# Formulas for “drop”

This post was inspired by me having watched too many flat Earth debunking videos on YouTube. It’s my second such post, though I didn’t advertise that fact in my first one: How fast does the Sun move?

While I’m pretty sure the Earth isn’t flat, I sometimes wish the debunkers — the round Earthers — would be a little more careful about getting the technical details right in their arguments.

## The scenario

Here’s the issue I’m nitpicking this time.

Flat Earther says: “Round Earthers tell us that the formula for the Earth’s curvature is 8 inches per mile squared”. They then present an observation that seems to be inconsistent with that formula.

Round Earther replies: “No, we don’t say that. For one thing, that formula is for drop, which is not what you probably think of as curvature. For another thing, it’s not correct. It’s just an approximation.” The round Earther often then points out that the formula is that of a parabola (or paraboloid), whereas the Earth is actually a spheroid. The round Earther rarely offers to correct the formula.

## My investigation

Discussion of drop only requires two dimensions, so I’m just going to consider a cross-section of the Earth, and use words like “circle”, instead of “sphere”.

What is drop? Imagine drawing a line tangent to the Earth’s surface, touching the ground at your feet. Then imagine a point on the ground some distance $x$ away from you. Drop (I’ll use the symbol $y$) is the height of the tangent line above that point. At least, roughly speaking.

The “8 inches” formula is worded confusingly. What is meant is 8 inches times the square of the number of miles: ${\rm drop}(x {\rm \ miles}) = 8 {\rm \ inches} \times x^2$.

Before going any further, I’ll convert the formula to more convenient units. I want everything to be in terms of the radius of the Earth. The “8 inches” formula is a parabola that best approximates a circle of radius 3960 miles. That’s remarkably close to the Earth’s mean radius of 3958.8 miles. Since I’m more interested in the abstract math functions than in the precision of the measurements, I’ll simplify things by assuming the Earth is a sphere 3960 miles in radius. Now let $r$ be the radius of the Earth. The “8 inches” formula then becomes $y = {x^2 \over {2r}}$.

I wanted to try to derive the correct formula, but I realized I wasn’t sure exactly what drop was. A few web searches later, and some digging through the fine print on Earth-curve-calculator websites, and I’m still not completely sure. I think I know, but in any case, what I really wanted to do anyway was to compare a few different variations of the definition.

As I see it, there are two sensible things that $x$ could be:

• The arc-length distance along the surface of the Earth
• The distance along the tangent line

And there are two sensible things that $y$ could be:

• The distance to/from the tangent line, perpendicular to the tangent line
• The distance to/from the tangent line, along a line going through the center of the Earth

That gives us four possible definitions. Here are the first two:

• $y_1$: $x$ units along surface, then up, perpendicular to tangent line
• $y_2$: $x$ units along surface, then up, away from center of Earth

(In these diagrams, the protagonist is at point A. The distance AB is the input to the drop formula.)

And the next two:

• $y_3$: $x$ units along tangent, then down, perpendicular to tangent line
• $y_4$: $x$ units along tangent, then down, toward center of Earth

All these ways measuring it will give very similar numbers, for distances that are small by comparison to the radius of the Earth.

If I’ve done the math correctly, the corresponding formulas are as follows:

• $y_1 = r \cdot (1-\cos{x_1 \over r})$
• $y_2 = r \cdot (\sec{x_1 \over r}-1)$
• $y_3 = r-\sqrt{r^2 - {x_2}^2}$
• $y_4 = \sqrt{r^2 + {x_2}^2}-r$

I think the first of these, $y_1$, is the correct one.

If you want to use lesser-known trig. functions, you can simplify the first two formulas:

• $y_1 = r \cdot {\rm versine}({ x_1 \over r })$
• $y_2 = r \cdot {\rm exsecant}({ x_1 \over r })$

Now I’ll graph them all together, along with the “8 inches” formula, to see how they compare. Admittedly, it’s a little misleading to put them together like this, since they’re measuring different things.

Zoomed in:

So we have a a sine curve, a secant curve, a semicircle, a hyperbola, and a parabola.

All of these, except “8 inches”, imply a round Earth, but only one is a circle. The others are distorted because the coordinate system has been changed.

Whatever the correct formula is, it isn’t a parabola. So I suppose round Earthers are right to criticize the “8 inches” formula for being a parabola. But if $y_1$ is the correct one, as I suspect, then they’re wrong if they say it should be a circle. It should actually be a sine wave. When the shape of the Earth is at stake, we should really try to get it right.