It’s April

This is really old news, but I have to announce that it’s April. I still don’t really have anything useful to add to the blog other than just advertising that I’m still occasionally streaming on Twitch. These streams are regularly uploaded to my YouTube channel for those interesting in watching my streams after the fact.

Is this where I ask people to “liek coment adn subscrib”?

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Fallout  7đź’©

There’s this one game that’s been out for a while, and I’ve heard increasingly bad news about it. Can someone please tell me what the hell is wrong with Bethesda and their shitty, sub-par, pseudo-Fallout-4.5 game, “Fallout 76”?

Let’s start with the basics: Why is it called “Fallout 76”? I honestly don’t know. Is it because, in Fallout lore, the bombs dropped in the year 2076? That’s not even the year they dropped. So, it’s not a more refi– Jesus fucking Christ, I can’t say write that with a straight face. Actually, that’s not true, but are you saying it takes place in the pre-war Fallout universe and is not a more refined version of “60 Seconds!”?

Actually, that’s not even true; the true story is that the story is set in 2102 in an alternate fucking timeline and that the player is a resident of Vault 76. That’s how it is, then? Fine. So, if I’m getting this straight, this is essentially a multiplayer, pseudo-battle-royale pile of feces with minimal plot, extremely dry gameplay and an immensely unstable engine to boot. What a fucking joke. It’s an uninspired cash-grab probably created with the same Gamebryo-successor trash heap as “Fallout 4”. Seriously, if you’re having trouble creating as simple a feature as the possibility of uninstalling the damn game, you’re doing something seriously wrong; this is quite a lot worse than that one time I ranted about “Fallout 4” not having proper out-of-disk-space exception handling and not even being able to handle a simple change of the save file location. The latter is just a string of characters, for fuck’s sake. Did they not learn about this during “Programming 101”?

This is why I have stopped jumping on any hype train for anything at all; “Guild Wars 2“, “WildStar” and “No Man’s Sky“, although I never bought the lattermost, have all desensitized me to any kind of hype, and color me surprised if it isn’t one of those times again. Kiss my fucking ass, Bethesda.

I won’t even ramble on about “Diablo: Immortal”; that was just an utter trainwreck and three quarters, and I have had just about enough trainwrecks for one month. Hold on: is it spelled with or without the colon? Actually, never mind; I don’t care. Also, my semi-colons are all over the place today. That’s quite alright.

New desktop PC

Here’s a blog post straight out of the blue, if ever there was one: I’m getting a new desktop soon – I guess it’s what you’d call a gaming PC. Hopefully, the parts actually fit together; I don’t actually know how to build a computer, as the last computer was pre-built by the company I bought it at, but I’m also not a computer-illiterate moron, hence why I’m currently attending computer science.

Anyway, my greatest concern is getting everything to fit in the case; there are a lot of screws to screw, you know, and the power supply may take up too much space. I’m pretty sure I investigated every specification and compared them in such a thorough manner that there won’t be any issues on this front, but time will tell. There’s also the matter of not blowing the whole thing to pieces during the assembly process, and the operating system – Windows 10, of course; to hell with Linux – has to work properly.

I guess I’ll post further updates. As for videos: I don’t know if I’ll be making any, but we’ll see what I decide to do during Christmas break. I probably should make some Let’s Plays soon; do you have any idea how much more expensive electronic devices are over here? It’s important that I make as much out of my purchase as possible.

A Plain Way of Thinking

There is something I have noticed over the past few months: a phenomenon has spread across YouTube. I have no idea whether this is the case elsewhere on the Internet, but I have noticed this mostly on YouTube.

You know how they call our current age the “Information Age”? I find this funny because there are in fact people who still believe the Earth is fucking flat. One of the most common complaints these imbeciles constantly throw at the spherically inclined crowd is the “fact” that distant objects are not actually obscured by the Earth once they pass the horizon. Rather, the object shrinks in perceived size simply by virtue of the increasing distance between the observer and the object. This, of course, is absolute bollocks; taking great care to make a proper attempt to observe a ship disappear in the distance, for instance, would utterly destroy this notion.

Since these morons seem completely incapable of forming a coherent mathematical argument as to why we should believe their nonsense, I have taken it upon myself to form my own. Please note that, while I have skills beyond a beginner’s in such matters, I am anything but an expert and thus liable to make mistakes. I will accept any corrections sent my way. Additionally, do note that, to the extent of my knowledge, WordPress.com does not allow for text editing beyond a few different titling styles, so there are going to be a few compromises made.

All this out of the way, let’s get on with it.

Problem statement

The problem we would like to solve is the following: if an observer is located at a certain point a given height H above the surface of the Earth, here modeled as a sphere, and a surface distance A away from an observed object, how much of the object would be obscured by the Earth’s surface?

It would be easier if we had a sketch of the problem. I took care to make a rough one on my tablet:

A crude sketch of the obscured-height-problem.

A crude sketch over the problem. Two new angles α and β have been created. These will come in handy later.

What we are interested in is the height h obscured by the horizon as a function of the distance A between the observer and the observed object only; the height is treated as a constant. To solve this problem, I will be using trigonometric functions and the Pythagorean theorem. While this probably sounds scary to the uninitiated plainthinker, the calculations are fairly straightforward once one manages to figure out how to solve this problem.

Deriving the solution

Let’s first figure out how we would describe the two horizon distances in the above images:

d_1 = \\sqrt(H(H+2R))

d_2 = \\sqrt(h(h+2R))

These equations are derived from the fact that both are catheti in each respective right triangle. Deriving equations for both is a matter of using the Pythagorean theorem.

Next, we must find equations for each of the angles α and β. This requires the use of trigonometric functions. In this case, I will be using the sine function. Recall the definition of the sine function:

\\sin\\left(angle\\right)=\\frac{opposite}{hypotenuse}

Using this definition in each right triangle, we can derive the following:

  • For the triangle of angle α:
    sin alpha something something square root and fraction stuffs
    alpha = \\sin^{-1}{\\left(\\frac{\\sqrt{H(H+2R)}}{H+R}\\right)}
  • For the triangle of angle β:
    okay, i actually fixed the beta now. anyway, still something something trigonometry and pythagoras and angles and stuff.
    \\beta = \\sin^{-1}{\\left(\\frac{\\sqrt{h(h+2R)}}{h+R}\\right)}

Here, we simply use the fact that taking the inverse of a function on the function itself returns the parameter of the function. In this case, the inverse sine of the sine function is just the angle – here α and β – supplied to the sine function.

I will now define two new values a_1 and a_2 to be the arc lengths subtended by α and β, respectively. Thus, we have the following:

a_1+a_2=A

How does this help us? Recall that, when we use radians as our unit of angle measurement, we get the following:

  • For a_1:
    \\frac{a_1}{2\\pi R}=\\frac{\\alpha}{2\\pi}
    a_1=\\alpha R
  • For a_2:
    \\frac{a_2}{2\\pi R}=\\frac{\\beta}{2\\pi}
    a_2=\\beta R

This works because the ratio between an arc length and the entire circle is linearly proportional to the ratio between an angle and two times pi.

Next, we can use the previous sum of the two arcs that we defined to be equal to A and plug in the results through substituting the yucky a_1 and a_2 with their respective equivalencies:

A = a_1 + a_2 = \\alpha R + \\beta R = \\left(\\alpha + \\beta\\right)R \\\\ = \\left(\\alpha + \\sin^{-1}\\left(\\frac{\\sqrt{h\\left(h+2R\\right)}}{h+R}\\right)\\right)R

Now, by doing a bit of algebraic manipulation, we can solve for the inverse sine function:

\\sin^{-1}\\left(\\frac{\\sqrt{h\\left(h+2R\\right)}}{h+R}\\right) = \\frac{A}{R} - \\alpha

Using the same trick as we did earlier, we can eradicate ourselves of the inverse sine function and get an equation for the internal fraction, which contains the variable h we’re after:

\\frac{\\sqrt{h\\left(h+2R\\right)}}{h+R} = \\sin\\left(\\frac{A}{R} - \\alpha\\right)

All that remains is to solve for h using simple algebraic manipulations of the single equation above. Simple knowledge on solving irrational and quadratic equations is required, but this is a really simple equation. I will be going through this step by step so as not to lose anyone, here, although getting lost at this point is admittedly a bit odd since this is probably the easiest step. Regardless, let’s just get on with it:

Before working on solving the equation itself, I will be using the following fact:

\\sin\\left(\\frac{A}{R}-\\alpha\\right) = \\sin\\beta

I will let it be up to the reader to determine that this is in fact true. Indeed, I could have skipped the whole part with a_1 and a_2 entirely and just gone straight to this point after determining the sines of the angles, but I decided to keep it this way because that’s how I roll, although, looking back, I really should have done just that so that I wouldn’t have had to deal with the stupid absence of any super- or subscript features in this text editor.

Regardless, let’s simplify things a little:

  1. Get rid of the denominator on the left side through multiplying by it on both sides:
    \\sqrt{h\\left(h+2R\\right)} = \\left(h+R\\right)\\sin\\beta
  2. Square both sides to get rid of the radical:
    \\phantom{X} \\\\ \\sqrt{h\\left(h+2R\\right)} = \\left(h+R\\right)\\sin\\beta
  3. Expand and/or distribute both sides:
    h^2+2Rh = h^2\\sin^2\\beta+2Rh\\sin^2\\beta+R^2\\sin^2\\beta
  4. Subtract the entire right side from both sides:
    h^2-h^2\\sin^2\\beta+2Rh-2Rh\\sin^2\\beta-R^2\\sin^2\\beta = 0
  5. Factorize wherever feasible:
    \\left(1-\\sin^2\\beta\\right)h^2+2R\\left(1-\\sin^2\\beta\\right)h -R^2\\sin^2\\beta = 0
  6. Use the following trigonometric identity to simplify:
    \\implies h^2\\cos^2\\beta+2Rh\\cos^2\\beta-R^2\\sin^2\\beta = 0This implies that:
    h^2\\cos^2\\beta+2Rh\\cos^\\2beta-R^2\\sin^2\\beta = 0
  7. Divide both sides by the squared cosine of β:
    h^2+2Rh-\\frac{R^2\\sin^2\\beta}{\\cos^2\\beta} = h^2+2Rh-R^2\\tan^2\\beta = 0Notice that this is a quadratic equation, which we can solve using the well-known quadratic formula:
    x = \\frac{-b\\pm\\sqrt{b^2-4ac}}{2a}In our case, x is h.
  8. Determine the coefficients to insert into the formula above:
    a = 1,\\qquad b = 2R,\\qquad c = -R^2\\tan^2\\beta
  9. Insert the coefficients into the quadratic formula:
    h = \\frac{-2R\\pm\\sqrt{4R^2-4*1*\\left(-R^2\\tan^2\\beta\\right)}}{2*1}
  10. Simplify:
    h = \\frac{-2R\\pm\\sqrt{4R^2+4R^2\\tan^2\\beta}}{2}
  11. Factorize the expression inside the square root:
    h = \\frac{-2R\\pm 2R\\sqrt{1+\\tan^2\\beta}}{2}
  12. Cancel the common factor 2 and factorize R out:
    h = \\left(-1\\pm\\sqrt{1+\\tan^2\\beta}\\right)R
  13. Convert the expression inside the square root into something simpler so that the square root can be removed:
    1+\\tan^2\\beta=1+\\frac{\\sin^2\\beta}{\\cos^2\\beta}=\\frac{\\cos^2\\beta+\\sin^2\\beta}{\\cos^2\\beta},\\\\\\cos^2\\beta+\\sin^2\\beta=1\\1implies\\frac{\\cos^2\\beta+\\sin^2\\beta}{\\cos^2\\beta}=\\frac{1}{\\cos^2\\beta}=\\sec^2\\beta
  14. Insert the simplified expression:
    h = \\left(-1\\pm\\sqrt{\\sec^2\\beta}\\right)R = \\left(-1\\pm\\sec\\beta\\right)R

We are, for the most part, done. All that remains is to realize that the height h cannot be negative in practice, even though that would be a perfectly valid mathematical solution to the quadratic equation. Realize also that the secant function is always greater than or equal to one. Expressed mathematically, this could be written as follows:

\\sec\\beta\\geq 1\\implies h = \\left(\\sec\\beta-1\\right)R\\geq 0

Thus, we can conclude that the function that best describes the height h obscured by the Earth is the following:

h = left(secbeta-1right)R

If we substitute the value for β found way earlier, we get our final result:

\\beta=\\frac{A}{R}-\\alpha=\\frac{A}{R}-\\sin^{-1}\\left(\\frac{\\sqrt{H\\left(H+2R\\right)}}{H+R}\\right)\\implies h=\\left[\\sec\\left(\\frac{A}{R}-\\sin^{-1}\\left(\\frac{\\sqrt{H\\left(H+2R\\right)}}{H+R}\\right)\\right)-1\\right]R

Conditions

Let’s plot this function; below is a selection of plots, where the height h is defined as a function of the surface distance A. Other values are implemented as sliders:

It appears that the observer’s height H shifts the graph along the A-axis and that the spherical body’s radius R widens it. However, more importantly, notice that there is a certain point where the obscured height equals zero; it touches the A-axis at exactly a single point. The value of A at this point is the surface distance to the horizon from the observer. As the surface distance A goes lower than the distance to the horizon, which happens to be a_1, the height obscured by the horizon must clearly be zero. Logically, it follows from our problem sketch that the angle β has to be greater than or equal to zero; the following condition must, therefore, hold:

\beta ≥ 0\iff\frac{A}{R}-\alpha ≥ 0\iff A ≥ \alpha R\iff A ≥ R\sin^{-1}\left(\frac{\sqrt{H\left(H+2R\right)}}{H+R}\right)

None of the values in this problem are negative, so we needn’t worry about the inequality’s direction.

We have established a lower bound for the domain of our function. There is yet an upper bound to be found: the value of A that leads to the height h approaching infinity. It actually requires no mathematical calculations to conclude what this upper bound is, as a simple sketch can illustrate what happens as the angle β grows larger:

more circular sketchiness...

Another crude sketch. This time, it’s to help us determine the domain of our function h(A), i.e. the range of values for A that would produce valid results.

Notice that the observer’s line of sight and the altitudinal line of the object are parallel once β approaches 90 degrees. At this point, the object can never be seen, assuming its width is negligible compared to the astronomical body upon which both the observer and object are located. We can, therefore, come to the following conclusion:

Once again, the absence of any negative values unnecessitates the need to consider the possibility of inequality inversion.

Conclusion

We have now shown that, to some observer located a certain height H above a reference level, the height h of an object obscured by the spherical body upon which both the observer and object are located is given as follows:

h=\\left[\\sec\\left(\\frac{A}{R}-\\sin^{-1}\\left(\\frac{\\sqrt{H\\left(H+2R\\right)}}{H+R}\\right)\\right)-1\\right]R

However, this equation requires that the following condition is met:

R\\sin^{-1}\\left(\\frac{\\sqrt{H\\left(H+2R\\right)}}{H+R}\\right)\\le A < \\left(\\frac{1}{2}\\pi+\\sin^{-1}\\left(\\frac{\\sqrt{H\\left(H+2R\\right)}}{H+R}\\right)\\right)R

Along with practical measurements, an argument of this quality magnitude or greater would be required to defeat the notion that the Earth is, in fact, a rounded, spheroidal object. Don’t sit there and pretend as though pointing a camcorder at the ocean and spouting that the oceanic horizon appears completely level. Remember: you’re basing this off technology that may or may not be capturing the horizon accurately; rely on more accurate tools of measurement instead of a deprecated camcorder potato and then present your findings in a less-than-rude way that doesn’t involve tearing books you haven’t even bought yet to pieces, you vandalous bastard. Spend your time looking at the derived formula, use a damn calculator for once, and take your time confirming that the above formula gives a relatively good estimate for the height h of an object obscured by the horizon and stop spouting the “eight inches per square mile” bullshit – that’s just an approximation to the recession of the terrestrial surface over smaller distances, not an actual steadfast rule. I’ll have to derive a proper formula for that one, too, I guess, although that would be very similar to this calculation.

Also, as a side note, I think some of my formulae aren’t transparent .jpg files, even though they totally are. I don’t know why; I checked the damn transparency box on the tool I used to create them, and they’re transparent when I view them on my PC. Oh, I suppose I ought to credit the tool I used, too. Right. Here:

All formulae used in this post were created using Roger Cortesi’s Roger’s Online Equation Editor found here: http://rogercortesi.com/eqn/. Additionally, I used Desmos to sketch the function.

Post scripta

I realize now, long after most of this post has been written, that I could have made this calculation much easier through not even having to solve a quadratic equation. I’m not going to update this post in depth, but I did scribble down the solution. It’s just like me to notice shit like that after I’ve already written a wall of text taller than the Burj Khalifa, isn’t it? At least it confirmed my earlier calculations, though, so there’s that.

Before anyone asks, I wrote this post over the period of a few days because I was bored. Also, I was tired of seeing plainthinkers and conspiritards everywhere, so I decided to add another counterargument to the mix. Plus, I like doing stuff like this, so I just did it.

Undeadifying myself

No, I’m not dead. Yes, I did watch SGDQ 2018. I don’t need a reminder that I used to post about every other GDQ lately but not this one. I suppose there’s nobody around to remind me, though, is there?

Anyway, I felt I should write something on this here blog instead of just letting it sit around dead. I haven’t forgotten about it; I have merely not had anything to write about. Generally, I like to jot down my thoughts as to whether I like or dislike a game I have recently played, but there have been no games that have managed to pique my interest as of late. It is, therefore, my duty to inform the masses of this very conundrum, except it’s not a conundrum at all, but simply my inane ramblings and pretentious choice of wording that block the way for anything even remotely resembling a coherent thought.

So, what have I been doing as far as games go? Not a lot, to be frank; there has been a little playing Grim Dawn, but I found myself uninterested after a couple hours, as the grind just got too grindy for me. The vast majority of my time playing games has been spent grinding through Euro Truck Simulator 2. Honestly, it’s the only game that has managed to keep me playing. Maybe I’m getting too old for this shit, but all the nonsense games these days just don’t hold up for me.

Seriously, what the fuck is up with all the battle royale games? Can someone please start a new trend, already? I thought trends came and went, but this battle royale garbage has clogged the system a bit too long, now. Everybody is out there playing Fortnite, PLAYERUNKNOWN’S BATTLEGROUNDS (ALL-CAPS!) and whatever other shitty battle royale games there are. It’s like neither the players nor the developers actually know that there are other genres to pick from, but, no, let’s just go with the flow and make something completely unoriginal that has no point nor purpose and which has absolutely no substance to the gameplay; it doesn’t matter, because it makes us money thanks to those fucking normies!

Never minding what I just blurted scribbled out, I’m trying to grind all the achievements for Euro Truck Simulator 2, because I need to have at least one game all the achievements of which I have completed*, and it appears to me that it is the most likely and least tedious candidate of the games I have anything resembling an interest in to one-hundred-percent, so that’s what I intend to do, assuming I can ever get that God damn Pathfinder achievement that’s taking an eternity. Fucking modular intersections! Why couldn’t the new ones just automatically fill out whenever you pass them so that I wouldn’t have to keep driving back and forth to fill them in!

Whatever; end of post.

JUST KIDDING! HERE IS A POST SCRIPTUM!

* That was a very clumsily written sentence, to be honest. Whatever. It’s not grammatically incorrect, as far as I’m aware, so I don’t give a flying shit!