I am also interested in knowing if there are any good reasons to pick one ordering of polygons over some other ordering.
I'll post what little I know, but I'm really hoping someone else will post something much better.
I have heard of 3 "standards" for ordering these polygons, but as far as I can tell they were arbitrarily picked:
Your icosahedron-based method,
the quadrilateralized spherical cube, and
Tegmark's icosahedron-based method for pixelizing the celestial sphere.
All three are examples of geodesic grids.
Ordering polygons on a polyhedron
icosahedron-based method for pixelizing the celestial sphere
"What is the best way to pixelize a sphere?" by Max Tegmark
gives a format that appears extremely similar to your proposal:
Tegmark starts with an icosahedron, just as you do,
and recursively subdivides it, just as you do.
So I find it hard to believe that his method is "computationally and geometrically far more complex than what [you] are proposing" or that it has significantly "greater distortion in either area or geometry".
Given some specified resolution (the same as your depth of recursion)
and some point on a sphere, Tegmark's system
assigns that point to a "pixel number" (the number indicating what major icosahedral face it is on, and which particular minor sub-triangle on that face that point is in).
quadrilateralized spherical cube
Wikipedia: "quadrilateralized spherical cube"
The all-sky, skyward looking, unfolded cube is reassembled by
arranging the faces in a sideways T with the bar on the right side:
0
4 3 2 1
5
where square face 0 is centered on the North pole,
square face 5 is centered on the South pole,
the vernal equinox -- at latitude=0, longitude=0 --
lies at the center of square face 1.
Ecliptic longitude increases from face 1 to face 4.
other polyhedra
Tegmark mentions that "it is clearly desirable to use as small faces as possible".
Subdividing each face of a (30-faced) rhombic triacontahedron into two nearly-equilateral isosceles triangles gives the (60-faced) pentakis dodecahedron, as you mentioned,
which one might expect to give less error than starting with a (20-faced) icosahedron.
Further subdividing each of those faces into two irregular triangles gives the (120-faced) disdyakis triacontahedron (dual to the truncated icosidodecahedron).
I've been told that the strictly convex polyhedron with the maximum number of identical faces (exactly equal-shaped and exactly equal-area) is that disdyakis triacontahedron.
However, apparently none of these shapes gives any improvement over your icosahedron proposal.
Tegmark's code immediately breaks up each equilateral triangle of the 20-sided icosahedron into 6 identical irregular triangles.
The resulting 20*6 = 120 identical triangles are the same triangles as the faces of the disdyakis triacontahedron, right?
Ordering polygons on a recursively subdivided polygon
The quadrilateralized spherical cube recursively subdivides each square -- each division of 4 by area adding 2 more bits of localization information -- using a Morton code (Z-order curve).
The Maidenhead Locator System uses a system similar to the Morton code.
You might also want to look at the military grid reference system.
Other ways of ordering such subdivided square (i.e., drawing a path that hits the center of each square exactly once) include the
Sierpiński curve, the Hilbert curve,
the Peano curve,
the Moore curve, etc.
There exist other space-filling curves based on recursively subdividing a triangle -- each division of 4 by area adding 2 more bits of localization area.
(Do any of them have popular names?)
There are 2 popular ways of subdividing squares and triangles:
( a, b )
"Class I"/"alternate" method:
each of the original triangles (or squares) is replaced with n^2 triangles (or squares) that exactly fit inside the original triangle (or square).
(n=1 gives the original base shape, n=2 gives a shape with 4 times as many faces).
The small triangles have edges (very close to) parallel to the original big triangles.
"Class II"/"triacon" method, employed on almost every major geodesic dome project in the 1950s ( a p. 30 ),
each of the original triangles is replaced with 3*n^2 triangles.
The small triangles have edges (very close to) right angles to the original big triangles.
(Unreliable sources tell me the triacon method can be applied to the cube or rhombic solids,
each of the original squares or rhombuses is replaced with 2*n^2 rhombuses).
(n=1 converts the icosahedron to the pentakis dodecahedron I mentioned earlier; n=8 converts the icosahedron to an approximation of the Epcot "Spaceship Earth"; n=1 converts the cube into the rhombic dodecahedron, etc.).
EDIT:
The most common application of numbering the sides of a platonic solid is while making dice.
One particular way of putting numbers of the sides of an icosahedron is shown on DiceCollector's page, but I don't know if that is particularly "standard".
Have you considered asking
"What is the proper way to number the sides of a 1d20 die?" or
"What is the proper way to number the sides of a 1d120 die?",
or both,
on the https://rpg.stackexchange.com/ ?
Best Answer
The math in lantius' answer is correct, and your comment underneath that is spot on: The difference between lantius' answer and yours arises from the fact that your solution assumes a circle, which contains an arc between two teeth instead of the straight line of the $n$-gon. Since you presumably measured the distance of $12.75$mm between the tips of the teeth along a straight line, the $n$-gon gives the correct solution. As lantius already pointed out, the difference this makes in the circumference is quite small in this case, only $.005$ inches, compared to the $.05$ inches discrepancy with the value calculated from the diameter. Since the error in your measurement of the distance between adjacent teeth is multiplied by $53$, a discrepancy of only $.05$ inches is very good; it means the error in your distance measurement for the teeth was only about $.001$ inches, or about $.025$ mm, which seems almost a bit too good to be true.
The difference between the $1/2$ inch chain standard and your $12.75$mm measurement is due to the fact that the centres of the chain pins sit slightly below the tips of the teeth, and the $1/2$ inch standard refers to the distance between the centres of adjacent chain pins (see here and here).
There's actually a further potential source of error that hasn't been mentioned yet in the other thread. You didn't specify how you measured the diameter. Since it's a 53-tooth chain ring, there are no opposite teeth to measure it at. Assuming that you measured between the tip of one tooth and the tip of either or both of the teeth almost but not quite opposite to it, what you would then actually measure is not the diameter, but slightly less. By the law of cosines, this distance is
$$\sqrt{r^2+r^2-2rr\cos\frac{52}{53}\pi}=r\sqrt{2-2\cos\frac{52}{53}\pi}=2r\sin\frac{26\pi}{53}\approx0.99956d\;,$$
where $r$ is the radius and $d$ is the diameter. So that makes a difference of around one twentieth of a percent in the diameter, or about $.1$mm, which is about one tenth of the discrepancy of $.05$ inches. However, this goes in the "wrong" direction; to correct for it, you'd have to add that difference to the measured diameter, which would move it very slightly away from the one calculated from the tooth tip distance.
Here's something fun you could do: For each number $k$ between $1$ and $26$, measure the distance between the tips of a tooth and another tooth $k$ teeth along the ring. By the law of cosines, the values you get should be
$$a_k=\sqrt{r^2+r^2-2rr\cos\frac{2k\pi}{53}}=r\sqrt{2-2\cos\frac{2k\pi}{53}}=d\sin\frac{k\pi}{53}\;,$$
so
$$d=\frac{a_k}{\sin\frac{k\pi}{53}}$$
in each case. Note that this establishes a connection between lantius' calculation of the diameter from a measurement of the distance between adjacent tooth tips ($k=1$) and my calculation of the diameter from a measurement of the distance between almost opposite tooth tips ($k=26$); they're both special cases of the same thing. You can make all $26$ measurements and see how the values scatter. You should be able to see that the scatter decreases with increasing $k$, since the measurement error is being magnified less. If you want to go the whole cog, you can do a least-squares fit on these measurements to get a more accurate estimate of the diameter. In doing that, you shouldn't just take the average, but weight the measurements according to their precision (see here for an explanation), so your best estimate for the diameter would be
$$d\approx\frac{\sum_{k=1}^{26}d_k\sin^2\frac{k\pi}{53}}{\sum_{k=1}^{26}\sin^2\frac{k\pi}{53}}=\frac4{53}\sum_{k=1}^{26}\frac{a_k}{\sin\frac{k\pi}{53}}\sin^2\frac{k\pi}{53}=\frac4{53}\sum_{k=1}^{26}a_k\sin\frac{k\pi}{53} \;. $$
[Edit in response to the comment:]
I'll break down some of the things I'm guessing you might be having trouble with; let me know specifically if you want anything else explained.
The law of cosines states that if $a$, $b$ and $c$ are the sides of a triangle and $\gamma$ is the angle opposite $c$, then
$$c^2=a^2+b^2-2ab\cos\gamma\;.$$
I applied this to triangles formed by two tips and the centre; two sides $a$ and $b$ are the distance from the tips to the centre, which is the radius of the circle, and the third side $c$, the distance between the two tips, is determined by taking the square root of the above equation. When the teeth are $k$ teeth apart, the angle is $k$ times one $n$-th of a full circle, where $n$ is the number of teeth; since a full circle is $2\pi$, this is $2\pi k/n$.
I then used the trigonometric identity $\sqrt{2-2\cos\gamma}=2\left|\sin\frac\gamma2\right|$; since all our angles are positive, I dropped the absolute value.
At the end, I simplified the denominator in the least-squares result as follows:
$$ \begin{eqnarray} \sum_{k=1}^m\sin^2\frac{2k\pi}{2m+1} &=& \sum_{k=1}^m\left(\frac1{2\mathrm i}\left(\mathrm e^{\mathrm i\frac{2k\pi}{2m+1}}-\mathrm e^{-\mathrm i\frac{2k\pi}{2m+1}}\right)\right)^2 \\ &=& -\frac14\sum_{k=1}^m\left(\mathrm e^{2\mathrm i\frac{2k\pi}{2m+1}}+\mathrm e^{-2\mathrm i\frac{2k\pi}{2m+1}}-2\right) \\ &=& -\frac14\left(\sum_{k=1}^m\left(\mathrm e^{2\mathrm i\frac{2k\pi}{2m+1}}+\mathrm e^{-2\mathrm i\frac{2k\pi}{2m+1}}\right)-\sum_{k=1}^m2\right) \\ &=& -\frac14(-1-2m) \\ &=& \frac{2m+1}4\;, \end{eqnarray} $$
where $n=2m+1$ is the odd number of teeth, $53$, so $m=26$, and the left-hand sum evaluates to $-1$ because the $(2m+1)$-th roots of unity sum to $0$ and the sum contains all of them exactly once, except $1$. I suspect you might not understand that last part; alternatively, you can just ask Wolfram|Alpha.
Regarding your measurement of the diameter: Yes, that does affect the conclusions. The distance between a tooth and the almost opposite teeth is already a bit shorter than the diameter; the distance from a tooth to the line between the almost opposite teeth is a bit shorter yet; the difference is roughly twice as much. The distance you measured is one radius plus the distance from the centre to the line connecting two teeth, which is the cosine of half the angle between two teeth times the radius, so you measured
$$r+\cos\frac\pi{53}r=\left(1+\cos\frac\pi{53}\right)r=\frac{1+\cos\frac\pi{53}}2d\approx0.99912d\;,$$
compared to $0.99956d$ above, so this would add about $.2$mm instead of $.1$mm to the discrepancy.