The Martian Tripod Problem and Transcendental Functions

I thought this week about a problem I originally considered about ten years ago. I imagined a source of a laser beam, mounted high up above the ground, shining straight down, and allowed to rotate upwards at a constant rate until it was shining horizontally. The point at which the laser beam touched the ground would travel from directly below the source to the horizon.

 

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The Martian Tripod Problem: What is the location where the beam strikes the ground as a function of time?

The question is, if I know where the source of the laser is, and how quickly it is rotating, can I know exactly where the beam strikes the ground over time?

The problem came about, I’m sure, as I was listening to Jeff Wayne’s Musical Version of the War of the Worlds, imagining beams of Martian heat rays mounted to towering tripods sweeping across the hull of the Thunder Child.

Trigonometry: What’s the length of a side of a right triangle with a constant height?

This is not so bad of a problem when only the geometry is considered. For now let’s call the angle that the laser makes with the “straight down” direction (the vertical) “omega-t”: \omega t. With t a length of time since the laser started shining, we can see that \omega is a sort of speed — when I multiply it by a length of time t, it gives a total angle, which is like a distance. The product \omega t works the same way that 30 miles per hour times 2 hours is 60 miles. In physics we’d call this speed of rotation \omega the angular speed, or angular velocity if you’re considering rotations in all three dimensions. For now, it doesn’t matter what the value of this speed of rotation is.

Setting up the laser at an angle \omega t from the vertical and a height H from the ground, we find it shines at a point H \sec(\omega t) away, and a horizontal distance H \tan(\omega t) to the right.

martian1.png
A laser at height H, at angle \omega t with the vertical, shines at a point H\tan(\omega t) to the right and a full distance H\sec(\omega t) away.

If you’re not so used to working with trig functions, you could get to the image above by first setting up the “classic” trig diagram, with the point a distance (hypotenuse) H away:

martian2.png
A scaled version of the previous image. Divide all lengths by cosine to get a constant height of H.

The above image has all the right ratios, but keeps the hypotenuse constant, not the height. Divide all the lengths by \cos(\omega t), and remember that secant = 1/cosine (by definition).

We’re already done. The point at which the laser beam hits the ground is

\bigg( H \tan (\omega t) , 0 \bigg)

Tangent “blows up” to infinity at $\pi/2$, which corresponds to the laser shining parallel to the ground. It intersects the flat ground an infinite distance away, at the horizon. Hopefully this agrees with your expectations; tangent is defined to act this way.

The Tripod Problem: Incorporating the speed of light

So that’s not super interesting. The real “tripod problem” was this: Where is the point of intersection if the speed of light isn’t infinite? If it takes some time for the laser beam to travel from the source to the ground, and the laser continues to rotate, then the location where the beam strikes will lag behind the orientation of the laser emitter.

This results in a “floppy” trajectory of the laser beam, drooping down to the ground

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A rough estimate of the shape of the laser beam given a very slow speed of light, a very quickly rotating heat ray, or a very tall tripod. Directly underneath the source, the movement of the intersection of the trajectory and the ground is dominated by the rotation of the laser. Far away from the source, the motion of the point on the ground is dominated by the speed of light.

The behavior of the point where the laser strikes the ground is very different with the speed of light restriction. It never reaches the horizon in finite time — the behavior for long lengths of time, and far distances, is totally dominated by the speed of light. It should travel along the ground at a speed approaching c.

The way the pointer location moves depends little on the speed of light directly under the source, where the distance to the ground doesn’t depend strongly on the angle, and a wider angle of the laser covers a small length. As the laser approaches the horizontal, the length covered by each small change in angle increases. The light that will eventually strike very far distances looks more like a point source, since the small angles will be covered in a very short length of time. The trajectory of the beam itself, while it’s still in the air, will look more and more like an expanding circle with time.

When I first mentioned the tripod problem to a friend recently, he had the insight of saying that a laser’s point could definitely travel faster than the speed of light. He could shine a laser at one side of the moon, and then the other. A quick enough rotation on a far enough canvas could result in a pointer appearing to travel faster than the speed of light. Remember, this doesn’t violate anything in relativity. No object is traveling faster than light, rather, a series of events in which different photons strike the distant moon are occurring. This situation is very much like that when the tripod laser is pointed nearly downwards. The speed of the pointer is dominated not by the speed of light, but by the rate of the laser rotation because the distance the light has to travel doesn’t change much when the laser is pointed directly downwards (or from one side of the moon to the other).

This suggests that the speed of the pointer, at very far distances from the tripod, would approach from slower speeds if the laser were rotating slowly. But, if the laser were rotating quickly enough, could counterintuitively approach from faster speeds.

Anyway, let’s try to deal with the problem. Take a look at our trig diagram again.

martian1
The laser beam has to travel a distance H\sec(\omega t).

When a certain portion of the laser beam is emitted at a time t (and an angle \omega t), it has to travel a distance H\sec(\omega t). Traveling this distance at speed c takes a length of time equal to

\cfrac{H}{c}\ \sec(\omega t).

Any portion of the heat ray travels in a straight line. Although the beam as a whole is curved, we’re still assuming it’s always traveling directly away from the source (no diffraction, etc.). The laser travels in the same direction and therefore strikes the ground at the point

\big( H\tan(\omega t),0 \big)

at a later time

t^\prime = t +  \cfrac{H}{c}\ \sec(\omega t)

This seems great. We have the basis for a complete understanding of the position of the laser pointer (or toasted Edwardian human) given some time. A portion of the laser, emitted at time t, will strike the ground a horizontal distance H\tan(\omega t) away not at t but at t^\prime above. This allows us to find the location corresponding to any time of emission in the interval

0 \leq t < \cfrac{\pi}{2\omega}.

If we were satisfied with this, the game plan would be to pick a time of emission t, determine how long that portion of the beam traveled, and then pair up the resulting t^\prime with the distance H\tan(\omega t).

I’m not satisfied, though. I’d like to get a trajectory of the laser pointer: a location as a function of the actual time t^\prime rather than the time that portion of the laser was emitted, t. In order to do this, we’d need to replace the t in the tangent function with an equivalent function of t^\prime. In order to do that, we’d need to solve

t^\prime = t +  \cfrac{H}{c}\ \sec(\omega t)

for t. Good luck.

What we’ve got above is a transcendental equation. This means it is not composed of a finite number of additions, subtractions, multiplications and divisions of our variable t and the constants, as well as rational powers of these. In most cases, and I’m pretty sure in this one, we can’t solve a transcendental equation exactly for the input variable. We cannot write t in terms of t^\prime.

It seems like the best we could do, if we wanted to create an animation with a step by step progression of the position of the pointer, is to prepare ahead of time. Pick an emission time t, find the value of the tangent function to find the distance, find the value of the strike time t^\prime, and record that pair. Then do this many, many times to create a table with more values than we expect someone to ask us for. We could find the position as a function of time with as much precision as we wanted, supposing we were willing to put the effort in.

I wanted a closed form solution to the problem, a trajectory x(t^\prime), and it seems more than out of reach. This annoyed me, until a friend (hey, there, buddo!) reminded me that “closed-form” is just a matter of what I’m allowing as a definition. In fact, like I mentioned in the last post, all of the trig functions are themselves transcendental: They can be written as Taylor polynomials, but these are polynomials of infinite length. The secant in the equation above can be estimated using

1 + \cfrac{1}{2}\ x^2 + \cfrac{5}{24}\ x^4 + \cfrac{61}{720}\ x^6 + \cfrac{277}{8064}\ x^8 +\dots

The problem with this, though, is that this isn’t much better. It would still take an infinite amount of time to achieve the exact value of secant given most x’s. The only reason I’m more comfortable using this and the other trig functions is because I’ve been trained to use this name for them, and rely on calculators or tables to give me the values whenever I need them. Anyone using a trigonometric function table is benefiting from someone else’s hard work to overprepare. When we use a calculator, we are relying on an estimation that is as precise as the manufacturer (or sometimes the user) dictates. One could make this estimation with a Taylor series, or with a more efficient method, but the calculator still wont give an exact decimal value.

Any single irrational number, whether it is the solution to an algebraic or a transcendental equation, is another instance of this. I’ve gotten used writing things like \sqrt{2} or \pi as representations of numbers with clear definitions. These numbers have exact values, but it’s hopeless for me to try writing them down. In a very definite sense, these numbers elude us. I could write or use them to any finite precision I wanted, with millions and millions of digits, so long as I were willing to come up with and use an efficient algorithm to find them, or if I were were willing to wait or work a very long time, or both. But, I still wouldn’t have the “exact” value, just one that was plenty good enough for whatever application I had in mind.

These examples remind me: it’s convenient to have named functions like “Cosine” to cover a mathematical idea, but we can’t let this name cover up the meaning of that idea. There are an infinite number of angles whose cosine is a transcendental value. We’re able to use cosine because we can always (right?) reach a higher precision than is necessary in a physical application. I’ve gotten used to working with cosine, and mentally separated it from the solution to the tripod problem, because someone gave it a name that I’ve adopted.

So, I guess I should name the solution. Let’s call the composed function

x(t^\prime) = H\tan\big(\omega t(t^\prime)\big)

where t and t^\prime are related by

t^\prime = t +  \cfrac{H}{c}\ \sec(\omega t)

the heat ray function. We could create a huge table for x(t), someone could come up with an efficient algorithm for calculating values of x, and in the future we could use these to invade infinite planes with laser pointers more effectively.

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Tangent of angles approaching 90 degrees

Last week a colleague came to me with a puzzle. He asked me to punch in the tangent of 89 degrees into a nearby TI-83 calculator.

\tan(89^\circ) = 57.28996163

He asked me what was surprising about this number. I wasn’t surprised. I didn’t have an answer for him, although in retrospect I probably should have. He had to tell me this was the number of degrees in a radian. Oh! So it is.

Even further, he said, try punching in tan(89.9), or tan(89.99), etc.

\begin{array}{rl} \tan(89^\circ) &= 57.28996163\\ \tan(89.9^\circ)&=572.9572134\\ \tan(89.99^\circ) &= 5729.577893\\ \tan(89.999^\circ) &= 57295.77951\end{array}

Each one is (about) ten times the previous. (With the TI-83, replacing the last result with each new one, I didn’t see the “about” until later.) This is kinda neat! His question to me was: WHY is this true?

Tangent is a function that accepts an angle and spits out a ratio of lengths. It seems weird that the answer for tan(89) looks like a number of degrees. It is a unitless output, though, and ~57 degrees per radian is also unitless, so I suppose this isn’t much of an issue. The question is, why does it appear that

\tan(89^\circ) = \cfrac{180^\circ}{\pi \text{ rad}}\text{ ?}

Ditching the degree measure,

\tan\left(\cfrac{\pi}{2}-\cfrac{\pi}{180}\right) \stackrel{?}{=} \cfrac{180}{\pi}

My problem was in trying to first tackle this question visually.

desmos-graph1.png
tan(89 degrees) is the length of the vertical line lying between an extended radius of a unit circle drawn 89 degrees from the horizontal and the right side of the circle. The numbers above make it seem like it is 180/\pi.

An equivalent image:

desmos-graph2.png
A triangle and circle \pi times larger have the same relative lengths.

I tried to explain this by imagining rolling the circle over the tangent line, wrapping the line around the circle, etc. I didn’t get anywhere.

I also tried considering the fact that there’s nothing particularly special about degree measure, except for the fact that 360 is an easy to divide number. Does this happen with other angle units? For example, what about a unit that was, instead of 1/360 of a circle, a larger 1/100 of a circle? We could instead take the equation above and ask,

\tan\left(\cfrac{\pi}{2}-\cfrac{\pi}{100}\right) \stackrel{?}{=} \cfrac{100}{\pi},

Is the tangent of one one-hundredth of a circle short of \pi/2 equal to the number of hundredths of a circle in a single radian? It looks to be true!

\begin{array}{rl} \tan\left(\cfrac{\pi}{2}-\cfrac{\pi}{100}\right)&= 31.82051595\dots\\\\ \cfrac{100}{\pi} &=31.83098862\dots \end{array}

But this is only approximate. We could extend this to any fractional unit 1/n of a circle:

\tan\left(\cfrac{\pi}{2}-\cfrac{2\pi}{n}\right) \approx \cfrac{n}{\pi}

Using this different unit, where the approximation is less accurate, I was able to see that the degree version wasn’t exactly true, either. It definitely looks like dividing the circle into a larger number (360, rather than 100) yields a closer approximation:

https://www.desmos.com/calculator/vqbu4zni5p?embed

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In the above, y= \tan\left(\cfrac{\pi}{2}-\cfrac{2\pi}{n}\right) (red) and y=\cfrac{n}{2\pi} (blue) converge for larger n (horizontal).

I was comfortable in concluding now that this wasn’t just a coincidence that relied on degree measure, and could extend this to include using 89.9, 89.99 etc degrees as well. In fact, tacking on .9s to the 1/nths of a circle units works. Just plugging in a bunch of numbers, it looks like

\tan\left(\cfrac{\pi}{2}-\cfrac{2\pi}{n}\ 10^{-a}\right) \approx \cfrac{n}{\pi}\ 10^a

works for any n, and also extends to any power a, not just the integers.

https://www.desmos.com/calculator/ctcicykylj?embed

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y=\tan\left(\left(\frac{n}{4}-10^{-x}\right)\frac{2\pi}{n}\right) (red) and y=\frac{n}{2\pi}\cdot10^x (green) lie almost on top of each other for positive x (horizontal). In the link you can see this works for any n by fiddling with a slider.

The question remained, why is this true? Now that I saw it’s only an approximation, I realized that I should be going about this algebraically from the start.

A trick, called the small angle approximation, is used in physics often to get rid of pesky sines and tangents when you’d rather just have an expression with the angle inside.

\begin{array}{rl} \sin x &\approx x\\ \tan x& \approx x \qquad\text{when }x\ll1 \end{array}

This behavior is clear when the functions are written in their Taylor series form:

\begin{array}{rl} \sin x &= x - \cfrac{x^3}{6} + \cfrac{x^5}{120} - \cfrac{x^7}{5040} +\dots \\\\ \tan x &= x +\cfrac{x^3}{3} + \cfrac{2x^5}{15} +\cfrac{17x^7}{315}+\dots \end{array}

When x is real small, all the higher power terms get super small, and the approximation becomes more accurate.

This approximation was my first thought, but there’s a problem: it works for small angles, but my colleague’s puzzle was about angles near 90 degrees. In fact, we can’t even fudge the Taylor series of tangent near here, because there is no Taylor series around 90 degrees. (This is a consequence of the fact that tan(x) blows up to infinity at 90 degrees.)

The problem is solved by noting that working with tangent near 90 degrees is the same as working with another trig function, cotangent, near 0 degrees.

\tan\left(\cfrac{\pi}{2}-x\right) = \cot(x) = \cfrac{1}{\tan(x)}.

Setting everything up:

\begin{array}{rl} \tan\left(\cfrac{\pi}{2}-\cfrac{2\pi}{n}\ 10^{-a}\right) &=  \left(\tan\left(\cfrac{2\pi}{n}\ 10^{-a} \right) \right)^{-1} \\ & = \left(\left(\cfrac{2\pi}{n}\ 10^{-a} \right) +\cfrac{1}{3}\left(\cfrac{2\pi}{n}\ 10^{-a} \right)^3 +\dots \right)^{-1}\\ &\approx  \left(\cfrac{2\pi}{n}\ 10^{-a} \right)^{-1} \qquad \text{(the approximation)}\\ &= \cfrac{n}{2\pi}\ 10^a \end{array}

Done! Having the 10^a instead of any old number is unneccessary — this works for any multiple. However, integer a‘s makes the trick of having the same digits show up in tan(89), tan(89.9), etc. work.

So, we can show this algebraically. I just wish I had a nice geometric argument.