Whatever explanation is the least exciting is the most likely to be true.
THE DAILY GUANO
The political right wing of the UK wins dramatic vote against the political right wing of the UK
Today, the electorate of the United Kingdom took part in a dramatic vote, favouring the right wing over the right wing.
In an historic referendum, the entire populous of a small shitty island off the coast of Europe decided to reject the arguments of the controlling right wing party, in favour of the arguments of the controlling right wing party.
In a decision that has dumbfounded everyone from pollsters to bookmakers to stock market investors to everyone on the globe, the population of the United Kingdom has astonishingly decided that the arguments of the right wing parties were more compelling than the arguments of the right wing parties.
Apparently there were people in the UK who identified as left wing, including an entire political party who apparently are the second largest party in the country, but their response to the entire referendum pre-, during and post- referendum has been “Ummmmmmmmmmmmmmmmm….. Blurgle wurgle, fugrle, turgle?”, before bashing their head against a wall whilst dribbling into a cup of tea and feeling guilty about how the tea trade signifies the greatest expansion of British imperialism in the history of the kingdom.
The political right in the small pissy island have been unsurprisingly unabashed in their celebrations of their victory against the political right, saying “NER NER! I told you we’d win no matter how the vote went!”
The entire populous has been said to feel both ecstatic and entirely betrayed across the board. I mean, every single person we spoke to said that they were both ecstatic and felt betrayed.
We tried to get a member of parliament from any political party to comment on the referendum, but the best we could find at 8am was a drunken old man collapsed outside the houses of parliament, who was both singing “Rule Britannia” whilst simultaneously crying into a pint of Stella Artois that they said “will probably cost even more from now on, but at least it won’t be subject to those fucking EU rules that it’s going to be subjected to anyway!”
We’re unable to confirm which party they belonged to, but it looks like it doesn’t really matter any way.
We also tried to get someone from the political left in the UK to comment, but they were surprisingly nowhere to be found.
I mean, literally – we kept asking people if they were left wing, and they kept showing us their middle finger.
We have no idea what to make of this. At all.
THE DAILY GUANO
“NOT HAVING CHILDREN IS SELFISH”, SAYS MAN WHO TOOK A VOW NEVER TO HAVE CHILDREN
The entire planet was left agasp in a fit of irony, as Pope Francis – the head of the Catholic church, which requires all its clergy to take a vow of celibacy, and who himself has taken this vow and will never have children – declared that not having children was a selfish act.
In what some members of the Catholic priesthood are calling an “impassioned sermon, right up there with anything the big JC ever said, innit”, the old white man who controls the seat of power of a religion populated by women and men who vow never to have sex, in order to make sure they aren’t distracted from their own quest for personal salvation, stated that people who don’t have children are being self-centered and creating a “depressed society”, as the entire world tried not to hold up a mirror the size of St Peter’s Square whilst pissing themselves.
Then the Pope began listing everything that the priesthood does, says and stands for as “selfish”, without even a hint of irony in his eyes.
Several clergy members had to jump on stage and wrestle the Pope to the floor in a dramatic scene, as he began spouting that acquiring lots of wealth and land, advising people not to follow scientific medical advice, and persecuting people were tearing society apart – and that spending all your time immersed in a cult centered around an imaginary friend who you think you can talk to, was not a substitute for going out and making the world a better place.
He was finally dragged away from the microphone and tranquilized just as he began to state that those caught having sex with children should feel the full force of justice.We tried to get someone from the Vatican to comment, but we ended up getting tranquilized and sold into slavery.
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”.
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.
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.
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.
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”.
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.
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.
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.
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.
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.
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.
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.
THE DAILY GUANO
HUMAN BODY NOT DESIGNED FOR BREATHING, SAYS MEXICAN CHURCH
Since proclaiming that the human body is not designed for homosexuality, earlier in the week, the Catholic webshite Desde la fe (Latin for “we’re a bunch of scientifically illiterate, out of date, bigoted fuckwits who frankly are past our use by date and should have done the decent thing and stopped stealing oxygen by now”. Wonderful language, Latin – it’s amazing how much can be expressed in such a few words) has declared that the human body is also not designed for breathing.
In an editorial titled, “Fuck me, the cardinal’s really losing his marbles now lads”, the church stated that oxygen has been linked to the aging process and bodily decay, and that this is proof that breathing is damaging to both the individual and society as a whole, urging its adherents to both protest any law that allows people to breathe, as well as abstaining from breathing themselves.
Shortly after this statement, one of Mexico’s leading bishops went live on air to announce that the Catholic church was taking a strong stand against sunlight, declaring the human body is not designed for living on a planet that orbited a massive star radiating trillions of trillions of joules of energy every day, citing the damage caused by sunburn and the extreme risk of cancer.
In what many unenlightened Catholics are calling a “landmark statement”, the bishop stated that all Catholics should go live underground, and urged the Vatican to begin building a “star destroyer” to eradicate the threat to human society once and for all, and told followers to petition their elected representatives to take firm actions against the offending star.
Things have finally escalated this evening, as Cardinal Norberto Rivera declared that the human body was not designed for existing in this universe, since it seems that everything is pretty much out to kill us or do us harm of some form.
He has urged all Catholics to refrain from existing in this universe, and suggested that they should start petitioning the government to stop passing legislation that is designed to make it in any way easier for people to exist in this universe.
We attempted to get hold of a spokesperson from the church to comment on the issue, but everyone was too busy running away and screaming at everything they saw.
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?
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.
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.”
“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.
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….
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.
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.
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.
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….
THE DAILY GUANO
USE OF HYPERBOLIC LANGUAGE “MURDERING OUR CHILDREN WHILE THEY SLEEP”, SAYS SENIOR GOVERNMENT FIGURE
Today, somebody who does something in the government, that we can report is very important and makes them “senior”, has called out the growing use of hyperbolic language in society and the damaging effects it is clearly having.
Sir Somebody-or-other (who gives a fuck?) said, “It began with children exaggerating in the playground about how strong their parents are, and slowly progressed to fishermen talking about the size of their catch and/or penises. Before long it found its way into our media, with every new film being called ‘the greatest piece of shit you’ll ever see in your life, ever’. Clearly, the use of hyperbolic language has become a pandemic. It is highly contagious. I can’t understate this – it is more contagious than cancer and AIDS. I know that cancer isn’t contagious, and AIDS isn’t contagious but is a syndrome brought about by a virus that itself is contagious, but what we’re basically dealing with here is cancer times AIDS squared, plus Hemorrhoids.”
The senior source went on, “I don’t think it is overdoing it to say that this pandemic is holding a pillow over our children’s mouths whilst they are sleeping, and will bring about the death of civilization as we know it. Something has to be done!”
We tried to get a spokesperson for No 10 to speak to us about the issue, but we were informed that they were too busy working on what is set to be “the greatest motherfucking piece of legislation this motherfucking world has ever seen. Like, seriously, this is like the Civil rights acts meet the Emancipation Declaration, which then all have sex with the Clean Air Act, and splices the offspring with the Representation of the People (Equal Franchise) Act”
We can confirm that the source was talking about a new act to lower the fees paid by people who moor their canal boats, by 20 pence.
I apologize right now for the horrendous amount of geometry involved in this post.
Somebody recently asked me to check their work relating to the curvature of the Earth, which prompted me to make this post – not just in answer to them, but also because I’ve seen a lot of people talk about the curvature of the Earth, most of whom don’t really get the whole picture correct. So I thought I’d show how the geometry works.
Also, I’ve seen many flat Earth proponents completely muck up this whole question, which isn’t surprising, considering their ability to grasp basic geometry is below that of a pre-schooler.
I’ll use the original question I was given, which asked how far below the horizon an object’s base would be when it is 10 miles away from us. In going through this, we’ll look at the height of an observer above the surface of the Earth, the height of the shortest object someone can see at this height (assuming a perfectly spherical Earth and discounting atmospheric refraction – very important points), and both the amount the Earth has curved away from the observer at this distance and how far below the horizon the base of the object is (which, we will see, are 2 different questions).
The original question looked at someone standing 6 feet above the surface of the Earth (for the sake of this post, we’re going to make 6 feet the altitude of the observer’s eyes), and how far an object 10 miles away over the surface of Earth would be below their horizon.
Firstly, I want to bring up the problem of precision.
We should give our results to only 1 or 2 decimal places at most, especially where distance values are concerned. This is important, because we have to acknowledge that we’re dealing with approximations here. The Earth isn’t a perfect sphere, and even the radius we’re using is an approximate value, no matter how precise we try to be (indeed, because it’s not a perfect sphere, there is a limit as to how precise we can be anyway).
Now, the thing you have to remember is that the 10 miles in this question is the arc length over the surface of the Earth, and not the length of the tangent from the surface of the Earth to the top of the object (the distance between our eyes and the top of the object), nor is it the distance between our eyes and the base of the object – as it is treated to be when we just use the Pythagorean theorem to calculate the value.
Thanks to the Earth’s massive curvature, the difference between these values is negligible over short distances when we are standing close to the ground, but if we fall for over precision we’re going to give an answer that is unwittingly inaccurate when we think it’s precise.
For short distances over the Earth’s surface, this approximation (just using the Pythagorean theorem involving the radius of the Earth plus the height of our eyes and the distance over the surface of the Earth pretending it’s these same as the distance between our eyes and the object we’re looking at) is fine – but if we make the answer over-precise, and we try to extrapolate this result over larger distances, we fall into the trap of making larger errors as the distances involved increase.
And again, I can’t overstate this fact: Even when we use the more precise formulas, we need to guard against giving a false impression of precision, as we will easily forget that the Earth isn’t a perfect sphere, and also that the radius of the Earth we’re using is an approximation.
To go through the geometry better, I’m going to try to make the problem visual:
Now when we talk about how much the Earth curves, there are 2 ways of looking at it.
The first is to think of ourselves on the top of a hill, and to measure how much the Earth drops from the tangent line that touches the point on the Earth where we are standing.
The second is to think of the horizon as being the peak of a hill, and to ask how much the Earth drops beneath the tangent line at the horizon, which is where the original question is going.
In this diagram, “x” is the height of our eyes above the surface of the Earth, and “y” is the height of the shortest object we can see the top of over the distance involved (10 miles).
Now, again, the first thing we have to note is that the 10 miles in question is an arc length over the surface of the Earth, given the Earth’s radius “r”. We’ll call this arc length “c”.
The distance between the spot on the Earth we’re standing on and the spot the object in question is standing on, is the arc width “n”. To figure this out, we need to first calculate the angle “C” by finding the circumference (2Pi*r) and then dividing the arc length by this circumference and multiplying the result by 360.
Let me just mention the Sine rule very quickly. It states that a/sin(A)=b/sin(B)=c/sin(C), where a, b and c are the sides of the triangle opposite the angles A, B and C, respectively. It is also therefore true that sin(A)/a=sin(B)/b=sin(C)/c.
This means that if we know A, b and B, we can find “a” by rearranging the equation a/sin(A)=b/(sinB) to a=b*sin(A)/sin(B).
Equally, if we know a, b and B, we can find “A” by rearranging the equation sin(A)/a=sin(B)/b to A=arcsin((a*sin(B))/b).
Importantly, the Sine of 90 degrees is 1, which becomes very useful when working with right angled triangles, because we can discount any terms of sin(90), because multiplying or dividing by 1 is a redundant process.
Now, we know that the triangle formed by the 2 radii and the arc width is an isosceles triangle by definition. 2 of its angles will have the same value, because 2 of its sides have the same value. Every triangle we make inside a circle, where 2 of its sides are defined by the radius of the circle will always, by definition, be an isosceles triangle.
We know that we can cut these triangles perfectly in half and end up with 2 right angled triangles. By cutting the angle “C” in half, we’ll put a line through this triangle that sits perpendicular to the line “n” and cuts it perfectly in half, that’s all we need to know here.
This means we can figure out what the value of n/2 is, using the Sine rule:
Remember that sin(90)=1, so we can rearrange this equation to find:
n/2 = sin(C/2)*r, so n = 2*sin(C/2)*r.
With this information, we can figure out how much the Earth’s surface drops from our position, which is the distance “d”.
Hopefully you’ve realised that the length “m” is the same length as the other dark blue line. This is useful, because we can find the value of m with the equation sin(C)*r, again thanks to the Sine rule.
We now have the hypotenuse and one of the other sides for the triangle “mnd”, so we can find the value of “d” (the amount the Earth has “dropped” away from the point where we are standing) by rearranging Pythagoras’ theorem: d=sqrt(n^2-m^2).
(remember that Pythagoras’ theorem states that the square of the hypotenuse is equal to the sum of the squares of the other 2 sides – which in this case means n^2=m^2+d^2. This means that d^2=n^2-m^2, so d=sqrt(n^2-m^2).)
Nice and simple.
Now we want to ask the question relating to how far below the horizon the base of the object is. This is the distance “h”.
There are 2 ways of approaching this problem. We’re going to look at the way to approach it in terms of an object we can see at least the top of over the horizon, first, so we can talk about it in terms of the height of the object itself.
Then we’re going to use the more general (and simpler) version, which will describe all objects – even those we can’t and can never see over the horizon – at the bottom of the post.
Let’s go through the first method – finding the distance below the horizon the base of an object is when we can see its top.
To do this we first want to find the height “y”, which is the shortest height this object can be for us to see its top over the horizon from this distance.
We begin by finding the distance “e” (the distance from our eyes to the horizon), which is sqrt((r+x)^2-r^2). Again this is just rearranging the Pythagorean theorem.
Now we’ve found “e”, we can find the angle “A”, which is arcsin(e/(r+x)), since the Sine of “A” is defined as the ratio of the opposite divided by the hypotenuse (in this case e/(r+y)).
Now we have the angle “A”, we can get the angle “B” as simply C-A.
With the angle “B”, we can now find the angle “T” as 90-B, and the angle “S” as 90-A.
We can again use the Sine rule to find the value “r+y”, which will equal (r+x)*sin(S)/sin(T). To find the value of “y”, we just take the radius away from this result (giving us ((r+x)*sin(S)/sin(T))-r), and we have the shortest height this object can be for us to be able to see its top over the horizon (actually the height at which its top disappears with the horizon).
Now we have the angle of “T”, we can see that the triangle that “h” is part of is just a similar triangle to the larger triangle (the scale is different, but the angles, and thus the ratios of the lengths of the sides, are all the same. When shapes have the same angles as each other, but are different scales, they are called “similar”, because the ratios of their sides and their angles will remain the same).
We can find the value of “h” by simply multiplying “r” by the ratio of y/(r+y), or we can use the Sine rule again, and find that h = sin(T)*y, which is exactly the same thing. Hopefully that makes sense. Remember that sin(T) = r/(r+y), so sin(T)*y is y*r/(r+y).
Again, even though this is the actual geometry behind the question(s), I hope the reader will notice 2 important things:
Firstly, we are still going to fall for the trap of over precision when we try to act as if the Earth is a perfect sphere, when it isn’t, and when we’re dealing with arbitrary distances as opposed to talking algebraically. Pick a number of significant figures and stick to using that and rounding to it. If the final result is going to be given in feet, then convert all other distances to feet (such as the radius of the Earth and the arc length of the distance between you and the object), as this will prevent us from getting larger errors due to rounding. I normally work in either meters or centimeters, but I did this in feet for our American friends 😉
Secondly, and most importantly, I hope the reader can see that the lower the value of “x”, and the lower the arc length “c”, then the closer the values of “h”, “d” and “y” come to each other – which is why using the approximation with Pythagoras is OK over short distances, but even more highlights why you have to acknowledge the approximation of the result. What we shall see, as well, is that even over a distance of 10 miles, the differences between the values of “d” and “h” and “y” are quite significant, though the difference between the values of “h” and “y” and insignificant.
Now. Let’s go through it and show what the answer would be if the Earth was a perfect sphere with a radius of 3963 miles, and the ground distance between the observer and the object was 10 miles (and discounting atmospheric refraction), and how we get the results.
NOTE: The illustrations below are not to scale, as the scales involved will be impossible to adequately portray. As such, the illustrations are purely schematic.
Firstly, I’m going to convert the miles to feet, so I’m only working in those units.
3963 miles = 20924640 feet, which is the radius “r”.
10 miles = 52800 feet, which is the arc length “c”.
I’m also going to round to 2 decimal places (more for the angles, though I shouldn’t to be honest, but it will help highlight something further on ), because that’s accurate enough for our needs.
Now I have the arc length “c” and the radius “r”, assuming a perfectly spherical Earth.
The circumference of the Earth is then 2pi*20924640 = 131473390.61 feet.
The arc length of 10 miles then relates to an angle of 52800/131473390.61*360 = 0.14458 degrees
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.
Hopefully, something has stood out to those of you paying attention. The arc length “c” should be longer than either of the lengths “m” or “n”.The reason they are larger is due to the fact that we have rounded the values for the angles.As we’ll see further below, the difference is insignificant (both values are only out by about 2 feet, putting the value for “d” out by less than 2 feet), and the problem is just one of rounding when it comes to the angles themselves.
Now for how much it is “below the horizon”, viewing the horizon like the peak of a hill.
The distance “e” is sqrt(20924646^2-20924640^2) = 15846 feet.
The angle “A” is then arcsin(15846/20924646) = 0.04339 degrees.
The angle “B” is 0.14458-0.04339 = 0.10119 degrees.
The angle “T” is 90-0.10119 = 89.89881 degrees.
The angle “S” is 90-0.04339 = 89.95661 degrees.
The distance “r+y” is therefore:
(20924646)*sin(89.95661)/sin(89.89881) = 20924672.63 feet.
So the distance “y” (the shortest height the object can be in order for us to see its top over the horizon), is 20924672.63-20924640 = 32.63 feet.
Finally, we can find the value “h” (the distance the base of the object “drops” below the “hill” of the horizon):
sin(89.89881)*32.63 = 32.63 feet (the Sine of 89.89881 degrees is so close to 1 that this is exactly what we expect it to be).
We can check this with the ratio method: 20924640*32.63/20924672.63 = 32.63 feet.
So, there we go.
The amount the base of the object is “below” the point of the surface of the Earth we’re standing on is around 65 feet, in this example, and the amount the base of the object is “below” the horizon is about 33 feet.
The lowest height this object can be so that we can see its top is about 33 feet.
Remember though, that we live on a planet which is not a perfect sphere and which has an atmosphere, and that light refracts as it moves through any region of space that isn’t a vacuum, which affects the distance we can see things, and allows us to see objects that are geometrically over the horizon. If the Earth was a perfect sphere given these dimensions and had no atmosphere, or atmospheric refraction didn’t exist, then this geometry would define exactly how far we can see.
A lot of imprecision is brought in the more steps you make, because with each step you are rounding the result.
The more astute will realise that the final equations can be expanded and written solely in terms of the values we’re given at the start of the exercise, namely “x”, “c” and “r” (or our height above the surface of the Earth, the radius of the Earth and the ground distance between ourselves and the object).
In this way, the amount a perfectly spherical Earth “drops” as you move across the surface (“d”) is:
and the amount the base of an object “drops” below the horizon is:
Plugging in the numbers for the example we have into the first equation to find “d”, we get:
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:
Which gives us 32.63 feet to 2 decimal places. Exactly the same result as above.
Now, the more astute among you will notice that the second equation breaks down at the points where C=180-S and C=180+S. This is because B=90 at these points, and as T=90-B, then T=0. The sine of 0 is 0, which means the equation to find “y” ends up with you dividing by 0.
This happens because the “y” term in the second equation – the shortest height an object can be in order for you to see the top of it over the horizon) is only meaningful for when C180+S.
In the region between C/=180-S, no matter what height the object is, the vector describing its height is divergent from your line of sight – which is described by the tangent line defined by extending “e” indefinitely. This means you will never see the top of the object, no matter how tall the object is.
At the points C=180-S and C=180+S the vector describing the height of any object is always parallel to your line of sight – the tangent line defined by extending “e” indefinitely – so again you will never see its top no matter how tall the object is.
At the points C=180-S and C=180+S, the “h” term loses its meaning in the context of the “y” term in this manner, and the “h” term becomes the value “r”, for reasons that should be obvious.
This is because the angle between the radius to the base of the object and the radius to the point of your horizon (the angle “B”) as we’ve pointed out is 90 degrees.
That means they form a right angled isosceles triangle with the arc width. Right angled isosceles triangles are what you get when you cut a perfect square diagonally in half – the 2 sides are equal and to non-right angles are both 45 degrees.
Using the same method as in the first equation to find “d”, this time to find “h”, we can easily see that “h” is equal to “r” (though this should be obvious anyway).
Of course, this demonstrates that the second equation is a short hand method for finding the solution, which breaks down at 2 specific points.
So, we’re going to go through the general method for answering this question in terms of any object, not just ones that we will be able to see above the horizon.
This method for finding the solution to the second problem is pretty much the same method we used to solve the first problem, but with a small adjustment.
This time, when we’ve found the angle “B”, we calculate the arc width of the angle, which is 2*sin(B/2)*r. We’ll call this “f”
Next, we want to calculate the length “g”, which is of course sin(B)*r.
With these we can find that h=sqrt(f^2-g^2).
We can check this result with the example figures we’re already given:
The angle “B” = 0.10119
The arc width “f” is 2*sin(0.10119/2)*20924640 = 36954.974 feet
The length “g” is sin(0.10119)*20924640 = 36954.96 feet.
The distance below the horizon “h” is sqrt(36954.974^2-36954.96^2) = 32.61 feet
Yes, this is the simpler version, but the first version expresses how far below the horizon an object you can see is.
(Also, though, I had to round the distances involved to 3 decimal places this time, as the rounding errors become more pronounced. Rounding to 2 decimal places gave the result for “h” as sqrt(36954.97^2-36954.96^2) = 27.19 feet.)
But what does this method look like when C>180-S?
Well, as with before, when we get the value of the angle “B”, we can find the value of the arc width “f” to be 2*sin(B/2)*r.
This time, the length “g” is still equal to sin(B)r, but why? This seems strange. We could see why this was the case in an example where C<180-S, because the lengths “g” and “r” formed part of a triangle that the angle “B” was inside of.
This time though, the lengths “g” and “r” form a triangle that seems to cut through the angle “B”.
So what’s going on?
Well, the angle I want to draw your attention to is the angle “K”, at the bottom.
Surely “g” should now equal sin(K)r? Well, it does.
What you should notice is that this angle is the supplementary angle to “B” – it has the exact same value as the angle that supplements “B” along this line (the angle “A+J”).
Well, the sines of supplementary angles are the same, and so sin(K) = sin(B). So yes, the value of “g” is sin(K)r which is exactly equal to sin(B)r.
There is another important reason why I wanted to go through the first method, though.
I’ve said above that the “y” term is only meaningful when C180+S. So what does that mean for the region between C>180-S and C<180+S (discounting the points where C=180=S and C=180-S)?
In this region, the building, no matter how tall it is, will always be diverging from the tangent line that defines your line of sight. So why does the equation still work in this region? What does “y” end up meaning?
Well, we can illustrate what happens very simply:
By extending the vector describing the height of the object, through the radius of the Earth and indefinitely onwards, we can see this line intersects the tangent line describing our line of sight at a point behind us. This intersection now defines the value of “T”, and also the value of “y”
Since the angle “T” is measured counter-clockwise from the tangent line (as you can see in the first example), we can see the angle “T” is now clockwise from the tangent line, giving it a negative value. The sine of an angle with a value of -n degrees is the same as the value of an angle of 360-n degrees (and an angle of 180+n degrees). In effect, as we can see, if we measure the angle “T” counter-clockwise as we have in the first example, we get the value 360-T.
OK, that’s how it defines “T”, but how does it define “y”?
Well, just as the angle “T” was measured counter-clockwise from the tangent line, making its value become negative when it became a clockwise angle, so to is the value of “y” positive when it is below the tangent line and negative when above it.
This is why the second equation works, even when C>180-S, as long as C doesn’t equal 180+S or 180-S.
THE DAILY GUANO
NIGEL FARAGE BLAMES THE UK’S BIASED LEFT-WING MEDIA FOR NOT ANNOUNCING HIS PARTY’S STUNNING ELECTION VICTORY
As polls came in thick and fast last night, showing a landslide victory for everyone who wasn’t a UKIP candidate, Nigel
Far Right Farage laid into the entire media industry of the UK for it’s blatant left-wing bias in not pronouncing him the next Prime Minister of the United Kingdom.
After being laughed out of the counting hall, Mr
Fat Rage Farage was seen mumbling to himself, shouting “Commie scum!” at anyone with a camera, and occasionally screaming at a passing cat.
Gathering his supporters in a nearby hotel, Mr
Fartage Farage finally came out to speak to reporters, saying “I just can’t f***ing believe you lot! I’ve clearly won this election, and yet you keep spreading your left wing lies to the British people. You must all be immigrants.”
When asked why the election results clearly showed that the vast majority of people in the UK didn’t even vote for Mr
Fistage Farage’s mob of brown shirts and racist simpletons political party, he said, “Well of course you’d tell people we didn’t win the election, because you hate white people and love immigrants coming over here and having sex with your jobs!”
Unable to accept the fact that he clearly didn’t win the election, and adamant that he had been elected into the country’s highest high office, he continued, “It’s all this left-wing, hippie bias in the BBC. I’m your f***ing Prime Minister, you simpering ingrates! ME! Worship me! Get on your knees and bow to your fuehrer!”
After which he began spasming wildly and foaming at the mouth, and had to be restrained from trying to bite an old lady walking past in the street.
We asked people in South Thanet and across the country why they didn’t vote for UKIP, and every single one of them said they couldn’t vote for someone who looked like they’d been kicked out of the muppets for inciting racial hatred.
We tried to contact a spokesperson for UKIP to comment on the result and Mr
Garbage Farage’s mental state, but they all just started goosestepping and throwing Nazi salutes everywhere.