Fred summarized: Necessary for planarity is the polynomial equation f_k(a, b, c, ...) = 0
[see "Horrid trig identity"], where A,B,C,... denote face-angles, a = sin^2(A) etc, and k varies depending on the number of faces at the corner under inspection. When f_k = 0, we are assured that, for some choice of the signs, A (+/-) B (+/-) C (+/-) ... = 0 mod \pi .
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The two branches cross where q = 0.674013, h = 1.285291, g = 1; Somehow it seems that this case should have an interesting property, though it's not clear what. This wasn't very observant: besides the two obvious right-angles at a U/V corner, there is a third belonging to the UWW face; and at a W corner, in the flat net there are two less obvious right-angles between UW and VW edges.
Hmmm --- awful lot of (1/2)\pi angles there --- this "cross-over" case really deserves to be renamed "right-angled" ...
Anyway pressing on, since it satisfies both constraints, we have for free a point on the second curve corresponding to a planar polytore; now by continuity, there should be an entire second family of planar polytors satisfying cusp = 0.
But unfortunately the "cusp" branch doesn't actually keep working; the signs turn out to be wrong. Those right angles Fred mentions are exactly the reason: at that point the angles around the degree-6 vertex can be partitioned into two parts (2 and 4 angles) that each sum to pi. So while the intersection point happens to satisfy the angular constraint with all + signs, continuity gets you bupkis. With Fred's exercise out of the way, what I really wanted to ask was about this other fact: There are two classes of polytore corner, "U/V" with 5 faces and "W" with 6.
For either class, we find f_k has two distinct quartic factors, christened "oval" and "cusp" after the shapes of their plane curves when g = 1 is fixed. The first class is characterised by the factor (due originally to Fred Helenius) oval(q,h,g) = 2(q - 2)^2 h^2 + (q^2 - 2)(q^2 - 4q + 2g^2 + 4) ; the second by the factor (whose author now prefers to remain anonymous) cusp(q,h,g) = 2(q^2 - 4q + 2)h^2 + q^2(q^2 - 4q + 2g^2 + 2) . The constraint oval(q,h,g) = 0 appears to be sufficient, as has been verified numerically, and by building models of examples.
The fact that f_5 and f_6 both turn out to be 0 at the same time is at first glance not surprising: Gauss-Bonnet says the total angular defect is 0, so if either type of vertex is flat, the other must be too. But the "cusp" branch *isn't* guaranteeing flatness, we now know; it's instead guaranteeing a 0 of the angle sum with some other combination of signs. So why on earth does this cusp branch show up as a factor of both f_5 and f_6? What else is forcing these to be simultaneously zero? (Pity it happens, whatever it is -- otherwise we could eliminate stray branches by taking the gcd of f_5 and f_6!) --Michael -- Forewarned is worth an octopus in the bush.