Debunking Flat Earthers 8 inches/mile squared - Irrelevant formula that both sides get wrong 

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

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)

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? 😆🐶

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.

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.

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.

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.

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.

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.

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.

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.

me too

U mean pseudoscience

@@flatearth7814 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.

@@flatearth7814he is quite good at clearly explaining the arguments of flat earthers and thus makes it easy to understand why they make no logical sense

​@@flatearth7814do a video trying to debunk anything he says. But you won't, because you cant. At all.

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.

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.

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..

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.

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.

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

Look at the comments sorted by new.

I don't i love debunking yall theoretical physics and pseudoscience along with millions of others

@@flatearth7814 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

@@flatearth7814 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

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 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

Nathan “things defract into the distance” Oakley wouldn’t know facts if it hit his child.

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.

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?

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.

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.

You Ankmorpork lot are always getting confused.

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.

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.

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? 😆

Also tides

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.

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.

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

well, i'll tell you one thing that's certainly true. the usage 8 inches per mile squared is a common MISUSE of the correct syntax. correct is (8 inches) TIMES (distance in miles) squared. the difference is NOT subtle or semantic. using incorrect syntax would give you demerits on any homework or exam in a physics course. however, the parabolic approximation is quite good, even out to 400 miles.

It's because of his dog, lol. :-)

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.

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.

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.

Exactly. In a flat Earth, we shouldn't see any horizon, ever. In every case, a blur caused by the atmosphere.

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.

While you can extract that constant from the equations, it doesn't make much sense on its own; to "convert to a consistent set of units" you need an area, and why should that area be a square with our distance as the side length? Consider the Taylor series for the trigonometric drop formula R*(1-cos(x/R)): x²/(2!*R) - x⁴/(4!*R³) + x⁶/(6!*R⁵) ... Here the dimension of R matches up with the dimension of x, restoring each term to be length. All uses of x came from x/R, which is a dimensionless angle. Furthermore, while x/R is small we'll have a good approximation with just the first term, as factorial grows rapidly. Now we divide the first term drop ≈ x²/(2!*R) by x² to make it a constant: 1/2R, and find a unit where it's conveniently round: drop/x² ≈ 1 / (2 × EarthRadius) ≈ 8.008 in/mi² ≈ 7.853 cm/km² It is accurate that this is half the curvature. I'm not certain this observation helps connect any dots, though it could be a useful mnemonic. Both "half" and "squared" were specific to the first non-zero term of the series, and the constant does not communicate the part of the squaring that's been moved to the other side of the equation. Multiply x² back in, round a bit more, and you get drop ≈ (8 in/mi²) * x² = 8in*(x/mi)². This reunites the squaring of x/mi which makes dimensional sense as a ratio of lengths. Of course, something that confuses me could well seem intuitive to someone else.

@@0LoneTech I agree with most of what you write after the first sentence. The first non-zero term in the Taylor series is equal to 1/(2R) which is ½κ, where κ=1/R is the curvature (and R is the radius of curvature of the curve: in this case a cross-section of Earth's surface). Using two different units, inches and miles, for length to define the same constant is unhelpful, and we can write 8''/mi²=[8/(12x3x1760)] mi/mi²= 1/(7956 mi) to put it into a consistent set of units. In mathematics, it is usual to measure the deviation of a curve at a point in terms of curvature (or radius of curvature if you must) rather than a local drop formula which depends on setting up a local coordinate system (see the Frenet-Serret formulae). [A quick aside: The drop formula can be truncated not because of the largeness of the growth of the factorials, but because we restrict its application to when x is sufficiently small. Had the formula not involved the cosine, but was R/(1-(x/2R)²) = R+(x²/2R)-(x⁴/8R³)+... (or any other series with a convergent series expansion) we would still have been able to apply it for sufficiently small x. Obviously the drop formula does involve a cosine, bt many other formulae do not and the truncations of their Taylor series are good provided x (the perturbation) is small.] I find your first sentence confusing when talking about an area. We do not talk about an area of time when discussing acceleration and it would unhelpful in doing so, why should this second order derivative be treated any differently? In any case, he above conversion was simply converting inches into miles.

@@0cgw I agree it's unhelpful to consider this squared length an area, because the length is cancelled in the very argument to cosine, and was entirely irrelevant to the derivative which produced the exponent. You consider it obvious it's part of the derivative, where both sides coexist; but the only link to derivatives was calling the expression a curvature. I believe you're right that Taylor series, like polynomials in general, should be truncatable for small values; I used the coefficient series to get me some idea of how small. Like any tool, you need some idea of where to apply it. Many will only half-remember the steps, not the context.

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.

Also in order for air to have water dissolved in it, its temperature needs to be a bit warm, which decreases density. As soon as the air cools it becomes too dense to have dissolved water and it precipitates out.

Seconds*

​@@justduro1527lmao

They are always hilarious.

Globe cultists are even more hilarious as they think their ridiculous theories are demonstrable in any way shape or form

​@@Level_No_CurveThe theories are also demonstrable and have, in fact, been demonstrated many, many, many times.

@@Level_No_Curve And flat earth is a hoax not even a theory

@@michaelmcgovern1561 unlike flat earth