It is very hard to say from that what you may expect without a lot more information.  Clearly while current is flowing normally there are small voltage drops all along the line, and when there is a far end short, the current increases and all the voltage drops rise pro-rata until the sum of all the drops across all the impedances in the fault loop exactly equals the original line voltage.
But how this plays out depends on a lot of detail we probably do not know - the source impedance of the generators while rotating is one, the resistances and inductances of the transformer are another - and then of course all the cable lengths and cross-sections that interconnect them.
However as a rule of thumb - if the far end voltage drop on normal load is a percentage of the open circuit voltage - say it droops 5% on full load, relative to no load, then the current that flows for a short circuit at that point, would be equivalent to 100% drop at that point- so 20 times the full -load current for that 5% example. Then the full-load fault condition voltage drops at all other points on the line will then also be 20 times what they would have been in normal operation. .

This is only roughly right, as a real fault is not a perfect short with no voltage drop - it was there would be no heat, no flash and no damage.
I bet there was.

Actually as an aside, the most dangerous fault, for anyone standing near it, is not  a perfect short, but one where the maximum power is dissipated in the fault. That occurs when the power dissipated along the line and in the fault are equal, and the energy available to do damage can be estimated from how fast such a fault would be disconnected. The results of such sums on HV systems are almost as unsettling as looking at the equipment that has been blown apart by such events. Large faults need short trip times - including for faults that are not always perfect shorts.
Mike.

It is very hard to say from that what you may expect without a lot more information.  Clearly while current is flowing normally there are small voltage drops all along the line, and when there is a far end short, the current increases and all the voltage drops rise pro-rata until the sum of all the drops across all the impedances in the fault loop exactly equals the original line voltage.
But how this plays out depends on a lot of detail we probably do not know - the source impedance of the generators while rotating is one, the resistances and inductances of the transformer are another - and then of course all the cable lengths and cross-sections that interconnect them.
However as a rule of thumb - if the far end voltage drop on normal load is a percentage of the open circuit voltage - say it droops 5% on full load, relative to no load, then the current that flows for a short circuit at that point, would be equivalent to 100% drop at that point- so 20 times the full -load current for that 5% example. Then the full-load fault condition voltage drops at all other points on the line will then also be 20 times what they would have been in normal operation. .

This is only roughly right, as a real fault is not a perfect short with no voltage drop - it was there would be no heat, no flash and no damage.
I bet there was.

Actually as an aside, the most dangerous fault, for anyone standing near it, is not  a perfect short, but one where the maximum power is dissipated in the fault. That occurs when the power dissipated along the line and in the fault are equal, and the energy available to do damage can be estimated from how fast such a fault would be disconnected. The results of such sums on HV systems are almost as unsettling as looking at the equipment that has been blown apart by such events. Large faults need short trip times - including for faults that are not always perfect shorts.
Mike.

If the generator tripped on Woodward that is good news as otherwise the unbalanced rotor would be crashing around in the housing possibly hitting the stator.  I know that it is annoying for a protection engineer as grading the relays even with low instantaneous setting on overcurrent fail to trip in time to beat the Woodward. 

Running several generators in parallel will help. Have you looked at the milli-second printout data of what tripped first yet?