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The 53-56KE Wakehomer will now cripple a Harper's Ferry LSD with a single hit, two will sink it. (I think it used to take four).
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Molon, are you sure?
Unless the modification to doctrine causes the damage to be modelled differently which is unlikely (I was thinking perhaps that a delayed torpedo detonation on a ship may be somehow modelled as the missing "under keel detonation" but that's farfetched I think), the number of torpedos needed should be the same, as both the warhead of the torpedo and the survivability of the Harper's Ferry have been left unaltered. It should take three 53-56k(e) to sink the HF LSD.
