Note that the inductance of bifilar reduces to that of the transmission line thus formed, shorted at one end, plus the wire resistance of course. It's much less than the full helix, but also a lot more than zero. For a parallel pair, it will be on the order of 0.6 uH/m.
Ayrton-Perry likewise has at least the linear (body length) inductance, plus some amount due to the nonuniformity of that conductor -- it's made of thin wires in a sort of anti-parallel zig-zag, not a solid conductor. Or, say you mounted the resistor inside or around a tube: you wouldn't have a coaxial structure, you'd have an approximation of one with the one conductor being this net structure.
Or if the two layers are insulated (not making contact at crossing points), then, it's simply the cancellation of two single-layer windings, at whatever relative dimensions they have. The coupling factor will be large (>0.9?) but still comfortably less than 1, and so the cancellation imperfect, and the residual calculated easily from that.
Note if the two layers were wound the same way (offset by half the pitch), A-P reduces to the first case (bifilar). To the extent we can approximate a helical winding as a smooth cylinder, then it doesn't depend on orientation (direction or alignment). But clearly it will.
Consider a single turn of both layers, containing a pair of crossing points (which connect). From the start (the proceeding crossing point), the two windings arc out apart, wrapping around the former; they overlap and connect on the opposite side. This forms two opposing half-turns. From the far side, they arc out and return again to the starting side, forming two more opposing half-turns. We can ignore the extended field from these half-turns by noting that all the current going up one half-turn, is opposed by the same current going down the half-turn below it (which is the diagonal opposite wire, but by symmetry, we don't care if it was the same wire or not; indeed this remains true if the crossings are insulated, but connected crossings should help enforce balance). Thus we have a parallel-wire transmission line, and the inductance per turn will be the parallel combination of two half-turns (so, about a quarter-turn's length), times the number of turns.
Ah, which should outperform the bifilar structure then, BUT: A-P probably can't be wound at as tight a pitch as bifilar, so the inductance of that parallel line (and they're not really parallel, there's an arc-rhombic shape to the spacing, think latitudes on a globe) will be higher to start with. Well, that would still seem to outperform, by somewhat less than 50% then, which is nice.
Hm, now I want to build and measure a few (both styles), hah.
Mind this also ignores proximity effects: except at the ends, each pair is adjacent to another pair, and so on. Though this effect cancels out between both, i.e. they both use periodic counter-currents in their windings, so will be in error from the ideal case (a parallel-wire TL in free space) by the same amount.
As for yet other types -- back in those days, constantan, maybe nichrome, wire was all they had; nowadays, with cermet glazes and evaporated/sputtered metals being standard, tubular (solid cylindrical element, spiral cut, etc.) shapes are available, giving inductance as low as the body length itself (or not many times more). Also flat film resistors in standard packages, etc.
The quirk of the latter, as I understand it, is the external field isn't always trivial. So if you need very flat resistance (say for calibrated flat voltage/current pulses), you still need a noninductive winding, not so much this time for reasons of self-inductance, but in terms of avoiding induction and thus eddy current tails into the heatsink! Or, you need to use a ceramic heatsink (which, surprisingly, was a thing that existed -- briefly, alas*).
*Oh! It was the CA series they dropped. Shame, that's the bigger one. They still have the WC series:
https://www.digikey.com/en/products/detail/ohmite/WC-T2X-38E/6677110Tim
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