Length of Wakes of Liberty and an Egyptian Destroyer
By Joseph Graziano (November 14, 2004)
Calculating the length of USS Liberty's wake.
There is a relationship between a ship's submerged surface area and drag caused by a vessel's movement. The rule of thumb is the area changes with the cube root of the increase in displacement squared. In other words, if displacement increases by a factor of 8, the surface area increases by 4 times.
Since Liberty's displacement was 4 times greater than a Z-class destroyer or El Quseir, the latter's submerged surface area was about 4/10ths as great as Liberty's.
Another factor governing drag or turbulence is a ship's beam to length ratio. Liberty's was roughly 1 to 7.5. That of a Z class destroyer 1 to 10. This means the destroyer's hull was about 3/4 Liberty's beam. By multiplying our difference in submerged surface area by the change in beam to length ratio we have 4/10 x 3/4 = 12/40 or 3/10. Inverting this we get 3.3 as a multiplication factor. This tells us that at any given speed Liberty's hull would produce 3.3 times the drag as the destroyer hull.
However, drag increases as the velocity squared. Therefore a destroyer making 28 knots generates approximately 30 times the drag or turbulence as one doing 5 knots. Dividing this by our factor of 3.3 we get 9 as the total increase in turbulence on our destroyer hull at 28 knots as opposed to that which would be produced by Liberty's hull at 5 knots. By multiplying that by the increase in speed we get 50+ as the increase in turbulence per unit of time and distance. In other words, while Liberty's wake at 5 knots would be a fraction of her length, that of a Z class destroyer at 28 knots would be several times the length of that ship. We would find the same difference between a rowboat's wake and that of a speedboat's.
It is surprising that the aircraft pilots didn't fly past Liberty in their hunt for a fast-moving ship that actually had guns to bombard El Arish.