From the Pythagorean theorem we have the relationship: a² + b² = c²

In our case we will let c = R+δ; b=R; a=d; where R is the radius of the Earth (we will use 3959 miles), d is our distance, and δ which is our unknown quantity (the drop height) so we can put these into the equation as follows:

(R+δ)² = R² + d²

We first want to solve for δ which gives us:

δ = √ [R² + d²] - R

Even this formula assumes a perfect sphere, so it is also an approximation, but if you plug in the right values for R you'll get a pretty good answer for the drop height at some distance and it remains close to the actual value over the full range of values; and we will show that [8"×d²] does not.

Below is the geometry for this equation. To understand this, imagine that you are at loc₁ and you go straight out for 3959 miles where you are at loc₀, and then, from THAT point, you draw a line back (from where ever you are, doesn't matter how far) to the center of the Earth - the length of that line, minus 3959 miles is your "Drop Height".

You can understand from this that there is NO distance (the length of 'a') from which you could NOT draw a line back to the center of the Earth. What happens is the further, and further out you go - the length of (a) VERY SLOWLY gets closer and closer to R+δ but no matter how big (a) becomes, it will always be just a hair smaller.

Some people seem to think that you can give this equation a distance value that is "too big" (bigger than one radius usually), this is clearly wrong. ANY VALUE for distance is perfectly fine, you just need to understand WHAT the equation is giving you back. Drop height CAN be far far greater than the radius of the Earth once you understand what you are measuring. I'll deal with why people THINK this is the case below and I'll show that they are just completely wrong.

From here, what the originator of [8"×d²] apparently did was to take our fairly accurate formula (which, as a reminder, is)

δ = √ [R² + d²] - R

and approximate it using a short-cut by using a truncated version of the Taylor series expansion assuming d is small, which gives you:

(√ R² - R) is zero so they ignored that, they kept the second term, and then ignored all terms after that, making this approximately equal to:

δ = [1/(2×R)] d²

This is why [8"×d²] is not accurate for longer distances, they must have assumed distance would be small to get a simple formula.

*NOTE: I was incorrect in which mathematical approach was used, see Addendum below for the similar method Rowbotham documented, but I doubt his was the original method. I'll let this stand as it is the same fundamental approach and formula both ways and this is a nice way to let Wolfram|Alpha do the work for us so we can see how this term comes out of the mathematics.*

So now we just need to find 1/(2×R) and then apply a scaling factor to convert miles to inches and since there are 63360 inches per mile we just multiply [1/(2×R)] by 63360, which gives us:

δ = [1/(2×R) × 63360] d²

Now we can plug in various values as R and see what we get:

For measurements along Earth's Equatorial radius (3965.1906 miles) you get: 7.989527…

For measurements along Earth's Common radius (3959.0000 miles) you get: 8.00202...

For measurements along Earth's Average radius (3956.5467 miles) you get: 8.006982…

For measurements along Earth's Polar radius (3949.9028 miles) you get: 8.020451…

You can just see someone thinking.. "meh, 8 inches is in there somewhere".

### Why 8" per mile squared is wrong

Now we can compare our better formula with this [8"×d²] estimation formula, I'll compare for Earth radius of 3959 miles:

Earth Radius | 3959 | ||

Distance (miles) | Actual Drop (miles) | Estimated Drop (miles) | ERROR (miles) |

1 | 0.000126 | 0.000126 | 0.000000 |

10 | 0.012629 | 0.012626 | 0.000003 |

100 | 1.262744 | 1.262626 | 0.000118 |

1000 | 124.341891 | 126.262626 | -1.920735 |

2000 | 476.502339 | 505.050505 | -28.548166 |

3000 | 1008.260915 | 1136.363636 | -128.102721 |

3959 | 1639.871493 | 1979.000126 | -339.128633 |

4000 | 1668.937544 | 2020.202020 | -351.264476 |

So my advice is to gently inform those using it that it's only fairly accurate to about 100 miles but don't fret a few inches here and there when discussing the 'curvature' of the Earth.

The more important issue to address is...

### Why It's Even More Wrong Than That

But more importantly, this [8"×d²] formula only gives you this somewhat wrong value for Drop Height, when what we actually care about, almost universally, is the height of a distant object that would be obscured for an observer at some elevation (h₀). I derive the correct formula to use for (h₁) height hidden by curvature at some distance in my other blog post but here is the short version:

The first one (which is the most common mistake) would be to imagine that we go directly out some distance along our tangent and then drop down perpendicular to our original plumb line. Like this:

But, if this were the case, when we go out a distance of one Earth radius (3959 miles) then the drop would be exactly 3959 miles also. This is CLEARLY nowhere even CLOSE to [8"×d²] which gives us just 1979 miles. Impossibly wrong. This geometry is the one where you couldn't have a distance greater than the radius - but I have NEVER seen anyone actually write this equation and it would be fairly complex because you have to find the intersection of a perpendicular line on a circle. This would be absurd on a spheroid anyway because nobody ever cares about A-B.

So we can clearly eliminate this as the intended equation modeled by [8"×d²] - it doesn't make geometric sense and it's not even in the BALLPARK mathematically.

The final one is what most people seem to actually do without intending to. They measure the distance using something like Google Earth which is giving you the Great Circle distance along the curve. This is NOT the same as the straight-line distance! And then they input this distance in the WRONG equation [8"×d²], which just compounds the error. You can see the difference between 'a' and curveDistance in this image below - however, for distances under 50 miles it doesn't make much difference at all. I see absolutely no use in giving a derivation for computing a drop height based off this value so I'll leave it here.

### Addendum

Since originally writing this derivation, I found the text below in Rowbotham's book 'Zetetic' (the Flat Earthers Bible).Rowbotham apparently cited a more geometric method from Britannica here, to the same end result. I hereby admit my error in the method although it is based on the same reasoning and the end result is identical and I find both approaches yield insights so I'm leaving the original for now.

More importantly my research was intended to confirm my geometrical assumptions (that CD would not be parallel to AB but rather form ADC with D being the intersecting point on the sphere) about the meaning of the distance (segment BC) and this 100% confirms it.

I also found a fun additional tidbit just following this part in which Rowbotham very clearly is aware of refraction -- which makes his 'Bedford Level Experiment' a complete sham. He also could not possibly have not understood that the viewers elevation would have mattered in how far one could see objects past the viewers horizon point.

Edited: added more explanatory text, addressed two common misconceptions, added & improved images, cleaned up some text, added Wolfram|Alpha links, made the taylor series much more clear, switched to just use 3959 miles for R everywhere (it doesn't really matter what value you use).

ReplyDeleteThe formula and data are still the same.

Rowbotham is quoting from Encyclopedia Brittanica in that section (thus the quote mark at the end). Letting the part about refraction stand without comment is, I think, telling, as is having someone else get into the waters of the Old Bedford Level instead of doing so himself (he mentions an observer, never in the first person); any magician is aware that a trick is much more convincing when someone else is brought in to verify the results.

ReplyDeleteI updated the note to include that it was from Britannica, thanks!

DeleteWhen a misperception walks hand-in-hand with misinformation what follows? It is human nature to argue from ignorance.

ReplyDeleteCan yopu give your opinion on this video?:

ReplyDeletehttps://www.facebook.com/100000927499200/videos/1301626756544895

Seriously!? LOL -- commented posted on video

Delete#1 at NO point do they say it REQUIRES a DIRECT line of sight to fire the weapon at something - they REGULARLY hit tanks on the OTHER SIDE OF MOUNTAINS by firing over the mountain using all kinds of technology with FOs. Have you never seen a >45 degree firing solution?

#2 at NO point do they say it has 125 mile range WHEN USED from sea-level -- it says that is the effective range with zero details.

#3 you realize the Navy also uses aircraft right? You guys realize they don't ONLY fire guns from sea-level right? If you have some altitude then you can SEE further. At 10K feet you would have a DIRECT line of sight to a target 125 miles away. *gasp*

But you don't even need that because ALL PROJECTILES follow a parabolic curve due to gravity so you are firing IN A CURVE.

#4 Not even ONE of you clowns thought to think about what effect the acceleration of gravity would have on an ballistic object airborne for ONE HUNDRED SECONDS!? Even if you deny 'gravity' (whatever the fuck that means) you can't deny that thrown objects curve in a parabola -- YES EVEN BULLETS ( https://www.youtube.com/watch?v=SmDDPqvcttk )

Let's assume it's fired at some angle and hits peak half-way -- that's 50 seconds up, 50 seconds down:

1/2*9.8*(50^2)*0.000621371 = 7.6

SEVEN POINT SIX FUCKING MILES

It's not symmetrical due to atmospheric drag but that just it makes worse for you because it's going to descend LONGER than it climbed which means it has to fall even further so it's more like 40 seconds up 60 seconds down -- which makes it almost ELEVEN miles up.

This is the video referred to by the above link:

ReplyDeletehttps://youtu.be/QmURWXZ9ssU

Thanks for the link -- I do not follow facebook links.

DeleteI posted a comment on the video and copied it here to our 'unknown'.

The caption to the above youtube video (on Facebook) reads:

ReplyDelete"Anybody? Somebody? Anyone?

How?

8 inches per mile squared. This means 125 miles the curvature of the Earth would be an astonishing 10,416.6 feet.

This means that they were able to line their sights up on their target when the target was over 10,000 feet below the horizon.

How is this possible? The only way that could happen is if the 8 inch per mile squared is bullcrap and the Earth is flat.

Thanks very much for your response.

ReplyDeleteCan you please answer this question from a flat earther on Facebook so he can be silenced forever? If you answer it I will post it in the flat earther's comment section with citation:

ReplyDelete"Here's your chance to silence me forever.

I'll shut up forever about flat earth if someone can explain to me how this can be seen if the earth is a globe.

This is clearly seen from 18 miles away off shore. 18 miles equals a 216 foot drop BELOW the horizon.

How is it possible to see something 200 feet below the horizon?"

http://testingtheglobe.com/images/Anacapa%20Arch%20Spherical%20Earth%20Notes.jpg

(If the above link to the photo does not work you can see the photo "Anacopa Arch and Spherical Earth" on the youtube video

https://youtu.be/w2UP6HrK6AQ at the 40:23 to 44:14 mark > Ricky)

Hoping for your kind consideration.

Thanks.

P.S. What is the 7.6 miles in your answer that you posted to our "Unknown"? Thanks.

The problem with this, if it is indeed 18 miles, is that the camera at the Ventura fairgrounds could easily be higher than ground level, and they're not telling you. The real challenge is when you can take a shot of the Toronto skyline from across the water, and a good portion of the base of the buildings is not visible. You can't cheat that way. You can't go below ground level to support your theory. Can you?

Delete..I am currently asking authorities if it is true that you can see Anacapa Island from the mainland and if you have to go to an elevated area to see it. Thanks.

ReplyDeleteWhat's needed is a clear picture from a known location that can be established (eg containing local features in the foreground) so we can establish a range for viewer height. The picture at the website you gave was very likely not from 18 miles away and is clearly from either a large boat or helicopter or the foreground water wouldn't appear so far down.

DeleteLet me know what you find

Any luck?

Delete...I will see what I can do (see for your late response).

ReplyDelete...I meant "sorry for my late response."

ReplyDeleteFantastic post! Nice use of the Geogebra plugin, too! :-)

ReplyDeleteIn the "Additional Notes" section you write, "but I have NEVER seen anyone actually write this equation and it would be fairly complex because you have to find the intersection of a perpendicular line on a circle."

It's the equation of a circle: x^2+y^2=r^2, but you want to place the bottom of the circle at the origin (raise the center up r units):

x^2+(y-r)^2=r^2

Solving for y: y= +/- sqrt(r^2-x^2)+r By inspection we see we want the negative square root, because we want it to cancel r when x=0:

y= -sqrt(r^2-x^2)+r

This results in the "perpendicular curvature drop" where the drop is measured perpendicular the tangent to the curve (at the observer's feet).

I'm going to have fun looking at the other resources on this blog...

Cheers!

Charles Breiling

aka FlatEarthMath on YouTube

Cool, thanks!

DeleteI am not an academic and I have not been following FE for long, My simple observation from all this is I think you are saying that if we see a tower say 100 miles away it would be leaning backward. I may be wrong but to my knowledge, nothing has ever been seen to be falling over when viewed from a distance except that is of course the leaning tower of Pizza [hope that is spelt correct]

ReplyDeleteYes, it would be but only very slightly and not observable or easily measurable at 100 miles away when you can barely see the entire giant building. Have you seen what things look like 100 miles away?

DeleteUse the calculator to find the approximate lean:

https://flatearthinsanity.blogspot.ca/p/fei-horizon-calculator.html?o=100

Tilt ∠ t=o/R or approximately 1.4° (1° 26' 50")

And since that is directly away from you how could you possibly detect it from 100 miles away? This is barely detectable 100 meters away from something. I have images with 1° lean -- I'll post them sometime.

The 8 inch per mile square is such a failure.

ReplyDelete0.2% of failure, I guess that means the whole flat Earth theory as been rendered useless by that 0.2%.

I mean, suddenly, that 0.2% make it totally logical for object to not be hidden by the curve of the Earth when they mathematically should be by the pathetic 8 inch per mile squared BS.

No -- the FAILURE of 8" x miles^2 is that it IGNORES OBSERVER HEIGHT.

DeleteRIGHT HERE:

Why It's Even More Wrong Than That

But more importantly, this [8"×d²] formula only gives you this somewhat wrong value for Drop Height, when what we actually care about, almost universally, is the height of a distant object that would be obscured for an observer at some elevation (h₀).

Are you lying on purpose or did you miss that somehow -- like not reading the article before commenting?

Thanks Dark Star - If refraction potentially increases visibility around the curve by 20%, that would explain a lot. But, by the same token, could it not decrease visibility by 20%?

ReplyDelete