I am a River Pilot and drive ships for a living. These ships are very large and range up to 160,000 Metric Tons. I am trying to figure out how to calculate the distance to stopping. I have a basic understanding of the physics 101 equations, but I think this is a little more complicated. The reason is because it takes less time for a ship to go from 15kts to 10kts than it does for it to go from 10kts to 5kts. The faster you are going, the quicker the speed comes off due to water pressure. When you get around 1-2kts the ship will float an extremely long distance. That same 1-2 kts came off the 15kts very quickly. I can calculate the negative acceleration rate, but it is different depending on how fast you are going. In the upper speeds the negative acceleration is greater than in the lower speeds. At this point, I would have to take the change in acceleration divided by the change in time, which I have read is known in the physics world as "jerk." So far I have been using $V_f = V_i + AT$ and $dX = \frac{1}{2}AT^2 + V_iT$ , however, I don't know how to calculate the distance and time to 0 kts using an equation that takes into account change in acceleration (jerk). As far as known variables are concerned, every 30 seconds I know the time and the speed. Anyone know how to calculate the total distance to 0kts?
Thank You!
Best Answer
When stopping the engines, how quickly will a ship lose its speed, and how far will it go?
Newton's law tells us the change in the ship's momentum equals the drag force:
$$M \frac{dv}{dt} = - F_{drag}$$
Here $M$ is the ship's mass, and $v$ is its speed. For ships with a large areal cross section $A$ under the water line and a speed $v$ such that $\sqrt{v^2 A} >> \nu$ with $\nu$ the kinematic viscosity (momentum diffusion constant) of the water, the drag force is given by:
$$F_{drag}= \frac{1}{2} C_D \rho v^2 A$$
Here, $\rho$ is the density of the water, and $C_D$ the drag coefficient, a dimensionless constant typically in the range 0.1 - 0.5, depending on the shape of the ship.
This is all you need. The rest is straightforward math. Substituting the equation for the drag force into Newton's law, one readily obtains
$$\frac{dv}{dt}= \frac{-1}{L}v^2$$
With $\frac{1}{L} = \frac{C_D \rho A}{2M}$. The solution to this equation is $v = L/(t+t_0)$ with $t_0$ chosen such that the ratio $L/t_0$ matches the initial speed of the ship.
Clearly, although the ship will shed its speed rapidly at early times, at later times the speed loss slows down considerably. The distance travelled is the integral over $v(t)$:
$$x(t) = L \ln{\frac{t+t_0}{t_0}}$$
Some specific results:
If it takes a time $t_0$ and a distance $(\ln 2) L \ = \ 0.693 L$ to half the ship's speed, it will take an additional time $2t_0$ and an additional distance $0.693 L$ to again half the speed. The total time to reduce the speed by 90% is $9t_0$. During that time period the ship will travel a distance of $2.30 L$
Estimation of the parameter $L$ and $t_0$ from velocity vs time data is easy: $t_0$ is the time it takes to reduce the initial speed $v_0$ to half the value, and $L_0$ is the product $v_0 t_0$.
Note that the derived results are valid up to times $t$ at which $v(t)\sqrt{A} >> \nu$ or $t+t_0 << L \sqrt{A}/\nu$.