Lake Balaton Laser experiment to determine the curvature of the Earth, if any.

[Sandor Szekely]

Update by Mick West

This thread documents the progress of planning, execution, and analysis of an Experiment conducted by Sandor Szekely (a believer in the "Flat Earth" theory) on Lake Balaton in Hungary. The goal of the experiment was to measure the curvature of the surface of a lake by shining a laser across it, and measuring the height of the laser at multiple points across the lake. If the lake was curved then plotting those heights on a graph should form a curve. If it was flat, then it would form a straight line.

The experiment was extensively documented with photos and video.

The results of the experiment were some very poor quality data. Poor initial calibration of the laser mean that the it was pointing slightly downwards instead of the intended level or slightly upwards. The laser target used was positioned so that the laser very quickly rose above it and could no longer be measured, and the laser beam diverged rapidly so as to be impossible to locate.

Since the laser seems to have been pointing downwards, the subsequent rise of the laser above the target is most consistent with a curved surface. However the very poor quality of the measurements makes this impossible to quantify.

Here re some pre-experiment key posts that were discussed:

Since the laser cannot be leveled exactly, multiple readings should be taken to calculate both the curve of the lake and the slope of the laser
https://www.metabunk.org/lake-balat...vature-of-the-earth-if-any.t7780/#post-186843

If the laser is pointing down a bit, then it will initially match the curve of the earth, and so appear to be level for the first mile or two.
https://www.metabunk.org/lake-balat...-of-the-earth-if-any.t7780/page-2#post-186880

A much more accurate "Wallace" method was suggested several times:
https://www.metabunk.org/lake-balat...-of-the-earth-if-any.t7780/page-3#post-187601

Initial results from the experiment started to be discussed here:
https://www.metabunk.org/lake-balat...-of-the-earth-if-any.t7780/page-4#post-188309

The experiment at night was unusable due to refraction of the beam, but gave some excellent photos of this refraction and inferior mirage. This led to some interesting discussion.
https://www.metabunk.org/lake-balat...-of-the-earth-if-any.t7780/page-4#post-188365

The refraction was due to the water temperature being warmer than the air temperature
https://www.metabunk.org/lake-balat...-of-the-earth-if-any.t7780/page-5#post-188421

The discussion of the results published on YouTube starts here:
https://www.metabunk.org/lake-balat...of-the-earth-if-any.t7780/page-13#post-190061

The following is a list of posts that identify issues with the experiment.

Calibration of the laser height was done with a tape measure at an angle of 20° from vertical, leading to it being lower than intended.
https://www.metabunk.org/lake-balat...of-the-earth-if-any.t7780/page-19#post-190631

Using the 1m edge of the board as a guide the height of the target tape is measured at under 1.20m, but was claimed to be 1.30m. Since the laser itself was at 1.25m, this means the laser was pointing downwards.
https://www.metabunk.org/lake-balat...of-the-earth-if-any.t7780/page-19#post-190631

Some measurements were taken with the boat stationary, some with it underway, resulting in an approximately 5cm variation.
https://www.metabunk.org/lake-balat...of-the-earth-if-any.t7780/page-19#post-190668

Times on the photo comparisons do not match, being several minutes off in some cases:
https://www.metabunk.org/lake-balat...of-the-earth-if-any.t7780/page-16#post-190262

After the laser left the target, subsequent sightings were only from a retroreflective patch on the back of the pilot's jacket, and from reflections off the camera glass.
https://www.metabunk.org/lake-balat...of-the-earth-if-any.t7780/page-18#post-190576

The motion of these reflections through the beam indicate that it eventually diverged to several feet wide, from the camera glass reflection
https://www.metabunk.org/lake-balat...of-the-earth-if-any.t7780/page-18#post-190574
And from the earlier jacket retroreflective patch
https://www.metabunk.org/lake-balat...of-the-earth-if-any.t7780/page-19#post-190623

Some later measurements were claimed to be from a "direct hit" in a camera on the boat, however these "direct hits" lasted for over a minute while the boat moved significant distances perpendicular to the beam. This indicated a "direct hit" could be had anywhere within a large cone, meaning height estimates were possibly catching the bottom of the cone.
https://www.metabunk.org/lake-balat...of-the-earth-if-any.t7780/page-18#post-190419

It was suggested geoid variations might be a factor, but did not seem to be significant
https://www.metabunk.org/lake-balat...of-the-earth-if-any.t7780/page-14#post-190169

Sandor claimed that a LIDAR study of the lake found it was flat, and that he had got confirmation of this. The lead author of the study refuted this.
https://www.metabunk.org/lake-balat...of-the-earth-if-any.t7780/page-23#post-190877

Sandor started a thread on scienceforums.net, and his OP seemed to suggest he was part of the LIDAR study, when he was not.
http://www.scienceforums.net/topic/98386-laser-curvature-test-on-lake-balaton/

Sandor was banned for a cumulation of violations of the posting guidelines, ending with his blatant misrepresentation of this paper (which actually proved the lake surface was curved exactly as would be expected)

The following is the original post that started the discussion
----------------------------------------------------------------
Water is absolutely FLAT
earth surface is 2/3 of water, so how can earth be a globe?

We have 33 meters (over 100 feet) missing target hidden height in our upcoming experiemnt!

The 14.3 miles laser experiment will be in August 15th - 16th
Please visit us if you are nearby Hungary and lake Balaton!

Now we will have the 0.03 mRad collimator ready to keep laser beam divergence at 4 inch (10cms)diameter on the 14.3 miles (23kms) distance. With this exact pinpoint laser beam we can measure and calculate the refraction volume. We will measure ALL the distances in every 500 meters to get 30 to 50 points of measurement on each distance.

some pictures of our pre-test video that debunks Stephen Hoaxking's similar 3 miles curvature experiment with his ridiculous result of over 6 feet curvature drop on this distance, please watch here:


Source: https://www.youtube.com/watch?v=cNEUOnlcIAQ

 

[Rory]

Thanks for the pointer to the Stephen Hawking experiment, Sandor, I hadn't seen that before. Very cool!


Content: A man on the shore of a lake uses a telescope to watch a helicopter hovering over the opposite shore 6 miles away. The helicopter descends towards the ground and disappears from his view while still in the air. It ascends again, becomes visible to the observer, and reports an altitude above the lake of 24 feet at the point that it appears 'level' with the horizon.

Interestingly, 8 inches x miles squared returns exactly 24 feet of obscured height at six miles. Or using the earth curvature calculator, 24 feet and 1/164th of an inch.

PS Are you aka Dr Zack?

 

[Z.W. Wolf]

No. "Dr. Zack" is someone else and lives in Spain; but they are collaborating on this.

There's also this thread: https://www.metabunk.org/upcoming-laser-experiment.t7664/

 

[Z.W. Wolf]

Water is absolutely FLAT
earth surface is 2/3 of water, so how can earth be a globe?

We have 33 meters (over 100 feet) missing target hidden height in our upcoming experiemnt!

The 14.3 miles laser experiment will be in August 15th - 16th
Please visit us if you are nearby Hungary and lake Balaton!

Now we will have the 0.03 mRad collimator ready to keep laser beam divergence at 4 inch (10cms)diameter on the 14.3 miles (23kms) distance. With this exact pinpoint laser beam we can measure and calculate the refraction volume. We will measure ALL the distances in every 500 meters to get 30 to 50 points of measurement on each distance.

some pictures of our pre-test video that debunks Stephen Hoaxking's similar 3 miles curvature experiment with his ridiculous result of over 6 feet curvature drop on this distance, please watch here:


Source: https://www.youtube.com/watch?v=cNEUOnlcIAQ

We've already had some exchange of ideas. Here are the mistakes you made this time

-Wrong type of laser equipment. This is a laser stage projector. (I've turned the photo upside down so that the writing is legible.)

Similar to this one: http://www.aliexpress.com/store/pro...-Disco-Bar-Stage-House/438375_1583509930.html

These units have a secondary low power laser as you demonstrate at 5:46 in your video. In the case of your unit it's a blue laser. The main laser, green in the case of your unit, is much more powerful which is why it was "more visible" in your test. Both lasers do not shine directly out of the unit. There is a device described here:

http://www.laserist.org/guide-to-laser-shows.htm#producing-laser-graphics

To make a laser graphic, two tiny computer-controlled mirrors aim the beam at a screen. The beam bounces first off of one mirror moving horizontally, then off another at right angles, moving vertically.

The computer literally "connects the dots", aiming the mirrors from one place to another fast enough that the viewer sees a single outline drawing. This process is called "scanning". The computer-controlled mirrors are galvanometer "scanners".

One can see at 5:46 in your video that the beam is not a narrow shaft that can be seen from a laser pointer, but a broad beam.

This is an illusion caused by the method described above. Why did you use this type of laser?

(I understand that you will use a different unit in your promised test. And you've promised to use a tight beam with a collimator. I think you'll find some different results if and when that happens.)

-Your method of leveling the laser is wholly inadequate. A tape measure on a bobbing boat. Attempting to determine the center of a broad beam. You will need a much better method.

-You claim that the boat was 5 Km from shore at maximum but the center of the beam was still 50 cm above the water surface. You don't say how you know the boat was at this distance. I have to assume that this was an estimate based solely on naked eye observation. From what I can SEE in your video the boat was no more than a few hundred meters from shore even at 11:51 in your video. You need to measure distances with an instrument and report your methods. GPS would be the simplest way.

(Given the amount of beam divergence and the modest power of this green laser I don't find it credible that you would even be able to see the laser on your target at 5Km. The beam would be very wide and attenuated.)


- At 11:51 in your video, it can be clearly seen that the beam has diverged to a number of meters wide and high. It's not possible to tell the exact number of meters. But it is clearly exceeding the size of your target by a large amount. The bottom of the laser "beam" is intercepting the water. Any attempt to measure the height of the "center" of the beam is not valid.


My conclusion is that you have not debunked the Hawking demonstration.

 

[Mick West]

A more accurate way of performing this experiment would be to

A) have all the elements on the ground.
B) have the laser as high as possible to reduce refraction effects
C) perform the test at dawn, to reduce refraction effects.
D) have a middle target between the laser and the longest range target

It should not really have to be that long a distance either. Three miles will result in a horizontal laser beam being six feet higher (as demonstrated on Hawkins Genius).

The difficulty is getting the laser beam horizontal. If you are out by 1" at 20 feet, then at three miles (15840 feet) then the error then becomes 66 feet.

To reduce the error to a reasonable 1 foot, you need an accuracy at 20 feet of 1/66th of an inch (0.38mm). Since you clearly don't have that on your bobbing boat, the results are irrelevant and prove nothing.

However, if you have a middle target, and everything on the ground, then you can measure the height of the beam at both points, and eliminate any calibration error (it likely won't be perfectly horizontal, but you can do the math from this to demonstrate the curve, assuming the laser beam is narrow enough to measure)

For example:

 

[Mick West]

Presumably you took some photos of the near opposite shore from this position (50cm above the waterline?). These photos should show a few feet of obscuration of the far side (over 7 feet ignoring refraction). Perhaps you could post a couple?

 

[Z.W. Wolf]

In the promised experiment both the laser and the target will be on land on different parts of the lake shoreline 23Km apart. The target will be a camera.

It's all explained here: https://www.metabunk.org/upcoming-laser-experiment.t7664/

Note: Sandor Szekely (aka SzS, aka DaMan) is the experimenter in Hungary. The narrator of these videos is "Dr. Zack" who lives in Spain.

Szekely has stated many times that he can see the opposite shoreline of the lake down to water level perfectly no matter how far away it is.

 

[Rory]

You claim that the boat was 5 Km from shore at maximum but the center of the beam was still 50 cm above the water surface. You don't say how you know the boat was at this distance. I have to assume that this was an estimate based solely on naked eye observation. From what I can SEE in your video the boat was no more than a few hundred meters from shore. You need to measure distances with an instrument and report your methods. GPS would be the simplest way.

Second that. A few days ago I kayaked to a platform 1km out to sea. From the shore, there was a buoy that appeared to be about half-way there. But I pretty soon realised it wasn't. And checking maps later I saw the buoy was only 15% of the way.

Conclusion: the naked eye is not reliable!

 

[Sandor Szekely]

We've already had some exchange of ideas. Here are the mistakes you made this time

-Wrong type of laser equipment. This is a laser stage projector. (I've turned the photo upside down so that the writing is legible.)

THANKS, you explain correctly our equippment used in the pre test. The mirrors of the scanner cause bigger divergence as you can see in the pictures.

This is the correct answer too:
(I understand that you will use a different unit in your promised test. And you've promised to use a tight beam with a collimator. I think you'll find some different results if and when that happens.)
Yes we will use only green laser with a 0.03 mRad (milli radius) collimator that we did not have ready at the pre test

-Your method of leveling the laser is wholly inadequate. A tape measure on a bobbing boat. Attempting to determine the center of a broad beam. You will need a much better method.
Yes this is the only way to level the laser on 2 kms distance to get a 0.01 degree accuracy


-You claim that the boat was 5 Km from shore at maximum but the center of the beam was still 50 cm above the water surface. You don't say how you know the boat was at this distance. I have to assume that this was an estimate based solely on naked eye observation. From what I can SEE in your video the boat was no more than a few hundred meters from shore even at 11:51 in your video. You need to measure distances with an instrument and report your methods. GPS would be the simplest way.
YES! we will have a GPS for water, the one we had on the pre test was not showing the coordinates on water surface


(Given the amount of beam divergence and the modest power of this green laser I don't find it credible that you would even be able to see the laser on your target at 5Km. The beam would be very wide and attenuated.)


- At 11:51 in your video, it can be clearly seen that the beam has diverged to a number of meters wide and high. It's not possible to tell the exact number of meters. But it is clearly exceeding the size of your target by a large amount. The bottom of the laser "beam" is intercepting the water. Any attempt to measure the height of the "center" of the beam is not valid.
The beam divergence was exactly 1 meter diameter at the 5 kms distance, just touching the water and the top of the white board
that was the reason we turned back

My conclusion is that you have not debunked the Hawking demonstration.

Hawking said it is 6.5 feet that is over 2 meters! WHERE IS IT? LOL

 

[Sandor Szekely]

A more accurate way of performing this experiment would be to

A) have all the elements on the ground.
PLEASE define what elements?

B) have the laser as high as possible to reduce refraction effects
We will make a 2 meter high above water level measurement as well.
ALSO we will repeat the measurements at daytime, sunset, night time and sunrise to collect more data on the volume of horizontal refraction

C) perform the test at dawn, to reduce refraction effects.
YES!

D) have a middle target between the laser and the longest range target
this is a good idea, we will put this into the experiment!
THANKS! these are the comments I am looking for I am really doing this in a good faith to prove the TRUTH
of course I made my own observations and therefore I can not pretent that I think we live on a globe
I am not an actor, but a researcher


It should not really have to be that long a distance either. Three miles will result in a horizontal laser beam being six feet higher (as demonstrated on Hawkins Genius).
We will make different distances and measure the laser height all accross the distance to prove more points on the line. The laser can not be parallel all accross the 14.3 miles distance on the globe model

The difficulty is getting the laser beam horizontal. If you are out by 1" at 20 feet, then at three miles (15840 feet) then the error then becomes 66 feet.
We can set 0.01 degree accuracy at the 2 kms distance

To reduce the error to a reasonable 1 foot, you need an accuracy at 20 feet of 1/66th of an inch (0.38mm). Since you clearly don't have that on your bobbing boat, the results are irrelevant and prove nothing.

However, if you have a middle target, and everything on the ground, then you can measure the height of the beam at both points, and eliminate any calibration error (it likely won't be perfectly horizontal, but you can do the math from this to demonstrate the curve, assuming the laser beam is narrow enough to measure)

For example:
View attachment 20170

 

[Sandor Szekely]

I took these photos to the direction of the 14.3 miles distance:
the last one is with a 165x zoom telescope

 

[Mick West]

Hawking said it is 6.5 feet that is over 2 meters! WHERE IS IT? LOL

It's only 6 feet if your beam actually is level, and within 1mm at 20 feet. How did you determine this?

 

[Z.W. Wolf]

Hawking said it is 6.5 feet that is over 2 meters! WHERE IS IT? LOL

You aren't answering any of the questions raised in my post. Let's talk about one. You say the boat was 5 km from shore. I think it was much closer.

How did you determine that the boat was 5 km distant? Please answer clearly.

If it was not 5 km from the laser, the 6 foot figure would not be expected.

 

[Sandor Szekely]

It's only 6 feet if your beam actually is level, and within 1mm at 20 feet. How did you determine this?

NOPE!

Hawkings setup:

laser at 3 feet? maybe
distance to the boat 3.1 miles (5kms)

there should be NO target hidden height!

curvature measured ON water surface is 6.4 feet!

they just mixed up the different values!

https://dizzib.github.io/earth/curve-calc/?d0=3.106855961185&h0=3&unit=imperial


our pre-test setup:
laser at 1.6 feet
distance to the boat 3.1 miles

target hidden height: 1.6 feet


please watch Jeranism's perfect explanation on the same Hawking subject:
HE is a member in Metabunk, how can we invite him into the talking here?


Source: https://www.youtube.com/watch?v=C1SQZVEg6Hw

 

[Sandor Szekely]

Images are automatically resized to 1600 pixels. Crop a region, or post a link.

Even with the little fata morgana you have at the horizon, you can clearly see that some building are obscured by the water, hence it's curved.
View attachment 20186

thanks, I will do so (crop a region)

"some building are obscured by the water, hence it's curved."

That building is the Balaton Sound BEER TENT at Siofok

NO, curvature shall be way much more! so it is NOT obscured because of curvature!
do not mix: the phenomena does not define an explanation, there may be more and even cotradicting explanations! we have to discard any explanation that is not possible
that is curvature for example, these building should be hidden under 14 feet of curvature! BUT they are NOT

So we can discard curvature if there is no debate on refraction
here is the data: https://dizzib.github.io/earth/curve-calc/?d0=6.21371192237&h0=1.6404&unit=imperial

but you know it is hard to prove refraction on a photo so therefore we make the laser measurement

 

[Sandor Szekely]

You aren't answering any of the questions raised in my post. Let's talk about one. You say the boat was 5 km from shore. I think it was much closer.

How did you determine that the boat was 5 km distant? Please answer clearly.

If it was not 5 km from the laser, the 6 foot figure would not be expected.

Sorry Wolf, I am just getting familiar here, and I'll visit your post soon

The 6 foot was the measurement from Hawking - we had 1.6 feet all the way.

The other side of the lake the boat was heading to is 10kms away. The captain of the boat TOLD US that they are over the half way when the bottom of the laser beam hit the water surface and the upper part hit the top of the white board.
That is 1 meter divergence and we could not make the measurement any further. The captain knows the lake very well as he is a rescue ship operator. SORRY, our GPS was not recording the actual position over the lake surface (BAD GPS) next time the captain will bring HIS professional GPS to locate the exact positions.
So fo this time you can check the distance like: I gave the parameters in the description and here above about the optics -you may see the sizes and relative distances and may give an idea
this is an 1120mm optical teleobjective

YOU understand what Hawking measured? 6.5 FEET? that is the CURVATURE of the water surface NOT the target hidden height. PLS take a look into Jeranism's video on the same subject

In the upcoming measurement we will record all possible data, so PLS list me here what datas you would like to see and how shall I prove them correct!
That would be a GREAT help for me THX

I thought of ambient temperature and humidity at the level of the laser beam and maybe 2 meter up?
measuring water level at both sides (at a bay with an official water level) and the temperature
taking the GPS with the boat, with more cameras to record everything (will you belive it is ONE SAME recording? pls give advices 4 this)
AND leveling the laser is still a debate I am waiting for ideas: but make sure it is 0.01 degrees accurate

 

[Sandor Szekely]

Can we invite Mr @Ray Von Geezer to the conversation here?

from Zombie Wolf's post:
"From my experience of aligning Free Space Optical laser network devices at distances of much less than that, in my opinion that was extremely unlikely, especially using the jury-rigged mountings and a target 1cm across. FSO equipment has targeting optics, digital alignment tools, stable mounts adjustable by fractions of degrees, and very forgiving receivers.

23Km I'd say virtually impossible without sophisticated alignment equipment, the adjustments needed are absolutely tiny. That's unless the beam diverges so much that it renders the whole experiment void anyway.

Ray Von"

He is pointing out well the problem of leveling the laser. This 0.01 degree is not possible with short bubble meter or gyroscope. This is why we figured out to use the lake itself to level the laser to the 0.01 accuracy at 2kms distance.
We place the laser on the ground on a rubber sheet and thin plastic sheets to align leveling. This may sound funny, but this way we can make sure to get the right accuracy.

We have changed the setup a bit from that first teaser video - thanks to comments there - and we use a boat to go all the way to the other side with the laser beam recorded on the white board all accross the distance measured in every 500 meters. When we reach the other side of the lake at 14.3 miles we will record the direct beam hit on the shore. We will have cameras in position A at the laser and at position B where the boat is heading to.

Ray Von Geezer is pointing out the MOST IMPORTANT subject!
"unless the beam diverges so much that it renders the whole experiment void anyway"

We DID NOT have the collimator ready at the time of the pre-test, but we will use it with the upcoming experiment!

this is a masterpiece! a 0.03 mRad collimator lens set designed for this experiment!
that is 0.03 milli radius

we will have a laser beam divergence of 3.9xx inch (about 10 cms) at the 14.3 miles (23kms) distance!
I think that is a pretty awsome divergence - not?

 

[Rory]

Personally, I look forward to your experiment very much, and feel that you're being much more thorough and 'scientific' than pretty much any flat earth believer I've come across. Particularly, I'm sure, if you take on board some of the tips from the guys here, which I think it's great that you're open to.

Just one thing - pointing us to Jeranism videos probably isn't going to get very far as one of his was looked at recently - a new formula for discerning curvature - and found to be somewhat lacking in accuracy.

Thread on that here:

https://www.metabunk.org/attempt-at...nt-by-flat-earth-believers.t7710/#post-185729

Good luck on the test. If it's done right I'm sure we'll all be super interested in the results.

 

[Chew]

a 0.03 mRad collimator lens set designed for this experiment!
that is 0.03 milli radius

we will have a laser beam divergence of 3.9xx inch (about 10 cms) at the 14.3 miles (23kms) distance!

How did you calculate 3.9 inches?

 

[Z.W. Wolf]

NOPE!

Hawkings setup:

laser at 3 feet? maybe
distance to the boat 3.1 miles (5kms)

there should be NO target hidden height!

curvature measured ON water surface is 6.4 feet!

they just mixed up the different values!

https://dizzib.github.io/earth/curve-calc/?d0=3.106855961185&h0=3&unit=imperial


our pre-test setup:
laser at 1.6 feet
distance to the boat 3.1 miles

target hidden height: 1.6 feet


please watch Jeranism's perfect explanation on the same Hawking subject:
HE is a member in Metabunk, how can we invite him into the talking here?


Source: https://www.youtube.com/watch?v=C1SQZVEg6Hw

You seem to remember me from the YT comments section as you call me Zombie Woof, which is what I call myself on YT. (Zomby Woof, actually. A Frank Zappa song.)

This video you have cited gives an incorrect analysis. I'll leave an explanation of that to someone else, as I have very little time just now.

Weeks ago on YouTube, I told you how to use the Earth's Curve, Horizon and Hidden Calculator in this situation and my logic behind it, and you agreed, but now I'm not sure you understand.

The calculator was designed to find the hidden height of an object beyond the observer's horizon, but in this case we can adapt it to find the "drop." The drop is what Samuel Rowbotham's formula was meant to calculate - 8 inches per mile squared.

In this case we enter the observer's height as zero.

In this case the "hidden height" can be read as "drop."

I'll use David Ridlen's illustration to explain.

I use this diagram to clearly illustrate their mistake without going into the trig they have no patience for, which has been working wonders for getting thru to them.

The Earth's Curve, Horizon and Hidden Calculator was designed to find the value for line B. Rowbotham's formula was meant to find the value for line D. In the case of these laser experiments we are trying to find the value for line D - how much the earth's surface "drops" beneath a line (the laser). In this illustration, you can think of the red line that starts at the man's FEET as the laser beam. So when we enter zero for observer's height we can "trick" the calculator into finding the value for line D for the laser.

 

[Sandor Szekely]

You seem to remember me from the YT comments section as you call me Zombie Woof, which is what I call myself on YT. (Zomby Woof, actually. A Frank Zappa song.)

This video you have cited gives an incorrect analysis. I'll leave an explanation of that to someone else, as I have very little time just now.

Weeks ago on YouTube, I told you how to use the Earth's Curve, Horizon and Hidden Calculator in this situation and my logic behind it, and you agreed, but now I'm not sure you understand.

The calculator was designed to find the hidden height of an object beyond the observer's horizon, but in this case we can adapt it to find the "drop." The drop is what Samuel Rowbotham's formula was meant to calculate - 8 inches per mile squared.

In this case we enter the observer's height as zero.

In this case the "hidden height" can be read as "drop."

I'll use David Ridlen's illustration to explain.



The Earth's Curve, Horizon and Hidden Calculator was designed to find the value for line B. Rowbotham's formula was meant to find the value for line D. In the case of these laser experiments we are trying to find the value for line D - how much the earth's surface "drops" beneath a line (the laser). In this illustration, you can think of the red line that starts at the man's FEET as the laser beam. So when we enter zero for observer's height we can "trick" the calculator into finding the value for line D for the laser.

HI Zomby Wolf, yes of course I remember You!

I don't remember this target hidden height conversation, but the good news is that I know this exactly - this is the same that I wrote up here.
David Ridlen's illustrations is okay too.

here is OUR DRAWING WITH Dr ZACK:
exact calculation on the 14.3 miles distance

 

[Sandor Szekely]

Personally, I look forward to your experiment very much, and feel that you're being much more thorough and 'scientific' than pretty much any flat earth believer I've come across. Particularly, I'm sure, if you take on board some of the tips from the guys here, which I think it's great that you're open to.

Just one thing - pointing us to Jeranism videos probably isn't going to get very far as one of his was looked at recently - a new formula for discerning curvature - and found to be somewhat lacking in accuracy.

Thread on that here:

https://www.metabunk.org/attempt-at...nt-by-flat-earth-believers.t7710/#post-185729

Good luck on the test. If it's done right I'm sure we'll all be super interested in the results.

Thanks Rory I am very happy to hear that!

and also happy to talk about the experiment with real sceptics here

 

[Sandor Szekely]

How did you calculate 3.9 inches?

This calculation is from our laserist, we will include the full documentation and calculation in the final video, and of course here as well.

The expected laser beam divergence is calculated from the collimation of 0.03 milli radius - and it is something like 3.9xx on the 14.3 miles distance. The exact value will be given on the exact distances. We can calculate the refraction volume as well from this value and our measured value at the experiment (with each measrement).

 

[Chew]

0.03 milli radius

What is a "milli radius"?

 

[Spectrar Ghost]

I think he means milliradian.

 

[Sandor Szekely]

I think he means milliradian.

Spectar Ghost tahnks for correcting me, yes I mean milliradian.

sorry for my English

SO how does it sound? that supposed to do the job well on long distance

 

[Mick West]

NOPE!

Hawkings setup:

laser at 3 feet? maybe
distance to the boat 3.1 miles (5kms)

there should be NO target hidden height!

curvature measured ON water surface is 6.4 feet!

they just mixed up the different values!

https://dizzib.github.io/earth/curve-calc/?d0=3.106855961185&h0=3&unit=imperial


our pre-test setup:
laser at 1.6 feet
distance to the boat 3.1 miles

target hidden height: 1.6 feet

You seem to be making the same mistake as @jeranism. For a level laser, the laser spot on the target (relative to the water surface) should be 6.4 feet higher at 3.1 miles, regardless of the laser height. The hidden height is entirely irrelevant to where the dot is.

I've asked @jeranism to correct this in his video, so far he has not replied.

 

[Sandor Szekely]

You seem to be making the same mistake as @jeranism. For a level laser, the laser spot on the target (relative to the water surface) should be 6.4 feet higher at 3.1 miles, regardless of the laser height. The hidden height is entirely irrelevant to where the dot is.

I've asked @jeranism to correct this in his video, so far he has not replied.

I understand what you mean with the leveled laser, but than you have to add up the height of the laser too.
Hawking did not prove how the laser was leveled it could be any angle. Surely not leveled, than it should be 6.4+ laserheight.

Check out my calculation on the experiment:

I should get the laser point at 140 feet when the laser is perfectly leveled on the 14.3 miles distance

I should get the laser point at 97 feet when the laser is tangent to the water surface on the 14.3 miles distance

IF I get a laser point well below 97 feet on the 14.3 miles distance than earth can not have this radius right?

 

[Chew]

SO how does it sound?

It sounds like your laserist can't do simple trigonometry.

 

[Spectrar Ghost]

It sounds like your laserist can't do simple trigonometry.

Basically this. You'll recall a radian is a segment of a circle with arc length equal to the circle's radius. If you have a radius of 23,000m (23km) and beam divergence of .00003 Rad (.03 mRad), your spread is the product of the two: 69cm/27".

 

[Sandor Szekely]

So you accept @jeranism was incorrect with the "Hawking Blunder"?

Jeranism talks about target hidden height 1 feet

You are taking about drop + laser height : 6 feet + 2 feet 7 inches = 8 feet 7 inches

The guy is saying (not measuring) 6 feet
forgot to add the laser height

 

[Sandor Szekely]

Basically this. You'll recall a radian is a segment of a circle with arc length equal to the circle's radius. If you have a radius of 23,000m (23km) and beam divergence of .00003 Rad (.03 mRad), your spread is the product of the two: 69cm/27".

SORRY, I wanted to say 0.003 mRad collimator lenses. this EXACT data will be given to us be the laserist when they calibrated the new collimator lens set unit. These are more lenses that are fixed in a special frame and calibrated by optical company. This is ONE OF A KIND in the world especially made for this experiment!
BUT all data will be given to reproduce the same or similar. Hawking's collimator is a simple one.

Thanks for pointing me out to the correct number Spectar Ghost

 

[Spectrar Ghost]

SORRY, I wanted to say 0.003 mRad collimator lenses. this EXACT data will be given to us be the laserist when they calibrated the new collimator lens set unit. These are more lenses that are fixed in a special frame and calibrated by optical company. This is ONE OF A KIND in the world especially made for this experiment!
BUT all data will be given to reproduce the same or similar. Hawking's collimator is a simple one.

Thanks for pointing me out to the correct number Spectar Ghost

I think your laserist is lying to you.

There is a fundamental limit to the collimation of a laser due to diffraction. Assuming the laser beam profile is a uniform disk it will be diffracted to an Airy disk at large distances, and the angular spread is approximately given by:


θ≈1.22λdθ1.22λd


where dd is the beam diameter. Assuming a diameter of 1 mm, which seems a reasonable estimate for most lasers I've seen, you get an angular divergence of about 0.6 milliradians for 500nm light.

http://physics.stackexchange.com/questions/70415/lasers-and-collimation

The most common Green lasers are about 532nm, so you've been sold something that works two orders of magnitude better than the physical limit to collimation. Using 532nm wavelength and 1mm beam diameter, it appears the lower limit to beam diameter at 23km is about 15m.

Even "correcting" the (repeatedly) claimed beam divergence value doesn't make your stated numbers work, BTW.

 

[Sandor Szekely]

I think your laserist is lying to you.

There is a fundamental limit to the collimation of a laser due to diffraction. Assuming the laser beam profile is a uniform disk it will be diffracted to an Airy disk at large distances, and the angular spread is approximately given by:


θ≈1.22λdθ1.22λd


where dd is the beam diameter. Assuming a diameter of 1 mm, which seems a reasonable estimate for most lasers I've seen, you get an angular divergence of about 0.6 milliradians for 500nm light.

http://physics.stackexchange.com/questions/70415/lasers-and-collimation

The most common Green lasers are about 532nm, so you've been sold something that works two orders of magnitude better than the physical limit to collimation. Using 532nm wavelength and 1mm beam diameter, it appears the lower limit to beam diameter at 23km is about 15m.

Even "correcting" the (repeatedly) claimed beam divergence value doesn't make your stated numbers work, BTW.

this is the complete quote from John Rennie

"Airy disk at large distances, and the angular spread is approximately given by:

θ≈1.22λdθ≈1.22λd

where ( ... ) Assuming a diameter of 1 mm, which seems a reasonable estimate for most lasers I've seen, you get an angular divergence of about 0.6 milliradians for 500nm light.

I know next to nothing about the design of lasers, but Wikipedia reports the divergence ascommonly less than 1 milliradian, which fits with the above estimate."

So you quote someone who knows nothing about lasers...

and this is explained to him by an other commenter:

"If you were talking about the device and not the beam in the answer, then what are you trying to say? Yes, "Lasers" the devices are not 100% efficient, and the beams do scatter due to air particles. But you said "it's fundamentally impossible to create a perfectly-collimated beam of light". If you add qualifiers like "maintain", "keep", "over all space", or "indefinitely", then I am OK. But as it stands the statement is misleading." Luke Burgess


I DON'T THINK my laserist is LYING to me, I think you have no idea about what you are arguing here.

"The most common Green lasers are about 532nm, so you've been sold something that works two orders of magnitude better than the physical limit to collimation. Using 532nm wavelength and 1mm beam diameter, it appears the lower limit to beam diameter at 23km is about 15m."

OHHH yeah? PROVE IT! BTW

I am not going into such a tone of debate on our laser technology. We have a SET OF LENSES that will be calibrated and the exact parameters will be shown ( and APPROVED)
I am not going into such childish debates... sorry..

I am an inventor with a state reward not a kid from the block

 

[Spectrar Ghost]

"Less than 1 mRad" is fundamentally not the same as .003 mRad. The Airy disk describes the limit of focus for a perfect mirror or lens at a given aperture size and wavelength. You're going to need to be more specific about how you've circumvented this limit...

 

[Sandor Szekely]

"Less than 1 mRad" is fundamentally not the same as .003 mRad. The Airy disk describes the limit of focus for a perfect mirror or lens at a given aperture size and wavelength. You're going to need to be more specific about how you've circumvented this limit...

I will be very specific and provide the official calibration documentation of the collimator (lens set built in the frame)

let's assume for now that we will have an awsome 4 inch - at least something reasonable - laser beam to measure on the different distances at different times of the day and night.
Please advise on the experiment, especially the "drop" and "target hidden height" issues.
Is that a valid statement, that :
if we can record the laser beam at a distance of over 10 miles in a height less than 10 feet (depending on our laser beam parallel precision) above water, than we can conlude that the water surface is perfectly level on this distance? NO curvature drop?
Or exact arguments?

 

[Mick West]

Jeranism talks about target hidden height 1 feet

And he's wrong in saying that's a mistake. Is he not?

You are taking about drop + laser height : 6 feet + 2 feet 7 inches = 8 feet 7 inches

The guy is saying (not measuring) 6 feet
forgot to add the laser height

That's a different issue, and also seems to be incorrect, the middle of the laser appears to be about six feet above where they first marked it.

However this experiment does suffer from the same problems yours has. i.e. it's very difficult to calibrate the laser to level. They say they know it is level, but don't explain exactly how they know.

The best way of doing it would be to assume the laser is not level, and that it has a slope. Then take sufficient intermediate measurement so you can calculate both the slope of the laser, and the curve of the earth.

We know we can approximate the curve over a few miles as 8*d^2 (8 inches per mile squared, we are approximating a segment of a circle with a segments of a parabola). Let's generalize this to a curve value E (where E=8 on the standard model of the earth). So height = E*d^2

Then we add in the slop of the Laser, assume a constant deviation from horizontal of L inches per mile.

So we get a value of the drop (p), relative to the original laser height, of:

p = E*d^2 + L*d

Now we can just take two readings of p at different values of d. If they are sufficiently different then since we know p and d, it's a trivial simultaneous equation to calculate both L and E. Ideally you would take multiple well spaced readings to verify exactly what is happening.

That way you don't need to calibrate your laser to the horizontal, you just need to be able to take reasonably accurate measurements of the height at, say, 1,2,3,4 miles

 

[Mick West]

If you like you can use the equation of a circle rather than a parabola, which should give the same result, but allow to to return a value for r.

p = r-sqrt(r^2 - d^2) + L*d

so if you have two values for p and d (p1,p2,d1,d2)

p1 = r-sqrt(r^2 - d1^2) + L*d1 (1)
p2 = r-sqrt(r^2 - d2^2) + L*d2 (2)

Multiply (1) by d2
d2*p1 = d2*(r-sqrt(r^2 - d1^2)) + L*d1*d2 (3)

and (2) by d1
d1*p2 = d1*(r-sqrt(r^2 - d2^2)) + L*d1*d2 (4)

(3)-(4)

d2*p1-d1*p2 = d2*(r-sqrt(r^2 - d1^2)) - d1*(r-sqrt(r^2 - d2^2))

That's a bit fiddly to directly solve for r, but you can use Wolfram alpha, for example:

d1 = 2 miles, p1 = 4.7 feet
d2 = 3 miles, p2 = 9.0 feet

so, (converting feet to miles)

3*4.7/5280 - 2*9.0/5280 = 3*(r-sqrt(r^2-2^2)) - 2*(r-sqrt(r^2 - 3^2))

Ask Wolfram to solve this for you:
https://www.wolframalpha.com/input/...=+3*(r-sqrt(r^2-2^2))+-+2*(r-sqrt(r^2+-+3^2))

Then you can substitute this into (1) or (2) to find L, just to see how flat your laser actually was.

p2 = r-sqrt(r^2 - d2^2) + L*d2 (2)
9.0/5280= 4061-sqrt(4061^2 - 3^2) + L*3
L(feet) = (9.0/5280- (4061-sqrt(4061^2 - 3^2)))/3 * 5280 ~= 1 foot/mile laser slope.

 

[Sandor Szekely]

And he's wrong in saying that's a mistake. Is he not?



That's a different issue, and also seems to be incorrect, the middle of the laser appears to be about six feet above where they first marked it.
View attachment 20252

However this experiment does suffer from the same problems yours has. i.e. it's very difficult to calibrate the laser to level. They say they know it is level, but don't explain exactly how they know.

The best way of doing it would be to assume the laser is not level, and that it has a slope. Then take sufficient intermediate measurement so you can calculate both the slope of the laser, and the curve of the earth.

We know we can approximate the curve over a few miles as 8*d^2 (8 inches per mile squared, we are approximating a segment of a circle with a segments of a parabola). Let's generalize this to a curve value E (where E=8 on the standard model of the earth). So height = E*d^2

Then we add in the slop of the Laser, assume a constant deviation from horizontal of L inches per mile.

So we get a value of the drop (p), relative to the original laser height, of:

p = E*d^2 + L*d

Now we can just take two readings of p at different values of d. If they are sufficiently different then since we know p and d, it's a trivial simultaneous equation to calculate both L and E. Ideally you would take multiple well spaced readings to verify exactly what is happening.

That way you don't need to calibrate your laser to the horizontal, you just need to be able to take reasonably accurate measurements of the height at, say, 1,2,3,4 miles

"And he's wrong in saying that's a mistake. Is he not?"

Jeran is talking about target hidden height because his experiment was about the tangent on water surface. We can only presume that Hawking's laser was horizontally leveled, so it can be ANY given value of beam measurement from 1 feet to almost 9 feet (or higher when the laser is upwards).

"it's very difficult to calibrate the laser to level." as you said

The laser beam marking and the guessing insted of the well proven measurement is ridiculous!
The guy is clearly saying: 6 feet above water!

The view of the picture is also tricky, the guy is actually bending towards the camera and not reaching upwards.
The laser point is NOT where he marks it!

"The best way of doing it would be to assume the laser is not level, and that it has a slope. Then take sufficient intermediate measurement so you can calculate both the slope of the laser, and the curve of the earth."

PERFECT! this is exactly what I am talking about too

"Ideally you would take multiple well spaced readings to verify exactly what is happening.
That way you don't need to calibrate your laser to the horizontal, you just need to be able to take reasonably accurate measurements of the height at, say, 1,2,3,4 miles"

YES! I am planning to make measurements like this:
at 500 meters I calibrate the laser beam to the same height as the unit on the shore. waves may cause +-5cms difference in reading.
at 1000 meters I calibrate the laser the same way and record the value before and after the setting.
at 1500 meters again, and at 2000 meters away I do the FINAL calibration of the beam to the exact height as the unit on the shore.

I do not touch the laser beam from here on* // *unless it is necessary - but then I will make a beam height recording before and after the calibration

SO from here on I take a READING only every 500 meters about the height of the laser beam above the water - all the way to the 3 different distances that are planned.

This way we will get 30 to 50 points / measurement to calculate the laser beam height drop from the original position. This way we can define as well if the concave theory is possible or not.

 

[Sandor Szekely]

If you like you can use the equation of a circle rather than a parabola, which should give the same result, but allow to to return a value for r.

p = r-sqrt(r^2 - d^2) + L*d

so if you have two values for p and d (p1,p2,d1,d2)

p1 = r-sqrt(r^2 - d1^2) + L*d1 (1)
p2 = r-sqrt(r^2 - d2^2) + L*d2 (2)

Multiply (1) by d2
d2*p1 = d2*(r-sqrt(r^2 - d1^2)) + L*d1*d2 (3)

and (2) by d1
d1*p2 = d1*(r-sqrt(r^2 - d2^2)) + L*d1*d2 (4)

(3)-(4)

d2*p1-d1*p2 = d2*(r-sqrt(r^2 - d1^2)) - d1*(r-sqrt(r^2 - d2^2))

That's a bit fiddly to directly solve for r, but you can use Wolfram alpha, for example:

d1 = 2 miles, p1 = 4.7 feet
d2 = 3 miles, p2 = 9.0 feet

so, (converting feet to miles)

3*4.7/5280 - 2*9.0/5280 = 3*(r-sqrt(r^2-2^2)) - 2*(r-sqrt(r^2 - 3^2))

Ask Wolfram to solve this for you:
https://www.wolframalpha.com/input/?i=3*4.7/5280+-+2*9.0/5280+=+3*(r-sqrt(r^2-2^2))+-+2*(r-sqrt(r^2+-+3^2))
View attachment 20256

Then you can substitute this into (1) or (2) to find L, just to see how flat your laser actually was.

p2 = r-sqrt(r^2 - d2^2) + L*d2 (2)
9.0/5280= 4061-sqrt(4061^2 - 3^2) + L*3
L(feet) = (9.0/5280- (4061-sqrt(4061^2 - 3^2)))/3 * 5280 ~= 1 foot/mile laser slope.

Thanks for the calculation. We will provide the raw data to YOU in first hand to match our findings.
We are making it in AutoCad and showing you the results as well.

"1 foot/mile laser slope"
By this you mean that the laser should RISE about 1 foot / mile on the distance? roughly?
Can you make an expected laser beam height above water surface calculation at 5, 10 and 15 miles distance?

I am happy that we agree on the subject of leveling the laser! VERY important.