MrMaker Posted September 12, 2016 Posted September 12, 2016 ... We had the laser at 1.25 meters that is 4.1 feet that is NOT close to the water level and outside the NUDTZ non uniform density transition zone What is a NUDTZ? I googled it and couldn't find any definition other than a small 3 lines Wiki article. Is it like refraction?
John Cuthber Posted September 12, 2016 Posted September 12, 2016 Can we start by confirming the obvious- the circumstances where the beam bends upwards are (as my mathematical modelling friends describe it) "unphysical"? If I put a slab of glass on a hot plate and cool the too of it so I get a temperature gradient through the glass it will bend a ray of light that enters the glass horizontally and traverses it. It will bend towards the "cold" side of the glass- so that's upwards if the bottom of the glass is hotter. The same would, in principle, be true for a body of air that was hot near the ground, and cool further up. You could set that experiment up in zero gravity and check it. However if you heat the "slab" of air from underneath- for example- but a warm lake surface- you set the air in motion. You can't get a stable system with the warm air under the cold air. So you can't get a stable system where the ray bends downward. Never mind that this effect is small- the problem is that someone is pretending that it goes in the opposite direction.
granpa Posted September 12, 2016 Posted September 12, 2016 (edited) h = d^2/2r r = radius of earth d = 77km h = (77km)^2/(2*6378km)=464meters Roughly 1km at 100km Edited September 12, 2016 by granpa
Mordred Posted September 12, 2016 Posted September 12, 2016 Can we start by confirming the obvious- the circumstances where the beam bends upwards are (as my mathematical modelling friends describe it) "unphysical"? Never mind that this effect is small- the problem is that someone is pretending that it goes in the opposite direction. That pretty covers it...
swansont Posted September 12, 2016 Posted September 12, 2016 ! Moderator Note I have and will continue to remove posts that state or imply that data were falsified, fabricated or otherwise improperly processed without evidence to support the claim.
Mordred Posted September 12, 2016 Posted September 12, 2016 (edited) In a way its become something of a challenge. I'm currently using the k-model with true elevation corrections. Starting from k=4/3. effective Earth radius of a signal without refraction [latex]h(\theta),r=\sqrt{r^2+2ra sin\theta+a^2-a}[/latex] with the following refraction corrections to the above formula [latex]a_e=\frac{a}{1+\frac{dn}{dh}}[/latex] More or less just in the curiosity of what refraction we may expect. (lol that or I'm bored enough to be curious) To be honest the approach Studiot is taking with the others is probably more revealing. Edited September 12, 2016 by Mordred
DarkStar66 Posted September 13, 2016 Posted September 13, 2016 (edited) So how much inversion would be required to overcome the natural tendency for the laser to follow the geodesic curvature?? http://aty.sdsu.edu/~aty/explain/atmos_refr/bending.html cites Wegener as 0.114 °K/m (inversion) to match Earth curvature (so any temperature data would need to be very accurate) -- keeping in mind that this would only need to hold for the narrow band where the laser was passing. Standard temp gradient isn't sufficient to prevent light from curving down towards Earth, so it doesn't have to be an inversion to have a significant amount of 'downwards' refraction. I also recommend anyone interested in this 'experiment' and the history to read the Metabunk thread which started before the experiment and has a lot of data following... https://www.metabunk.org/lake-balaton-laser-experiment-to-determine-the-curvature-of-the-earth-if-any.t7780/ Where you'll find that most of the readings were made by the method of 'we can see the laser in the camera' from a moving boat - which we found to not be very reliable since there is clear evidence that this is not sufficient is cited in the metabunk thread. There is no control or measurement for laser beam divergence, which was originally promised to be impossibly fine (0.003 mRad, here) but clearly shows up as much larger through-out the 'experiment'. No controls for refraction, no fine-grained temperature data at or near the level of the laser, no clear data on laser divergence past about 700m, leaning board (correction, likely significant as it impacted the 'leveling' measurements by perhaps 10-11cm), some measurements made from still boat others while moving, inaccurate method for height measurements past the 700m mark (camera hits not established to be reliable), laser too low (they were warned about this repeatedly and ignored it), the photo timestamps and gps data do not align in some cases raising concerns about the data accuracy. They did not take multiple short sightings, they did not have loop closure, error calculation, or use any surveying technology, techniques, or methodology. Most of Sandor's first post here is from the Lake Balaton paper ( http://publik.tuwien.ac.at/files/PubDat_228814.pdf ) which is not his work. They really only have the video and some stills and their table of measurements. My post showing clear evidence of tons of refraction: https://www.metabunk.org/lake-balaton-laser-experiment-to-determine-the-curvature-of-the-earth-if-any.t7780/page-13#post-190072 Post with plot of their Excel data: https://www.metabunk.org/lake-balaton-laser-experiment-to-determine-the-curvature-of-the-earth-if-any.t7780/page-16#post-190250 HOWEVER, their data is interesting -- who knows, maybe the water on this lake is flatter than it should be. Hope this is helpful Update: Mick created two clips showing the beam divergence issue and why hits on the back of the guys jacket reflector and from camera lens do not likely indicate 'center of beam'. This combined with possible measurement error for '130 cm' on the board actually being closer to 1.185cm likely explain a significant portion of the observations. Edited September 13, 2016 by DarkStar66
Mordred Posted September 13, 2016 Posted September 13, 2016 (edited) thanks for the info, I had already looked through that thread a few days ago. Some excellent details over there in particular the inverse mirage scenario. (at least for my particular interest on the test). Also thanks for pointing out that the one publication is not his paper... The similarity in names could catch readers here. Edited September 13, 2016 by Mordred
michel123456 Posted September 13, 2016 Posted September 13, 2016 I don't know. What I know is that each country has its own reference system which may not coincide with the ellipsoid. In Greece where I am the national system is the ΕΓΣΑ 87. The Greek system is shifted in height by 246.62m (from the same link in Wiki) Although HGRS87 uses the GRS80 ellipsoid, the origin is shifted relative to the GRS80 geocenter, so that the ellipsoidal surface is best for Greece.[1] The specified offsets relative to WGS84 (WGS84-HGRS87) are: Δx = -199.87 m, Δy = 74.79 m, Δz = 246.62 m. What is a NUDTZ? I googled it and couldn't find any definition other than a small 3 lines Wiki article. Is it like refraction? It is written exactly after the NUDTZ in the same post NUDTZ non uniform density transition zone ! Moderator Note I have and will continue to remove posts that state or imply that data were falsified, fabricated or otherwise improperly processed without evidence to support the claim. What a relief! I couldn't sleep all night. I thought I was hit by Alzheimer. Please next time put your warning immediately. Thank you.
Mordred Posted September 13, 2016 Posted September 13, 2016 (edited) 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. Edited September 13, 2016 by Mordred
Sensei Posted September 13, 2016 Posted September 13, 2016 (edited) The laser was leveled using the boat so at 720 meters on the GE model, the beam should have risen from the surface from 1.25 starting height to 1.29. The ended up setting it to 1.32 because moving it in .01 adjustments proved difficult without an appropriate adjustable mount. What they were looking for was either a curve in the data plot against distance or an updward curve in the height data. What the test showed was a straight ascending line. The video shows the leveling at 9 minutes. https://youtu.be/GBhDFO4NMrw?t=9m I watched the all 27 minutes of the video yesterday. And honestly must say, you spend the more time flying helicopter and playing volleyball than thinking and preparing for gathering data during the experiment. From your's precalculations of globe curvature there is visible that from 1.25m it will go to 4.32m, and what size of "target" you were using?! That's has barely 2m width and 1m height. It makes completely no sense to me at all.. ?! (If I would be expecting melting some metal at 1000 C, would I use container that has melting temperature at 200 C?) Additionally target was mounted at angle. IMHO you should mount it as perpendicular to the ground of boat as possible using spirit level.. If you would be seriously going to take this experiment, you should print grid on paper with 1 cm x 1 cm spacing, with some 10cm x 10 cm with slightly different background color (and/or chessboard), with printed numbers on side. With size 2m width, and 5m height. To be able catch the all laser positions on it. Mount it steadily to the boat using spirit level to be sure angles. Mount steadily camera (better 4K,60 FPS+) in the front of target, and filming it all the time, without breaks, and as fast as possible, to have weather influence as small as possible. (target, boat and recording camera should be each other not moveable during the entire experiment) Captain should be controlling boat looking at laser spot, and trying to keep it in the center of target. Don't do sharp movements. Reading results do in hotel room, looking at where is spot on target with grid, on the video recorded by camera. Nobody serious is doing measurements in the middle of experiment, instead store data by electronics devices. (The guy using tape measure, I could not believe my eyes.. completely unprepared people for experiment.. you looked like the first time in life doing physical experiment) And obviously REPEAT it several times. Additionally this "flat earth" mentioned everywhere in the video. It will obviously put you in "crackpot zone" if you will be going to discuss experiment with other scientists. I suppose so it's just there to catch attention to YouTube video, to earn money on advertisements from Google. Scientists seeking for the truth don't behave like that. At least should not. You didn't confirm that Earth is flat, but that photons from laser flied curved trajectory, at most. And identify source causing bending of the beam. Flat Earth + photons going straight line will have the same (or similar) results as curved globe Earth + photons going curved path. And obviously REPEAT it several times. With different conditions, with different time of day, and different time of year.. Edited September 13, 2016 by Sensei 5
Mordred Posted September 13, 2016 Posted September 13, 2016 Repeat measurements would have certainly helped in different weather conditions. At least then you can average out optical refractions somewhat. Relying on lasers isn't precise, too much potential for error.
studiot Posted September 13, 2016 Posted September 13, 2016 (edited) Thank you sensei for some good thoughts, particularly as it allowed me to locate and watch the video this morning. (I did not see it before the original link was removed which is why I kept asking for method statements) +1 My thoughts on the 'experiment' are 1) Safety - No one was wearing eye protectors, yet those in the boat often looked into the laser beam. 2) It was never clear what the vertical angle of the laser was or if there was a vertical angular readout. It was however stated that one some occasions the laser was 'tilted down' to compensate for the curvature of the Earth. Perhaps this is why no extended board was need. 3) How the distances to stations C1, C2 C3 etc were obtained was not clear. 4) Some attempt was made to find the cente of the beam by measuring the upper and lower limits of the light spot and splitting the difference. This is correct procedure and corresponds to standing theodolite instructions (which are better) to read both the upper and lower stadia lines to avoid misreading a staff. 5) All this was really showmanship and quite unnecessary. Better results could have been obtained by standard tacheometry or just reading vertical angles, sighted onto the same point set low on the boat and using a purely optical theodolite or a digital station. Why try to measure up and down a scale jigging about on a small boat when you can stand firmly planted on shore and measure the scale there as an angle sighted to a particular spot on the boat as it moves around the lake? Edited September 13, 2016 by studiot 2
michel123456 Posted September 13, 2016 Posted September 13, 2016 HOWEVER, their data is interesting -- who knows, maybe the water on this lake is flatter than it should be. (...) That makes me remember the sheep in scotland joke.
Sensei Posted September 13, 2016 Posted September 13, 2016 (edited) Studiot, you gave me idea to reverse experiment. As a yet another variation. Instead of having laser on the ground and reading spot location on the boat. It's possible to mount steadily laser on the boat, and have large target (with grid of course) on the ground. And camera recording where is spot. It will be jumping back and forth as boat swimming on the waves. Certainly easier to have 10m height target on the ground, than swimming with 5m height target on the boat. Additionally catamaran could be used to even further reduce influence of waves and winds. Edited September 13, 2016 by Sensei
studiot Posted September 13, 2016 Posted September 13, 2016 (edited) Studiot, you gave me idea to reverse experiment. As a yet another variation. Instead of having laser on the ground and reading spot location on the boat. It's possible to mount steadily laser on the boat, and have large target on the ground. And camera recording where is spot. It will be jumping back and forth as boat swimming on the waves. Certainly easier to have 10m height target on the ground, than swimming with 5m height target on the boat. Wouldn't work directly. The experiment relies on using the laser line as a reference to measure from, and that reference line is supposedly set horizontal. A laser in a boat would not be horizontal without some very sophisticated mechanical mountings. I am suggesting something quite different. Edited September 13, 2016 by studiot 2
Sandor Szekely Posted September 13, 2016 Author Posted September 13, 2016 It would appear that the OP is indeed another attempt to push the misguided flat earth dogma. However instead of ridicule or minor arguments against his results we should all consider this. Sandor's results, if the survey was properly conducted, will be correct. The water surface departs from following the Earth's curvature; it is essentially flat. Note my question in post#6 that has gone unanswered Lake Balaton is an important lake because of its size and topography. Only 12 metres deep at max and much less on average as my early question shows there is a basic issue with surveying the water/air surface. [aside]Although I have not been there since 1956 and at that time I was too young to be interested in geomorphological history, my daughter had a wonderful holiday there earlier this year. I can definitely recommend it, it is one of Europe's Geopark sites.[/aside] As such it has attracted much august study about its formation which reveals the clue to the conundrum. Lake Balaton is a rift lake. That is it lies in the bottom of a rift valley. Rift valleys, such as the African and Red Sea rifts, are noted for their flat bottoms, which can run for great distances. Rifts are no great respectors of the Earth's curvature. So the underlying topography is, well, flat. And a very shallow lake has formed with sediment deposits on the rift valley floor giving a slight slope to the bottom. But since the average depth is only about 3 m How can the water surface not be as flat as the ground on which it is resting? This is not support for the flat earth view, just modern science sorting the fact from the fiction and properly acknowledging real facts. Here is a sketch geo history of the rift. lakeBalaton-098.jpg Dear Stuiot and the other members I am sorry I have not been here for a day, I will catch up now with the comments and answer them. I am researching the GE vs FE theory since years now but here I would like to focus on the lake curvature experiment in the thread. "Sandor's results, if the survey was properly conducted, will be correct. The water surface departs from following the Earth's curvature; it is essentially flat." our 16th of august experiment was the first measurement we have done so far with the laser so it has some things to reconsider and make better in the upcoming measurements. We will work toghether with a Hungarian university on the study so the results will be more precise and definite too. Studiot, I have learned about the bottom of the lake from your information here! Sorry, for your question at post #6 I have to gather information from the university as well. "How can the water surface not be as flat as the ground on which it is resting? This is not support for the flat earth view, just modern science sorting the fact from the fiction and properly acknowledging real facts." This shall be our motto for the thread! Exactly right question I can notice from your video in Youtube that your measuring white board is around 1.35m high. How did you take your measures starting in row 15 in your table as those are all higher than 1.35m? After 12min in your video the laser is not hitting the white board anymore ... I also noticed the beam diameter in the white board getting bigger and bigger. At around 14min the circle is nice and small. At around 19:10, the circle is barely recognizable. How did this "spread" impact your measures? How did you estimate the correct beam height as it was not nice and small anymore? Great structure you put there by the way. Hello Mr Marker I have uploaded all the measurement evaluation pictures here: https://drive.google.com/drive/folders/0B2gyF12ygRBjU0NSSmIxbXU4bVk 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. The collimator on the laser beam is 0.08 mRad in our measurements. Good point, +1 Variation of spot size is an important point when finding the centre of a laser beam. Are you aware that a collimation error of 0.001 degrees corresponds to 1.3 metres at 77km range? By the way, .001 degrees is nearly 4 seconds of arc and pretty poor accuracy for this sort of work. Standard geodetic work at this range would be at least two orders of magnitude finer and really accurate work better still. Studiot, please note: our collimator was 0.08 mRad this time (on the longest range measurement it will be probably 0.003 mRad) and I say we need an accuracy of 0.001 degrees for laser leveling.
studiot Posted September 13, 2016 Posted September 13, 2016 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. 5
MrMaker Posted September 13, 2016 Posted September 13, 2016 Hello Mr Marker I have uploaded all the measurement evaluation pictures here: https://drive.google.com/drive/folders/0B2gyF12ygRBjU0NSSmIxbXU4bVk C18.png 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. Studiot, please note: our collimator was 0.08 mRad this time (on the longest range measurement it will be probably 0.003 mRad) and I say we need an accuracy of 0.001 degrees for laser leveling. 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.
snaphat Posted September 13, 2016 Posted September 13, 2016 I initially was not going to post but since the discussion from the other forum was referenced here including a plot of the data from nominal values vs sandor's results, I've decided to post regarding the data itself. A week or so ago I plotted the difference between the results obtained and flat earth nominal calculations for the expected laser heights for each point of his data. I noticed a lack of random noise that I would typically expect in uncontrolled datasets (due to uncontrolled variables like boat movement, drift, currents, measurement inaccuracy, beam divergence and lack of centering, etc.). What I noticed were what looked like patterns in the differences instead that I would not expect in such a data set. Plot of differences in Sandor's results and expected flat earth laser heights at each point along the path: Last night, I took another look at the data. I decided to plot the change between points along the path for both Sandor's data and the flat earth nominal heights. To say that differently, the difference between point C1 & C2; C2 & C3, so on so forth. This is shown in the dRes and dNom columns below. Then I computed the difference of those two values as shown in the dNom - dRes column below. What I noticed is that in for the second half of the data that the change in height always deviated by 0.00 or 0.01 from the expected change between corresponding points in the path. This would show a slight divergence getting larger and larger if the curves were plotted. What was more interesting though were the results for the first half of the data points. I noticed that for the first half, expected changes between points did occur (as in matched or only deviated by 0.01) but that it was fanned out over multiple points. I attempted to highlight this in red and green below. For example the changes of 0.02 + 0.04 shown in red for dRes match the change of 0.06 shown in green for dNom. It almost looks to me that there was an attempt to compensate for the difference between the obtained results and nominal values and that this resulted in patterns occurring in the results; as opposed to what one would expect: random noise. In the coalesced column, I show that if you treat these deviations as single additive points (E.g. Adding C1 and C2 differences together) what you end up with is a net difference of only 0.01 maximum along any point of the path from expected values. This does not all appear to be realistic to me for a data set obtained with such uncontrolled variables. What I would have expected to see was simple random deviation from nominal values and under and overshooting randomly at various points along the path. Instead what we see is either the perfectly expected change, or under shooting of the change that ultimately results in the curves beginning to linearly diverge. Deltas between point N and N+1 for all points along the path for the obtained results (dRes) and nominal expected flat earth results (dNom): Graph of curves diverging (from Boxer): 1
michel123456 Posted September 13, 2016 Posted September 13, 2016 (edited) This is a picture of the lake frozen - this should have a 465 meters curve on the surface: balaton-1.jpg (...) 465m Well that sounds huge and made my eyebrows raise. So I took some time to check it with Autocad as you did. Here below in sketch (a print from Autocad gives only overlapping lines unless printed at huge scale and still not very much understandable) What you did corresponds to sketch 1. 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. -------------------------- I made the same for 6km transverse, I get a bulge of 70cm. (0,7063m from Autocad) IOW at normal height (human eye approx 150cm from ground) the other side of the lake is entirely observable. **edit added the word horizontally which means perpendicular to the radius at this point. Edited September 13, 2016 by michel123456 1
MrMaker Posted September 13, 2016 Posted September 13, 2016 ... I made the same for 6km transverse, I get a bulge of 70cm. (0,7063m from Autocad) IOW at normal height (human eye approx 150cm from ground) the other side of the lake is entirely observable. That's very interesting. So, disregarding refraction, with the point of view at 1.25m, everything higher than 55cm would be visible on the other side of the lake. Oh the power of math
studiot Posted September 13, 2016 Posted September 13, 2016 (edited) 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. Edited September 13, 2016 by studiot
michel123456 Posted September 13, 2016 Posted September 13, 2016 (edited) That's very interesting. So, disregarding refraction, with the point of view at 1.25m, everything higher than 55cm would be visible on the other side of the lake. Oh the power of math And the power of Autocad. And some experience on how to cut pieces (of wood or marble) in order to make a circle. Edited September 13, 2016 by michel123456
MrMaker Posted September 13, 2016 Posted September 13, 2016 It is written exactly after the NUDTZ in the same post NUDTZ non uniform density transition zone 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.
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