1

Flat Earth Debate / Re: Cavendish's Experiment

« on: March 29, 2011, 11:24:15 AM »

20% is a huge number in this context.

The fact that is nonzero is what is important for this discussion, not the exact value.

2

Flat Earth Debate / Re: Industrial flat glass, needs a flat earth

« on: March 23, 2011, 04:23:31 PM »

If it helps anyone think of the bending of glass if needs be, in true "personal experience BS" style:
The Young's modulus of a piece of glass in lab was measured to be (8.06ą0.36)x10^10 Pa using Cornu's method.
Noticeable bending can easily be observed in glass by putting a flat piece of glass on top of it and lighting it from below.

3

Flat Earth Debate / Re: Cavendish's Experiment

« on: March 23, 2011, 04:01:38 PM »

What was the result, GD? I did it but i was 20% under the expected value for G.

I don't recall the exact figure, but I know it was closer than 20%.

Yeah my fist uni experiment was hardly going to be accurate, might reanalyse it and see if it looks any better or if i just screwed it up. I'm guessing it was screwed up because my correction for cross-attraction made the result worse, not better.

4

Flat Earth Debate / Re: Cavendish's Experiment

« on: March 23, 2011, 07:41:09 AM »

What was the result, GD? I did it but i was 20% under the expected value for G.

5

Technology, Science & Alt Science / Re: A QUESTION TO ROUND EARTH BELIEVERS

« on: March 22, 2011, 09:25:54 AM »

Have you personally performed or seen performed the The Cavendish Experiment?

I'm going to be the answer is no but for some reason..... you still think it's true..

Yes. My error margin in finding G was pretty poor, about 20% (measured value 5.43x10^-11 Nm^2kg^-2). It was my first experiment at Manchester, and such is not as good as some of my later experiments. I have my logbook in my room if you have any questions.

6

Flat Earth Q&A / Re: On the Notion of a John Davis Comment

« on: March 22, 2011, 05:59:02 AM »

Not really, the question of the Earth's sphericity has two conceivable answers. It either is or it isn't (it isn't).

You could say the exact same about any scientific experiment with a yes/no answer (and for a continuous measurement, by extension we get an infinite series of possibilities). The point of falsifiability is to remove bias from your measurement. Hence your criticism of the scientific method is moot.

7

Flat Earth Debate / Re: Cavendish's Experiment

« on: March 20, 2011, 02:13:30 PM »

Has anyone who has been bashing this experiment done it themselves?

8

Flat Earth Debate / Re: Sinking Ship in ENaG

« on: March 20, 2011, 01:43:36 PM »

Astronomers have not measured the speed of the earth through the universe. It is not possible to get a speed reading by looking at the slight color differences of the stars. Doppler Shift does not tell one the speed of a body.

If you are going to make a claim that no physicist in the world agrees with, at least give some evidence to the contrary.

Quite the claim with no backing or evidence yourself.  Several physicists disagree that the doppler shift in stars is necessarily indicative of their relative speeds.

Correct, but you are missing the point. I have used the Doppler shifts of 21cm hydrogen lines to measure the velocity of the Andromeda galaxy (otherwise known as M31) relative to the galactic centre. The Doppler shift for M31 does not directly give the velocity relative to earth. You have to correct for Hubble and the relative velocity through expanding space to get the correct answer. My answer was (320ą30) billion solar masses at 30kpc from the galactic centre (the distance is important due to "dark matter"), which is in agreement with other measurements at that sort of distance from the galactic centre. Data was obtained from Jodrell Bank.

In short, at long distances (at least megaparsecs) but not too far (where Hubble swamps the relative velocity), the Doppler shift is NOT indicative of the relative velocity, you must correct for Hubble.

9

Flat Earth General / Re: Heat Below The Surface of Earth?

« on: March 17, 2011, 03:33:40 PM »

Dark energy is probably bollocks and dark matter is explained fairly well using MOND (modified newtonian dynamics, its an emipircal theroy that works quite well for galaxy rotation curves), however there is no real basis to completely reject the concepts of dark matter and energy. Dark matter energy, to me (and i am studying physics at uni) seems to have the following form:
1:unexplained phenomenon
2:make something up to explain it
3:
4.PROFIT!!

For heat generated by radioactive stuff, consider the vast amount of stuff (6x10^24kg) that makes up the earth, if there is even a small density of energy release, there is still a lot of power down there.

10

Flat Earth General / Re: UK Flat Earth Meeting

« on: March 17, 2011, 03:23:54 PM »

This is just a thread of trolls trolling trolls isn't it?

11

Flat Earth Debate / Re: Coriolis Effect Proves Earth's Rotation

« on: March 17, 2011, 02:45:35 PM »

Let us consider a flat earth map with the north pole in the middle (quite common) and assume that the Coriolis effect is as measured. In order for the flat map to move as is measured requires discontinuities in the rotation of earth or changes in the physics of rotational motion. The former is not observed and the latter is not justified.

12

Flat Earth Debate / Re: The Correct Formula for the Displacement Expected by RE's Curvature

« on: March 14, 2011, 07:07:06 PM »

@ClockTower
Was this really worth another thread? Are the 12 pages and counting, of the 'Perfectly flat, flaw-free' glass thread not enough?

And despite all the moaning about numbers, you are still agreeing that there would be some curvature? Which then contradicts the very definition of Perfectly flat, flaw-free glass. So after 12 pages, a second thread and despite me having given you the RE answer twice before in other threads, you are still unable to show how the glass could be flat on a round earth?

I will say this to you one final time.

It is Perfectly Flat Flaw-Free F*****g Glass! Ahhhhhhhh! Ahhhhhhhhhhh! AAAAAAHHHHHH!!!!!!!!!!!!!!  

Define Perfect.
Define Flat.
Define Flaw-Free.

What does this tell you about the glass? Is it curved? Are there any flaws that might allow for it to be curved? What do we know about this glass. Oh yeah, that's right. Its Perfectly flat flaw-free glass.

It isn't marketing hyperbole. It isn't that you can cut into the earth to make the liquid flat or use a level. It isn't because they only make tiny pieces. Its not down to tolerances or any of the other things I have told you a million times. Altogether now ... What is it?

Perfectly Flat Flaw-Free Glass!

well, becuase it is made on a liquid, and liquids have surface tension, it curves towards the walls (like the meniscus in the graduated cylinder) and you can make it couteract the gravitational pull that causes the  curve in that way

and hence, the curved liquid has now become perfectl flat, on a round earth

Orders of magnitude, Vhu.

?

Since you aren't going to research this yourself, consider the model that you have of a flat earth, this would generate a flat ocean. If we consider the vast weight of evidence to the contrary ie sinking ships, we see that gravity has an effect on the earth's shape. A spherical earth implies a spherical (locally) ocean. Reading really is fundamental.
This thread is miles off topic. Well done FES for helping to derail a maths thread.
To answer the "?" specifically, all your argument suggests is that the earth appears flat (due to its size).
lrn2calculus.

13

Flat Earth Debate / Re: The Correct Formula for the Displacement Expected by RE's Curvature

« on: March 03, 2011, 12:30:00 PM »

@ClockTower
Was this really worth another thread? Are the 12 pages and counting, of the 'Perfectly flat, flaw-free' glass thread not enough?

And despite all the moaning about numbers, you are still agreeing that there would be some curvature? Which then contradicts the very definition of Perfectly flat, flaw-free glass. So after 12 pages, a second thread and despite me having given you the RE answer twice before in other threads, you are still unable to show how the glass could be flat on a round earth?

I will say this to you one final time.

It is Perfectly Flat Flaw-Free F*****g Glass! Ahhhhhhhh! Ahhhhhhhhhhh! AAAAAAHHHHHH!!!!!!!!!!!!!!   

Define Perfect.
Define Flat.
Define Flaw-Free.

What does this tell you about the glass? Is it curved? Are there any flaws that might allow for it to be curved? What do we know about this glass. Oh yeah, that's right. Its Perfectly flat flaw-free glass.

It isn't marketing hyperbole. It isn't that you can cut into the earth to make the liquid flat or use a level. It isn't because they only make tiny pieces. Its not down to tolerances or any of the other things I have told you a million times. Altogether now ... What is it?

Perfectly Flat Flaw-Free Glass!

well, becuase it is made on a liquid, and liquids have surface tension, it curves towards the walls (like the meniscus in the graduated cylinder) and you can make it couteract the gravitational pull that causes the  curve in that way

and hence, the curved liquid has now become perfectl flat, on a round earth

Orders of magnitude, Vhu.

14

Flat Earth Debate / Re: The Correct Formula for the Displacement Expected by RE's Curvature

« on: January 26, 2011, 06:46:36 PM »

The solution to the puzzle is:
d=(r+h)(1-cos(s/2r))
which, if we are talking about miles rather than intercontinental distances, would approx as
d=(r+h)(sē/4rē)
d= declination on central pole
s= distance from first post to last post around the earth
h= height of poles (all have the same height in this formula)
r= earth radius.
The cosine is in radians, not degrees. If you want to use degrees use:
d=(r+h)(1-cos((90*s)/(pi*r))
The equation was derived from:
Consider a right angled triangle with hypotenuse = r+h, and the angle opposite to the right angle being theta/2. By simple trig/vectors, we can find the adjacent side.
The projection of this adjacent side will give the equation:
r+h-d=(r+h)cos(theta/2)
from which finding d is no problem.
The approximate equation uses the approximation:
cos x = 1 - xē
The reason we use radians is the simple relation s=r*theta. Using degrees gives us unnecessary constants. From this information, the equation should be self evident.

15

Flat Earth Debate / Re: The Correct Formula for the Displacement Expected by RE's Curvature

« on: January 25, 2011, 07:37:11 PM »

There is a more elegant solution to the maths puzzle, any takers? Will post solution tomorrow.

16

Philosophy, Religion & Society / Re: Money - The Great Challenge Of Our Time

« on: January 23, 2011, 04:57:26 PM »

As if removing the device that allows a modern economy to operate would allow that same economy to operate more efficiently. 

Did anyone advocate that?

What else is the argument of the alleged evil of modern monetary systems?  The abuse of the system at the cost of depositors and/or taxpayers, that's bad...but our very way of life is made possible by the system that people complain about. 

Without a fractional reserve banking system, how would the non-wealthy acquire the capital necessary to create a business?  Without central banking, entire economies would be dominated by private interests.

Fractional reserve banking and central banking are completely different. Under a full-reserve banking system with responsible expansion of the money supply, governments can provide services and tax cuts that will help people get onto the ladder and will help them out of debt.

17

Technology, Science & Alt Science / Re: Clarification on Gravity

« on: January 23, 2011, 04:50:04 PM »

gravity travels at the speed of light

Do you have any evidence for this outlandish claim?

http://adsabs.harvard.edu/abs/2006FoPh...36.1244K

Abstracts are not accepted on this site as evidence.

Any good book on GR will derive it.

18

Flat Earth Debate / Re: Noether's Theorem

« on: December 28, 2010, 06:01:16 PM »

Momentum isn't conserved along the vertical direction.

Because there is a force acting. I mentioned the need to compensate for it.

19

Flat Earth Q&A / Re: Davis Model, Gauss's law and gravitation.

« on: December 28, 2010, 05:42:18 PM »

I used the equation I wrote before, assuming that the density is equal to the RE value for a mean earth as 5.52kgm^-3 and the mass distribution is uniform, and the distance works out to be 2.12x10^9m, ie 2 million km. That's pretty big, however they are probably still big assumptions to make and would need to be accounted for.

20

Flat Earth Q&A / Re: Ok, admitted skeptics with questions about your theory.

« on: December 28, 2010, 05:08:40 PM »

I was under the impression the three laws were incorrect anyway.

I'm pretty sure they are right.
1st law: things have a constant velocity without forces. This is pretty much stating the obvious.
2nd law: definition of force as the rate of change of momentum. Remember that in relativity, we much correct the momentum, so we get different equations. The second law is more a definition than a physical law.
We can verify the first law with the second by arguing that if the rate of change of momentum wrt time is zero, the velocity must be constant.
3rd law: This is a statement of the principle of conservation of momentum

21

Flat Earth Q&A / Re: Davis Model, Gauss's law and gravitation.

« on: December 28, 2010, 04:50:56 PM »

I think you should try using Gauss's law again.

Well spotted. well I'm a silly bugger, the field is a constant 4*pi*G*[area density (units of mass per area)]. Regardless of that correction, the argument that an addition of layers of mass to approximate a flat, infinite earth would result in an infinite field, not because of the infinite earth but because of the infinite number of layers (each having a finite contribution) would all have to be accounted for.

22

Philosophy, Religion & Society / Re: Money - The Great Challenge Of Our Time

« on: December 28, 2010, 03:09:47 PM »

Money corrupts, absolute money corrupts absolutely.

Spot on there. The money system is incredibly corrupt.
"Of all the many ways of organising banking, the worst is the one we have today." - Mervyn King, Governor of the Bank of England

23

Flat Earth Debate / Re: Distances on RE and FE consistent thanks to bendy light.

« on: December 27, 2010, 07:30:48 PM »

http://www.astronomyknowhow.com/setting-circles.htm

Ninian shows you. He knows what hes talking about.

I do not have Windows Media Player, so I cannot launch it to watch the video.

Troll "Bendy light specialist" is troll.

24

Flat Earth Debate / Re: The Correct Formula for the Displacement Expected by RE's Curvature

« on: December 27, 2010, 07:21:48 PM »

the brute force solution is:
d = h + r(1-cos (s/2r) ą 0.5sqrt((-2h-2r+2rcos(s/2r))ē - 4[hēcosē(s/2r) + 2h + rē - 2rhcos(s/2r) - 2rēcos(s/2r) + rēcosē(s/2r)])
Can anyone see any useful simplification here?

25

Flat Earth Debate / Re: The Correct Formula for the Displacement Expected by RE's Curvature

« on: December 27, 2010, 05:45:57 PM »

Maths challenge (I'm going to do it too):
hēsinē(s/2r) = hē - 2hd + dē + 2hr - 2rd +rē -2rhcos(s/2r) + 2drcos(s/2r) - 2rēcos(s/2r) + rēcosē(s/2r)
Solve for d.
In this equation, I am modeling the declination by using three collinear flag poles, at a height h above still water. A laser is used to point from the top of the first pole to the top of the last. d is the distance between the top of the middle pole and the point where the laser light points. r is the radius of the earth, s is the co-ordinate distance. You can get this by drawing it how I explained it hopefully, as well everyone makes mistakes, especially during a lot of derivation lol. Again, sin and cos need radians here, as i cannot be bothered with stupid factors of 180/pi tbh.
The equation can be brute-forced by completing the square/quadratic formula, however elegance is always good in proofs. Since there appears to be two roots due to the quadratic nature of the problem, any solutions below zero or above h are unphysical. If anyone finds two roots between zero and h, tell me because I have made a mistake.

26

Flat Earth Q&A / Davis Model, Gauss's law and gravitation.

« on: December 27, 2010, 03:43:40 PM »

From the FAQ, there is reference to a Davis Model for a flat earth. The FAQ states that this model has an infinite earth, yet a finite gravitational pull. If we use Gauss's law, using a cylindrical symmetry on an infinite plane with infinitesimal thickness, we get a finite field proportional to (1/altitude). If we then extend our model by integrating through the thickness of the model (the earth is NOT of infinitesimal thickness, I think we can all agree on that), we find that the gravitational field would be infinite, not finite.
Do you have any comment on this, John? I have a feeling that the FAQ needs an update and we need to collate material on theories so they can be tested properly.

27

Flat Earth Debate / Re: Next Assaults on FE

« on: December 27, 2010, 03:09:33 PM »

There is only one pole point on the FE surface drawn. Two are observed in reality.

Vertical field lines in the arctic and antarctic circles are observed in reality.

If your diagram is true, we would observe that the magnetic field strength is much stronger at the north pole compared with the south. This is not observed. You are correct that a compass would not work for either pole. I should have been more specific. A flat surface with one dipole cannot have two poles of equal field density on the same surface.

28

Flat Earth Q&A / Re: How do the FEers know that other planets are round?

« on: December 27, 2010, 04:03:30 AM »

So if someone asks:

"Hey Billy! Doing fine? How's your mom?"

and Billy answers:

"She's just died"


Do you infer that Billy is fine because it is in the question?

Of course not. Billy's answer doesn't confirm whether or not he is fine. It doesn't even answer how his mom is, but you can infer that his mother isn't fine because she is dead.

The same thing with the FAQ question. The answer doesn't answer the question directly, it states that the other planets are different from Earth, from which you can infer that even if the other planets are round or not, it doesn't make it so Earth HAS to be round.

Missing the point here. The answer in the FAQ question has no indication that the answer is contrary to the question, so the inference is valid.

29

Flat Earth Debate / Re: Sunset model

« on: December 26, 2010, 02:18:21 PM »

Big problem. Refractive index reduces with increasing altitude (as you mentioned this is in the RE model). We can infer that by thinking about optical density. Optical density of a some stuff (ie air) is related to concentration/pressure of that stuff. Therefore optical density is positively correlated (may be directly proportional is I's expect but do the experiment). We can infer that pressure decreases by observing the change in the boiling point of water with altitude. On Mt Everest, water boils around 70 deg C, so the pressure is obviously lower.

30

Flat Earth Debate / Re: The Correct Formula for the Displacement Expected by RE's Curvature

« on: December 26, 2010, 01:32:06 PM »

I say it's given by the following:
http://img825.imageshack.us/img825/6573/16863268.jpg

I disagree here. I used vectors instead, using the same diagram.
I make it:
d=r(1-cos(s/r))
where d is declination
r is radius
s is arc distance.
RADIANS FOR THE COSINE. n00bs use degrees.
For small (s<<r) distances, we can taylor-expand the cosine and we get
d=sē/r

From this formula, we can easily see that Rowbotham cocked his table up in his book.

For us less mathematically incline may you include some examples - liker perhaps how Rowboatem cocked up his table?

Berny
Ugh Math

table (using miles for distance, declination in feet, using earth radius of 4000 miles)
distance       declination
1                 0.73
2                 2.90
3                 6.53
4                 11.60
5                 18.13
6                 26.01
7                 35.53
8                 46.40
9                 58.73
10               72.50


http://www.sacred-texts.com/earth/za/za05.htm
{ //bollocks, he's about right there
Rowbotham is out by about a factor of 2 on the whole table.
Another incorrect treatment is his experiments is that he assumes that his eye is at water level, which it is clearly not.
}
EDIT: Oops... Major error. Earth radius is about 4000, not 8000 miles. Schoolboy error on my part. It seems Rowbotham is right with the curvature, however he is still using the wrong equations for his experiments, due to the longer horizon resulting from more height. Will find the formula ASAP for the general case, in which the observer is at nonzero altitude. This is an important formula. table fixed