Following Takahashi (*"On the concrete construction of hyperbolic structure of 3-manifolds"*), I was able to construct the Euclidean cusp cross-section for the 5_2 knot complement (please see http://kias.dyndns.org/topogeoimages/5_2.cusp.png), and determine the vertex invariants
```
x = 0.12256116687665364 + 0.7448617666197441 i
y = 0.6623589786223728 + 0.5622795120623012 i and
z = 0.7849201454990264 + 1.3071412786820453 i.
```
I would like to be able to compute the cusp shape from x, y and z, but as a retired physicist and learning this on my own, I have not been able to come up with an algorithm to do this. I have been able to compute the cusp shape following Yokota ("On the cusp shape of hyperbolic knots"), but this is from the knot, not from the cusp cross-section. I feel this should be easy, but the SnapPy documentation (and code, for that matter) is not entirely clear to me, nor is Coulson et. al. ("Computing Arithmetic Invariants of 3-Manifolds").

If I could just see the cusp shape in terms of x1, x2, x3, y1, y2, y3, z1, z2 and z3, I am reasonably sure I can work backwards to an algorithm I can use. Thanks in advance.

In reference to Sam Nead's answer:

I believe that the image in the link ("5.2.cusp.png") is topologically the correct tiling. Using Mathematica to compute the angles around each edge in the cusp tiling, and then to draw the associated rays along each angle, I pieced these together to make an approximate geometrical sketch of the tiling (please see http://kias.dyndns.org/topogeoimages/5_2.geometric.png). This matches closely with Takahashi's figure 25.
Then I did what any physicist would do: I took measurements of the sides of the bounding parallelogram. The ratio of those lengths (arbitrarily in pixels, but since similar parallelograms have equal side ratios, that should not be a problem) is approximately 452/153 = 2.9425. The acute angle of the bounding parallelogram (corresponding to x1, for instance) is 80.6562 degrees or 1.40772 radians. As per Adams, Hildebrand and Weeks ("Hyperbolic invariants of knots and links"), I should obtain the homological cusp shape as

(longitude/meridian) . Exp(Pi.I.angle_radians)

For this I get 0.479646 + 2.91505 I.
In checking this with SnapPy (which gives a cusp shape of -2.4902446675 + 2.9794470665 I), I noticed that, while I got the same tetrahedra shapes as Takahashi, SnapPy disagrees with those (my y is the difference between SnapPy's first two shapes).
Further, https://knotinfo.math.indiana.edu gives translations yielding a cusp shape of 2.49024 - 2.97945 I. I am assuming this is simply a choice of orientation (the real part of the meridian differs by a sign). When I followed Yokota, I got SnapPy's result.
I know that there can exist inequivalent cusp tilings, but it seems to me that the tetrahedra shapes should be invariant, and certainly the cusp shape is an invariant. I am now pretty much thoroughly confused.

Thank you!

Now, with some judicious use of the law of sines, I find for the magnitude of the cusp shape

Csc[Arg[y2]] Csc[Arg[z3]] Sin[Arg[y3]] Sin[Arg[z1]] + 
  Csc[Arg[z1]] Sin[Arg[z2]] + 
  Csc[Arg[x2]] (Sin[Arg[x3]] + 
     Csc[Arg[z3]] Sin[Arg[x1]] Sin[Arg[z2]]) = 3.01951

which with the complement of the acute angle (Arg[z1]+Arg[x3]) = 99.3438 degrees, gives precisely the result of the shape of m015 from SnapPy.