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@@UncleKennysPlace I've sent it off but awaiting the results Although my partner used their DNA facility a year or so ago and found some relatives in America
@@DaveMcKeeganMistake at 9:08 Water vapor is actually lighter than air, that is why it rises up. Water has two Hydrogen atoms and one Oxygen. Oxygen has two oxygens and Nitrogen two nitrogens both heavier than water vapor
It's like he's saying "I didn't tell you to stop". I've seen a similar video of this little monkey who kept grabbing a woman's hand and putting it back on its head when she'd stop petting him
It's like 1 meter = 3 feet rule. It works well enough for rough calculation of small measurements, but not for longer measurements and it's meant to simply be an easy and quick way to convert meters to feet to understand size or distance.
This has nothing to do with your point but ... who ever does 1m=3ft? Its barely more difficult to do 1m=3.3ft. Just add another 10% if you're going from meters to feet, or subtract 10% if going feet to meters. You'll be accurate to within about half a percent. Rounding will probably cause a bigger error.
@@dgthe3 I can't speak for everybody, but I've found the "3ft = 1m" is used more for a simple comparison rather than for measuring. "What's 100 meters?" "About the length of a (US) football field." While 100 meters is quite a few feet longer, you can convey the idea to somebody easily. (edit to fix the glaring error as outlined in replies below)
@@dgthe3 To be more specific it's "a meter = a yard" but a yard is 3 feet, so same thing and we better understand small measurements in feet. It's meant to just get a rough idea, for example if something is 50m long I can quickly go "over 150ft" or using yards it's "over half a football field", both of which are far more useful to me than 50m. Now when we are dealing with 10s (10, 100, 1000 etc) then I would use 3.3, or heck even 3.28, since it's just moving the decimal and adding zeros. It's for quick, doesn't matter much, in your head math.
@@dgthe3 Im from a metric country and have always used metrics... but I often use the 3ft to a meter when doing something that doesn't require accuracy... sure over 10meters you've suddenly lost a meter but hey, when measuring a few meters for something in the garden or telling someone something in terms of 'it was this big', 20-30cms isnt gonna matter much (3x a foot thats 30cms, the size of a school ruler, is easier to visualize and approximate with your hands than a meter i find).
@@Well...Darn. I have NO IDEA how long a football field is. Nor an Olympic swimming pool. Why can't commentators just use comprehensible units like feet, yards or metres? As a rough Imperial to Metric equivalent, 3 metres = 10 feet is probably better and quite close to accurate.
I love how flat earthers think a floating ball in space is craaaaaazy, but a flat disc surrounded by an ice wall, guarded by the evil government and inside a giant upside down salad bowl sounds completely reasonable.
Not just "the government" but *every* government, military, space agency, airline, shipping company, aircraft manufacturer, surveyor, weather agency, and on and on - including their ex-employees - for _at least_ the past several centuries. Conservatives, liberals, socialists, Communists, devoutly religious people and atheists, scrupulously honest people and con-men, quiet truth-seekers and blatant attention-seekers... Somehow, every person who holds any position of responsibility is captured by this conspiracy. Makes Scientology look like a clique of preschoolers.
As usual, the "ballers" have it all wrong AGAIN. The flat disc surrounded by an ice wall, inside a giant upside-down salad bowl isn't guarded by THE government. It's guarded by ALL the governments on the planet. And has been for all of time. (And, no, don't ask me how the disc got here.) This was the case even when there was no means of communication between the governments (which was 99.9% of that time). Wait, when you say it like that, it really does sound crazy, doesn't it?
Excellent video as always. No insults, no yelling, just cold hard facts, science, and logical reasoning. You're at this point the best debunker I've seen, keep up the good work!
This is exactly why I watch this channel. No bombastic insults, no degrading, no hyperbole. Just facts, clearly expressed, in a gentle tone, while giving a very good dog some love. He isn't insulting flat earthers, just explaining how their conclusions are inconsistent.
I much prefer Professor Dave awesome debunkings exactly because he is not afraid to tell Flattards what they are: retards who slander scientists and engineers and everyone else. M Keegan is nice but not my favorite. He respects too much the flattards
@@mikepictor Very true. However, I will say that the majority of flat Earthers aren't likely to be swayed by facts given their absurd adherence to pseudoscience and faulty argumentation, so the lack of insults probably won't do much to help them realise their errors. Great video though as always.
@@ZoogleasI agree that they won't easily be convinced that the earth isn't flat, but if they were insulted for that it would be even more unlikely for them to realise that.
Honestly, the fact that Dave video feels more like a simple fact explanation than a rage debunking makes it 100 times more enjoyable for everyone. Keep going ! (And flat earther, for God sake, please go sailing.)
Absolutely... if the point is to help borderline flateathers and/or help flat-earthers see better... Dave's presentation is so much less arrogant, angry debunking, which nearly always produces the backfire effect. As one who looked deep into the anecdotal evidence of a flat earth (finding it compelling), the clean, calm, and clear presentations of real experiment and objective observations, is so much better at keeping a mind on the facts rather than the emotion. Which is so unlike Prf Dave who's snark and arrogance is a real turn off for people needing facts, experiment, and objectivism. Just be cordial, clear, and respectful and you will be able to deliver truth to those who need it.
He would actually make a great professor. I'd call him "Dave the legit professor" Edit: if he became a professor, then I would call him "Dave the legit professor". I didn't think it would be necessary to clarify that 🤷♂️😄
@@steevo101The only credit I’ll give Professor Dave is that a lot of his annoyance and frustration is directed at those who have directly attacked him or frankly are scummy enough to deserve it. If you look at some of his earliest Flat Earth content that are simply just informational debunks, they are a lot more tame.
This is similar to the difference between the line of sight of a rifle scope, and the line of a rifle bore. The ballistic arc of a fired projectile appears to “rise up” through the line of sight, at the first point of aim, and then drop back to intersect the line of sight again at the second point of aim. In reality the projectile is dropping the second it leaves the barrel. The apparent “rise” is a product of the angular relationship, and the offset, between the scope and the bore.
Hehehe... there is a video game series "Sniper Elite" that I've enjoyed. One of the neat physics features is that if I can figure out exactly where the 'bad guy' is and he's just below an intervening obstacle at some distance (say, 200m), I can shoot OVER the obstacle and hit them. The game correctly calculates the difference between scope sight and rifle bore, and the shallow arc the bullet travels over the obstacle, dropping down to hit the 'hidden' target. :)
@@mikefochtman7164 I know the series, and played it on the PlayStation 2. It’s one of the only games that models the external ballistics and sight picture relationship well. I particularly enjoyed it, because I grew up in the middle of nowhere, and shot regularly until my mid 20s. I used to hunt, periodically, mostly rabbits because my grandmother made an awesome rabbit stew, but that was using a point blank aim setup - where the trajectory is fairly flat, and the arc between the 1st and 2nd point of aim is contained within the kill zone. Which is only about an inch for rabbits. But what I really enjoyed was long range target shooting for fun. I used to nerd over ballistics calculators, making shooting tables (of minute of angle corrections), range finding and scope set up. I live in a much more built up area, these days. I kinda miss not having anywhere suitable to shoot.
@@lucasdrudi7231 Yeah, the basic concept of the relationship between line of sight, and a ballistic trajectory, holds true with any form of shooting. Ironically, I also did archery regularly from age 11-16, bare-bow with a recurve, and a light 45lb compound bow with sights. But yeah, the principle is basically the same.
This doesn't make complete sense though. If you're firing at a significant distance (200 yards or more) then you have to tilt the gun upwards slightly. This is extremely obvious using iron sights. As such the bullet isn't always dropping the moment it leaves the barrel, there is a period where it will be moving upwards slightly before it starts dropping. The only time this wouldn't happen is if the gun is completely level with the ground, or obviously pointing downwards.
Noticed one mistake. Humid air is less dense, not more dense. Higher the humidity, lower the density. So the least dense air would be hot and humid. Cheers
@@notgonnahappen7899Except that you do sometimes get clouds at ground level - that's what fog is, a cloud which has fallen down to the ground. This is related to the classic question of "how do clouds stay up in the air", or "why don't clouds fall down to the ground", or similar: Clouds ARE constantly falling - it's usually thermal air currents which keep them up in the air, fog happens when those thermals dissipate and the cloud falls back to earth before it can evaporate. Which is why fog often happens in the early morning, after the daytime thermals have dissipated overnight, and the fog clears later in the day when enough energy is supplied by the sun to evaporate it, or the thermal air currents regenerate and lift the water vapour back up into the air. Density differences are of course a factor: Thermals take warm air (which is less dense) upwards and the moisture goes with them as humidity, since warmer air can hold more water vapour. Clouds form when that rising air cools down and can no longer hold the moisture as humidity, so water droplets condense out, forming a visible cloud. Which can often continue rising or at least not fall, depending on the strength of the thermal supporting it.
Wow, I never thought about that and at first it confused me, but it absolutely makes sense: If one assumes, the air pressure is approximately equal inside and around the cloud, by ideal gas law (pV = nRT) the particle density is equal inside and outside. Since H2O is lighter than O2 or N2, the density of humid air is lower. Thanks for the the hint.
Saturated moist air density at 15°C is 0.9936 times the density of dry air. Moist air is maximal 0.64% less dense than dry air at 15°C. But Note: not the absolute density creates atmospheric refraction, but the vertical density _gradient._ So if humidity is constant, it does not contribute to the density gradient and hence to refraction at all, only to the absolute density. The influence on the density gradient is in practice negligible. So refraction is practically the same for try and moist air, except humidity has a very strong gradient, which is limited because humidity can only vary between 0% and 100%. The stronger the humidity gradient, the smaller the layer it influences.
After Dave explained why simple Earth's curvature calculators are not so accurate with what we observe I was afraid that he won't mention Wolter Bislins work. That guy is a legend. Thanks to Dave it wasn't a case. ;)
Thanks Martin for introducing me to Walter Bislins calculator. I have always wanted to see graphically the relation between observer height and horizon distance. Thanks to you, now I got it. Thanks again.
Bislin's Earth Curve Calculator is great. Shows the difference between hidden and drop. Factors in refraction. ...A great many in depth articles on his sight as well. I especially enjoyed the detailed Rainy Lake Advanced Level Bedford Experiment.
8" x Miles squared is a rounded off version by Samuel Rowbotham based on an approximation given by the 1860's Encyclopedia Britannica in their article on Levelling. Samuel included it in his first small edition but left out the article in the larger later edition. The article included the entire mathematical derivation and is quite interesting. It also included a very explicit warning against the effects of refraction that Samuel completely ignored.
This matches well with the [Wikipedia: Bedford level experiment] article on Rowbotham's canal experiment, but the encyclopedia sure gave me a merry chase! Levelling: See surveying. Surveying: See trigonometry. I believe the formula can be arrived at using a Taylor series for 1-cos(x), which produces roughly x²/2 (constant terms cancelled, every other term becomes 0, and further terms scale with an inverse factorial, so it converges - at least for x
@@0LoneTech Why not check out the first edition of Zetetic Astronomy by Parallax. The preview available at Google Books has the quote starting on the 4th page of text after the contents list. The maths used in the quote is super basic. No Taylor series.
I have derived the formula here byggvir.de/2023/08/01/curvature-of-the-earth-eight-inches-per-mile-squared/ The approximation is good enough for eye levels below 10,000 m. The error is less than 2 % and less than 0.04% if you use 7.8481 cm / km squared. The refraction raduis in air is between 40.000 to 50.000 km. That is approximately 1 cm per km squarred. That adds an error < 15 %. In most flat earth evidence, this error can be neglected because if something is hidden, regardless of how much, the earth can't be flat. Flat earth mostly porve that something can be seen, but not how much is seen and how much is hidden.
While yes it was used by Samuel, the roots of this old rough approximation is older than Samuel and had been around before him. Samuel did not invent it, he was just merely smart enough to likely understand that represents a parabola but used it as the con man he was. It is used by surveyors to do quick calculations as it works fairly well for short distance. Especially when talking about single digit miles like surveyors typically deal with in line of sight. And was used back in the days for approximations without doing complex maths. Any real calculation of Earth curvature _must_ include the radius as part of the calculation. There is no radius accounted for in 8" per mile squared.
This flat earther in the beginning says "with 8 inches per mile squared, the curve would most certainly be visible, but it's not." Then when you show flat earthers the Lake Pontchartrain pylons photo taken by Soundly, they say "I'm seeing too much curve." How do we simultaneously see not enough curve *and* too much curve?
Well in this case too much is the fact you can see it at all. They live trying to make everything black and white and use things like the approximations intended to be close enough before the days of computers when everything had to be done by hand.
@@Isolder74 They also love to parrot that it isn't the "geometric horizon" so that makes it invalid. They like to ignore the fact they can't calculate to the "geometric horizon" on a flat earth because they don't even have a model of a flat earth.
To add another point about atmospheric refraction, it can result in hot air and cold air mirages. Just as the plane appears reflected on a tarmac, so too can the sky appear reflected in the desert, giving the illusion that there is a source of water ahead. Cold weather mirages had a significant impact upon the tragedy of the Titanic as well. It was a moonless night surrounded by Icebergs. The refraction made the horizon appear higher than it was, hiding the iceberg until immediately before it struck the Titanic, it also made the Titanic appear like a completely different ship to the surrounding vessels, and when the Titanic's crew tried to used the Titanic's lights to send a distress signal, the cold air caused to to flicker like starlight, scrambling the messages.
I have a hard time understanding the laws of physics, you see, I never studied law.🐰 Thanks for the clear and concise explanations Dave, I really enjoy them.
I hate the expression "laws of physics." The laws of physics are things like "don't leave radioactive materials laying around on benches", "turn off the lights if you're the last one out of the lab" and "grad students should have the same number of limbs and digits at graduation that they had when they started." The laws are the laws of nature as discovered and described by the science of physics.
@@d614gakadoug9 I think it's called the rules of physics Class not the Laws of Physics because Rules of Physics Class make a lot more sense than seemingly breaking the Law if you don't follow them(Which luckily is impossible)
@@FlatEarth-q1f He's also very good at communicating pseudoscience, as in explaining the flat earth position, and then using real science to explain why it's all bunk. He's generally just good at communicating both sides, steel-manning his interlocutors and then clearly and thoroughly debunking the nonsense flat earthers espouse.
@@FlatEarth-q1fhe is quite good at clearly explaining the arguments of flat earthers and thus makes it easy to understand why they make no logical sense
Most flat Earth UA-camrs have dug themselves in so deep with their beliefs that they have no way of getting out without destroying their online reputation. It's because of this that I firmly believe that even if a flat Earther was gifted a seat on Blue Origin, they would never report about it or they would come up with some extravagant conspiracy that they were drugged and plugged into a computer simulation or something crazy like that.
I just love how the water level absolutely destroys the horizon is always level argument. I get that it can be hard for someone to imagine the scale of things, we spend all our lives close to the ground, but I don't get how they can ignore the tons of evidence.
Cognitive dissonance and lack of education is my best guess. There's a very conspiratorial tinge to most flat Earthers though I do find. Has there ever been one flat Earther who didn't think the government/NASA was lying to them about the shape of the Earth? I've yet to find one! Not that NASA has even been the authority on what shape the Earth is, so even if they lied about all their missions, that still wouldn't debunk the shape of the Earth lol.
I think you are mistaken here. The water level demonstration does NOT destroy the horizon is always level argument. It destroys the argument that the horizon always rises to eye level!
I only ran across the modern flat Earth stuff a few months ago, but I've followed blogs of a number of science communicators, mostly in matters related to biology such as evolution and medicine, for many years. When I "discovered" flat Earth I quickly came to realize that flerfs are very much like others who deny science and the knowledge we get therefrom. They are often remarkably arrogant and they are almost all quite profoundly ignorant of the basics of the topics whereof they speak. They'll assert that they are "critical thinkers." Fine, but you can't critically think your way through something about which you know almost nothing. Actually, and real critical thinker with limited knowledge would conclude that when they are arguing against a large body of evidence from a multitude of highly qualified scientists, it is the odd man out who is likely to be wrong. Of course sometimes the odd man out turns out eventually to be correct, but it is pretty darned rare these days in science.
@@FlatEarth-q1f theoretical? Oh you mean the proven and reputable sources and evidence rather than the guy in his moms basement who thinks he’s proven the worlds smartest people wrong. Keep trying, keep failing and I’ll keep laughing
@@FlatEarth-q1f I see that you have a new script that you don't understand. "Theoretical Physics! *SQUAWK!* Theoretical physics!" You don't know any theoretical physics. You have never even touched a physics book. 🤦♀
3:06 It amuses me how flat earthers look at that picture of Chicago (for example) and suggest that it's a proof of flat earth and yet ignore the ELEPHANT in the room that is the missing bottom half of the buildings. No matter how or what formula is used there is no way that image should be like that on a flat earth.
well, our globe has been mapped every millimeter... Yet flatards can't even agree of a map of their fairytale Frisbee, as none of the distances matches reality within hundreds of miles 😂
That city being visible from across the shore is funny because it only happens under specific conditions. If it was proof that the Earth is flat then it would be visible every day of the year, but it isn't.
The only reasonable explanation is that THEY dug a big hole to hide Chicago, but every now and then, freedom fighters manage to release enough hot air to make the buildings float up and give the game away. Nothing else fits /s
@@thearmouredpenguin7148Correct. Didn't stop Rob Skiba (RIP) from using this as his argument that the Earth is flat. Funny thing is, and this is true of a lot of flat Earthers that I've seen, he seems overall like a fairly intelligent guy but his religious leanings have just turned that part of his brain off.
One important detail about those photos of the Chicago skyline, and why refraction doesn't explain the observed shifts: they're taken from ~100 feet above water level, atop very large sand dunes that are common on the Lake Michigan coasts across from Chicago.
Occasionally you can see looming of the French beaches from the high cliffs of Dover. Most of the time France is visible but only cliff tops. Diffraction makes a big difference. Head over to the Bislin advanced earth calculator and use the refraction slider to change levels.
Great video. This is a topic that has been the bane of my existence. My brother does not believe refraction can cause as much as I would think based on the videos. But with every rebuttal, he points to another TikTok FE video, and ignores anything I say. I also just started researching this topic, and found another video that when through the math to show how this formula works, so it was good to see another person mention it as well.
Be careful arguing with siblings who've fallen into the Flerf cult, my brother got so deep in the rabbit hole he eventually had to be sectioned. He's out now, but since he's started bringing up other conspiracies recently I speak to him less and less (his current favourite is mudflood). On the plus side, he never mentions flat earth now, but when he was sectioned the doctors explained how arguing with him about it would have made him worse and I kind of see their point. Sadly it means we rarely ever speak, as I don't want to contribute to another more lengthy stay in a care facility.
@@smiffsoft One thing that I have noticed about conspiracy people is they never bring up the real stuff. There is the story of how Target knew a 15 year old girl was pregnant before her father did. In the county I grew up in there is a conservation authority that had a lot of power. It was not really part of the government but still the guy in charge could block construction in some places. And there are other real life examples of people and groups manipulating people but the conspiracy people never talk about them. I'm not sure if any of the people complaining about shrinkflation are conspiracy believers.
Something I would love to see explained by the Flat Earthers: if the flat earth is the reason why Chicago skyline is visible in these images... why isn't it visible all the time? Or at least, most of the time, without heavy cloud cover or fog obstructing the view? I have asked Flerfs about this a number of times... but they all chose to ignore the question instead of making something up as usual.
Always a fun way to pick apart their arguments. "Hurr durr this thing is visible!" Yes. It's visible in this instance, but if the Earth was flat, then it would be visible *all* the time instead of waiting for specific conditions to make that thing visible. So why do photographers have to wait for specific conditions to get the picture?
@@Tabearnack A flat earth wouldn't negate refraction, but the density stratification would be horizontal, not curved, so the visual effects would be very different from what we see in reality.
@@5peciesunkn0wnEver seen the fake horizon above the real horizon?. Maybe go outside and observe a sunset or sunrise over the ocean for a few days and you'll see..
It can not be stated enough that your explanations are thorough, clear, and straightforward. A person could not care at all about the “debate” (if it can even be called that) and still learn a lot. It’s also refreshing that a video about this isn’t just heavy handed with condescension insults. Refreshing to watch.
The approximation "8 inches per mile squared" IS accurate out to about 750 miles. If you want the "HUMP HEIGHT" of the earth in between the two endpoints, you must use HALF the distance between the two endpoints. If you want "DROP FROM A TANGENT" then you use the full distance. The Flerfs don't know the difference, so they always calculate "Drop from a Tangent" when they really want "Hump Height". SO, it's no wonder they can never find the curve they are looking for because they are always looking FOR TOO MUCH! There is a video on my YT channel that explains the math in detail.
As a hump height calculator it is useful for trying to explain to flerfs why Earth's curvature can usually be ignored in short-distance surveying, as in laying out the foundation for even fairly large buildings, or explaining why trying to measure the hump in the water in an Olympic-size swimming pool would be a daunting task. I've used it a few times as a check to make sure I've used an on-line circular arc/chord calculator correctly. I don't use such a thing much, so I don't have a good "feel" for what to expect and therefore like a check to make sure I haven't erred due to specifying the wrong unit or some other blunder. A lot of people don't seem to take a moment to ask themselves "does this calculated value look reasonable?" and wind up with something like being 9 orders of magnitude off because they confused "milli" with "mega."
@@d614gakadoug9 "confused 'milli' with 'mega'." is the same thing as flerfers dumping water on a basket ball and asking why it doesn't stick to the surface of the ball lol.
They don't want the hump, they want hidden height, but do not regard observer height. But that is easy to do (when you know math): You reverse the 8 inches formula, then you can determine the distance to the (geometrical) horizon for the observer height. I.e., Distance = sqrt(8"/height)
In that distance your line of sight is tangential with the earth surface.
Then you subtract the distance to the horizon from the distance to the object and then can use the 8 inches for that rest, getting the geometrical drop..
8" per Mile^2 was a standard approximation used by surveyors and civil engineers, before the availability of computers, since at least the early 19thC. The earliest reference I have found is in "A Treatise on Surveying and Civil Engineering, Wherein Everything That is Useful and Curious is Demonstrated from its First Principles", by P.A O'Shaughnessy (p30). Published 1848 in New York.
The formula gives a slightly too small approximation for short distances. It is most accurate at 400 km. Until 550 km or 342 mi the error stays within ∓0.032%. Until 800 km or 497 mi the error is less than 0.1%.
Surveyors then and now used it for an approximation of the drop from the horizontal line of their theodolite, starting from the surveying point. It works well enough for typical surveying distances but was never supposed to be used for drop beyond the horizon.
Love this kind of stuff Dave. Once upon a time, I was taught and became quite good at working out what the Brit Army called "Intervisiblity" ( or did, when I were a lad!) Instead of spending huge amounts of time tromping up and down mountains to check what could be seen from higher up - we used a trig formula ( which I've completely forgotten!) applied to the contour height and distance from and to - this was very handy if you wanted to observe say a valley C from mountain A but with big hill B in between - the formula would be used to find the best vantage point. Don't suppose anyone is bothered now that "drones" are available.
I agree with all. The one major point that you missed is that no flat earther ever uses 8 inches per mile squared to measure curvature. They always use it to say, "you shouldn't see this building, tower, or mountain range" when once you calculate that the hill you're standing on is 500 feet above sea level and that tower is 200 feet high... you totally should see that tower.
Something of a tangent (perhaps somewhat appropriately), but this reminds me of reading about something from World War II. USS _Atlanta_ was Shanghaied into helping USS _North Carolina_ (I think. She might have been helping _Washington_ instead) with an offset gunnery exercise off Guadalcanal. _North Carolina_ took aim at _Atlanta_ at maximum range, offset her rangefinder by a degree or two, and opened fire. _Atlanta's_ crew could only see the very top of _North Carolina's_ mast, where her big rangefinder is. One small puff of black smoke and about 30 seconds of flight time later and _Atlanta's_ crew were treated to the disconcerting experience of 9 16 inch shells smashing down in her wake. If not for the offset, they'd have hit dead center. _North Carolina_ proceeded to do this over and over and over again, never once properly coming into view or up over the horizon.
The 'offset' may have just been the target's course and speed. I was amazed at the analog fire-control system of WW II battleships. If the target is steaming at 20 knots, it'll have moved over 300 yds by the time the shell gets there so an estimate of target course and speed was one of many inputs. To be as accurate as they were is a testament to the engineering of those systems.
i am tempted to fall back to my simplistic answer: why does someone on a boat, at the top of a mast, spots land before the ones on decK? On a flat earth, everyone on the boat (with the same eyesight, of course) would spot land at the same time. There would be no need to put someone in the crow's nest, as mariners have been doing for thousands of years. Flerfers jus trigger me.
The reason they try to get up s high as they can it to be able to look over the ocean waves. Of course you probably don't know how high waves can get if you've never been in the middle of the ocean.
Not always Ocean level is not constant Just to tell you that simple disturbance of the sea can give you a 15 meter difference A big disturbance like a big storm is more than 20,30 meters. And i mean not like in the movies Just a quiet and up and down elevator feeling. So no even if the earth was flat you cannot see farther than a few miles. Local fishermen use the view of the opposite city as a reference. If the city across the shore is visible the weather is good I mean they can see the shoreline If they cannot the water is too high the weather is bad. It has nothing to do with curvarure.
Formulas 8"/mi^2 you want to graph y = -x^2/7920 + 3950 This is just converting 8" into 1/7920 of a mile. The + 3950 is to shift the parabola up, so the vertex aligns with the north pole which is 3950mi above the center of the earth at the origin. Alternatively you could shift the elipse downward 3950 and have the north pole and parabola vertex at the origin. To graph the earth in miles you want to graph an elipse with major axis 3963mi and minor axis 3950mi. So you get (x/3963)^2 + (y/3950)^2 = 1 The 8"/mi^2 is actually quite an accurate formula. You can calculate the discrepancy between the two curves by graphing Y = (-x^2/7920 + 3950) - 3950sqrt(1 - (x/3963)^2) If you do this you find the formula is quite accurate in absolute distances up to 550mi. The discrepancy is never more than 0.0323mi = 170.5ft in that range. Which is pretty good considering that occurs in conjunction with an actual drop of 15.94mi or roughly 0.2%. Considering percentage errors it's pretty accurate up to about 700mi. Over that range the error never exceeds 0.4%. Ranges of accuracy err < 0.1% : 436mi-560mi err < 0.2% : 357mi-614mi err < 0.3% : 255mi-661mi err < 0.4% : 47mi-706mi err < 0.5% : 0mi-748mi err < 1.0% : 0mi-928mi err < 5% : 0mi-1751mi err < 10% : 0mi-2319mi
Great video. One thig that struck me is that it is a prime example of the frustration on how to address simplistic arguments. To fully explain the whys and hows, science (and real life in general) requires a detailed explanation and requires the listener being willing to learn. On the other hand, the simplistic argument latches on to a simple easy to repeat phrase that they can repeat ad nauseum. It may be wrong, but the debunking of it can require a detailed argument that can go over many people's heads. Thank you Dave for making straight forward videos that explain and debunk the arguments without descending into ridicule.
Flat earthers seems to actually believe that there is a real debate about the shape of the earth. They don't seem to understand that flat earthers are just entertainment for smarter people.
The fact that you can maintain a straight face, while your dog demands pats and scratches, astonishes me. The way the dog puts its paw on your arm just cracks me up.
the fact that he can have a straight face while lying through his teeth and find time to shoot ,produce, draw and edit these videos in minimum time really astonishes me....
5:19 That is one I learned long ago... If you stand on the beach and watch for the sun rising. The moment you see the sun breaking the horizon, lay down on the beach and you will not see the sun. So there's the curve, anyways this is just another way grifters/conmen marketing for those Patreon Dollar!
Honestly though Dave, as much sense as your video makes and even people with remedial education can actually grasp these concepts, remember….and I have stated this countless times, the flat earth individual is suffering from an illness. I suspect it is on par with an addiction illness or a chronic lack of the ability to realize and understand that you were wrong. No different than a thief being shown a video of them committing the crime and insisting that it is not them. The hypothetical person who caught this individual has them dead to rights….and it could be a crystal clear picture of their face and an accurate height of the person can be established along with witness testimony, yet they will until their dying day claim that it is not them in the video. Dave, it’s like trying to change someone’s religion or political ideology. It is practically impossible. So maybe one day a pill will be developed that will help these poor souls but until then…I truly give you credit for dealing with the childish behavior and viewpoints that are culminated in the realms of the flat earthers mind.
And also this formulas for the curvature are meant to measure perfect spheres, which the earth is not, even it can have large amounts of land actually flat but in the full scale it’s basically a sphere
My most recent dog would do that stuff. He would even put his head to my hand and move his head so he got his scratches. Reilly was a good boy and I miss him.
The "it's a parabola, not a circle" thing to me is less about accuracy and more about the fact that so many flerfers constantly claim that "8 inches per mile squared" is *the* formula that "all scientists" say is how you must calculate this stuff, when *it is just obviously not* because _it doesn't even represent the right shape_ and any _real_ scientist could immediately see that to be the case. Yes, it can be used as an approximation over short distances, but _that's never what the flerfers are actually claiming_ it's for. And the fact that they can't even understand that that formula is not _actually_ the formula for a spherical earth, because it mathematically cannot be, and it also _is not the formula that any scientist or earth curve calculator app actually uses,_ or claims to be accurate, just shows how little they understand about the reality of how any of this actually works.
Agreed. That's why it's important to argue that point, because that's Cult 101: Lie about the premise, then attack that premise. "This is their formula and it's obviously wrong, look!" Well, no, it's *not* our formula, but they'll keep repeating that lie because they desperately require their converts to be misinformed. Flat earth doesn't spread without misinformation and misunderstanding.
One of the key things I learned early on in engineering school is how useful approximations can be, and also to respect their limitations. There's plenty of useful approximations that engineers use everyday, the thin walled stress approximation for cylinders, the coherence of electromagnetic waves after 10 wavelengths, etc. these aren't reality, but they are damn close for almost every use. But we scientists and engineers must still acknowledge and justify our use of approximations. 8 in/mi/mi is a great approximation, if and only if you respect its limits. The fact that flerfs don't acknowledge that it's an approximation, don't acknowledge the limits and justify the use of this approximation shows their deep ignorance of science, math, engineering, etc.
@@phillyphakename1255 The 8" per mile² comes from surveying text books in the USA. It is referred to as Cc, curvature correction. Actually it is not 8" per mile it is 8*miles² and the result is in inches. The eight is a constant and it could equally be 0.66666*miles² and the answer is in feet. The metric equivalent used in the rest of the world is - Curvature correction: Cc = -0.07849 * D² (D is in kilometres) answer is in metres. It is used by surveyors along with a refraction correction Rc to get readings from a surveyor's level over the relatively small distances a surveyor will normally be working at. Goes like this, an assistant holds a surveying staff on the ground 1000m away and the surveyor looks through their level and reads off a height from the staff. The surveyor takes that reading, subtracts the height of their level from the ground it is on, subtracts Cc and adds Rc to come up with the figure for the height difference of the surveyed point with the observer's position. If you want to know why it is a parabolic value which does not match the surface of a sphere consider this. A circle on a cartesian graph centred at x0, y0 has the formula r²=x²+y² (Thanks go to Pythagoras). For the earth r is the earth's radius, a pretty big number in miles or kilometres. At the top of the circle at x=0 y=r so y²=r². If you move along the x axis the amount likely to be used in surveying x is going to be a small distance compared to r or y so x² is small enough to be ignored and you just look at the changes to y due to the small angular change. This is why until x becomes significant the shape of a parabola is very close to a section of a circle, another approximation that can be useful but has limitations. Another approximation is that over these small distances the surveying staff can be considered parallel to a plumb line at the surveyor's level.
Thank you! Every time I hear that formula, my brain, which is terrible at math and math concepts, still, my brain said: wait, but, that's not enough? To describe our experience with seeing the globe, and sure enough! The height! The formula as described is like a floppy string with no anchor point, no reference, until you anchor the viewer's height: thank you. :)
Also, if you put a coin on the ground in front of you. What is the difference between that coin and eye level? Probably about 5.5 feet give or take. If you move that coin 50 feet away from you. What is the difference between your eye level, and the level at which the coin sits in the ground? Still 5.5 feet and it will remain at that level untill it is moved far enough away that it drops behind the horizon. So the horizon does not rise to eye level
Think about this...even if the Earth were flat, the horizon would NEVER rise to the level of your eyes! The difference in height between your eye level and the coin/floor will ALWAYS be 5.5 feet no matter how far away the coin is moved. Flat Earthers are just morons.
@@mangojulie123 Not only that, but flat earthers believe that the earth is disc shaped right. So it's still round in their idea. Which means the horizon would still appear curved. The only way the horizon would appear flat, is if the earth was square
@@righty-o3585 You and I think very much alike. That's why they had to also bring the conspiracy of space travel...we can't go into space to take a picture of their pancake earth 🤣. And even if the Earth were infinitely flat as some flerfs claim, the horizon would still appear curved because the radius of the limit of vision of our eyes would carve out perfectly curve.
@@mangojulie123 Bro, I remember like 8 years ago, when this new era of flat earth just started gaining some attention. Somebody had commented about how because of gravity in space, the default shape for anything planet sized, is a globe, or sphere. And I replied..... DON'T LET EM HEAR YOU SAY THAT, THEY'LL START CLAIMING GRAVITY IS FAKE.... Sure as shit like 2 and a half years later. Wish I could have screen shotted lol
Rememeber that time, when a flat-earther, instead of going out of his way to deny the globe, actually came up with evidence to prove flat-earth ?? Neither do I...
here let me use maths to explain why approximating a circle with a parable works: circle: y= srqt(r²-x²) parabola: y=ax²+b we care about the difference between these functions: ->Δy= circle - parabola= sqrt(r²-x²) - (ax²+b) we want to see what happens for small x, so xΔy≈ r - r= 0 so in conclusion for small x the function values of circle and parabola are almost identical. that means for a few dozen km the approximation works absolutely fine.
For the sake of mathematical accuracy, I want to point out that this is only a first order approximation, i.e. you've only shown that the tangent lines are equal. This is not surprising and it's also not something flat-earthers would dispute because a flat earth is also tangent to a circle (cause the tangent is flat). This can be easily seen by using a constant function y = r instead of the equation for a parabola. That means your equation for Δy would look like: Δy = circle - constant ≈ sqrt(r²-0) - r = r - r = 0 You haven't taken curvature into account at all, unfortunately. The real reason a parabola is sufficient for approximating the curvature of a circle is because curvature only depends on the second derivative, and a parabola is the first polynomial to have a non-zero second derivative. The curvature of a circle and a parabola at its vertex are well known, and fairly simple to derive if you know how, but I won't go into it here. A circle's curvature is just 1 / r and a parabola's curvature at its vertex is 2a, so the two curvatures are exactly equal at the parabola's vertex when a = 1 / 2r. Here's a desmos graph demonstrating this principle: www.desmos.com/calculator/mdywqupmws Plugging in the radius of the Earth for r and multiplying by 63360 inches per mile gives 8.0024 inches per mile²
I always tell flerfs to graph this shit. I also tell people who know we live on a globe to graph it too, because while it is a parabola, its actually oretty accurate for an observer with an eye height of 0 out to around 175 miles. I also always use the part where the fact that they do fit quite closely for a while as evidence that flerfs haven't done the math, because that would be the natural response to "it's a parabola"
Another issue most people forget, is the "squared". Miles squared is not a linear unit of distance, but a planer unit. When Flat Earthers' say "8 inches per mile squared", they mean "8 inches per linear mile".
stuart has it right; it describes the formula, not a conversion factor. Here's an example translated into Qalculate as a function called paradrop. > angle(x) := x arcmin/NauticalMile > EarthRadius = radian * NauticalMile / arcmin > drop(x) := (1-cos (x arcmin/NauticalMile)) * EarthRadius > paradrop(x) := 8in*(x/mile)^2 > compare(x) := transpose (vector(drop x, paradrop x, drop x - paradrop x, (paradrop x-dropx)/drop x)) > [compare(1km), compare(10km), compare(100km), compare(1000km), compare(3000km)] [compare(1 kilometer) compare(10 kilometers) compare(100 kilometers) compare(1000 kilometers) compare(3000 kilometers)] ≈ (78.53 mm) (7.853 m) (785.3 m) (78.37 km) (693.8 km) (78.46 mm) (7.846 m) (784.6 m) (78.46 km) (706.1 km) (77.57 μm) (7.756 mm) (759.6 mm) (−83.74 m) (−12.28 km) −9.878E-4 −9.876E-4 −9.673E-4 0.001069 0.01770 That's a relative error under 2% at 3Mm. I divided by the mile unit to get "number of miles" before squaring, which is dimensionless. Side note, here I'm assuming it's surface distance, not horizontal distance, that is the argument for the function. They're nearly identical for small angles anyway. The formula has nothing to do with what's hidden, it only describes how far off horizontal will be from level in the distance.
I've heard a couple of flerfs say "eight inches per square mile" and I'm pretty sure it wasn't just a slip of the tongue in either case. Similarly I've heard one refer to the "inverse squares [plural] law."
@@d614gakadoug9I just worked it out from the Taylor series again, and one form is "drop / distance² ≈ 8 inches / mile²". The squarings match because it's derived from (distance/radius of Earth), which is an angle (dimensionless). If you solve it for drop, you get "drop ≈ 8 inches * (distance/mile)²", which I prefer because it doesn't split the square and keeps all operations on one side, but spoken aloud it often gets ambiguous; "square mile" is "mile²" is "mile squared", but where did the squared distance go? Meanwhile there's an entirely unrelated inverse square law, which has to do with the spread of e.g. light or gravitational pull over a surface with distance. And when that's applied to a reflection, like the light we see from the moon, you get squared squared inverses.
The BIG question about this springs to mind..... Will Slappy show up still lying about being a pilot or will he be unable to resist showing off the lovely plumage of his Norwegian Blue, i mean Black Swan?
Love your videos, Dave. However, as a meteorologist for the past 43 years, you got one point wrong. At 9:00 in the video, you state that higher humidity air due to increased water vapor is more dense. That's incorrect. Clouds rise because water vapor is lighter or less dense than air. The molecular weight of air is 28.96 g/mol, while the molecular weight of water vapor is 18.02 g/mol. Adding water vapor to air makes the air less dense. Temperature is a very big factor, though.
Hi Dave, really enjoying your channel and content pal. I watched a video yesterday of a parachutist doing a HALO jump from 41000ft and you could clearly see the earths curvature as he exited the plane. Would love to see you explain the camera work on a similar video (anything that pains the flerfs 😂)
Glober here! If you use 6.7 in.mi2, it take into account standard refraction. And if you reverse it with observer altitude to get distance to horizon that you subtract to observer to target distance, then you use 6.7 in.mi2 to get hidden height. It is quite good under 100 mi observations! Still it is a rough estimate. Also notice that Rawbotham do not indicate the drop perpendicular to the tangente but angled toward earth center. Anyway the difference is negligible for short distance!
@@duncanmcneill7088 Refraction correction can be estimated at 1/7th of curvature correction. In the surveying text I have from my freshman year the curvature and refraction correction can be calculated together at 6.75cm per km^2.
@@duncanmcneill7088 8in.mi2 is an empiric formula, so unit goes to trash! Normally result should be a volume! It is a point I throw to flerf face as well.
@@NicoLeDahut More specifically it is an engineering equation, so all the unit conversions were bundled into the constant and are implied by the equation. Half the EIT exam is knowing what unit conversions are bundled into the equation sheet and what unit conversions are not.
@@Lassisvulgaris My rational is at 12, usually 7th grade, is when kids start to have self-awareness and an overall understanding of the world around them. But yeah you're probably right.
I know this is the point of the video and all but the formula works like this: 8"/mi² means for every mile away you ADD 8" of drop. That is to say, 1 mile = 8", 2mi = 16", 3mi = 24" and so on. The reason, from your eye level, you cant see 8 inches of drop from 1mi away is because: 1) it does drop but its hard to tell because youre tall so it isnt "hidden" 2) you start at x inches tall To find where the horizon should be you take your height, in inches, and divide it by 8 and youll get the miles to the horizon If youre 6ft = 72" = 9mi. That means, from your eye level (this is assume your eyes are on top of your head which is ridiculous but simple) an object will not be occluded by the horizon by 8" UNTIL you get NO LESS THAN 10 MILES AWAY. So a city, 60 miles away, is effectively 51 miles away. 51 × 8 = 408" / 12" = 34ft. The buildings 60 miles away from you should only APPEAR to have the lowest 34ft of their hieght occluded by the horizon. That would mean that you could still see the upper 66ft of a 100ft building. So the problem is flerfs are dorks that dont know what numbers mean and how to use them.
It’s nice you can actually admit you were wrong/not completely accurate in the past, while flerfs stick to their statements no matter how wrong they are, even if they probably secretly deep down inside know they are wrong, but just too scared/ashamed, stubborn or maybe even stupid to admit it. To any flerf out there reading this; it’s okay to admit you were wrong, we won’t think less of you, it’s not the end of the world.
This atmospheric refraction effect has to be calculated in optical design of high-magnification telescope instrumentation. I found this out a couple years ago designing a camera for installation at the Keck Observatory. When the telescope is pointed 60° off zenith (30° above the horizon), light between 550 nm and 950 nm will refract differently with a dispersion of one arc second. Doesn’t sound like a big deal unless the pixels on your camera are scaled at 6.7 mili-arc seconds (an effective focal length of 350 m)! Corrective zero deviation prisms pairs (with matching indexes, but differing dispersive powers) need to be included in the instrument design to correct for this if observing more than a single narrow waveband. Otherwise you would get each star turned into a spectrum ~ a millimeter long.
Land surveyors used this formula (or rather one that actually *did* account for refraction) to calculate the drop of the horizontal in times when personal computers, laptops, pocket calculators and embedded computers weren't a thing. Nowadays the equipment has the calculating power to use the actual curvature for these calculations (I don't know if the old formula is still taught for when the equipment does not cooperate ;-)). It is really easy to understand, so it takes a lot of willpower by the flerfs to misunderstand and misrepresent it. Or maybe they are really that stupid that they are completely unable to understand it. Maybe both.
The formula y = a×x^2 is a parabola, but that's not really where the squared term comes from. Using trigonometry you get y = a×( 1 - cos(r) ) where r is the angle in radians round the earth's circumference. I.e A very tiny angle in practice. The power series for cos(r) is 1 - r^2÷2 + r^4÷24 - ... . With r so small you can get the reasonable approximation y = a×(1 - 1 + r^2. ) which is y = a×r^2. Then r radians has to be translated into the (small) distance from the observer giving 8 inches per mile squared.
I explain the below concisely here, "How to Use 8 Inches Per Mile Squared to Estimate the Hump Correctly" The issue with 8 inches per mile squared is that it is the drop etc etc, yet flerthers think it's the sagitta - the "hump" of Earth between them and some distant point. Check out how it was derived, and you quickly see that it IS a sagitta, but not the one they think it is. It's the one they're standing on top of, since the drop and the sagitta are the same thing for that segment, and what they think is the arc length is only half the arc length. The other half of the arc is behind them. To estimate the sagitta they're on top of, you double the distance between them and the point they're observing when calculating the sagitta geometrically. That is the real arc length for that sagitta, and it corresponds relatively reasonably to the 8 inches per mile squared drop calculated estimate that they gave you, but for clarity it is the sagitta they're on top of, not the sagitta that describes the "hump" between them and some point they're observing. To work out the sagitta of the "hump" between them and some distant point, they have to move half away along that arc to be on top of that sagitta, aka "the hump." Thus, you don't use 8 inches per mile square at all. You use 8 inches per miles halved squared. 8*(0.5d)^2 The formula for a sagitta is s = r +/- the square root of r^2 - l^2, where l is r * sin(a/2r) using radians R = radius L = half the chord length A = the arc length You can ignore the + in the +/- because we're only interested in the short sagitta, not the long one. You can also ignore calculating half the chord length for observations less than 250 miles, since the difference between the arc length and the chord length is trivial, so half the arc length will be approximately the same as half the chord length. For l, just substitute 0.5a instead. It'll have a 0.001 mile variance at 3 d.p. If you do go ahead and calculate half the chord length, the variance doubles to 0.002 mi. s = r - sqrt (r^2 - l^2) =3963 - sqrt (250^2 - 125^2) = 1.972 mi s = r - sqrt (r^2 - l^2) = 3963 -sqrt (3963^2 - (3963 * sin (250/(2*3963)))^2) = 1.971 mi 8 in per halved miles squared = (8 * (0.5 * 250)^2)/(5280*12) = 1.973 mi I compared the results out to 5000 miles, and at that distance using the correction above, there was about a 27 mile variance between the methods. At typical observation ranges that the flerfers bring up, it's perfectly fine - within a few feet. Examples: 1 mile, 0.167 ft sagitta, 0.167 ft "hump," 0.67 ft "drop" 10 miles, 16.65 ft sag, 16.67 ft "hump," 66.67 ft "drop" 50 miles, 416.35 ft sag, 416.67 ft "hump," 1667.67 ft "drop" 1000 miles, 31.5 mi sag, 31.57 mi "hump," 126.26 mi "drop" 5000 miles, 762.74 mi sag, 789.14 mi "hump," 3156.57 mi "drop" As you see, using 8 inches per miles halved squared for the "hump" tracks very well with a sagitta calculation for a 1000 miles, after which the divergence becomes noticeable. For the Chicago example, the distance given is 52 miles which you round up to 60 miles. Half of 60 is 30, so 8*30^2 gives us that hump between sea level where the observer is and sea level where Chicago is. = 7200 inches = 600 feet
Glad to see you point out the biggest error people make when trying to use 8 inches per miles squared: they get the angle of a person's view wrong or don't account for it altogether.
Flerfs will utter their arguments even if you will take them to space, they are hopeless. They're deaf, they're illogical, they're lying and they're spreading.
I've always regarded the 8"/mi^2 as the simpleton's baseline RATE calculator, from a viewpoint at the surface. The main problem is not that observer height is unaccounted for, but that the parabolic curve can crudely fit a circle only over a limited range.
Not so crudely and not over such a short range. The difference between the “8 inches” and the correct “cosine formula” is less than 0.5% over 1000 miles.
@@betaorionis2164 ...and most _errr_ "discussions", for lack of a better word, involving flerfs are talking about distances of less than a few hundred miles.
I was trying to explain this to a viewer of I Can Science That a few weeks ago about a lake in his vicinity. I hope he's watching this video and listening well because you explain it much better than I did.
IIRC the 8" equation comes from surveying. Because surveying is all about tangent lines at ground level and not about observer height, it was perfect for the job: a correction factor that 19th century surveyors could do in their heads. They knew it was an approximation, but it had less error than their instruments, so it was acceptable for short distances. I wish Dave had confirmed or debunked this. Maybe tack on one minute at the end . . . ?
Another excellent demonstration of refraction is a notorious pair of photos of two oil rigs where one has a large amount of refraction, making straight objects appear bent, and the other has a more standard amount of refraction, causing the water to occlude the bottom of one of the oil rigs. The less refracted photo can be referred to as a "black swan" because even if you assume the horizon will never bulge and occlude your view to an object, just one demonstration of that happening is enough to invalidate that assumption.
I can’t believe how quickly your channel has grown!! Well, actually I can believe it; your explanations are well laid out and easy to understand. I’ve been here since your first flat earth video. I’ll be cheering you on to that 100K subscriber achievement. Well done.
Have flat Earthers ever explained why you can see clouds lit from underneath at sunrise and sunset and why it always has a more red component? I've never seen a model made by them that would account for it.
Especially mammatus clouds after a severe thunderstorm. They're wavy lumpy texture produce shadows that make it so obvious they're being lit from the underside. I really curious about bringing this up.
Dave, I figured out how to use that formula correctly. It’s not 100%, but it’s not far off. It’s within 7% out 2000 miles. At least when I use the Pythagorean Theorem and assume everything is taking place on a perfect sphere with a 4000 mile radius and the horizon is always at sea level.
Liquid water is denser than air, but the molecular weight of H20 as vapour is lighter than the nitrogen-oxygen mix of the air. So while it may seem counterintuitive, one should expect refraction is reduced when humidity is high.
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Good debunk! Those flatards will not understand.
Did you do the DNA as well on MH?
@@UncleKennysPlace I've sent it off but awaiting the results
Although my partner used their DNA facility a year or so ago and found some relatives in America
Win the top prize on a major lottery, you'll find all the relatives you never knew or had!
@@DaveMcKeeganMistake at 9:08 Water vapor is actually lighter than air, that is why it rises up. Water has two Hydrogen atoms and one Oxygen. Oxygen has two oxygens and Nitrogen two nitrogens both heavier than water vapor
My favorite part of these videos is watching Dave's doggo grab his hand every time he pulls his hand away! XD
Same
I've always thought that Rusty is the real star of the videos.
Fr he's so adorable ❤
It's like he's saying "I didn't tell you to stop". I've seen a similar video of this little monkey who kept grabbing a woman's hand and putting it back on its head when she'd stop petting him
@@0cgwwhat do you think we are here for? 😆🐶
It's like 1 meter = 3 feet rule. It works well enough for rough calculation of small measurements, but not for longer measurements and it's meant to simply be an easy and quick way to convert meters to feet to understand size or distance.
This has nothing to do with your point but ... who ever does 1m=3ft? Its barely more difficult to do 1m=3.3ft. Just add another 10% if you're going from meters to feet, or subtract 10% if going feet to meters. You'll be accurate to within about half a percent. Rounding will probably cause a bigger error.
@@dgthe3 I can't speak for everybody, but I've found the "3ft = 1m" is used more for a simple comparison rather than for measuring. "What's 100 meters?" "About the length of a (US) football field." While 100 meters is quite a few feet longer, you can convey the idea to somebody easily.
(edit to fix the glaring error as outlined in replies below)
@@dgthe3 To be more specific it's "a meter = a yard" but a yard is 3 feet, so same thing and we better understand small measurements in feet.
It's meant to just get a rough idea, for example if something is 50m long I can quickly go "over 150ft" or using yards it's "over half a football field", both of which are far more useful to me than 50m. Now when we are dealing with 10s (10, 100, 1000 etc) then I would use 3.3, or heck even 3.28, since it's just moving the decimal and adding zeros.
It's for quick, doesn't matter much, in your head math.
@@dgthe3 Im from a metric country and have always used metrics... but I often use the 3ft to a meter when doing something that doesn't require accuracy... sure over 10meters you've suddenly lost a meter but hey, when measuring a few meters for something in the garden or telling someone something in terms of 'it was this big', 20-30cms isnt gonna matter much (3x a foot thats 30cms, the size of a school ruler, is easier to visualize and approximate with your hands than a meter i find).
@@Well...Darn. I have NO IDEA how long a football field is. Nor an Olympic swimming pool. Why can't commentators just use comprehensible units like feet, yards or metres?
As a rough Imperial to Metric equivalent, 3 metres = 10 feet is probably better and quite close to accurate.
I love how flat earthers think a floating ball in space is craaaaaazy, but a flat disc surrounded by an ice wall, guarded by the evil government and inside a giant upside down salad bowl sounds completely reasonable.
Not just "the government" but *every* government, military, space agency, airline, shipping company, aircraft manufacturer, surveyor, weather agency, and on and on - including their ex-employees - for _at least_ the past several centuries.
Conservatives, liberals, socialists, Communists, devoutly religious people and atheists, scrupulously honest people and con-men, quiet truth-seekers and blatant attention-seekers...
Somehow, every person who holds any position of responsibility is captured by this conspiracy.
Makes Scientology look like a clique of preschoolers.
@@larrywest42 for *at least* the whole history of mankind.
Some of flerfs are going way before 5000 BC.
As usual, the "ballers" have it all wrong AGAIN. The flat disc surrounded by an ice wall, inside a giant upside-down salad bowl isn't guarded by THE government. It's guarded by ALL the governments on the planet. And has been for all of time. (And, no, don't ask me how the disc got here.) This was the case even when there was no means of communication between the governments (which was 99.9% of that time). Wait, when you say it like that, it really does sound crazy, doesn't it?
Just the fact that they think humans can keep secrets is ridiculous enough.
@@ReValveiT_01 It would be hilarious to see the CEO of NASA stuttering and swearing when asked for proof if the earth is round(I'm not a flat earther)
Excellent video as always. No insults, no yelling, just cold hard facts, science, and logical reasoning. You're at this point the best debunker I've seen, keep up the good work!
This is exactly why I watch this channel. No bombastic insults, no degrading, no hyperbole. Just facts, clearly expressed, in a gentle tone, while giving a very good dog some love. He isn't insulting flat earthers, just explaining how their conclusions are inconsistent.
I much prefer Professor Dave awesome debunkings exactly because he is not afraid to tell Flattards what they are: retards who slander scientists and engineers and everyone else.
M Keegan is nice but not my favorite. He respects too much the flattards
@@mikepictor Very true. However, I will say that the majority of flat Earthers aren't likely to be swayed by facts given their absurd adherence to pseudoscience and faulty argumentation, so the lack of insults probably won't do much to help them realise their errors. Great video though as always.
@@ZoogleasI agree that they won't easily be convinced that the earth isn't flat, but if they were insulted for that it would be even more unlikely for them to realise that.
I would have to agree. As much as I enjoy scimandan I think dave edges him for me, for the reason stated by the OP.
Honestly, the fact that Dave video feels more like a simple fact explanation than a rage debunking makes it 100 times more enjoyable for everyone.
Keep going !
(And flat earther, for God sake, please go sailing.)
Cannot agree more.
I agree every time I watch a debunking vid and have to listen to flat earther, I get annoyed.
Absolutely... if the point is to help borderline flateathers and/or help flat-earthers see better... Dave's presentation is so much less arrogant, angry debunking, which nearly always produces the backfire effect. As one who looked deep into the anecdotal evidence of a flat earth (finding it compelling), the clean, calm, and clear presentations of real experiment and objective observations, is so much better at keeping a mind on the facts rather than the emotion. Which is so unlike Prf Dave who's snark and arrogance is a real turn off for people needing facts, experiment, and objectivism. Just be cordial, clear, and respectful and you will be able to deliver truth to those who need it.
He would actually make a great professor. I'd call him "Dave the legit professor"
Edit: if he became a professor, then I would call him "Dave the legit professor". I didn't think it would be necessary to clarify that 🤷♂️😄
@@steevo101The only credit I’ll give Professor Dave is that a lot of his annoyance and frustration is directed at those who have directly attacked him or frankly are scummy enough to deserve it. If you look at some of his earliest Flat Earth content that are simply just informational debunks, they are a lot more tame.
This is similar to the difference between the line of sight of a rifle scope, and the line of a rifle bore. The ballistic arc of a fired projectile appears to “rise up” through the line of sight, at the first point of aim, and then drop back to intersect the line of sight again at the second point of aim. In reality the projectile is dropping the second it leaves the barrel. The apparent “rise” is a product of the angular relationship, and the offset, between the scope and the bore.
Hehehe... there is a video game series "Sniper Elite" that I've enjoyed. One of the neat physics features is that if I can figure out exactly where the 'bad guy' is and he's just below an intervening obstacle at some distance (say, 200m), I can shoot OVER the obstacle and hit them. The game correctly calculates the difference between scope sight and rifle bore, and the shallow arc the bullet travels over the obstacle, dropping down to hit the 'hidden' target. :)
Oh so that's what happens with the arrows in Skyrim!
@@mikefochtman7164 I know the series, and played it on the PlayStation 2. It’s one of the only games that models the external ballistics and sight picture relationship well. I particularly enjoyed it, because I grew up in the middle of nowhere, and shot regularly until my mid 20s. I used to hunt, periodically, mostly rabbits because my grandmother made an awesome rabbit stew, but that was using a point blank aim setup - where the trajectory is fairly flat, and the arc between the 1st and 2nd point of aim is contained within the kill zone. Which is only about an inch for rabbits. But what I really enjoyed was long range target shooting for fun. I used to nerd over ballistics calculators, making shooting tables (of minute of angle corrections), range finding and scope set up. I live in a much more built up area, these days. I kinda miss not having anywhere suitable to shoot.
@@lucasdrudi7231 Yeah, the basic concept of the relationship between line of sight, and a ballistic trajectory, holds true with any form of shooting. Ironically, I also did archery regularly from age 11-16, bare-bow with a recurve, and a light 45lb compound bow with sights. But yeah, the principle is basically the same.
This doesn't make complete sense though. If you're firing at a significant distance (200 yards or more) then you have to tilt the gun upwards slightly. This is extremely obvious using iron sights. As such the bullet isn't always dropping the moment it leaves the barrel, there is a period where it will be moving upwards slightly before it starts dropping. The only time this wouldn't happen is if the gun is completely level with the ground, or obviously pointing downwards.
Noticed one mistake. Humid air is less dense, not more dense. Higher the humidity, lower the density. So the least dense air would be hot and humid. Cheers
Thanks Mike 👍
Correct. It's literally how clouds and rain work. If moist air was more dense, rain clouds would be at ground level.
@@notgonnahappen7899Except that you do sometimes get clouds at ground level - that's what fog is, a cloud which has fallen down to the ground. This is related to the classic question of "how do clouds stay up in the air", or "why don't clouds fall down to the ground", or similar: Clouds ARE constantly falling - it's usually thermal air currents which keep them up in the air, fog happens when those thermals dissipate and the cloud falls back to earth before it can evaporate. Which is why fog often happens in the early morning, after the daytime thermals have dissipated overnight, and the fog clears later in the day when enough energy is supplied by the sun to evaporate it, or the thermal air currents regenerate and lift the water vapour back up into the air.
Density differences are of course a factor: Thermals take warm air (which is less dense) upwards and the moisture goes with them as humidity, since warmer air can hold more water vapour. Clouds form when that rising air cools down and can no longer hold the moisture as humidity, so water droplets condense out, forming a visible cloud. Which can often continue rising or at least not fall, depending on the strength of the thermal supporting it.
Wow, I never thought about that and at first it confused me, but it absolutely makes sense:
If one assumes, the air pressure is approximately equal inside and around the cloud, by ideal gas law (pV = nRT) the particle density is equal inside and outside.
Since H2O is lighter than O2 or N2, the density of humid air is lower.
Thanks for the the hint.
Saturated moist air density at 15°C is 0.9936 times the density of dry air. Moist air is maximal 0.64% less dense than dry air at 15°C.
But Note: not the absolute density creates atmospheric refraction, but the vertical density _gradient._ So if humidity is constant, it does not contribute to the density gradient and hence to refraction at all, only to the absolute density. The influence on the density gradient is in practice negligible.
So refraction is practically the same for try and moist air, except humidity has a very strong gradient, which is limited because humidity can only vary between 0% and 100%. The stronger the humidity gradient, the smaller the layer it influences.
After Dave explained why simple Earth's curvature calculators are not so accurate with what we observe I was afraid that he won't mention Wolter Bislins work. That guy is a legend. Thanks to Dave it wasn't a case. ;)
heh yeah that and his refraction simulator are great.
Thanks Martin for introducing me to Walter Bislins calculator. I have always wanted to see graphically the relation between observer height and horizon distance. Thanks to you, now I got it. Thanks again.
@@gaetanoroccuzzo You should thank Dave. But it was my pleasure to mention Walter Bislin. That guy is a godsend. :)
Bislin's Earth Curve Calculator is great. Shows the difference between hidden and drop. Factors in refraction.
...A great many in depth articles on his sight as well. I especially enjoyed the detailed Rainy Lake Advanced Level Bedford Experiment.
You're expecting flerfers to understand refraction when they can't even understand up and down.
Nathan Oakley 3 minutes after this video goes up be like : "Dave Refraction Mckeegan!"
Nathan “things defract into the distance” Oakley wouldn’t know facts if it hit his child.
8" x Miles squared is a rounded off version by Samuel Rowbotham based on an approximation given by the 1860's Encyclopedia Britannica in their article on Levelling. Samuel included it in his first small edition but left out the article in the larger later edition. The article included the entire mathematical derivation and is quite interesting. It also included a very explicit warning against the effects of refraction that Samuel completely ignored.
This matches well with the [Wikipedia: Bedford level experiment] article on Rowbotham's canal experiment, but the encyclopedia sure gave me a merry chase!
Levelling: See surveying.
Surveying: See trigonometry.
I believe the formula can be arrived at using a Taylor series for 1-cos(x), which produces roughly x²/2 (constant terms cancelled, every other term becomes 0, and further terms scale with an inverse factorial, so it converges - at least for x
I thank both of you for the history lesson and the math.
@@0LoneTech Why not check out the first edition of Zetetic Astronomy by Parallax. The preview available at Google Books has the quote starting on the 4th page of text after the contents list. The maths used in the quote is super basic. No Taylor series.
I have derived the formula here
byggvir.de/2023/08/01/curvature-of-the-earth-eight-inches-per-mile-squared/
The approximation is good enough for eye levels below 10,000 m. The error is less than 2 % and less than 0.04% if you use 7.8481 cm / km squared.
The refraction raduis in air is between 40.000 to 50.000 km. That is approximately 1 cm per km squarred. That adds an error < 15 %. In most flat earth evidence, this error can be neglected because if something is hidden, regardless of how much, the earth can't be flat. Flat earth mostly porve that something can be seen, but not how much is seen and how much is hidden.
While yes it was used by Samuel, the roots of this old rough approximation is older than Samuel and had been around before him. Samuel did not invent it, he was just merely smart enough to likely understand that represents a parabola but used it as the con man he was. It is used by surveyors to do quick calculations as it works fairly well for short distance. Especially when talking about single digit miles like surveyors typically deal with in line of sight. And was used back in the days for approximations without doing complex maths.
Any real calculation of Earth curvature _must_ include the radius as part of the calculation. There is no radius accounted for in 8" per mile squared.
This flat earther in the beginning says "with 8 inches per mile squared, the curve would most certainly be visible, but it's not." Then when you show flat earthers the Lake Pontchartrain pylons photo taken by Soundly, they say "I'm seeing too much curve." How do we simultaneously see not enough curve *and* too much curve?
Well in this case too much is the fact you can see it at all. They live trying to make everything black and white and use things like the approximations intended to be close enough before the days of computers when everything had to be done by hand.
Simple: stupidity.
@@Isolder74 They also love to parrot that it isn't the "geometric horizon" so that makes it invalid. They like to ignore the fact they can't calculate to the "geometric horizon" on a flat earth because they don't even have a model of a flat earth.
@@Tsudico They never both to say why that matters just playing word games so they can just ignore anything they don’t wish to address.
The Earth must be flat because everyone on it can see Cori Celesti - er, I mean the Himalayas - from their back garden. Oh, no, hang on a minute...
You Ankmorpork lot are always getting confused.
To add another point about atmospheric refraction, it can result in hot air and cold air mirages. Just as the plane appears reflected on a tarmac, so too can the sky appear reflected in the desert, giving the illusion that there is a source of water ahead.
Cold weather mirages had a significant impact upon the tragedy of the Titanic as well. It was a moonless night surrounded by Icebergs. The refraction made the horizon appear higher than it was, hiding the iceberg until immediately before it struck the Titanic, it also made the Titanic appear like a completely different ship to the surrounding vessels, and when the Titanic's crew tried to used the Titanic's lights to send a distress signal, the cold air caused to to flicker like starlight, scrambling the messages.
You only need to ask a flerfer "why is the mile squared?" to see that they have no idea what that formula means.
I have a hard time understanding the laws of physics, you see, I never studied law.🐰 Thanks for the clear and concise explanations Dave, I really enjoy them.
😂😂
I hate the expression "laws of physics." The laws of physics are things like "don't leave radioactive materials laying around on benches", "turn off the lights if you're the last one out of the lab" and "grad students should have the same number of limbs and digits at graduation that they had when they started."
The laws are the laws of nature as discovered and described by the science of physics.
@@d614gakadoug9 I think it's called the rules of physics Class not the Laws of Physics because Rules of Physics Class make a lot more sense than seemingly breaking the Law if you don't follow them(Which luckily is impossible)
@@d614gakadoug9 Lmao, I hope this was sarcastic
Dave, you're a truly skilled science communicator. I'm glad you've chosen to do what you're doing.
me too
U mean pseudoscience
@@FlatEarth-q1f He's also very good at communicating pseudoscience, as in explaining the flat earth position, and then using real science to explain why it's all bunk. He's generally just good at communicating both sides, steel-manning his interlocutors and then clearly and thoroughly debunking the nonsense flat earthers espouse.
@@FlatEarth-q1fhe is quite good at clearly explaining the arguments of flat earthers and thus makes it easy to understand why they make no logical sense
@@FlatEarth-q1fdo a video trying to debunk anything he says. But you won't, because you cant. At all.
Most flat Earth UA-camrs have dug themselves in so deep with their beliefs that they have no way of getting out without destroying their online reputation. It's because of this that I firmly believe that even if a flat Earther was gifted a seat on Blue Origin, they would never report about it or they would come up with some extravagant conspiracy that they were drugged and plugged into a computer simulation or something crazy like that.
It would be fun to watch them "debunk" the entire video of them walking onto the ship and looking out the windows though
@@lyndafjellman3315 Agreed.
@@lyndafjellman3315They must have hacked my optical nerves! Those damn NASA lizard people reeeeeeeeeee!
That's why I respect Ranty. He admitted he was wrong, and Owned it. That took guts and maturity.
Technically we haven't it gives more fuel that yall have to guess with pseudoscience and theoretical physics
I just love how the water level absolutely destroys the horizon is always level argument. I get that it can be hard for someone to imagine the scale of things, we spend all our lives close to the ground, but I don't get how they can ignore the tons of evidence.
Cognitive dissonance and lack of education is my best guess. There's a very conspiratorial tinge to most flat Earthers though I do find. Has there ever been one flat Earther who didn't think the government/NASA was lying to them about the shape of the Earth? I've yet to find one! Not that NASA has even been the authority on what shape the Earth is, so even if they lied about all their missions, that still wouldn't debunk the shape of the Earth lol.
I think you are mistaken here. The water level demonstration does NOT destroy the horizon is always level argument. It destroys the argument that the horizon always rises to eye level!
It needs showing to every flat earther that says 'water finds its own level'.
@@jdmjesus6103 Well yeah, it does find it's own level around the globe - sea level.
@@teebosaurusyouwhich also happens to vary across the globe
"Flat Earthers have a misconception of how _____write something -anything____ works."
I only ran across the modern flat Earth stuff a few months ago, but I've followed blogs of a number of science communicators, mostly in matters related to biology such as evolution and medicine, for many years.
When I "discovered" flat Earth I quickly came to realize that flerfs are very much like others who deny science and the knowledge we get therefrom. They are often remarkably arrogant and they are almost all quite profoundly ignorant of the basics of the topics whereof they speak. They'll assert that they are "critical thinkers." Fine, but you can't critically think your way through something about which you know almost nothing. Actually, and real critical thinker with limited knowledge would conclude that when they are arguing against a large body of evidence from a multitude of highly qualified scientists, it is the odd man out who is likely to be wrong. Of course sometimes the odd man out turns out eventually to be correct, but it is pretty darned rare these days in science.
LMAOOOO I laughed loud at this
They don't know how work works. Which is why they live off their mums and spend all day on flat earth discords.
Spheres. Flat Earthers don't know how spheres work. They keep talking about the top and bottom.
@@Thirdbase9 So many think north is up and south is down.
I bet flat earthers really hate your videos. You're calm and precise. They don't know how to handle it. Well done again Dave.
Look at the comments sorted by new.
I don't i love debunking yall theoretical physics and pseudoscience along with millions of others
@@FlatEarth-q1f theoretical? Oh you mean the proven and reputable sources and evidence rather than the guy in his moms basement who thinks he’s proven the worlds smartest people wrong. Keep trying, keep failing and I’ll keep laughing
@@FlatEarth-q1f I see that you have a new script that you don't understand. "Theoretical Physics! *SQUAWK!* Theoretical physics!" You don't know any theoretical physics. You have never even touched a physics book. 🤦♀
@@The_Beer_Hunter u mean yall playing the guessing games
Reasons for watching video:
10% knowledge
20% laughing at flat earthers.
70% doggy
3:06 It amuses me how flat earthers look at that picture of Chicago (for example) and suggest that it's a proof of flat earth and yet ignore the ELEPHANT in the room that is the missing bottom half of the buildings. No matter how or what formula is used there is no way that image should be like that on a flat earth.
Most hilarious of all is that this is even still a debate.
well, our globe has been mapped every millimeter...
Yet flatards can't even agree of a map of their fairytale Frisbee, as none of the distances matches reality within hundreds of miles 😂
It’s not really a debate though, is it. It’s sensible people trying to edu take drooling morons.
The flat earthers aren’t listening as they only listen to each other
That city being visible from across the shore is funny because it only happens under specific conditions. If it was proof that the Earth is flat then it would be visible every day of the year, but it isn't.
In fact it happens so infrequently that when it occurs it makes the news.
The only reasonable explanation is that THEY dug a big hole to hide Chicago, but every now and then, freedom fighters manage to release enough hot air to make the buildings float up and give the game away.
Nothing else fits /s
Also, on a flat earth you would see the streets also, not just the highest buildings. With pictures like that they proof the world is a globe.
@@thearmouredpenguin7148Correct. Didn't stop Rob Skiba (RIP) from using this as his argument that the Earth is flat. Funny thing is, and this is true of a lot of flat Earthers that I've seen, he seems overall like a fairly intelligent guy but his religious leanings have just turned that part of his brain off.
Simplest explanation is that the Earth is flat, but only when nobody is checking.
That's how it works, right? 😆
This always does my effing head in when i see them say the 8 inches thing, the fact that they cant even grasp the idea of the curve in the first place
One important detail about those photos of the Chicago skyline, and why refraction doesn't explain the observed shifts: they're taken from ~100 feet above water level, atop very large sand dunes that are common on the Lake Michigan coasts across from Chicago.
Occasionally you can see looming of the French beaches from the high cliffs of Dover. Most of the time France is visible but only cliff tops. Diffraction makes a big difference. Head over to the Bislin advanced earth calculator and use the refraction slider to change levels.
Great video. This is a topic that has been the bane of my existence. My brother does not believe refraction can cause as much as I would think based on the videos. But with every rebuttal, he points to another TikTok FE video, and ignores anything I say. I also just started researching this topic, and found another video that when through the math to show how this formula works, so it was good to see another person mention it as well.
Be careful arguing with siblings who've fallen into the Flerf cult, my brother got so deep in the rabbit hole he eventually had to be sectioned. He's out now, but since he's started bringing up other conspiracies recently I speak to him less and less (his current favourite is mudflood). On the plus side, he never mentions flat earth now, but when he was sectioned the doctors explained how arguing with him about it would have made him worse and I kind of see their point. Sadly it means we rarely ever speak, as I don't want to contribute to another more lengthy stay in a care facility.
Anyone who gets science from Tik-Tok. . .could probably turn lead into gold with all that "education."
@@smiffsoft
One thing that I have noticed about conspiracy people is they never bring up the real stuff.
There is the story of how Target knew a 15 year old girl was pregnant before her father did.
In the county I grew up in there is a conservation authority that had a lot of power. It was not really part of the government but still the guy in charge could block construction in some places.
And there are other real life examples of people and groups manipulating people but the conspiracy people never talk about them.
I'm not sure if any of the people complaining about shrinkflation are conspiracy believers.
Something I would love to see explained by the Flat Earthers: if the flat earth is the reason why Chicago skyline is visible in these images... why isn't it visible all the time? Or at least, most of the time, without heavy cloud cover or fog obstructing the view?
I have asked Flerfs about this a number of times... but they all chose to ignore the question instead of making something up as usual.
Always a fun way to pick apart their arguments. "Hurr durr this thing is visible!" Yes. It's visible in this instance, but if the Earth was flat, then it would be visible *all* the time instead of waiting for specific conditions to make that thing visible. So why do photographers have to wait for specific conditions to get the picture?
Why would a flat earth negate atmospheric refraction? You making that shit up bro?
@@Tabearnack A flat earth wouldn't negate refraction, but the density stratification would be horizontal, not curved, so the visual effects would be very different from what we see in reality.
@@5peciesunkn0wnEver seen the fake horizon above the real horizon?. Maybe go outside and observe a sunset or sunrise over the ocean for a few days and you'll see..
@@stanlee4217 I don't believe I've seen one yet. Sadly nowhere near the ocean. Would it work on the Great Lakes?
It can not be stated enough that your explanations are thorough, clear, and straightforward. A person could not care at all about the “debate” (if it can even be called that) and still learn a lot. It’s also refreshing that a video about this isn’t just heavy handed with condescension insults. Refreshing to watch.
The approximation "8 inches per mile squared" IS accurate out to about 750 miles.
If you want the "HUMP HEIGHT" of the earth in between the two endpoints, you must use HALF the distance between the two endpoints.
If you want "DROP FROM A TANGENT" then you use the full distance.
The Flerfs don't know the difference, so they always calculate "Drop from a Tangent" when they really want "Hump Height".
SO, it's no wonder they can never find the curve they are looking for because they are always looking FOR TOO MUCH!
There is a video on my YT channel that explains the math in detail.
I always thought that they couldn't see the curve because they can't see past their noses.
@@JavaBum Can't see past their ears more like.
As a hump height calculator it is useful for trying to explain to flerfs why Earth's curvature can usually be ignored in short-distance surveying, as in laying out the foundation for even fairly large buildings, or explaining why trying to measure the hump in the water in an Olympic-size swimming pool would be a daunting task.
I've used it a few times as a check to make sure I've used an on-line circular arc/chord calculator correctly. I don't use such a thing much, so I don't have a good "feel" for what to expect and therefore like a check to make sure I haven't erred due to specifying the wrong unit or some other blunder. A lot of people don't seem to take a moment to ask themselves "does this calculated value look reasonable?" and wind up with something like being 9 orders of magnitude off because they confused "milli" with "mega."
@@d614gakadoug9 "confused 'milli' with 'mega'." is the same thing as flerfers dumping water on a basket ball and asking why it doesn't stick to the surface of the ball lol.
They don't want the hump, they want hidden height, but do not regard observer height. But that is easy to do (when you know math):
You reverse the 8 inches formula, then you can determine the distance to the (geometrical) horizon for the observer height.
I.e., Distance = sqrt(8"/height)
In that distance your line of sight is tangential with the earth surface.
Then you subtract the distance to the horizon from the distance to the object and then can use the 8 inches for that rest, getting the geometrical drop..
8" per Mile^2 was a standard approximation used by surveyors and civil engineers, before the availability of computers, since at least the early 19thC. The earliest reference I have found is in "A Treatise on Surveying and Civil Engineering, Wherein Everything That is Useful and Curious is Demonstrated from its First Principles", by P.A O'Shaughnessy (p30). Published 1848 in New York.
The formula gives a slightly too small approximation for short distances. It is most accurate at 400 km. Until 550 km or 342 mi the error stays within ∓0.032%. Until 800 km or 497 mi the error is less than 0.1%.
Surveyors then and now used it for an approximation of the drop from the horizontal line of their theodolite, starting from the surveying point. It works well enough for typical surveying distances but was never supposed to be used for drop beyond the horizon.
Love this kind of stuff Dave. Once upon a time, I was taught and became quite good at working out what the Brit Army called "Intervisiblity" ( or did, when I were a lad!) Instead of spending huge amounts of time tromping up and down mountains to check what could be seen from higher up - we used a trig formula ( which I've completely forgotten!) applied to the contour height and distance from and to - this was very handy if you wanted to observe say a valley C from mountain A but with big hill B in between - the formula would be used to find the best vantage point. Don't suppose anyone is bothered now that "drones" are available.
I agree with all. The one major point that you missed is that no flat earther ever uses 8 inches per mile squared to measure curvature. They always use it to say, "you shouldn't see this building, tower, or mountain range" when once you calculate that the hill you're standing on is 500 feet above sea level and that tower is 200 feet high... you totally should see that tower.
Something of a tangent (perhaps somewhat appropriately), but this reminds me of reading about something from World War II. USS _Atlanta_ was Shanghaied into helping USS _North Carolina_ (I think. She might have been helping _Washington_ instead) with an offset gunnery exercise off Guadalcanal. _North Carolina_ took aim at _Atlanta_ at maximum range, offset her rangefinder by a degree or two, and opened fire. _Atlanta's_ crew could only see the very top of _North Carolina's_ mast, where her big rangefinder is. One small puff of black smoke and about 30 seconds of flight time later and _Atlanta's_ crew were treated to the disconcerting experience of 9 16 inch shells smashing down in her wake. If not for the offset, they'd have hit dead center. _North Carolina_ proceeded to do this over and over and over again, never once properly coming into view or up over the horizon.
It was USS Washington and the crack shot Adm Willis Lee.
Edit: if added to a computer game he’d be called too OP!
The 'offset' may have just been the target's course and speed. I was amazed at the analog fire-control system of WW II battleships. If the target is steaming at 20 knots, it'll have moved over 300 yds by the time the shell gets there so an estimate of target course and speed was one of many inputs. To be as accurate as they were is a testament to the engineering of those systems.
Love how "water finds its own level" disproved "horizon rises to our eye level."
i am tempted to fall back to my simplistic answer: why does someone on a boat, at the top of a mast, spots land before the ones on decK? On a flat earth, everyone on the boat (with the same eyesight, of course) would spot land at the same time. There would be no need to put someone in the crow's nest, as mariners have been doing for thousands of years. Flerfers jus trigger me.
Refraction? wonder why you are getting triggered? Cause MAYBE you've been lied to your whole life.. and you are falling for it...!
Land Ho!
The reason they try to get up s high as they can it to be able to look over the ocean waves. Of course you probably don't know how high waves can get if you've never been in the middle of the ocean.
Not always
Ocean level is not constant
Just to tell you that simple disturbance of the sea can give you a 15 meter difference
A big disturbance like a big storm is more than 20,30 meters.
And i mean not like in the movies
Just a quiet and up and down elevator feeling.
So no even if the earth was flat you cannot see farther than a few miles.
Local fishermen use the view of the opposite city as a reference.
If the city across the shore is visible the weather is good
I mean they can see the shoreline
If they cannot the water is too high the weather is bad.
It has nothing to do with curvarure.
Formulas 8"/mi^2 you want to graph
y = -x^2/7920 + 3950
This is just converting 8" into 1/7920 of a mile. The + 3950 is to shift the parabola up, so the vertex aligns with the north pole which is 3950mi above the center of the earth at the origin. Alternatively you could shift the elipse downward 3950 and have the north pole and parabola vertex at the origin.
To graph the earth in miles you want to graph an elipse with major axis 3963mi and minor axis 3950mi. So you get
(x/3963)^2 + (y/3950)^2 = 1
The 8"/mi^2 is actually quite an accurate formula. You can calculate the discrepancy between the two curves by graphing
Y = (-x^2/7920 + 3950) - 3950sqrt(1 - (x/3963)^2)
If you do this you find the formula is quite accurate in absolute distances up to 550mi. The discrepancy is never more than 0.0323mi = 170.5ft in that range. Which is pretty good considering that occurs in conjunction with an actual drop of 15.94mi or roughly 0.2%.
Considering percentage errors it's pretty accurate up to about 700mi. Over that range the error never exceeds 0.4%. Ranges of accuracy
err < 0.1% : 436mi-560mi
err < 0.2% : 357mi-614mi
err < 0.3% : 255mi-661mi
err < 0.4% : 47mi-706mi
err < 0.5% : 0mi-748mi
err < 1.0% : 0mi-928mi
err < 5% : 0mi-1751mi
err < 10% : 0mi-2319mi
Great video. One thig that struck me is that it is a prime example of the frustration on how to address simplistic arguments. To fully explain the whys and hows, science (and real life in general) requires a detailed explanation and requires the listener being willing to learn.
On the other hand, the simplistic argument latches on to a simple easy to repeat phrase that they can repeat ad nauseum. It may be wrong, but the debunking of it can require a detailed argument that can go over many people's heads.
Thank you Dave for making straight forward videos that explain and debunk the arguments without descending into ridicule.
Flat earthers seems to actually believe that there is a real debate about the shape of the earth.
They don't seem to understand that flat earthers are just entertainment for smarter people.
uh oh you discussed math so now the flerfs won't comment here unless they didn't watch the video and try to recite their scripts
The fact that you can maintain a straight face, while your dog demands pats and scratches, astonishes me. The way the dog puts its paw on your arm just cracks me up.
It's not a real dog, all dogs are fake just like the spherical Earth!!
the fact that he can have a straight face while lying through his teeth and find time to shoot ,produce, draw and edit these videos in minimum time really astonishes me....
@@stanlee4217Awww, poor diddums. You need someone to pet/massage your ego to make you feel special too.
Rusty does seem to be firmly of the opinion that if Dave is going to wave his hands about they should be used to pet Rusty.
@stanlee4217 What lies has he said? Just out of curiosity, I'm not the brightest.
5:19 That is one I learned long ago... If you stand on the beach and watch for the sun rising. The moment you see the sun breaking the horizon, lay down on the beach and you will not see the sun. So there's the curve, anyways this is just another way grifters/conmen marketing for those Patreon Dollar!
Honestly though Dave, as much sense as your video makes and even people with remedial education can actually grasp these concepts, remember….and I have stated this countless times, the flat earth individual is suffering from an illness. I suspect it is on par with an addiction illness or a chronic lack of the ability to realize and understand that you were wrong.
No different than a thief being shown a video of them committing the crime and insisting that it is not them. The hypothetical person who caught this individual has them dead to rights….and it could be a crystal clear picture of their face and an accurate height of the person can be established along with witness testimony, yet they will until their dying day claim that it is not them in the video. Dave, it’s like trying to change someone’s religion or political ideology. It is practically impossible. So maybe one day a pill will be developed that will help these poor souls but until then…I truly give you credit for dealing with the childish behavior and viewpoints that are culminated in the realms of the flat earthers mind.
Dave your segways to sponsor are legendary mate😜
So 8" per mile^2 only works if you lie flat on a beach after the atmosphere has been stripped away ... and you can't use metric ...
basically, yeah
And also this formulas for the curvature are meant to measure perfect spheres, which the earth is not, even it can have large amounts of land actually flat but in the full scale it’s basically a sphere
Also tides
5:05 Get back to scratches, human
My most recent dog would do that stuff. He would even put his head to my hand and move his head so he got his scratches.
Reilly was a good boy and I miss him.
Always calm, collected and pleasantly explains things without resorting to insults, name calling or getting angry/triggered. Flerfers take note.
Thanks for shining a light on this subject. I had never understood how viewing height above the ground changed how the drop should be calculated.
The "it's a parabola, not a circle" thing to me is less about accuracy and more about the fact that so many flerfers constantly claim that "8 inches per mile squared" is *the* formula that "all scientists" say is how you must calculate this stuff, when *it is just obviously not* because _it doesn't even represent the right shape_ and any _real_ scientist could immediately see that to be the case. Yes, it can be used as an approximation over short distances, but _that's never what the flerfers are actually claiming_ it's for. And the fact that they can't even understand that that formula is not _actually_ the formula for a spherical earth, because it mathematically cannot be, and it also _is not the formula that any scientist or earth curve calculator app actually uses,_ or claims to be accurate, just shows how little they understand about the reality of how any of this actually works.
Agreed. That's why it's important to argue that point, because that's Cult 101: Lie about the premise, then attack that premise. "This is their formula and it's obviously wrong, look!" Well, no, it's *not* our formula, but they'll keep repeating that lie because they desperately require their converts to be misinformed. Flat earth doesn't spread without misinformation and misunderstanding.
One of the key things I learned early on in engineering school is how useful approximations can be, and also to respect their limitations.
There's plenty of useful approximations that engineers use everyday, the thin walled stress approximation for cylinders, the coherence of electromagnetic waves after 10 wavelengths, etc. these aren't reality, but they are damn close for almost every use.
But we scientists and engineers must still acknowledge and justify our use of approximations. 8 in/mi/mi is a great approximation, if and only if you respect its limits. The fact that flerfs don't acknowledge that it's an approximation, don't acknowledge the limits and justify the use of this approximation shows their deep ignorance of science, math, engineering, etc.
@@phillyphakename1255 The 8" per mile² comes from surveying text books in the USA. It is referred to as Cc, curvature correction. Actually it is not 8" per mile it is 8*miles² and the result is in inches. The eight is a constant and it could equally be 0.66666*miles² and the answer is in feet. The metric equivalent used in the rest of the world is - Curvature correction: Cc = -0.07849 * D² (D is in kilometres) answer is in metres. It is used by surveyors along with a refraction correction Rc to get readings from a surveyor's level over the relatively small distances a surveyor will normally be working at. Goes like this, an assistant holds a surveying staff on the ground 1000m away and the surveyor looks through their level and reads off a height from the staff. The surveyor takes that reading, subtracts the height of their level from the ground it is on, subtracts Cc and adds Rc to come up with the figure for the height difference of the surveyed point with the observer's position. If you want to know why it is a parabolic value which does not match the surface of a sphere consider this. A circle on a cartesian graph centred at x0, y0 has the formula r²=x²+y² (Thanks go to Pythagoras). For the earth r is the earth's radius, a pretty big number in miles or kilometres. At the top of the circle at x=0 y=r so y²=r². If you move along the x axis the amount likely to be used in surveying x is going to be a small distance compared to r or y so x² is small enough to be ignored and you just look at the changes to y due to the small angular change. This is why until x becomes significant the shape of a parabola is very close to a section of a circle, another approximation that can be useful but has limitations. Another approximation is that over these small distances the surveying staff can be considered parallel to a plumb line at the surveyor's level.
@phillyphakename1255
And is usefull for engineering jokes. 😊
Google joke close enough for all practical purposes.
Thank you!
Every time I hear that formula, my brain, which is terrible at math and math concepts, still, my brain said: wait, but, that's not enough?
To describe our experience with seeing the globe, and sure enough!
The height!
The formula as described is like a floppy string with no anchor point, no reference, until you anchor the viewer's height: thank you. :)
Your segues into your advertisements never ceases to amuse.
The most surprising lesson I learned in this video was that Rusty actually does have back legs.
Also, if you put a coin on the ground in front of you. What is the difference between that coin and eye level? Probably about 5.5 feet give or take. If you move that coin 50 feet away from you. What is the difference between your eye level, and the level at which the coin sits in the ground? Still 5.5 feet and it will remain at that level untill it is moved far enough away that it drops behind the horizon. So the horizon does not rise to eye level
Think about this...even if the Earth were flat, the horizon would NEVER rise to the level of your eyes! The difference in height between your eye level and the coin/floor will ALWAYS be 5.5 feet no matter how far away the coin is moved. Flat Earthers are just morons.
@@mangojulie123 Exactly my point 😁
@@mangojulie123 Not only that, but flat earthers believe that the earth is disc shaped right. So it's still round in their idea. Which means the horizon would still appear curved. The only way the horizon would appear flat, is if the earth was square
@@righty-o3585 You and I think very much alike. That's why they had to also bring the conspiracy of space travel...we can't go into space to take a picture of their pancake earth 🤣. And even if the Earth were infinitely flat as some flerfs claim, the horizon would still appear curved because the radius of the limit of vision of our eyes would carve out perfectly curve.
@@mangojulie123 Bro, I remember like 8 years ago, when this new era of flat earth just started gaining some attention. Somebody had commented about how because of gravity in space, the default shape for anything planet sized, is a globe, or sphere. And I replied..... DON'T LET EM HEAR YOU SAY THAT, THEY'LL START CLAIMING GRAVITY IS FAKE.... Sure as shit like 2 and a half years later. Wish I could have screen shotted lol
Rememeber that time, when a flat-earther, instead of going out of his way to deny the globe, actually came up with evidence to prove flat-earth ??
Neither do I...
here let me use maths to explain why approximating a circle with a parable works:
circle: y= srqt(r²-x²)
parabola: y=ax²+b
we care about the difference between these functions:
->Δy= circle - parabola= sqrt(r²-x²) - (ax²+b)
we want to see what happens for small x, so xΔy≈ r - r= 0
so in conclusion for small x the function values of circle and parabola are almost identical.
that means for a few dozen km the approximation works absolutely fine.
For the sake of mathematical accuracy, I want to point out that this is only a first order approximation, i.e. you've only shown that the tangent lines are equal. This is not surprising and it's also not something flat-earthers would dispute because a flat earth is also tangent to a circle (cause the tangent is flat). This can be easily seen by using a constant function y = r instead of the equation for a parabola.
That means your equation for Δy would look like:
Δy = circle - constant ≈ sqrt(r²-0) - r = r - r = 0
You haven't taken curvature into account at all, unfortunately. The real reason a parabola is sufficient for approximating the curvature of a circle is because curvature only depends on the second derivative, and a parabola is the first polynomial to have a non-zero second derivative. The curvature of a circle and a parabola at its vertex are well known, and fairly simple to derive if you know how, but I won't go into it here. A circle's curvature is just 1 / r and a parabola's curvature at its vertex is 2a, so the two curvatures are exactly equal at the parabola's vertex when a = 1 / 2r. Here's a desmos graph demonstrating this principle: www.desmos.com/calculator/mdywqupmws
Plugging in the radius of the Earth for r and multiplying by 63360 inches per mile gives 8.0024 inches per mile²
@@APaleDot dude i wanted to keep it simple
@@S1nwar
Nothing complicated about 1/r and 2a.
The problem is that your math doesn't show why a parabola is any better than a straight line.
I don't get it, even if the Earth was flat, the horizon still wouldn't rise to eye level. Why do they use this as an argument?
flat earthers all copy/paste off a script written by fraudsters to trick low-grade morons.
no need for any of it to make sense
One theory suggest about infinite flat earth
Lol they don't even have an unified theory
@@SpectreGR396 An infinite disc's horizon still wouldn't be at eye level if you were standing on it.
@@mrosskne yeah idk its their logic
I think it is due to their visualising Earth as a few miles in circumference and the FE as endless.
At this point I watch these videos primarily to see if Dave will ever defeat his dog in their constant battle of hand-to-paw combat.
I always tell flerfs to graph this shit.
I also tell people who know we live on a globe to graph it too, because while it is a parabola, its actually oretty accurate for an observer with an eye height of 0 out to around 175 miles.
I also always use the part where the fact that they do fit quite closely for a while as evidence that flerfs haven't done the math, because that would be the natural response to "it's a parabola"
They don't even understand a simple concept like a tangent. We cannot expect them to grasp that a parabola can approximate a circle in 2D.
It depends on exactly how you measure the distance (tangential distance or along the curve) but it's within 1% at around 750miles.
You can do the math with the corde instead of the circumference.
…I think we need a video of your puppy sleeping in your chair…that would be very soothing
Another good reason to use this approximation in the 'good old days' of engineering is that it's really easy to calculate on a slide rule.
Another issue most people forget, is the "squared". Miles squared is not a linear unit of distance, but a planer unit. When Flat Earthers' say "8 inches per mile squared", they mean "8 inches per linear mile".
NO that is not what is meant. What is meant IS: "8 inches times the number of miles squared"
What?
stuart has it right; it describes the formula, not a conversion factor. Here's an example translated into Qalculate as a function called paradrop.
> angle(x) := x arcmin/NauticalMile
> EarthRadius = radian * NauticalMile / arcmin
> drop(x) := (1-cos (x arcmin/NauticalMile)) * EarthRadius
> paradrop(x) := 8in*(x/mile)^2
> compare(x) := transpose (vector(drop x, paradrop x, drop x - paradrop x, (paradrop x-dropx)/drop x))
> [compare(1km), compare(10km), compare(100km), compare(1000km), compare(3000km)]
[compare(1 kilometer) compare(10 kilometers) compare(100 kilometers) compare(1000 kilometers) compare(3000 kilometers)] ≈
(78.53 mm) (7.853 m) (785.3 m) (78.37 km) (693.8 km)
(78.46 mm) (7.846 m) (784.6 m) (78.46 km) (706.1 km)
(77.57 μm) (7.756 mm) (759.6 mm) (−83.74 m) (−12.28 km)
−9.878E-4 −9.876E-4 −9.673E-4 0.001069 0.01770
That's a relative error under 2% at 3Mm. I divided by the mile unit to get "number of miles" before squaring, which is dimensionless. Side note, here I'm assuming it's surface distance, not horizontal distance, that is the argument for the function. They're nearly identical for small angles anyway.
The formula has nothing to do with what's hidden, it only describes how far off horizontal will be from level in the distance.
I've heard a couple of flerfs say "eight inches per square mile" and I'm pretty sure it wasn't just a slip of the tongue in either case. Similarly I've heard one refer to the "inverse squares [plural] law."
@@d614gakadoug9I just worked it out from the Taylor series again, and one form is "drop / distance² ≈ 8 inches / mile²". The squarings match because it's derived from (distance/radius of Earth), which is an angle (dimensionless). If you solve it for drop, you get "drop ≈ 8 inches * (distance/mile)²", which I prefer because it doesn't split the square and keeps all operations on one side, but spoken aloud it often gets ambiguous; "square mile" is "mile²" is "mile squared", but where did the squared distance go?
Meanwhile there's an entirely unrelated inverse square law, which has to do with the spread of e.g. light or gravitational pull over a surface with distance. And when that's applied to a reflection, like the light we see from the moon, you get squared squared inverses.
The BIG question about this springs to mind.....
Will Slappy show up still lying about being a pilot or will he be unable to resist showing off the lovely plumage of his Norwegian Blue, i mean Black Swan?
I do love that Dave corrects both sides in this video. Very well made, I actually learned a few things in this video.
Love your videos, Dave. However, as a meteorologist for the past 43 years, you got one point wrong. At 9:00 in the video, you state that higher humidity air due to increased water vapor is more dense. That's incorrect. Clouds rise because water vapor is lighter or less dense than air. The molecular weight of air is 28.96 g/mol, while the molecular weight of water vapor is 18.02 g/mol. Adding water vapor to air makes the air less dense. Temperature is a very big factor, though.
Hi Dave, really enjoying your channel and content pal. I watched a video yesterday of a parachutist doing a HALO jump from 41000ft and you could clearly see the earths curvature as he exited the plane. Would love to see you explain the camera work on a similar video (anything that pains the flerfs 😂)
Dave, great video and very well explained! Keep up the good work!
The very fact that the formula exists proves that the Earth is curving
The mere fact that they use the imperial system already discredits them entirely
Yeah, too many words. Flearths aren't going to be able to pay attention to more than the first couple of minutes.
Seconds*
@@justduro1527lmao
Glober here! If you use 6.7 in.mi2, it take into account standard refraction. And if you reverse it with observer altitude to get distance to horizon that you subtract to observer to target distance, then you use 6.7 in.mi2 to get hidden height. It is quite good under 100 mi observations! Still it is a rough estimate. Also notice that Rawbotham do not indicate the drop perpendicular to the tangente but angled toward earth center. Anyway the difference is negligible for short distance!
But what is that in PROPER units?
e.g. 7.85cm per km^2
@@duncanmcneill7088 Refraction correction can be estimated at 1/7th of curvature correction. In the surveying text I have from my freshman year the curvature and refraction correction can be calculated together at 6.75cm per km^2.
@@duncanmcneill7088 8in.mi2 is an empiric formula, so unit goes to trash! Normally result should be a volume! It is a point I throw to flerf face as well.
No forgetting it multiple in by mi by mi to result in in. Flerf magic!
@@NicoLeDahut More specifically it is an engineering equation, so all the unit conversions were bundled into the constant and are implied by the equation. Half the EIT exam is knowing what unit conversions are bundled into the equation sheet and what unit conversions are not.
When flat earthers only have 6th grade education and try using math...
Do they even reach 6th grade...?
@@Lassisvulgaris My rational is at 12, usually 7th grade, is when kids start to have self-awareness and an overall understanding of the world around them. But yeah you're probably right.
I know this is the point of the video and all but the formula works like this:
8"/mi² means for every mile away you ADD 8" of drop. That is to say, 1 mile = 8", 2mi = 16", 3mi = 24" and so on.
The reason, from your eye level, you cant see 8 inches of drop from 1mi away is because:
1) it does drop but its hard to tell because youre tall so it isnt "hidden"
2) you start at x inches tall
To find where the horizon should be you take your height, in inches, and divide it by 8 and youll get the miles to the horizon
If youre 6ft = 72" = 9mi. That means, from your eye level (this is assume your eyes are on top of your head which is ridiculous but simple) an object will not be occluded by the horizon by 8" UNTIL you get NO LESS THAN 10 MILES AWAY.
So a city, 60 miles away, is effectively 51 miles away. 51 × 8 = 408" / 12" = 34ft.
The buildings 60 miles away from you should only APPEAR to have the lowest 34ft of their hieght occluded by the horizon. That would mean that you could still see the upper 66ft of a 100ft building.
So the problem is flerfs are dorks that dont know what numbers mean and how to use them.
It’s nice you can actually admit you were wrong/not completely accurate in the past, while flerfs stick to their statements no matter how wrong they are, even if they probably secretly deep down inside know they are wrong, but just too scared/ashamed, stubborn or maybe even stupid to admit it.
To any flerf out there reading this; it’s okay to admit you were wrong, we won’t think less of you, it’s not the end of the world.
What you will NEVER see is a Flat Earther using FLAT PLANAR Geometry to prove the Flat Earth.
I'll be honest, I got confused, but at least I do understand that the world is a ball shape and NOT flat.
This atmospheric refraction effect has to be calculated in optical design of high-magnification telescope instrumentation. I found this out a couple years ago designing a camera for installation at the Keck Observatory. When the telescope is pointed 60° off zenith (30° above the horizon), light between 550 nm and 950 nm will refract differently with a dispersion of one arc second. Doesn’t sound like a big deal unless the pixels on your camera are scaled at 6.7 mili-arc seconds (an effective focal length of 350 m)! Corrective zero deviation prisms pairs (with matching indexes, but differing dispersive powers) need to be included in the instrument design to correct for this if observing more than a single narrow waveband. Otherwise you would get each star turned into a spectrum ~ a millimeter long.
Excellent explanation Dave! You always nail these things.
Land surveyors used this formula (or rather one that actually *did* account for refraction) to calculate the drop of the horizontal in times when personal computers, laptops, pocket calculators and embedded computers weren't a thing. Nowadays the equipment has the calculating power to use the actual curvature for these calculations (I don't know if the old formula is still taught for when the equipment does not cooperate ;-)). It is really easy to understand, so it takes a lot of willpower by the flerfs to misunderstand and misrepresent it. Or maybe they are really that stupid that they are completely unable to understand it. Maybe both.
The formula y = a×x^2 is a parabola, but that's not really where the squared term comes from. Using trigonometry you get y = a×( 1 - cos(r) ) where r is the angle in radians round the earth's circumference. I.e A very tiny angle in practice. The power series for cos(r) is 1 - r^2÷2 + r^4÷24 - ... . With r so small you can get the reasonable approximation y = a×(1 - 1 + r^2. ) which is y = a×r^2. Then r radians has to be translated into the (small) distance from the observer giving 8 inches per mile squared.
I explain the below concisely here, "How to Use 8 Inches Per Mile Squared to Estimate the Hump Correctly"
The issue with 8 inches per mile squared is that it is the drop etc etc, yet flerthers think it's the sagitta - the "hump" of Earth between them and some distant point.
Check out how it was derived, and you quickly see that it IS a sagitta, but not the one they think it is. It's the one they're standing on top of, since the drop and the sagitta are the same thing for that segment, and what they think is the arc length is only half the arc length. The other half of the arc is behind them. To estimate the sagitta they're on top of, you double the distance between them and the point they're observing when calculating the sagitta geometrically. That is the real arc length for that sagitta, and it corresponds relatively reasonably to the 8 inches per mile squared drop calculated estimate that they gave you, but for clarity it is the sagitta they're on top of, not the sagitta that describes the "hump" between them and some point they're observing.
To work out the sagitta of the "hump" between them and some distant point, they have to move half away along that arc to be on top of that sagitta, aka "the hump."
Thus, you don't use 8 inches per mile square at all.
You use 8 inches per miles halved squared.
8*(0.5d)^2
The formula for a sagitta is s = r +/- the square root of r^2 - l^2, where l is r * sin(a/2r) using radians
R = radius
L = half the chord length
A = the arc length
You can ignore the + in the +/- because we're only interested in the short sagitta, not the long one.
You can also ignore calculating half the chord length for observations less than 250 miles, since the difference between the arc length and the chord length is trivial, so half the arc length will be approximately the same as half the chord length. For l, just substitute 0.5a instead. It'll have a 0.001 mile variance at 3 d.p. If you do go ahead and calculate half the chord length, the variance doubles to 0.002 mi.
s = r - sqrt (r^2 - l^2)
=3963 - sqrt (250^2 - 125^2)
= 1.972 mi
s = r - sqrt (r^2 - l^2)
= 3963 -sqrt (3963^2 - (3963 * sin (250/(2*3963)))^2)
= 1.971 mi
8 in per halved miles squared
= (8 * (0.5 * 250)^2)/(5280*12)
= 1.973 mi
I compared the results out to 5000 miles, and at that distance using the correction above, there was about a 27 mile variance between the methods. At typical observation ranges that the flerfers bring up, it's perfectly fine - within a few feet.
Examples:
1 mile, 0.167 ft sagitta, 0.167 ft "hump," 0.67 ft "drop"
10 miles, 16.65 ft sag, 16.67 ft "hump," 66.67 ft "drop"
50 miles, 416.35 ft sag, 416.67 ft "hump," 1667.67 ft "drop"
1000 miles, 31.5 mi sag, 31.57 mi "hump," 126.26 mi "drop"
5000 miles, 762.74 mi sag, 789.14 mi "hump," 3156.57 mi "drop"
As you see, using 8 inches per miles halved squared for the "hump" tracks very well with a sagitta calculation for a 1000 miles, after which the divergence becomes noticeable.
For the Chicago example, the distance given is 52 miles which you round up to 60 miles.
Half of 60 is 30, so 8*30^2 gives us that hump between sea level where the observer is and sea level where Chicago is.
= 7200 inches
= 600 feet
Hilarious how refraction (the excuse flerfers use for almost everything that doesnt fit their "model") is what defeats this particular pseudologic
The horizon IS the curvature of the Earth. No matter what your height is, the horizon is where the Earth curves away out of view. Period.
Exactly. In a flat Earth, we shouldn't see any horizon, ever. In every case, a blur caused by the atmosphere.
Glad to see you point out the biggest error people make when trying to use 8 inches per miles squared: they get the angle of a person's view wrong or don't account for it altogether.
Teach them Dave, teach these dudes and teach us too...I hope I never hear 8 inches/miles²💆
Flerfs will utter their arguments even if you will take them to space, they are hopeless.
They're deaf, they're illogical, they're lying and they're spreading.
I've always regarded the 8"/mi^2 as the simpleton's baseline RATE calculator, from a viewpoint at the surface. The main problem is not that observer height is unaccounted for, but that the parabolic curve can crudely fit a circle only over a limited range.
Not so crudely and not over such a short range. The difference between the “8 inches” and the correct “cosine formula” is less than 0.5% over 1000 miles.
Actually that's not true. The 8"/m squared is surprisingly accurate over any distance likely to be involved in any relevant observation.
@@betaorionis2164 ...and most _errr_ "discussions", for lack of a better word, involving flerfs are talking about distances of less than a few hundred miles.
Good job, if we cant regulate our own group we would be as bad as they.
I was trying to explain this to a viewer of I Can Science That a few weeks ago about a lake in his vicinity. I hope he's watching this video and listening well because you explain it much better than I did.
My favourite fact about this topic is the sunset illusion: when you see the Sun touch the horizon, it has already set.
IIRC the 8" equation comes from surveying. Because surveying is all about tangent lines at ground level and not about observer height, it was perfect for the job: a correction factor that 19th century surveyors could do in their heads. They knew it was an approximation, but it had less error than their instruments, so it was acceptable for short distances. I wish Dave had confirmed or debunked this. Maybe tack on one minute at the end . . . ?
And that's why they all parrot it, because it was an approximation in use when Rowbotham was running his flat erf grift.
As an approximation of the actual curvature of the earth, It's accurate to within a fraction of a percent out to around 500 miles.
Another excellent demonstration of refraction is a notorious pair of photos of two oil rigs where one has a large amount of refraction, making straight objects appear bent, and the other has a more standard amount of refraction, causing the water to occlude the bottom of one of the oil rigs.
The less refracted photo can be referred to as a "black swan" because even if you assume the horizon will never bulge and occlude your view to an object, just one demonstration of that happening is enough to invalidate that assumption.
Dave...you rule.
That’s why witsit conveniently hid the parameters of the photos taken
I can’t believe how quickly your channel has grown!! Well, actually I can believe it; your explanations are well laid out and easy to understand. I’ve been here since your first flat earth video. I’ll be cheering you on to that 100K subscriber achievement. Well done.
It's because of his dog, lol. :-)
Have flat Earthers ever explained why you can see clouds lit from underneath at sunrise and sunset and why it always has a more red component? I've never seen a model made by them that would account for it.
Especially mammatus clouds after a severe thunderstorm. They're wavy lumpy texture produce shadows that make it so obvious they're being lit from the underside. I really curious about bringing this up.
Dave, I figured out how to use that formula correctly. It’s not 100%, but it’s not far off. It’s within 7% out 2000 miles. At least when I use the Pythagorean Theorem and assume everything is taking place on a perfect sphere with a 4000 mile radius and the horizon is always at sea level.
Liquid water is denser than air, but the molecular weight of H20 as vapour is lighter than the nitrogen-oxygen mix of the air. So while it may seem counterintuitive, one should expect refraction is reduced when humidity is high.