The Flat Earth Society
Flat Earth Discussion Boards => Flat Earth Investigations => Topic started by: Zack Bimmel on December 20, 2019, 09:19:29 PM

Hello everybody,
thought you might be interested in some science concerning measurements of distances and heights in the lower atmosphere which has been applied in the real world for decades now. References given at the end of this post span the time from 1979 to 2016. Hyperlinks to the original articles are provided.
Bedford Level Experiment https://wiki.tfes.org/Bedford_Level_Experiment
(https://wiki.tfes.org/Bedford_Level_Experiment)
It is wellknown that the measurements in surveying over distances of some miles can be affected by atmospheric refraction due to the change in index of refraction of air with height. Such changes are due to change in temperature and relative humidity with height above ground. I do not conѕider the latter in this post.
References [1], [2] and [3] (at the bottom of this post) present this phenomena in terms of a coefficient of refraction, k, defined as :
k = R / r
where R is the radius of earth and r the radius of the circle describing the path of the refracted light beam. Hence k=0 represents the case when the light is not refracted and follows a straight line from the target to be investigated to the observer. For k=1 the light would follow exactly the curvature of a round earth of radius R = 6370 km. Negative values for k correspond to a case with the light path curving upwards.
In the above mentioned references a mathematical expression for the value of k is presented :
k = 503*p*(0.0343 + dT/dh)/(T*T)
p = atmospheric pressure in mbar (1015 mb = 14.7 psi)
T = absolute temperature in degree Kelvin (288 K = 59 Fahrenheit)
dT/dh = change of temperature with height in Kelvin/meter.
The values for p and T are mine representing an average atmospheric condition.
If the temperature does not change with height then dT/dh = 0 and the value of k becomes k=0.21 with the above numbers for pressure and temperature.
Close the ground dT/dh can be expected to be positive if the ground is cooler than the air above. Higher up we observe in general that the temperature decreases with height and therefore a negative dT/dh is common in that region.
In order to get a feeling for what numbers we are talking about, let's assume the air temperature changes by +0.13 deg kelvin ( = 0.234 Fahrenheit) per vertical meter ( about a yard ). We get k=1.01. In that case, on a round earth, any object a few miles away would still appear to sit on the horizon in full view regardless of distance. An observer who is not familiar with refraction will therefore conclude that the earth is flat.
Reference [2] cites a variety of typical ranges for k depending on at which height above ground geodesic measurements are taken.
In the region of 100 m (330 ft) and above the temperature gradient is fairly constant at dT/dh = −0.006 K/m resulting in values of k around 0.17 .
For heights between 2030 m up to a 100m dT/dh = −0.01 K/m resulting in k=0.15. For these values of k the bending of light due to refraction would be still small in comparison to the curvature of an earth with a radius of 6370 km and lineofsight measurements would give pretty conclusive evidence about the flatness of earth's surface. But, both, target and observer and anything in between has to be more than 20m ( 65 ft ) above ground. Even if that is the case, a really good experiment will be accompanied by precise measurements of the temperature gradient at that height.
Below 2030m the temperature gradient in the air is subject to the thermal properties of the surface underneath. [2] cites several experimental studies with values for the coefficient of refraction, k, ranging from 14 to +18 as distance to the ground decreases to below 10m (33 ft). Based on that and my calculations with dT/dh=0.13 K/m giving already a value of k=1.01 it is clear that lineofsight experiments conducted close to the ground must be accompanied by very precise measurements of the temperature gradient along the path of light.
In summary : the accuracy of the measurement of the temperature gradient must better than, let's say, a few hundreds of a degree per vertical meter in order for an experiment to prove or disprove conclusively the flatness of earth's surface.
[1] D. Gaifillia, D et.al.
"Empirical Modelling of Refraction Error in Trigonometric Heighting Using Meteorological Parameters"
Journal of Geosciences and Geomatics, Vol. 4, No. 1, 2016, pp 814
http://pubs.sciepub.com/jgg/4/1/2/index.html
(http://pubs.sciepub.com/jgg/4/1/2/index.html)
[2] Hirt, Christian et.al.
"Monitoring of the refraction coefficient in the lower atmosphere using a controlled setup of simultaneous reciprocal vertical angle measurements"
Journal of Geophysical Research: Atmosphere, Volume 115, Issue D21; Nov 2010
https://agupubs.onlinelibrary.wiley.com/doi/full/10.1029/2010JD014067 (https://agupubs.onlinelibrary.wiley.com/doi/full/10.1029/2010JD014067)
[3] Fraser, C.S
"Atmospheric Refraction Compensation in terrestrial Photogrammetry"
Photogrammatic Engineering and Remote Sensing, Vol 45, No.9, September 1979, pp.12811288
https://www.asprs.org/wpcontent/uploads/pers/1979journal/sep/1979_sep_12811288.pdf
(https://www.asprs.org/wpcontent/uploads/pers/1979journal/sep/1979_sep_12811288.pdf)

The whole bumph depends on the assumption that earth is a sphere where R = 6370km
so use of an imaginary equation K=R/r to quantify any observation falls into the category of pseudoscience .

somerled,
sorry to say, you are wrong. R=6370km is merely used as a convenient reference for those surveyors who believe in a round earth. R has no influence on the amount by which a light beam gets refracted in terms of feet (or whatever units you prefer) for given atmospheric conditions and distance between target and observer. Personally, I would have chosen to not have R appear because it leads to confusion as it happened to you. It was more important to me to represent the equations in a form given by the authors.
The main point remains, experiments like the Bedford level experiment need to determine the temperature gradient with a very high accuracy and document carefully their measurements. Do you know of any such experiment ?

It is wellknown that the measurements in surveying over distances of some miles can be affected by atmospheric refraction due to the change in index of refraction of air with height.
Is that because they see an earth which doesn't match up with theory?

somerled,
sorry to say, you are wrong. R=6370km is merely used as a convenient reference for those surveyors who believe in a round earth. R has no influence on the amount by which a light beam gets refracted in terms of feet (or whatever units you prefer) for given atmospheric conditions and distance between target and observer. Personally, I would have chosen to not have R appear because it leads to confusion as it happened to you. It was more important to me to represent the equations in a form given by the authors.
The main point remains, experiments like the Bedford level experiment need to determine the temperature gradient with a very high accuracy and document carefully their measurements. Do you know of any such experiment ?
Don't talk bollards. The imaginary R is the basis for the coefficient k as stated in all those experiments so k is imaginary and this is then used in an equation with real atmospheric temperature measurements to give imaginary results.
Now if these scientists had surveyed the curvature across the land over which they took their pressure , temperature and altitude measurements then we might have had a real experiment rather than bumph designed to shore up the their imaginary model globe of 6370 imaginary km big R.
Still, tis the pantomime season I suppose . Have a merry Christmas and good new year .

It is wellknown that the measurements in surveying over distances of some miles can be affected by atmospheric refraction due to the change in index of refraction of air with height.
Is that because they see an earth which doesn't match up with theory?
Tom your observation is unfair as Rowbotham himself keeps referring to a supposed "round earth theory" in his ENAG. For example, in https://www.sacredtexts.com/earth/za/za07.htm he states:
If the earth is a globe, the series of flags in the last experiment would have had the form and produced the results represented in the diagram, Fig. 5
but we don't know what kind of globe he was referring to. He observes that his experiment didn't match his own expectation of the globe, but this doesn't prove the globe wrong if his expectation of how a globe would work were wrong in the first place. I think Rowbotham interprets how light would travel in a globe in a purely geometrical way, ignoring any possible effect of refraction or any other of the many effects we now know exist on earth independently to its shape. And let me add that yes, afaik if the Bedford level experiment was conducted on a ideal world where light travels in perfect vacuum without being refracted or modified in any possible way, that would be just simple logic that earth was flat, light being a placeholder of a quite long horizontal level. But why Rowbotham never consider the possibility of refraction? Wasn't he aware of the existence of that?

Rowbotham does reference some tests with barometers in Experiment 9. (https://www.sacredtexts.com/earth/za/za14.htm)
Experiment 2 (https://www.sacredtexts.com/earth/za/za07.htm) that you linked is also probably intended to be a refraction experiment, as multiple point tests give greater confidence for the path of light. In that experiment the five foot tall flags were spaced a mile apart and the tops of the flags were observed to be lined up in a straight line.
(https://www.sacredtexts.com/earth/za/img/fig04.jpg)
It would be a pretty big refraction coincidence to put the top of the poles in a line like that. The top of the first flag would have to have been projected 8 inches into the air, the second flag 2.67 feet, the third flag 6 feet, the fourth flag 10.6 feet, the fifth flag 14.29 feet, and the sixth flag 24.01 feet into the air, when the later flags should be below the horizon.

someled
Sorry that I was not careful enough in explaining the appearance of R in the equation in my submission from Dec 20. You were absolutely correct when you stated that what people perceive as the radius of a round earth should in no way affect the calculation of the amount of refraction of light in the atmosphere.
So let me try again :
As stated :
k = 503*p*(0.0343 + dT/dh)/(T*T)
divide that equation 6370,000 then by definition of k=R/r you get :
1/r = 79*10^{6}*p*(0.0343 + dT/dh)/(T*T)
with r now in meters. The value of R does not appear anymore. You can verify the validity of this division by inspecting the equations (8) through (10) in reference [3]. Feel free to calculate values for r for various values of dT/dh, the vertical temperature gradient, assuming standard atmospheric conditions.
Now, this equation and similar ones incoporating humidity, wave length of the light (think of lasers) and other effects have been used for decades. It seems to me somebody would have caught on to significant errors on the part of surveyors.
All this quarrel about refraction can of course be avoided by conducting a laserbased experiment inside a vacuumfilled, long tube. Maybe, if we were to look around, somebody has built already such a tube ? Maybe they did it for another purpose, but with the condition that the laser beam coincides with the tube's central axis all the way from one end to the other ? Did they consider earth's supposed curvature ?
Happy Holidays to you and everybody else ... Zack

Rowbotham does reference some tests with barometers in Experiment 9. (https://www.sacredtexts.com/earth/za/za14.htm)
There he writes that
Refraction can only exist when the medium surrounding the observer is different to that in which the object is placed. As long as the shilling in the basin is surrounded with air, and the observer is in the same air, there is no refraction; but whilst the observer remains in the air, and the shilling is placed in water, refraction exists
and that being moisture equal at the two ends of the bank,
In a short time afterwards the two sets of observers met each other about midway on the northern bank of the canal, when the notes were compared, and found to be precisely alikethe temperature, density, and moisture of the air did not differ at the two stations at the time the experiment with the telescope and flagstaff was made. Hence it was concluded that refraction had not played any part in the observation, and could not be allowed for, nor permitted to influence, in any way whatever, the general result.
refraction *should* not have played (in his view) a role, because source and observer are in the same medium. But this is obvious and Rowbotham doesn't mention the other refraction due to gradual changes of the air in height (as the OP) and this happens to be exactly the explaination given to the Bedford Experiment by REs. So there you have from two different theories of how a roundearthwithatmosphere should work two different conclusions deriving from the same experiment.
It would be a pretty big refraction coincidence to put the top of the poles in a line like that. The top of the first flag would have to have been projected 8 inches into the air, the second flag 2.67 feet, the third flag 6 feet, the fourth flag 10.6 feet, the fifth flag 14.29 feet, and the sixth flag 24.01 feet into the air, when the later flags should be below the horizon.
Earth curve calculator (https://dizzib.github.io/earth/curvecalc/?d0=6&h0=5&unit=imperial) gives me a smaller number. Anyway refraction makes it look like a curved surface is straight, so everything in between looks like on the same line.

someled
Sorry that I was not careful enough in explaining the appearance of R in the equation in my submission from Dec 20. You were absolutely correct when you stated that what people perceive as the radius of a round earth should in no way affect the calculation of the amount of refraction of light in the atmosphere.
So let me try again :
As stated :
k = 503*p*(0.0343 + dT/dh)/(T*T)
divide that equation 6370,000 then by definition of k=R/r you get :
1/r = 79*10^{6}*p*(0.0343 + dT/dh)/(T*T)
with r now in meters. The value of R does not appear anymore. You can verify the validity of this division by inspecting the equations (8) through (10) in reference [3]. Feel free to calculate values for r for various values of dT/dh, the vertical temperature gradient, assuming standard atmospheric conditions.
Now, this equation and similar ones incoporating humidity, wave length of the light (think of lasers) and other effects have been used for decades. It seems to me somebody would have caught on to significant errors on the part of surveyors.
All this quarrel about refraction can of course be avoided by conducting a laserbased experiment inside a vacuumfilled, long tube. Maybe, if we were to look around, somebody has built already such a tube ? Maybe they did it for another purpose, but with the condition that the laser beam coincides with the tube's central axis all the way from one end to the other ? Did they consider earth's supposed curvature ?
Happy Holidays to you and everybody else ... Zack
I will make the point again , k is the coefficient of refraction deduced from the assumed (imaginary) curvature of a globe earth of R = 6370. Hence the silliness of using it anywhere in calculations . If these scientists wanted to carry out real experiments then the curvature , or lack of , would have to be accurately surveyed by good old proper measurement methods using real precision instruments and there would be no need to use imaginary values.
I mean the earth is either pear shaped or squashed orange shape , depending on which greengrocer you believe so why use this R = 6370 since it doesn't exist . It's interesting that in the case of k=1 light follows the curvature ,or lack of , of the earth . If Rowbotham had used a better telescope he'd have seen his own arris since the light was clearly following the " curve " on globeworld.
No one disputes that there are atmospheric effects but putting these down to refraction is pseudoscience . Refraction takes place at distinct boundaries dependent upon angle of incidence and differing density of medium involved . The atmosphere diffuses ,absorbs , diffracts etc.
Refraction calculated using this mathematical trickery is used to cover the fact that there is no curvature and the trick always provides an answer although it is nothing to do with reality .

Dear somerled,
thanks for response. Unfortunately a few improvements to it are in order. First of all, nobody has claimed nor is deducing the amount of refraction from the curvature of earth. To assert publicly that anybody ever would do that is akin to spreading false information. Please clarify that to yourself by reading the equations in ref.[3] I had mentioned in a previous post. If you can't do that by yourself ask somebody for help but do not classify Algebra 101 matters as "mathematical trickery". That makes FET look really bad.
Secondly, you start off correctly by stating that "refraction takes place at distinct boundaries dependent of angle of incidence and differing density of medium involved." The latter part is not exactly correct. You can have two media with equal density but different refractive index and vice versa. More importantly, your statement should be appended to include the fact that refraction ALSO takes place inside media where the index of refraction varies continuously. In connection with our discussion, feel free to google for "refraction thermal gradient" and "refraction humidity gradient". You will find ample links to these topics. Some of the linkedto articles derive the equations of how light (or in general electromagnetic waves) refract in the presence of temperature and humidity gradients in some detail. For those articles understanding calculus is often prerequisite.
Happy New Year

Do you not read the articles you link . Ref 3 introduces big imaginary R in equation 10 and combines with equation 7 to give what can only be an imaginary answer . Big R is defined as 6370km which apparently is a mean value which tells us that it is not a measured value .
Why bother measuring to the nth degree the pressure ,altitude , humidity etc over the surface of an area of land and then not measure the curve of the land over which the experiment is carried out.
Again the real value of big R is there to be measured but is ignored and a mean value ,which cannot be accurate over any part of the surface , inserted which invalidates the results. Pseudoscience . And mathematical trickery in plain language . Who, and when , carried out the measurements of our pearshaped squashed orange which allowed a mean value of 6370 km to be deduced . A curvature we cannot find or measure .
All any scientist has to do to find the shape of the earth is carry out a physical survey .
Insults are thrown when debates are lost .

somerled
I certainly admire and appreciate your tenacity in our discussion. Let me try to guide you a little bit through the math involved.
First, let's see that we find a common starting point and then go from there. Allow me to suggest eq.(7) you mentioned as a starting point. It does not contain R=6370km in any way shape or form as you so rightfully demanded.
Look at that equation in connection with Fig.1 and the subsequent list of variables to see that S stands for chord length and Delta_beta is the refraction angle. If you are not quite sure what I mean by Delta_beta look up the capital version of the Greek letter delta and the lower case version of the Greek letter beta and write them next to each other.
For the next step we need to know that in mathematics angles are measured in radians (some times abbreviated as rad). Look that up as well if you are not familiar with that. But first let's agree on the starting point and address any questions you might have about it before we take the next step.

Pretty equations but as the saying goes , you can't polish a turd. I see you are trying to distance yourself from the imaginary k = R/r so prevalent in your original post in which these imaginary results are used in your "real world"
Please forget about that you say . But k , that imaginary coefficient with no basis in reality , is central to all your linked experiments . Hope you've learned a bit more about the silly pear shaped squashed orange model you think bears a resemblance to reality .
Also ,angles can be measured using something called "degrees" . Look it up along with the word " assumption " which is used several times in your not so empirical experiment links .

somerled
You are mistaken when you assume I am distancing myself from k, the opposite is the case. I will lead you to it and will also show you how R=6370km appears.
You have to be a little bit patient with me though. Math is sometimes not easy and sometimes multiple steps have to be executed flawlessly. Just by way of preliminaries : of course I know that most people measure angle in degrees ( 360 for the full circle ). I think I learned that in grade school. But then later on when we got to trigonometry radians popped up as an alternative. The relationship between the two is easy (PI=3.1415...) :
radians = PI/180 * degree
Hence, 360 deg = 2*PI , 180 deg = PI , 90 deg = PI/2 etc.
The reason why I bring this up is that later on I will make use of the fact that :
sin(Delta_beta) is in very good approximation equal to Delta_beta (in radians) if Delta_beta is small. Example :
Delta_beta = 10 degrees = 0.1745 rad ; sin(10 degree) = 0.1736
Luckely, as far refraction of light in air is concerned, Delta_beta is much smaller than 10 degrees and the error between it and its sin() becomes even smaller than the 1% error in my example.
Let me know whether what I brought so far is OK with you. Feel free to ask any question you might have and with a little bit of patience we will get to eq.(10) of Ref.[3] and see how k and R arrive on the scene. May I recommend that in your response you just comment on the math I brought in this post. Talking about the silly pear shaped orange model is as the saying goes, balking up the wrong tree.

Then lead away , I am interested to see the logical steps .
I would like to point out again that R is assumed to be equal to 6370km , a value that cannot be arrived at by empirical measurement since it cannot exist if the earth is an oblate spheroid . This is why it will always be quoted as an assumption when used in experiment .

Then lead away , I am interested to see the logical steps .
Let me try that:
Fact 1: Atmosomething refraction due to temperature gradient exists, proof: mirages as in the following figure:
(http://hyperphysics.phyastr.gsu.edu/hbase/atmos/imgatm/mirageinf.png)
Fact 2: Atmosomething refraction can indeed be modeled with Math formulas (we may not agree on the exact formulas, but those formulas do exist).
Fact 3: In a FE temperature gradient would go upwards in layers that are horizontal, but in a RE they would be concentric spherical shells.
Fact 4: Given the Math formulas believed by REs, and assuming Fact 3, and a temperature gradient, from the RET point of view the Bedford Level experiment would give the same exact results both on FE and RE.
Conclusion: Rowbotham concluded FE from an observation that was enough for him, but not enough for those using a certain set of Math formulas (aka REs). For the latters, temperature gradient should have been taken into account. Now somerled you don't trust REs Math formulas, but this doesn't disprove Fact 1 and also doesn't exempt FE experimenters to not consider the possibility that visual results on land survey could be due to atmosomething refraction.

Like your sensible conclusion BP .
I don't have a problem with formulae . Math is a good tool to help describe reality but once assumption is built in to a formula then it no longer describes reality .
The logical way to avoid that would be to actually survey across the land . Use precise physical , optical and laser measuring equipment  would show us the true shape .

Hi,
Don’t mean to hijack this conversation, but I do have two comments.
I agree with somerleds position that if you have a flat earth model, refraction doesn’t make sense as the line of sight is always perpendicular with the thermal gradient, so refraction would always be zero.
But you do have two problems otherwise.
1. In the Wiki, it is stated that one of the reasons that ships vanish below the horizon, which is basically the same thing as this Bedford experiment, is refraction. You can have it both ways, either refraction interferes with the path of light in the flat earth model or it doesn’t.
2. Whenever I have seen these “Rowbotham effecst” demonstrated, it is pretty clear that the experimenter always puts the observation point and the target very close to the ground. (including Rowbotham). The reason seem pretty obvious, when you are close to the ground, the thermal gradient is the highest and therefore the refraction is the highest and you get the illusion that the earth is curved more than it actually is. If you see others perform the experiment, they always make sure to do it well above the surface which minimizes refraction and shows that the earth is curved.

But you do have two problems otherwise.
1. In the Wiki, it is stated that one of the reasons that ships vanish below the horizon, which is the same thing as this Bedford experiment, is refraction. You can have it both ways; either refraction interferes with the path of light in the flat earth model or it doesn’t.
Based on the wiki, the sinking ship effect may have many causes. I don't know if FEs consider one to be the most frequent, but in that case, I believe it would be the FE theory of Optical Resolution, which places the vanishing point of perspective at no more than 7 miles in front of you.
2. Whenever I have seen these “Rowbotham effects” demonstrated, it is pretty clear that the experimenter always puts the observation point and the target very close to the ground. (including Rowbotham). The reason seems pretty obvious when you are close to the ground, the thermal gradient is the highest and therefore the refraction is the highest and you get the illusion that the earth is curved more than it is. If you see others experiment, they always make sure to do it well above the surface, which minimises refraction and shows that the earth is curved.
The primary sin of these experiments is that they are visual. So considering that refraction creates optical illusions, they both don't prove anything either way. The OP stated we'd need a vacuum tube and laser, and I think he's right.

Your experiment has already been done: check out the 'LIGO Experiment'.
It's a laser in a vacuum. The arms are long enough so the concrete supports for the
pipes had to be adjusted to compensate for the earth's curvature.
Maybe they should have consulted the wiki on here and saved themselves a bunch of money!

Hi someled
Here is my last attempt to convey to you that indeed, in the references I provided the occurrence of the radius of the earth, R, does not indicate that anybody suggests that the mechanism of the refraction of light is influenced by R. It also should become clear to you why it pops up in the equations in the references. If that seems to you a contradiction in terms than please study the following with care to see the contradiction vanish in thin air (pun intended).
(http://)
For the following see my attached image, I have it available only on my home computer and didn't know how to imbed into this post.
This image is a my counterpart to Fig. 1 in Ref.[3]. As in that figure we have the cordlength S stretching from point 1 to 2 with point 3 in the middle. The circular arc going from point 1 to 2 represents a beam of light under influence of refraction with point 2 being the target and point 1 the observer. Line 14 is tangential to the circular arc at point 1 and points in the direction the observer at point 1 perceives the target to be. Again, Delta_beta is the angle of refraction. Perpendicular to this tagent is the radius line 10. By way of similarily of right triangles you can prove that the angle Deltabeta occurs again between the lines 01 an 03.
Some trigonometry :Or with extremely small errors (less than 1%) according to the info in my previous post :
Delta_beta = S/(2*r)
Now use this relationship to eliminate Delta_beta from Eq.(7) in Ref.[3]. The resulting equation is :
S/(2*r) = S/2 * (dN/dh) * 1.e6
I assumed that the cordlength, line 13, is running horizontally. Hence cos(beta) = 1
The factor S/2 appears on both sides of the equation and therefore cancels out.
1/r =  (dN/dh) * 1.e6
This equation still does not contain R=6370km which is what you rightfully demand to be. So, how does R=6370km come into being in eq.(10) in Ref.[3] you might ask.
Well, here is what scientist often do. In order to converse easily among each other it is better to present r with respect to some reference length, let's call that L. One can argue endlessly as to how big the value of L might be. If all the involved scientists were to live in the UK one (usually it is the one who publishes first on the subject at hand) might suggest the distance between London and Glasgow. If subsequent publishers in the field like that choice because they think that distance is relevant to the refraction of light in the context of surveying the landscape or the building of large structures the original suggestion for L will pervail, otherwise somebody else comes up with a different idea.
I suggest that L = distance between London and Glasgow is really not a good idea but maybe using the radius of the earth might not be a bad choice for two reasons. a) Everybody knows at least a good approximate value of that ( 6370km or the equivalent in miles or whatever units one prefers). b) r in above equation often comes out to be in the neighborhood of thousands of kilometers under common atmospheric conditions.
Last step to get to Eq.(10) in Ref.[3].
Multiply my last equation by R on both sides :
R/r =  R*(dN/dh) * 1.e6
and abbreviate : k = R/r to obtain eq.(10) in Ref.[3]. Bingo.
And this is my last post trying to explain to you that in all the references I provided the radius of earth is simply used as a convenient way to display and communicate the effect of refraction of light due to temperature gradient in the atmosphere. Thank you for your time. Zack
sin(Delta_beta) = S/(2*r)

Your experiment has already been done: check out the 'LIGO Experiment'.
It's a laser in a vacuum. The arms are long enough so the concrete supports for the
pipes had to be adjusted to compensate for the earth's curvature.
Maybe they should have consulted the wiki on here and saved themselves a bunch of money!
Congratulations, but you really took away my punch line. ;D

Hi,
Don’t mean to hijack this conversation, but I do have two comments.
I agree with somerleds position that if you have a flat earth model, refraction doesn’t make sense as the line of sight is always perpendicular with the thermal gradient, so refraction would always be zero.
But you do have two problems otherwise.
1. In the Wiki, it is stated that one of the reasons that ships vanish below the horizon, which is basically the same thing as this Bedford experiment, is refraction. You can have it both ways, either refraction interferes with the path of light in the flat earth model or it doesn’t.
2. Whenever I have seen these “Rowbotham effecst” demonstrated, it is pretty clear that the experimenter always puts the observation point and the target very close to the ground. (including Rowbotham). The reason seem pretty obvious, when you are close to the ground, the thermal gradient is the highest and therefore the refraction is the highest and you get the illusion that the earth is curved more than it actually is. If you see others perform the experiment, they always make sure to do it well above the surface which minimizes refraction and shows that the earth is curved.
DaveP : sorry to contradict you a little bit. Refraction of light takes place of cause when light enters a medium with a n index of refraction different from that medium the light beam comes from. In addition to that, light is also refracted when light beam and temperature gradient are perpendicular to each other. Or for that matter by gradients of absolute humidity.
But, you are correct with you last paragraph, you can use refraction to make points in favor or against the flat earth model. Unless, of course, you measure the temperature gradient exactly along the line of sight during your experiment and then correct accordingly. I am not aware of anybody in the FE community or otherwise has done that nor published about his/her/their experimental evidence.
There is at least one experiment (conducted for totally unrelated purposes) called LIGO of which we have two examples in the US and one in Italy. It is based on a laser beam traveling down a long straight tube ( 4km) and being reflected back to where it came from. No air or other gases on the inside, just a plain, highquality vacuum. They had to take earth's curvature into account increasing the cost of a rather expensive experiment (hundreds of millions of dollars). LIGO is doing well continuously listening to the "sounds" of the universe.
It seems to me there might be other such instance where creating large structures might have to consider earth's curvature. Maybe underground subway system where two tubes cross each other; one on top of the other with little clearance in between. I am thinking of those instances where these tunnels were "dug" by those huge machine which grind their way for miles on end never to see the light of day why doing so.

When you construct a microwave link, it's essential that you consider the curvature of the earth.
The engineering plan must consider the height of the antenna at each end, the height of the terrain in between,
and the curvature of the earth. The height of any object or a rise of terrain near the middle of the path could be
a game changer causing the costly construction of taller towers at one or both ends.
If it weren't for the curvature of the earth than a microwave transmitter on top of the Sears Tower in Chicago could easily
communicate with a tower in Omaha. In the real world, that's impossible all because of the nasty curvature of the earth.
The same goes for TV towers. If it weren't for the earth's curvature you wouldn't be able to put a TV transmitter on any of
the same channels as are used for the ones in Chicago anywhere in the USA. Mutual interference would be the result.
That doesn't happen for the vast majority of the time. Yes, I used to sometimes watch TV from a distant city usually on the
lower TV channels 60 years ago. Quirky weather conditions would allow the TV signals to bounce back to the earth, but those
kind of things only happened less than 1% of the time. All that means is that, yes, you can indeed receive those signals from a
longer distance but for the most part it's the earth's curvature that keeps that distant TV signal from interfering with your local
TV station and keeping all the viewers happy.

In the LIGO experiment which model of earth curvature (oblate spheroid or pearoid) did the building bods account for ? Did they just account for the fictional R = 6370km in such a delicate experiment ?

In the LIGO experiment which model of earth curvature (oblate spheroid or pearoid) did the building bods account for ? Did they just account for the fictional R = 6370km in such a delicate experiment ?
The LIGO setup is not floating on water, so I don't think it would prove anything. Maybe the closest thing to a visual proof (that REs would take as a disproof of their theory blablabla) would be to check the Bedford Canal supposed flatness at an increased height, where the temperature gradient wouldn't affect light so much.

Funny you should mention that......
Rowbotham repeated his experiments several times over the years but his claims received little attention until, in 1870, a supporter by the name of John Hampden offered a wager that he could show, by repeating Rowbotham's experiment, that the earth was flat. The noted naturalist and qualified surveyor Alfred Russel Wallace accepted the wager. Wallace, by virtue of his surveyor's training and knowledge of physics, avoided the errors of the preceding experiments and won the bet.[5][6] The crucial steps were to
Set a sight line 13 feet (4 m) above the water, and thereby reduce the effects of atmospheric refraction, and
Add a pole in the middle that could be used to see the "bump" caused by the curvature of the earth between the two end points.[1]
Despite Hampden initially refusing to accept the demonstration, Wallace was awarded the bet by the referee, editor of The Field sports magazine. Hampden subsequently published a pamphlet alleging that Wallace had cheated and sued for his money. Several protracted court cases ensued, with the result that Hampden was imprisoned for threatening to kill Wallace[7] and for libel.[8][9] The same court ruled that the wager had been invalid because Hampden retracted the bet and required that Wallace return the money to Hampden. Wallace, who had been unaware of Rowbotham's earlier experiments, was criticized by his peers for "his 'injudicious' involvement in a bet to 'decide' the most fundamental and established of scientific facts".[1]
https://en.wikipedia.org/wiki/Bedford_Level_experiment (https://en.wikipedia.org/wiki/Bedford_Level_experiment)

Had a quick look on YouTube as I figured someone must have reproduced the experiment. Couldn't find anything but I did find this
https://www.youtube.com/watch?v=CnrjdD08dWg
It's a similar experiment over a 9 mile stretch of water with no significant waves.
I'll be fair and say they picked some of the crazier flat earthers but it was unsurprising to see results which you'd expect on a globe being declared by Mark Sargent as exactly what he'd expect on a flat earth.

AllAroundTheWorld (also mentioning RonJ's post)
thanks for posting that video. But, it really doesn't help the discussion about flatness or not because the experiment again does not address the impact of refraction in a satisfactory manner. One of the FE persons tries to talk about it ( naming it mirage effect or something like that) and the "science" guy just smiles condescendingly. At the end, everybody, on both side of the debate, felt confirmed in their opinion. Come to think of it, that was utterly foreseeable and one has to blame the "science" guy.
As I said before, lineofsight experiments like the Bedford level and Lady Bount experiments are only really useful when the atmospheric conditions including temperature and humidity gradients have been measured accurately or when the atmospheric conditions do not come into play in the first place like with LIGO with its vacuum tubes. Also, the post by RonJ about establishing microwave links must be taken into account. We have those links all over the world and they are working and they take earth's curvature into account.
Does anybody know of more things being built which have to take earth's curvature into account ?
Interestingly enough, comments from the FE side have not been definite in this thread although they should be shaking to their core. Guess I am rattling the cage here hoping that something constructive falls out.

AllAroundTheWorld (also mentioning RonJ's post)
thanks for posting that video. But, it really doesn't help the discussion about flatness or not because the experiment again does not address the impact of refraction in a satisfactory manner. One of the FE persons tries to talk about it ( naming it mirage effect or something like that) and the "science" guy just smiles condescendingly. At the end, everybody, on both side of the debate, felt confirmed in their opinion. Come to think of it, that was utterly foreseeable and one has to blame the "science" guy.
As I said before, lineofsight experiments like the Bedford level and Lady Bount experiments are only really useful when the atmospheric conditions including temperature and humidity gradients have been measured accurately or when the atmospheric conditions do not come into play in the first place like with LIGO with its vacuum tubes. Also, the post by RonJ about establishing microwave links must be taken into account. We have those links all over the world and they are working and they take earth's curvature into account.
Does anybody know of more things being built which have to take earth's curvature into account ?
Interestingly enough, comments from the FE side have not been definite in this thread although they should be shaking to their core. Guess I am rattling the cage here hoping that something constructive falls out.
The WGS84 model is the proven shape.

As King Arthur's horse once said  It's only a model . https://www.esri.com/news/arcuser/0703/geoid1of3.html

The WGS84 model is the proven shape.
Your WGS84 geographic syestem is a flat map management system:
http://www.boshamlife.co.uk/wpcontent/uploads/2013/11/PrimeMeridian.pdf
“ By 1911, the Greenwich meridian had been accepted as the prime meridian for the whole world. However, relating the maps of an individual country or region to a standard system of latitude and longitude is not only difficult, it is nearly impossible. The earth is approximately spherical, but maps are flat. They are fitted as closely as possible to the surface of the earth in one region, but when fitting them to a standard system of latitude and longitude, there are bound to be slight discrepancies. The differences between the coordinate systems used by different maps really didn’t matter until recently. When the GPS system was introduced in the 1980s, it was realised that having dozens of ‘local’ systems of latitude and longitude for different countries wasn’t going to work. A single coordinate system had to be devised, which would give the best results for every part of the world. It is known as WGS 84 (World Geodetic System 1984). ”

The WGS84 model is the proven shape.
Your WGS84 geographic syestem is a flat map management system:
http://www.boshamlife.co.uk/wpcontent/uploads/2013/11/PrimeMeridian.pdf
“ By 1911, the Greenwich meridian had been accepted as the prime meridian for the whole world. However, relating the maps of an individual country or region to a standard system of latitude and longitude is not only difficult, it is nearly impossible. The earth is approximately spherical, but maps are flat. They are fitted as closely as possible to the surface of the earth in one region, but when fitting them to a standard system of latitude and longitude, there are bound to be slight discrepancies. The differences between the coordinate systems used by different maps really didn’t matter until recently. When the GPS system was introduced in the 1980s, it was realised that having dozens of ‘local’ systems of latitude and longitude for different countries wasn’t going to work. A single coordinate system had to be devised, which would give the best results for every part of the world. It is known as WGS 84 (World Geodetic System 1984). ”
And:
WGS 84 is an Earthcentered, Earthfixed terrestrial reference system and geodetic datum. WGS 84 is based on a consistent set of constants and model parameters that describe the Earth's size, shape, and gravity and geomagnetic fields. WGS 84 is the standard U.S. Department of Defense definition of a global reference system for geospatial information and is the reference system for the Global Positioning System (GPS).
https://gisgeography.com/wgs84worldgeodeticsystem/
The Global Positioning System uses the World Geodetic System (WGS84) as its reference coordinate system. It comprises of a reference ellipsoid, a standard coordinate system, altitude data and a geoid.
Do you understand how the various GPNSS systems work  GPS, Glonass, Beidou, Galileo?
Very strange you think WGS84 is a flat map management system. We look forward to your reply.

Tom bishop is just a black belt in cherry picking. Maps are flat, based on a globe map still. There’s no use picking words out of an article that has already stated the earths shape being spherical in an attempt to prove they’re saying the earth is flat... Surely you can see that the overall paragraph Isn’t saying the war is flat, Tom?

Tom bishop is just a black belt in cherry picking. Maps are flat, based on a globe map still. There’s no use picking words out of an article that has already stated the earths shape being spherical in an attempt to prove they’re saying the earth is flat... Surely you can see that the overall paragraph Isn’t saying the war is flat, Tom?
I've read WGS84 related posts from Tom before and really don't understand his reasoning.
If you want to map the Earth using modern technology then you need some sort of mathematical reference shape to base your coordinates on. This needs to be an approximation since it would be impractical to model every single pile of rock or lake bed in a usable mathematical model.
If you start with the assumption the Earth is a globe you could pick a sphere as your reference shape, but an ellipsoid (spheroid) is a better approximation. There is no such thing as a best reference ellipsoid, all of them are compromises, they fit better in some places than others and you choose the reference ellipsoid which best fits your location. In North America you might decide to use NAD83, but in Australia maybe AGD84 (Australian Geodetic Datum 1984).
Of course if you are using GPS/GNSS then you don't have a location as such, it's a global system, so you need a really generic, best fit everywhere ellipsoid and that's what WGS84 gives you. It's a bit like choosing an off the peg suit designed for an average person of a certain size (WGS84) vs an expensive tailor made suit just for you (NAD83  if you live in North America).
Once you have recorded all your features in your WGS84 or NAD83 (or whatever) coordinate system, you need to display the results. If you happen to have a computer to hand you can do this more or less directly, but if you want a 2D map, you have to mathematically project your 3D ellipsoidal model onto a 2D surface and you have to accept a compromise here. If you want accurate distances for example, then your directions and areas and shapes are compromised. There are many different projections, and you choose one which best fits your need and location, so a GA pilot in North America needs a different projection to an Australian trucker for example.
Every description you'll ever read about WGS84 or NAD83 will talk about a reference ellipsoid, so I just don't get why Tom keeps saying WGS84 is a flat map system.

Trying to getting back to the OP:
1  Refraction does exist for FEs and it may cause the sinking ship effect, see https://wiki.tfes.org/Sinking_Ship_Effect_Caused_by_Refraction
2  Rowbotham acknowledges the existence of the Sinking Ship effect, but adding that it's unpredictable and it doesn't prove RE.
3  Rowbotham shows he doesn't understand air refraction and that, by the same token as 2, also the Bedford Level Experiment wouldn't necessarily prove FE.
4  FET is still unable to provide formulas for that, and maybe it could be a totally unpredictable effect for FEs. Even though I'd be curious to know how in the mentioned link in 1 it's boldly stated:
Firstly, the reader should note that, if that curvature seen the photograph were actually the curvature of the earth, the image would suggest that earth is very small.
without any reference to any formula. The same sentence somehow shows that FET does indeed acknowledge that seeing that the earth is round is really difficult because it's so big.

Wgs84 is a mathematical defined surface that approximates the geoid which itself is a model based on the assumption that earth is some sort of pear shape or oblate spheroid.It's use is to map the earth as some sort of globe . Tom is correct imo .
This globe is then flattened to produce workable maps and to allow gps to function.
The flat earth geocentric model is used in all endeavors that involve navigation or survey . Geodesy is system of measure of earth based on the assumption of it's shape.
The shape of earth is easily measured by survey that doesn't have to include any atmospheric refraction or assumption of shape . Surveyors take no account of ( cannot find ) any curvature over any area of 100sq. miles  because we live on a plane .
Refractive distortion provides a smoke screen for the non existence of the globe . That is why science uses a coefficient of atmospheric refraction based on figures that cannot be accurate.

Wgs84 is a mathematical defined surface that approximates the geoid which itself is a model based on the assumption that earth is some sort of pear shape or oblate spheroid.It's use is to map the earth as some sort of globe . Tom is correct imo .
This globe is then flattened to produce workable maps and to allow gps to function.
The flat earth geocentric model is used in all endeavors that involve navigation or survey . Geodesy is system of measure of earth based on the assumption of it's shape.
The shape of earth is easily measured by survey that doesn't have to include any atmospheric refraction or assumption of shape . Surveyors take no account of ( cannot find ) any curvature over any area of 100sq. miles  because we live on a plane .
Refractive distortion provides a smoke screen for the non existence of the globe . That is why science uses a coefficient of atmospheric refraction based on figures that cannot be accurate.
There is no assumption of shape, we accurately know it. A sphere.

Wgs84 is a mathematical defined surface that approximates the geoid which itself is a model based on the assumption that earth is some sort of pear shape or oblate spheroid.It's use is to map the earth as some sort of globe . Tom is correct imo .
Tom says WGS84 is "a flat map management system". You say "it's use is to map the earth as some sort of globe". These two statements sound to me like more or less opposite views. Yet you say Tom is right? I'm very confused.
This globe is then flattened to produce workable maps and to allow gps to function.
The flat earth geocentric model is used in all endeavors that involve navigation or survey . Geodesy is system of measure of earth based on the assumption of it's shape.
Surely geodetic surveying does not use a flat earth model and heliocentrism is irrelevant to any kind of map or survey. Navigation involving ocean voyages and long distance flights are both based on following great circle routes which only make sense if you assume you are on a globe. If all navigation is based on a flat earth, why are the routes curved?
The shape of earth is easily measured by survey that doesn't have to include any atmospheric refraction or assumption of shape . Surveyors take no account of ( cannot find ) any curvature over any area of 100sq. miles  because we live on a plane .
Refractive distortion provides a smoke screen for the non existence of the globe . That is why science uses a coefficient of atmospheric refraction based on figures that cannot be accurate.

The shape of earth is easily measured by survey that doesn't have to include any atmospheric refraction or assumption of shape . Surveyors take no account of ( cannot find ) any curvature over any area of 100sq. miles  because we live on a plane
Incorrect
https://www.ligo.caltech.edu/page/facts
LIGO’s arms are long enough that the curvature of the Earth was a factor in their construction. Over the 4 km length of each arm, the Earth curves away by nearly a meter! Precision concrete pouring of the path upon which the beamtube is installed was required to counteract this curvature.

A further quote from the LIGO curvature compensation:
https://books.google.com/books?id=6mMlDwAAQBAJ&lpg=PA168&ots=fhaBCNVwoX&dq=ligo%20curvature&pg=PA168#v=onepage&q=ligo%20curvature&f=false
"The ends of each arm are actually situated several feet higher off the
ground than their starting point at the center station. That’s to compensate
for the Earth’s curvature."
It says that the ends of each arms are situated several feet off the ground than the center station to "account for the curvature of the earth."
Would it work if the earth was flat? Yes, it is possible to point a laser slightly downwards or upwards on a Flat Earth.
LIGO diagram:
(https://www.mathworks.com/company/newsletters/articles/confirmingthefirsteverdetectionofgravitationalwavesbyanalyzinglaserinterferometerdata/_jcr_content/mainParsys/image_1545705967.adapt.full.high.jpg/1523628233006.jpg)
They had to take steps to aim and align the components at some point. To hit the opposite marker they would have aimed the laser beam at the destination, or would have aligned the intermediate components so that the laser beam would hit the destination. The set up does not appear to be a physical obstacle on a FE.

That’s the second time recently you’ve posted a quote which says the earth is a globe. Have you finally seen the light?
I was simply responding to the point that they don’t ever need to compensate for the earth’s curve in engineering projects. They sometimes do in larger scale ones, thanks for backing up my point.
Were the earth flat they wouldn’t need to. Yes, they could have made a sloped tunnel and pointed the laser slightly upwards, but why would they?

Comment 1 :
The pipes of the LIGO experiments are not only supported at either end, some 4 km apart, but at many intermediary support piers. Compensating for earth's curvature means that the height of theses piers would have to change quadratically with distance if measured with respect to the water level inside a 4 km long water trough running parallel to LIGO pipes. FE would result in a linear dependency.
Comment 2 :
Some of the more sophisticated GPS system measure to within a few centimeters the altitude with respect to an ellipsoid model of the earth. The value for the altitude can be positive or negative depending on whether the elevation you wish to measure is above or below the ellipsoid. Again, the elevation of various points along the length of the pipe should change quadratically. (P.S. Of course one could base a GPS system by approximating earth by a sphere. The principle is the same)
Comment 3 :
All GPS measurements ultimately lead back to having (preferably) four satellites above the horizon at the moment of measurement. All GPS related satellites orbit the round earth at an altitude of about 20,000 km (12,427 miles) and complete two full orbits every day.
Comment 4 :
In addition to millions of earthlings, NASA, NOAA and the entire military complex of NATO uses GPS rather successfully. Russia has its own GPS version (Glonass) and so do China (Beidou ?) and Europe (Galileo). All these systems are satellite based.

1. LEGO experiment link states vaguely that several feet of curvature is accounted for . Which model , pearshape or squashed ball or imaginary R= 6370km is used in this curvature correction?
Could just be allowance for topography .
2. Yes , as stated before WSG is a mathematical ellipsoid surface used to model a globe ,with all vertices converging at earth's supposed centre of gravity . Where is that in whichever ellipsoid model used?
3. High altitude craft , balloon satellites , signals reflected of the dome, triangulation masts.
4. It's easy to brainwash young people . That's what schooling is about . Critical thinking is not part of the curriculum hence people cannot question what they are told they know .
i

1. LEGO experiment link states vaguely that several feet of curvature is accounted for . Which model , pearshape or squashed ball or imaginary R= 6370km is used in this curvature correction?
Could just be allowance for topography .
2. Yes , as stated before WSG is a mathematical ellipsoid surface used to model a globe ,with all vertices converging at earth's supposed centre of gravity . Where is that in whichever ellipsoid model used?
3. High altitude craft , balloon satellites , signals reflected of the dome, triangulation masts.
4. It's easy to brainwash young people . That's what schooling is about . Critical thinking is not part of the curriculum hence people cannot question what they are told they know .
i
3. Not according to the documentation and information provided by receivers. Do China and Russia have balloons over the USA?
4. It's not just about young people, engineers and scientists understand, design, build and operate GPNSS systems in many countries.
Conclusion  GPNSS operates exactly as it says on the tin.

Then which globe model does it say on the tin?

Then which globe model does it say on the tin?
WGS84

But WGS84 is a reference coordinate system system , not a globe models. Clearly explained here.
https://gisgeography.com/wgs84worldgeodeticsystem/
OP's original post shows how surveyors who believe the earth is a sphere ( he kindly quotes this in his 2nd post) use a set of equations including a constant of refraction k calculated with a value of R = 6370km . This is valid is on a perfect sphere only . We know this is not true whichever model we choose to believe .
WGS84 is an ellipsoid reference coordinate system applied to one of many geoids which in the words of that site leads to Geodesists to " believe the error is less than 2cm which is better than NAD83."
The use of that word " believe " leads me to conclude that we are encouraged to have faith in whatever they tell us. Science now has the appearance of religion in which no one is allowed to question the high priests and the faithful have no need for truth .

1. LEGO experiment link states vaguely that several feet of curvature is accounted for . Which model , pearshape or squashed ball or imaginary R= 6370km is used in this curvature correction? Could just be allowance for topography .
Sigh.
And this is the problem with debating anything with someone like you. You said
Surveyors take no account of ( cannot find ) any curvature over any area of 100sq. miles  because we live on a plane
You provided no basis for that claim or no supporting evidence, you just asserted it.
So I found a link about a large scale project which I imagined would probably have had to take account of the earth's curve and, sure enough, found some information on their website where they said they did have to do that.
This is the people who build the sodding thing saying they did indeed have to take the earth's curve into account because of the scale of it.
This is the point where you're supposed to concede the point and see the error of your ways but instead you're just saying "well, maybe it's this, maybe it's that". Again, with no basis or supporting evidence. It's easy to prove yourself right if you ignore or dismiss all evidence showing you to be wrong...

But WGS84 is a reference coordinate system system , not a globe models. Clearly explained here.
https://gisgeography.com/wgs84worldgeodeticsystem/
OP's original post shows how surveyors who believe the earth is a sphere ( he kindly quotes this in his 2nd post) use a set of equations including a constant of refraction k calculated with a value of R = 6370km . This is valid is on a perfect sphere only . We know this is not true whichever model we choose to believe .
WGS84 is an ellipsoid reference coordinate system applied to one of many geoids which in the words of that site leads to Geodesists to " believe the error is less than 2cm which is better than NAD83."
The use of that word " believe " leads me to conclude that we are encouraged to have faith in whatever they tell us. Science now has the appearance of religion in which no one is allowed to question the high priests and the faithful have no need for truth .
Dear somerled,
I find it amazing that you still think about R=6370km used by surveyors when displaying and communicating with each about refraction of light in the presence of thermal and humidity gradients. For a last time : they could have used as well the distance between Glasgow and London as a reference value or any other distance. The choice of a reference variable does NOT ever influence the calculations of the amount of refraction when expressed in terms of lengths and angles .
I realize that it is hard for you to accept that the lineofsight experiments conducted in the past without taking thermal and humidity ' gradients into account are inconclusive. Therefore, true to the method Zeteticism, we have to strive for more uptodate experiments, like a laser in a vaccum tube.
And for the record : I did not state that surveyors believe that the earth is a sphere. I used the term "round" which can be a sphere or an ellipsoid or a pearshaped entity. But, as far as earth is concerned the naked eye is not good enough to discern the difference. The model ellipsoid of GPS is a pretty good approximation to earth with a semimajor and semiminor axis of 6378137 meter and 6356752 meter, respectively. Choosing 6370 km as a reference length does seem to me a reasonable choice.
An ellipsoid, like the sphere, has no vertices. But you are correct in that the ellipsoid we are talking about is centered on the earth's center of gravity. You might want to study this web site https://www.state.nj.us/transportation/eng/documents/survey/Chapter4.shtm (https://www.state.nj.us/transportation/eng/documents/survey/Chapter4.shtm) . I also encourage you to have somebody who has a GPS receiver with the required accuracy for vertical distance, verify the Bedford Level Experiment. Wouldn't it be worthwhile to find the truth ?

1. LEGO experiment link states vaguely that several feet of curvature is accounted for . Which model , pearshape or squashed ball or imaginary R= 6370km is used in this curvature correction? Could just be allowance for topography .
Sigh.
And this is the problem with debating anything with someone like you. You said
Surveyors take no account of ( cannot find ) any curvature over any area of 100sq. miles  because we live on a plane
You provided no basis for that claim or no supporting evidence, you just asserted it.
So I found a link about a large scale project which I imagined would probably have had to take account of the earth's curve and, sure enough, found some information on their website where they said they did have to do that.
This is the people who build the sodding thing saying they did indeed have to take the earth's curve into account because of the scale of it.
This is the point where you're supposed to concede the point and see the error of your ways but instead you're just saying "well, maybe it's this, maybe it's that". Again, with no basis or supporting evidence. It's easy to prove yourself right if you ignore or dismiss all evidence showing you to be wrong...
You could have wiki'd that  https://en.wikipedia.org/wiki/Surveying
Go down to the part about " plane and geodetic surveying " . 100 square miles .
Then you can check out some survey books  the older real explanatory types are better . You will find that plane survey is used in areas up to 100 square miles because no curvature is found that does not fall within error limits of precision survey equipment . That's alot of missing curvature over a circular area of 100 square miles  80 feet or so .
So how much curvature over a few square miles did bob the ligo builder account for? , in exact terms . Vague statements mean nothing .
Zack  all the scientists have to do is measure , as exact as possible, the amount of curvature over the distance they use for their experiment . Measuring gradients of pressure , heat etc then applying imaginary values into their equations turns everything into mere ramblings.

You could have wiki'd that  https://en.wikipedia.org/wiki/Surveying
Go down to the part about " plane and geodetic surveying " . 100 square miles .
OK. It says this:
In geodetic surveying the curvature of the earth is taken into account while calculating reduced levels, angles, bearings and distances. This type of surveying is usually employed for large survey works. Survey works up to 100 square miles (260 square kilometers ) are treated as plane and beyond that are treated as geodetic
Right. So what this says that if the area being surveyed is less than 100 square miles  that's 10x10, not 100x100  they treat the area as a plane. Not because it is a plane, simply because over this distance the error is so small that it makes no difference for their purposes. 10 miles is 0.04% of the earth's circumference. The curve of 0.04% of a circle is negligible. Have a go at this:
https://www.mathsisfun.com/geometry/polygonsinteractive.html
You'll note that if you ramp up to a 20 sided polygon you're already getting to the point where it's starting to look a bit like a circle. Were the earth a regular polygon of 10 mile segments it would be a 2490 sided polygon. The angle between each side would be 0.145 degrees. It would be in all practical senses a sphere
Over lager distances it does start to make a difference so they use geodetic surveying. If the earth is flat then why is geodetic surveying a thing?
So how much curvature over a few square miles did bob the ligo builder account for? , in exact terms . Vague statements mean nothing.
1 meter over a length of 4km. Because the tunnel has to be straight but the earth curves. Again, were the earth flat then no accounting for this would have to be made.
TL;DR  it depends what you're doing. If you're making maps or whatever then over short spans you can think of the earth as flat, it doesn't make enough difference to cause errors. If you're drilling narrow tunnels and you need clear line of sight from one end to the other then you need to take the earth's curve into account.
It's weird that you provided the link that explains this and are pretending it says something else. ???

Only you mention an area of 10,000 square miles  I take it you are not familiar with mathematical terms describing areas .
You can use simple maths to work out the diameter of a circle of area 100 square miles ( not 100 miles square  they are different ) . Then you can use any curvature calculator to see that's about 80 feet of curvature that can be ignored across that diameter of 11 miles .Such a small amount can be ignored in survey. Fantastic .
Of course the 100 square mile limit is arbitrary as far as I can find out . A vague limit matched in vagueness by the LIGO "allowance" for curvature , 1m or several feet . In such a scientific undertaking I think everything deserves measurement to as exact a degree as possible .
There again since earth is flat that is possible .

Only you mention an area of 10,000 square miles  I take it you are not familiar with mathematical terms describing areas.
I was making sure that you are, I was just making explicit that 100 square miles is an area of 10 miles x 10 miles.
I explicitly said it was NOT 100x100 just in case you had confused "square miles" with "miles square".
You can use simple maths to work out the diameter of a circle of area 100 square miles ( not 100 miles square  they are different ) . Then you can use any curvature calculator to see that's about 80 feet of curvature that can be ignored across that diameter of 11 miles .Such a small amount can be ignored in survey. Fantastic.
Well, it depends what you're doing. If you're making a map then yes, the error is negligible and you can treat the earth as a plane on that scale.
For some engineering projects like LIGO then even over smaller distances the earth's curve has to be accounted for.
If the earth were flat then it wouldn't ever have to be considered and geodetic surveying wouldn't exist as it wouldn't need to.

Some good articles about the "true shape" of the earth and the problems of modelling this .
https://www.quora.com/Whatistheexactshapeoftheearth?redirected_qid=26177289&share=1
https://zeilon.squarespace.com/wellgroundedblog/2016/6/28/theshapeoftheearth
https://techinabottle.wordpress.com/2017/01/23/whatistherealshapeofearth/
Plenty of PhD in there  that's Phoctor of Dilosophy to you and I .
Begs the question of which model did bob the builder make use of use when allowing for curvature in LIGO. Interesting . Will we ever know?

Begs the question of which model did bob the builder make use of use when allowing for curvature in LIGO. Interesting . Will we ever know?
Here’s probably everything you need to know regarding the engineering, methodology, and ellipsoid reference points used in constructing LIGO:
Precision alignment of the LIGO 4 km arms using dualfrequency differential GPS
"The curvature of the Earth will cause the Earth's surface to deviate from the straight line propagated by light in vacuum by 1.25 meters over a 4 km path if the line starts out level with the surface. The alignment was, therefore, not the same as that for a level highway or pipeline.”
“The fundamental coordinate system for the alignment was the Earth ellipsoidal model WGS84"
https://dcc.ligo.org/public/0072/P000006/000/P000006A.pdf

Thanks for that link Stack . A lot to get through but it basically says the tubes are perfectly straight and level. Stand for correction though not read it all .
So far.
Uses GPS based ellipsoid model WSG84 coordinate system  converts to plane survey along with plumb line and bubble level equipment , nice to see that in there. Computer software is always susceptible to error .
The 1.25 m allowance for curvature is based on the ellipsoid model WSG84 coordinate system  which does not represent the true shape of earth as shown in previous links .
Will give a proper opinion when I've studied it  which may be some time ha .

Thanks for that link Stack . A lot to get through but it basically says the tubes are perfectly straight and level. Stand for correction though not read it all .
So far.
Uses GPS based ellipsoid model WSG84 coordinate system  converts to plane survey along with plumb line and bubble level equipment , nice to see that in there. Computer software is always susceptible to error .
The 1.25 m allowance for curvature is based on the ellipsoid model WSG84 coordinate system  which does not represent the true shape of earth as shown in previous links .
Will give a proper opinion when I've studied it  which may be some time ha .
I'm not sure what you mean by, "...WSG84 coordinate system  which does not represent the true shape of earth as shown in previous links." The point is, you were wondering what reference model was used by the LIGO engineers to account for the curvature of the earth. As stated in the paper, they relied on the WSG84 coordinate system which is a Spherical/Global/Ellipsoidal model, hence the accounting for the curvature of the earth.
If you claim they shouldn't have accounted for the curve of the earth and that they were essentially wasting their time in doing so, well, then that's between you and the LIGO engineers. But fact of the matter, they chose to use a spherical earth model to measure and construct the installations. That alone certainly doesn't prove the 'true shape of the earth', but I'm pretty sure their intent wasn't to do so in the first place anyway.

Did LIGO account for curvature as expected from the pear shaped earth or the oblate spheroid sphere ? That is what I am trying to find out . Which shape did Bob the builder account for ? Apparently neither since he used a ellipsoid model constructed on a mythical centre of the earth.
We know that WSG84 is nothing but a mathematical construct in which earths centre of gravity is the starting reference point for the coordinate system .
I mean read this quote from the conclusion on this link https://gisgeography.com/wgs84worldgeodeticsystem/
It states "Never before have we’ve been able to estimate the ellipsoid with such precision." Ho ho  estimate with precision .
Where is that centre of gravity on the pear shape /oblate model . Read the links I posted and you will see that is a big problem . There is no centre of gravity .
LIGO constructed its arms plane and level according to normal plane surveying techniques . That's my view .

The whole bumph depends on the assumption that earth is a sphere where R = 6370km
so use of an imaginary equation K=R/r to quantify any observation falls into the category of pseudoscience .
The introduction of 'K' is clearly just to compare the actual meaningful value 'r' to the globe model's 'R' in order to make the results more humanreadable.
You could say instead that when the light isn't refracted, r=infinity, and when it's refracted the same amount as the RE globe curves, it's r=6,371km.
No information is gained or lost here  it's just easier for a person to parse a value of K that's usually 10<K<10 rather than using r, which is [infinity]>r>~0

Did LIGO account for curvature as expected from the pear shaped earth or the oblate spheroid sphere ? That is what I am trying to find out . Which shape did Bob the builder account for ? Apparently neither since he used a ellipsoid model constructed on a mythical centre of the earth.
An oblate spheroid is a surface of revolution obtained by rotating an ellipse about its minor axis, so it is a form of ellipsoid and indeed it is sometimes called an oblate ellipsoid. They mean the same thing in this case.
We know that WSG84 is nothing but a mathematical construct in which earths centre of gravity is the starting reference point for the coordinate system .
Yes of course. It's the closest approximation we have for the true shape of the Earth which is also a simply defined mathematical shape. Its accuracy varies depending on where you are, but is perfectly acceptable for GPS etc. For many purposes, a simple sphere would do, but this is just a better approximation.
I mean read this quote from the conclusion on this link https://gisgeography.com/wgs84worldgeodeticsystem/
It states "Never before have we’ve been able to estimate the ellipsoid with such precision." Ho ho  estimate with precision .
Every physical real world measurement is an approximation within acceptable tolerances, so yeah, estimate with precision. Parts for racing engines are made with high precision, but they're never exact, they are either within tolerance or not.
Where is that centre of gravity on the pear shape /oblate model . Read the links I posted and you will see that is a big problem . There is no centre of gravity .
LIGO constructed its arms plane and level according to normal plane surveying techniques . That's my view .
Well except that the people who built it are telling you they used WGS84. You have the evidence, you quote from the source. Do you accept it?

Did LIGO account for curvature as expected from the pear shaped earth or the oblate spheroid sphere ? That is what I am trying to find out . Which shape did Bob the builder account for ? Apparently neither since he used a ellipsoid model constructed on a mythical centre of the earth.
We know that WSG84 is nothing but a mathematical construct in which earths centre of gravity is the starting reference point for the coordinate system .
I mean read this quote from the conclusion on this link https://gisgeography.com/wgs84worldgeodeticsystem/
It states "Never before have we’ve been able to estimate the ellipsoid with such precision." Ho ho  estimate with precision .
Where is that centre of gravity on the pear shape /oblate model . Read the links I posted and you will see that is a big problem . There is no centre of gravity .
LIGO constructed its arms plane and level according to normal plane surveying techniques . That's my view .
This may be your view but it is not the view of the Engineers that designed/built LIGO as spelled out in the paper:
"A straight line in space varies in ellipsoidal height by ~1.25 m over a 4 km baseline. At each of the fiducial points, the design ellipsoidal height of the beam tube centerline was calculated using the WGS84 model with the latitude and longitude as inputs. These heights were used to perform preliminary alignment of the tube sections as the supports were installed during beam tube fabrication on the slab."
So you can basically say one of the following:
 They lied and didn't really survey and construct LIGO taking into consideration a curvature of the earth because there is none
 They took into account a supposed curvature of the earth when they didn't have to because there is none

I still don't get how FEs interpret refraction and how they model it. Rowbotham clearly ignored it, but he doesn't represent all FEs. As far as I can't see the current state of this discussion is that the current theory of refraction assumes a Globe Earth in it, even though those assumptions could be "theoretically" ignored. But still this doesn't rule out that the Bedford canal apparent flatness could be well due to refraction, and this fits nicely with the common idea of FEs to ignore visual proofs.