Working with the curvature of a spherical Earth

I apologize right now for the horrendous amount of geometry involved in this post.

Somebody recently asked me to check their work relating to the curvature of the Earth, which prompted me to make this post – not just in answer to them, but also because I’ve seen a lot of people talk about the curvature of the Earth, most of whom don’t really get the whole picture correct. So I thought I’d show how the geometry works.

Also, I’ve seen many flat Earth proponents completely muck up this whole question, which isn’t surprising, considering their ability to grasp basic geometry is below that of a pre-schooler.

I’ll use the original question I was given, which asked how far below the horizon an object’s base would be when it is 10 miles away from us. In going through this, we’ll look at the height of an observer above the surface of the Earth, the height of the shortest object someone can see at this height (assuming a perfectly spherical Earth and discounting atmospheric refraction – very important points), and both the amount the Earth has curved away from the observer at this distance and how far below the horizon the base of the object is (which, we will see, are 2 different questions).

The original question looked at someone standing 6 feet above the surface of the Earth (for the sake of this post, we’re going to make 6 feet the altitude of the observer’s eyes), and how far an object 10 miles away over the surface of Earth would be below their horizon.

Firstly, I want to bring up the problem of precision.

We should give our results to only 1 or 2 decimal places at most, especially where distance values are concerned. This is important, because we have to acknowledge that we’re dealing with approximations here. The Earth isn’t a perfect sphere, and even the radius we’re using is an approximate value, no matter how precise we try to be (indeed, because it’s not a perfect sphere, there is a limit as to how precise we can be anyway).
Now, the thing you have to remember is that the 10 miles in this question is the arc length over the surface of the Earth, and not the length of the tangent from the surface of the Earth to the top of the object (the distance between our eyes and the top of the object), nor is it the distance between our eyes and the base of the object – as it is treated to be when we just use the Pythagorean theorem to calculate the value.
Thanks to the Earth’s massive curvature, the difference between these values is negligible over short distances when we are standing close to the ground, but if we fall for over precision we’re going to give an answer that is unwittingly inaccurate when we think it’s precise.
For short distances over the Earth’s surface, this approximation (just using the Pythagorean theorem involving the radius of the Earth plus the height of our eyes and the distance over the surface of the Earth pretending it’s these same as the distance between our eyes and the object we’re looking at) is fine – but if we make the answer over-precise, and we try to extrapolate this result over larger distances, we fall into the trap of making larger errors as the distances involved increase.

And again, I can’t overstate this fact: Even when we use the more precise formulas, we need to guard against giving a false impression of precision, as we will easily forget that the Earth isn’t a perfect sphere, and also that the radius of the Earth we’re using is an approximation.

To go through the geometry better, I’m going to try to make the problem visual:

Now when we talk about how much the Earth curves, there are 2 ways of looking at it.
The first is to think of ourselves on the top of a hill, and to measure how much the Earth drops from the tangent line that touches the point on the Earth where we are standing.
The second is to think of the horizon as being the peak of a hill, and to ask how much the Earth drops beneath the tangent line at the horizon, which is where the original question is going.

In this diagram, “x” is the height of our eyes above the surface of the Earth, and “y” is the height of the shortest object we can see the top of over the distance involved (10 miles).

Now, again, the first thing we have to note is that the 10 miles in question is an arc length over the surface of the Earth, given the Earth’s radius “r”. We’ll call this arc length “c”.
The distance between the spot on the Earth we’re standing on and the spot the object in question is standing on, is the arc width “n”. To figure this out, we need to first calculate the angle “C” by finding the circumference (2Pi*r) and then dividing the arc length by this circumference and multiplying the result by 360.

Let me just mention the Sine rule very quickly. It states that a/sin(A)=b/sin(B)=c/sin(C), where a, b and c are the sides of the triangle opposite the angles A, B and C, respectively. It is also therefore true that sin(A)/a=sin(B)/b=sin(C)/c.

This means that if we know A, b and B, we can find “a” by rearranging the equation a/sin(A)=b/(sinB) to a=b*sin(A)/sin(B).

Equally, if we know a, b and B, we can find “A” by rearranging the equation sin(A)/a=sin(B)/b to A=arcsin((a*sin(B))/b).

Importantly, the Sine of 90 degrees is 1, which becomes very useful when working with right angled triangles, because we can discount any terms of sin(90), because multiplying or dividing by 1 is a redundant process.
Now, we know that the triangle formed by the 2 radii and the arc width is an isosceles triangle by definition. 2 of its angles will have the same value, because 2 of its sides have the same value. Every triangle we make inside a circle, where 2 of its sides are defined by the radius of the circle will always, by definition, be an isosceles triangle.
We know that we can cut these triangles perfectly in half and end up with 2 right angled triangles. By cutting the angle “C” in half, we’ll put a line through this triangle that sits perpendicular to the line “n” and cuts it perfectly in half, that’s all we need to know here.

This means we can figure out what the value of n/2 is, using the Sine rule:
(n/2)/sin(C/2)=r/sin(90).
Remember that sin(90)=1, so we can rearrange this equation to find:

n/2 = sin(C/2)*r, so n = 2*sin(C/2)*r.

With this information, we can figure out how much the Earth’s surface drops from our position, which is the distance “d”.
Hopefully you’ve realised that the length “m” is the same length as the other dark blue line. This is useful, because we can find the value of m with the equation sin(C)*r, again thanks to the Sine rule.

We now have the hypotenuse and one of the other sides for the triangle “mnd”, so we can find the value of “d” (the amount the Earth has “dropped” away from the point where we are standing) by rearranging Pythagoras’ theorem: d=sqrt(n^2-m^2).

(remember that Pythagoras’ theorem states that the square of the hypotenuse is equal to the sum of the squares of the other 2 sides – which in this case means n^2=m^2+d^2. This means that d^2=n^2-m^2, so d=sqrt(n^2-m^2).)
Nice and simple.

Now we want to ask the question relating to how far below the horizon the base of the object is. This is the distance “h”.

There are 2 ways of approaching this problem. We’re going to look at the way to approach it in terms of an object we can see at least the top of over the horizon, first, so we can talk about it in terms of the height of the object itself.

Then we’re going to use the more general (and simpler) version, which will describe all objects – even those we can’t and can never see over the horizon – at the bottom of the post.

Let’s go through the first method – finding the distance below the horizon the base of an object is when we can see its top.
To do this we first want to find the height “y”, which is the shortest height this object can be for us to see its top over the horizon from this distance.

We begin by finding the distance “e” (the distance from our eyes to the horizon), which is sqrt((r+x)^2-r^2). Again this is just rearranging the Pythagorean theorem.
Now we’ve found “e”, we can find the angle “A”, which is arcsin(e/(r+x)), since the Sine of “A” is defined as the ratio of the opposite divided by the hypotenuse (in this case e/(r+y)).

Now we have the angle “A”, we can get the angle “B” as simply C-A.
With the angle “B”, we can now find the angle “T” as 90-B, and the angle “S” as 90-A.
We can again use the Sine rule to find the value “r+y”, which will equal (r+x)*sin(S)/sin(T). To find the value of “y”, we just take the radius away from this result (giving us ((r+x)*sin(S)/sin(T))-r), and we have the shortest height this object can be for us to be able to see its top over the horizon (actually the height at which its top disappears with the horizon).

Now we have the angle of “T”, we can see that the triangle that “h” is part of is just a similar triangle to the larger triangle (the scale is different, but the angles, and thus the ratios of the lengths of the sides, are all the same. When shapes have the same angles as each other, but are different scales, they are called “similar”, because the ratios of their sides and their angles will remain the same).
We can find the value of “h” by simply multiplying “r” by the ratio of y/(r+y), or we can use the Sine rule again, and find that h = sin(T)*y, which is exactly the same thing. Hopefully that makes sense. Remember that sin(T) = r/(r+y), so sin(T)*y is y*r/(r+y).

Again, even though this is the actual geometry behind the question(s), I hope the reader will notice 2 important things:
Firstly, we are still going to fall for the trap of over precision when we try to act as if the Earth is a perfect sphere, when it isn’t, and when we’re dealing with arbitrary distances as opposed to talking algebraically. Pick a number of significant figures and stick to using that and rounding to it. If the final result is going to be given in feet, then convert all other distances to feet (such as the radius of the Earth and the arc length of the distance between you and the object), as this will prevent us from getting larger errors due to rounding. I normally work in either meters or centimeters, but I did this in feet for our American friends 😉
Secondly, and most importantly, I hope the reader can see that the lower the value of “x”, and the lower the arc length “c”, then the closer the values of “h”, “d” and “y” come to each other – which is why using the approximation with Pythagoras is OK over short distances, but even more highlights why you have to acknowledge the approximation of the result. What we shall see, as well, is that even over a distance of 10 miles, the differences between the values of “d” and “h” and “y” are quite significant, though the difference between the values of “h” and “y” and insignificant.

Now. Let’s go through it and show what the answer would be if the Earth was a perfect sphere with a radius of 3963 miles, and the ground distance between the observer and the object was 10 miles (and discounting atmospheric refraction), and how we get the results.

NOTE: The illustrations below are not to scale, as the scales involved will be impossible to adequately portray. As such, the illustrations are purely schematic.
Firstly, I’m going to convert the miles to feet, so I’m only working in those units.
3963 miles = 20924640 feet, which is the radius “r”.
10 miles = 52800 feet, which is the arc length “c”.
I’m also going to round to 2 decimal places (more for the angles, though I shouldn’t to be honest, but it will help highlight something further on ), because that’s accurate enough for our needs.

Now I have the arc length “c” and the radius “r”, assuming a perfectly spherical Earth.
The circumference of the Earth is then 2pi*20924640 = 131473390.61 feet.
The arc length of 10 miles then relates to an angle of 52800/131473390.61*360 = 0.14458 degrees

The arc width “n” is therefore 2*sin(0.14458/2)*20924640 = 52801.16 feet.

Please note that the discrepancy making the chord value longer than the arc length is due to rounding, specifically rounding the angle “C”, which is actually 0.1445767840350298281 degrees, making the arc width (the chord) actually 52799.986 feet.

The take away message here is that on such a scale with a spheroidal object with a radius of over 20 million feet, when dealing with distances of just 52800 feet, the arc width and arc length  are essentially the same, which should provide a big clue to anyone who isn’t utterly devoid of intellect as to why the earth around them looks flat at these relatively small scales, as it proves the fact that curvature increases and decreases inversely with scale.

But then, surely even a 5 year old could tell you that, let alone a grown adult. Or so you’d think.

So, on with the show.

The length “m” is sin(0.14458)*20924640 = 52801.12 feet.
The length “d” is therefore sqrt(52801.16^2-52801.12^2) = 64.99 feet.

Hopefully, something has stood out to those of you paying attention. The arc length “c” should be longer than either of the lengths “m” or “n”.The reason they are larger is due to the fact that we have rounded the values for the angles.As we’ll see further below, the difference is insignificant (both values are only out by about 2 feet, putting the value for “d” out by less than 2 feet), and the problem is just one of rounding when it comes to the angles themselves.
Now for how much it is “below the horizon”, viewing the horizon like the peak of a hill.
The distance “e” is sqrt(20924646^2-20924640^2) = 15846 feet.
The angle “A” is then arcsin(15846/20924646) = 0.04339 degrees.
The angle “B” is 0.14458-0.04339 = 0.10119 degrees.
The angle “T” is 90-0.10119 = 89.89881 degrees.
The angle “S” is 90-0.04339 = 89.95661 degrees.

The distance “r+y” is therefore:
(20924646)*sin(89.95661)/sin(89.89881) = 20924672.63 feet.
So the distance “y” (the shortest height the object can be in order for us to see its top over the horizon), is 20924672.63-20924640 = 32.63 feet.
Finally, we can find the value “h” (the distance the base of the object “drops” below the “hill” of the horizon):
sin(89.89881)*32.63 = 32.63 feet (the Sine of 89.89881 degrees is so close to 1 that this is exactly what we expect it to be).
We can check this with the ratio method: 20924640*32.63/20924672.63 = 32.63 feet.

So, there we go.
The amount the base of the object is “below” the point of the surface of the Earth we’re standing on is around 65 feet, in this example, and the amount the base of the object is “below” the horizon is about 33 feet.
The lowest height this object can be so that we can see its top is about 33 feet.

Remember though, that we live on a planet which is not a perfect sphere and which has an atmosphere, and that light refracts as it moves through any region of space that isn’t a vacuum, which affects the distance we can see things, and allows us to see objects that are geometrically over the horizon. If the Earth was a perfect sphere given these dimensions and had no atmosphere, or atmospheric refraction didn’t exist, then this geometry would define exactly how far we can see.

A lot of imprecision is brought in the more steps you make, because with each step you are rounding the result.

The more astute will realise that the final equations can be expanded and written solely in terms of the values we’re given at the start of the exercise, namely “x”, “c” and “r” (or our height above the surface of the Earth, the radius of the Earth and the ground distance between ourselves and the object).

In this way, the amount a perfectly spherical Earth “drops” as you move across the surface (“d”) is:

sqrt((2*sin(360(c)/(2pi*r)/2)r)^2-(sin(360(c)/(2pi*r))*r)^2)

and the amount the base of an object “drops” below the horizon is:

sin(90-(360*(c/(2pi*r))-arcsin(sqrt((r+x)^2-r^2)/(r+x))))*((((r+x)sin(90-arcsin(sqrt((r+x)^2-r^2)/(r+x))))/(sin(90-(360*(c/(2pi*r))-arcsin(sqrt((r+x)^2-r^2)/(r+x))))))-r)

Plugging in the numbers for the example we have into the first equation to find “d”, we get:

sqrt((2*sin(360(52800)/(2pi*20924640)/2)20924640)^2-(sin(360(52800)/(2pi*20924640))*20924640)^2)

Which gives us 66.61 feet to 2 decimal places.

The reason for the discrepancy here is due to our rounding the values for the angles in the equations above to 5 decimal places.

The error is just around 1.62 feet, though, which is insignificant for the scales we’re talking about.

Plugging in the numbers into the second equation to find “h”, we get:

sin(90-(360*(52800/(2pi*20924640))-arcsin(sqrt((20924640+6)^2-20924640^2)/(20924640+6))))*((((20924640+6)sin(90-arcsin(sqrt((20924640+6)^2-20924640^2)/(20924640+6))))/(sin(90-(360*(52800/(2pi*20924640))-arcsin(sqrt((20924640+6)^2-20924640^2)/(20924640+6))))))-20924640)

Which gives us 32.63 feet to 2 decimal places. Exactly the same result as above.

Now, the more astute among you will notice that the  second equation breaks down at the points where C=180-S and C=180+S. This is because B=90 at these points, and as T=90-B, then T=0. The sine of 0 is 0, which means the equation to find “y” ends up with you dividing by 0.

This happens because the “y” term in the second equation – the shortest height an object can be in order for you to see the top of it over the horizon) is only meaningful for when C180+S.

In the region between C/=180-S, no matter what height the object is, the vector describing its height is divergent from your line of sight – which is described by the tangent line defined by extending “e” indefinitely. This means you will never see the top of the object, no matter how tall the object is.

At the points C=180-S and C=180+S the vector describing the height of any object is always parallel to your line of sight – the tangent line defined by extending “e” indefinitely – so again you will never see its top no matter how tall the object is.

At the points C=180-S and C=180+S, the “h” term loses its meaning in the context of the “y” term in this manner, and the “h” term becomes the value “r”, for reasons that should be obvious.

This is because the angle between the radius to the base of the object and the radius to the point of your horizon (the angle “B”) as we’ve pointed out is 90 degrees.

That means they form a right angled isosceles triangle with the arc width. Right angled isosceles triangles are what you get when you cut a perfect square diagonally in half – the 2 sides are equal and to non-right angles are both 45 degrees.

Using the same method as in the first equation to find “d”, this time to find “h”, we can easily see that “h” is equal to “r” (though this should be obvious anyway).

Of course, this demonstrates that the second equation is a short hand method for finding the solution, which breaks down at 2 specific points.

So, we’re going to go through the general method for answering this question in terms of any object, not just ones that we will be able to see above the horizon.

This method for finding the solution to the second problem is pretty much the same method we used to solve the first problem, but with a small adjustment.

This time, when we’ve found the angle “B”, we calculate the arc width of the angle, which is 2*sin(B/2)*r. We’ll call this “f”

Next, we want to calculate the length “g”, which is of course sin(B)*r.

With these we can find that h=sqrt(f^2-g^2).

We can check this result with the example figures we’re already given:

The angle “B” = 0.10119

The arc width “f” is 2*sin(0.10119/2)*20924640 = 36954.974 feet

The length “g” is sin(0.10119)*20924640 = 36954.96 feet.

The distance below the horizon “h” is sqrt(36954.974^2-36954.96^2) = 32.61 feet

Yes, this is the simpler version, but the first version expresses how far below the horizon an object you can see is.

(Also, though, I had to round the distances involved to 3 decimal places this time, as the rounding errors become more pronounced. Rounding to 2 decimal places gave the result for “h” as sqrt(36954.97^2-36954.96^2) = 27.19 feet.)

But what does this method look like when C>180-S?

Well, as with before, when we get the value of the angle “B”, we can find the value of the arc width “f” to be 2*sin(B/2)*r.

This time, the length “g” is still equal to sin(B)r, but why? This seems strange. We could see why this was the case in an example where C<180-S, because the lengths “g” and “r” formed part of a triangle that the angle “B” was inside of.

This time though, the lengths “g” and “r” form a triangle that seems to cut through the angle “B”.

So what’s going on?

Well, the angle I want to draw your attention to is the angle “K”, at the bottom.

Surely “g” should now equal sin(K)r? Well, it does.

What you should notice is that this angle is the supplementary angle to “B” – it has the exact same value as the angle that supplements “B” along this line (the angle “A+J”).

Well, the sines of supplementary angles are the same, and so sin(K) = sin(B). So yes, the value of “g” is sin(K)r which is exactly equal to sin(B)r.

There is another important reason why I wanted to go through the first method, though.

I’ve said above that the “y” term is only meaningful when C180+S. So what does that mean for the region between C>180-S and C<180+S (discounting the points where C=180=S and C=180-S)?

In this region, the building, no matter how tall it is, will always be diverging from the tangent line that defines your line of sight. So why does the equation still work in this region? What does “y” end up meaning?

Well, we can illustrate what happens very simply:

By extending the vector describing the height of the object, through the radius of the Earth and indefinitely onwards, we can see this line intersects the tangent line describing our line of sight at a point behind us. This intersection now defines the value of “T”, and also the value of “y”

Since the angle “T” is measured counter-clockwise from the tangent line (as you can see in the first example), we can see the angle “T” is now clockwise from the tangent line, giving it a negative value. The sine of an angle with a value of -n degrees is the same as the value of an angle of 360-n degrees (and an angle of 180+n degrees). In effect, as we can see, if we measure the angle “T” counter-clockwise as we have in the first example, we get the value 360-T.

OK, that’s how it defines “T”, but how does it define “y”?

Well, just as the angle “T” was measured counter-clockwise from the tangent line, making its value become negative when it became a clockwise angle, so to is the value of “y” positive when it is below the tangent line and negative when above it.

As you can see, the value of “y” now becomes the value of this length above the tangent line behind us, intersected by a line perpendicular to the tangent line equal in length to “h”.

This is why the second equation works, even when C>180-S, as long as C doesn’t equal 180+S or 180-S.