### Eric Dubay's Flat Earth Maths Calculations Have Too Little Curvature In Them. Officially Taught Curvature Appears to Have Too Much. The Truth Must Be Somewhere Between The Two

Henry Curteis of Tapblog Investigates:

What do Tap readers make of this list of reasons why the earth has to be flat? It’s got some highly persuasive points to make. Who can bust this and prove the opposite?

He keeps repeating a complex mathematical formula for the amount of expected fall on a round earth of 25,000 miles in circumference, eight inches per mile squared. He never goes into how that figure is arrived at in detail. If you divide 25,000 miles by 360 degrees, each degree of the circle is 69.4 miles approximately. At the surface of the earth, the fall over that distance would surely be imperceptible.

You can, however, see ships disappearing over the horizon as they travel away from you, or appearing over the horizon.

Yet why are aircraft never routed over Antarctica, and why is it so much colder in the Antarctic there than the North Pole?

Why is dawn and dusk instantaneous in places like the Philippines, but drawn out over places like England?

Why is the sun visible for 72 hours without a break at the summer solstice at locations near the north pole?

This should be easy to clear up, one way or the other.

If the earth curves 8 inches at 1 mile away, you would think that it would be nine miles to drop 6 feet. But this, it is claimed, is not so. You only need three miles to drop six feet, if the earth is a 25,000 mile sphere, claims the video. The example below explains the Pythagorean method used to calculate this. Keep thinking. To my mind the simple eight inches is right per mile, and the pythagorean explanation is not calculating the rate of curve of a sphere, but comparing a round earth with a flat earth. The sum seems wrong. The amount of curvature increases very slowly at first and then dramatically, so that it would be impossible to see a point 90 degrees round the earth no matter how high you are. To see a 45 degree span of the earth, you’d need to be more than 1000 miles high. Yet you can see tall items like hills or towers thirty miles away in any direction from six feet high. What’s the right way to calculate the curve?

I found this example, and then did my own afterwards.

It is true that Harley showed quite correctly that the earth curves approximately 8 inches in one mile. The solution presented then goes on to find out over how many miles does the earth curve 72 inches or 6 feet. Then this distance is doubled which is required (and would be easy to forget), because each man looks this distance to his horizon, where the line of sight is tangent to the earth’s surface midway between them. All of this reasoning is correct so far.

#### However, it turns out that while the earth does curve 8 inches in one mile, it does not take 9 miles to curve 72 inches. To show this, let us return to the Pythagorean Theorem method used by Harley, but using 6 feet for the curvature. Here is a copy of Harley’s diagram with the 1 in the diagram replaced by x, since in this case the distance is unknown.

#### Again, using the theorem of Pythagoras

a^{2}= 3963

^{2}+ x

^{2}= 15705369 + x

^{2}

#### Solving for x,

x^{2}= a

^{2}– 15705369

####
a must be 3963 miles + 6 feet (Let’s say the men are actually 6’3″, so their eyes are six feet above ground.). Thus

a = 3963.001136 miles

x^{2}= 15705378 – 15705369 = 9

x = 3 miles

#### Now, remember that each man looks 3 miles to the horizon, giving their distance from each other as 6 miles.

#### This shows that at eye level of 6 ft. the horizon is 3 miles (at sea or on a level plain).

#### A rule-of-thumb for line of sight problems such as this, where the distance is small in comparison to the size of the earth is

####
c = (2/3) times x^{2}, where x is distance in miles and c is curvature in feet.

####
For the problem at hand, we then have x^{2} = (3/2)c

x^{2}= (3/2) 6 = 9

x = 3

#### This is the same result that the more lengthy solution yielded.

#### –Jerry

#### ---------------------------------------------------------------------------------

####
Now my own thoughts.

If the first mile lowers the horizon 8 inches, the 6250th mile (one quarter of 25,000) lowers the horizon one mile (less a miniscule amount of curvature). There must be a formula which expresses the accumulation of curve the further you travel round the globe, between one and 6,250 miles. The curve starts off very slowly indeed, and the rate of tip builds progressively. A mile is 1760 yards, 5280 feet and 63, 360 inches. The 8 inches tip at one mile becomes a 63,360 inches tip at the 6250th mile.

What is the relationship between the two? I divided 63,360 inches by eight giving 7920. That figure is not too far far off the 6250, which would suggest the rate of tip of the earth’s curvature is an arithmetic progression, not a squared one or geometric progression, as is claimed in the video and by other flat earthers. You tip 8 inches in the first mile, 16 inches more in the second, and 24 inches more in the third, all the way to 63,360 inches in the 6250th. You have to accumulate these totals of inches to calculate the rate of tip at a given total of miles.

**At three miles, using the assumed figures, the accumulated tip is 48 inches (8 + 16 + 24), which most adults can see to. Another mile more and the tip is 80 inches (8+16+24+32), a distance most adults cannot see to, unless they climb a hill.**

The simple progression, counting by 8 goes as follows – 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, 104, 112, 120, 128, 136, 144, 152, 160, 168, 174, 182, 190, 198, 206, 214, 222, 230, 238

But this progression needs to be accumulated at each step, as follows – 8, 24, 48, 80, 120 (5 miles), 168, 224, 288, 360, 440 (ten miles), 528, 624, 728, 840, 960 (15 miles), 1088, 1224, 1368, 1520, 1680 (20 miles), 1848, 2022, 2204, 2394, 2592 (25 miles), 2798, 3012, 3234, 3464, 3702

**At twenty miles, the tip would be 1680 inches or 140 feet.**

The lighthouses would be visible at twenty miles if they are 250 feet high.

**At thirty miles, the tip would be 3702 inches or 308.5 feet.**

The lighthouse at 250 feet would disappear at about 27 miles – if my very rough assumptions are any good.

(I don’t seem to be far off, do I, which is encouraging. Hopefully a real mathematician will show up, and confirm my workings.

**At 64 miles, the tip is 1213.5 feet.**

**At 100 miles, the tip is 3102 feet or thereabouts.**

The flat earthers are saying the earth’s tip is 8 inches times the number of miles distant squared. That’s not true. They are approximately doubling the actual rate of tip, and then saying that they can see islands or lighthouses which should not be visible. Then they ignore the fact that the identified islands have mountains on them. The rate of tip is not a squaring formula like the one they are using, but a simple multiplying one. You take each mile of the distance under consideration and add 1 for each extra mile, then multiply the total of all of them by 8 inches.

So, for example, to calculate the earth’s tip at five miles, you add 1+2+3+4+5 = 15, and multiply that by 8 to get 120 inches, or 10 feet. Taking nearly every example in the video or seen elsewhere on other flat earth videos, you can calculate that the apparently impossibly visible islands should indeed be visible. Of course some of the examples are of the kind ‘people travelling from A to B are saying that they can see X or Y’. Why not drop off a video to prove these assertions? I would estimate that their maths is in error in every case, and a formula such as I suggest above solves the problem. The key is getting rid of the square of the distance times 8 inches formula which seems to be taking a hold of the flat earth internet, despite its being completely wrong.

As to why there are no commercial flights over Antarctica, now that is a very good question.

**TAP**

UPDATE –

If you take Dubays’ formula of the distance travelled squared times 8 inches, one quarter of the earth is 6250 miles (one quarter of 25,000 miles). The square of that is 39 million plus a bit times 8/12 is 26 million feet, or 26000000/5280 = 4924 miles. Yet the radius of the earth is only 3980 odd miles. So the formula from Dubays appears wrong, with too much curvature.
My formula drops the earth not enough at 6250 miles travelled . I did all the workings to 69.5 miles, which is 1/360 of the earth, or 1/90th of a quarter section. And then used the same formula up to 90 degrees, or 6250 miles, and only dropped the earth 1500 miles, so while Dubays has too much curvature, I have too little. More work required.
The squaring formula would correct at 6250 miles by reducing the inches per mile to 6.5 from 8. That would give a fall of 1950 feet at sixty miles, which would still leave The Isle Of Man visible from Blackpool Pier, for example, as the Isle of Man has peaks of 2000 feet, and the pier must be 100 feet high.
Corsica is 100 miles from Genoa, which would be 5400 feet lower, but there are peaks at over 6000 feet, not to mention that Genoa has hills. Dropping Dubays from 8 inches to 6.5 inches per mile brings all the impossible distances back into the possible. Is that they key to understanding this?

If you take Dubays’ formula of the distance travelled squared times 8 inches, one quarter of the earth is 6250 miles (one quarter of 25,000 miles). The square of that is 39 million plus a bit times 8/12 is 26 million feet, or 26000000/5280 = 4924 miles. Yet the radius of the earth is only 3980 odd miles. So the formula from Dubays appears wrong, with too much curvature.

My formula drops the earth not enough at 6250 miles travelled . I did all the workings to 69.5 miles, which is 1/360 of the earth, or 1/90th of a quarter section. And then used the same formula up to 90 degrees, or 6250 miles, and only dropped the earth 1500 miles, so while Dubays has too much curvature, I have too little. More work required.

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