[Zack Bimmel]

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


It is well-known 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 20-30 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 line-of-sight 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 20-30m 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 line-of-sight 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 8-14
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

[3] Fraser, C.S
"Atmospheric Refraction Compensation in terrestrial Photogrammetry"
Photogrammatic Engineering and Remote Sensing, Vol 45, No.9, September 1979, pp.1281-1288
https://www.asprs.org/wp-content/uploads/pers/1979journal/sep/1979_sep_1281-1288.pdf

[somerled]

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 .

[Zack Bimmel]

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 ?

[Tom Bishop]

It is well-known 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]

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 .

[Bikini Polaris]

It is well-known 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.sacred-texts.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?

[Tom Bishop]

Rowbotham does reference some tests with barometers in Experiment 9.

Experiment 2 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.



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.

[Zack Bimmel]

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 ( 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 laser-based experiment inside a vacuum-filled, 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

[Bikini Polaris]

Rowbotham does reference some tests with barometers in Experiment 9.

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 alike--the temperature, density, and moisture of the air did not differ at the two stations at the time the experiment with the telescope and flag-staff 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 round-earth-with-atmosphere 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/curve-calc/?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.

[somerled]

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 ( 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 laser-based experiment inside a vacuum-filled, 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 . 



     

[Zack Bimmel]

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 linked-to 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

[somerled]

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 .

[Zack Bimmel]

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.

[somerled]

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 .

[Zack Bimmel]

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.

[somerled]

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 .

[Bikini Polaris]

Then lead away , I am interested to see the logical steps .

Let me try that:
 
Fact 1: Atmo-something refraction due to temperature gradient exists, proof: mirages as in the following figure:



Fact 2: Atmo-something 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 atmo-something refraction.

[somerled]

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 .

[DaveP]

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. 

[Bikini Polaris]

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.