the Earth. I think the experiments are simply false rather than the canals being built flat.

The canal would be built flat because humans would build it flat. That way the depth would always be know, and if a barge didn't ground at one end it won't ground in the middle or at the other. See my topic in Debate and Discussion, when I put it there.

That's true but how would they dig the canal in such a way that it didn't follow the curvature of the Earth. I think the experiments are simply false rather than the canals being built flat.

The canal would be built flat because humans would build it flat. That way the depth would always be know, and if a barge didn't ground at one end it won't ground in the middle or at the other. See my topic in Debate and Discussion, when I put it there.

I think the north-south elliptical curve of the earth (which is quite significant as I have shown), local irregularities in the curve of the earth due to gravitational variations, a bit of refraction/mirages, and the poor quality of telescopes of the day might adequately explain the Rowbotham experiments within the RE model.  But, then again, I have never personally conducted the experiments, seen the exact circumstances in which they were performed, or seen photographs.

The Bedford level experiment was not a controlled experiment.

The canal would be built flat because humans would build it flat. That way the depth would always be know, and if a barge didn't ground at one end it won't ground in the middle or at the other. See my topic in Debate and Discussion, when I put it there.

So they rise out of the ground the closer you are to each end or do they simply follow the curvature of the Earth.

Quote from: Conspiracy Agent on February 20, 2008, 11:25:32 AM

The canal would be built flat because humans would build it flat. That way the depth would always be know, and if a barge didn't ground at one end it won't ground in the middle or at the other. See my topic in Debate and Discussion, when I put it there.

So they rise out of the ground the closer you are to each end or do they simply follow the curvature of the Earth.

What?

Well if you built them flat then eventually they would start to rise out of the ground due to the Earth being curved.

Aren't canals built flat but with locks seperating the different levels?

Very thought provoking post, Tom.  But, I think there might be an explanation for your observations in the RE model.  You see, the earth is not a perfect sphere but rather an oblate spheroid.  This means that a cross section of the earth is an ellipse through a north-south plane.  Your math uses a perfectly circular earth.  However, your observations were along a north-south line, so we must take the elliptical curve of the earth into account.

The radius of the earth increases as one moves further south.  The radius of the earth at the poles is b=3949.901 mi and a=3963.189 at the equator.

The formula for finding the radius of the earth at any latitude is



I looked up the latitude of the two locations you mentioned, and they are 36 deg 38 min and 36 deg and 57 min. I plugged in the numbers.  The radii of the earth at the two locations are 3958.4860 miles and 3958.4157 miles.  So the difference in the radius of the earth is (3958.4860-3958.4157)=.0703 mi=371.1 feet.  As you can see this is quite significant compared to the drop you calculated with a perfect sphere.  This should really be taken into account.  Also, I looked up the straight line distance between your two locations and it is more like 22 miles not 33.  These two factors may very well account for your observations in the RE model.  Get back to me; I'll try to use these numbers and some more math to see how far you could possible see in the RE model.

A few questions:  What kind of telescope did you use?  With my telescope, I have difficulty resolving people at a few miles much less 33 miles.  Do you have any pictures?  Are you sure you were precisely 20 inches above water level?  It seems that the surf would get in the way at that height.  Any more details would be greatly appreciated.

I think the north-south elliptical curve of the earth (which is quite significant as I have shown), local irregularities in the curve of the earth due to gravitational variations, a bit of refraction/mirages, and the poor quality of telescopes of the day might adequately explain the Rowbotham experiments within the RE model.  But, then again, I have never personally conducted the experiments, seen the exact circumstances in which they were performed, or seen photographs.

Tom, would like to share your thoughts here?  I'm pretty much claiming that your experiments prove nothing about the shape of the earth.

Tom, would like to share your thoughts here?  I'm pretty much claiming that your experiments prove nothing about the shape of the earth.

No matter how you distort the math, there should still be visible curvature when looking across the Monterey Bay. Even with your revisions there should still be a wall of water 300 feet high blocking the opposite beach.

There isn't.

No matter how you distort the math, there should still be visible curvature when looking across the Monterey Bay. Even with your revisions there should still be a wall of water 300 feet high blocking the opposite beach.

No, if the distance is only 22 miles as I have looked up, the drop would only be 51.85ft. 

3958.4860 - 3958.4152/cos[(36 deg 57 min) - (36 deg 38 min)] = .00982 mi = 51.85 ft

This drop might be less noticeable, or a little variation of the curve the earth may eliminate it completely.  The main point to get from this is that the curve of the earth is far from uniformly circular over these distance scales and is prone to variation. 

Also, would you please share the details of the experiment I asked for?

No, if the distance is only 22 miles as I have looked up, the drop would only be 51.85ft.

I didn't see a 50 foot tall wall of water either. 

No matter how you distort correct the math...

Fixed

I didn't see a 50 foot tall wall of water either.

50ft is well within the margin of error for something as large as the earth, Tom.  But, we could clarify this a lot if you would give me the details I asked for.

Also, wouldn't this elliptical curve of the earth put the other FE experiments in doubt?  They rely on deviations from the curve of the  earth on the order of 10-100ft from a perfect sphere.  But, I have shown that the deviations of the elliptical curve of the earth from a perfect circle is on the same order as this.

Will you please answer my questions Tom?  Keep in mind, I am only trying to determine if your experiments really are conclusive evidence for a flat earth, not prove you wrong or vindicate RE theory.  So far, I am unconvinced.  Provide the details I asked for, and we can see if your observations are possible in the RE model or really do show a flat earth.  Everyone will benefit.  So again:

A few questions:  What kind of telescope did you use?  With my telescope, I have difficulty resolving people at a few miles much less 33 miles.  Do you have any pictures?  Are you sure you were precisely 20 inches above water level?  It seems that the surf would get in the way at that height.  Any more details would be greatly appreciated.