Flat Earth Follies: How to derive 8" per mile squared and why it's wrong

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

From this is it easy to see that [8"×d²] is somewhat accurate for distances up to about 100 miles and then it is absolutely terrible after that.  And now we know why - they threw out all the little corrections in the Taylor series that keep it accurate in order to make a simple "rule" that was "good enough" for what they needed at the time.

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:

Additional Notes

There are TWO other ways to potentially measure Drop Height.

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.