Archive | April 2014

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.

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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.

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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.

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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.

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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.

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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.

My first blog

Well, I’m finally breaking my blogging virginity – and I expect the experience to be just as awkward and leave me with just as much of an anti-climax, mixed with disorientation and confusion about what’s just happened when it’s over, as when I lost my actual virginity.

What to expect? Hmmmm…. I’m not wanting to overexcite the reader with incredible promises of lyrical delight, only to leave them deflated like a child’s balloon that’s been mistakenly taken by the pet dog as a sex toy.

All I’ll say is that there will be short stories, occasional comedy (disclaimer: this is solely contingent on a subjective interpretation of “comedy”), political musings, bits of what I think counts as news and editorial, and a look at some of the more egregious claims of pseudo-scientists.

Oh, and lot’s of long sentences that would be better served by being broken up with full stops – leaving the reader breathless as they try and keep up. I promise nothing less than an endurance test, and warn smokers and people suffering from asthma not to try and read along.

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