Just for s$%^ and giggles, estimate the wetted area as an ellipsoid.
a=draft, 55 ft
b=length/2, 1000 ft
c=beam/2, 131 ft
From wikipedia, the area of an ellipsoid can be approximated as this nasty thing:
Where p is a factor of about 1.6.
Our wetted area is half of that, though there's some fudge because of course it's a bit bigger than that because it's not a semi-ellipsoid, and by observation of the hull form and an ellipsoid I think it'll have more area. Say...50% more? Wetted area is thus 66,761 m^2.
From this paper, coefficient of drag using the wetted area should be something like 4*10^-3, and thus using the density of water (1000 kg/m^3) we can solve for drag via good old D=(0.5*density*v^2)*(area)*(drag coefficient). Power for steady speed is drag times the velocity, so I can calculate power requirements. Say that the prop is 80% efficient in turning shaft power into power in the water, and we can find the required engine output.
Results for sample top speeds:
Code:
Speed Drag Power Req'd
20 knots 14 MN 243,718 HP
22 knots 17 MN 324,388 HP
25 knots 22 MN 476,011 HP
27 knots 26 MN 599,637 HP
30 knots 31 MN 822,547 HP
So...you know. To make 30 knots, it'd only take the thrust equivalent of the Saturn V, and about 613 MW of power. This would be about 6 A4W reactors, or twin A1B reactors (based on
this page listing 104 MW shaft power for the A4W and
this one saying the A1B has about triple that). Or about 8 of the largest diesels existing in our world:
Well, that was fun mental floss. People can point out where I screwed up, and I'll deal with it in the morning.