# Thrust generated by a ship

While reading about airplane engines on the internet, it is common to find figures of their thrust, in kilo Newton. However, in the case of ships and naval vessels, when talking about the engine, the only parameter mentioned is the horsepower of the power plant driving the propeller.

So, my question is, what is the typical thrust generated by a large naval ship (such as the nuclear-powered USS Enterprise aircraft carrier, with a power of 210 MW)? How can this be calculated?

• Thrust of a naval screw (don't call it a propellor unless you want to walk the plank) is strongly dependent on avoiding cavitation. Apr 24, 2017 at 13:49
• Thrust of a naval screw is strongly dependent on avoiding cavitation this explains why the ships are slower than air planes. Apr 24, 2017 at 15:23
• @Fennekin I'm sure that's just one of many.
– JMac
Apr 24, 2017 at 16:09
• For tug boats, this parameter is called "bollard pull", i.e. the force that can be applied via a tow rope when the tug is stationary. The usual range is about 1.2 to 1.5 tons per 100 HP (sorry about the non-SI units!) Whether that order of magnitude applies to an aircraft carrier, I have no idea - tugs are not designed to operate at high speeds, for example. Apr 24, 2017 at 16:13

If you know the Froude efficiency $\eta_{\text{Fr}}$ and assume a constant velocity $u$ for the ship. Then you can calculate the thrust $F_{\text{T}}$ for given power at the shaft $P_{\text{S}}$ as $$F_{\text{T}}=\eta_{\text{Fr}}\frac{P_{\text{S}}}{u}.$$
There is also a dimensionless thrust coefficient $C_{\text{T}}$ which is defined as:
$$C_{\text{T}}=\frac{F_{\text{T}}}{1/2\rho v^2 (D/2)^2}.$$
In which $\rho$ is the density of the fluid (here: water) and $D$ is the diameter of the screw. $v = u(1-w)$ in which $w$ is the so-called wake fraction.