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<!DOCTYPE html>
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<title>A Suspicion of Plausible Answers - Comfortably Numbered</title>
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<a href="/"><span class="left-word">Comfortably</span> <span class="right-word">Numbered</span></a>
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<section>
<h1>A Suspicion of Plausible Answers</h1>
<center><em><p>Investigating the physics of <em>Rendezvous with Rama</em></p>
</em></center>
<h4>Friday, August 14, 2015 · 3 min read</h4>
<blockquote>
<p>He [Norton] had a suspicion of plausible answers; they were so often wrong.</p>
<p>— <em>Rendezvous with Rama</em>, Arthur C. Clarke</p>
</blockquote>
<p>Clarke’s <em>Rendezvous with Rama</em> describes the exploration of a giant spaceship
called “Rama”. If you haven’t read the book yet, go read it and come back,
because the rest of this post is a really big spoiler. I’ll be waiting; don’t
worry.</p>
<hr>
<p>Welcome back.</p>
<p>To me, the charm of <em>Rendezvous with Rama</em> is the way Clarke introduces parts
of the spaceship, lets you guess what they’re for, and then reveals their
purpose in a series of intertwined narratives. The book is a guessing game.</p>
<p>One of the central mysteries, of course, is the Southern Cliff of the
Cylindrical Sea: why is it so much higher than the Northern Cliff?</p>
<p>In a flash of inspiration, the exobiologist Dr. Perera realizes that when Rama
accelerates (to the north), the Sea would rise against the Southern shore; the
Cliff is a barrier to prevent a great flood. In his own words:</p>
<blockquote>
<p>“The Cylindrical Sea is enclosed between two cliffs, which completely circle
the interior of Rama. The one on the north is only fifty meters high. The
southern one, on the other hand, is almost half a kilometer high. Why the big
difference? No one’s been able to think of a sensible reason.</p>
<p>“But suppose Rama <em>is</em> able to propel itself—accelerating so that the
northern end is forward. The water in the sea would tend to move back; the
level at the south would rise, perhaps hundreds of meters. Hence the cliff.”</p>
<p>— <em>Rendezvous with Rama</em>, Arthur C. Clarke</p>
</blockquote>
<p>On a roll now, Perera goes on to predict—with not more than twenty seconds of
thought and scribbling—the maximum possible acceleration of Rama based on the
height of the Cliff (500 meters). His result, 0.02g (2% of Earth’s
gravitational acceleration), is confirmed at the end of the book.</p>
<p>How did he do it? Let’s investigate.</p>
<hr>
<p>Here’s a diagram of Rama’s cross-section to help you follow along. It’s taken
from the cover art of a video game based on the novel; the image was posted to
<a href="http://www.foundation3d.com/forums/archive/index.php/t-551.html">a forum
thread</a> by
someone named DELTA.</p>
<p><img src="static/rama.jpg" alt="Rama cross section"></p>
<p>The first page of <a href="http://ebruneton.free.fr/rama3/rama.pdf">this paper</a>
contains a simpler schematic as well as some pretty pictures.</p>
<p>First, some raw data scraped by scouring the novel.</p>
<p>We have a couple of ways of determining the unaffected “gravity” (centrifugal
effect) on Rama. The centripetal acceleration is proportional to the square of
the angular velocity and the radius. Knowing that the Plains are 8km from the
axis about which Rama rotates, and that it rotates at 0.25rpm (a rotation every
4 minutes), we calculate that Rama’s gravity is 0.6 that of Earth’s:</p>
<p>\[ g_\text{rama} = \omega^2r = \left(\frac{2\pi}{4\times60\,\text{sec}}\right)^2\times(8000\;\text{m}) = 5.483 \text{m}/\text{s}^2 \approx 0.6\, g_\text{earth} \]</p>
<p>For those without MathJax, that said</p>
<pre><code>g_rama = w^2 r
= (2pi / (4*60s))^2 * 8000m
~= 0.6 g_earth
</code></pre><p>This result agrees with a statement by one of the explorers, Mercer: when he
was less than 2km down the stairway, he said his weight was around a tenth of
what it would be on Earth.</p>
<hr>
<p>We know the width of the sea: it’s 10km across. Unfortunately, we know very
little about the depth of the sea. We do know that the seafloor is not uniform;
it’s ridged to disperse large waves. We even have a lower bound on the deepest
portion: at one point, an anchor is lowered 30 meters into the sea. However,
these facts were discovered <em>after</em> Perera’s calculation, so we could be
justified in assuming that the seafloor is uniformly “flat”. We also have an
upper bound of 2km because that’s the difference between Rama’s inner and outer
radii.</p>
<p>Let’s assume that the sea is at least 0.5 kilometers deep*. As long as it’s at
least that deep, we don’t need to know exactly how deep it is, because the sea
surface will never intersect with the seafloor.</p>
<p>Here’s what it looks like<sup>†</sup>:</p>
<p><img src="static/rama-geom-1.png" alt="A diagram."></p>
<p>Based on the force diagram, a bit of similar triangles magic shows that for the
water to reach a height of 500 meters, the ship must accelerate northwards at
6% of Earth’s gravity. This is arguably pretty close to 0.02 g, considering all
the approximations and eyeballed measurements involved. (At least, it’s the
correct order of magnitude, which is supposedly all astronomers really care
about.)</p>
<hr>
<p>*Interestingly, if you assume that the sea is shallower than 0.5 km, you can
try to reverse-engineer the exact depth Perera must have assumed. It turns out
that there are no real solutions for depth which yield 0.02g as the
acceleration. Here’s the diagram that corresponds to the scenario—the key
insight is that the area of the rectangle and the triangle must be the same
since the volume of liquid is the same.</p>
<p><img src="static/rama-geom-2.png" alt="Another diagram."></p>
<p>[<sup>†</sup>Apologies for the Paleolithic-era hand-drawn diagrams which
look like they were scanned in the ‘80s. I would appreciate it if someone could
whip up a nice computerized image in a vector format…]</p>
<hr>
<p>Overall, I’m impressed with the accuracy of Clarke’s physics. I suppose
attention to detail like that is what makes these novels so fascinating.</p>
<p>I would love to see other explanations of Perera’s result. Maybe I missed
something. Let me know if you make some discoveries.</p>
</section>
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