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. http://www.pbs.org/video/2365763828/ 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.