Drag

The speed a vehicle can attain is a function of the available power and the total drag imposed by the vehicle and tether. This is characterized by the equation:

Drag = 1/2 x s AV2 Cd

where:

s = density of sea water/gravitational acceleration

density of seawater = 64 lb/ft3 (1,025 kg/m3)
gravitational acceleration = 32.2 ft/sec2 (9.8 m/sec2)

A = Characteristic area on which Cd is nondimensionalized. For vehicles, it is usually the cross sectional area of the front or the volume of the vehicle to the 2/3 power. For cables, it is the diameter of the cable in inches divided by 12 times the length perpendicular to the flow. For ships, it is the wetted surface.

V = Velocity in feet per second

(1 knot) = 1.689 feet/second
               = 0.51 meters/second

 

Cd = Nondimensional drag coefficient. Cd is in the range of 0.8 to 1 based on the cross sectional area for most vehicles. Cd is in the range of 1.2 for unfaired cables, 0.5 to 0.6 for hair faired cables and 0.1 to 0.2 for faired cables.

NOTE: Calculations will be the same using metric units provided units are consistent. Do not mix meters and centimeters.

 

The power absorbed is characterized by;

where 550 is a constant, which converts feet/pounds/seconds to horsepower. Thus, the power is proportional to the velocity cubed (recall that drag is proportional to velocity squared). Simply stated, because the power absorbed is proportional to the velocity cubed, a vehicle will require (3/2)³ = 3.4 times as much power to go 3 knots as 2 knots. This means that if the power to weight ratio is constant, the propulsion system on a 3-knot vehicle will weigh 3.4 times that of a 2-knot vehicle. This does not turn out exactly this way because components come in discrete sizes. Nonetheless, it is clear that a requirement for higher speed has a dramatic impact on power, which in turn has the same sort of effect on system weight. A rule of thumb is that you can get about 35 to 40 lb (15.9 to 18.1 kg) of thrust per horsepower available.

The vehicle drag is only one part of the equation as the tether usually dominates the vehicle-tether combination. This can be best illustrated by an example for a vehicle cable system.

Drag = 1/2 s Av V2 Cdv + 1/2 s Au Vu2 Cdu
(v = vehicle; u = umbilical)

As an example, suppose a vehicle is being live boated at 1 knot (1.9 km/hr) in 1,000 ft (305 m) of water. Suppose further that the cable is hanging straight down and there is a float on the surface and a weight on the bottom of the umbilical. Assume further that the umbilical drag from the ship to the float is small and the drag on the umbilical from the weight to the vehicle is small. Other data:

Umbilical diameter = 1 inch (2.54 cm)
The frontal projected area of the vehicle = 16 ft2 (1.5 m2)

Then:

Vehicle drag = 1/2 X 64/32.2 x 16 x (1.689)2 x 0.8 = 36 Ib (16.3 kg)

Umbilical drag = 1/2 x 64/32.2 x (1/12 x 1000) x (1.689)2 x 1.2 = 284 Ib (129 kg)

This simple example shows why improvements in vehicle geometry do not make significant changes to system performance.