do you divide the R1+R2 values by 4

Dividing by 4 only works for ring circuits when you start with r1 & r2 numbers - rather R1 & R2. The difference between "r" and "R" is significant. r signifies the  resistance of a conductor all the way around the ring - while R signifies the effective resistance from the origin to the point of the fault under consideration (usually the furthest point from the origin, measured along the cable runs - i.e. half way around for a simple ring). Comparing r with R, it's halved because the furthest point is only half way around the ring so only half the length of the cable is involved to that point, and then halved again because of having two conductors in parallel - hence dividing by 4.

when calculating thermal constraints

Be careful here - faults (by their very nature) may occur at any point on the circuit, not necessarily at the furthest point. For a simple single-conductors radial fed by a fuse, the worst case was normally when the fault current was lowest (i.e. when Zs highest) - i.e. at the furthest point. When circuit breakers are involved the worst case is usually when the fault current is largest - i.e. for faults close to the start of the circuit. Having parallel conductors makes things a lot more complicated - a fault might involve just one of the paralleled cores - so there would be no equal sharing of the fault current between the parallel conductors as there would be for load currents - in the worst case one of the cores might have to carry practically the entire fault current on its own. Where you have many conductors in parallel, which each relatively thin compared to the circuit's overall rating, you might even need individual fault protective devices on both ends of each conductor (see appendix 10 of BS 7671). The whole situation is made more complicated as faults currents can flow from both ends of each conductor, not just from source towards the fault.

A common fault (especially on small circuits) is a wire becoming detached from its terminal and coming into contact with something at a different potential (e.g. a c.p.c. snapping off and touching a L terminal, or a the end of a live wire touching a back box). If such a fault occurred close to the origin of a circuit with parallel conductors, the fault would flow the 'long way;' around - up the other conductor and then backwards down the broken conductor to the fault. In effect that would double the resistance for the faulty conductor. Often that's compensated for because the other half  of the fault path is correspondingly shorter - but it's only a precise balance if both L1 & c.p.c. are of the same resistance per metre - reduced c.p.c.s can increase the overall resistance for errant c.p.c.s, while lower c.p.c.s resistances (e.g. steel conduit) man mean the doubling of R1 isn't fully compensated for for such faults on L conductors.

   - Andy.

Thank you for the indepth reply it explains the path the fault current would take a lot better 

Thank you for the indepth reply it explains the path the fault current would take a lot better