### 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 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.

#### Again, using the theorem of Pythagoras

a2 = 39632 + x2 = 15705369 + x2

#### Solving for x,

x2 = a2 – 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

x2 = 15705378 – 15705369 = 9
x = 3 miles

#### For the problem at hand, we then have x2 = (3/2)c

x2 = (3/2) 6 = 9
x = 3