The Correct Formula for the Displacement Expected by RE's Curvature

In fact from the same thread just a few posts earlier EG writes

Quote from: EnglshGentleman on October 15, 2010, 09:16:54 PM

Quote from: Thork on September 17, 2010, 05:14:03 PM

*I'm not gonna post the whole thing here since it is huge, but you can easily follow the link.*
-EG

Despite your objections, you were incapable of proving Thork's maths wrong. In fact, zork even agreed his maths are correct, just to start with his claim liquids won't follow the Earth's curvature.

There is nothing in that link to support your claim. Your failure to understand the math doesn't make Thork's maths right. Indeed, I believe that even Thork understand his error now, after I explained it in 'baby steps' for him. Are there smaller steps I can take for you? What do you think the displacement formula is for the RE?

Quote

2. ClockTower's equations were shown to be wrong on that same page. Do note, just because some posts something against FE, does not mean that it is instantly correct.

Yep, your equations. Not my equations. What you subsequently do with my work, is up to you. But after you butcher it to make it fit your argument, do not then complain about the methodology you had as a starting point.

Your just pissy because Clocktower's math doesn't suck. Also becoming a grammar Nazi on these forums indicates Tunnel Vision, please see a doctor. My point that you should leave stands. OR you could rationalize your math :snicker:.

Quote from: Thork on October 16, 2010, 06:59:51 PM

Quote

I believe that no one has made a significant criticism of the formula, after discount Parsifal's stupid comments.

Well I don't want to criticise the formula. I wrote it. Its perfect. Surely criticising it, is your job?

So you agree with the formula then? EG will be so disappointed.

I never disagreed with Thorks calculations, only yours.

Your just pissy because Clocktower's math doesn't suck. Also becoming a grammar Nazi on these forums indicates Tunnel Vision, please see a doctor. My point that you should leave stands. OR you could rationalize your math :snicker:.

What are you even doing here? All you are doing is being a cheerleader for ClockTower. How about instead of just saying, "I'm right, your wrong, so ha!" you actually prove it.

You don't understand what Emission Spectroscopy is, and it had to be proven to you that liquids will follow the curvature of the Earth regardless of their container. These are extremely basic concepts. How old are you? Like 14 years old in 9th grade? You don't seem to understand anything at all.

Quote from: zork on October 16, 2010, 07:48:58 AM

Quote from: Parsifal on October 16, 2010, 07:40:45 AM

Mountains, valleys, trees, rivers and so forth exist. Therefore, the Earth cannot be modelled as having a perfectly circular cross-section on arbitrarily small scales, and certainly not on the scale that you are proposing to model it as such.

I guess you have never seen the sea or lake.

So you admit that in RET there is curvature to sitting bodies of water? Good to know.

No, I am amazed that you claim that there are no 6m area on the earth which we can assume to be flat and the earth is only filled with mountains, valleys, trees, rivers and so forth by every 6 meter or less distance . Sure, the sea is not so good example because it is rarely calm(in bays still can happen sometimes). In current context the earth can quite well modeled as perfectly circular.

I believe that no one has made a significant criticism of the formula, after discount Parsifal's stupid comments.

I didn't criticise your formula, since it is only a statement of Pythagoras's Theorem anyway. I criticised your derivation and your application of it, and you failing to provide a valid refutation does not make my comments stupid.

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 guess that we are even because you are also unable to show why the glass should have curvature on the round earth. Relying blindly on marketing sentence is pathetic.

I will say this to you one final time.
It is Perfectly Flat Flaw-Free F*****g Glass! Ahhhhhhhh! Ahhhhhhhhhhh! AAAAAAHHHHHH!!!!!!!!!!!!!!   

It is a marketing slogan and you just have failed to show that it is really so perfect and flat. Where is your evidence other that this marketing slogan?

Quote from: ClockTower on October 16, 2010, 07:02:05 PM

Quote from: Thork on October 16, 2010, 06:59:51 PM

Quote

I believe that no one has made a significant criticism of the formula, after discount Parsifal's stupid comments.

Well I don't want to criticise the formula. I wrote it. Its perfect. Surely criticising it, is your job?

So you agree with the formula then? EG will be so disappointed.

I never disagreed with Thorks calculations, only yours.

Where did you disagree with my calculations? And do tell us how mine are wrong. Even Parsifal agrees with the formula.

Quote from: ClockTower on October 16, 2010, 07:02:05 PM

Quote from: Thork on October 16, 2010, 06:59:51 PM

Quote

I believe that no one has made a significant criticism of the formula, after discount Parsifal's stupid comments.

Well I don't want to criticise the formula. I wrote it. Its perfect. Surely criticising it, is your job?

So you agree with the formula then? EG will be so disappointed.

I never disagreed with Thorks calculations, only yours.

Quote from: Danukenator123 on October 16, 2010, 07:19:21 PM

Your just pissy because Clocktower's math doesn't suck. Also becoming a grammar Nazi on these forums indicates Tunnel Vision, please see a doctor. My point that you should leave stands. OR you could rationalize your math :snicker:.

What are you even doing here? All you are doing is being a cheerleader for ClockTower. How about instead of just saying, "I'm right, your wrong, so ha!" you actually prove it.

You don't understand what Emission Spectroscopy is, and it had to be proven to you that liquids will follow the curvature of the Earth regardless of their container. These are extremely basic concepts. How old are you? Like 14 years old in 9th grade? You don't seem to understand anything at all.

I don't understand it? You think you can prove life on the cloud tops of Venus. Also all you and Thork have done on this forum so far is pat each other on the back. I'm just going by the example set by the FE'ers.

Quote from: Thork on October 16, 2010, 04:15:05 PM

I will say this to you one final time.
It is Perfectly Flat Flaw-Free F*****g Glass! Ahhhhhhhh! Ahhhhhhhhhhh! AAAAAAHHHHHH!!!!!!!!!!!!!!   

It is a marketing slogan and you just have failed to show that it is really so perfect and flat. Where is your evidence other that this marketing slogan?

Where is your evidence that it is a marketing slogan? Unless you have any, I have no reason to think it is. You are going the wrong way. You are starting with the assumption that it is a marketing slogan and than demanding we prove otherwise. You forget that it's evidence then conclusion. For shame!

I don't understand it? You think you can prove life on the cloud tops of Venus.

When did I ever say this?  Once again you are putting words in my mouth which is both outrageous and intellectually dishonest. Typical RE'er.

I never said I could prove that life exists on the cloud tops of Venus. I said that it has been suggested that life could survive there since the clouds' composition is similar to that of sulfur springs here on Earth which bear life. When asked how they can determine what the cloud tops are similar to the sulfur springs I stated they can use emission spectroscopy to find this out.

You lose. I suggest you stop trying to pick up the scraps of your failure.

Where is your evidence that it is a marketing slogan?

Here's a hint for you: It contains the word, "perfectly". Duh.

Quote from: EnglshGentleman on October 17, 2010, 10:19:50 AM

Where is your evidence that it is a marketing slogan?

Here's a hint for you: It contains the word, "perfectly". Duh.

Sounds like a bunch of conjecture to me. Where is your proof?

Quote from: ClockTower on October 17, 2010, 10:21:32 AM

Quote from: EnglshGentleman on October 17, 2010, 10:19:50 AM

Where is your evidence that it is a marketing slogan?

Here's a hint for you: It contains the word, "perfectly". Duh.

Sounds like a bunch of conjecture to me. Where is your proof?

Please ask for your proof in the proper thread.

Quote from: EnglshGentleman on October 17, 2010, 10:33:15 AM

Quote from: ClockTower on October 17, 2010, 10:21:32 AM

Quote from: EnglshGentleman on October 17, 2010, 10:19:50 AM

Where is your evidence that it is a marketing slogan?

Here's a hint for you: It contains the word, "perfectly". Duh.

Sounds like a bunch of conjecture to me. Where is your proof?

Please ask for your proof in the proper thread.

Please leave your conjectures out of this forum.

Quote from: ClockTower on October 17, 2010, 10:37:29 AM

Quote from: EnglshGentleman on October 17, 2010, 10:33:15 AM

Quote from: ClockTower on October 17, 2010, 10:21:32 AM

Quote from: EnglshGentleman on October 17, 2010, 10:19:50 AM

Where is your evidence that it is a marketing slogan?

Here's a hint for you: It contains the word, "perfectly". Duh.

Sounds like a bunch of conjecture to me. Where is your proof?

Please ask for your proof in the proper thread.

Please leave your conjectures out of this forum.

Typical FE tactic... Now please stay on topic. Do you agree that my formula is correct or can you point to a problem with it?

Do you agree that my formula is correct or can you point to a problem with it?

The formula isn't incorrect, but as pointed out by Parsifal, the application of it is. It is as if you used the Ideal Gas Law to measure the gravitation force exerted on an object.

I didn't criticize your formula, since it is only a statement of Pythagoras's Theorem anyway. I criticised your derivation and your application of it, and you failing to provide a valid refutation does not make my comments stupid.

Quote from: ClockTower on October 17, 2010, 10:45:38 AM

Do you agree that my formula is correct or can you point to a problem with it?

The formula isn't incorrect, but as pointed out by Parsifal, the application of it is. It is as if you used the Ideal Gas Law to measure the gravitation force exerted on an object.

Quote from: Parsifal on October 17, 2010, 02:52:51 AM

I didn't criticize your formula, since it is only a statement of Pythagoras's Theorem anyway. I criticised your derivation and your application of it, and you failing to provide a valid refutation does not make my comments stupid.

How does my application differ from Thork's that you accept?

So - do we have a working formula for the Correct Displacement Expected by a Spheroid fitting the dimensions of the RE's expected Curvature.  I would think that it would be rather simple for someone who understands maths.  Therefore I await someone elses answer because of the obvious.

Berny
Really?  Intelligent Pulling?

 Seems to have some relevance here. Document named "Experiments using a helium-neon laser" . page 41, Experiment 19. Measuring the Curvature of the Earth

I say it's given by the following:

I disagree here. I used vectors instead, using the same diagram.
I make it:
d=r(1-cos(a/r))
where d is declination
r is radius
a is arc distance.
d, r and a have the same units.
RADIANS FOR THE COSINE. n00bs use degrees.
For small (s<<r) distances, we can taylor-expand the cosine and we get
d=a²/2r
{
From this formula, we can easily see that Rowbotham cocked his table up in his book.
}
EDIT: I found the difference between the formulae. My formula calculates the distance from the surface to the tangent ray in the vertical direction, not the direction of the radius vector. For small distances they are the same so I stand by my formula for this sort of treatment.
more edit. Taylor expansion fail in approx, cos x is roughly 1-(x²)/2 so there is a factor of two. It can be used to generate the table in the other post, so is a suitable approximation for the declination observed by an observer level with a perfect sphere.

Quote from: ClockTower on October 15, 2010, 11:17:16 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=s²/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

I say it's given by the following:

I disagree, the earth is not a sphere, even in RET.  You need to take your own model into account.

Quote from: ClockTower on October 15, 2010, 11:17:16 PM

I say it's given by the following:

I disagree, the earth is not a sphere, even in RET.  You need to take your own model into account.

The difference between the sphere and the oblate spheroid is so small it is not worth thinking about. Are you talking about global effects ie the oblateness or local effects ie geography? The geography can be added in later, for example you can't see through mountains.

Quote from: hahahaidiots on December 24, 2010, 08:17:04 AM

Quote from: ClockTower on October 15, 2010, 11:17:16 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=s²/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

Maths challenge (I'm going to do it too):
h²sin²(s/2r) = h² - 2hd + d² + 2hr - 2rd +r² -2rhcos(s/2r) + 2drcos(s/2r) - 2r²cos(s/2r) + r²cos²(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.

the brute force solution is:
d = h + r(1-cos (s/2r) ± 0.5sqrt((-2h-2r+2rcos(s/2r))² - 4[h²cos²(s/2r) + 2h + r² - 2rhcos(s/2r) - 2r²cos(s/2r) + r²cos²(s/2r)])
Can anyone see any useful simplification here?

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

IDK, but I've already shown the transformation that light would have to "bend" would be that of the secant.
this is also describes the height the earth falls per mile, so yes this is the most elegant solution I know:
y=r(sec(x/r)-1). where x is the distance along the earth. y is the height of the line segment from point x to the corresponding point on the top line (this segment is made to be perpendicular to the earth) r is the radius of the earth
EDIT I realize that I mistranscribed my formula, and edited it here, plus simplified

Derivation.

c=circumference of Earth. r=radius of earth. theta=interior angle. x=distance along earth's surface. y=altitude of beam.
Take the right triangle in CT's post. one leg is the light beam, one leg is the radius of the earth, the hypotenuse is
the radius of the earth plus the altitude of the light beam. from this we gather that:

cos(theta) = r/(r+y)                    Eq1 [definition of cosine]

From the wedge of the circle, we know that:

theta/2pi =x /c                            Eq2 [proportions]

For convenience, we note:

c = 2pi*r                                      Eq3 [circumference of a circle]

Substituting Eq3 into Eq2:

theta/2pi = x/2pi*r

Solving for theta and simplifying:

theta = x/r                                                   Eq4

Substituting Eq 4 into Eq1:

cos(x/r) = r/(r+y)

Reciprocal:

1/cos(x/r) = (r+y)/r

Trigonometric manipulation and splitting the fraction:

sec(x/r) = 1 + y/r

Solving for y:

y = r(sec(x/r)-1)

IDK, but I've already shown the transformation that light would have to "bend" would be that of the secant.
this is also describes the height the earth falls per mile, so yes this is the most elegant solution I know:
y=r(sec(2pi*x/r)-1). where x is the distance along the earth. y is the height of the line segment from point x to the corresponding point on the top line (this segment is made to be perpendicular to the earth) r is the radius of the earth.

Can you put in a working example please?

Berny
Likes examples.

Quote from: Thevoiceofreason on January 26, 2011, 04:46:27 AM

IDK, but I've already shown the transformation that light would have to "bend" would be that of the secant.
this is also describes the height the earth falls per mile, so yes this is the most elegant solution I know:
y=r(sec(2pi*x/r)-1). where x is the distance along the earth. y is the height of the line segment from point x to the corresponding point on the top line (this segment is made to be perpendicular to the earth) r is the radius of the earth.

Can you put in a working example please?

Berny
Likes examples.

example?
imagine that you were to have a laser beam tangential to the earth.
and then you were to walk one kilometer (the x in the equation) along the earth's surface in the general direction of the laser.if you were to look up at that point, you'd see that the laser was about 8cm (the y in the equation) above your head.

Also one mile away lead to a drop of about 8 inches, which has been confirmed elsewhere on the site.

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)(s²/4r²)
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 - x²
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.