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.
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 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:
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:
Using this definition in each right triangle, we can derive the following:
- For the triangle of angle α:
- For the triangle of angle β:
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:
How does this help us? Recall that, when we use radians as our unit of angle measurement, we get the following:
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:
Now, by doing a bit of algebraic manipulation, we can solve for the inverse sine function:
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:
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:
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:
- Get rid of the denominator on the left side through multiplying by it on both sides:
- Square both sides to get rid of the radical:
- Expand and/or distribute both sides:
- Subtract the entire right side from both sides:
- Factorize wherever feasible:
- Use the following trigonometric identity to simplify:
This implies that:
- Divide both sides by the squared cosine of β:
Notice that this is a quadratic equation, which we can solve using the well-known quadratic formula:
In our case, x is h.
- Determine the coefficients to insert into the formula above:
- Insert the coefficients into the quadratic formula:
- Factorize the expression inside the square root:
- Cancel the common factor 2 and factorize R out:
- Convert the expression inside the square root into something simpler so that the square root can be removed:
- Insert the simplified expression:
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:
Thus, we can conclude that the function that best describes the height h obscured by the Earth is the following:
If we substitute the value for β found way earlier, we get our final result:
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:
H = 100, R = 6750
H = 500, R = 6750
H = 900, R = 6750
H = 100, R = 2800
H = 100, R = 8700
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:
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:
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.
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:
However, this equation requires that the following condition is met:
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.
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.