Some of those figures might be valid for a circuit with 100% of the load at the end of the circuit.
The guide was to consider two sockets fully loaded to 20 amp at the centre or end or ring or radial, and then any left over i.e. 12 amp with a 32 amp overload even spread, using that clearly with a 20 amp radial the load is simply 20 amp.
I will agree the whole idea of working out the volt drop based on a 26 amp load (Design current for circuit Ib) can be questioned, one could consider 13 amp at furthest point and 19 amp even spread, so Ib = 22.5 amp with a 32 amp supply, so with a 16 amp radial it would be 14.5 amp as Ib instead of 16 amp, but 42 meter to 47.5 meters is hardly worth writing about. I am including the Correction factor (Ct) in my calculations.
But there is nothing so annoying to find after laying a cable, that the volt drop is too high, only had it once, and lucky for me I had expressed my doubts about the volt drop before installing, and my foreman was unlucky that the shrink wrap machine was so voltage dependent, but lucky we could hid the expense of replacing the 4 mm² cable with 10 mm² so no one who could cause a problem was aware of our error.
I wrote the program in java script as I was worried with the PIR as it was then, now called an EICR that some one would find out if under size cable was used, volt drop wise, but the program pointed out that a loop impedance meter showing 1.42 Ω is not really that accurate, repeat the measurement and one can get 1.40 to 1.44 Ω the same at the origin so 0.33 to 0.37 Ω so once one adds the +/- into the calculations one really needs to be way out before some one can really point the finger.
So most electricians who have measured the prospective short circuit current or line - neutral loop impedance on a regular basis get to know how much they can get away with without working it all out first, it is the DIY person who has the problem, as they don't have the experience. So they need to work it out first, and can't really have a reasonable guess as to what the likely result will be.
But again today most equipment is designed to work on around 200 to 254 volt, and in the main our supply is near the higher mark, although officially 230 volt, more like 240 volt in most homes, so we can have a 40 volt volt drop and equipment still work, where the limit is 11.5 volt.
However easy for an electrician even if potential volt drop is over the limit to disregard it, as the EIC does not have a place to record the line - neutral loop impedance at the end of the circuit, so any error will not show on the paperwork. As long as he gets 1.38 Ω so trip will operate on the magnetic part, with a B32, he's OK, even if with a 0.35 Ω incomer he should show 0.94 Ω line - neutral centre of the ring to be within the volt drop, since he does not need to record it on the form, they get away with excessive volt drop.
However getting away with it, does not make it correct.