# Using calculus to find stopping distance of object towed from a vessel

• Jun 29th 2009, 07:02 AM
Stewart Fraser
Using calculus to find stopping distance of object towed from a vessel
Hello

I studied this thirty years ago but it is all very rusty. The problem is to find out how far a towed object will travel if the boat that is pulling it suddenly stops. (We will be working in ice, and so might possibly stop suddenly ... we don't want to towed object to get caught in the propellers).

The towed object is called a "steamer", a plastic-coated oil-filled cylinder with a 50 mm diameter and usually about 5 km long. It is filled with transducers (hydrophones) and is used in conducting seismic surveys.

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Assuming 5 knots vessel speed and a 5 km Sercel streamer.

Initial velocity : v(t=0) = 5 kts = 5 x 1852 m / 3600 sec = 2.57 m/sec

Mass of one filled 150 m section is 320 kg, so total mass of streamer = (5000/150) x 320 kg = 10,667 kg

Tension at shooting speed is around 1000 kg so this is the drag force at 5 knots (around 10,000 Newtons)

Now ... force = mass x acceleration
So ... acceleration = force / mass = 10,000 / 10,667 = 0.9375 m/sec/sec ... actually deceleration in this case, of course.

Now we can draw a diagram of velocity against time. Velocity is the "y" axis, time is the "x" axis.

The streamer velocity will cut the "y" axis at 2.57 m/sec and cut the "x" axis at 2.74 seconds.

The distance travelled is the area under the line = 2.57 x 2.74 / 2 = 3.521 metres

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However this is wrong as the drag force is not constant, it is proportional to velocity through the water squared (I think) so calculus must be used.

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Could somebody solve this one for me ... please.
• Jun 29th 2009, 08:23 AM
skeeter
initial drag force equals the initial tension ...
$T_i = D_i = -kv_i^2$ ... solve for k using the initial tension and velocity.

when the boat stops, the net force on the object will be drag ...

$m\frac{dv}{dt} = -kv^2$

separate variables ...

$\frac{dv}{v^2} = -\frac{k}{m} \, dt$

integrate ...

$-\frac{1}{v} = -\frac{k}{m} \, t + C_1$

$v = \frac{m}{kt + C_2}$

solve for $C_2$ using the given initial conditions.
• Jun 29th 2009, 05:33 PM
Stewart Fraser
Thanks skeeter

Plugging in the constants I get ....

v = 10667 / (1514t + 4159)

Now to get the distance that it will take to stop I guess we have to integrate v with respect to t. I have looked for the formula on the web. To integrate t^-1 is ln t. But I can not find a formula that would take care of the 4159 value.
• Jun 30th 2009, 04:35 AM
CaptainBlack
Quote:

Originally Posted by skeeter
initial drag force equals the initial tension ...
$T_i = D_i = -kv_i^2$ ... solve for k using the initial tension and velocity.

Valid for high Reynolds numbers, but the TA is 5km long. at 5kts is this still in the high Reynolds number regime? (probably, though will need to check the calculation)

CB
• Jun 30th 2009, 06:18 AM
Stewart Fraser
Thanks CaptainBlack

Yes that was something I was not sure about. But now I have read up on Reynold's number a bit in Wikipedia.

So it should be drag is proportional to the speed through the water (more or less). OK I think this situation is a bit easier mathematically.

I have made a stab at it myself, see notes below;-
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Assuming 5 knots shooting speed and a 5 km Sercel streamer.

Initial velocity : v(t=0) = 5 kts = 5 x 1852 m / 3600 sec = 2.57 m/sec

Mass of one filled 150 m section is 320 kg, so total mass of streamer = (5000/150) x 320 kg = 10,667 kg

Tension at shooting speed is around 1000 kg so this is the drag force at 5 knots (10,000 Newtons)

force (drag) = - constant x velocity .... plugging in the force and speed at time=0 gives

force (drag) = - 3891 x velocity

Using Newton’s law ( Force = mass x acceleration) we get the differential equation of motion we must solve

mass x acceleration = Force = - 3891 x v

mass dv/dt = - 3891 v

=> 10,667 dv/dt = - 3891 v

=> 2.74 dv/dt = - v

Separating the variables

=> 2.74 dv/v = - dt

And now integrating both sides

2.74ln v = - t + constant

Inputting the initial speed and t=0 to determine the constant

2.74 ln v = - t + 2.586

ln v =( 2.586 - t) / 2.74

v(t) = e to the power ( 2.586 - t) / 2.74 ...... [where e is 2.71828....]

or v(t) = 2.57 x e to the power (-0.36477t)

Now we must integrate the above with respect to t to find the distance travelled.

And that is

2.57 x 1/0.36477 = 7.0455 metres (I think)
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Any comments on the above? As I said before I am very rusty when it comes to calculus.