Sandor Szekely

Laser curvature test on lake Balaton

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Dear Visitors

we would like to introduce you our laser curvature experiments.

We are looking for the curvature on the surface of the lake Balaton in Hungary with special laser technology to detect the supposed curvature on the water surface.

How much is the curvature of earth on the distance of 77kms? (47.8 miles)
465 meters! (1526 feet)
that means that an object this tall in one side of the lake must not be seen from the other side.

Online curvature calculator:
https://dizzib.github.io/earth/curve-calc/?d0=47.845581802249&h0=0&unit=imperial

But is that assumption of GE curved water surface model true?

We have conducted a series of experiments and surveyed Lake Balaton in western Hungary, which is a shallow (average depth 3.3 m), large (597 km2 ) area. To what extent does the lake resemble a level surface?

The airborn LIDAR laser scanner measurement in 2013

post-120902-0-23645600-1473549296_thumb.jpg

Figure 2. (a) Map of lidar-measured ellipsoidal water surface heights (inside flight strip outlines) and modeled gravity potential isosurface height (background raster) with the same elevation color scheme. Inset shows detailed height distribution of water surface measurements, histogram refers to lidar measurements (not to gravity potential isosurface model). (b) Map of lidar-derived normal water surface heights, obtained by correction with local deviation of LMLL and local quasi-geoid height. Histogram refers to water surface normal heights.

On Lake Balaton, currents are driven by wind forcing and by seiche while thermal convection remains hardly observable (Muszkalay, 1973). The flow of water from the tributaries towards the outlet is also very weak, because significantly more water leaves the lake through surface evaporation than through the outflow.

The rule of thumb for air pressure response is 1 cm decrease in water level for 1 mbar increase in air pressure (Ponte and Gaspar, 1999). However, this only applies to the open ocean and after a period of 1–2 weeks. Since the lake was very shallow, most of the air pressure is directly forwarded to the lake bottom. Therefore we assume that air pressure changes could not have influenced the local lake level

Lake tides are known to have amplitudes of up to 10 cm in larger lakes (Trebitz, 2006). In case of Lake Balaton, the shallow depth and the relatively low water volume of the lake suggest this effect would be even smaller.

Due to its fresh water, shallow depth (3.3 m avg), limited in- flow and small water volume (2 km3 ) compared to its surface (600 km2 ), Lake Balaton experiences very limited variations in water density. There is no thermocline, warming and cooling can only create maximum vertical temperature gradients of up to 1.1 ◦C m−2 , and horizontal differences in surface temperature between basins have been documented to remain less than 3 ◦C (Virág, 1998). The salinity of the water is also quite constant over its area and over time, so no major changes in density are expected. The height effect of local warming or cooling can dissipate relatively quickly on this scale, therefore no contribution of density change to lake topography can be expected.

Both in the cases of the 2013 LIDAR experiment and the 2016 laser experiment confirmed that standing water has a truly level surface.


The result of the new 6kms laser measurement are shown here in the data sheet.

Note: that the GE model curved water surface expected laser beam height does not match the observed height.
The accuracy of the test is negligable to the difference of the calculated and measured value.

post-120902-0-37034200-1473550937_thumb.png

Balaton-laser-exp-816-data-sheet-4th-measurement final.pdf



OUR NEW 6kms laser measurement:

 

youtube links deleted

 

The airborn LIDAR measurement
http://publik.tuwien.ac.at/files/PubDat_228814.pdf

 

The pre-test video


post-120902-0-21770400-1473543698_thumb.jpg


Here is an other new video from Mr Thrive and Survive youtube channel - talking about our new 6kms laser measurement:


Please ask questions

Edited by Sandor Szekely

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Rule 2.7 forbids you from posting to advertise your youtube channel or post without sufficient supporting text.

 

Advertising and spam is prohibited. We don't mind if you put a link to your noncommercial site (e.g. a blog) in your signature and/or profile, but don't go around making threads to advertise it. Links, pictures and videos in posts should be relevant to the discussion, and members should be able to participate in the discussion without clicking any links or watching any videos. Videos and pictures should be accompanied by enough text to set the tone for the discussion, and should not be posted alone.

 

You have failed to provide sufficient information to accompany the videos. The links have been removed. Post the details of the experiment, and then you can post a video to accompany them

 

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Seems like an interesting subject, but I fear that this thread will soon be closed as it is in breach of forum rules.

 

This would be a shame as it is a subject I have personal experience of.

 

So please consult with the moderators about making this a regular thread for discussion.

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Seems like an interesting subject, but I fear that this thread will soon be closed as it is in breach of forum rules.

 

This would be a shame as it is a subject I have personal experience of.

 

So please consult with the moderators about making this a regular thread for discussion.

Thank you Studiot I will do so as the links were taken out of the thread.

I will include now the text version to the thread - can I do that?

 

 

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Thank you Studiot I will do so as the links were taken out of the thread.

I will include now the text version to the thread - can I do that?

I thought I was clear: you must do that.

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Please run me through your curvature calculation, as my rough and ready calculator makes it under 150 metres.

 

Here are my calcs.

 

Grazing height = deviation of a circular curve from straight = (distance along straight)2 / (2 * Radius)

 

Grazing height will occur at semi distance = 77/2 = 38.5km

Radius = 6400km

 

So grazing height = 38.52 / 12800 = 0.116km = 116m

 

Edit correct typing in calcs.

 

***************************************************************************************************************************************************

 

Moderators

 

Why don't I have permission to download the pdf?

 

Is this another piece of windows 10 s__t ?

 

********************************************************************************************************************************************************************

 

 

Thank you for extending the post to include some figures.

 

If the lake is less than 5m average depth then it is a very thin layer of water in comparison to the curvature of the Earth which is up to two orders of magnitude greater.

 

In other words the lake is a surface film.

 

Furthermore the bottom must also follow this profile.

 

Since the underlying land is nearly planar, why would you expect a surface layer of water to be anything else?

Edited by studiot

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Please run me through your curvature calculation, as my rough and ready calculator makes it under 150 metres.

 

Here are my calcs.

 

Grazing height = deviation of a circular curve from straight = (distance along straight)2 / (2 * Radius)

 

Grazing height will occur at semi distance = 77/2 = 38.5km

Radius = 6400km

 

So grazing height = 38.52 / 12800 = 0.116km = 116m

 

Edit correct typing in calcs.

 

 

 

Moderators

 

Why don't I have permission to download the pdf?

 

Is this another piece of windows 10 s__t ?

Studiot, your calculation is the saggita as I see.

 

The question here is: what is the drop of curvature over the 77kms distance. This means how tall object is in level to the original starting position on the curved water surface model.

 

You gave answer to the how high should the water bumb be in the middle. In an other way how deep the laser should go into the water to reach that height at the other shore.

 

We use autocad to calculate the exact drop over distance, but most of the online calculators give the same result.

I have a math from the Hungarian ELTE university:

 

h = (0.28 * s)'2

where

h = height in meters

s = distance in kilometers

 

We made this calculator for imperial units:

 

post-120902-0-89629100-1473551719_thumb.jpg

 

Please run me through your curvature calculation, as my rough and ready calculator makes it under 150 metres.

 

Here are my calcs.

 

Grazing height = deviation of a circular curve from straight = (distance along straight)2 / (2 * Radius)

 

Grazing height will occur at semi distance = 77/2 = 38.5km

Radius = 6400km

 

So grazing height = 38.52 / 12800 = 0.116km = 116m

 

Edit correct typing in calcs.

 

***************************************************************************************************************************************************

 

Moderators

 

Why don't I have permission to download the pdf?

 

Is this another piece of windows 10 s__t ?

 

********************************************************************************************************************************************************************

 

 

Thank you for extending the post to include some figures.

 

If the lake is less than 5m average depth then it is a very thin layer of water in comparison to the curvature of the Earth which is up to two orders of magnitude greater.

 

In other words the lake is a surface film.

 

Furthermore the bottom must also follow this profile.

 

Since the underlying land is nearly planar, why would you expect a surface layer of water to be anything else?

You are right by saying that the lake is indeed a thin surface film, and therefore it should follow the curvature of the earth as it has no depth to compensate with as the water takes the form of it's container.

 

We should be measuring the expected curvature on the water surface.

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Yes my grazing line was centered between the two shores, as should any be reference straight line.

 

Referring to the intervisibility of points on opposite shores is in my opinion a bit of theatrical dramatics since it assumes one observers eyes are in his toes (ie at ground level).

 

But yes, such an observer would need a 400+ metre giant to see on the other side.

 

 

 

However I have struggled with those idiot programmers to add another more pertinent comment at the end of my last post concerning the depth of water.

This is really what we need to discuss since it means that the land surface cannot have the required curvature either.

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Yes my grazing line was centered between the two shores, as should any be reference straight line.

 

Referring to the intervisibility of points on opposite shores is in my opinion a bit of theatrical dramatics since it assumes one observers eyes are in his toes (ie at ground level).

 

But yes, such an observer would need a 400+ metre giant to see on the other side.

 

 

 

However I have struggled with those idiot programmers to add another more pertinent comment at the end of my last post concerning the depth of water.

This is really what we need to discuss since it means that the land surface cannot have the required curvature either.

Well it's not our intention to use "a bit of theatrical dramatics" :) but this is the correct way to calculate the expected height of the laser beam on the opposite shore.

 

We calculate the expected laser height as:

 

Drop over distance + original height of the laser.

 

We use a slope correction method at the leveling distance of 720 meters to make a true horizontal level to a 0.001 degree accuracy at the position of the laser on the shore. We do tht by adding a 4 cms to the original 1.25 cms laser height to set the beam level on the boat, as the GE curved surface model has a 4cms drop in this 720 meters distance.

 

This is the method and setup that we used at the 4th measurement of the 2016 aug 16 experiment.

 

Here I show an example of our 14.3 miles distance calculation (upcoming experiment)

 

The expected height of the laser beam shall be at 139.7 feet on the GE curved surface model.

On the GE curved surface model the beam must not be detectable below 97 feet in any possible laser leveling method.

 

 

post-120902-0-93367100-1473554203_thumb.jpg

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Well it's not our intention to use "a bit of theatrical dramatics" :) but this is the correct way to calculate the expected height of the laser beam on the opposite shore.

 

We calculate the expected laser height as:

 

Drop over distance + original height of the laser.

 

We use a slope correction method at the leveling distance of 720 meters to make a true horizontal level to a 0.001 degree accuracy at the position of the laser on the shore. We do tht by adding a 4 cms to the original 1.25 cms laser height to set the beam level on the boat, as the GE curved surface model has a 4cms drop in this 720 meters distance.

 

This is the method and setup that we used at the 4th measurement of the 2016 aug 16 experiment.

 

Here I show an example of our 14.3 miles distance calculation (upcoming experiment)

 

The expected height of the laser beam shall be at 139.7 feet on the GE curved surface model.

On the GE curved surface model the beam must not be detectable below 97 feet in any possible laser leveling method.

 

 

attachicon.gif13517593_1154676367952041_9076900752493008852_o.jpg

From your diagram I get the feeling that Heights & distances are not exactly in the correct direction.

For example, the 3.3 ft on the right should be on the earth's radius. I wonder for other distances (as the 14.3 miles for example, maybe is is an arc and not a straight line, or a straight line but not in this direction).

Guessing.

Also the point of tangent is very difficult to catch in practice. The lake surface is a mirror.

-------------

edit: and your sketch is too much unscaled. If the height on the right is 3.3 ft and on the left 97.2 ft, it means the sketch is not symmetric, the point of tangent is very much on the right and therefore will produce a large measurement error.

------------------

edit2. Notice that the errors (the red numbers in your table post#1) start approx at the distance to the tangent. Before that point everything seems OK.

Edited by michel123456

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As far as I can tell from a brief skim, nobody in this thread has mentioned the variation of refractive index with height.

That make light bend down slightly as it goes through the air and throws the calculation out slightly.

 

https://en.wikipedia.org/wiki/Bedford_Level_experiment#Refraction

 

It seems you are in danger of repeating a 19th century mistake.

Edited by John Cuthber

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Refraction in this case would have curved the beam upwards.


Michael, the 3.3ft on the right is correctly perpendicular to the surface. 14.3 looks like a straight line to me. Why do you think it's curved?

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Refraction in this case would have curved the beam upwards.

 

 

Can you show how that is calculated? And how large the effect is.

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Thank you for your post#9 which begins to describe your methodology, but more is needed.

 

As I understand it you are extrapolating, rather than interpolating your corrections.

 

I further understand you are applying standard corrections at the end of a 720m observation line, although I am not familiar with the GE model.

 

Standard close to the ground (water) surface observations like these are subject to curvature and refraction corrections.

 

Yes I agree with 4cm for the curvature (I make it 41 mm).

 

John is correct in noting that there is also a refraction correction, that I was coming to as it is more minor.

 

The standard refraction correction, is equal to one seventh of the standard curvature correction ie 6mm.

 

This make a total correction over the observed distance of 34mm.

 

Both of these would be eliminated by suitable interpolation methodology.

 

To return to the implications I mentioned earlier,

 

Using standard curvature and refraction corrections it would imply a bulge (see attachment) in the Earth land surface in this area, which should be detectable by land surveying techniques, as opposed to hydrographic ones.

This is entirely feasible since most of the lake is only a couple of metres deep old fashioned 10 foot survey towers could even be used.

 

Alternatively

(1) There could be an undisclosed error in the methodology

 

(2) There could have been freak atmospheric effects causing effect at the time of measurement.

 

post-74263-0-19717400-1473592277_thumb.jpg

Edited by studiot

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I can't show the calculation but from the way I understand it, refraction goes in the direction of higher density and with the measured lake temp higher than the air above, means it would have curved up since cooler air is denser. Sandor will probably post the calculation.

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Refraction in this case would have curved the beam upwards.

How?

If the air nearer the ground was warmer it would rise rather than sit there covered by cold air. The same is true for humid air which is less dense than dry air.

 

If there were strong thermal currents when you made the measurements, that's another complication.

 

Were there air temperature measurements at different heights?

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From your diagram I get the feeling that Heights & distances are not exactly in the correct direction.

For example, the 3.3 ft on the right should be on the earth's radius. I wonder for other distances (as the 14.3 miles for example, maybe is is an arc and not a straight line, or a straight line but not in this direction).

Guessing.

Also the point of tangent is very difficult to catch in practice. The lake surface is a mirror.

-------------

edit: and your sketch is too much unscaled. If the height on the right is 3.3 ft and on the left 97.2 ft, it means the sketch is not symmetric, the point of tangent is very much on the right and therefore will produce a large measurement error.

------------------

edit2. Notice that the errors (the red numbers in your table post#1) start approx at the distance to the tangent. Before that point everything seems OK.

Hi Michel,

I can make a quick video in AutoCAD showing you how to calculate the curvature. But I highly recommend watching the Laser Test carefully, There is scene where the curvature's calculations were explained. Take a look at them if they don't match your calculations, I can make a video about that and show you how correct they are.

 

Thanks

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It should be remembered that the observation lines we are talking about are not tens of kilometres long with associated hundres of metres of correction.

 

This as I said, was the overdramatic headline stuff.

 

The real lines appear to be a distributed network of much shorter lines about 1 metre from the surface and the deviations from this are a few tens of millimetres.

 

(This by the way is the right way to go about this surveying exercise)

 

But we need more information about this methodology and also about any actual land surface survey.

Edited by studiot

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Thank you for your post#9 which begins to describe your methodology, but more is needed.

 

As I understand it you are extrapolating, rather than interpolating your corrections.

 

I further understand you are applying standard corrections at the end of a 720m observation line, although I am not familiar with the GE model.

 

Standard close to the ground (water) surface observations like these are subject to curvature and refraction corrections.

 

Yes I agree with 4cm for the curvature (I make it 41 mm).

 

John is correct in noting that there is also a refraction correction, that I was coming to as it is more minor.

 

The standard refraction correction, is equal to one seventh of the standard curvature correction ie 6mm.

 

This make a total correction over the observed distance of 34mm.

 

Both of these would be eliminated by suitable interpolation methodology.

 

To return to the implications I mentioned earlier,

 

Using standard curvature and refraction corrections it would imply a bulge (see attachment) in the Earth land surface in this area, which should be detectable by land surveying techniques, as opposed to hydrographic ones.

This is entirely feasible since most of the lake is only a couple of metres deep old fashioned 10 foot survey towers could even be used.

 

Alternatively

(1) There could be an undisclosed error in the methodology

 

(2) There could have been freak atmospheric effects causing effect at the time of measurement.

 

attachicon.gifobs1.jpg

 

Hello John nice to meet you

Actually we made a complete video explanation of our measurements that includes all the correlating subjects - like refraction - but not allowed to quote it here.
This makes me explain my points much more harder and I have this post only (max 5 replies) - well kinda hard to debate the subject.
I will try to answer the upcoming questions by editing this reply.

Jon: "That make light bend down slightly as it goes through the air and throws the calculation out slightly"
Refraction is not bending light downwards in our conditions, but UPwards - as the lake temperature is warmer (22C) than the air above (17C).

We choose all our measurement timeframes in such ambient conditions where the refraction effect is bending the laser beam upwards, so this would make the refraction corrected height even worse for the GE model.
Indeed we did not experience any significant refraction of the laser beam in any of our 4 measurements at the last experiment.

The standard value of terresterial refraction in normal conditions (daytime) is 0.4 millimeter / kilometer in geodezy.

The refraction of the beam was not detected and it must have been within our measurement accuracy. (waves, measurement precision)
The GE curved water surface and the FE flat water surface model has a much more significant difference in expected height difference than our measurement accuracy.

post-120902-0-30140000-1473595778_thumb.png


post-120902-0-54852600-1473599173_thumb.jpg

John: "It seems you are in danger of repeating a 19th century mistake."
Are you refering here to the Robowham or the Wallace experiment?
We had our laser at 1.25 meter (4.1 feet) at the last measurement - that supposed to be over the N.U.D.T.Z. ( non uniform density transition zone)


Studiot: "As I understand it you are extrapolating, rather than interpolating your corrections."
"Both of these would be eliminated by suitable interpolation methodology."
Please explain this to me in more details. We are continuing our measurements over the lake and land surface as well, so we can fine tune the measurements in the future. Now we had our measurements taken by photographic evidence and evaluated by photo analysis. This gives us a pretty accurate reading for us way within the accuracy needed to make a conclusive measurement.

This time we did not have the possibility to use geodezic equipment as you suggested, but we will do in the upcoming ones.

 

"Using standard curvature and refraction corrections it would imply a bulge "
As I wrote to John, the refraction of the laser beam would be a reversed effect to curve the beam upwards, and not downwards. In that case we should have even a greater difference in the models and would make the GE calculation worse.
We had the laser beam at 1.8 meters height above water surface at the distance of 6kms.
If refraction would account for let's say 0.3 meters upwards curve than that would make the measurement 1.5 meters above water level.

This was not the case here, the FE flat water model just fits perfectly with the 0.005 degree slope correction applied to the laser leveling. We had the laser beam height just as expected on the truly level water surface.

This is exactly the same outcome as the airborn LIDAR experiment.


Studiot:"It should be remembered that the observation lines we are talking about are not tens of kilometres long with associated hundres of metres of correction.

This as I said, was the overdaramtic headline stuff."

The hypothesis of GE curved surface water model states that the curvature drop is 465 meters on the 77kms total width of the lake Balaton.
We will be probably able to measure this complete distance in the winter time when the lake is frozen.

This is a picture of the lake frozen - this should have a 465 meters curve on the surface:


post-120902-0-81878200-1473599307_thumb.jpg



AS a new member I have a limit of 5 comments in the first 48 hours... still bout 12 hours to go!
(this is insane by the way... not to be able to comment correctly on my own thread)

Sorry I had a lost message just before, when I restored it from saved it messed up the quote signs maybe... I had to retype the half of the message again.

Studiot: "I am interested in you methodology because of your apparent claim that the water surface is flat over a distance of 77km, yet is only a couple of metres deep.

Therefore the lake bed under the water must also exhibit this characteristic.

We have both agree that the deviation from a spherical surface over this distance is of the order of several hundred metres.

Since a 3m layer of water apparently exhibits this, the lake bed underneath it must also exhibit this.

Hence the bulge.

I have repeatedly asked you to clarify this point and you have repeatedly avoided it."

Studiot please clarify what should I clarify here? :)
The hypothesis of GE curved water surface model is 465 meters over the 77kms distance as we agree.
The measurements were taken upto 6kms distance and we found no curvature drop yet.
The LIDAR experiment had the same outcome: "
the two data systems confirmed that standing water has a truly level surface" (page 8/15 in LIDAR pdf)

Studiot: "
Arguing about whether a minor correction of a few millimeters to several hundred metres is upwards or downwards is irrelevant."
I aggree with this absolutely. Refraction of the laser beam played no significant role in the measurement.

Sorry to everyone, until I do not have my ban lifted I can not answer any comments.
It may be the problem with my account by I had this message yesterday:

AS I am also not alllowed to edit my comment here many times - I have to wat 12 hours from now.
Admin, if you can please do something about it.


post-120902-0-75915200-1473600909_thumb.png

Edited by Sandor Szekely

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This makes me explain my points much more harder and I have this post only (max 5 replies) - well kinda hard to debate the subject.

Studiot: "As I understand it you are extrapolating, rather than interpolating your corrections."

"Both of these would be eliminated by suitable interpolation methodology."

Please explain this to me in more details. We are continuing our measurements over the lake and land surface as well, so we can fine tune the measurements in the future. Now we had our measurements taken by photographic evidence and evaluated by photo analysis. This gives us a pretty accurate reading for us way within the accuracy needed to make a conclusive measurement.

This time we did not have the possibility to use geodezic equipment as you suggested, but we will do in the upcoming ones.

 

"Using standard curvature and refraction corrections it would imply a bulge "

As I wrote to John, the refraction of the laser beam would be a reversed effect to curve the beam upwards, and not downwards. In that case we should have even a greater difference in the models and would make the GE calculation worse.

We had the laser beam at 1.8 meters height above water surface at the distance of 6kms.

If refraction would account for let's say 0.3 meters upwards curve than that would make the measurement 1.5 meters above water level.

 

This makes me explain my points much more harder and I have this post only (max 5 replies) - well kinda hard to debate the subject. This was not the case here, the FE flat water model just fits perfectly with the 0.005 degree slope correction applied to the laser leveling. We had the laser beam height just as expected on the truly level water surface.

This is exactly the same outcome as the airborn LIDAR experiment.

 

Please separate quotes and answers as here when replying.

It is much more difficult for all otherwise.

 

As far as I know there is no limit on the number of (sensible within the rules) replies you may make on this site, certainly it is far in excess of 5.

 

I am interested in you methodology because of your apparent claim that the water surface is flat over a distance of 77km, yet is only a couple of metres deep.

 

Therefore the lake bed under the water must also exhibit this characteristic.

 

We have both agree that the deviation from a spherical surface over this distance is of the order of several hundred metres.

 

Since a 3m layer of water apparently exhibits this, the lake bed underneath it must also exhibit this.

 

Hence the bulge.

 

I have repeatedly asked you to clarify this point and you have repeatedly avoided it.

 

Arguing about whether a minor correction of a few millimeters to several hundred metres is upwards or downwards is irrelevant.

Edited by studiot

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Actually we made a complete video explanation of our measurements that includes all the correlating subjects - like refraction - but not allowed to quote it here.

!

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No, that's not true. You have been asked to quote it. What you can't do is just link to it with nothing else posted.

 

This makes me explain my points much more harder and I have this post only (max 5 replies) - well kinda hard to debate the subject.

I will try to answer the upcoming questions by editing this reply.

!

Moderator Note

I don't know why you think you can only make 5 replies. That's not a restriction of this forum. Your ability to edit will expire after a short time.

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Sandor, you truly are a magician.

 

You managed to quote my post#20 in your post#19

 

:eek:

 

It really is wonderful that you can answer my questions, before I ask them.

 

Nevertheless I thought that both my diagram in post#14 and subsequent repeats of the question were plain enough.

 

What is the shape of a surface 3m below one you claim is dead flat, if it is not also dead flat?

Edited by studiot

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Thank you for your post#9 which begins to describe your methodology, but more is needed.

 

 

There is a large button, clearly labelled "Quote" beneath each post. You can use that to make it clear when you are quoting someone else (as here). Otherwise it becomes nearly impossible to follow the discussion.

 

I assume (as it has been completely obvious for many millennia that the world is round) that you are simply trying to understand why your experiment produces the wrong results.

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There is a large button, clearly labelled "Quote" beneath each post. You can use that to make it clear when you are quoting someone else (as here). Otherwise it becomes nearly impossible to follow the discussion.

 

I assume (as it has been completely obvious for many millennia that the world is round) that you are simply trying to understand why your experiment produces the wrong results.

I begin to wonder.

 

post-19758-0-28031200-1473604820_thumb.jpg

What is that last column about?(enhanced in red) If the Earth was flat, why are the values diverging? If the laser beam is send horizontally from 1.26m high, all the values should be the same 1.26m.

--------------------

(edit)

Now I don't wonder anymore after googling "Lake Balaton laser experiment", it appears that you have tried this on another forum.

Who is behind this?

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I begin to wonder.

 

attachicon.gifpost-120902-0-37034200-1473550937.jpg

What is that last column about?(enhanced in red) If the Earth was flat, why are the values diverging? If the laser beam is send horizontally from 1.26m high, all the values should be the same 1.26m.

--------------------

(edit)

Now I don't wonder anymore after googling "Lake Balaton laser experiment", it appears that you have tried this on another forum.

Who is behind this?

The laser was leveled using the boat so at 720 meters on the GE model, the beam should have risen from the surface from 1.25 starting height to 1.29. The ended up setting it to 1.32 because moving it in .01 adjustments proved difficult without an appropriate adjustable mount. What they were looking for was either a curve in the data plot against distance or an updward curve in the height data. What the test showed was a straight ascending line. The video shows the leveling at 9 minutes. https://youtu.be/GBhDFO4NMrw?t=9m

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