Upcoming Laser Experiment

Z.W. Wolf

Senior Member.
In approximately two weeks these researchers are going to place a laser 1 meter above the water surface on the shore of Lake Balaton in Hungary. They plan to shine the laser 23 kilometers across the surface of the lake into the lens of a camera sitting on another part of the shore. The object is to test whether the surface of the lake - and the earth - is flat or curved.

Full explanation here:

They are trying to shine a laser into the lens of a camera 23 kilometers away. How do you do that? I was going to try to calculate the angular size of a 40mm diameter camera lens at 23 kilometers, but I realized... why bother? It would be a fraction of an arc second. But what would that figure mean on an intuitive level? It would mean that it would be the size of a 40mm lens at 23 kilometers! A tiny target. How do you position a target that small so that a laser beam would hit it at that distance? Are they going to hunt the camera around until suddenly they see a bright point of light? How many days would it take using that method?

Granted there is the question of the beam divergence of their particular laser. I don't know what they are going to use but it's safe to assume the beam at 23 Km is going to be more than a meter in diameter. But would that help? And that brings up the question of how bright that beam is actually going to be after divergence and after passing through 23 Km of air.

One way of finding the beam would be to see it shining on a target; such as a large white board. That would help a bit. Much more surface area than a camera lens. But if you found it a on a target... why would you have to place the lens in the beam? You could just measure the height of the center of the beam on the target. But is the laser really bright enough at 23 Km for you to see it on a target? Even at night? Certainly not in daylight.

You could use a powerful telescope with crosshairs to sight the laser in on a particular area. That would help. But then why would you need the laser? Just use the telescope to sight on a target. If the target is visible simply lower the target until it is cut off by the horizon. If it is never cut off, then the surface is flat.

I think these researchers haven't thought this out completely. They've been hypnotized by the idea of measuring something straight with something straight just above the straight surface. In other words... "A laser beam is straight, right? So let's measure the flat surface with a straight laser just above the water surface and do it across a big distance to make really sure."

What I would suggest is one of two things.

1. Decrease the distance:
Three miles is plenty of distance. If you position the laser exactly parallel to the water surface at the shore - (tricky) - then shine the laser onto a target 3 miles from the laser, the beam will hit the target 6 feet above the level of the laser. In other words 3.3 feet (1 meter) plus 6 feet = 9.3 feet above the surface of the water. Much more practical.

This is in fact the way it was done in the recent episode of Genius by Stephen Hawking.

Episode 6. It's viewable by anyone.

Problem: Is the laser really parallel to the surface of the water at its position? That has to be tested carefully. The question of fraud can always be raised.

2. At 23 Km, rather than a laser, use a target and a telescope. In daylight, I would suggest a large piece of Masonite painted orange. A cheap 6 inch dobsonian telescope should be able to resolve a square 2 meters on a side at 23 Km pretty easily. Move the target until it is just flush with the horizon and measure the distance above the surface of the water. If there is no convenient way to move the target up and down that distance (no convenient building, ferry boat, cliff, etc.) - how about using a 6 foot diameter helium balloon?

At night just use a powerful omnidirectional light source. Or illuminate the balloon with a powerful spotlight. No laser necessary.

I would suggest placing the telescope 10 feet above the surface of the water

A. To avoid refraction effects above water (to satisfy Sphere Earthers)
B. To satisfy Flat Earthers that waves aren't getting in the way. That's an explanation they've used to explain why distant objects appear to be cut off by the horizon.

At 23 Km - 14.3 miles - using the Metabunk earth curvature calculator, a camera at 10 feet would see a target flush with the horizon at 72.5 feet above the water surface on its side of the lake. I expect there would be some rationalization about why the target is cut off by the horizon when it is above the water surface. There's always some rationalization. But we're making a prediction about exactly how many feet it would be: 72.5 feet. If that prediction turns out to be true, is that a coincidence?

Note: Be sure the camera is 10 feet above the surface of the water. Not just 10 feet above the surface of the beach or the jetty or pier it's sitting on. That's a problem I've seen several times.

A potential problem I imagine, is that at night these researchers are going to see a trace of color in the sky and conclude that they are seeing the laser in a direct line. One can see the beam of a lighthouse that's distant enough to be below the horizon, simply because the beam is shining on clouds or even haze high above the surface of the sea. You see some light, even though there is no direct line of sight. I don't know if a laser would have the brightness to produce the same effect or not. But they might even just be seeing some light the same approximate color as their laser and wrongly conclude they are seeing their laser.

Certainly they should interrupt the laser with an object so that there so that there is an off-on pattern that can be identified. In other words wave a hat in front of it in a definite pattern.
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I've found that only one of these men will actually be doing this experiment (with the help of this company: http://goldlaser.hu/?lang=en). "Dr. Zack" is in Spain. It's "SzS" in Hungary who will be on the spot. Here's a test he has done - shining a laser directly into the lens of a camera.

The laser in this test is only a short distance from the camera. Has "SzS realistically planned for a distance of 23 Km from the camera?

This is a screenshot from the first video. This is the laser:
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There was a previous thread on someone who claimed to have done a 2km experiment using a stick as a target, a emitter attached to a pole with wing nuts, aligning it by hand, using instructions passed via mobile 'phone. Oddly, they had no evidence that the laser had reached the other side, just said that it had.

Unfortunately the thread got closed, so a few of us continued the conversation.

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

I don't have anything useful to contribute on the results, rather the claim that this is a simple experiment that anyone could recreate, since I have experience with Free Space Optical comms links which work in almost exactly the same way as this experiment.

FSO uses a narrow beam laser (also may use LEDs) transceiver at both ends to link networks where there's line-of-sight between two remote locations. We use them to link sites where installation of underground or aerial cables is impossible/impractical, or third parties are unwilling (or make unreasonable demands) to grant wayleave.

The furthest I've dealt with is about 450-500m, and I'd have to say that even at that distance alignment is not trivial. FSO transceivers usually have an in-built monocular sight, or a mount for a calibrated "sniper scope", to allow course manual alignment - basically you just line the centre of the remote transceiver up in the crosshairs.

Accurate aiming is then done with screws which allow micro-fine adjustments, using software (or an LED readout on some cheaper units) to gauge the optimal position.

Mounting is extremely important, and the units either need to be fixed to a solid surface like a wall, or on a stable mount like a tripod or a pole with guy wires. To improve reliability some units use multiple lasers.

The author ("Ar Pi"?) mentions using two wing nuts to secure the laser. With the FSO transceiver the fine adjust screws are required because accurately securing it is impossible with just the mounting bolts - torqueing up the bolts will always cause some movement, even if ever so slight. The poles would also need to be plumb otherwise any X or Y adjustment of the laser could also introduce movement in the other plane.

While it's possible that the author is simplifying the process they used for the sake of brevity, and they actually did plumb the poles, have suitable mounts and fixed aiming optics, I really don't think it's realistic to suggest that just anyone can recreate the experiment as seems to be claimed here.

Bearing in mind the target area on a transceiver is usually +4 inches across, and that even over 500m the tiniest movement is enough to misalign (before the transceiver is securely mounted the vibration from a passing lorry or bus is enough), experience tells me that the odds that someone could essentially hammer a stake in the ground, jury rig a powerful laser pointer to it and then use a 'phone and binoculars to hand align it with cm accuracy over 2Km are essentially zero.

I think a great way to demonstrate the difficulty here would be to get a fairly modest laser and try and align it accurately over a couple of hundred meters using the method described. If I had such a pointer I'd give it a go.


The method of mounting the laser is not adequately addressed - not only does it need to be very, very stable it also needs to have some method of finely adjusting the aim.

If my memory (and admittedly poor maths!) serve, using the 1 in 60 rule of thumb a divergence of 1 degree is about 90ft off target at a mile distance, 1/10 of a degree is ~9ft of target, 1/100th of a degree is ~2.5cm off target.

The author is claiming 1cm accuracy, at least vertically, so needs a method of aligning the laser in <1/100th of a degree increments, and keeping it fixed there.

That's not going to be achieved with stakes, wing nut fixings, hand adjustments and binoculars.


Actually, it's 1/1000 of a degree for 2.5cm, if the rule holds true.

There, I said my maths was iffy :)
Content from External Source
The laser unit they show in the picture seems to be this one - "High Power 3W/5W/8W/10W/20W Landmark Green Laser Light, Outdoor Fat Beam Sky Laser Landmark" , so looks like they're just renting a laser display unit.

Not sure whether the pictures on that page are from that unit, but the divergence looks to be considerable. Although the unit is computer controllable, I doubt it'll have the level of accuracy required.

Ray Von