[TomInAustin]

Check out this video, a leveled green laser is shot across a lake and what would you guess happened when the height of the beam was measured?

Plus a recreation of the bishop experiment to boot.

Plus they had a flat earth guy.

[Tontogary]

Very interesting. Perspective perhaps???

Or is it a fake?

Abnormal refraction? (Although refraction makes an object look higher)

How about some “waves”?

[AATW]

This has been dealt with in this thread

https://forum.tfes.org/index.php?topic=8832.0

Tom declared it a fraud, predictably, but showed in doing so he didn't actually understand the experiment - although amusingly he claimed it was me who didn't understand it. I thought he finally did understand it after I did a diagram and then Stargiri did a better version of it but he then started talking about boat tilt so I'm still not sure if he really has understood it. And he said this:

They say that the laser (at your higher height) reaches the boat where it would be if the height of the laser was at zero altitude. Look it up. They are not comparing anything. They flatly say that the higher laser reaches the boat where it should not reach.

which is just a flat out lie, they absolutely do not say that.

[TomInAustin]

Very interesting. Perspective perhaps???

Or is it a fake?

Abnormal refraction? (Although refraction makes an object look higher)

How about some “waves”?

Or the obvious, curvature?

[Tom Bishop]

The results of both experiments are identical to what you would get if you put the distance into an earth curve calculator and set the height of the observer to zero. This proves that the experiments are fraudulent.

---

Experiment 1

The first experiment was looking at a boat 3 miles away. The height of the laser is 2 feet.

In the experiment they say that their results showed that the boat went below the horizon by 6 feet. On the boat they show the laser hitting it at 8 feet above the water (+2 to account for the height of the laser) But "6 feet hidden" is the value you would get if you used an earth curve calculator to calculate how much would be hidden if you set the distance to 3 miles and the height to 0 feet.

https://www.metabunk.org/curve/

Set the distance to 3 miles in the calculator, and the height to zero, and you see that 6 feet are hidden.

The height of the laser is at a height 2 feet, so this doesn't make sense. Only 1.07 feet are hidden if you make that adjustment.

When we discussed this AllAround had a ridiculous explanation that, even though the laser is seeing a boat where 1.07 feet are hidden (sunken), and that the laser is 2 feet in altitude, that the laser beam would hit a spot on the boat that is 8 feet above the surface of the water. That does not really make sense. A straight line shot would make a mark that is 3.07 feet above the sunken area (1.07 feet hidden + 2 foot high laser).

When the boat was close to the laser the dot was 2 feet above the water on it, reflecting the height of the laser. The boat only dropped to a point where 1.07 feet was hidden of the boat when it went out to 3 miles, but for some reason, which I still cannot figure out, AllAround says that the laser would arrive at a spot on the boat that is 8 feet above the water.

---

Experiment 2

The second experiment was 6 miles away. Set the distance to 6 miles in the calculator, and the height to zero, and you get 24 feet hidden.

Set the height to 2 feet and the number changes to 12.15 feet.

Take a guess what results they got on this experiment? They said that they got a result where exactly 24 feet were hidden, which matches perfectly to the earth curve calculator if the height were zero (the telescope is not zero, it is about 2 or 2.5 feet).

Funny how their results always exactly match up with what an earth curve calculator gives if the height is zero. This is a clear example of fraud.

[TomInAustin]

The results of both experiments are identical to what you would get if you put the distance into an earth curve calculator and set the height of the observer to zero. This proves that the experiments are fraudulent.

---

Experiment 1

The first experiment was looking at a boat 3 miles away. The height of the laser is 2 feet.

In the experiment they say that their results showed that the boat went below the horizon by 6 feet. On the boat they show the laser hitting it at 8 feet above the water (+2 to account for the height of the laser) But "6 feet hidden" is the value you would get if you used an earth curve calculator to calculate how much would be hidden if you set the distance to 3 miles and the height to 0 feet.

https://www.metabunk.org/curve/

Set the distance to 3 miles in the calculator, and the height to zero, and you see that 6 feet are hidden.

The height of the laser is at a height 2 feet, so this doesn't make sense. Only 1.07 feet are hidden if you make that adjustment.

When we discussed this AllAround had a ridiculous explanation that, even though the laser is seeing a boat where 1.07 feet are hidden (sunken), and that the laser is 2 feet in altitude, that the laser beam would hit a spot on the boat that is 8 feet above the surface of the water. That does not really make sense. A straight line shot would make a mark that is 3.07 feet above the sunken area (1.07 feet hidden + 2 foot high laser).

When the boat was close to the laser the dot was 2 feet above the water on it, reflecting the height of the laser. The boat only dropped to a point where 1.07 feet was hidden of the boat when it went out to 3 miles, but for some reason, which I still cannot figure out, AllAround says that the laser would arrive at a spot on the boat that is 8 feet above the water.

AllAround, please read the previous two sentences slowly, several times in your head, and then try to make a very clear explanation, with very clear words, expressing very clear ideas, on why the beam should be 8 feet above the water.

---

Experiment 2

The second experiment was 6 miles away. Set the distance to 6 miles in the calculator, and the height to zero, and you get 24 feet hidden.

Set the height to 2 feet and the number changes to 12.15 feet.

Take a guess what results they got on this experiment? They said that they got a result where exactly 24 feet were hidden, which matches perfectly to the earth curve calculator if the height were zero (the telescope is not zero, it is about 2 or 2.5 feet).

Funny how their results always exactly match up with what an earth curve calculator gives if the height is zero. This is a clear example of fraud.

You tried to sell this BS in the other thread.   With a level laser, the effective height is zero and you know it.  Still waiting for the proof of the so called Bishop Experiment.

[Curious Squirrel]

The results of both experiments are identical to what you would get if you put the distance into an earth curve calculator and set the height of the observer to zero. This proves that the experiments are fraudulent.

---

Experiment 1

The first experiment was looking at a boat 3 miles away. The height of the laser is 2 feet.

In the experiment they say that their results showed that the boat went below the horizon by 6 feet. On the boat they show the laser hitting it at 8 feet above the water (+2 to account for the height of the laser) But "6 feet hidden" is the value you would get if you used an earth curve calculator to calculate how much would be hidden if you set the distance to 3 miles and the height to 0 feet.

https://www.metabunk.org/curve/

Set the distance to 3 miles in the calculator, and the height to zero, and you see that 6 feet are hidden.

The height of the laser is at a height 2 feet, so this doesn't make sense. Only 1.07 feet are hidden if you make that adjustment.

When we discussed this AllAround had a ridiculous explanation that, even though the laser is seeing a boat where 1.07 feet are hidden (sunken), and that the laser is 2 feet in altitude, that the laser beam would hit a spot on the boat that is 8 feet above the surface of the water. That does not really make sense. A straight line shot would make a mark that is 3.07 feet above the sunken area (1.07 feet hidden + 2 foot high laser).

When the boat was close to the laser the dot was 2 feet above the water on it, reflecting the height of the laser. The boat only dropped to a point where 1.07 feet was hidden of the boat when it went out to 3 miles, but for some reason, which I still cannot figure out, AllAround says that the laser would arrive at a spot on the boat that is 8 feet above the water.

AllAround, please read the previous two sentences slowly, several times in your head, and then try to make a very clear explanation, with very clear words, expressing very clear ideas, on why the beam should be 8 feet above the water.

Not AllAround, but let's see if we can spell this out well enough for even you to grasp, seeing as the diagrams apparently did nothing.

The boat was 50 feet away in the test part, correct? Not enough to make a difference in drop. Let's put two lasers on the ground instead of one. Both running parallel to one another. Laser G(the green one from the video) and Laser I(our new imaginary one) where laser G is at 2 feet, and laser I as on the ground. Laser G hits the boat at about 2 feet up. Laser I hits the boat at the water line. What happens to both of these points when we move the boat 3 miles away?

Option 1) The boat has now 'fallen' by 6 feet. Laser I hits the boat 6 feet up from the water line. Laser G hits the boat 8 feet up from the water line. They have remained parallel, just like they were at the start.

Option 2) The boat has now 'fallen' by 6 feet. Laser I hits the boat 6 feet up from the water line. Laser G hits the boat...3 feet up from the water line? How? The lasers are no longer parallel with each other. This however is the scenario you are suggesting should happen.

I don't know how to make this any more clear.

[Tom Bishop]

Not AllAround, but let's see if we can spell this out well enough for even you to grasp, seeing as the diagrams apparently did nothing.

The boat was 50 feet away in the test part, correct? Not enough to make a difference in drop. Let's put two lasers on the ground instead of one. Both running parallel to one another. Laser G(the green one from the video) and Laser I(our new imaginary one) where laser G is at 2 feet, and laser I as on the ground. Laser G hits the boat at about 2 feet up. Laser I hits the boat at the water line. What happens to both of these points when we move the boat 3 miles away?

Option 1) The boat has now 'fallen' by 6 feet. Laser I hits the boat 6 feet up from the water line. Laser G hits the boat 8 feet up from the water line. They have remained parallel, just like they were at the start.

Option 2) The boat has now 'fallen' by 6 feet. Laser I hits the boat 6 feet up from the water line. Laser G hits the boat...3 feet up from the water line? How? The lasers are no longer parallel with each other. This however is the scenario you are suggesting should happen.

I don't know how to make this any more clear.

Your explanation does makes more sense than when I was first trying to decipher AllAround's explanation. I will abandon my point. If we are talking about how much boat we should see at 3 miles, then it doesn't drop that much. But we are talking about how much a theoretical earth should drop, then it might make more sense. It might be 6 feet at either height.

However, this still doesn't explain the results in Experiment 2 which got the exact values from the earth drop calculator at zero height. There are no lasers pointing out into hypothetically level directions in this experiment.

[AATW]

I really don't know how else to explain this



The green line is the laser, it hits the boat at 2 feet and 8 feet.
The red line is what you can see from 3 miles at a viewer height of 2 feet, only one foot is hidden.

The green line starts parallel with the earth but the earth slopes away. In the red line you look down (horizon dip) and over the curve of the earth.
Obviously the amount of cure and dip are exaggerated in that diagram.

The second experiment I have agreed that it coming to the exact amount was fishy, given the viewer height. But you can still clearly see the helicopter disappear behind and rise from behind the hill of water which Rowbotham said didn't happen. There is no flat earth explanation for that, nor have you suggested any motive for fraud.

[Tom Bishop]

I concede my point on that subject. You are correct. If we put in 3 miles and 2 feet in height into the metabunk calculator, the drop is 6 feet. It says that 1.07 is hidden to the observer. This is not a normal earth curvature experiment that is based on the appearance of hidden bodies, and so the assumptions are different when we account for a hypothetically level laser.

The second experiment is a regular earth drop experiment that is based on vision rather than lasers, however.

It is very odd that they got exactly 24 feet hidden -- which is the value if we set 6 miles and 0 feet oberver height in the earth curve calculator. If we set the height to 2 feet or 2.5 feet, only 11 or 12 feet are hidden.

There doesn't have to be a motive for fraud. Its just laziness on account of the production staff. They knew that the earth was round, and used these calculators to determine what the results should be in their presentation.

[TomInAustin]

I concede my point on that subject. You are correct. If we put in 3 miles and 2 feet in height into the metabunk calculator, the drop is 6 feet. It says that 1.07 is hidden to the observer. This is not a normal earth curvature experiment that is based on the appearance of hidden bodies, and so the assumptions are different when we account for a hypothetically level laser.

The second experiment is a regular earth drop experiment that is based on vision rather than lasers, however.

It is very odd that they got exactly 24 feet hidden -- which is the value if we set 6 miles and 0 feet oberver height in the earth curve calculator. If we set the height to 2 feet or 2.5 feet, only 11 or 12 feet are hidden.

There doesn't have to be a motive for fraud. Its just laziness on account of the production staff. They knew that the earth was round, and used these calculators to determine what the results should be in their presentation.

But more importantly, they showed the lake was not flat.    So much show they shocked the participants.

Also, I am impressed that you conceded the point.  We don't hear that much around here.

[Tom Bishop]

But more importantly, they showed the lake was not flat.    So much show they shocked the participants.

Also, I am impressed that you conceded the point.  We don't hear that much around here.

There are plenty of refraction effects that could present the helicopter disappearing.

But there are not many things which can explain how they got a value which matches exactly what an earth curve calculator says if the height is zero. What a coincidence that refraction added enough height to exactly match the calculator result at zero height. We are interested in truth, and they are not being truthful when they give phony results.

[TomInAustin]

I concede my point on that subject. You are correct. If we put in 3 miles and 2 feet in height into the metabunk calculator, the drop is 6 feet. It says that 1.07 is hidden to the observer. This is not a normal earth curvature experiment that is based on the appearance of hidden bodies, and so the assumptions are different when we account for a hypothetically level laser.

The second experiment is a regular earth drop experiment that is based on vision rather than lasers, however.

It is very odd that they got exactly 24 feet hidden -- which is the value if we set 6 miles and 0 feet oberver height in the earth curve calculator. If we set the height to 2 feet or 2.5 feet, only 11 or 12 feet are hidden.

There doesn't have to be a motive for fraud. Its just laziness on account of the production staff. They knew that the earth was round, and used these calculators to determine what the results should be in their presentation.

Tom,  I am curious what you would accept as a valid experiment?   I happen to own a 300 milliwatt green laser with a stupid bright beam.   I live near Lake Travis Texas and while the lake is quite twisted there are stretches with a clear 3 miles.   What I propose is to set up a similar test on a clear night using the beam, levels,  and white targets.   Would have to establish how much beam spread there is but that would be pretty simple.  What would make it better is if any flat earther live around here and would participate? 

[TomInAustin]

But more importantly, they showed the lake was not flat.    So much show they shocked the participants.

Also, I am impressed that you conceded the point.  We don't hear that much around here.

There are plenty of refraction effects that could present the helicopter disappearing.

But there are not many things which can explain how they got a value which matches exactly what an earth curve calculator says if the height is zero. What a coincidence that refraction added enough height to exactly match the calculator result at zero height. We are interested in truth, and they are not being truthful when they give phony results.

Looked a pretty clear calm day to me. You have no proof they gave phony results.  They did not give exact numbers anyway so the point is moot.   What they demonstrated was the beam was much higher over distances as would be expected.  If the lake was flat the beam would have been roughly the same height.

[Curious Squirrel]

But more importantly, they showed the lake was not flat.    So much show they shocked the participants.

Also, I am impressed that you conceded the point.  We don't hear that much around here.

There are plenty of refraction effects that could present the helicopter disappearing.

But there are not many things which can explain how they got a value which matches exactly what an earth curve calculator says if the height is zero. What a coincidence that refraction added enough height to exactly match the calculator result at zero height. We are interested in truth, and they are not being truthful when they give phony results.

Yeah, the height there was a bit weird/suspicious. Buuut, that's definitely not a very important part of the video. Both the laser AND the helicopter do things they should not do on a flat Earth. On a flat Earth the laser shouldn't have moved from it's 2 foot height, the helicopter shouldn't have been able to vanish from his sight as we clearly see. How does the flat Earth hypothesis explain either of these phenomena? Both are results that should not be possible to get on a flat plane.

[HorstFue]

Tom,  I am curious what you would accept as a valid experiment?   I happen to own a 300 milliwatt green laser with a stupid bright beam.   I live near Lake Travis Texas and while the lake is quite twisted there are stretches with a clear 3 miles.   What I propose is to set up a similar test on a clear night using the beam, levels,  and white targets.   Would have to establish how much beam spread there is but that would be pretty simple.  What would make it better is if any flat earther live around here and would participate?

I can't follow the preference, to use lasers. It's a tedious job, to align them to hit the target. Learn from lighthouses and do it the other way round: Use a bright spotlight as target, a floodlight, maybe even a single bright LED is visible over this distance. The eye or camera is selecting the one and only light ray, that goes from the target to the eye/camera. Find the elevations for target and observer, for which the target just becomes visible, or just vanishes.
You could also use light sources with different colors, which would give a hint for the conditions for atmospheric refraction during the experiment, as refraction also depends on wavelength.
Maybe your spotlight has a "blinking" mode, so it could be better identified, if other light sources are near by.

[AATW]

There are plenty of refraction effects that could present the helicopter disappearing.

Can you elaborate on this?

[TomInAustin]

Tom,  I am curious what you would accept as a valid experiment?   I happen to own a 300 milliwatt green laser with a stupid bright beam.   I live near Lake Travis Texas and while the lake is quite twisted there are stretches with a clear 3 miles.   What I propose is to set up a similar test on a clear night using the beam, levels,  and white targets.   Would have to establish how much beam spread there is but that would be pretty simple.  What would make it better is if any flat earther live around here and would participate?

I can't follow the preference, to use lasers. It's a tedious job, to align them to hit the target. Learn from lighthouses and do it the other way round: Use a bright spotlight as target, a floodlight, maybe even a single bright LED is visible over this distance. The eye or camera is selecting the one and only light ray, that goes from the target to the eye/camera. Find the elevations for target and observer, for which the target just becomes visible, or just vanishes.
You could also use light sources with different colors, which would give a hint for the conditions for atmospheric refraction during the experiment, as refraction also depends on wavelength.
Maybe your spotlight has a "blinking" mode, so it could be better identified, if other light sources are near by.

It comes down to being able to see the beam.  It is obviously straight and not curving or bending due to any refraction.   I have shot it off the stern of a cruise ship in the middle of the night and the beam is straight as far as you can see it.  No difference from shooting it up in the air.

[SiDawg]

It is very odd that they got exactly 24 feet hidden -- which is the value if we set 6 miles and 0 feet oberver height in the earth curve calculator. If we set the height to 2 feet or 2.5 feet, only 11 or 12 feet are hidden.

There doesn't have to be a motive for fraud. Its just laziness on account of the production staff. They knew that the earth was round, and used these calculators to determine what the results should be in their presentation.

This is a good example of "something is wrong therefore the entire thing can be dismissed"... The second experiment is a very odd they get 24 feet... i suspect the height was calculated later and then edited to seem live. They incorrectly set the telescope height to 0 feet in the same way they could with the laser experiment, and added post production the reading of 24 feet.

So that's a bit lame/stupid/devious of them... but you still haven't explained how the helicopter can disappear from site. Just because an element of an experiment is wrong, why does that discredit the entire thing? You think cause they "lied" or made a mistake in measuring then they also just tricked us in to thinking the helicopter disappeared?

There's hundreds if not thousands of videos of objects disappearing below the horizon, yet they seem to all get dismissed just because someone finds 0.001% of videos that show either refraction or a calculation error... Highly irrational.

[JohnAdams1145]

From a practical point of view, you're going to have a hard time performing the laser experiment at distances on the order of thousands of meters.

I don't know the perceived brightness of a 2W blue laser vs a 300mW green laser (although I suspect the green one is brighter) and whether you'll even be able to see it in daylight, but that's irrelevant. I don't think you'll be able to see much other than a big blob of light at the other end. Lasers typically have a beam divergence of about 0.5mrad to 1mrad. If you're going to shine it at a distance of 1000 meters, you're looking at anywhere between a 0.5m and 1m wide blob of light.