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!

  • $\begingroup$ Drag should be the Square of the speed. Freeboard catches wind and introduces another variable. Experience is the best teacher. The mind receiving feedback will adjust and correct for unknown variables. $\endgroup$ Jul 28, 2013 at 9:01
  • $\begingroup$ @Optionparty, there's only so much the mind can do. That's why we have to rely on physics! $\endgroup$
    – udiboy1209
    Jul 28, 2013 at 11:59
  • $\begingroup$ My first guess would be, as @Optionparty said, to let drag be proportional to speed squared. Then write the differential equation and solve it, numerically or analytically. Don't get hung up on "jerk". It's just a name for a derivative. $\endgroup$ Jul 29, 2013 at 11:23
  • $\begingroup$ Thanks, but I don't really now know how to do what you are saying. $\endgroup$ Jul 30, 2013 at 11:27

5 Answers 5


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 $t+t_0 << L \sqrt{A}/\nu$.

  • 1
    $\begingroup$ Does this indicate then a ship will come to a stop quicker in salt-water than in fresh-water? $\endgroup$
    – Everyone
    Jul 29, 2013 at 13:08
  • 5
    $\begingroup$ @Everyone - Salt water has a higher density than fresh water and therefore the $\rho$ in the drag force will be higher. However, this is counterbalanced by the fact that the higher density of salt water would also mean that it has a higher buoyancy effect and therefore the cross sectional area $A$ will decrease. The net effect (the value of the product $\rho A$) will depend on the shape of the hull. $\endgroup$
    – Johannes
    Jul 29, 2013 at 13:14
  • $\begingroup$ @Everyone - and if the cross section $A$ changes, there will also be an effect on $C_D$. Whether the drag in salt water will be higher than in fresh water will depend on the combined effects on $\rho A C_D$. $\endgroup$
    – Johannes
    Jul 29, 2013 at 13:54
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    $\begingroup$ Thank you for your detailed answer. The main problem is that I do not know the block coefficient of the ship. All I have is velocities and times. I was thinking that maybe figuring out the change in acceleration, jerk, would help me estimate the distance. I guess the question is will the change in acceleration be the same from 15-10kts and 5-0kts? Do you think this would work? Please remember I have limited physics knowledge. Thanks! $\endgroup$ Jul 29, 2013 at 14:53
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    $\begingroup$ @Trimok - The model is intended to model the slowing down of the ship under turbulent flow conditions (as mentioned: $v >> \nu/\sqrt{A}$). The last fraction of a percent of the slowing down of the ship's velocity is not modeled (and hardly relevant for the problem at hand). $\endgroup$
    – Johannes
    Jul 29, 2013 at 16:37

Nice to see some ship questions around here, I'm a naval engineer!

So, you're looking for a simple, ball-park number for a question that is in reality pretty complicated. Johannes answer might give reasonable results since it's constantly updating the number; but I want to point out some assumptions made here which might affect the result's accuracy.

Background Info: First is that Johannen's $C_d$ (which in Naval Arch is usually named $C_t$ since it would correspond to the total drag coefficient) is actually described as $C_t = f_1(\frac{V^2}{gL})+f_2(\frac{VL}{v})$, where $f_1$ and $f_2$ represent the wave-making (residuary) resistance coefficient ($C_r$) and frictional resistance coefficient ($C_f$) accordingly, $V$ is the ship's speed, $L$ is the ship's length, $g$ is gravitational acceleration, and $v$ is water's viscosity. As you can see, it's far from constant and changes from ship to ship, strongly dependent on their length. So to have an accurate result for your computer algorithm, you would need the chart for the boat's $C_t$. But even if you had this, it would still be off (but on the conservative side) since ships fouling strongly affect $C_f$.

Answering your Question: If your speed readings updated a little quiker, you could approximate the "instantanous" $C_t$ by approximating it with a Taylor's expansion, and then setting a system of equations with Johannes third equation. However, even with a first order approximation, you would need 3 samples or 1.5 minutes to get your first reading. And this might mean that the your "accuracy" might be lagging by the same amount. So, it might be that without any prior information of the ships (and no fancy smart/learning algorithms saving/estimating information of the ships from past data), the best you could do is Johannes approach, with some few modifications so that you can get the information you are asking for:

Quick-and-Dirty Method: First (sorry for any Kosher mathematicians out there), consider that:

$$\frac{\partial^2 x}{\partial t^2} = \frac{\partial }{\partial x}\left(\frac{\partial t}{\partial x} \right ) = \frac{\partial V}{\partial t}\left ( \frac{\partial x}{\partial x} \right ) = V\left ( \frac{\partial V}{\partial x} \right )$$

Substituting this to Johannes third equation, and integrating using separation of variables (let's assume that Johannes $L$ is actually constant, and let's name it $\alpha$) with limits of integration $(0-x_{end})$ and $(V_0-\delta )$ for the $x$ and $V$ accordingly, we get:

$$x_{end} = \alpha\ln\left(\frac{V_0}{\delta}\right)$$

where $V_0$ would be your initial speed (in your case, your current speed), $\delta$ is the speed your going to end at, and $\alpha$ you assume to be a constant (but in reality you'll updated at each time step). You mentioned you want $\delta$ to be zero, but as you can see this is not possible, your result would be infinity (classic example of Zeno's Paradox, as Johannes result more clearly illustrates).

You have many options to estimate $\alpha$. If you get erratic results with the most basic option I'm going to present here, I recommend you look into derivative smoothing. The most basic option would be to use a numerical derivative in Johannes third equation, $$\frac{V_t-V_{t-1}}{\Delta t} = \frac{-1}{\alpha_t} V_t^2$$ Solving for $\alpha$, $$\alpha_t = \frac{V_t^2\Delta t }{V_{t-1}-V_t}$$ To make this obvious, you'll calculate at each time step $\alpha_t$, and apply it on $$x_{end} = \alpha_t\ln\left(\frac{V_t}{\delta}\right)$$ Now $\delta$ would have to be a speed you'll reach when you're at $x_{end}$ (this result will be very ballpark, for the reasons I commented above). You mentioned zero, so a speed you'll consider small enough to be zero... perhaps 0.02 knots? But let's be real, in a river you'll have currents so you'll never really get to zero unless you're going upstream or you're facing some strong winds. So you'll have to play around with $\delta$ until you get results that seem useful to you (and probably conservative as well).


It appears given your other post that you have already determined something to this information you have requested.

However, I do not think, though, that your answer is going to be found in determining the jerk. You really have a drag force acting on your boat. Rather than a higher-order time component, you have an extra velocity component with your total acceleration: $$ v_f =v_i + a_{tot}t = v_i + \left(a_{ship} + k v^\alpha \right)t $$ where $k$ is some constant and $\alpha$ is a power (usually either 1 or 2).

  • $\begingroup$ Well, I have limited physics knowledge. I was thinking that since the acceleration is changing, I must determine the rate at which it is changing. The problem is, I only know velocities and times for the ship. I do not know the effect the shape of the hull has on the drag or its block coefficient. I'm not sure if the jerk equation will work, but I was hoping to test it and see. Thanks $\endgroup$ Jul 29, 2013 at 14:50

You could begin with a law :

$$\dot v = - a(1+bv^2)$$ where $a$ and $b$ are positive constants.

The integration of this gives, (see this), the formula :

$$v(t) = \frac{v_0 - \large \frac{\tan (a \large \sqrt{b}t)}{\sqrt{b}}}{1+\sqrt{b}v_0 \tan(a \sqrt{b}t)}$$

The ship stops at time :

$$t_{stop} = \frac{\tan^{-1}(\sqrt{b} v_0)}{a \sqrt{b}}$$

The equation for $x(t)$ is, (see this):

$$x(t) = - \frac{\log (\sec^2(a \sqrt{b}t)) - 2 \log(\sqrt{b} v_0 \tan(a \sqrt{b}t)+1)}{2ab}$$

Plugging $t_{stop}$ in this equation, we get :

$$x_{stop} = \frac{\log(1 + b v_o^2)}{2ab}$$

  • 2
    $\begingroup$ where does the law come from? for v=0, your $\dot{v}$ = -a, that's clearly wrong. $\endgroup$
    – mart
    Jul 29, 2013 at 11:18
  • $\begingroup$ I think the first equation should read $\dot v = - a v(1+bv)$. $\endgroup$
    – Johannes
    Jul 29, 2013 at 14:14
  • $\begingroup$ @mart : It is a model. A non-null $a$ is required if you want that the ship stops. $\endgroup$
    – Trimok
    Jul 29, 2013 at 15:19
  • $\begingroup$ @Johannes : A non-null $a$ is required if you want that the ship stops. With the model you propose in your comment , you will have a decreasing exponential speed. , so the ship does not stop in a finite time, and the distance before stopping is infinite, (this is the case too for the model that you propose in your answer) $\endgroup$
    – Trimok
    Jul 29, 2013 at 15:23
  • $\begingroup$ That's right, the ship never really comes to rest. It just keeps reducing it's speed. $\endgroup$
    – Johannes
    Jul 29, 2013 at 15:47

As you know the wave making resistance will reduced by reducing the ship speed. but this resistance does not decrease linearly. When the ship moves at a slow speed its value is very low. And non-linearly with increasing speed wave making resistance will increase. Thus reducing the speed at high speeds faster will happen

  • $\begingroup$ Increasing the occurences of the capital cases and markings would help people to understand your posts. $\endgroup$
    – peterh
    Dec 25, 2016 at 20:47

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