iwire,
As I understand Smart$'s example, the loads on the two pole breakers do not involve a neutral. The current through the pole in slot 9 must therefore be equal to the current through the pole in slot 11, similarly for 13 and 15.
But the load added to the _feeder_ by this current flow is _not_ equal to the current flowing through the breaker. This is simply because we are adding a load with a phase angle that is not necessarily the same as the phase angle of the current already flowing on the feeder.
Smart$'s table is IMHO confusing, because it appears to list the net increase in current flow caused by each load, skipping the intermediate step of showing the phase angles involved.
Circuit 3 puts a 16A, 120 degree load on leg B
Circuit 9,11 puts a 16A, 90 degree load on leg B
Circuit 13,15 puts a 16A, 150 degree load on leg B
Circuit 17,19,21 puts a 16A, 120 degree load on leg B
The total current ends up being 59.7A, 120 degrees on the feeder for leg B, in agreement with Smart$.
Circuit 13,15 is also placing a 16A, 330 degree load on leg A.
Circuit 9,11 is putting a 16A, 270 degree load on leg C.
Thus I disagree with Smart$'s totals for legs A and C.
For leg A, I get 62.4A with a phase angle of 7 degrees. (unity power factor would be 0 degrees)
For leg C, I get 46.55A with a phase angle of 230 degrees. (I'm rushing the math on this, and may have reversed a sign on my A and C phase angles.)
The same current flows through both poles of the breakers in a given circuit. But the effect on the total phase load is _different_ because of the different phase angles already on the legs.
In all honesty, this is an intellectually interesting analysis, but IMHO it exhibits an apparent precision that would not really be justified by real world loads. IMHO it is accurate enough to simply assign VA to each leg and divide by the leg to neutral voltage.
-Jon