Step 4 of the Hellbore damage procedure (4K3d) says of the second damage number, 'The defending player [who owns the ship] applies one point of damage to each of the other shields until all shields have taken one damage point....if any points remain, repeat the process.
Which process? The whole damage process [i.e. the extra damage begins on the weaker shield] or the five-shield sharing out process?
Hellbore shield damage
Moderators: mjwest, Albiegamer
Yes, I think it means "repeat the process of applying damage to the reamining five shields one point at a time, until you run out of hits". So say for example you had a normal hellbore hit a D7 at range 2, and the D7 has 5 points of damage on the #6 shield and no other damage, the damage would be 9+(8). The Klingon D7 would take the 9 points on the #6 shield as this it the weakest, leaving the remaining 8 damage points to be spread between 5 shields. So this must be one point on each of those 5 shields, plus 3 other damage points between the 5 shields. For example, the Klingon player could assign one of the remaining 3 points to the #1 shield, one to the #2 shield and one more to the #3 shield.
This would make the total damage allocation from the Hellbore hit 9 points on shield #6, 2 points on shield #1, 2 on the #2, 2 on the #3, and 1 damage point on each of shields #4 and #5.
This would make the total damage allocation from the Hellbore hit 9 points on shield #6, 2 points on shield #1, 2 on the #2, 2 on the #3, and 1 damage point on each of shields #4 and #5.
