Laser curvature test on lake Balaton

[michel123456]

Michel, please be advised that intervisibility is more complicated than this.

In general it depends where the obstruction is along the line.

In the second case you show, you would need to remove the entire bulge, not just part of it, to allow intervisibility.

So this would be appropriate if the model of the Earth was a faceted solid,.

But we seem to be debating a flat Earth which is different.

Yes it is complicated.

But if you replace the bulge with a wall 70 cm high exactly in the middle, it gives the same effect.

Which makes me think that from 1.50 height, even the shore is visible. 1.40m is the limit.

From 1.25m, it must be calculated.

 

So, it's like non-uniform-micro-refraction-artifacts that would cause the beam to change chaotically, making it impossible to predict the real direction of the beam?

 

I think this concept is being used to try to explain why the beam is apparently bending upwards.

 

It's interesting and would be a paper in itself. They should also explore that in the next experiment.

Bizarre. That would make the beam go left & right, not only up & down.

[MrMaker]

 

Bizarre. That would make the beam go left & right, not only up & down.

 

Maybe it's happening? I don't know ....

 

And I am silly, I always read NUDTZ and giggles because it looks like NUDEZ ...

 

It's probably not helping my image here in this forum, but I am what I am. ;-)

[DarkStar66]

so you can check them individually and see how I marked the laser hits.

We had laser direct hits on the reflective jacket and as well in the camera optics too.

 

 

MVI_7124-Sail-Far-Away---loop.gif

 

The collimator on the laser beam is 0.08 mRad in our measurements.

 

This image perfectly shows that the laser beam divergence is fairly large as you are clearly getting reflections over several meters wide as the boat slews across the view. These clips are found in the video starting at https://youtu.be/GBhDFO4NMrw?t=1111 (severalbroken clips from that point) and you can clearly see the boat is slewing very much side-to-side, it is not moving straight away from the laser, and you can see these glints across a very wide swath.

 

So all your showing is that a very divergent laser beam is hitting the reflective patch on the back of the white jacket and sometimes glinting off the camera lens pointing back at the dock.

 

These are not accurate measurement methodologies and do not indicate the center of the laser beam.

 

This, combined with the ~10cm error in your 'slope corrected leveling process' mean this experiment does not give us accurate results.

 

Here is my summary with l inks to the main metabunk posts that support them (credit to Mick for the analysis here, I'm just summarizing):

 

1) measuring height along the slanted board, likely resulting in ~9% error in 'slope corrected leveling' method [link1,link2] - confirmed by observing measuring methodology used and also by photo analysis. This likely resulted in a slight initial downward slope on the laser.

2) laser beam divergence is clearly much greater than they are admitting, creating false hits (shown in #3 & #5)

3) reflective patch on back of jacket and glints off camera lens giving false 'hits' [link1, link2]

4) 'direct hit in camera' is also demonstrated as not reliable (combined with #2 & #3) [link]

5) evidence that the laser is initially pointed slightly down (data at C5 & C8) [link]

 

If these are wrong please post the images that refute them. Here is kind of a guide as to what I think would be needed to address these points...

 

1) you would need images/video to show that the 130cm marking tape was placed accurately. We can see them using inaccurate methodology (slanted measurements) in other parts of the video but this isn't shown. So you can refute this by showing that the same methodology was not used when placing the 130cm tape mark but that it was made very plumb to the surface of the lake and not made by measuring the distance on the slanted board.

2) you would need to show the laser fully hitting the board at each point past ~C5 showing at least 1/2 extents of the beam divergence. All the evidence I can see past C4 is an increasingly divergent beam. Hits on the reflective tape and lens of the camera do not count to show that the laser divergence is small when we can plainly see in the video that this is not the case (something you seem to have a problem accepting).

3) the reflective patch and camera lens glints are pretty much self explanatory as being invalid. I don't see how you can show otherwise but I'm open to any further evidence presented.

4) 'direct hit on the camera' methodology is just busted - you can see the same effect in the camera even when the laser spot is small on the board in the closer measurements.

5) demonstrating #1 would call this into question depending on the results of that further analysis. But already at C8 we can see the beam is so diverged that it's difficult to see the actual extent of the beam hitting the board.

 

Do you have *anything* that shows the beam extents very clearly past C8 or not?

465m Well that sounds huge and made my eyebrows raise. So I took some time to check it with Autocad as you did.

...

The lake is 77km long.

You stand on one side of the lake and send a laser beam horizontally** through the 77km and measure the level on the other side. It is indeed 465m on the basis of Earth radius 6371 km.

But on sketch 2, with the exact same dimensions, if you draw a straight line between the sides and measure the bulge, it gives a value of 116m. That is the eight of "hill", the height that is an obstacle for viewing the other side, not 465m.

 

 

You might appreciate this calculator: https://www.metabunk.org/curve/?d=77&h=3&r=6371&u=m&a=n&fd=60&fp=3264

 

For 77km and an observer height of 3 meters it gives us the following:

 

Distance = 77 km (77000 m), View Height = 3 meters Radius = 6371 km (6371000 m)

Horizon = 6.18 km (6182.72 m)

Bulge = 116.33 meters

Drop = 465.33 meters

Hidden= 393.57 meters

 

With Standard Refraction 7/6*r, radius = 7432.83 km (7432833.33 m)

Refracted Horizon = 6.68 km (6678.1 m)

Refracted Drop= 398.85 meters

Refracted Hidden= 332.65 meters

 

Tilt Angle = 0.692 Degrees, (0.0121 Radians)

Horizon Dip Angle = 0.056 Degrees, (0.0010 Radians)

Most of these are pretty obvious and the full results explain some of the math used but very quickly...

Horizon = distance from observer to horizon point (assuming no obstacles and level ground)

Bulge = 116.33 meters -- that's the hump in the middle as you found

Drop = 465.33 meters -- the 'drop from a tangent line', as you found

Hidden= 393.57 meters -- how much of an object at that distance the observer would be unable to see due to the Bulge blocking the view (again assuming everything is level and no obstacles)

These all assume a Spherical Earth approximation but you can tweak the radius at your location if you want (but it only changes the results by micrometers for these smaller distances)

Tilt angle is how much an object at that distance would be 'tilted' away from the observer (not very much)

Horizon Dip Angle is more for higher elevations and over open waters, but gives the angle down to the horizon assuming no obstacles.

[Mordred]

Here is a food for thought conjecture I've been considering. If one must use lasers. Could we not use two or more lasers at different elevations.

 

Essentailly several parallel beams, then use a similar technique to look for curvature deviation via deviations on parallel transport.

 

Much the same manner as one goes about calculating geodesics.

 

I've been tinkering with this idea so still mulling it over...

Edited September 13, 2016 by Mordred

[michel123456]

 

 

You might appreciate this calculator: https://www.metabunk.org/curve/?d=77&h=3&r=6371&u=m&a=n&fd=60&fp=3264

 

For 77km and an observer height of 3 meters it gives us the following:

 

Distance = 77 km (77000 m), View Height = 3 meters Radius = 6371 km (6371000 m)

Horizon = 6.18 km (6182.72 m)

Bulge = 116.33 meters

Drop = 465.33 meters

Hidden= 393.57 meters

 

With Standard Refraction 7/6*r, radius = 7432.83 km (7432833.33 m)

Refracted Horizon = 6.68 km (6678.1 m)

Refracted Drop= 398.85 meters

Refracted Hidden= 332.65 meters

 

Tilt Angle = 0.692 Degrees, (0.0121 Radians)

Horizon Dip Angle = 0.056 Degrees, (0.0010 Radians)

Most of these are pretty obvious and the full results explain some of the math used but very quickly...

Horizon = distance from observer to horizon point (assuming no obstacles and level ground)

Bulge = 116.33 meters -- that's the hump in the middle as you found

Drop = 465.33 meters -- the 'drop from a tangent line', as you found

Hidden= 393.57 meters -- how much of an object at that distance the observer would be unable to see due to the Bulge blocking the view (again assuming everything is level and no obstacles)

These all assume a Spherical Earth approximation but you can tweak the radius at your location if you want (but it only changes the results by micrometers for these smaller distances)

Tilt angle is how much an object at that distance would be 'tilted' away from the observer (not very much)

Horizon Dip Angle is more for higher elevations and over open waters, but gives the angle down to the horizon assuming no obstacles.

 

Nice! +1 (to transfer also to Mick)

with 6km input, as in the experiment, and 1.25 height, it gives 32 cm hidden.

And i was wrong that with 1.50m height the entire shore is visible, you need 2.60m height for a null hidden result.

Here is a food for thought conjecture I've been considering. If one must use lasers. Could we not use two or more lasers at different elevations.

 

Essentailly several parallel beams, then use a similar technique to look for curvature deviation via deviations on parallel transport.

 

Much the same manner as one goes about calculating geodesics.

 

I've been tinkering with this idea so still mulling it over...

That is a lot of work to prove what we already know (that the Earth is round) and I am pretty sure that the precision will not be much better than that of Eratosthenes 2000 years ago. Each cm of lack of precision will have a large impact on the estimated Earth radius. Maybe you'd need another version of the calculator to show that.

[Sandor Szekely]

Thank you again, Boxer for the partial explanation.

 

But still why present the data as a histogram?

 

If we regard this as a statistical sampling of the surface water then it seems to point to the Wiki number of 105 being the mean of the histogram.

 

But this information is useless without knowing what part of the water surface was sampled.

All the flight paths were along the edge.

 

Like the rest of the information provided it requires a proper method statement and declaration of variables to accompany it to make any true sense of it.

 

Further none of this alters the possibility of local flatness of the geoid and ground surface (ie the lake bed) supporting a flat water surface since the geoid 'corrections' are not shown.

Studiot

 

"All the flight paths were along the edge."

 

Correct, this was confirmed by the leader of the LIDAR experiment today. I will get soon the flight path measurement data on the exact same route as our boat measurement. I will share for review.

 

I fully agree with everything else you wrote here.

[Mordred]

My conjecture is looking for deviations of the lasers whether through refraction or curvature. It should help filter out refraction due to turbulence and possibly the lower regions directly above the lake.

That is a lot of work to prove what we already know (that the Earth is round) and I am pretty sure that the precision will not be much better than that of Eratosthenes 2000 years ago. Each cm of lack of precision will have a large impact on the estimated Earth radius. Maybe you'd need another version of the calculator to show that.

The purpose of the experiment isn't to assume curvature or flat...

 

Its pretty clear their were too many errors on his test. Were looking for ways to improve accuracy.

Edited September 13, 2016 by Mordred

[DarkStar66]

Studiot

 

"All the flight paths were along the edge."

 

Correct, this was confirmed by the leader of the LIDAR experiment today. I will get soon the flight path measurement data on the exact same route as our boat measurement. I will share for review.

 

Please note that Mick had emailed the PI of the referenced paper [pdf] who replied as follows:

 

 

from: Zlinszky András <XXX>

to: Mick West <YYY>

date: Mon, Sep 12, 2016 at 4:23 AM

 

[...]

 

As first author of the publication "Observation of a local gravity potential isosurface by airborne LIDAR of Lake Balaton, Hungary", published in Solid Earth 5 355-369, 2014, I clarify the following:

 

with the statement "As far as the resolution of the geoid model allowed, the close correlation of the two data systems confirmed that standing water has a truly level surface" we do NOT state that the water surface would be planar or "flat". This means that it is "level" in terms of following a gravity potential isosurface at a constant orthometric height above the geoid. The geoid is a curved surface, resembling an ellipsoid but with considerable deviations from rotational symmetry ("geoid undulation"). Only over a very small surface can this be approximated as flat. 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.

 

I am not in collaboration with Sándor Székely, nor has he ever contacted me. In the cited paper, we have a co-author, dr. Balázs Székely, who is NOT to be confused with Sándor Székely.

 

Best regards,

 

András Zlinszky

 

[Mordred]

by the way Dark thanks for this link.

 

http://aty.sdsu.edu/~aty/explain/atmos_refr/bending.html

[studiot]

Mordred

Here is a food for thought conjecture I've been considering. If one must use lasers. Could we not use two or more lasers at different elevations

 

Isn't one poorly managed laser moredred than enough?

 

I noted the safety implications earlier.

Any laser with enough punch to produce a visible light by day at several km range is potentially dangerous.

Yet none of the crew were wearing eye protection and several were seen looking into the beam in the video.

Perhaps Mr Maker's comments about beam spread were an underestimate - this would lead to a safer beam.

 

Further the measurement of vertical angles by ordinary means as would allow direct calculation avoiding the built in corrections for curvature available in more sophisticated equipment.

 

Dark star

 

Thank you for your comments, I note your post#108 and others

 

I see nothing in the composite fig2 posted in post#1 to substantiate or refute the statement that the lidar survey shs the water surface to closely followed the local curve of the geoid, although I would expect this to be the case.

There is just not enough information there to make such a judgement.

 

Of course there would still be a geoid curve if the Earth was flat and the water would still more or less follow it, as would the underlying lakebed.

 

Sandor

 

Thank you for your recent replies, it would be good if you would also reply to my most recent posts as they contain genuine well intentioned advice to help your group proceed.

[Mordred]

 

Isn't one poorly managed laser moredred than enough?

 

I noted the safety implications earlier.

Any laser with enough punch to produce a visible light by day at several km range is potentially dangerous.

Yet none of the crew were wearing eye protection and several were seen looking into the beam in the video.

Perhaps Mr Maker's comments about beam spread were an underestimate - this would lead to a safer beam.

 

Further the measurement of vertical angles by ordinary means as would allow direct calculation avoiding the built in corrections for curvature available

Lol a fixed mount of the additional lasers would help. Quite frankly the safety aspects should have been addressed in the first place....

 

Including anyone on the shoreline and other boaters

[Sandor Szekely]

Sandor, in response to your post#92, my thoughts now are.

 

1) I am pleased if you wish to turn this into a proper scientific experiment.

 

2) Go to the university and find a proper surveyor. Buy her a cup of coffee (take her out to dinner?) and run my post#88 and the rest of these thoughts past her.

 

3) Send the laser back to the fairground and save the money.

 

4) Depending upon your budget hire a shore base tacheometer (Wild RDS etc) or a Total station.

 

5) Fix an outrigger pontoon on the boat and paint a large target at water level.

 

6) Sight on this target with your new equipment. It will give you sufficiently accurate horizontal distance and heighting over the ranges you are actually using.

 

7) Track the boat + pontoon on a series of traverses across the bay during daylight hours. This leaves the night for dancing etc.

 

8) Good luck in your mission, should you choose to accept it.

 

9) This post will self destruct in 9,000 years.

1. Thanks that is my intention!

 

2. Yeah I did find the beast possible geodezists and geophysisists to work with (not "she" but all "him"). I will name my partners after we have signed our cooperation agreement.

 

3. I love that laser and we are improving it. We will have a much finer increment adjusting of the beam and as well we will calibrate it in a longer distance.

 

4- We will surely use geodezy equipments in the next experiment with professional stuff.

 

5.-7. I am thankfull if you write your ideas of the proper or best possible method of making the measurements. We will model and make real test of the best ones.

 

8. Thank you again! I am commited to find the truth.

 

9. Time is an illusion

(think about the one hour you spend at the dentist - or your loved ones)

 

Yes it is complicated.

But if you replace the bulge with a wall 70 cm high exactly in the middle, it gives the same effect.

Which makes me think that from 1.50 height, even the shore is visible. 1.40m is the limit.

From 1.25m, it must be calculated.

Bizarre. That would make the beam go left & right, not only up & down.

I am answering in time order (to the comments I can) but this comment caught my eye.

 

"Bizarre. That would make the beam go left & right, "

 

that is exactly what happened in our 2nd experiment! It was before midnight, and the beam was like waving left to right in a very noticable amount viewing from the boat. I even called the laserist on the shore if they are moving the laser? she said of course not the laser is not touched.

The effect was exactly as you described it - a huge (like 10 meters) swing looking at the beam from the boat.

We saw this effect at the midnight measurement only.

 

Yes, I would also say : "bizarre"

 

Thanks for the reply, Sandor.

 

I checked your online folder. In C2.jpg and C3.jpg you can see almost the whole circumference of the beam spot. I would estimate the diameter at around 20cm, maybe 30 cm at least as the top of the spot is not visible. That would mean a 20 cm minimum spread at 400m to 600m distance. Please let me know if I am wrong here as this is the basis for my whole analysis

 

Image below is C2.jpg for reference.

 

https://drive.google.com/file/d/0B2gyF12ygRBjZUhnMWRPZXhza2M/view?usp=sharing

 

 

With a 0.08 mRad collimator, the expected diameter should be around 7cm to 8cm at 400m (I am estimating a 4cm initial beam size, again, let me know if I am wrong), but I can see it's a little bigger than that. I think the refraction is disrupting your beam a lot, or maybe the laser was not well calibrated.

 

In the C18.jpg image you attached, the beam is not hitting the white board anymore, so we need to work with math only. With a 0.08mRad, at 5.038m distance, the spot diameter would be around 44cm. But at 400m, it looked like the spot was already at 20cm, so at 5km distance I would estimate it to be 205cm, or 2.05m. It's consistent with the animated gif you posted, as we can see the beam hitting the reflective jacket several times as the boat moves horizontally. The laser could be hitting the jacket anywhere in the 2m spot I think.

 

With all that said, I am happy that you are planning to get a 0.003mRad collimated laser in the second experiment, but you need to make sure that it's well calibrated, as the 0.08 mRad you used in this first experiment was not doing well. Or if the disruption was caused by refraction only, this also needs to be taken care of.

 

Well let's start with the board. It is plastic with 0.5 cm inner square tubing (sold in Praktiker garden shops)

I am not sure this is a good solution for the board as it leads the lights internally well so the beam looks bigger.

Next time I would use a normal plastic sheet or a retroreflective material - not sure what would be the best.

 

Our opinion is that the beam was not spreading out very much over the long distance.

Here I attach my comparison on the beam divergence at 2 distances.

 

 

Fine calibration of the laser beam can be an issue, next time we do it on longer distance. (it was now calibrated on about 100 meters distance)

 

"With a 0.08mRad, at 5.038m distance, the spot diameter would be around 44cm."

 

this could be realistic in my opinion.

 

"But at 400m, it looked like the spot was already at 20cm, so at 5km distance I would estimate it to be 205cm, or 2.05m."

 

Then the beam center would be about 1 meter (radius) so the beam should be visible at a height of minimum 2.45 meters (3.45 - 1).

The beam was recorded at 1.7 meters so it is still a big difference (0.75 meter).

I think our beam was well collimated. The motion gif picture is a high speed very distorted image (1200mm optics) that shows the reflexion in the camera lens when we are pointing it directly into the laser.

 

here I attach the corresponding dat, C18 at the last row:

 

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

 

 

[Mordred]

Well glad to see the approach to eliminate systematic errors. I've decided to pick up a couple of cheap lasers for myself lol.

 

I want to test my idea...

 

(curiosity killed the cat...)

Edited September 14, 2016 by Mordred

[Sandor Szekely]

 

 

This image perfectly shows that the laser beam divergence is fairly large as you are clearly getting reflections over several meters wide as the boat slews across the view. These clips are found in the video starting at https://youtu.be/GBhDFO4NMrw?t=1111 (severalbroken clips from that point) and you can clearly see the boat is slewing very much side-to-side, it is not moving straight away from the laser, and you can see these glints across a very wide swath.

 

So all your showing is that a very divergent laser beam is hitting the reflective patch on the back of the white jacket and sometimes glinting off the camera lens pointing back at the dock.

 

These are not accurate measurement methodologies and do not indicate the center of the laser beam.

 

This, combined with the ~10cm error in your 'slope corrected leveling process' mean this experiment does not give us accurate results.

 

Here is my summary with l inks to the main metabunk posts that support them (credit to Mick for the analysis here, I'm just summarizing):

 

1) measuring height along the slanted board, likely resulting in ~9% error in 'slope corrected leveling' method [link1,link2] - confirmed by observing measuring methodology used and also by photo analysis. This likely resulted in a slight initial downward slope on the laser.

2) laser beam divergence is clearly much greater than they are admitting, creating false hits (shown in #3 & #5)

3) reflective patch on back of jacket and glints off camera lens giving false 'hits' [link1, link2]

4) 'direct hit in camera' is also demonstrated as not reliable (combined with #2 & #3) [link]

5) evidence that the laser is initially pointed slightly down (data at C5 & C8) [link]

 

If these are wrong please post the images that refute them. Here is kind of a guide as to what I think would be needed to address these points...

 

1) you would need images/video to show that the 130cm marking tape was placed accurately. We can see them using inaccurate methodology (slanted measurements) in other parts of the video but this isn't shown. So you can refute this by showing that the same methodology was not used when placing the 130cm tape mark but that it was made very plumb to the surface of the lake and not made by measuring the distance on the slanted board.

2) you would need to show the laser fully hitting the board at each point past ~C5 showing at least 1/2 extents of the beam divergence. All the evidence I can see past C4 is an increasingly divergent beam. Hits on the reflective tape and lens of the camera do not count to show that the laser divergence is small when we can plainly see in the video that this is not the case (something you seem to have a problem accepting).

3) the reflective patch and camera lens glints are pretty much self explanatory as being invalid. I don't see how you can show otherwise but I'm open to any further evidence presented.

4) 'direct hit on the camera' methodology is just busted - you can see the same effect in the camera even when the laser spot is small on the board in the closer measurements.

5) demonstrating #1 would call this into question depending on the results of that further analysis. But already at C8 we can see the beam is so diverged that it's difficult to see the actual extent of the beam hitting the board.

 

Do you have *anything* that shows the beam extents very clearly past C8 or not?

 

You might appreciate this calculator: https://www.metabunk.org/curve/?d=77&h=3&r=6371&u=m&a=n&fd=60&fp=3264

 

For 77km and an observer height of 3 meters it gives us the following:

 

Distance = 77 km (77000 m), View Height = 3 meters Radius = 6371 km (6371000 m)

Horizon = 6.18 km (6182.72 m)

Bulge = 116.33 meters

Drop = 465.33 meters

Hidden= 393.57 meters

 

With Standard Refraction 7/6*r, radius = 7432.83 km (7432833.33 m)

Refracted Horizon = 6.68 km (6678.1 m)

Refracted Drop= 398.85 meters

Refracted Hidden= 332.65 meters

 

Tilt Angle = 0.692 Degrees, (0.0121 Radians)

Horizon Dip Angle = 0.056 Degrees, (0.0010 Radians)

Most of these are pretty obvious and the full results explain some of the math used but very quickly...

Horizon = distance from observer to horizon point (assuming no obstacles and level ground)

Bulge = 116.33 meters -- that's the hump in the middle as you found

Drop = 465.33 meters -- the 'drop from a tangent line', as you found

Hidden= 393.57 meters -- how much of an object at that distance the observer would be unable to see due to the Bulge blocking the view (again assuming everything is level and no obstacles)

These all assume a Spherical Earth approximation but you can tweak the radius at your location if you want (but it only changes the results by micrometers for these smaller distances)

Tilt angle is how much an object at that distance would be 'tilted' away from the observer (not very much)

Horizon Dip Angle is more for higher elevations and over open waters, but gives the angle down to the horizon assuming no obstacles.

 

Hello Darkstar

 

I don't agree with your assumption that the laser beam is spread out like 2 meters at that distance in the boat loop video.

The laser center height should be over 3 meters so therefore that slice of the circle - that is 2 meters wide - would suggest a huge laser beam like 10 meters. In this case we would see the laser hitting the board too, but it is not.

SO I think that the 1200mm optics is tricking us here.

 

WE have no error at the slope correction laser leveling. Mick was wrong and instead of admitting it he banned me from metabunk (for "trolling").

We proved in the autocad video that his 118.5 cms leveling theory will result a 3.19 meters beam height instead of the 3.45 meters (not a significant difference).

 

I answered to Mick's comments until I was not kicked from meta. I will not answer them again, as you copy pasted them here, PLS copy paste my answers too.

 

Thanks for sharing the meta calculator, but I am banned from meta. (reason: I am trolling... can someone explain me how I can be trolling on my own thread?)

 

so Mick case is closed, I am not debating with him. (unless he is here)

 

3) the reflective patch and camera lens glints are pretty much self explanatory as being invalid.

 

I don't see them invalid.

 

the refraction calculator is a joke...

refraction is DUE to CHANGE. so how can we estimate a "standard" constant change?

 

where can I put : time of the day, temperatures, humidity values and so on into that calculator?

what is the change here? only distance? this calculator was made to give the near same value as curvature. but this is not how terresterial refraction is calculated.

 

The sun is not where we see it, we see an apparent sun (exept when the sun is directly 90 degrees above us) - that is a huge atmospheric refraction. That is constant due to atmosphere conditions. But the terresterial refraction caused by local differences that curve light, it can not be calcualted with that calculator...

Well glad to see the approach to eliminate systematic errors. I've decided to pick up a couple of cheap lasers for myself lol.

 

I want to test my idea...

 

(curiosity killed the cat...)

Cool if you are trying your self too!

 

Please share your findings, and as well your difficulties with making such an experiment, like leveling with this accuracy.

 

I forgot to write that you can get pretty good collimators cheap for hand held lasers too, that is much more important than the power of the unit.

Lol a fixed mount of the additional lasers would help. Quite frankly the safety aspects should have been addressed in the first place....

 

Including anyone on the shoreline and other boaters

 

We have discussed the safety aspects of the laser with my laserist well ahead of the experiments. As this it was used as a 3W version it was safe to look into with the naked eye from over a kilometer distance. Actually we tried the sunglasses but at night time let's say it is impossible to wear in a boat.

A camera CCD is more sensitive to the laser beam but none of the cameras had any problem eighter.

 

We used green and blue dual lasers at the pre-test but we had collimator only for the green now. It would be good to use the blue to see the refraction (as the 2 colours have different refraction).

 

Our laserist and the Goldlaser company we work with is the biggest Hungarian outdoor laser show (and laser theater). They do events every week and well aware of the safety aspects.

Nudtz isn't a standard terminology. You won't find that terminology used in any textbooks. In point of detail the attempt to add it to Wiki has recently been placed on the deletion list.

 

http://deletedwikipedia.gawker-labs.com/wiki/N.U.D.T.Z.

 

I would just ignore this term as invalid

 

Particularly when the video states they sink into the NUDTZ ????

 

Either way there was far greater actual data presented on the other site to work with than was presented by the OP here. There is far too many systematic errors involved. The major issues posted by darkstar above.

We named the zone that we

non uniform density (that is the water vapour above the lake surface)

transition zone (meaning the zone - height above the surface)

 

I see this is the best explanation to what we experience.

 

why do you think it is invalid?

 

"far too many systematic errors" is not someting I can argue about, pls nem that you think is not accurate and pls define your point how much it effects our outcome.

 

I don't like this: Mick said this already at meta... first of all I can not see those posts, second the opposion claims should be explained here not referenced.

To explain my concern, on refractive index lets consider an example. The refractive index is typically highest the closer to the surface. Lets assign value N_1. At higher elevations temperature and pressure drop. Giving refractive index N_2.

 

Snells law [latex]\frac {sin\theta_1}{sin\theta_2}= \frac {N_2}{N_1}[/latex]

 

So as you increase elevation the ray will curve downward. This occurs regardless of the curvature of the Earth.

 

The humidity along the water surface will be higher than the ambient air. This gets into the potential of ducting...

 

The question is can we determine that the OP has the right conditions for sub-refraction? Which is possible but I don't see sub-refraction with the environment conditions mentioned in the OP..

Too bad the OP can't provide the proof that sub-refraction and not ducting is occurring. The quoted section occurs more readily over deserts with temperature inversion. Its a different situation with humidity of 71%.

 

[latex]N=77.6\frac{P}{T}+3.73*10^5\frac {e}{T^2}[/latex]

 

first term is dry air second term moist air. Key note the temperatures may invert but the pressure does not. In point of detail the moist air off the lake increases the pressure near the surface

 

That quoted section is just too convenient as "JUST THE RIGHT conditions to get the results I want...

 

I've been tinkering with the numbers all night. I can't find a combination that has a refractive curve upward from the Earths surface. Not within reasonable temperature variations. There is an extent in range of the signal due to reduced curvature but in all cases there is still a slight downward curve.

 

Anyone willing to show the calcs of upward refraction?

Mordred, please explain more about refraction conditions that could have occured at the experiment.

(or anyone who is expert in refraction)

 

This is an interesting thought " I can't find a combination that has a refractive curve upward from the Earths surface."

I assume that with inverted conditions from day to night the direction of refraction should change too. Of course there are many other factors to consider too. But this would really suprise me that refraction is always curving downwards.

Actually I am not sure if there is a possiblity to gather all ambient data - like we should measure the temp / hum factor all the way too.

So I think we should find at least the best possible timing of the experiment. We have to concider that a strong sunlight makes the laser invisible.

 

Please share advices on timing and weather conditions to look for in the upcoming experiment.

Edited September 14, 2016 by Sandor Szekely

[Mordred]

We named the zone that we

non uniform density (that is the water vapour above the lake surface)

transition zone (meaning the zone - height above the surface)

 

I see this is the best explanation to what we experience.

 

why do you think it is invalid?

Physics is all about what you can mathematically define. In this case the thermodynamic state of this "self named region" requires a metric modelling. As well as the corresponding thermodynamic degrees of freedom of the system state being described.

 

Hence it is an undefined region, until you can properly define it

 

Mordred, please explain more about refraction conditions that could have occured at the experiment.

(or anyone who is expert in refraction)

 

This is an interesting thought " I can't find a combination that has a refractive curve upward from the Earths surface."

I assume that with inverted conditions from day to night the direction of refraction should change too. Of course there are many other factors to consider too. But this would really suprise me that refraction is always curving downwards.

Actually I am not sure if there is a possiblity to gather all ambient data - like we should measure the temp / hum factor all the way too.

So I think we should find at least the best possible timing of the experiment. We have to concider that a strong sunlight makes the laser invisible.

 

Please share advices on timing and weather conditions to look for in the upcoming experiment.

I wouldn't call myself an expert in atmospheric refraction. It's been awhile since I last practiced those particular equations.

 

To be 100% honest with you, I honestly feel learning how to calculate the best times yourself and show the refraction corrections on your paper is an important step in eliminating error margins.

 

I can certainly help explain how those equations work. (once you understand them, they can easily be programmed)

 

There are calcs available but if you don't understand how pressure density and temperature affects refraction and how to mathematically model such in an atmosphere they won't do much good.

 

Most calculators are typically designed with the basic formulas. They don't often have elevation corrections etc.

I'll post some metric examples when I get a chance. Takes a bit of time to post.

Edited September 14, 2016 by Mordred

[DarkStar66]

I don't agree with your assumption that the laser beam is spread out like 2 meters at that distance in the boat loop video.

The laser center height should be over 3 meters so therefore that slice of the circle - that is 2 meters wide - would suggest a huge laser beam like 10 meters. In this case we would see the laser hitting the board too, but it is not.

SO I think that the 1200mm optics is tricking us here.

 

The horizontal distance the boat travels is very clear in the video. Your white board is 2 meters wide so you can easily mark that width on the video and see the extents showing how far to the left and right the reflective patch and camera lens are showing reflections. You keep denying this is the case but that doesn't make the clear evidence go away.

 

WE have no error at the slope correction laser leveling. Mick was wrong and instead of admitting it he banned me from metabunk (for "trolling").

We proved in the autocad video that his 118.5 cms leveling theory will result a 3.19 meters beam height instead of the 3.45 meters (not a significant difference).

 

I answered to Mick's comments until I was not kicked from meta. I will not answer them again, as you copy pasted them here, PLS copy paste my answers too.

 

Thanks for sharing the meta calculator, but I am banned from meta. (reason: I am trolling... can someone explain me how I can be trolling on my own thread?)

 

so Mick case is closed, I am not debating with him. (unless he is here)

You don't have to debate anything - the information is here for others to evaluate and review.

I linked to the evidence showing the photographic analysis showing the 130cm mark is incorrect.

The analysis was that the 1m tall board was ~452 pixels (which gives 452 pixels/meter).

Total vertical height to the top of tape marked "1.3m" = 536 pixels.

This is closer to ~1.186m than 1.3m

Sandor Szekely wrote: "On the photos you can tell the height of the laser by calculating the number of pixels above the board and comparing them with the board size. If you use a high resolution photoshop you can calculate quite precisely."

Isn't that exactly what you did in all of your photos past about C8? Because you don't have a board even, you have a reflection off a special reflector on one guys coat and off the camera lens and that's it. But you called all of those slightly increasing heights by counting pixels - correct?

This is exactly what Mick did and anyone can replicate it as he provided the details in his video.

They also estimated that your board was slanted at ~24° and we see the people on the boat repeatedly putting the measuring tape along side the slanted board.

This puts your measurements at ~1.3*cos(24°) off -- that's about .91 so that puts you at about 1.188 -- which agrees with the photographic analysis.

So there are two data points that suggest this error is real.

You can easily refute this -- please show the photo or place in the video where this tape is marked at 1.3m and it was done accurately with the tape measure fairly plumb and not slanted at ~24° along the board.

it's that simple really. Maybe we just missed it.

But continuing to assert that everything is perfect doesn't make your case. There is very good reason here to believe that this mark was placed incorrectly. That throws everything else about your calculations off.

You also did not measure or demonstrate in any sufficient way the beam divergence past about C8 -- and you also have not accounted for the beam divergence in your calculations. Assertions do not make the data reliable.

Again - your evidence is only needed if you want to convince others, or you can continue to ignore this -- it's really up to you.

3) the reflective patch and camera lens glints are pretty much self explanatory as being invalid.

 

I don't see them invalid.

I gave the evidence supporting my argument, perhaps that could be addressed?

I know that's painful to hear but these are absolutely critical and deep flaws in the methodology and measurements here. As I'm sure you are aware.

=========== the rest of this is mostly irrelevant commentary on the aforementioned calculator ==================

the refraction calculator is a joke...

First of all, NOTHING about the refraction portion here was mentioned in the context of your experiment. I was showing another member that the calculator exists and is very handy and the results matched their numbers which gives us confidence in both. It simply uses a "surveyor standard refraction" as a GUIDE to what one would expect under fairly common conditions (see below).

This, in no way, makes it a 'joke'. One must use information appropriately, as always.

refraction is DUE to CHANGE. so how can we estimate a "standard" constant change?

If by change you mean the medium contains components with different refractive indices then yes. But refraction doesn't require some kind of dynamic 'change' to make it happen. Everything could be entirely stable. In practice it isn't but that's a different store as to what refraction is "DUE to".

It is just a tool -- you are free to go write your own refraction calculator. Mick & I have discussed our desire to do exactly that but it is an extremely complex and large project. 7/6*R is very easy and gives you a ballpark/starting point to consider. Perhaps it would be wiser to consider it since every surveyor knows that you must account for refraction when shooting longer sightlines.

this calculator was made to give the near same value as curvature. but this is not how terresterial refraction is calculated.

Note even remotely, it uses 7/6*R (see next bit)

The sun is not where we see it, we see an apparent sun (exept when the sun is directly 90 degrees above us) - that is a huge atmospheric refraction. That is constant due to atmosphere conditions. But the terresterial refraction caused by local differences that curve light, it can not be calcualted with that calculator...

 

It's not really all that 'huge' -- on the horizon, for the Sun, the "average amount of atmospheric refraction" is given as ~34 arcminutes (slightly more than one apparent solar disc). That's going through the entire atmosphere the long way.

 

The calculator uses the approximation given by ATY based on a surveying rule of thumb:

 

Usually, the air is densest at the surface, so the rays of light are concave toward the surface; see the bending page for details.

...

This assumption is made so often that it's conventional in surveying and geodesy to use a “refraction constant” that's just the ratio of the two curvatures. A typical value of the ratio is about 1/7; that is, the ray curves about 1/7 as much as the Earth does (or, equivalently, the radius of curvature of the ray is about 7 times that of the Earth's surface).

Using this “typical” value means we should just use the formula given above, but use a value R′ instead of R for the effective radius of the Earth, where

1/R′ = 1/R − 1/(7R) = 6/(7R) ,

so that

R′ = R × 7/6 .

 

 

[michel123456]

(...)

Sandor Szekely wrote: "On the photos you can tell the height of the laser by calculating the number of pixels above the board and comparing them with the board size. If you use a high resolution photoshop you can calculate quite precisely."

Isn't that exactly what you did in all of your photos past about C8? Because you don't have a board even, you have a reflection off a special reflector on one guys coat and off the camera lens and that's it. But you called all of those slightly increasing heights by counting pixels - correct?

This is exactly what Mick did and anyone can replicate it as he provided the

.

They also estimated that your board was slanted at ~24° and we see the people on the boat repeatedly putting the measuring tape along side the slanted board.

 

If the board was slanted at ~24° then No, you cannot "tell the height of the laser by calculating the number of pixels above the board and comparing them with the board size".

 

In any case, even with a vertical board* it is unsafe to measure on a photo. From my experience, every time there is a significant lack of precision.

 

*not even vertical, it should be perpendicular to the laser beam, and the photo taken from the origin of the laser beam, which is quite difficult (it should be installed like a rifle scope)

[Sandor Szekely]

Physics is all about what you can mathematically define. In this case the thermodynamic state of this "self named region" requires a metric modelling. As well as the corresponding thermodynamic degrees of freedom of the system state being described.

 

Hence it is an undefined region, until you can properly define it

I wouldn't call myself an expert in atmospheric refraction. It's been awhile since I last practiced those particular equations.

 

To be 100% honest with you, I honestly feel learning how to calculate the best times yourself and show the refraction corrections on your paper is an important step in eliminating error margins.

 

I can certainly help explain how those equations work. (once you understand them, they can easily be programmed)

 

There are calcs available but if you don't understand how pressure density and temperature affects refraction and how to mathematically model such in an atmosphere they won't do much good.

 

Most calculators are typically designed with the basic formulas. They don't often have elevation corrections etc.

I'll post some metric examples when I get a chance. Takes a bit of time to post.

We did do approximation on the refraction at different times of the day that is how we came to the conclusion to do the experiments at night time and dawn. It is very hard to model (actually impossible) to model the terresterial refraction as it is due to change in small local conditions.

Like for example a turbulence is not possible to aviod if present.

We should try to define the best possible timing. As daytime has more changes in ambient conditions (more sunlight energy, wind factor) and we can see mirage only at daytime and laser is not well seen in daylight: I figured out that this is not the good time for the laser test.

 

Thank you for posting some examples so we can evaluate the data for the best possible timing of the measurement.

[michel123456]

I went back to Mike's calculator

https://www.metabunk.org/curve/?d=77&h=3&r=6371&u=m&a=n&fd=60&fp=3264

 

A small remark: (I did the same mistake)

The input distance should not be the straight line (the chord) but the length of the arc

 

I did the same mistake because in Autocad there is no way (at my knowledge) to draw an arc of a specific length. So I choose the facility and draw a straight line.

But the distances on Earth (as the length of the lake) are geodesics, they are not straight lines.

[Mordred]

We did do approximation on the refraction at different times of the day that is how we came to the conclusion to do the experiments at night time and dawn. It is very hard to model (actually impossible) to model the terresterial refraction as it is due to change in small local conditions.

Like for example a turbulence is not possible to aviod if present.

We should try to define the best possible timing. As daytime has more changes in ambient conditions (more sunlight energy, wind factor) and we can see mirage only at daytime and laser is not well seen in daylight: I figured out that this is not the good time for the laser test.

 

Thank you for posting some examples so we can evaluate the data for the best possible timing of the measurement.

Actually there may be an aspect to consider here. Over large bodies of water there is a specific type of Ducting.

 

Evaperation duct. I'm still looking into this but if I'm right your chosen height of the laser may have been too low...

 

Did you perchance recall the wind speed in knots? the evaperation duct height calculation requires that detail.

 

The back of envelope naval chart I'm looking at gives roughly 3 metres in height for the water temp/air temp in the video at zero knots.

 

Best time of day for sampling

 

when the temperature of the atmosphere most closely matches the surface temperature of the water. Reduces the potential evaperation duct height due to humidity gradient.

 

Bigger boat and higher height above water surface highly recommended to get to a more uniform elevation distribution. (with as little wind as possible)

Edited September 14, 2016 by Mordred

[Sandor Szekely]

 

The horizontal distance the boat travels is very clear in the video. Your white board is 2 meters wide so you can easily mark that width on the video and see the extents showing how far to the left and right the reflective patch and camera lens are showing reflections. You keep denying this is the case but that doesn't make the clear evidence go away.

 

You don't have to debate anything - the information is here for others to evaluate and review.

I linked to the evidence showing the photographic analysis showing the 130cm mark is incorrect.

The analysis was that the 1m tall board was ~452 pixels (which gives 452 pixels/meter).

Total vertical height to the top of tape marked "1.3m" = 536 pixels.

This is closer to ~1.186m than 1.3m

Sandor Szekely wrote: "On the photos you can tell the height of the laser by calculating the number of pixels above the board and comparing them with the board size. If you use a high resolution photoshop you can calculate quite precisely."

Isn't that exactly what you did in all of your photos past about C8? Because you don't have a board even, you have a reflection off a special reflector on one guys coat and off the camera lens and that's it. But you called all of those slightly increasing heights by counting pixels - correct?

This is exactly what Mick did and anyone can replicate it as he provided the details in his video.

They also estimated that your board was slanted at ~24° and we see the people on the boat repeatedly putting the measuring tape along side the slanted board.

This puts your measurements at ~1.3*cos(24°) off -- that's about .91 so that puts you at about 1.188 -- which agrees with the photographic analysis.

So there are two data points that suggest this error is real.

You can easily refute this -- please show the photo or place in the video where this tape is marked at 1.3m and it was done accurately with the tape measure fairly plumb and not slanted at ~24° along the board.

it's that simple really. Maybe we just missed it.

But continuing to assert that everything is perfect doesn't make your case. There is very good reason here to believe that this mark was placed incorrectly. That throws everything else about your calculations off.

You also did not measure or demonstrate in any sufficient way the beam divergence past about C8 -- and you also have not accounted for the beam divergence in your calculations. Assertions do not make the data reliable.

Again - your evidence is only needed if you want to convince others, or you can continue to ignore this -- it's really up to you.

I gave the evidence supporting my argument, perhaps that could be addressed?

I know that's painful to hear but these are absolutely critical and deep flaws in the methodology and measurements here. As I'm sure you are aware.

=========== the rest of this is mostly irrelevant commentary on the aforementioned calculator ==================

First of all, NOTHING about the refraction portion here was mentioned in the context of your experiment. I was showing another member that the calculator exists and is very handy and the results matched their numbers which gives us confidence in both. It simply uses a "surveyor standard refraction" as a GUIDE to what one would expect under fairly common conditions (see below).

This, in no way, makes it a 'joke'. One must use information appropriately, as always.

If by change you mean the medium contains components with different refractive indices then yes. But refraction doesn't require some kind of dynamic 'change' to make it happen. Everything could be entirely stable. In practice it isn't but that's a different store as to what refraction is "DUE to".

It is just a tool -- you are free to go write your own refraction calculator. Mick & I have discussed our desire to do exactly that but it is an extremely complex and large project. 7/6*R is very easy and gives you a ballpark/starting point to consider. Perhaps it would be wiser to consider it since every surveyor knows that you must account for refraction when shooting longer sightlines.

Note even remotely, it uses 7/6*R (see next bit)

 

It's not really all that 'huge' -- on the horizon, for the Sun, the "average amount of atmospheric refraction" is given as ~34 arcminutes (slightly more than one apparent solar disc). That's going through the entire atmosphere the long way.

 

The calculator uses the approximation given by ATY based on a surveying rule of thumb:

 

 

Please model the beam size on the boat clip. I explained that it is not possible to have 2 meter of the laser at the bottom of the beam, because that would make the beam like 10 meters diameter. Make a drawing that shows that the beam is 2 meters at the 1.8 meter height and it is not extending to the white board, and the beam center is at 3.4 meters.

This is not possible.

 

on the 118.5 cms / 132 cms debate I attached a video that was deleted from meta. Zack clearly explains the math with autocad: the beam should be at 3.19 meters (instead of the 4.32 meters)

That makes marginal difference in the outcome as the beam was recorded at 1.8 m - still a huge 1.4 m difference

 

here I lost all the huge sript I made so far so I have to rewrite it again. does this usually happen on the website? no saved copy

 

I attach here a picture: the first hit of the loop and the last direct hit of the loop.

 

 

 

My question: if the beam was spread out to more meters as you suggest, how come that the beam is not seen on the retroreflective surface few inches below, but only in the camera lenses?

 

 

 

I have explained that NONE of the measurements were taken with the tape. At C1, C2 and C3 leveling process Dave used a tape to tell my laserist a number to calculate the increment of raising the laser. It has nothing to do with the measurements! Dave could as well say only "raise up".

 

We have explained in 2 video already that the photoshop measurements are inaccurate. I can not upload them here. Admin - any suggestions?

 

1st video: explaining the GE and FE calculated expected beam height. The 3.19 meters beam center is still not possible on the GE model.

2nd video: presenting on a real capture of a box why the picture can not be evaluated like Mick did.

 

Your calculation on the board height from angle is inaccurate.

 

"There is very good reason here to believe that this mark was placed incorrectly. That throws everything else about your calculations off."

 

So you believe in something that throws my calcualtion off...

 

" Assertions do not make the data reliable."

 

Please provide me your unrefutable datat that shows that my divergence was not as I said with the 0.08 mRad collimator. Unless you make a model of beam divergence that is coherent with the photo evidences, your claim is only an opinion. Check the above pictures and my concerns on your guess of beam divergence.

 

I see no evidences supporting your argument, you have not even modeled it if it possible.

 

Please refrain from comments like "I know that's painful to hear"

 

You are telling your opinion, do not make it look like you are the teacher here...

 

I lost a lot of script here on the explanation of the terresterial refraction.

 

so I conclude here:

 

You wrote: "But refraction doesn't require some kind of dynamic 'change' to make it happen. Everything could be entirely stable."

this is wrong, refraction is caused by CHANGE in the medium light travels through:

In a constant density, pressure , temperature medium there is no refraction - only retardation.

 

atmospheric refraction is a near constant value depending on the sun angles as sunlight coming from near vacuum and entering more dense, humid, and warmer air with more pressure and CO2

this should be - but not - taken into account when measureing the sun shadow lenght

 

terresterial refraction can not be calculated like this as local conditions effect the outcome very much. A local turbulance or a mirrage effect can never be calculated like this, so this calcualtor is wrong.

 

How come, that we make a good understandable definition of what we experience: the NUDTZ non uniform density transition zone - and it is criticized.

 

How come you suggest a calculator without any references to it's accuracy? I think this calculator is deceptive - I ask other people here not to take that into concideration.

Especially terresterial refraction is impossible to calculate like that on the many different aspects of modifying properties.

 

Standard refraction used in geodezical terrain measurements is 0.4 millimeter / kilometer. Dubble run leveling in 1999 by Swiss authors I have to look for that link)

 

"In order to obtain a hypothesis-free reduction of refraction influences, it is necessary to determine the refraction influences integrally along the propagation path. In the scope of this research work, the evaluation of the measured signal only is assumed to provide a satisfying estimation of the refraction influences. This estimation is based on image processing techniques which use the image data from built-in geodetic image sensors. Additionally, to estimate the refraction influences, the image processing techniques must be combined"

 

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

 

please read section 7 on the laser refraction experiment.

 

your comments on the avarage amount of refraction and the refraction calculator are wrong.

If the board was slanted at ~24° then No, you cannot "tell the height of the laser by calculating the number of pixels above the board and comparing them with the board size".

 

In any case, even with a vertical board* it is unsafe to measure on a photo. From my experience, every time there is a significant lack of precision.

 

*not even vertical, it should be perpendicular to the laser beam, and the photo taken from the origin of the laser beam, which is quite difficult (it should be installed like a rifle scope)

I agree with you that the pixel calculation is not an exact measurement. the best way is to mark all the sublines on the board too. In my measurements this gives some inaccuracy over long distance. in Mick's calculation it gives a greater difference.

I went back to Mike's calculator

https://www.metabunk.org/curve/?d=77&h=3&r=6371&u=m&a=n&fd=60&fp=3264

 

A small remark: (I did the same mistake)

The input distance should not be the straight line (the chord) but the length of the arc

 

I did the same mistake because in Autocad there is no way (at my knowledge) to draw an arc of a specific length. So I choose the facility and draw a straight line.

But the distances on Earth (as the length of the lake) are geodesics, they are not straight lines.

Of course AutoCad can calculate the lenght of an arc

Actually there may be an aspect to consider here. Over large bodies of water there is a specific type of Ducting.

 

Evaperation duct. I'm still looking into this but if I'm right your chosen height of the laser may have been too low...

 

Did you perchance recall the wind speed in knots? the evaperation duct height calculation requires that detail.

 

The back of envelope naval chart I'm looking at gives roughly 3 metres in height for the water temp/air temp in the video at zero knots.

 

Best time of day for sampling

 

when the temperature of the atmosphere most closely matches the surface temperature of the water. Reduces the potential evaperation duct height due to humidity gradient.

 

Bigger boat and higher height above water surface highly recommended to get to a more uniform elevation distribution. (with as little wind as possible)

Yes I think the ducting made some strange effects like the beam swinging sideways horizontally in our measurements from sunset to midnight as the laser was at 50cms (4.1 feet). This effect was not so much at the 4AM measurement this effect was reduced, no horizontal swinging of the laser beam.

 

We agree that the best timing is after sunrise when all temperature parameters are similar. We agree that raising the laser to higher position reduces this problem, we thought of raising the laser to 3 meters.

After sunset the water until the morning is very calm, low or no waves and low wind factor (on the 16th of Aug).

[michel123456]

Of course AutoCad can calculate the lenght of an arc

Sure. But you cannot input the length in the command line(not in my version of Autocad). You draw an arc and then measure it.

[MrMaker]

 

Well let's start with the board. It is plastic with 0.5 cm inner square tubing (sold in Praktiker garden shops)

I am not sure this is a good solution for the board as it leads the lights internally well so the beam looks bigger.

Next time I would use a normal plastic sheet or a retroreflective material - not sure what would be the best.

 

Our opinion is that the beam was not spreading out very much over the long distance.

Here I attach my comparison on the beam divergence at 2 distances.

 

laser beam divergence.png

 

Fine calibration of the laser beam can be an issue, next time we do it on longer distance. (it was now calibrated on about 100 meters distance)

 

"With a 0.08mRad, at 5.038m distance, the spot diameter would be around 44cm."

 

this could be realistic in my opinion.

 

"But at 400m, it looked like the spot was already at 20cm, so at 5km distance I would estimate it to be 205cm, or 2.05m."

 

Then the beam center would be about 1 meter (radius) so the beam should be visible at a height of minimum 2.45 meters (3.45 - 1).

The beam was recorded at 1.7 meters so it is still a big difference (0.75 meter).

I think our beam was well collimated. The motion gif picture is a high speed very distorted image (1200mm optics) that shows the reflexion in the camera lens when we are pointing it directly into the laser.

 

here I attach the corresponding dat, C18 at the last row:

 

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

 

 

Thanks again for the clarifications, Sandor.

 

It's good to know that you want to enhance the experiment, that's always a good practice.

 

Regarding the board, I wouldn't use a semi-transparent material anymore. This may be causing extra refraction and disrupting your data. I would suggest using an opaque material, taking the measures where the laser hits, and not on the other side. I would also strongly suggest you to not use any reflective material as a mean to capture the data. You want the light to be absorbed, not reflected. The reflective one is probably a good idea to help finding the beam, but not measuring it.

 

I agree with your point that 3.45m - 1m is also not looking good for the GE model, but I was intrigued by the "moving laser" effect you mentioned. If a laser can move up to 2m left and right, I think it's possible it may be also moving up and down by the same amount. A well collimated laser (lets say 40cm at 5km) may cause the top and bottom limits of the laser to be 2.4m off, and a not well collimated laser (let's say 2m at 5km) may cause the top and bottom limits to be up to 4m off. I had no idea that this NUDTZ effect, like you call it, could have such a big impact. Very interesting.

 

Only a quick correction, the beam center is always ~0 nm, not 1m. . But anyway, you need to be 100% sure about what the beam size and beam dispersion are. From the pictures you sent, I was making a wild estimation of 2.05m at 5km, but it could easily be 40cm or 4m, there is no way to tell. The images, as you mentioned, are from too far away.

 

My suggestions, some of them already mentioned in the thread:

• Put your laser higher than 1.25m. The refraction at this height is causing too much variation to the data.
• Use a different board material, preferably opaque.
• Definitely a higher board. Do you think you can make it at least 5m? Can you get a bigger boat so that bigger structures can be mounted there? 10m would be cool
• Validate the beam dispersion. Measure the top and bottom of the laser spot at 10m, 100m, 500m, 1km, 2km. After that, you go back and start the experiment. This is only a calibration round.
• Same water and air temperature
• Perpendicular board as it's easier to measure heights. If possible, pre-mark the heights there so that you don't need to measure it during the experiment.
• More measuring: Top and Bottom of the laser spots every 500m. This will give you 24 measurements every time you run the experiment.
• Don't switch measuring tools (from board to jacket).

 

I know it's too much to ask, but you have an uphill battle ahead of you.

Edited September 14, 2016 by MrMaker

[DarkStar66]

Our opinion is that the beam was not spreading out very much over the long distance.

Here I attach my comparison on the beam divergence at 2 distances.

 

laser beam divergence.png

 

Fine calibration of the laser beam can be an issue, next time we do it on longer distance. (it was now calibrated on about 100 meters distance)

 

"With a 0.08mRad, at 5.038m distance, the spot diameter would be around 44cm."

 

this could be realistic in my opinion.

 

"But at 400m, it looked like the spot was already at 20cm, so at 5km distance I would estimate it to be 205cm, or 2.05m."

 

Then the beam center would be about 1 meter (radius) so the beam should be visible at a height of minimum 2.45 meters (3.45 - 1).

The beam was recorded at 1.7 meters so it is still a big difference (0.75 meter).

I think our beam was well collimated. The motion gif picture is a high speed very distorted image (1200mm optics) that shows the reflexion in the camera lens when we are pointing it directly into the laser.

 

Here is my analysis of the C5 image, the actual, observed, beam spread here is nowhere near 0.08 mRad - whatever you want to attribute the cause, perhaps it's all atmospheric, but it is obviously very large.

 

The spot on the board is almost 70cm across at just 870 meters. That's huge and it gets worse with distance as you clearly see by looking at C2 on up until it's off the board. In C11 you can see it almost covering the whole board but yet, you mark the EXACT SAME specular reflection off the metal bar as the height which you claim is "1.47m" even though it's the EXACT SAME SPOT as in C5 which you called 1.32m in your spreadsheet. How did you get a 15cm rise when the spot on the boat is clearly exactly the same between these two images (C5 and C11)?

 

 

The smaller spot which you suggested as indicative of the actual beam diameter in C5 is only the specular reflection off the metal bar and doesn't match actual spot size seen clearly in the left side of photo.

 

Every photo past this point only shows the laser reflecting off the patch, metal bar, or camera lens and there are no images that show a tight laser spot on the board but a continuous spreading from C2 onwards. But even though you have only reflections from those 3 spots you continue to mark the "measured beam height" with higher and higher values.

 

Annotated version of C5:

 

 

 

If the board was slanted at ~24° then No, you cannot "tell the height of the laser by calculating the number of pixels above the board and comparing them with the board size".

 

In any case, even with a vertical board* it is unsafe to measure on a photo. From my experience, every time there is a significant lack of precision.

 

*not even vertical, it should be perpendicular to the laser beam, and the photo taken from the origin of the laser beam, which is quite difficult (it should be installed like a rifle scope)

 

I agree that you cannot be 100 percent accurate here but you can get an idea.

 

And remember that the analysis *Mick* was doing NOT looking along the slanted board but looking at it edge on and the marking is nowhere near appearing 130cm above the water. Sure, it COULD be perspective distortions. So let Sandor show how the tape marking was placed and verify it is 130cm above the water.

 

If you watch the video Mick very carefully marks out the top of the tape mark and measures the pixels more accurately than here, this was just the rough swag but it shows what I'm talking about:

 

 

I don't think it's unreasonable to ask for confirmation of how this specific and critical marking was made to confirm that it wasn't done by placing the tape measure along the slanted angle (but was accurately a measure of that point above the water) when you can very clearly see that do this at two other places in the video - shown below. Since this was their 'slope corrected leveling' mark it determines if the laser is pointing downwards or upwards slightly. An inaccurate measurement at this point renders their overall data inaccurate. Although other methodological errors swamp even this error.

 

And if you reject the pixel counting method then you have to reject all their data past ~C5 because that is all they did - and worse, each of those cases they are counting the pixels to either the reflection off the metal bar, the reflective patch on the jacket, or the camera -- at no point past C5 do they have anything that would mark the actual center of the laser beam.

 

I even showed above where they are calling the exact same reflective spot on the boat as being two different heights.

 

 

[Sandor Szekely]

 

Thanks again for the clarifications, Sandor.

 

It's good to know that you want to enhance the experiment, that's always a good practice.

 

Regarding the board, I wouldn't use a semi-transparent material anymore. This may be causing extra refraction and disrupting your data. I would suggest using an opaque material, taking the measures where the laser hits, and not on the other side. I would also strongly suggest you to not use any reflective material as a mean to capture the data. You want the light to be absorbed, not reflected. The reflective one is probably a good idea to help finding the beam, but not measuring it.

 

I agree with your point that 3.45m - 1m is also not looking good for the GE model, but I was intrigued by the "moving laser" effect you mentioned. If a laser can move up to 2m left and right, I think it's possible it may be also moving up and down by the same amount. A well collimated laser (lets say 40cm at 5km) may cause the top and bottom limits of the laser to be 2.4m off, and a not well collimated laser (let's say 2m at 5km) may cause the top and bottom limits to be up to 4m off. I had no idea that this NUDTZ effect, like you call it, could have such a big impact. Very interesting.

 

Only a quick correction, the beam center is always ~0 nm, not 1m. . But anyway, you need to be 100% sure about what the beam size and beam dispersion are. From the pictures you sent, I was making a wild estimation of 2.05m at 5km, but it could easily be 40cm or 4m, there is no way to tell. The images, as you mentioned, are from too far away.

 

My suggestions, some of them already mentioned in the thread:

• Put your laser higher than 1.25m. The refraction at this height is causing too much variation to the data.
• Use a different board material, preferably opaque.
• Definitely a higher board. Do you think you can make it at least 5m? Can you get a bigger boat so that bigger structures can be mounted there? 10m would be cool
• Validate the beam dispersion. Measure the top and bottom of the laser spot at 10m, 100m, 500m, 1km, 2km. After that, you go back and start the experiment. This is only a calibration round.
• Same water and air temperature
• Perpendicular board as it's easier to measure heights. If possible, pre-mark the heights there so that you don't need to measure it during the experiment.
• More measuring: Top and Bottom of the laser spots every 500m. This will give you 24 measurements every time you run the experiment.
• Don't switch measuring tools (from board to jacket).

 

I know it's too much to ask, but you have an uphill battle ahead of you.

I 'd like to make the curvature experiment perfect, that would probably take a few more attempts but I am very persistent in my works

 

Thanks for sharing the ideas to improve the measurements!

I am about to sign a cooperation agreement with the top geodezist and geophysics in Hungary and I will point them to this forum to review the suggested ideas.

"Regarding the board, I wouldn't use a semi-transparent material anymore."

"Regarding the board, I wouldn't use a semi-transparent material anymore.

I have only one concern: that the camera from the starting position is probably not going to be able to capture the beam on the boar all the way (we plan to do 10kms next time). In this case we can have a camera on the opposite shore, but that will not see the laser hit if the board is not transparent. We might as well use a camera in a boat that is following the measurement boat from the side.We will have a camera onboard facing the board to make the exact readings, but we can confirm the position only with timecode (that may not be well proved).

We have to check now in a test if an opal surface or a retroreflective surface (like a moviescreen) is better to capture the beam.

Please tell me your opinion on all this.

 

About the movement of the laser beam: it was noticed only at the measurements upto midnight.

The beam was moving ONLY sideways! It looked like someone was moving it on purpose, an unusually high horizontal vibration: quite high frequency and movement. I would like to attach a video here but I can't - can someone give me an idea how to?

What could be the purpose of the only sideway movement? (at this time the laser was at 50cms 4.1 feet)

 

We will definitely put the laser higher next time and use a big enough board to measure the beam. Bigger boat is not a problem, we can have powerboat, or a sailing boat with engine. The sail is a huge but not an ideal recording surface (in no wind conditions of course).

 

The beam dispersion can be automatically validated if the board is functioning well. The board would be the best indicator for refraction as well.

 

Water and air temperature measurements are quite hard to accomplish with the moving boat. I am not sure how we can make that accurate.

 

Perpendicular and measured board (the height of the board may be varied here to match the exact height from water surface - the exact height is still a question so it is 1.5 on the picture for now)

 

 

 

Edited September 14, 2016 by Sandor Szekely