TheNavalArch is pleased to launch a new article: How to use empirical formulas to estimate a ship’s resistance
The article talks about the popular ‘Holtrop-Mennen’ method of resistance estimation for ship-shaped vessels in detail.
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I like to tinker with hydrodynamics every now and then. The ship’s resistance is an important parameter in this. During the time of the Milano Bridge accident I did some crude calculating exercises on that ship which I will copy here.
The dimensions of the ship are 365x51.2x15.5 meters. The weight with a Cb of 0.77 is then 365x51.2x15.5x0.77 = 223041 ton. The deadweight is 146931.
The frictional resistance coefficient is related to the surface roughness of the hull and calculated according to the formula it is 0.001387 and the friction resistance Rf = 929 kN
In the formula rho = 1025, the wetted surface S = 24130 and V the service speed of 14.3 knots in m/s, that is 7.36 m/s.
The wetted surface for container ships (based on analysis of 38 vessels) can be calculated according to following formula :
Now we have to calculate the total resistance Rt which is more difficult but I have a workaround for that.
I use an older version of the Holtrop Mennen excel sheet with a maximum ship length of 300 m while is ship is 365 m. Filled in it gives the following result:
These very long ships have a rather ridiculous low Froude number which is responsible for a relatively low resistance. Here Fn = V/(gL)^0.5 = 7.36/ (9.81 * 365)^0.5 = 0.123. See the graph.
Rf is here, due to the 300 m limit, 760.4 kN instead of 939 kN. Therefore I multiply the Rt of 1260 kN with 929/760.4 = 1.22 and that is 1539 kN. Power = force x speed = 1539 x 7.36 = 11330 kW.
As the ship is travelling at a constant speed, there is no resultant force on the ship. In this case the forward force of the engine must have the same magnitude as the resistance forces, i.e. 1539 kN.
The brake power Pb is due to all the losses 2.42 x the effective power Pe. Therefore the Pe here is 2.42 x 11330 kW = 27419 kW that is 56% of the engine’s power of 48900 kW. Theoretically on full power the ship would run 17.4 knots. We can also calculate the acceleration factor a. As F = m*a then a = F/m = 1539/223041 = 0.0069 m/s^2.
In a second part I will do a few more calculations like stopping distance.
As the ship is travelling at a constant speed, there is no resultant force on the ship. In this case the forward force of the engine must have the same magnitude as the resistance forces, i.e. 1539 kN. The forward force is D(rive) and the resistant force is R thus D - R = 0
The acceleration formula is F or D = m*a Then a = F/m = 1539/223041 = 0.0069 m/s^2. This is a constant acceleration and as such it is a more specific type of accelerated motion. An object that travels with constant acceleration has a speed that changes by the same amount each second.
If the engine is stopped the Drive force is zero and only the Resistance force is present. The ship will loose speed from that moment on due to primarily hull friction and other friction factors. The hull friction amounts to 929/1539 = 0.6 or 60% of the total resistance.
We use the formula v = u + at to find the time till the ship is completely stopped, thus u = 0 and v = 7.36 m/s and because the acceleration is in this case not constant, due to the fact that the resistance force is getting smaller and smaller, we take the mean value between 0.0069 and 0, once stopped: 0.00345 m/s^2 -----> 7.36 = 0 + 0.00345 * t, the t = 7.36/0.00345 = 2133 sec = 35.55 minutes.
The distance travelled, also using the mean value of a, in that time is ------> s = 1/2 a t^2 = 0.5 * 0.00345 * 4549689 = 7848 m = 4.24 nm. All this with no wind and waves! The stopping distance is 13.33 times the ship length, just over the maximum of 13 ship lengths allowed by IMO requirements, but then my calculation is not that accurate. It is no wonder that with the enormous mass of the ship, the slender form, the low resistance as also reflected by the very low Froude number of 0.123, it drifts on and on…
Thanks for adding to the discussion!
