# 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.

### 23 responses to “Working with the curvature of a spherical Earth”

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- February 29, 2016 -
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Could you possibly plug this into an easy calculator? So I can say I’m at 6ft eye level and a tower is 26 miles away and is X tall, can I see it? Or how much is hidden? Something easy? I see all these flat earthers use Pyth but according to your formulas they use it incorrectly. The way I’m seeing your explanation is the ‘drop’ is from the angle of perpendicular to the surface where the object you want to see is, not directly straight out and down.

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Pythagoras can give you good approximations over short distances, but the main things to remember are that it falls apart over longer distances, and that flat Earthers routinely mistake the “drop” of the Earth away from you with the amount an object is “below” the horizon – as this article shows, they are 2 very different things.

Also, how far the ground at the base of a tower is “below” the horizon is not the same as how much of the tower is hidden by the horizon, because the tower will be at an angle to you.

Again, over short distances, these values are essentially the same, though.

There are very good calculators on line, such as Webcalc 2.0, which I recommend. You can literally just copy and paste the formulas from this article into the calculator and substitute your values for the variables (your height above the Earth, the height of the tower, the distance to the tower), and it will give you your result – it’s literally just a matter of copying and pasting from this article into the calculator.

Unfortunately, I haven’t seen any calculators that deal with advanced trig functions, such as the versed sine and exsecant, which is a shame, because it would make those formulas much shorter. It’s a shame that these functions have fallen out of fashion.

The “drop” of the Earth is literally just, if you drew a line tangent to the surface of the Earth where you are standing, and dropped a line perpendicular to that to the base of the target object, then it’s the value of that perpendicular line. It’s also the coversine (or co-versed sine) of the angle of the arc length from you to the target object – yet another trig function that would make everything really simple if people were still taught it.

The amount the base of the target object is below the horizon is simply the versed sine of the angle of the arc length from the horizon to the target object; and the amount of the target object that is “hidden” by the horizon is the exsecant of the angle of the arc length from the horizon to the object.

If only calculators had these functions on them today. Life would be easier.

The take home thing is that these functions – indeed, all trig functions – are not just arbitrary or esoteric and mystical things. They have actual geometric identities – you can draw them and show what they are. The have a physical existence and identity.

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Earth Curve Calculator

This app calculates how much a distant object is obscured by the earth’s curvature, and makes the following assumptions:

https://dizzib.github.io/earth/curve-calc/?d0=10&h0=6&unit=imperial

You can go here. There number match this sites answers. Just put in your numbers

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You seem to have missed the section explaining the fact that distances and angles have been rounded, which causes discrepancies such as this in the values.

Shall I repeat it for you?

“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.”

And just below that, you’ll even find the points that answer your ridiculously desperate grasping at straws, where I state:

“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 no. The chord length value being larger than the arc length is solely a result of rounding values so I don’t have angle values that are 20 freaking decimal places long. Plugging in the full non-rounded angle, and you can see that the formula provides a chord length that, whilst essentially the same as the arc length, is in fact shorter.

But you’d have known that, had you the mathematic literacy above that of a 2 year old, and plugged the values into the formulae yourself.

So, do tell me. Are the formulae wrong? Or is it just when I round values that such minor discrepancies in rounded values arise?

It’s not hard to figure out, is it?

Well, at least I wouldn’t have thought so.

Perhaps you need to spend less time reading your book of fairy tales and talking to your imaginary friend, and more time reading mathematics text books,k you geometrically illiterate fuckwit.

Just a thought.

Oh, and complaining about minor rounding errors is a big no-no. It shows you haven’t got the basic acuity to go through the formulae and figure out where the answers come from, mostly due to being a mathematically fucktarded bunglecunt.

Thanks for documenting your ignorance and general inability to read 😉 xx

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Lol my apologies man.. no need for the language though. Mind if we delete that comment I made.. haha i need to finish reading before I say anything. Sorry bout that. very quick response btw!

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No, worries. Apologies for confusing you with a flat earther. The language only reflects my exhausted patience from 4-5 years of replying to their stupidity and threats. My mistake for assuming.

Will delete shortly.

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They’ve threatened you?! thats crazy..

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Yes, I’ve received everything from puerile and vacuous whining to people threatening my life and well being, just because I go through geometry and it unfortunately proves them wrong.

After 4-5 years it has become extremely tiresome, hence my terseness with them.

Again, apologies for mistaking you. My bad.

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Hi,

The one thing I like about coming across the flat earth cult is that it has made me think about geometry and science that I normally ignore. In the comments section today I corrected a post from someone arguing against some cult member and a ridiculous “open letter to Neil DeGrasse Tyson”. One assertion was that over a 400 mile lake there would be a “hump” of 20 miles.The responder claimed it would be 1/20th of a mile. Both seemed wrong to me and I posted the following. I can’t see how it could be wrong but wondered if you could read it over so I’m not posting misinformation. Thanks for your time

+mezsh So this morning I did the calculation, both you and the cult members are totally off, the correct figure is roughly 5 miles. In fact the maths is very simple, you just need to know Pythagoras ( for a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides) The earth’s radius is roughly 4000 miles (actually 3995 average but I’m rounding to make it easy) So imagine an isosceles triangle going from the centre of the earth to two points 400 miles apart on the surface. Dissect the angle at the centre until it meets the 400 mile line in the centre. Now you have two right angled triangles with a hypotenuse of 4000 miles and one other side of 200 miles. So the difference between the lengths of the hypotenuse and the unknown side is how much the earth’s surface rises due to curvature (if you draw it this will be obvious). So 4000 squared – 200 squared= unknown side squared. This is 16000000- 40000=15960000. The square root of this is 3995 (rounded up).4000-3995=5 So the “hump” is roughly 5 miles.

So I was wrong, the cult’s favourite formulae (8″ x distance in miles squared) isn’t anywhere near accurate, and for people who are trying to debunk the curvature of the earth it is shocking that they cannot calculate it using Pythagoras.Unless they believe that Pythagoras, and all mathematics, and all triangles are “in on it”, probably. Also the notion that you could “see” this is ridiculous, all you would see it what we do see, a horizon. Still it’s just a cult having fun in an internet echo chamber”

.By the way I read your blog and wonder if we have met. I’m really rubbish at remembering stuff and people, but spend many years living on the road, my name’s Paddy and I used to play fiddle and sing for a band called the Tofu Love Frogs.

Paddy

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Your math is right, and there is little difference between your straight-line measurement between the points and measuring along the surface of the earth (<1%).

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n

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Would it be more accurate if, rather than using the mean radius for the earth, we used the actual radius? There are calculators online that give data for the distance from the center of the earth to any given point we enter coordinates for. It might not seem a lot but I have seen points less than 150 miles apart that had differences in radius measuring over 2000 feet.

Also, what about refraction?

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Thanks for your comment. I probably should have mentioned my reasons for this in the article.

The article assumes a perfectly spherical earth with no topography (smooth – no hills and valleys).

The main reason for this is because the geometry is simpler.

Using the actual radius, you then have to specify what direction you’re looking in, and unless it’s due East or due West you will have to calculate for an ellipse rather than a circle as the arc over which you are viewing.

Since the Earth is so close to spherical, I felt it doesn’t greatly affect the point being made – which was to counter the geometrically incorrect claims made by flat Earthers (particularly with respect to the “drop” of the earth away from you being the same as the amount an object is hidden by the horizon).

As for atmospheric refraction, I felt it would add a layer of complexity that a piece about just geometry didn’t need.

There are better resources than I could hope to achieve for explaining the calculations behind refraction, so I decided to stick purely to the geometric nature of the problem.

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Agreed. And given that Pythagoras is still too complicated for some flat earthers, wise to keep it simple. 🙂

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Man, the concept that a person would be looking “down” at the horizon on a spheroidal Earth is beyond most flatheads.

I kid you not.

Even Rowbotham knew this, yet his new acolytes seem completely oblivious to basic geometry, and complain that in this article the observer is looking “down” to the horizon.

But then, nota single flat Earther has been able to answer the question of what characteristics define the shape of any object, demonstrating just how geometrically illiterate they are – at the most basic level!

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I took a stab at deriving formula for height obscured by curvature of a distant object in 2 parts (full details, including a reference diagram for the geometry here: http://flatearthinsanity.blogspot.com/2016/07/derivation-for-height-of-distant.html )

I assume that elevation of observer and distance to object are known and we can factor in refraction, elevation bias, and approximate oblateness.

Where

h₀ = elevation of observer

d₀ = total distance to distant object

R = Earth Radius – given a latitude you can get a better estimate for R using this calculator: https://rechneronline.de/earth-radius/

You should also add any elevation bias into R (basically your lowest elevation point between two points)

distance to horizon at height h, and Earth radius R is given by: d₁ = √h₀ √[h₀ + 2 R]

and if this distance to horizon (d₁) is LESS THAN distance to object, then the height obscured by curvature will be:

+√[(d₀ – d₁)² + R²] – R

or putting the two together:

h₁ = √[(d₀ – [√h₀ √[h₀ + 2 R]])² + R²] – R

(otherwise the value is be negative and giving how much we could see if we dug away the ground – but usually not interesting at that point)

So that’s it! h₁ = √[(d₀ – [√h₀ √[h₀ + 2 R]])² + R²] – R

Now interestingly, as complex as refraction is ( http://emtoolbox.nist.gov/Wavelength/Documentation.asp ) we can treat the NET EFFECT along one sight-line as a simple percentage of Earth radius. Of course, each sight-line can have a different value but for this calculation we only care about the one that just grazes the horizon so all you have to do is multiply R by your refraction estimate (say 1.07 = 7%, 1.14 = 14% — these are common values used in surveying when actual value isn’t known).

I use wolfram|alpha to do the calculation work.

For example, you can look at the example I did for Toronto’s CN Tower: http://flatearthinsanity.blogspot.com/2016/07/toronto-cn-tower-from-olcott-ny.html

A quick example:

Latitude was 43.340372, elevation bias was 73 meters, giving R = 6368183 m

Distance was 63634 meters, observer height was 3.5 meters, 7% refraction (estimated)

which gives us estimated height obscured of Toronto:

√[(d – (√h √[h + 2 R]))² + R²] – R, d=63634, h=3.5, R=6368183*1.07 = ~236 m

If links work: [Wolfram|Alpha]

If not: http://www.wolframalpha.com/input/?i=%E2%88%9A%5B(d+-+(%E2%88%9Ah+%E2%88%9A%5Bh+%2B+2+R%5D))%C2%B2+%2B+R%C2%B2%5D+-+R,+d%3D63634,+h%3D3.5,+R%3D6368183*1.07

Which is right about where it should be for the images we see.

Hope this helps — I find this version of the equation pretty easy to work with and I can work in the refraction and oblateness so I get more accurate values.

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It’s important to note, however, that refraction is subject to changes in atmospheric composition (particularly humidity), density and temperature.

Whilst you can generally, under normal conditions, estimate it at around 7%, more extreme conditions (such as those responsible for superior mirages, Fata Morgana, etc) will produce a much higher refraction value.

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Could you show the curvature calculations for looking at the horizon from left to right instead of over the horizon.

Like if the width of the field of view is 10 miles would the curvature be apparent?

In this case would it not be roughly 8 inches per mile so 6 feet and not noticeable.

As you can tell, I do not have strong math skills just the basics.

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if the thing you’re trying to look at is 10 miles away and the circumference of the earth is roughly 25,000 miles then you are trying to see something through a distance of .04%(.0004) of the circumference of the earth. So shouldn’t the arc length, ‘C,’ be proportionate to the sphere being measured if the angles are to be correct? ‘C’ should be 0.04% of the whole circle, or 0.144 degrees of the circle. that is really really small, so that should change everything.

i don’t know, maybe i’m missing something.

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Don’t know what you’re getting at.

“C” is clearly stated as around 0.144 degrees, for the 10 mile example.

So what exactly is your objection?

You’ve literally just said “but if this thing is true then shouldn’t this value be exactly the value you’ve given?”

Why yes. Yes it should. That’s why it has that value in the article.

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