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Physics rule #54

Whatever explanation is the least exciting is the most likely to be true.

Star Wars and the Parsec Palaver: Even if George Lucas did mean to use “parsec” as a unit of distance, he’s STILL wrong.

Trigger warning: This article explains basic geometry and science, with the unfortunate downside of demonstrating that Star Wars is wrong about something, and that the fan boys don’t know what they’re on about. Expect heated comments from angry fans who lack the ability to understand reality, or have a sense of humour, below 😉

OK, it’s time to bury this nonsense once and for all.
Ever since Star Wars: A New Hope originally came out, there has been a minor scandal surrounding the scene where Han Solo declares that he did the Kessel Run in the Millennium Falcon in “less than 12 parsecs”.

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Now, it is absolutely obvious that originally George Lucas heard the word “Parsec”, and thought that because it involves the word “sec” (which he correctly thought was short for “second”) then this is a spacey/sciencey sounding unit of time that would be appropriate to drop into a sci-fi film.

Well, it wasn’t long before people explained that it isn’t a unit of time but a measure of distance, and even Neil DeGrasse Tyson weighed in.

Unable to accept a major cock-up in the script of their favourite film, the fan boys went into full mental gymnastic overdrive, trying to come up with why this line could still make sense – talking about a region of black holes that you had to pass through, and that only a fast ship could go a shorter route through this region of space. Therefore, the Millenium Falcon was fast enough to fight against the gravity in this region in order to allow it to take a shorter route that was less than 12 parsecs (or about 39 light years) long.

Sounds good, eh?

Well, actually, no.

Yet again, this may sound good to lay people with no real knowledge of astronomy or astrophysics, but to anyone who ACTUALLY knows what a parsec is, this is just as idiotic as the idea that a parsec is a unit of time, and the fan boys who use it only prove that they have no idea what a parsec even is.

So, let’s first explain what a parsec ACTUALLY is.
The word is an abbreviation, which comes from “parallax arc second”. In short, it is the distance an object is from you when it appears to shift its position by 1 arc second over half an arbitrary distance you or the object travel. It’s basically a measure of parallax and how it relates to distance.

Something should stand out to anyone who isn’t geometrically illiterate or desperately trying to save the integrity of their favourite film franchise – that the distance is dependent on the amount you or the object moves, and hence is not really a standard unit of distance (or rather, wouldn’t be used as one by a space-faring civilization or galactic empire). It is simply a RATIO that relates apparent motion to distance using trigonometry.

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It’s used today in respect to the distances of stars from the solar system, using the Earth’s orbit and the apparent shift in the star’s positions throughout the year.
It’s very simple.
Basic trigonometry states that the ratios of the sides of triangles are a function of the angles of the triangle (in fact, this is true of all shapes, and can be called one of the most basic laws of shapes – as it pretty much defines what “shape” means. From this fact we find how similar shapes are defined, because the ratios of their sides are always exactly the same for both shapes no matter what different sizes they are, as long as they have the same number of sides and values for angles).

If we have a right angled triangle, then the ratio between the opposite side and the adjacent side of an angle is a function of that angle, which we call the tangent. That is to say, if we know the baseline of the triangle and we know the angle opposite the baseline, then we know that the adjacent side must have a specific ratio to the baseline which will tell us its exact value.

Trigonometry_triangle.svg

So here’s how we use that information in astronomy to find the distances to certain stars (those close enough for our instruments to measure any parallax at all).
Measure the position of a star with respect to other background stars on one evening. Then wait until you’re on the other side of your parent star in your orbit and measure the same star’s position and how much it has shifted. Using the semi-major axis of the planet’s orbit as a base line (that is, cutting the diameter of the orbit in half), you can halve the angular shift in the position of the star that you measured and divide the semi-major axis of the planet’s orbit (or its distance to the parent Star) by the tangent of this angle (using radians, rather than degrees), and this gives you the distance to the star you are measuring.

Stellarparallax_parsec1.svg

Typically, we talk about it in relation to the Earth, because, well, that’s where all our observations of the universe are made.
So, we know the Earth is about 150,000,000 km from the Sun. We measure the position of a distant star in the night sky, then wait 6 months when the Earth is on the other side of the Sun and measure the same star’s position again.
We now have an isosceles triangle (or something close enough to an isosceles triangle, given the vast distances in space compared to the paltry diameter of the Earth’s orbit being about 300,000,000 km – in fact, time it right and you can make sure it’s exactly an isosceles triangle, but it’s not really that necessary given the immense distances involved). We can cut this isosceles triangle in half, drawing a line from the distant star to the Sun, making 2 right angled triangles.
Over the scales we’re talking, since the distance between the Earth and the sun is only a measly 150,000,000 km and the angles involved are around 1 arc second or even less, then the difference between the length of the hypotenuse of one of these right angled triangles (which is the distance from the Earth to the star), and the length of the opposite side (the distance between the Sun and the star) is essentially zero – they are as close to the same distance as makes no difference. Plus, when we’re talking about the distance to other stars, it only really makes sense to talk about it in terms of their distance from the Sun, or rather the barycenter of the solar system.
Hey presto, you have the distance to the star.

As the angle approaches 0 it’s tangent in radians approaches the same value as the angle itself. This is thanks to something called the “small angle approximation”. So when you convert 1 arc seconds into radians, the tangent of that angle is equal to 1 arc second in radians. This is very convenient because we can also define our distance to the sun as 1 AU (or astronomical unit), which gives us a distance of 1AU/1 arc second, or just 1.
“1 what?” I hear you ask.
Well, 1 parallax arcsecond, or “parsec”.

pimg1

Actually, I’ve kind of cheated there by having the tangent of 1 arcsecond equal 1, and whilst I still think it’s OK to think in terms of 1 Astronomical unit divided by 1 arc second equaling 1 parsec, I can already hear astronomers and geometry teachers having heart attacks around the world, so I better explain it better.
What actually happens is you convert the arc second into radians, which is 1/3600 x pi/180, which comes to pi/648000. You want to divide 1 AU by this value, which is the same as 648000/pi AU (Any number divided by a ratio is equal to that number multiplied by the inverse of the ratio – so 7/(3/4) is equal to 7 x 4/3). That means a parsec viewed from Earth is 206264.81 AU.
By calculating the distance to the sun as about 150,000,000 km and multiplying this by 206246.81, you can find that this parsec is equal to about 3.09 x 10^13 km. Since a light year is about 9.5 x 10^12 km, then a parsec is (3.09 x 10^13)/(9.5 x 10^12), or about 3.26 light years. Simples.

Well, that’s great, isn’t it? A Parsec equals 3.26 light years. There, we’ve numerically defined it as a unit of distance so we can use it in navigation, haven’t we?

Not really, and here’s why:
If I move to Jupiter, and then I have to measure the parallax motion of stars from Jupiter’s orbit, I’m now dealing with an orbit around 5 times as large as the Earth’s, which means that a star would have to be around 5 times further away from the solar system for me to be able to see it appear to move 1 arc second.
And that’s because the parsec is not really a unit of distance, but actually a ratio that relates distance to apparent motion.

Remember the basics of parallax: The further you move, the more something appears to move relative to you – so the further you move, the further away an object has to be to appear to move only 1 arc second, because things in the background appear to move less than things in the foreground.

It’s the exact same law of perspective that states that things look smaller the further they are from you.
But wait! We’ve defined the astronomical unit as the distance between the Earth and the Sun, so since the AU is an integral part of the Parsec, doesn’t that mean the Parsec IS a unit of distance?
I mean, it still works with Jupiter if we just put in the value 5AU, right?
After all, the International Astronomical Union (IAU – or “they who demoted Pluto”) recently defined the parsec as exactly 648000/pi AU, so that means that it has a defined numerical value, and that the galactic empire can use a parsec with a defined numerical value, right?

I’m afraid not, for many reasons.

badexplanation

First and foremost, in Star Wars we’re dealing with a galactic empire encompassing many different star systems inhabited by many different species. Each one is going to define the distances they measure the stars to be from them, if they use the parallax arc second, according to the semi-major axis of their own planet’s orbit. All those planets are not going to magically be the same distance from their parent star, and so the concept of a parallax arc second being a standard unit of distance is completely meaningless to them, because they will all measure a different parallax amount for the stars they see thanks to the differently sized orbits of their planets.
The IAU has only managed to define the parsec numerically because we only live on one planet and all our parallax measurements are taken with respect to Earth’s orbit, and setting it at this value just helps deal with having a standard definition of the ratio to help us deal with the problem of being over-precise when dealing with the slight differences in the Earth’s distance to the Sun over the course of a year. It wasn’t made a standard unit so that we can somehow use it when we start traveling among the stars. As soon as we start colonizing further afield – and definitely when we start colonizing other star systems – this standardized unit will have no meaning.

Secondly, nobody would try to navigate by parallax arc seconds, precisely because it is defined by the amount you or the target object has moved.
If you’re in your space ship and you travel 1 million kilometers and you measure the amount a distant star has appeared to move as 1 arc second, then someone in a ship next to you that has traveled 2 million kilometers will measure that same star to have appeared to move about 2 arc seconds.
You both plug in the values for the parallax motion you observed from the star into the parallax formula on your handy space calculators and find that the star is 1 parsec away from you and about 0.5 parsecs away from the other ship!
Even worse, you’ll measure the star to have a different parsec value when you yourself move different distances!
And even worse than that, how the hell do you use this unit to figure out the distance you’ve traveled?
Well, you find out your distance to different stars by measuring the difference in their position, which is dependent on the distance you moved, and given you don’t know the distance you’ve moved (because that’s what you want to figure out) and you don’t know the distance to the stars around you (because you need to know the distance you’ve moved in order to figure that out) you’re lost in a pathetic mathematical loop where both values you need to find out are completely dependent on you knowing the other meaning that you can’t know either, unless you depart from this silly “unit” of measurement and refer to a more standard unit of measurement for one of the values, begging the question WHY THE HELL DON’T YOU JUST USE THE FRIGGIN’ STANDARD UNIT OF MEASUREMENT?

It just makes no sense.

hmmm

You see, the problem with this argument isn’t just a case of assigning a standard numerical value to a Parsec. It’s with how you MEASURE it.
It’s like deciding to measure velocity in meters per heartbeat.
Sure, you could decided to standardize the “heartbeat” as a unit of time, by suggesting that the “standard” number of heartbeats in a human being in a minute is about 80, so a “heartbeat” is 3/4 of a second, and now you can use meters per heartbeat as a standard measure of velocity. But the problem is when it comes to measurement.
Everyone has different heart rates (and they can change with the level of activity and stress a person is subject to – and even worse, different species will have vastly different heart rates), so when someone tries to measure it using their heart rate, they will have a non-standard value.
So then they have to do some maths to find out how many heartbeats per second they were experiencing, and do some calculations to fit this value into your “standard” unit of measure, find out how many seconds passed and how many heartbeats they counted and how that relates to the amount of heartbeats in the “standard” unit…. but why? You’d just use meters per second.

Similarly, how are you going to measure a parsec? Well, you have to measure the parallax motion of the distant star – but that’s dependent on the motion of the observer.
There’s no standard way to measure it as a standard unit – and THAT’S the problem, and that’s why nobody in their right mind would suggest it as a set unit of distance for use in navigation.
Standardizing the numerical value of a parsec won’t help you, and the idea that you can standardize a numerical value for a trigonometric RATIO is possibly one of the most mathematically illiterate things you can suggest.

spock stupid

I love putting Star Trek memes into Star Wars related articles….

This is the crux of the problem. A parsec is a way of measuring the distance of objects FROM you, not distances you travel.

The idea of using the Parsec as some standard unit of distance is insanely stupid, and nobody who actually understands what a parsec is would ever think of it being used by a galactic empire or even for interstellar navigation as some standard unit of distance, because that isn’t what it is.
Only fan boys who can’t let the fact that their favourite film said something hilariously stupid would think that such a mind-bogglingly dumb idea would make any sense whatsoever.

If you have a galactic empire, or are navigating through interstellar space, the only standard unit of distance you’re going to use is the light year, because it is a STANDARD unit of distance that everyone can measure, and not a RATIO dependent on the distance the observer moves which will be different for everyone.

And who they hell is going to come up with a standard unit of distance that is 3.26 light years anyway? That’s just a silly idea. It’s so close to a light year that you’d just talk in light years. That’s like deciding to come up with an extra standard unit of distance in the metric system that equals 2.7 kilometers. Who the hell does that?
You’d have thought the fact that a parsec equated to such an odd numerical value would have made people think that maybe it’s not what they think it is….

It’s just a convenient measure of distance we can treat as a unit because the way we use it relates solely to the distances to stars when observed from our Earth, where we all live and which moves a pretty set amount throughout the year. Take away all those factors, and the parsec ceases to have any coherent meaning as a unit of distance.

now-thats-a-level-of-stupid

There is a final problem with the whole apologetic around the “parsec” line, and it has to do with hyperspace.
As the argument has been put in one blog, “Traveling at hyperspace is much more complicated than just pressing a button and going directly from point A to point B. A ship’s computer has to be programmed with a route to avoid the known obstacles along that route.”
This completely misunderstands what hyperspace is. You don’t travel at hyperspace, you travel through hyperspace.
In any n-dimensional space, a hyperspace is any n+1 (or more) dimensional space in which the original n-dimensional space is embedded. Think of a 2 dimensional space, like the surface of a piece of paper, and hyperspace is the 3 dimensional space it exists in.
Now think of 2 points on that piece of paper. You want to plot a course between them, but you’ve put some objects between those 2 points. If you want to travel between them through that 2 dimensional space, you have to avoid the obstacles.
But if you travel through the hyperspace of 3 dimensional space, you can plot a course in which you don’t have to think about avoiding those obstacles at all – so the idea of having to plot a course to avoid objects that exist only in a dimensional space that you aren’t going to be traveling through is completely nonsensical.

hyperspace

You can expand this idea to think about a 4 or more dimensional space in which our 3 dimensional space exists, and the same rules apply when you travel through that 4 or more dimensional hyperspace – no need to avoid obstacles that only exist in the 3 dimensional space.

So no, traveling through hyperspace literally IS as simple as “just pressing a button and going directly from point A to point B” – that’s the whole freakin’ point of hyperspace!
To claim it’s not that simple is to prove that you really have no idea what hyperspace is.

The thing is that, in Star Wars, hyperspace doesn’t mean hyperspace – it just means going really fast, which isn’t what hyperspace means at all.
(There’s also a topological definition of hyperspace, but that’s still nothing like the hyperspace of Star Wars.)

Seriously, Star Wars fans, I’d have stuck to accepting that Lucas made a dumb mistake thinking that a Parsec was a unit of time if I were you, because this new stupid idea where you try to appear clever just proves how ignorant you are of what a Parsec is, and even what geometry is.

yodafail

And in fact, if one of the fans had just decided to say that a “parsec” is a unit of time in the Star Wars galaxy, rather than engage in this incredible feat of mental gymnastics that fundamentally misunderstands what a parsec is in astronomy and astrophysics, I’d have been absolutely fine.
Really, I would have been completely OK with that explanation.
After all, they’re in a galaxy far, far away, right? They can have whatever units of time they want and call them whatever they want – so they can happily have a unit of time that just happens to be called a “parsec”. There’s absolutely nothing wrong with that.
Sure, we’d still all know that George Lucas originally wrote the line because he didn’t understand what a parsec is, but it would be perfectly acceptable as an answer. Not just because it’s OK for them to have whatever units of time they want, but because we know that Star Wars isn’t sci-fi – it’s fantasy.
Similarly with hyperspace. Stop pretending it relates in any way to the scientific concept of hyperspace, and just accept that this is a fantasy series where General Relativity and the rules of space-time geometry don’t exist, and which isn’t using the scientific terms in any way close to what they actually mean.
The problem only comes when you try to pretend you’re clever, and make it very obvious that you haven’t got a clue.

Stop pretending Star Wars is sci-fi, because when you do, you make up ridiculous apologetic arguments for it’s flaws that really just completely misunderstand and misrepresent science – and that’s why people have to write blog posts explaining what things like parsecs actually are, after you butcher their meaning to pretend your favourite fantasy has some scientific relevance.

(Please note: Yes, it is fun to bait Star Wars fans, but actually I hope this helps people understand some basic trigonometry and geometry and astrophysics in an entertaining manner.)

This article is also posted on my “Astro-gnome” blog at Trolling with Logic.

Thor’s hammer explained: Mjolnir-centrism!

Having a conversation with my friend, Wolf (yes, that is his name – I’m not just getting high and talking to wild animals, pretending they are conversing with me, though I can understand why someone may think that given my relationship with my dog), we got on to the inevitable subject that all Marvel comics fans do at least several times a year – How does Thor’s hammer work?

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My insight was spurred by this conversation and by a recollection of a conversation from the Big Bang theory, in which Penny, Bernadette and Amy were arguing over if you could pick up Thor’s hammer in space. Their conversation focused on the semantics of the word “up”, which is meaningless in space – and however amusing and insightful about space itself, this didn’t really answer the question.

thor hammer big bang

However, my mind began to race, thinking about the fact that only certain people can ever MOVE Thor’s hammer.

Now, go with me on this (famous last words).

If an object is in space and you somehow push it, Newton’s 3rd law states that the same force you exert pushing that object will also be exerted on you in the opposite direction.

As Newton states:

“To every action there is always opposed an equal reaction: or the mutual actions of two bodies upon each other are always equal, and directed to contrary parts.”

Or, as wikipedia summarizes it:

“When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body.”

And his first law states:

“Every body persists in its state of being at rest or of moving uniformly straight forward, except insofar as it is compelled to change its state by force impressed.

Or, again, according to wikipedia:

“When viewed in an inertial reference frame, an object either remains at rest or continues to move at a constant velocity, unless acted upon by an external force.

Now, this presents us with a problem. If I am in space and I push against Mjolnir (Thor’s hammer, for those who are inexplicably reading this but who don’t know what Mjolnir is), Newton’s laws state that I should move Mjolnir and that I should also be moved backwards. I may move it only very little, but I will move it.

However, Mjolnir, apparently, should not move at all.

This has a profound implication for physics. For Mjolnir must, therefore, represent THE preferred frame of reference for the universe – in fact, it must be the center of the universe. And there must be some kind of technology which enables only certain people (or a certain person) to allow Mjolnir to move the rest of the universe around it.

This makes sense, with everything that we observe it doing.

Think about it – we see Mjolnir appear too heavy to lift, no matter what force is applied to it, and yet it doesn’t go smashing through a glass coffee table when it is rested upon it, as such a heavy object should.

Now, we could say that there’s some convoluted technology that is close to magic, which allows it to have a perfectly balanced anti-gravity mechanism that perfectly allows it to adjust its apparent weight appropriately.

Tyson-Tweet-thor

OR we could say that it is the center of the universe, and that it isn’t falling through the coffee table because the coffee table is not moving relative to Mjolnir at that point in time.

Sounds great to me….

fry idea

And then we have the way Mjolnir appears to fly around finding Thor’s hand via strange trajectories through space at incredible velocities – and yet stopping it’s motion magically just in time for him to grab it, without him flying backwards with its momentum.

Well instead, Mjolnir, as the center of the universe, is moving the entire universe around itself in the most efficient trajectory until the point at which Thor’s hand occupies the same space, and it then stops the universe from moving around it. This is perfectly in keeping with Newton’s laws of motion, if Mjolnir is the center of the universe, and the universe is being made to move around it. It doesn’t send Thor flying backwards with any momentum, because essentially there is no momentum.

Then there is the incredible way that Thor wields Mjolnir – striking things all over the place, whilst the amount they appear to fly or have damage inflicted seems quite arbitrary to the momentum Thor would be imparting to the hammer with his swing.

Instead of Thor wielding the hammer like a mason, and giving the hammer momentum, Mjolnir is moving the universe around it appropriately enough to cause the exact damage it wants to the people or objects it strikes – by making them hit it with the required force and then fly away with the appropriate trajectory.

This would also explain how Mjolnir allows Thor to fly. He isn’t flying – he’s just moving the entire universe around him, as he’s holding Mjolnir.

Image credit: Thor Visionaries : Walt Simonson volume 1, taken from http://www.comicsrecommended.com/articles/marvel/thor-mighty-simonson-1.html

Image credit: Thor Visionaries : Walt Simonson volume 1, taken from http://www.comicsrecommended.com/articles/marvel/thor-mighty-simonson-1.html

There is, of course, the origin story of Mjolnir itself – that it was forged in a dying star.

Well, how about the people who forged Mjolnir found that there really was a center of the universe, and that this center occupied some space inside an old dying star – possibly the oldest star in the universe (cough cough, let it slide, cough cough)?

Upon finding this center of the universe, they create some technology that allows a person holding an object forged around this center to move the entire universe around it (or that allows the object itself to move the entire universe around this center).

Once forged, you would only need to activate the object and make it move the dying star away from this center, and move whatever planet or relative location in the universe to this center.

Image credit: ESA Herschel space telescope image, taken from http://frenchtribune.com/teneur/1317317-european-space-agency-captures-image-dying-star

Image credit: ESA Herschel space telescope image, taken from http://frenchtribune.com/teneur/1317317-european-space-agency-captures-image-dying-star

So, there we have it. Newton’s laws of motion seem to indicate that Mjolnir must occupy THE preferred reference frame in the universe, and since it must move the universe around itself, it therefore occupies (in the only way we could define it, I guess) its center.

Mjolnir-centrism.

Somebody tell the Geocentrists they were right all along – they just have the wrong thing in the center of the universe. And it only works in a comic book universe anyway, with an extreme butchering of physics….

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:

curvature b

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.

curvature02 b

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.

Sine rule

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.

curvature03

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.

curvature04 b

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).

curvature05 b

(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.

curvature06
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)).

curvature07 b
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).

curvature08

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).

curvature09

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.

curvature10

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

curvature11
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.

curvature12
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.

curvature13Hopefully, 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.

curvature14
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.

curvature15
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”

curvature16

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).

curvature17

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.

curvature16e

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.

curvature17

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

curvature16d

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:

curvature16b

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.

curvature16cAs 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.

Iovandrake And The First Basic Law Of Shapes (video)

After an interesting (?) conversation with a flat Earth proponent, who tried to claim that there is no difference between reality for either a flat Earth or a spherical one, I’ve decided to make a short video exploring this conjecture.

Join me as we explore the first basic law of shapes and discover – magically – that shapes have different properties, and that circles and spheres are MEASURABLY different.
Who’d have thunk it?

I’m truly amazed that, in the 21st century, I have to explain something that most children learn before they even begin to talk.
It appears that geometric and mathematical illiteracy is something of a prerequisite for belief in a flat Earth.

Apologies for the sound issues at the beginning. There also appears to be some flickering with the blender animations. I’m not sure if this is just my computer or a Lightworks issue. I hope to have it sorted for the next videos. The Lightworks render seems to have a bit of a nightmare as well, sometimes.
When I did my film degree – all those years back – we were still using celluloid. I can change a film can with my eyes closed and I’m a whizz in an old style editing suite, but you crazy kids have gone all digital these days, so it’s about time I had a go at it.

I’ve decided to upload it as it is, because I can’t be bothered with the headache with Lightworks at the minute, as it seems to like crashing whenever I move the mouse.
If anybody knows how to deal with these issues – or of another program as versatile as Lightworks, but without the annoying tendency to freeze up all the time – then I’d be glad to hear from them.

Still, I hope you enjoy the video.

Music Credits:

Kevin Macleod:

Truth Of The Legend
Exotic Battle
Doh De Oh
Gone Beyond
Flighty Theme
Investigations
Harlequin
Intended Force
Open Those Bright Eyes
Fluffing A Duck

Bach:
Brandenburg Concerto #5 In D, Allegro
Brandenburg Concerto #1 In F, Allegro.

Why I Can Say With Absolute Certainty That There Is NO POSSIBLE SCENARIO In Which Our Planet (At This Stage) Will Be Invaded By An Alien Civilisation Bent On Wiping Out Humanity.

Okay, so apart from their being no evidence yet for space faring ETs roaming around the place, it is apparently a scenario that gets explored not only in the imaginations of story-tellers, but also by many in the scientific community – and even some within government and military circles have reportedly looked into it.

As a massive fan boy of both the SETI project and science-fiction classics such as H G Wells’ War Of The Worlds, it is with great dismay that I burst this little paranoid bubble humanity likes to live in regarding such matters, by disabusing them of some fundamental flaws in their logic.

 

There are only a certain number of reasons why a sentient life form would want to invade Earth right now (and I intend to show, briefly, that they actually aren’t reasons – which makes us safe, as it were).
They are:
1) Resources
2) Enslavement
3) Our biosphere (as in, they need a knew one because theirs is under threat/lost)
4) Religious/dogmatic fanaticism

 

So, let’s explore these one at a time and evaluate them.

 

1) Resources:

To begin with, the resources you can find on earth – all the minerals, water, etc – can be found in abundance anywhere in the Galaxy, on many worlds, which don’t involve staging a costly war. Why waste money making weapons then mining equipment, when you can just make mining equipment and work on the several moons and planets even within our solar system – let alone the several star systems you passed to get here?

 

2) Enslavement:

Enslavement would be stupid. Any advanced civilisation that can make it here would be able to make machines that can do anything humans can do for them (and most likely much more besides), with the added advantage of not dealing with slaves who need feeding and who are prone to uprisings – not to forget that their uprisings could even spark sympathy revolutions within your own populous, or other slave populations.

When you become a space faring civilisation with the ability to build amazing machines, the concept of biological slaves is, frankly, meaningless.

 

3) Our Biosphere:

The biosphere question comes from the idea that an alien race loses – or is about to lose – it’s home planet, so comes to ours. There are a number of problems with this idea.

Our Biosphere is great (understatement #1). However, there’s more than likely many other worlds without a technologically developed civilisation on them, which you’d encounter on your way here. Which would be easier to take control of? The biosphere with microbes and simple plant and animal life, or the planet with a species that makes aircraft carriers and atomic weapons and has mastered the art of warfare (even if your weaponry is far in advance of theirs)?

War costs a lot. More importantly, going to war in a land you don’t know against people who do know it very well is a very dangerous thing to do – look at Vietnam for one (or even Afghanistan). Small Guerrilla armies with less advanced weaponry can all too often succeed in fighting off larger armies (or at least make the effort too costly for them, such that the larger army retreats), simply because they know the land and how to live/fight there.
When you want to colonise a planet, you have to look at what it would cost you – and when you have cheap planets around that would cost you nothing (but which are still just as useful), why waste time and energy making enemies that you don’t even need?


The argument doesn’t just stop there, though. Any civilisation that can get here has probably (and this isn’t a big ‘IF’) mastered terraformation. Unless you master faster than light travel (FTL), you will be traveling for some time before you get here. Keeping people in stasis takes a lot of energy (and may not even be that viable – there may be a biological impediment on keeping complex life forms almost indefinitely in a state of stasis). Not only do you need to master creating a floating habitat, but you may also need to stop off at places to refuel (at least), and these places will have to be made in some way habitable before you can move off again. If you can keep a floating habitat going, you can create a biosphere. Why start a war to take over another one, when you’ve been doing it anyway without having to start a messy war?
If you manage to create faster than light travel, then you surely must have terraforming capabilities, which would make your excursion here pointless. FTL is a lot more unlikely – and even if it is likely, a lot more difficult – than the idea of terraforming a planet/moon. Also, what possible advantages would you get from a planet that is unable to leave its own orbit, over a floating habitat that you can go anywhere you want in?


Also, maybe our biosphere isn’t that great. Maybe it’s very hostile to you. Who would happily sling themselves into an Ebola epidemic, just because they’ve run out of food where they are? Our microbial life has taken us several years (massive understatement #2) to get used to. We’ve even integrated with it. If it wasn’t for this integration, we mammals couldn’t digest food or even have children (we got that trick from a virus). You can’t just land on an alien world and expect to just be fine, even if you wiped out its number one sentient species. Integrating with a biosphere takes a long time. Just go on holiday to the tropics and see how badly your immune system copes compared to the people who live there – and remember, they at least have the same DNA as you!


The biggest argument I find against this, though, is that any life forms that can create a ship that can travel across the stars will not have just one planet. That level of technology doesn’t just come from one species suddenly having to leave their one home. That technology most likely comes from a species that has gone out (as we’re doing) and slowly colonised their surroundings. Losing their home planet would not be a problem for them (other than an emotional one). They could easily move the population left there to other worlds nearby.
Just think, we’re not even going to reach our closest neighbour, Alpha Centauri (yes, I’m forgetting Proxima Centauri, oops), for a long time – even though it’s a measly 4.5 light years away. Before we reach that, we will have colonised the moon, set up on Mars, made it to the outer solar system, etc. We won’t just suddenly jump into an untested ship and fling ourselves at the nearest habitable planet, which could be several hundreds of light years away – not just because that’s a stupid idea, but actually even more because in order to reach that level of technology we’d have gone through all the steps of colonising our neighbouring regions.


There’s also the final point, here, that they would have known about their home planet’s impending doom for quite some time (as have we) and they would have had a long time to come up with an easier colonisation/rehabituation plan than one that involves hurling off their entire population in an untested ship and invading a planet infested with a thriving intelligent species.

 

4) Religious/Dogmatic Fanaticism:

The last point regarding religious/dogmatic fanaticism I find very problematic (whereby our very existence offends them – a theme explored in the video game Halo). We can easily see from our own experiences – with the Abrahamic religions for instance and the resulting conflicts even within them (between the Catholics and Protestants, and Shiite and Sunni Muslims, etc) – that religious/dogmatic fanaticism can and does lead to dangerous schisms within your own populous. I’d argue that it stands to reason that any such fanatics would most likely be fighting amongst themselves either too much to care about us, or even so much that they’d wipe themselves out before getting very far.
Even though I don’t think a sufficiently technologically advanced civilisation would succumb to this fanaticism, due to how much reason and logic plays a part in technological and scientific advances – even if they did, I don’t feel they’d survive it in time to disrupt our way of life.

I did say ‘at this time’ for a very good reason.
I believe the only time we would find ourselves in a war with another species is when we fight over resources/territorial influence. That would typically be somewhere far away where our zones of influence meet. Certainly that war could spread and lead to Earth being invaded, but that’s way in the future – when we’ve also become a galactic player with many colonies and the invasion of a single planet becomes less significant in the grand scheme.
In all likelihood, as an aside, by the time that happens the human race will have evolved and will no longer exist – it may even have evolved into several different species, all of which could also end up at war with each other. That, however, is another story.

Our world holds no use (cost effective wise) to any aliens. To come and invade would be pointless, which makes me not just doubt but say with all certainty that ‘there is NO SCENARIO POSSIBLE where our planet at this stage will be invaded by an alien civilisation bent on wiping out humanity.’

Flat Earth “Theory” And The Pole Stars Conundum

About a month ago, that titan of modern cosmology and philosophy, Rory Cooper (Rorycoopervids), decided to take on what is possibly the greatest proof possible for anybody unable to leave the surface of the Earth, that the Earth is a spheroid.

We’ll call this “The Pole Stars Conundrum”.

Anybody able to use the photosensitive ganglion cells in their retinas, together with all the other highly evolved parts of their optic system enabling them to see, can look up at the night sky and notice that the stars they can see appear to revolve around a central point (with an exception mentioned below).

In the Northern Hemisphere, you will notice that the closest star to this central point is Polaris. In the Southern Hemisphere, you will notice that the closest star is Sigma Octantis.

Image

Rory Cooper’s frankly childish conjecture is that these are actually the same star, meaning that the Earth has only one pole star – which would certainly seem to confirm a flat Earth.

However, as usual, Rory runs into a few problems with his asinine approach to astronomy.

 

The first problems, as pointed out by JimSmithInChiapas, in his video No, Flat Earthers: Polaris ISN’T Sigma Octantis, are quite simple:

 

1) Polaris and Sigma Octantis have different apparent magnitudes – which is to say that they aren’t the same brightness.

Polaris is bright and clear in the night sky at magnitude 2, whereas Sigmas Octantis is significantly fainter at magnitude 5.42.

(A quick note: The lower the magnitude, the brighter the star.)

Flat Earthers can attempt to explain this away by stating that someone past the government-conspiracy-concocted line called the equator (that’s someone in the Southern Hemisphere for those of us still in full control of our critical faculties), would be further away from the pole star and that this would account for the difference in the observed brightness of the star.

The problem with this conjecture is that the brightness of Sigma Octantis doesn’t change for someone in Australia or someone near the equator. Neither does the apparent magnitude of Polaris change for someone in Scandinavia or someone in Kenya.

Image

Instead, the pole star must then magically jump from magnitude 5.42 to magnitude 2, as soon as you move a few kilometers either side of the equator – and yet stay at this magnitude no matter how much further away from the equator you travel.

Flat Earthers can provide no physics to explain this phenomenon.

ImageImageImage

2) The star fields surrounding Polaris and Sigma Octantis are completely different – they are surrounded by different constellations.

Again, Flat Earthers try to explain this away by arguing that when you move towards or away from the pole star, the sky would obviously change.

Here they run into several problems, though.

To begin with, just as last time, the star fields don’t change for someone in Australia or someone near the equator (or in the Northern Hemisphere, someone in Scandinavia and someone in Kenya), but somehow the stars magically jump to new positions once you travel just a short way across the equator.

Image

Another problem they find difficult to explain are the different distances and directions each star must move in order for us to see them in their new constellations.

There is still another large problem though, which is apparent to anyone who lives near the equator – because they can see that stars to the South seem to rotate around one central point, whilst stars in the North appear to rotate around another central point. Not only that, but they can confirm that the constellations in the Northern sky and those in the Southern sky are not the same. They can see Ursa Major (which the Plough is part of) AND the Southern Cross at the same time.

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Why is this significant?

Simply because Gacrux and Acrux (the “pointer stars” in the Southern Cross) point to Sigma Octantis, whereas Merak and Dubhe (the “pointer stars” in the Plough) point to Polaris.

If Polaris and Sigma Octantis were the same star – and the star fields around them consisted of the same stars – then it would be impossible to see the Plough and the Southern Cross, because these must just be the same constellation viewed from a different perspective.

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But perhaps the largest problems with any attempts Flat Earthers fling around to explain “The Pole Stars Conundrum” away, like distraught monkeys flinging their faeces at a passing lion, are the directions the stars lie in and their motions across the night sky.

 

To begin with, Polaris is visible when you look towards the North (in the Northern Hemisphere) and Sigma Octantis is visible when you look towards the South (in the Southern Hemisphere).

Somehow – and god only knows how – as you walk away from whatever pole star you can see, as you cross the equator, the pole star must jump ahead of you.

This is a remarkable celestial feat.

 

The other problem arises when you observe the motions of the 2 pole stars.

Polaris can be seen moving counter-clockwise around the night sky, whilst Sigma Octantis travels clockwise around its central point.

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This provides a massive problem to the Flat Earth conjecture and a F***ING MASSIVE CLUE about the shape of the Earth.

 

Flat Earthers of any denomination cannot account for the fact that anybody living “inward from that government-conspiracy-concocted line” called the equator would view Polaris moving in a counter-clockwise direction – and yet anybody living outside that same line would not see Polaris, but would see Sigma Octantis moving in the opposite direction.

When pressed on this, Flat Earthers draw an amusing blank.

 

The sky must, for Rory’s conjecture about Polaris and Sigma Octantis being the same star to be true, magically flip over when one crosses the equator.

The best, or rather funniest and only attempt at an answer that I have ever been provided with by a Flat Earth proponent, came from someone going by the name of “Can Attal”.

It was a simple video that involved 2 circles, one inside the other, both rotating in opposite directions. (I’d show it to you, but I haven’t worked out how to make wordpress not be a complete arse, yet).

The more astute among you will realise some important issues with this model.

Firstly, we should be able to see a shearing motion in the sky – a line at which the stars can be seen to move in completely opposite directions.

Oddly, nobody sees this or mentions it.

More importantly, though, the center of the sky will still be moving in one direction. That means that the pole star is still going to be moving in the same direction, so we still wouldn’t see Sigma Octantis and Polaris moving in different directions to each other.

 

I’ve got to be honest, here. I really tried to give Can Attal’s model every chance it had of explaining how the same star can be seen to move in 2 different directions across the sky, depending on where you observed it from on a flat Earth.

The very best I can conceivably come up with is if this is how they believe the Earth itself revolves.

That is to say, their model is only true if everything within that “government-conspiracy-concocted line” called the equator rotates one way and everything without it rotates in the other direction.

 

I’m frankly amazed how this hasn’t been reported in the news. Surely it would not escape the notice of someone in Southern Kenya, say, that part of their country disappeared every day, to be replaced by seas and parts of Asia, then more seas and parts of South America – all zooming past at break neck speed.

I’m pretty certain that I’d have seen this strange shearing effect when I flew to Tanzania in 1998.

And we’re not even touching on the problems that this would have on trade and communication.

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But here’s the crux of “The Pole Stars Conundrum”: It’s not that it is impossible for the stars in the sky to be seen to have 2 different points around which they revolve, if you live on a Flat Earth. That’s fine.

No, the point is that, on a Flat Earth of any kind (yes, there are more than one kind of Flat Earth model), it is impossible for one pole star to be visible to anyone living within a concentric circle of that Earth, whilst being invisible to anyone outside that concentric circle – and for everyone outside that concentric circle to be able to see a pole star that is invisible to everyone who lives within that concentric circle.

The only way that 2 pole stars can exist in a Flat Earth model is if someone at one end of your flat Earth sees a completely different pole star to someone at the other end – but people all over the Southern hemisphere see the same pole star.

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People standing at Points A and C can only see a pole star that moves clockwise around the sky and cannot see a pole star that moves counter-clockwise around the sky.

People standing at Point B can only see a pole star that moves counter-clockwise around the sky and cannot see a pole star that moves clockwise around the sky.

It is impossible for someone standing at Point A to be able to see the same star as someone standing at Point C without being able to see the same star as someone standing at Point B.

Every Flat Earth model is only possible if the people at Point A can only see one pole star moving in one direction and the people at Point C can only see a different pole star moving in the completely opposite direction – and people at Point B can see both stars.

 

In short, you can only have 2 pole stars that behave this way on an Earth that is not flat – and I mean not flat according to any denomination of Flat Earth models.

Finding the sizes of – and distance to – distant targets (or “Why Fernieboy100 is menatlly challenged”).

In this post, we’re going to discover just how easy it is to measure the distances and sizes of objects, without having to walk to them.

It’s an inverted triangulation method. Instead of triangulating your position, you can triangulate a chosen object.

YouTube user Fernieboy100, once asserted that we couldn’t measure the distances to objects using telescopes or binoculars, during one of his least mentally attuned attempts to debunk a video by CoolHardLogic.

CoolHardLogic‘s brilliant video response, on how we can measure the distance to the Sun, can be found here: https://www.youtube.com/watch?v=yQoz5iWPrbs&list=PLmWeueTF8l82THrHwihtcmhQdjcBQBXjT&index=8

Indeed, Fernieboy100 seems to believe that we can only measure distances by getting a tape measure and walking towards our targets.

He’s obviously never met anyone who works with maps – and must be living under the delusion that maps are made by people walking all over the countryside with a massive ruler.

However, we’ll just need 2 angles and a baseline and we can calculate everything else.

 

We need a pair of binoculars/telescope, a pencil, a ruler, a protractor, a pocket calculator (or very good knowledge of maths), a sheet of paper and a magnetic compass. We also need to know our binoculars’/telescope’s field of view (normally written on the binoculars/scope), so we can measure the angular size of whatever object we like (a mountain, a tower, whatever).

 

First, we take a bearing to the object we want to measure the size of and distance to. Let’s say we pick a house.

Next, we’re going to draw a simple schematic map (schematic means it’s just a representation, we don’t need to draw the terrain).

We’re going to draw an arrow on our map to indicate which direction North is, and draw a long line in the direction of the object from a point that represents our starting point. We need to write on the bearing to the object as well.

Let’s say our house lies on a bearing of 120°.

Now we’re going to walk in any direction we want (that isn’t directly towards or away from the object), covering any distance we choose. A vaguely lateral path is best.

We only need to walk 500m or 1km at most, or if our compass has mil-radians marked on it (and we’re good enough at taking a bearing), we could even just walk 100m or less.

Before we set off, we’ll make a note of the bearing we’re walking in and draw a line on our schematic representing this. We’re going to scale the map as well. We can make 1cm=100m or something appropriate for the distance we’re going to walk.

In this case, we’re going to walk 500m on a heading of 210°.

 

I know my pace over 100m on a flat surface is 60 “right feet” (adjusting accordingly for terrain), so I can count out this distance, using my compass to stay on a straight bearing. The “right feet” unit means that I only count when I step with my right foot, which makes it easier to count over long distances. You can find out what your own measure is beforehand, by simply walking 100m and counting.

 

When we’ve reached our 2nd observation spot, we take a bearing on the object and draw a second line on our schematic map along that bearing from where we are now.

We can now know the angle described by the vectors emanating from our starting point to where we are now and the object, which is just the difference between the bearing from our first point and the bearing we’ve walked on.

We can also find the angles described by the vectors emanating from the object to our 2 observation points, as it’s just the difference between the 2 bearings.

Since the angles in a triangle must add up to 180°, we can also figure out the final angle at the point we’re now standing.

 

In this exercise, let’s imagine the bearing to the object from our 2nd point is 110°.

We know that if we’ve walked on a bearing of 210° from our first point to here, so the angle at our starting point is simply the bearing we’ve walked minus the bearing to the object from our starting point. 210-120 = 90°.

Since our original bearing was 120°, we can subtract our new bearing from this to find the angle at the object we’re measuring to be 10°.

180-90-10 = 80, so the angle at the point we’re standing at now is 80°. We only really need this final angle if we want to calculate the distance from our starting point to the object.

 

Because of the distances involved, we’re probably not going to have a piece of paper big enough to deal with the scales we’ve used, otherwise we could have just measured the distance with a ruler.

We’ll probably end up with something that looks more like this:

 

Which is why we’ll use trigonometry.

 

Now we have 3 angles and a length, we can calculate our distance to the object using the law of sines:

a/sin(A) = b/sin(B) = c/sin(C)

Rearranging this equation to find the length we want, we use:

a = (sin(A) x b)/sin(B)

Plugging in the numbers we’ve got, our distance is (sin(90) x 500)/sin(10), which gives us about 2879.4m, or 2.88km.

We can even find the distance from our starting point, if we wish. It’s (sin(80) x 500)/sin(10), which is about 2835.6m, or 2.84km.

 

Hey presto! We have the distance, thanks to a bit of maths and some basic tools.

Now we have the distance, we can find the size of the object.

 

Looking through our binoculars/telescope from the second point, we measure the angular size of the house to be about 11 arc minutes (if we’re good enough, we can even calculate the arc seconds for a better precision). Dividing our arc minutes by 60, we find that that’s 0.1833°.

The equation for finding the size of an object from its angular size is:

S is the size of the object.

a is the angular size of the object.

D is the distance to the object.

Plugging our numbers in, we find tan(0.1833/2) x 5857.8 comes to about 9m.

 

So, we’ve managed to measure the distance to, and size of, any distant object on earth using just binoculars/telescopes and some very basic trigonometry.

 

Now that we have the size of the object, if it is an object of roughly standard size, we can find other such objects and reverse the angular size equation to find our distances to them, using:

 D = S/(2 x tan(a/2))

 

Now we can easily measure the distance to anything and anywhere, using simple trigonometry and tools, to an acceptable level of accuracy.

Yet more problems for Iovandrake

As we saw in the last post, there was a marked difference between the areas of our flat circles and our hemispheres.
Indeed, there were greater differences between the flat worlds and the spherical one.

We also know that for the land masses to be the same on both, they must have the same area.
(We’ll ignore the insurmountable problem of them needing the same dimensions – lengths of sides and angles – for the moment.)

The accepted land mass area of the earth is roughly 148,940,000 km^2. The accepted land mass area minus Antarctica is roughly 134,980,000 km^2.
The area of the seas is accepted as roughly 361,132,000 km^2.
Land, minus Antarctica, makes up about 26.5% of the surface area of the spherical earth.

Whacking the same land mass size (minus Antarctica) into the first hemisphere, with its total area of 255,036,000 km^2, we find that the land makes up nearly 53% of the total area.
This leaves the seas to take up a piddling 120,056,000 km^2.
Which means flat Earthers would expect us not to notice that 241,076,000 km^2 (almost 67%) of the water surface of earth is missing, from the spherical model to the more convenient (for them – or so they misguidedly believe) hemispherical one.

But it gets even worse.

On the flat earth, with its total area of 314,284,941 km^2, the land masses now occupy around 43% of the Earth’s surface.
This leaves only 179,304,941 km^2, for the water surface of their world.

This means that they expect us not to notice that 181,827,059 km^2 (over 50%) of the water surface on earth is missing.
Even giving them the benefit of the hemispherical model, they still expect nobody to notice a difference of 59,248,941 km^2 of water surface between the 2 models. To put it into perspective, that’s an area equivalent to over 6 times the size of the USA.
Which we’re not meant to notice.

But hang on, couldn’t Antarctica account for that area?
Well…. No.
That would mean that the radius from the center of the flat Earth to Antarctica was 9,012 km^2.
The accepted distance between South Africa and Antarctica is around 4000 km, leaving the Southern tip of Africa to be around 5000km from the center. However, Africa is 8000 km from its northernmost tip to its southernmost – so now not only does the southern sea have to squeeze into a space of 1000 km (1/4 it’s accepted size), it even has to share that available space with Europe to the north of Africa.
That is a remarkable feat.
The problem is that whatever area we give Antarctica on the rim of the flat Earth, this compresses the area left over inside for the rest of the Earth to take up. This means we would notice a difference in the meridional lengths of all the land masses.

 

Things get even funnier when we use the second hemisphere and flat circle from the last post, in order to stop this compression of the meridian lengths of the land masses.

Our land masses now comprise only just over 13% of the total surface area of the hemisphere and under 12% of the total surface area of the flat Earth. That translates to the seas taking up 884,020,000 km^2 of the hemisphere’s surface area and 999373721.55 km^2 of the flat Earth’s surface area – a difference of 115,353,721.55 km^2, or just under 12 times the area of the USA. Which, again, we’re not meant to notice somehow.

Further more, that’s a whopping difference between this flat Earth and the spherical earth of 638,241,721.55 km^2 – making the water surface area of this flat Earth over 2.7 times as large as it is on the spherical Earth. This is an area the size of 65 USA’s! Again, flat Earthers don’t think we’d notice the effect this extra surface water area would have on the distance between the land masses.

Clearly the distances between land masses will be significantly different between the 2 models, no matter how flat Earthers will try and swing it.

We shall continue to explore the problems with this claim of Iovandrake’s and go on later to show how we can take simple measurements to determine which model is correct.

Iovandrake & the First Basic Law of Shapes

Well, I recently had a fun conversation with a flat earth proponent – aren’t they always fun?

It began with the usual heaping on of the furtive fallacy, bizarre and unfounded ad hominem attacks, appeals to motive and the egregious claim that no good debunkings of the flat Earth theories exist (especially those espoused by that great modern philosopher, Rory Cooper – see his mind-effluent here: https://www.youtube.com/user/Rorycoopervids).

After a long process of providing argument after argument and evidence after evidence, the veritable titan of wisdom (that is, Iovandrake), produced the most humourous line I’ve ever read:Image

“The distance between any place on the earth is the same regardless of whether it is a flat circle or a sphere.”

 

Now, I shouldn’t have to work too hard to point out the major flaws in this argument. But let’s do it anyway….

 

Anybody whose understanding of shapes, when they were a child, went beyond just trying to put the funny looking blocks into their mouths and instead led to them trying to fit cubes into square holes, should have a simple grasp of geometry.

Indeed, anyone with a basic grasp of geometry should have learned an important fact, which we can happily call “The First Basic Law Of Shapes”: In order for the distances between ALL points on any 2 objects to be the same, those objects have to be congruent – that is to say, they have to be geometrically identical, which means the exact same size and shape. Don’t confuse this with genetically identical, which is what I’m betting Iovandrake’s parents were….

After pointing this out, and explaining that he must then believe circles and spheres are the exact same thing, I was told that this wasn’t what he meant…. before he went on to make the exact same argument. Repeatedly.

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In fact, what Iovandrake was proposing was that he’d found a way to do something that the entire human race has found to be impossible.

The example is simple. Look at a flat map of the world. Now look at a globe. You should be able to see a big difference, other than one being flat and the other being spherical. The land masses and the seas are not quite the same.

Cartographers and navigators have known about this for a long time. It is impossible to make a true flat representation of a spheroid.

Anybody who uses hiking maps will know about a little key on those maps, pointing to grid north, true north and magnetic north – and they will know that these maps differ in how big the angle between grid north and true north is. This is because, even though these are the closest representations of the surface of the Earth, if you stitched together every map of the same scale from around the world (if you could get ones for the oceans as well as lands, that is), you wouldn’t get a true representation of the surface of the Earth – or even of the distances as portrayed from one map to the next.

But Iovandrake thinks it is possible to take a sphere and a circle (hell, I’ll even give him just a hemisphere and a circle if he likes, it doesn’t change a thing) and place 2 maps upon them that perfectly agree with each other.

Iovandrake’s conjecture seems to be a common one among flat Earth proponents, eager to insist that whatever measurements we can make to supposedly prove a round Earth, can also fit a flat Earth. In this regard, they are not unlike Geocentrists, who also like to claim that any astronomical predictions made using a heliocentric model would be the same using a geocentric model, without backing that claim up by showing their workings.

 

What can I say?

Well, some basic geometry may help.

To begin with, we can demonstrate that on a sphere you can have 2 circumnavigational distances of the same length. You can take one circuit around the Earth tracing all points equidistant from one pole, and find it has the same length as a circuit around the Earth tracing all points the exact same distance from the other pole.

On a flat earth no 2 circumnavigational routes (tracing all points equidistant from the center) can be the same length. Simply put, anyone traveling around the world in the Southern hemisphere will find it takes them less time than traveling around the earth at the equator, and yet they can find a line of latitude in the northern hemisphere where their route takes the same amount of time. This is impossible on a flat Earth.

Now, some flat Earthers try to claim that nobody has traveled such routes, so how could we know – and some even seem to claim that we couldn’t tell if the antarctic actually lies at what we would call the equator, blocking our passage to what we would like to think of as the southern hemisphere.

As I will show below, none of that matters in order to prove Iovandrake’s conjecture to be wrong.

In order for the distances between all points on any 2 objects to be the same, they must necessarily have the same area. NB, we’re only talking about points on the surface of the shapes here, so volume doesn’t matter. Though it’d be worse if it did, because flat shapes have no volume by definition. They must also have the same dimensions – all their angles and edges must be equal, which in this case means that the circumference of the hemisphere and the circumference of the circle must be equal.

So, if we just assume a hemisphere for a moment (spheres get even more messy for Iovandrake’s conjecture), then we shall need to calculate the quarter circumference, which would be the same as the distance between any point on the hemisphere’s rim and its top. This will be analogous to the radius of the flat circle. The circumference is 2 x pi x r, just like a circle, so the quarter circumference will be 1/2 x pi x r.

Now, the area of a circle is pi x r^2 – as I hope we all remember.

The area of a hemisphere (not including it’s base), is naturally half the area of a circle (4 x pi x r^2), which makes it 2 x pi x r^2.

Hmmm, I can spot a problem coming up already. But let’s go with it.

Let’s give the Hemisphere a radius of 100, nice and simple (units don’t matter here, we could be talking about miles or millimeters for all we care). That gives it a circumference of 628.32.

Let’s take a quarter of that, which is 157.08. This is the distance along the surface from the top to any point on the rim. So, this distance must be equal to the distance from the center of the flat circle to any point on its rim, so we’ll make this our flat circle’s radius.

Spot the problem yet? It will become very clear soon.

Now, calculating the area of our flat circle, we find it has an area of 77516.05.

OK, so let’s calculate our hemisphere’s area (again, not counting the base), which will be half the sphere’s area. Remember, its radius is 100. This gives us an area for the sphere of 125663.706, which we halve to get the hemisphere’s area of 62831.853.

Hmmm, we have a problem, don’t we? Yep, the flat circle doesn’t have the same area as the hemisphere, which means that all the distances between all the points on the 2 shapes cannot be the same. In fact, apart from the distances between the centers and any points, none of the distances are the same – and we can prove it.

As we’ve calculated, the hemisphere has a circumference of 628.32, but the flat circle has a circumference of 986.96.

This means that all the distances between all the points are larger on a flat circle than they are on the hemisphere, except for the distances between the center and any points on either shape. In fact, as the circumference of the circle is over 1.5 times that of the hemisphere and the area of the circle is over 20% larger than the area of the hemisphere (and these ratios will be constant no matter what number we wish to begin with), the difference in the distances is going to be obvious and measurable.

Now, some may argue that the Earth is an oblate spheroid. Well, that’s OK, we just calculate for the oval of the polar circumference. The equatorial radius of the Earth is 6378.1 km. The polar radius is 6356.8 km. The more astute among you will realise from this alone that it’s not going to magically make this flat Earther’s claim true.

For one thing, the polar circumference is then 40008 km. Quarter this again for our distance from one pole to a point on the equator, and we get 10002 km – which must be the same as the flat Earth’s radius (the distance from the center to a point on the rim). But we have a problem straight away. The circumference of the flat earth is now 62844.42 km, whereas the circumference of the Earth with an equatorial radius of 6378.1 km is 40075 km.

Hmmm, we have a big problem, don’t we. The distances here are still not the same.

Neither are the areas.

The area of the flat Earth clocks in here at 314284941 km^2.

The given area of the Earth with the above dimensions is roughly 510,072,000km^2, again halving for the hemisphere area to give us 255,036,000 km^2. Still nowhere near.

For those interested, the equation for an oblate spheroid is:

2 x pi x a^2 (1+(1-e^2/e) x tanh^-1(e)), where e^2 = 1 – a^2/b^2.

a is the semimajor axis (the equatorial radius, here) and b is the semiminor axis (the polar radius).

And the equation for the circumference of an oval (the perimeter of an ellipse) is roughly:

pi x [3(a+b) – sqrt((3a+b)(a+3b))]

 

Now, even more people may complain that I’ve only taken half the earth here. Well, let’s see what happens when we take the accepted distance from the North Pole to Antarctica, and turn that into a hemisphere (as it would seem the flat Earthers want us to do) and also translate it to a flat Earth.

We’ll take a ring 1000 km from the South Pole to be the hemisphere’s and the circle’s rim.

This would give our hemisphere a polar radius of about 12713.6 km a quarter polar circumference of about 19,002 km, an equatorial radius of about 12756.2 km, an equatorial circumference of about 80149.57 km, and an area (excluding the base, again) of around 1,019,000,000 km^2.

Give our flat circle the quarter circumference of 19,002 for its radius (because the distance on the surface from the top of the hemisphere to the rim must equal the radius of the flat circle), and we find it has a circumference of 119,393.08km and an area of 1,134,353,721.55 km^2.

 

Well, what can we see? There’s still a difference of around 39,ooo km in their circumferences and there’s still a difference of over 115,000,000 km^2 between their areas – that’s an area over 11 times the size of the USA. And we wouldn’t notice that?

Now we can compare this flat Earth to the actual spherical Earth and find that the area is over twice as big – and the longest circumference on the Earth is less than half as long as the circumference of the flat Earth now.

In fact, we can also notice that there is a massive difference between the spherical Earth and the hemisphere. The reason for this is that hemispheres and spheres are not congruent either, thus we would find it easy to spot whether everywhere we’ve explored is the whole of the spherical Earth as opposed to just down to the equator – and it would be just as easy as discerning if we were on a flat or a round Earth.

 

So, we can safely assert that it is impossible for the statement “The distance between any place on the earth is the same regardless of whether it is a flat circle or a sphere” to be true. We can also safely assert that distances on land masses and the distances between them would be very different, and that we would be able to easily calculate the difference between these distances.  Not only that, but by measuring these distances, we should be able to determine which model is correct, without the hassle of going into space.

I shall endeavour to go further into this in a later post.

Suffice to say, for the moment, that we can easily state that when Iovandrake claims that “there is no difference between REALITY for either a flat earth or a spherical one”, he is clearly demonstrating that he hasn’t got even a child’s grasp of geometry and he has no idea about The First Basic Law Of Shapes.

This, unsurprisingly, seems to be a recurring theme in flat Earth claims.

Jack Flacco

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