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Laser curvature test on lake Balaton


Sandor Szekely
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equipment datasheets are definitely handy. I can also use them to look for some of these affects mentioned. The effect of the elevator duct varies depending on equipment frequencies. Glad to see someone that can help double check the equations involved for this application. I'm going down memory lane on everything involved, in terms of refraction, scatter etc.

 

 

The more I think of this test, the more I keep thinking that two emitters may be a good idea. I keep looking at the pros and cons involved. Assuming parallel transport placement. Safety issues addressed (lol). KEY NOTE use two distinct frequencies.

 

 

 

PROS.

 

1) detection in deviation in parallel transport can be used to find changes in refractive index.

2) if the two lasers are mounted in tandem vertically we can use geometry relations between the two beams to assist in leveling. Combined with a grid backboard we can break the leveling aspects down to straightforward trig. This will help correct leveling errors in 3D if done correctly

3) the changes in frequency and loss of parallel transport can be used to calculate the refraction index. (this in principle is used in equipment today. Usually through a known sample) however a known sample isn't needed just more complex. If not.

 

CONS

1) added equipment and complexity, adds to potential systematic errors

2) greater number of control point data to collect. (measurements between the two laser beams for example, as well as receiving frequencies.)

 

Even without worrying about refraction, etc the advantages in PRO 2 is in my opinion worth the added cons

 

The above may work better if the two lasers were aimed to a common point. Forming two sides of our triangle, this can help in leveling and distance from emitter calcs.

 

(KEY NOTE do not mount the two lasers onto the same mount backboard. We don't want to add material expansion/contraction problems)

Hello Mordred

 

We did think about using dual colour laser (and used it on the pre-test) to calculate reftraction by the difference of the blue / green beam. The cons are not a problem, indeed I'd like to add more types of equipments to improve the measurement accuracy.

 

Actually we have only one problem with this: we have only 1 collimator that is on the green laser. We might be able to get an other one too.

 

Could you please explain this in greater details?

 

"The above may work better if the two lasers were aimed to a common point. Forming two sides of our triangle, this can help in leveling and distance from emitter calcs.

(KEY NOTE do not mount the two lasers onto the same mount backboard. We don't want to add material expansion/contraction problems)"

 

 

Did someone say that this is one of those laser projecters used for creating vector images through a series of mirrors? Could it be that it is always being reflected and that one of the mirrors is wobbling/vibrating?

no, wedo not have laser scanner (mirrors) in this laser unit.

 

As I pointed out, the horizontal vibrating effect was due to the laser beam hitting the water surface at the 3rd measurement (reflexion)

 

I think these extracts show the confusion between a flat surface and a level surface.

 

A level surface is most definitely curved, a flat surface is essentially planar.

 

So how can a flat water model be consistent with laser or any other leveling?

I think there is a confusion on the meaning of "level". Here are some definitions, copy Nr1 here:

 

"Level \Lev"el\ (l[e^]v"[e^]l), a.

1. Even; flat; having no part higher than another; having, or

conforming to, the curvature which belongs to the

undisturbed liquid parts of the earth's surface; as, a

level field; level ground; the level surface of a pond or

lake.

[1913 Webster]"

 

http://onlinedictionary.datasegment.com/word/level+surface

I've not watched the video. I did fear this might be the case. You can get "bright" spots off of thr beam path, and you'll be killed by any beam divergence or laser wobble. I don't really think this experiment is within the reach of a hobbyist. Although it looks simple as soon as you start actually thinking about it there are so many subtleties that will make your results meaningless.

 

That's a terrible way to measure. You could do something similar with a camera sensor but where you measure the intensity from each pixel. But you'd need a large area CMOS or ccd and some clever way of removing ambient light.

Okay so you have NOT watched the video? then what are we arguing about?

 

"not within the reach of hobbist"... come on.. please be more serious in commenting the experiment!

I watched the all 27 minutes of the video yesterday.

And honestly must say, you spend the more time flying helicopter and playing volleyball than thinking and preparing for gathering data during the experiment.

From your's precalculations of globe curvature there is visible that from 1.25m it will go to 4.32m,

and what size of "target" you were using?! That's has barely 2m width and 1m height.

It makes completely no sense to me at all.. ?!

(If I would be expecting melting some metal at 1000 C, would I use container that has melting temperature at 200 C?)

Additionally target was mounted at angle.

IMHO you should mount it as perpendicular to the ground of boat as possible using spirit level..

 

If you would be seriously going to take this experiment, you should print grid on paper with 1 cm x 1 cm spacing,

with some 10cm x 10 cm with slightly different background color (and/or chessboard), with printed numbers on side.

With size 2m width, and 5m height. To be able catch the all laser positions on it.

Mount it steadily to the boat using spirit level to be sure angles.

Mount steadily camera (better 4K,60 FPS+) in the front of target, and filming it all the time, without breaks, and as fast as possible, to have weather influence as small as possible.

(target, boat and recording camera should be each other not moveable during the entire experiment)

Captain should be controlling boat looking at laser spot, and trying to keep it in the center of target. Don't do sharp movements.

Reading results do in hotel room, looking at where is spot on target with grid, on the video recorded by camera.

Nobody serious is doing measurements in the middle of experiment, instead store data by electronics devices.

(The guy using tape measure, I could not believe my eyes.. completely unprepared people for experiment.. you looked like the first time in life doing physical experiment)

 

And obviously REPEAT it several times.

 

Additionally this "flat earth" mentioned everywhere in the video.

It will obviously put you in "crackpot zone" if you will be going to discuss experiment with other scientists.

I suppose so it's just there to catch attention to YouTube video, to earn money on advertisements from Google.

Scientists seeking for the truth don't behave like that. At least should not.

You didn't confirm that Earth is flat, but that photons from laser flied curved trajectory, at most.

And identify source causing bending of the beam.

 

Flat Earth + photons going straight line will have the same (or similar) results as curved globe Earth + photons going curved path.

 

And obviously REPEAT it several times.

With different conditions, with different time of day, and different time of year..

Sensei, I did not answer this post because I found it unrelevant and unpolite.

you asked for it so I'll answer.

 

"And honestly must say, you spend the more time flying helicopter and playing volleyball than thinking and preparing for gathering data during the experiment.

 

what is this very first sentence? I'm usually not replying to intriques... helicopter was supposed to enter the measurement, but we had to cancel that due to the storm. On the 3rd (THIRD) day we made some test flights for the upcoming measurement over land.

We did not play volleyball... that is a videoclip from someone else... OMG... this is like facebook... I am here for scientific debate - I'll leave all coments like this unanswered in the future.

 

About the size and position of the board : I answered that already, so I will not repeat my self - look back in the comments here.

 

I am sorry, if you have not seen the GoPro mounted camera on the board... 4k is not need

 

Captain WAS doing just fine job. I see you have no idea about a real measurement...

 

I see you still do not understand that WE DID NOT USE the tape measure... it was at the leveling process and NOT the measurements.

I am bored to answer questions like this...

 

NEVER EVER AGAIN COMMENT ON MY BELIEFS!

AND

NEVER EVER mention like "It will obviously put you in "crackpot zone""

 

I give respect to people here and I DO WNAT THE SAME.

 

If it starts to be like the other forum I will leave you too...

 

 

"I suppose so it's just there to catch attention to YouTube video, to earn money on advertisements from Google.

Scientists seeking for the truth don't behave like that. "

 

REPEAT AGAIN: NEVER EVER TALK TO ME LIKE THAT! I AM NOT YOUR FRIEN, NOR A PARTNER. I AM GOING TO IGNORE YOU FROM NOW.

 

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Okay so you have NOT watched the video? then what are we arguing about?

!

Moderator Note

As has been previously pointed out, you need to present enough information for discussion here, so that people need not watch the video to participate.

 

I AM GOING TO IGNORE YOU FROM NOW.

!

Moderator Note

You can do that without the histrionics. But the price for using this platform to discuss your experiment is answering criticism about the experiment

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From the given data, it looks the Balaton lake is not level. It flows like a river. It has a level +150.55 at inflow (left on the picture) and 149.65 at outflow, a difference of 90cm from level.

attachicon.gifpost-120902-0-23645600-1473549296.jpg

(I have added labeling on the maps provided in the OP)

 

The experiment was taken at a point after a bottleneck, transverse to the lake. From the color legend, the lake shows a slight level difference transverse also.

If the experiment has been correctly conducted, then the surface lake is concave along the experiment. Eventually the concave of the lake surface and Earth curvature would cancel each other and gives an almost perfectly flat section.

But that would be a extraordinary coincidence that a F.E. knew about this and decided to take an experiment right at this spot and at the right time of the year.

Michael you are wrong.

Read through the lidar experiment again, Balaton lake DOES NOT flow like a river. It is a "trully level surface".

 

" It has a level +150.55 at inflow (left on the picture) and 149.65 at outflow, a difference of 90cm from level."

Very wrong and missleading.. NO the highest point of the lake is NOT at the inflow, and the lowest point is NOT at the outflow. (just to tell you for a less "misspresentation of the lidar experiment" I have been banned from an other forum...

 

amm... just to inform You all: the RIVER Danube has a 26 cms drop over the COMPLETE route in Hungary. You say seriously that the lake has 3x more on a distance of 1/10? lol

!

Moderator Note

As has been previously pointed out, you need to present enough information for discussion here, so that people need not watch the video to participate.

 

!

Moderator Note

You can do that without the histrionics. But the price for using this platform to discuss your experiment is answering criticism about the experiment

As a moderator I think you should be concerned to avoid discussions like this from the forum:

"I suppose so it's just there to catch attention to YouTube video, to earn money on advertisements from Google.

Scientists seeking for the truth don't behave like that. "

 

IS this personal or has to do with the experiment evaluation?!

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Michael you are wrong.

Read through the lidar experiment again, Balaton lake DOES NOT flow like a river. It is a "trully level surface".

 

" It has a level +150.55 at inflow (left on the picture) and 149.65 at outflow, a difference of 90cm from level."

Very wrong and missleading.. NO the highest point of the lake is NOT at the inflow, and the lowest point is NOT at the outflow. (just to tell you for a less "misspresentation of the lidar experiment" I have been banned from an other forum...

 

amm... just to inform You all: the RIVER Danube has a 26 cms drop over the COMPLETE route in Hungary. You say seriously that the lake has 3x more on a distance of 1/10? lol

As a moderator I think you should be concerned to avoid discussions like this from the forum:

"I suppose so it's just there to catch attention to YouTube video, to earn money on advertisements from Google.

Scientists seeking for the truth don't behave like that. "

 

IS this personal or has to do with the experiment evaluation?!

I may be wrong, but I was simply reading the map provided. If the lake was level, it should be represented on the map by a single color. However is not a single color over the lake. Or maybe the colors are representing the bottom of the lake?

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Here is my first stab at putting this into a spreadsheet:

 

https://drive.google.com/open?id=0B22sA7_PYGQvMFY5WWtIenU0ZVk

 

Assumptions: WGS84 ellilpsoid

 

f=1/298.257223560

a=6378137.000
Radius of Curvature at latitude θ ~ a(1-f sin²(θ))
That the top of the black tape mark is ~1.185m as estimated by Mick West's analysis.
The slope calculation is against a flat baseline calculated between the laser itself and measurement point C4, where they re-leveled the laser.
ΔGround[2.286] + laserHeight@C4[1.185] - laserHeight[1.25] / distance@C4[720] = ~0.003084 meters per meter
You can use column T to ignore the beam width or column U attempts to find where the bottom of the beam might be.
Beam divergence angle for C0, C1 is just set to same value as C2 as I don't have good values - the values in green are actual estimates, the values after that are extrapolated but conservative estimates from the green values only.
No refraction is taken into account.
Unfortunately the data is not clean enough to do a 'fit' analysis. But this does align well with the first few measurement points, accounts for the beam spread and later reflective hits.
Using simple curvature did NOT fit the observations well.

 

Thank you.

 

By 'Gaussian beam divergence' I just mean the amount of divergence expected from a laser beam given some degree of collimation, even in a vacuum. This beam is spreading at an increasing rate of divergence rather than some fixed divergence angle. So I think that atmospheric effects are dominating. Some of that is diffraction, some is refraction - but end result is that the spot we see at C1 and C2 has grown considerably by C4 and by C11 it's so spread out we can barely see it on the board.

 

If you assume an ~100mm aperture (hands width) at the start and try computing the divergence angle at each of C2, C3, C4, & C5 you get different answers for each distance -- so it's not a constant rate of divergence (or maybe the distances are wrong but I think it's too great for this to be the only cause). These are the measurements that I'm referring to and I posted a montage showing these in my earlier post:

 

426m shows ~25cm diameter of spread

631m shows ~31cm diameter of spread

720m shows ~48cm diameter of spread

870m shows ~72cm diameter of spread

 

I think your calculation is far off from reality. PLease explain that excel with words too:

 

you suggest, that at c37 6000meters: beam diameter is 11.75 meter?

you say, that the calculated bottom of the beam is at -2.97 (meter?)

How is that possible?

then it should be visible on the board and it shall hit the water

 

NONE of that is recorded...

 

Please clear your calculations.

your calculation is way wrong starting with a false assumption on the leveling.

C1 - C4 is NOT measurement points.

How do you come to this divergence calculation anyway? do you realize that the beam was reflected from the metal frame and therefore it spreads the beam?

 

"This beam is spreading at an increasing rate of divergence rather than some fixed divergence angle. " how did you come to that conclusion?

 

I think your calculation on beam divergence is way wrong

 

 

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You've failed to address any of my comments. Only emotionally responded to one part. Did you actual read or understand my concerns about beam divergence and measuring the power in a cross section of the beam?

 

Opps, whitch one I forgot to answer?

 

I've not watched the video. I did fear this might be the case. You can get "bright" spots off of thr beam path, and you'll be killed by any beam divergence or laser wobble. I don't really think this experiment is within the reach of a hobbyist. Although it looks simple as soon as you start actually thinking about it there are so many subtleties that will make your results meaningless.

 

so I should answer: no I can't get "bright spots" off the beam path. "killed by" what is that term?

"Although it looks simple as soon as you start actually thinking about it there are so many subtleties that will make your results meaningless."

what does this sentence mean? you say surface can't be measured with laser? well... this is your opinion and I have mine

 

That's a terrible way to measure. You could do something similar with a camera sensor but where you measure the intensity from each pixel. But you'd need a large area CMOS or ccd and some clever way of removing ambient light.

IS that a terrible way? LOL have you seen Stephen Hawking Genious video (attached here before)?

my favourite part:

But you'd need a large area CMOS or ccd

 

a digital camera IS MADE OF A CCD

 

please point me to your important suggestion, it looks like I did not find it

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As a moderator I think you should be concerned to avoid discussions like this from the forum:

"I suppose so it's just there to catch attention to YouTube video, to earn money on advertisements from Google.

Scientists seeking for the truth don't behave like that. "

 

IS this personal or has to do with the experiment evaluation?!

 

 

!

Moderator Note

 

I don't see how it's personal. It's about how you present your results. Do scientists do this on a regular basis or not?

 

Now, how about we refrain from further off-topic discussion and not respond to modnotes.

 

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Er No.

 

The problem is that I don't think you understand the surveying theory.

 

I seriously doubt that the Swiss authors were referring to a standard correction as we are discussing.

Since they mention double run leveling, they are talking about calculational adjustments to a leveling network.

 

I make the refraction correction 11mm not 0.4mm at an observational distance of 1000m

 

Here is a proper Swiss discussion, including a table which includes refraction and curvature corrections.

Note carefully their value at 1km. They make the refraction correction 9mm using a slightly lower value of refraction that the international standard constant of 0.07, but still very close to my calculation.

This is taken from the Swiss manufacturer's textbook

The Theodolite by Wild of Heerbrugg Switzerland.

 

attachicon.gifrefract1.jpg

 

This lack of surveying knowledge may also have lead to you abrupt response to my posts on reciprocal levelling (post#137) and the previous one (post#133) on survey control points.

 

This is why I keep recommending obtaining the services/advice of a professional surveyor.

Hello Studiot

 

I've been sick for the last days therefore had a lot of reading on the refraction subject. :)

 

Please note here at end of page 26

"Assuming 17 levelling setups along a levelling line of 1 km, the total systematic devia¬ tion caused by refraction is 0.75 mm/km. For comparison: the standard deviation of height differences obtained by precise levelling using digital levels is about 0.3 to 0.4 mm/km (double run levelling) as reported, e.g., in [iNGENSAND. 1995J. There¬ fore, the refraction influences should be taken into consideration."

 

"In order to obtain a hypothesis-free reduction of refraction influences, it is necessary to determine the refraction influences integrally along the propagation path"

Summary of page 29

 

http://e-collection.library.ethz.ch/eserv/eth:23778/eth-23778-02.pdf

 

Please explain in your referenced table what is the 2nd column and what it is it measured in? (meter?)

 

 

In general I think terresterial refraction can not be calculated with a formula, especially not correspondent to the earth curvature. In other words: refraction has nothing to do with curvature.

 

In my opinion terresterial refraction can be measured but not calculated, as terresterial refraction is due to LOCAL changes. Unlike atmospheric refraction where the ambient data can be estimated and the change of conditions are more alike.

 

Refraction is bending the light due to CHANGE of the medium.

I see no constant or presumable change on horizontal distance measurement. It is not like sunshine coming from near vacuum and entering into more dense, humidand warmer atmosphere - where we can estimate the effects roughly.

Terresterial refraction is caused by LOCAL changes and temperature/ pressure differences, like a hot spot. That is a bending point of the laser, but NOT a continous curve at all.

 

At the moment I see that refraction may be measured with different color of laser beams, and it can not be estimated nor calculated.

 

so we have to look for theautoconvective laps rate condition:

 

 

In my opinion the laser beam is going in a straight line between the bending points, therefore that is not a curve.

 

The suggested laser beam refraction is 1/7th of the curvature in this example on this page (or the horizon distance is +7%)

 

http://mathscinotes.com/2013/08/distance-to-the-horizon-assuming-refraction/

 

 

here some interesting data on convective stability

 

http://meteorologytraining.tpub.com/14312/css/14312_53.htm

 

 

 

A digital camera has a ccd or CMOS sensor but lots of other stuff too like lenses. That's not what you want. You want a sensor array with no optics to measure the beam power to try and find the beam centre.

Kalynos, I am a cameraman...

 

a ccd is a device that has lenses and other stuff too.. that is called the camera. All sensors have optics.

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That's not true. I have used many ccd and CMOS sensors during my time as a professional optical physicist with no attached lenses at all. The coolest was liquid nitrogen cooled to detect very low intensities.

 

You also need something with really accessible and understood electronics to actually measure the intensities. Not some processed output as you get from a camera.

 

Your set up as is is not good enough to find the centre of the beam.

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sandor szekely post#185

Please explain in your referenced table what is the 2nd column and what it is it measured in? (meter?)

 

There is really only one column, but it is split into 5.

Within each of the 5 'columns' the combined curvature and refraction correction, as calculated from the formula given on the preceding page, is tabulated on the right hand side against the sighting distance D.

D is in kilometers, the correction is in metres.

 

So for instance the first 'column runs from a distance of 100 metres (=0.1km) to 1000 metres (=1 km)

 

At a distance of 1km the correction is given as 0.07 metres = 70 mm.

 

 

sandor szekely post#185

I've been sick for the last days therefore had a lot of reading on the refraction subject.

 

I'm sorry to hear you were ill (not too much Bull's Blood I hope? :) )

 

Unfortunately you seem to have been reading in the wrong places about refraction.

 

Refraction affects both vertical, horizontal, and true distance measurement as well as angular measurements.

Further the effect depends upon the type of EDM and measurement employed.

Consequently there are many formulae, each for different purposes and you need to be something of an expert to chose the right one.

I do not intend protracted discussion about refraction here since it small in your case and largely irrelevant.

Further it can be eliminated by proper observational methods, as I keep saying.

 

One interesting point is that the variation of refraction with the frequency of the EDM leads to a phenomenon called dispersion which allows accurate measurement of the refraction over the whole observed line of sight with polychromatic light.

Monochromatic sources do not allow this but special two colour laser systems such as 'Georan' are available to exploit this.

 

Another irrelevancy is the CCD issue.

 

Charge coupled devices are much older than their use in solid state cameras.

They were originally introduced to facilitate digital signal processing for instance in filters and delay lines.

My brother-in-law did his final year project on Digital Filters in the late 1970s.

Optical applications came a decade later.

 

So, back to the project in hand.

Your objective is to measure the actual shape in space of the surface of the lake, is it not?

 

You should be aware that the surface of water bodies on Earth is rarely level, flat, equipotential etc.

In hydraulic engineering we consider something called the 'backwater curve', whcih is the shape of flowing water surface as it approaches an obstruction.

This curve can even include actual physical steps in the surface, known as hydraulic jumps.

The surfaces of main oceans of the Earth (Atlantic, Pacific, Indian and Southern) are not at the same 'level'.

This does not go as far as jumps or step changes but manifests itself as a region of turbulence where they meet.

The so called 'cape rollers' off South Africa are an example of this.

 

Returning to the actual shape (and size) of the water surface at Lake Balaton,

 

Do you fully understand the difference between actual on site measurements and the 'reduced measurements' you would plot on a spheroid or geoid or a map?

 

This is very important and would help greatly in your discussions with your surveyor.

 

Would you like me to explain the difference between the terms level, equipotential, spheroid, geoid etc ?

Edited by studiot
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That's not true. I have used many ccd and CMOS sensors during my time as a professional optical physicist with no attached lenses at all. The coolest was liquid nitrogen cooled to detect very low intensities.

 

You also need something with really accessible and understood electronics to actually measure the intensities. Not some processed output as you get from a camera.

 

Your set up as is is not good enough to find the centre of the beam.

We will use a board big enough to capture the whole beam in the upcoming experiment. Way more easy than to get the same size of any detector.

temperature is such that de

 

My question is : how can anyone estimate, or use a standard value for terresterial refraction?

 

Refraction is caused by change in ambient conditions like pressure, temperature, density or humidity (...)

 

In the case of atmospheric refraction, light coming from thousands of miles from vacuum through more dense, humid, warmer and pressured air it has a well definable "standard" value. Change is more-less the same.

 

But in the case of terresterial refraction like in our experiment the laser beam is going within a short height difference above water, conditions are nearly the same all the way. In this case the refraction can not be pre-determined as it depends on local changes.

Refraction is also not a curve but a bend at certain points (where conditions change) and connected with straight path of light.

 

So in my opinion terresterial refraction can only be measured, but not calculated with a standard formula.

What is your opinion on the best way to measure it?

(mine is to use different colours, but we did not have the collimator for the 2nd)

 

 

 

 

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You can't use the human eye as it's s nonlinear uncalibrated sending device. Cameras try and reproduce this so you can't use one of those either. You'll just make more poor measurements. You need a quantitative, reoroduceable method of measuring the beam centre.

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There is really only one column, but it is split into 5.

Within each of the 5 'columns' the combined curvature and refraction correction, as calculated from the formula given on the preceding page, is tabulated on the right hand side against the sighting distance D.

D is in kilometers, the correction is in metres.

 

So for instance the first 'column runs from a distance of 100 metres (=0.1km) to 1000 metres (=1 km)

 

At a distance of 1km the correction is given as 0.07 metres = 70 mm.

 

 

I'm sorry to hear you were ill (not too much Bull's Blood I hope? :) )

 

Unfortunately you seem to have been reading in the wrong places about refraction.

 

Refraction affects both vertical, horizontal, and true distance measurement as well as angular measurements.

Further the effect depends upon the type of EDM and measurement employed.

Consequently there are many formulae, each for different purposes and you need to be something of an expert to chose the right one.

I do not intend protracted discussion about refraction here since it small in your case and largely irrelevant.

Further it can be eliminated by proper observational methods, as I keep saying.

 

One interesting point is that the variation of refraction with the frequency of the EDM leads to a phenomenon called dispersion which allows accurate measurement of the refraction over the whole observed line of sight with polychromatic light.

Monochromatic sources do not allow this but special two colour laser systems such as 'Georan' are available to exploit this.

 

Another irrelevancy is the CCD issue.

 

Charge coupled devices are much older than their use in solid state cameras.

They were originally introduced to facilitate digital signal processing for instance in filters and delay lines.

My brother-in-law did his final year project on Digital Filters in the late 1970s.

Optical applications came a decade later.

 

So, back to the project in hand.

Your objective is to measure the actual shape in space of the surface of the lake, is it not?

 

You should be aware that the surface of water bodies on Earth is rarely level, flat, equipotential etc.

In hydraulic engineering we consider something called the 'backwater curve', whcih is the shape of flowing water surface as it approaches an obstruction.

This curve can even include actual physical steps in the surface, known as hydraulic jumps.

The surfaces of main oceans of the Earth (Atlantic, Pacific, Indian and Southern) are not at the same 'level'.

This does not go as far as jumps or step changes but manifests itself as a region of turbulence where they meet.

The so called 'cape rollers' off South Africa are an example of this.

 

Returning to the actual shape (and size) of the water surface at Lake Balaton,

 

Do you fully understand the difference between actual on site measurements and the 'reduced measurements' you would plot on a spheroid or geoid or a map?

 

This is very important and would help greatly in your discussions with your surveyor.

 

Would you like me to explain the difference between the terms level, equipotential, spheroid, geoid etc ?

 

I have a confusion here:

 

your chart says 4.9kms is 1.68 meter

the curvature of earth in the 4.9 kms distance is 1.88 meter.

 

so is the refraction here 0.2 meter or 1.68?

 

"I do not intend protracted discussion about refraction here since it small in your case and largely irrelevant."

in the case of 0.2 meter, maybe but I still don't see how it can be standardised as it deends on the local condition change.

 

"One interesting point is that the variation of refraction with the frequency of the EDM leads to a phenomenon called dispersion which allows accurate measurement of the refraction over the whole observed line of sight with polychromatic light.

Monochromatic sources do not allow this but special two colour laser systems such as 'Georan' are available to exploit this."

Please tell me more abut this, I think this is exactly what I figured out with the dual colour laser too. So as I understand there are eqiupment like this?

 

"Your objective is to measure the actual shape in space of the surface of the lake, is it not?

You should be aware that the surface of water bodies on Earth is rarely level, flat, equipotential etc."

My objective is to find out the shape of the water surface, if it is flat or curved.

I think there is a huge difference in the outcome of the 2 models already at a shorter distance of 10kms, and a huge difference on the 77kms of the lake. This should be significant enough to exclude one of the models. I am not aiming to make the most precise measurement just to the accuracy that we can determine the shape of the surface: is it flat or curved?

 

Is it possible to have 465 meters drop of curvature on the lake Balaton?

 

Yeah maybe some Bull's blood would be good for me :) but I never drink alcohol :)

 

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sandor szekely post#190

My objective is to find out the shape of the water surface, if it is flat or curved.

I think there is a huge difference in the outcome of the 2 models already at a shorter distance of 10kms, and a huge difference on the 77kms of the lake. This should be significant enough to exclude one of the models. I am not aiming to make the most precise measurement just to the accuracy that we can determine the shape of the surface: is it flat or curved?

 

 

Exactly, thank you.

 

Do you realise that geoids, spheroids or any other oids (droids, androids etc) are quite irrelevant to your work?

 

You would have to perform a deal of calculations to refer the actual measurements you take to any of these.

The first thing an engineer would have to do, if asked to place a giant curtain across the lake, just touching the water surface, would be to undo all those calculations to obtain real world measurements.

 

Please confirm some things before I answer the other parts of your post.

 

You did say that you have a degree in mathematics did you not?

 

So I have assumed you can easily substitute values into the formula given in the left hand page of my extract from Wild.

So you could generate your own table and check the calcs for yourself.

The derivation is simple enough it only requires first or second year high school geometry.

 

Please say if you are having trouble with the technical English (I thought yours was pretty good).

 

But yes, at 4.9 kilometers sight distance the combined addition for curvature and refraction is 1680 mm.

 

The curvature alone amounts to 1885mm.

 

But the question is, what do you intend to show with this value?

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I think your calculation is far off from reality. PLease explain that excel with words too:

 

you suggest, that at c37 6000meters: beam diameter is 11.75 meter?

you say, that the calculated bottom of the beam is at -2.97 (meter?)

How is that possible?

then it should be visible on the board and it shall hit the water

 

NONE of that is recorded...

 

Please clear your calculations.

your calculation is way wrong starting with a false assumption on the leveling.

C1 - C4 is NOT measurement points.

How do you come to this divergence calculation anyway? do you realize that the beam was reflected from the metal frame and therefore it spreads the beam?

 

"This beam is spreading at an increasing rate of divergence rather than some fixed divergence angle. " how did you come to that conclusion?

 

I think your calculation on beam divergence is way wrong

 

 

 

>>> you suggest, that at c37 6000meters: beam diameter is 11.75 meter?

>>> ...

>>> NONE of that is recorded...

Yes it is, look at your data more carefully, already at C11 it's all over the board and glinting off the water. It's just very dim because it's spread out so much.

 

This is easily 3 meters of spread already if not more. I've enhanced the color separation so you can see the green laser light very clearly.

 

post-121005-0-64657600-1474395183_thumb.png

 

And you can look at your beam diameter at C2, C3, C4, C5 and see it heading that direction. These earlier measurements show that the beam is spreading wildly out of control. I have no idea if it's 12 meter or 30 meter by C37 - that's why it's labeled extrapolation, however, it's irrelevant because I show that none of you measurements past about C8 are even close to reliable based on what we can see.

 

post-121005-0-99994900-1474386294_thumb.jpg

 

The real point here is that you have no way to show that I'm wrong. You have to present the data to validate your measurements. What I'm showing is that what you have shown is simply not reliable and I've shown why (in great detail which you have mostly ignored).

 

>>> do you realize that the beam was reflected from the metal frame and therefore it spreads the beam?

 

 

Since the beam is spreading both horizontally and vertically the metal support frame wouldn't be the cause -- that would only contribute to vertical reflections.

 

>>> C1 - C4 is NOT measurement points.

 

 

C4 obviously is but taking your intention here please note that I do not use C1, C2, or C3 as a height measurement point - In fact, I specifically point out that you MOVED the laser at these points and so they are invalid for this purpose - I calculate the slope based on 1.25m laser and point C4. They ARE however valid observations of the beam divergence and I use them for that purpose.

 

>>> "This beam is spreading at an increasing rate of divergence rather than some fixed divergence angle. " how did you come to that conclusion?

 

 

I showed and measured the obvious beam spread in the images. Why don't you take your images for C1, C2, C3, C4, C5 and you tell us how big your laser spot is at each one?

 

 

 

Sandor -- the most critical error so far, which was raised by maximillian12 on the 18th, is that the ellipsoid shows a 17m delta over the 6km distance. Ideally Zach should remodel your 'GE Model' using this ellipsoidal data -- as I did in my spreadsheet -- and then you guys need to admit that most of your data past about C5 is very poor due to mismeasuring reflective spots on the boat (as you've not yet responded to any of that, see linked post).

 

 

 

Please note that my spreadsheet is only PRELIMINARY and is certainly open to review - I don't claim they are perfect. If you find an error point it out and we can discuss.

 

But which one do you not understand specifically and I can discuss them in more detail?

 

Let me point out some of the data locations and overall calculations:

 

Ellipsoid calculations:

 

J3 = WGS84 ellipsoidal flattening (f)

J5 = WGS84 ellipsoidal equatorial radius (a)

J8 = equation (approx) used for Ellipsoidal Radius of Curvature at latitude θ: a(1-f sin²(θ))

 

Laser/beam spread calculations:

 

O7 = Laser Height (1.25m) AGL (above 'ground' level)

Q5 = equation for extrapolated beam width: c*e^(bx)

Q6 = value for c - from INDEX(LINEST(LN(Q12:Q15),LN(F12:F15)),1)

Q7 = value for b - from EXP(INDEX(LINEST(LN(Q12:Q15),LN(F12:F15)),2))

Where Q12-Q15 are the measured spread values, F12-F15 are the associated distances

 

R7 is your starting aperture, I've used 100mm

R8 is the equation used for 1/2 angle divergence (atan(g/d/2)), where g = beamwidth-aperture, r=distance(column F) (*1000 for mRad)

S8 gives the equation for 1/2 beam spread (d*tan(a)-mm/2), where d = distance(column F), a = 1/2 angle(column R), mm(aperture)=$R$7 -- but this is also JUST half the beam width, I did it the 'hard' way as a check on the math for divergence angle - so basically S=Q/2

 

As for where these formula come from, they come from basic trig:

 

post-121005-0-05831800-1474396818.png

 

tan(α) = a/b

 

This says that tan(α) gives you the slope of c, a=rise, b=run, slope=rise/run -- conversely atan() takes a slope and gives you the angle back.

 

We can apply atan() to both sides:

 

atan(tan(α)) = atan(a/b)

 

which simplifies to:

 

α = atan(a/b)

 

Now, in our case, a is 1/2 of the diameter -- so that's why you see atan(g/d/2) -- and we subtract out/add back in the starting aperture of 100mm (0.1m) (or 1/2 that for the 1/2 spread) -- essentially treating it as a point source in the middle. Feel free to give us your equations for beam spread.

 

 

P7 is the Slope calculation: (K14+N14-O7)/(F14) -- which is: (ΔGround@C4[2.286] + EstMid[1.185] - LaserHeight[1.25]) / Distance@C4[720]

 

Column O is not used, but shows the spherical curvature approximation over distance (column F), given r=6371000

 

We calculate the Ellipsoidal Radius for column J using the values f and a from WGS84 using your given latitude values in the spreadsheet. I use an approximation function (as stated in the spreadsheet: a(1-f sin^2(θ)). If you want to use the more accurate equation use this:

 

post-121005-0-19791700-1474388678.png

 

ΔGround is simply the total change to that point in the Ellipsoid radius in column J (=J10-$J$10) - this shows the ~17m change in radius of curvature over the 6km

 

>>> your calculation is way wrong starting with a false assumption on the leveling.

 

 

Then you need to show us the evidence you promised a while back. I have shown mine.

 

"est mid" at C4 is 1.185 which is from Mick's calculations BOTH on the pixels and looking at the slope of the board from numerous lines of evidence. You still haven't provided any evidence that your measurement of 1.3m for the top of the tape was correct. I also looked at the dimensions of your exact boat plus the height of the board at the observed angle. It all points to ~1.185m for the top of the tape, not 1.3m.

 

If you are hoping to approach scientific accuracy on this experiment then you need to take this seriously. You just keep denying that this is a problem despite the significant evidence against it and no evidence supporting your claim of 1.3m.

 

 

Est Beam" is just the estimate beam path, this is simply based on the slope, ΔGround, and laser height.

 

Calculated Bottom of Beam is beam path plus taking into account the radius of the beam width.

 

 

summary of issues:

 

need to use ellipsoidal model, not spherical

need to account for 1.3m height measurement - measurements must be dead accurate and supportable

need to account for the observed beam spread (C1, C2, C3, C4, C5, C11 show evidence of significant beam spread)

cannot use 'direct hit in camera' as a valid measurement (C31 example)

cannot use reflective spots on the support bar on boat as valid measurements (C5, C11, C19 are the same spot, but you marked at different heights)

cannot use reflections off jacket reflector as a valid measurement (C13, C14, C16, all the same jacket reflector, but again, all marked at different heights)

cannot use glints off the camera lens as valid measurements (C17, C18 are both camera lens glints but marked at different heights)

 

 

Additional notes on other comments:

 

>>> As I pointed out, the horizontal vibrating effect was due to the laser beam hitting the water surface at the 3rd measurement (reflexion)

 

 

I agree that in the night-time video the most likely explanation is that the laser was hitting the water and reflecting upwards and while very pretty & interesting in other ways, isn't relevant to the test results labeled Laser 4 or indicative of refraction effects.

 

 

>>> I think there is a confusion on the meaning of "level". Here are some definitions, copy Nr1 here:

 

The email from András Zlinszky makes it very clear that their results are in no way consistent with planar, flat lake.

 

The surface of the lake in our survey was confirmed to be close to hydrological equilibrium, that is, to closely follow the local curve of the geoid. Fig. 2 a of this paper shows how the water surface height deviates from the WGS 84 ellipsoid, it is by no means planar.

 

 

 

You keep using this sense of 'level' that is not used in the world of Geodesy. http://nptel.ac.in/courses/105107122/modules/module1/htmlpage/24.htm

Level surface : A level surface is the equipotential surface of the earth's gravity field. It is a curved surface and every element of which is normal to plumb line. A body of still water provides the best example of a level surface.

 

 

 

 

You also made this comment in the thread:

>>> amm... just to inform You all: the RIVER Danube has a 26 cms drop over the COMPLETE route in Hungary. You say seriously that the lake has 3x more on a distance of 1/10? lol

 

 

You seem to be confusing a gradient in the orthometric height of the terrain (which would cause water to flow) with the geoid -- these are not the same thing can cannot be compared in this manner. The geoid shape can vary and water will not flow, but will conform to the geoid shape as long as other forces are not acting upon it (tides, winds, etc). This shape is defined by the slightly lumpy gravity of the Earth and its spin (which cancels out a very small portion of the Gravity and varies with latitude, highest at the equator and zero at the poles).

 

When you give an elevation from a map, you are usually giving the Orthometic height or the closely related, MSL height. So a change in THIS height means water flows. But a 100m rise in the Geoid does NOT cause water to flow.

 

The shape of the gravity equipotential is what defines 'level' at every point. That's why, even though the South end of your path over the lake is 17 meters further from the center of the Earth than your starting point, every point along that line is 'level' even though the water is 17 meters higher. This change does NOT cause water to flow and you do not feel it or measure it as "going up hill" because it's a gravity equipotential.

 

It is this gravity equipotential shape that the water has conformed to. That shape just happens to be 17m higher on one end than the other.

 

 

Some definition of terms might help:

 

post-121005-0-12472600-1474400513.png

 

Geoid: The geoid is the shape that the surface of the oceans would take under the influence of Earth's gravitation and rotation alone, in the absence of other influences such as winds and tides. This surface is extended through the continents (such as with very narrow hypothetical canals). All points on the geoid have the same gravity potential energy (the sum of gravitational potential energy and centrifugal potential energy). The force of gravity acts everywhere perpendicular to the geoid, meaning that plumb lines point perpendicular and water levels parallel to the geoid.

 

Orthometric Height: The orthometric height of a point is the distance H along a plumb line from the point to the geoid.

 

 

In my spreadsheet I ignored the Geoid height because it was constant to 1cm and saw little point in adding it in only to subtract it back out. Essentially the shape of the water matches very closely to the Ellipsoidal shape so we just need to know how much THAT changes, the rest all cancels out to within 2cm.

 

If you want to be MAXIMALLY accurate, then you need to download the software and data sets for the Geoid, and Orthometic heights and get exact data for your measurements points (and measure actual depth of water since that would be a variable), that will get you an additional 1-2 cm of accuracy for your 'GE Model' (aka Globe Earth).

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Dark Star, considering all the work you have put in there I hate to be a wet blanket but your calculations are flawed for the reasons I have already given.

 

Nor are your hydraulic predictions any more reliable (are you a hydraulic engineer?) again for the reasons recently given.

 

Sandor does not need (in fact shouldn't be) calculating on the spheroid.

Your formulae need the measured distance to be reduced to a spheroidal distance, but I do not see any attempt in the calculations to do this.

 

I do, however, agree that the difference and significance of that difference between actual measurements on the surface of the Earth and mathematically modelled figures is very confusing if you are not a professional geodecist.

Edited by studiot
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Exactly, thank you.

 

Do you realise that geoids, spheroids or any other oids (droids, androids etc) are quite irrelevant to your work?

 

You would have to perform a deal of calculations to refer the actual measurements you take to any of these.

The first thing an engineer would have to do, if asked to place a giant curtain across the lake, just touching the water surface, would be to undo all those calculations to obtain real world measurements.

 

Please confirm some things before I answer the other parts of your post.

 

You did say that you have a degree in mathematics did you not?

 

So I have assumed you can easily substitute values into the formula given in the left hand page of my extract from Wild.

So you could generate your own table and check the calcs for yourself.

The derivation is simple enough it only requires first or second year high school geometry.

 

Please say if you are having trouble with the technical English (I thought yours was pretty good).

 

But yes, at 4.9 kilometers sight distance the combined addition for curvature and refraction is 1680 mm.

 

The curvature alone amounts to 1885mm.

 

But the question is, what do you intend to show with this value?

I don't have a degree in math, but I get along well :)

I worked as a cameraman long time ago and became an inventor 2 decades ago.

 

I get along with technical English, but I am thankfull if you correct me! I use dictionary to search terms that I am not familiar with, but I may use them incorrecty.

 

"the combined addition for curvature and refraction"

thanks this makes it perfectly clearly understandable :)

 

I'd like to calculate the amount of curvature that we are looking for, so we know what precision we have to do in our measurements. I see that it would be very hard to determine the difference of few centimeters on these distances, but looking for meters of difference should be very well measurable.

 

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Dark Star, considering all the work you have put in there I hate to be a wet blanket but your calculations are flawed for the reasons I have already given.

 

Nor are your hydraulic predictions any more reliable (are you a hydraulic engineer?) again for the reasons recently given.

 

Sandor does not need (in fact shouldn't be) calculating on the spheroid.

Your formulae need the measured distance to be reduced to a spheroidal distance, but I do not see any attempt in the calculations to do this.

 

I do, however, agree that the difference and significance of that difference between actual measurements on the surface of the Earth and mathematically modelled figures is very confusing if you are not a professional geodecist.

 

I didn't make any hydraulic predictions, I agree with your assessment on that front. I make no attempt to model those. I agree you need to actually make the measurements.

 

I would also note that I just did a couple of things -- I took Sandor's EXISTING spreadsheet (don't know if you realized this or not but that wasn't my original spreadsheet -- this is the one that Sandor had published with their 'experiment') and I replaced the spherical approximations they had used with the the ellipsoid approximation, and I tried to guesstimate how much the beam was actually spreading out over the distances. Sandor is the one building the model, I'm just trying to show their original had significant issues. Unfortunately, for all the reasons I mentioned previously, we really have no usable data from their 'experiment' to compare it against. I think that would be fun even if we had issues.

 

Even reading what you have written I'm not sure why you think the shape of the lake cannot be modeled accurately enough for the task at hand -- Sandor's goal is not to get an accurate survey but rather to prove that the lake is absolutely 100% a Flat Plane on a Flat Earth. And he believes (and has repeatedly stated in this thread despite the PI on the study explicitly stating that this is not the case) that the cited 2013 study claims the lake is exactly this 100% flat plane.

 

So the question is -- do you think we can model the curvature on a Lake "well enough" to fit some observations that show it is not flat? I think over a large enough distance with proper techniques this is easily done. But the real problem is their laser isn't up to the task even if we had a perfect model to compare it against. You can't make "hide nor hair" of the laser past C5 and since they pointed the laser on the board at C4 that's as good as having no data. And the laser spread at C5 at 870 meters is already all but unusable (I would say even that is unusable)... you tell me:

 

post-121005-0-83774300-1474435304_thumb.png

 

Ignore that the board is slanted -- can you read a reliable "height" from that? Even if everything else was known perfectly?

 

 

That said, the distances calculated ARE based on great circle distance calculations between lat/long points so they are already spherical distances. However, the measurement error for the lat/long readings FAR exceed the difference between "straight-line" and Great Circle distances over 6km. After showing their first data point at C1 wasn't 111 meters from the laser but 1.9 meters I have zero reason to trust their GPS data either.

 

Let's do a quick sample, assuming a circle. The straight-line distance given s=curved distance would be R*2*sin(s/R/2)

 

So for s=10km distance

 

6371*2*sin(10/6371/2) = 9.99999897347

 

So 10 km spherical distance is just 1.03mm off on the straight-line distance. We're not getting 1mm accurate GPS coordinates here, we're getting +/- several meters. :)

 

My main point has been that NONE of their measurements are valid for a multitude of reasons and I've tried to highlight some of them. There are so many it's hard to hit them all.

 

But from his comments Sandor still refuses to accept that the laser is spreading out.

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So the question is -- do you think we can model the curvature on a Lake "well enough" to fit some observations that show it is not flat? I think over a large enough distance with proper techniques this is easily done. But the real problem is their laser isn't up to the task even if we had a perfect model to compare it against.

I have zero reason to trust their GPS data either.

 

Let's do a quick sample, assuming a circle. The straight-line distance given s=curved distance would be R*2*sin(s/R/2)

 

So for s=10km distance

 

6371*2*sin(10/6371/2) = 9.99999897347

 

So 10 km spherical distance is just 1.03mm off on the straight-line distance. We're not getting 1mm accurate GPS coordinates here, we're getting +/- several meters

good point on required precision. A 10 km sample would be extremely difficult to measure the amount of curvature.

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To summarise the situation here at ScienceForums.

 

An OP has expressed an interest in measuring the actual water profile on a modest, but significant body of water and carried out a trial measurement which showed several methodological flaws.

 

Some of the flaws go right back to fundamental misunderstandings of the science involved, some to lackadaisical organisation of the actual trial measurement.

 

So the main outcome of the trial, as I see things, is a requirement to go 'back to the drawing board' for a complete rethink from the beginning.

 

I am not interested in mud slinging or raking over past mistakes but am prepared to help develop a proper new strategy.

 

1) This should start with a standard aims and objectives statement.

 

2) Then provide an estimate of the expected range of measurement values and accuracies required so suitable equipment and experimental methods can be devised.

 

3) Offer a system of control or reference points, established to a higher standard than the survey itself and preferably established by a different method, independent of the survey measurement technique.

 

4) Provide a proper method statement for the survey itself.

 

I think that is enough to be going on with.

The issue of (mis)understanding can be dealt with as we go along.

 

Sandar, your once worked in the film industry so you will be familiar with a storyboard.

This is the scientific and technical equivalent.

 

We have a basis for (1)

 

So to kick off (2) here is a hydraulic assessment of what we might expect to be found.

 

The surface of larger bodies of water is mostly influenced by gravitational forces.

Water collects in depressions in the land surface.

Because continental crust is about 2.5 times the density of water and oceanic crust 3 times as dense, the material in the depressions is less dense than the average.

This results in a depression of the water surface in the deep ocean and an elevation or bulge in shallow water.

Bottom or bed features such as sea mounts, ridges or trenches are therefore reflected in profile disturbances at the surface.

An ocean floor feature 1km high results in a surface depression of 2 metres.

 

If you like the water surface follows the bed surface more closely than you might think.

 

Here is a Seasat ocean surface heighting scan showing a fully recognisable bed shape in its own surface.

 

So I would expect the lake surface to show its bed shape in miniature, rather than following a 'truly level' surface, whatever that might mean (I propose to discuss the meaning of that phrase later).

 

post-74263-0-02888100-1474465305_thumb.jpg

 

Edit

 

I will just add some more hydraulic comments.

 

Lake Balaton has the following hydraulic characteristics.

It has a main inflow to the south west and a main outflow to the northeast.

These are both relatively narrow channels from the internet pictures, but otherwise unrestricted so at the inlet the inlet water will be slowing down.

 

Therefore there will be a local rise in water surface as the dynamic head (kinetic energy) of the inflow is exchanged for static head (potential energy) in the lake itself.

 

The outlet is flush with the lake so there is no draw down effect, but the necessary static head to propel the water along the outlet channel must be generated so there will be a build up of water around the outlet, again generating a surface rise. This is the backwater curve I mentioned before.

 

Through the main body of the lake the cross section is much larger so the flow rate will be much more sluggish, almost zero.

So the head will be essentially static.

 

However about midway there is a projecting peninsula, restricting whatever flow there is so the water will again pile up behind (to the west of) the peninsula reduce through the gap and return to normal lake levels to the east.

Edited by studiot
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So I would expect the lake surface to show its bed shape in miniature, rather than following a 'truly level' surface, whatever that might mean (I propose to discuss the meaning of that phrase later).

 

Maybe these will help...

 

Level doesn't mean flat:

 

post-121005-0-20799900-1474484232_thumb.jpg

 

And with some exaggeration in scale, showing that level also is lumpy:

 

post-121005-0-85982600-1474484610_thumb.jpg

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To summarise the situation here at ScienceForums.

 

An OP has expressed an interest in measuring the actual water profile on a modest, but significant body of water and carried out a trial measurement which showed several methodological flaws.

 

Some of the flaws go right back to fundamental misunderstandings of the science involved, some to lackadaisical organisation of the actual trial measurement.

 

So the main outcome of the trial, as I see things, is a requirement to go 'back to the drawing board' for a complete rethink from the beginning.

 

I am not interested in mud slinging or raking over past mistakes but am prepared to help develop a proper new strategy.

 

1) This should start with a standard aims and objectives statement.

 

2) Then provide an estimate of the expected range of measurement values and accuracies required so suitable equipment and experimental methods can be devised.

 

3) Offer a system of control or reference points, established to a higher standard than the survey itself and preferably established by a different method, independent of the survey measurement technique.

 

4) Provide a proper method statement for the survey itself.

 

I think that is enough to be going on with.

The issue of (mis)understanding can be dealt with as we go along.

 

Sandar, your once worked in the film industry so you will be familiar with a storyboard.

This is the scientific and technical equivalent.

 

We have a basis for (1)

 

So to kick off (2) here is a hydraulic assessment of what we might expect to be found.

 

The surface of larger bodies of water is mostly influenced by gravitational forces.

Water collects in depressions in the land surface.

Because continental crust is about 2.5 times the density of water and oceanic crust 3 times as dense, the material in the depressions is less dense than the average.

This results in a depression of the water surface in the deep ocean and an elevation or bulge in shallow water.

Bottom or bed features such as sea mounts, ridges or trenches are therefore reflected in profile disturbances at the surface.

An ocean floor feature 1km high results in a surface depression of 2 metres.

 

If you like the water surface follows the bed surface more closely than you might think.

 

Here is a Seasat ocean surface heighting scan showing a fully recognisable bed shape in its own surface.

 

So I would expect the lake surface to show its bed shape in miniature, rather than following a 'truly level' surface, whatever that might mean (I propose to discuss the meaning of that phrase later).

 

attachicon.gifwatersurface1.jpg

 

Edit

 

I will just add some more hydraulic comments.

 

Lake Balaton has the following hydraulic characteristics.

It has a main inflow to the south west and a main outflow to the northeast.

These are both relatively narrow channels from the internet pictures, but otherwise unrestricted so at the inlet the inlet water will be slowing down.

 

Therefore there will be a local rise in water surface as the dynamic head (kinetic energy) of the inflow is exchanged for static head (potential energy) in the lake itself.

 

The outlet is flush with the lake so there is no draw down effect, but the necessary static head to propel the water along the outlet channel must be generated so there will be a build up of water around the outlet, again generating a surface rise. This is the backwater curve I mentioned before.

 

Through the main body of the lake the cross section is much larger so the flow rate will be much more sluggish, almost zero.

So the head will be essentially static.

 

However about midway there is a projecting peninsula, restricting whatever flow there is so the water will again pile up behind (to the west of) the peninsula reduce through the gap and return to normal lake levels to the east.

That is new to me (the bold part).

References?

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