I struggle to believe that a bolt 3 phase short circuit between all phases could be smaller in magnitude than a fault between any phase and earth

I think I know where you're coming from. If you have a bolted 3-phase fault it should in theory be perfectly balanced - so no N (or PE) current. From the point of view of each phase it's like having a L-N fault onto an artificial N point right at the point of the fault - so while the L conductor contributes some impedance to the loop, the N (or PE) contributes nothing at all - so it's like a L-N fault with a perfect zero-Ohms N (or PE) conductor back to the source. Any number of parallel paths still isn't going to reduce R2 entirely to zero.

On the other hand say we had a reduced N in the supply - so let's say R1 = 0.8Ω and Rn = 1.2Ω (round numbers for ease of calculation rather than being realistic) - you might measure L-N on each phase as 2Ω/115A and double it to 230A. But in reality (unbeknown to you) the fault current on each phase for a bolted fault would be closer to 230V/0.8Ω = 287.5A and so R2 could be anything up to 0.2Ω and still give a higher fault current.

As has been said, "establish Ipf between lines and divide by 0.87" would be better but that does require an instrument that suitable for 400V use (or 460V if it happens to be a splt-phase supply) - which not all instruments are.

   - Andy.

I struggle to believe that a bolt 3 phase short circuit between all phases could be smaller in magnitude than a fault between any phase and earth

I think I know where you're coming from. If you have a bolted 3-phase fault it should in theory be perfectly balanced - so no N (or PE) current. From the point of view of each phase it's like having a L-N fault onto an artificial N point right at the point of the fault - so while the L conductor contributes some impedance to the loop, the N (or PE) contributes nothing at all - so it's like a L-N fault with a perfect zero-Ohms N (or PE) conductor back to the source. Any number of parallel paths still isn't going to reduce R2 entirely to zero.

On the other hand say we had a reduced N in the supply - so let's say R1 = 0.8Ω and Rn = 1.2Ω (round numbers for ease of calculation rather than being realistic) - you might measure L-N on each phase as 2Ω/115A and double it to 230A. But in reality (unbeknown to you) the fault current on each phase for a bolted fault would be closer to 230V/0.8Ω = 287.5A and so R2 could be anything up to 0.2Ω and still give a higher fault current.

As has been said, "establish Ipf between lines and divide by 0.87" would be better but that does require an instrument that suitable for 400V use (or 460V if it happens to be a splt-phase supply) - which not all instruments are.

   - Andy.