Speed after Stop - Mathematical Analysis

It is under discussion whether the STOP command was given immediate before the collision, or immediate after the warning of the lookout. It is not sure whether the time between the warning and the collision was 37 seconds or considerable less or even more.

To have a base to judge the various testimonies I will try to workout how the speed changes after the command STOP is given AND executed immediately.

Summary of Results

As long as the ship is moving through the water with constant speed, the force of the screws equals the resistance of the hull driven through the water. When the screws stop, the ship will slow down by the resistance. The force which slows down the ship decreases with the speed.

Because it takes a few seconds to stop the engines and the screws (there was no gear, so the screws stop in the same way as the engines did), we have distinguish two conditions:

In the first phase of the slowing down of the screws these will neither propel nor brake. The ship slows down by the resistance of the hull only.

The speed change is described by

v(t) = V[sub]0[/sub] * (1+t/T)[sup]-1[/sup]

where:

V[sub]0[/sub] = 23 knots, speed at the time of STOP command

T = constant, 5 minutes.

It is assumed that the screws need 30 seconds to stop. During the first 15 seconds they do neither propel nor brake. The speed after 15 seconds:

V(t) = 23 knots (1+15sec/5min) )[sup]-1[/sup] = 23 knots * 21.9 knots

After the first 15 seconds both the resistance of the hull and the screws will slow down the ship. The equation is the same, however with modified constants:

v(t) = V[sub]0[/sub] * (1+t/T)[sup]-1[/sup]

where:

V[sub]0[/sub] = 22 knots, speed at the time where the screws have stopped

T = constant, 2.9 minutes.

Elapsed time / speed

-------------------------------------------

00 sec . . . . 23.0 knots (78 rpm)

15 sec . . . . 22.0 knots (screws stopped)

30 sec . . . . 20.2 knots

45 sec . . . . 18.8 knots

1 min . . . . . 17.5 knots

2 min . . . . . 13.7 knots

3 min . . . . . 11.3 knots

Fig.1: Speed and elapsed time. For the first 15 seconds only the resistance of the hull has an effect on the speed.

Parks stated that the pistons of the reciprocating engines aboard Titanic would run a cycle or two after steam was redirected, then stop. After considering that I came to the conclusion that the screws indeed can stop quite fast.

Reason:

As long as the ship is going at full speed, the power of the screws

(power = torque * angular velocity ) is used to propel the ship. When the

steam is cut of to stop the ship, the same torque from now on will stop the screw, but not the whole ship. To stop the screw one has just to take the

kinetic rotation energy out of the shafts and the screws. The ship keeps on moving. So I provide an alternative table with the assumption that the ship

is stopped by the resistance of the hull and the not turning screws from the very beginning:

v(t) = V[sub]0[/sub] * (1+t/T)[sup]-1[/sup]

where

V[sub]0[/sub] = 23 knots, speed at the time where the screws have stopped

T = constant, 2.7 minutes.

Elapsed time / speed

--------------------------

00 sec . . . . 23.0 knots (78 rpm), screws stop almost immediately

15 sec . . . . 21.0 knots

30 sec . . . . 19.4 knots

45 sec . . . . 18.0 knots

1 min . . . . . 16.8 knots

2 min . . . . . 13.2 knots

3 min . . . . . 10.9 knots

Fig.2: Speed and elapsed time. It is assumed that the screws stop immediately. From the very beginning the resistance of the hull and the screws slow down the ship.

The main difference between these two versions is the decrease of speed during the first 15 seconds. In the first version the speed decreases 1 knot, in the second one two knots. The difference of speed after 45 seconds is rarely one knot.

I will go on to explain in detail how I found these results.

Coherence of Force, Velocity and acceleration

There are two coherences which we have to consider to describe the slow down:

1. Force and acceleration

F = M * a; Force = Mass * Acceleration ; (equation 1)

2. Force and speed

The force which is necessary to drive the ship through the water increases with the square of the velocity:

F = v[sup]2[/sup] * ½ A * rho * c[sub]w[/sub] ; (equation 2)

where

v: velocity

A: cross section area of ship's part under water

rho: density of water = 1025 kg/m[sup]3[/sup]

c[sub]w[/sub] : constant, depends of shape of the hull

The Equation for speed

1. Velocity and Acceleration

To connect equations (1) and (2) we need the coherence of speed and acceleration.

The acceleration is the derivation of the speed.

a = dv/dt = v´(t) ; (equation 3)

2. Force, Velocity and Acceleration

Equations (1) and (2) both have the force on one side. We replace the acceleration in equation (1) by v´(t):

F = M * a = M * v'(t)

F = v[sup]2[/sup] * ½ A * rho * c[sub]w[/sub]

These forces equal one another at any time:

M * v'(t) = - v[sup]2[/sup] * ½ A * rho * c[sub]w[/sub] ; (equation 4)

The force on the right side has a negative sign because it is directed against ship's course.

We need now a function v(t) with the property that the square of v(t) equals the derivation v'(t). This condition is fulfilled by

v(t) = V[sub]0[/sub] * (1+t/T)[sup]-1[/sup]

where V[sub]0[/sub] is the velocity at the beginning of the slow down.

v[sup]2[/sup](t) = (V[sub]0[/sub])[sup]2[/sup] * (1+t/T)[sup]-2[/sup]

v'(t) = V[sub]0[/sub] * (-1/T) * (1+t/T)[sup]-2[/sup]

We replace v(t) and v'(t) in equation (4):

- v'(t) = v[sup]2[/sup](t) * ½ A * rho * c[sub]w[/sub] / M

V[sub]0[/sub] * (1/T) * (1+t/T)[sup]-2[/sup] = (V[sub]0[/sub])[sup]2[/sup] * (1+t/T)[sup]-2[/sup] * ½ A * rho * c[sub]w[/sub] / M

With t = 0 we receive:

V[sub]0[/sub] * (1/T) = (V[sub]0[/sub])[sup]2[/sup] * ½ A * rho * c[sub]w[/sub] / M

(1/T) = V[sub]0[/sub] * ½ A * rho * c[sub]w[/sub] / M

T = M / ( V[sub]0[/sub] * ½ A * rho * c[sub]w[/sub] )

Determination of Constants T, A, c[sub]w[/sub]

To determine the T we need :

M: mass of the ship, 66000 * 10[sup]3[/sup] kg

A: cross section area = draught * width = 10 meters * 28 meters = 280m[sup]2[/sup]

rho: density of water, 1025 kg / m[sup]3[/sup]

c[sub]w[/sub]

To find the c[sub]w[/sub] we remember to equation (2):

F = v[sup]2[/sup] * ½ A * rho * c[sub]w[/sub]

When we know the force for any speed we can find the c[sub]w[/sub].

In Simon Mills Olympic we find the relevant data:

Power at 21 knots: 50000 IHP (indicated horsepowers)

The power at the screws is about 10% less, 45000 horsepowers.

Not all power of the screws is available to propel the ship. A part gets lost because the water is thrown back. The efficiency is estimated 70%.

Velocity in MKS-units: 21 knots = 21 * 1852 m/3600 sec = 10.8 m/sec

Power in MKS-units: 45000 Ps * 0.735 kW / Ps * 0.7 = 23.152 kW

Force = Power / Velocity = 23,152 kW / 10.8 m/sec = 2143 * 10[sup]3[/sup] Newton

F = v[sup]2[/sup] * ½ A * rho * c[sub]w[/sub]

c[sub]w[/sub] = F / ( v[sup]2[/sup] * ½ A * rho )

c[sub]w[/sub] = 2143 10[sup]3[/sup] Newton / ( (10.8 m/s)[sup]2[/sup] * ½ * 280 m[sup]2[/sup] * 1025 kg/m[sup]3[/sup] )

c[sub]w[/sub] = 0.128 Newton / (kg * m / s[sup]2[/sup]) = 0.128

Now we have all we need to determine the constant T:

T = M / ( V[sub]0[/sub] * ½ A * rho * c[sub]w[/sub] )

T = 66,000 10[sup]3[/sup] kg / ( 10.8 m/s * ½ 280 m[sup]2[/sup] * 1025 kg/m[sup]3[/sup] * 0.128 )

T = 333 sec = 5.5 min

V(t) = 21 knots (1+t/T)[sub]1[/sub], where t is the time, and T = constant 5.5 min.

The constant T depends on the speed at the beginning of the slow down.

T is reciprocal proportional to V. The speed immediate before the collision was about 23 knots through the water. When we enter with 23 knots we need

T = 333 sec * (21/23) = 304 sec = 5 minutes.

V(t) = 23 knots (1+t/T)[sub]1[/sub], where t is the time, and T = constant 5 min.

Using this formula the speed would change like this:

0 . . . . . 23 knots

1 min . . 19.2 knots

2 min . . 16,4 knots

3 min . . 14.4 knots

This formula applies as long as the ship is just slowed down by the resistance of the hull. As soon as the screws are stopped they will add an additional resistance.

We need to know the total area of the blades of the screw.

Additional Resistance of the Screws

The diameter of the port- and starboard screw is 7 meters. I took some measurements from pictures in Simon Mills Olympic and tried to specify the geometrical measures of one propeller:

Diameter of screw: 7 meters

Diameter of shaft: 1.6 meters

long diameter of blade from shaft to edge: 2.7 meters

short diameter of blade : 2.1 meters

area of blade: pi/4 (2.7*2.2) m[sup]2[/sup] = 4.7 m[sup]2[/sup]

area of screw with 3 blades: 14 m[sup]2[/sup]

The diameter of the centre screw is 5 meters only. The area of one blade is 25/49 (square of ratio 5/7), which is about half the area of a blade of the big screw. But the centre screw has four blades, whereas the side screws have only three blades.

Area of the centre screw = area of side screw * ( 25/49 * 4/3 ) = 0.7 * area of side screw.

Total area of port and centre and starboard screws:

Total area = Area of side screw * (1+1+0.7) = 14 m[sup]2[/sup] * 2.7 = 38 m[sup]2[/sup]

We need now an estimate of the c[sub]w[/sub] of a blade.

Examples for some c[sub]w[/sub]

---------------------------------------------------

streamlined body . . . . . . . . 0.056

hull of Titanic . . . . . . . . . . . 0.13

hemispere, opening back . . 0.34

car . . . . . . . . . . . . . . . . . . 0.35 … 0.5

round plate . . . . . . . . . . . . 1.1 … 1.3

hemispere, opening front . . 1.33

What is the blade of a ship's screw like ? While it is not turning it is an obstacle, something between a car and a round plate. The c[sub]w[/sub] is something between 0.5 and 1.1.

I chose 0.8.

(This decision is a typical one of those I made with my stomach. Would like to hear the opinion of an naval engineer about this proceeding).

Again we return to equation (2):

F = v[sup]2[/sup] * rho * ½ A * c[sub]w[/sub]

The term A * c[sub]w[/sub] has two components:

1. cross section area of the ship (part below the water line)

with c[sub]w[/sub] = 0.13

2. area of the screws with c[sub]w[/sub] = 0.8

F = v[sup]2[/sup] * rho * ½ ( 280 m[sup]2[/sup] * 0.13 + 38 m[sup]2[/sup] * 0.8 )

F = v[sup]2[/sup] * rho * ½ ( 36.4 m[sup]2[/sup] + 30.4 m[sup]2[/sup] )

The additional resistance by the screws not turning is nearly the same as the resistance by the hull of the ship. The force which makes the ship slow down increases 1.8fold with the screws being stopped!

So we have to determine again the T-constant in the equation for speed.

We chose V[sub]0[/sub]= 21 knots = 10.8 m/s.

T = M / ( V[sub]0[/sub] * rho * ½ A * c[sub]w[/sub] )

T = 66,000 10[sup]3[/sup] kg / ( 1025 kg/m[sup]3[/sup] * 10.8 m/s * ½ 66.4 m[sup]2[/sup] )

T = 179.6 sec = 3 minutes.

v(t) = 21 knots * (1+t/T)[sub]1[/sub], where T = 3 minutes

The constant T depends in the same way reciprocal to the velocity at start condition.

For 23 knots we have T = 3 min * (21/23) = 2.7 minutes

For 22 knots we have T = 3 min * (21/22) = 2.9 minutes

Frictional Forces were neglected !

Note: The ship will never come to stop using this formula alone:

In fact there are two forces:

1. A frictional force which increases linearly with the speed

2. resistance which results from the difference of pressure before and after the object moving through the water. This force increases with the square of the speed.

While considering high speed the frictional forces can be neglected. But these will finally make the ship stop.