Current division for fault on high side of substation

[minus3db]

Hi all,
I was reading IEEE 80-2013 (Guide for Safety in AC Substation Grounding) and came across Figure 30 which shows current division during a substation fault.
Please see the below snapshot.
I am having difficulty in following KCL to arrive at the numbers shown in the diagram. I can see and understand that the returning current at the remote source is simply 1048A + 444A, but I don't follow how IG=742A at the faulted structure is determined. I feel really dumb as I thought that I understood the fundamentals of how fault current returns both via the neutral / shield wire and via ground, but I am not connecting the dots here. Would appreciate any insight that you folks could offer. Thank you.

 

[waross]

Delta contributions.
Consider three transformers in delta fed from phases A, B and C.
If you lose B phase, the transformer fed fro A-C still has full voltage applied.
The other two transformers are now in series across phases A and C.
This causes a voltage unbalance at the wye side of the distribution substation transformer.
The delta winding at the load substation tries to correct the unbalance,
The result of that is neutral/ground currents at the wye points of the wye windings.
Forgive my poor explanation,but consider the effect of two phase windings being in series across a healhty phase.

 

[bacon4life]

I share your confusion at various aspects of the diagram. The phase angles on the currents are not shown, which makes it hard to do KCL on portions where the labeled currents are out of phase.