STOP engineorder command


Status
Not open for further replies.
Mar 3, 1998
2,745
261
358
There is enough corroborating testimony to assert that Murdoch sent an engine order down to the engine room just prior to or during the collision. Some take Boxhall's cue and assert that command was FULL ASTERN. I look at other testimony and conclude the order sent down was STOP (there may be some who remember my earlier theories about twisting the engines...I have debunked those myself after thorough consideration).

But why STOP? I am one who believes that Murdoch ordered the bow first turned to port, then to starboard, to clear the iceberg. Surely, he could have accomplished that maneuver at FULL AHEAD? Why stop the engines? I have tried to advance a few theories on this, but none sounded convincing, even to me.

There has been discussion about a possible loss of rudder effectiveness while speed is bleeding off the ship, but I have never felt that played a significant role in the collision. At an original speed of 22.5 knots through the water, there was sufficient water acting on the rudder to keep it effective during the avoidance maneuver, no matter if the screws were turning or not.

I was considering what the engines were doing at the time (as part of discussion on the "Reversing Engines" thread) when I remembered something from my shiphandling days. Modern naval ships (ones I have been on, at least) have a STOP SHAFTS command, in addition to the normal STOP command. Nowadays, STOP means to close the throttles, leaving the screws windmilling in the stream of water passing down the length of the ship. STOP SHAFTS is an emergency order that applies steam necessary to stop the rotation of the shafts and essentially lock the screws in one position. This order is used mostly to protect the propeller blades when the ship is passing close by an obstacle.

In my view, that clinched it. The pistons of the reciprocating engines aboard Titanic would run a cycle or two after steam was redirected, then stop. The wing propellor shafts, which were directly linked to the engines, would likewise stop and their respective props would halt in place (the centre prop would wind down with the turbine). Murdoch came from the Olympic, which had already thrown one blade after a collision with an underwater obstacle, so he might have been mindful of protecting Titanic's starboard prop (or, if he was aware of Dave's grounding theory, both wing props). With the obstacle close aboard, Murdoch knew he had enough momentum to complete his maneuver, so ordering STOP was not a speed consideration, as we've always assumed.

So, basically, I'm saying that Murdoch intended to avoid the iceberg with the rudder, while stopping the props to protect them from incidental damage. This makes practical sense to me...does it for you?

Parks
 

Inger Sheil

Member
Feb 9, 1999
5,343
67
398
Parks, you could feed me anything technical and I'd believe you! You know I'm a follower when it comes to helm orders...

On a slightly different but related note, Moni was out having another look at Smith's testmony at the Olympic/Hawke Inq again (she was mainly indulging in copying Thomas Andrews' testimony as it has some entertaining asides, but she did look at the rest of it as well). There was one phase of the testimony where the questions dealt with Smith's understanding of how the centre prop worked. I'm not even going to attempt to paraphrase here in case I bollix it up, but would be happy to snag a copy for you next time I'm out there. That and the testimony of the Olympic's engineers who were called to give evidence might be useful to you.

All the best,

Ing
 
Mar 3, 1998
2,745
261
358
Ing,

Yes, please...I'll take anything you come up with, even if it only concerns the colour of Lady Astor's bloomers. :)

Parks
 
M

Morgan Eric Ford

Guest
I'd like to see the Olympic/Hawke inquiry. Has it been published somewhere?

What is SOP for a modern multi-screw ship with a inoperative turbine on one shaft? Is the idle shaft allowed to pin wheel or is it locked in place? I read that the "Queen Mary" made her voyage to Long Beach on two shafts to save fuel.

Morgan
 
Mar 3, 1998
2,745
261
358
Morgan,

I'll let Inger tell you about the Olympic/Hawke inquiry. She's the one who stumbled across the unindexed boxes of materials in the PRO.

I can say that modern naval vessels allow idle shafts to windmill in the slipstream in order to minimise unwanted drag.

Parks
 

Bill Sauder

Member
Nov 14, 2000
230
20
263
By Morgan Eric Ford asks:

What is (the standard operating procedure) for a modern multi-screw ship with an inoperative turbine on one shaft? Is the idle shaft allowed to pin wheel or is it locked in place? I read that the "Queen Mary" made her voyage to Long Beach on two shafts to save fuel


Eric,

That depends entirely on the ship's design, the nature of the damage, and the chief engineer's solution. To offer a concrete example using the Queen Mary
each of the ship's 4 engines had 4 turbines (High Pressure, 1st Intermediate Pressure, 2nd Intermediate Pressure and Low Pressure) driving a single propeller through a reduction gear. The steam pipes were arranged in such a way that any turbine (except the 2nd IP) could be isolated and the engine run at 75% of its normal horse power.

If the damage is such that the entire engine is disabled (say teeth failed on a reduction gear) then my suggestion would be to disconnect the engine from the screw and allow the latter to windmill. This should be done because of problems with asymmetric drag, a waste of fuel to overcome the resistance of a dead propeller, and to help reduce pressure brought to bear on the propeller blades by the slip stream.

Again, in the case of the QM, I would lock the engine with the jacking screw and lock the propeller shaft separately with a friction break mounted in the shaft alley for that express purpose. Once the drive train was locked, there are about 18, 4 inch bolts that disconnect the bull gear drive shaft from the first segment of the propeller shaft. Once the bolts are out and the blanking flange removed, release the break on the shaft and allow the propeller to windmill. Thrust will be taken up by the main thrust block for that shaft which is "downstream" from the disconnect point. Of course, a single lube oil pump must remain on duty for that engine to keep the main thrust block lubed.

I have deliberately not addressed the question of how to turn the turbines over with steam applied to prevent warping.

BTW, the drag from the two dead screws on the QM's last voyage caused her to run late her entire "Great Last Cruise" and she wound up calling most of her stays in port short.

Bill Sauder
 

Inger Sheil

Member
Feb 9, 1999
5,343
67
398
Hallo, Morgan -

There are several boxes of evidence from the Hawke/Olympic inquiry held in the Public Records Office at Kew. They have not yet been microfilmed/fiched (but then, neither have the crew agreements for the Lusitania's last voyage, believe it or not).

One box contains a bound publication of testimony taken at the inquiry into the collision, which is formatted much the same as the Titanic inquiry (numbered questions etc.) First witness was EJ Smith on the 16/11/1911, then Bowyer. Second day had Olympic the officers Harold Holehouse, David Alexander, William Murdoch, Henry Wilde, Adolphus Tulloch, Robert Hume and QM Albert Haines, then Chief Engineer Fleming and four other engineers. The third day was primarily witnesses and expert testimony, including Thomas Andrews (who was rather reluctant to part with his plans until he had an assurance that they would not be copied!). Don't have notes on the Hawke witnesses, so can't tell you who testified on that side.

The other boxes consist of 'loose' evidence, and are arranged in no particular order. Blueprints, photos of the damage on both craft, letters from the public with comments...a real mixed bag. I keep trying to get Parks over here so he can have a snuffle through and sort out any material of significance.

All the best,

Inger
 
Mar 3, 1998
2,745
261
358
Ing,

All you have to do is provide me with two tickets (my wife will not allow me to fly to England without taking her) and a babysitter for the kids left behind while we're gone and I'm there!

Would the Olympic material find under a raincoat, do you think? :)

Parks
 

Cal Haines

Member
Nov 20, 2000
308
1
263
Tucson, AZ USA
Inger wrote:
... The other boxes consist of 'loose' evidence, and are arranged in no particular order. Blueprints, photos of the damage on both craft, letters from the public with comments...a real mixed bag. ...

Hi Inger,

If you get a chance next time you are there, could you see if the Olympic blueprints have titles on them? The Harland & Wolff drawings that I have seen have a title centered at the top of the sheet and/or vertically across one end. I would be very interested to know what those titles are.

re the Titanic Inquiry, have you seen any trace of the statements of the crew that were taken immediately upon their return to England?

Warm Regards,

Cal
 

Inger Sheil

Member
Feb 9, 1999
5,343
67
398
Hallo, Cal -

Blueprints I saw were, unfortunately, all from the Hawke (I could be wrong, but we did pay special attention for obvious reasons to these documents). Thomas Andrews was quite entertaining on the subject of his plans - he refused to even hand them over as evidence until he was assured that they would not be copied and would be returned to him.

I do have a copy of one of the crew statements, but it is not from a public archive. I haven't begun a concerted effort to find the others yet, although one would hope they still exist.

Will answer your email when I get back in just over a week :)

Parks, should that big lottery windfall come my way you and the whole family and the babysitters will be coming over here.

All the best,

Ing
 

Cal Haines

Member
Nov 20, 2000
308
1
263
Tucson, AZ USA
Hi Ing,

Thanks for the information. Those crew statements have been of intense interest to me since I learned the they had been taken.

Is it true that some of the Titanic documents are still under seal by the British government?

Enjoy your trip! I hope your Jet Lag isn't too bad.

Warm Regards and Holiday Wishes,

Cal
 

Erik Wood

Member
Aug 24, 2000
3,519
15
313
Dear Mr. Stephenson,

I would appear as though you and I may agree on one thing. Why did Murdoch stop his forward momentum in the middle of a evasive manuver? Or did he? Smith ordered full astern with the New York hopping the prop wash would clear the ship.

Erik
 
Mar 3, 1998
2,745
261
358
Erik,

I contend that Murdoch did not stop Titanic's forward momentum. He knew he was close aboard an obstacle, so he ordered STOP on the engines as a temporary condition in order to protect the propellers. Had Titanic not struck, I would assume that Murdoch would have immediately resumed speed.

The incident with the New York was in a restricted channel, not the open ocean...two entirely different scenarios.

Parks
 

Erik Wood

Member
Aug 24, 2000
3,519
15
313
Parks,

They were to different scenarios but they both used the stop manuver. I would think as stated in a different thread that Murdoch was attempting a crash stop but forgot about the helm needing to stay in a "central"position. He put the helm over and as the berg passed along the starboard side he also ordered a port rounding. But either way he has the engines at stop and he forward momentum is falling off. The rudder of Titanic was far to small for the ship one would assume that any removal of free water moving past it what have an effect on it's effectivness.

The New York was something completely different. Smith used the Stop or Full Astern (which escaptes me presently) to create a prop wash to clear the aft end away. Now obviously because of the confines of the channel this was easy to accomplish. Now that I think on it I don't know why I brought up the New York in this thread. I must be getting OLD.

Erik
 

Cal Haines

Member
Nov 20, 2000
308
1
263
Tucson, AZ USA
<FONT COLOR="119911">in subtopic "Reversing engines" Cal wrote:
Do you suppose Murdock's initial engine order was one that just alerted the engine room to standby? They might have had a prearranged signal to alert the engine crew, perhaps it was ringing "full-ahead" or "standby" followed immediately by "full-ahead"?

Parks wrote:
... I used to think, as Dr. Griffiths did, that STOP was rung down in preparation for another engine command that never came. However, I have convinced myself this past week that Murdoch rang STOP to protect the propellers from hitting an obstacle that he was about to pass by close aboard. ...

Hi Parks,

"Stop" to protect the propellers makes perfect sense to me--I'm with you until someone comes up with a better explanation. But if Murdock was intending to try to port around and wanted to keep power on the engines, at least on the wing propellers, it would make good sense to take some steps to alert the engine room that he was maneuvering and might need them to answer bells--particularly if he was anticipating the need to stop the props, as you suggest. If there was some sort of pre-arranged signal, the response from the engine room might well be to a) cut steam production (hence the "Stop" order sent to the boiler rooms} and, b) drop out the turbine. Just something to think about.

Oh my, look at the time! I'd better get started on my Christmas shopping.
happy.gif


Cal
 
Nov 5, 2000
245
0
261
65
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] * &frac12; 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&acute;(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&acute;(t):

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

F = v[sup]2[/sup] * &frac12; A * rho * c[sub]w[/sub]

These forces equal one another at any time:

M * v'(t) = - v[sup]2[/sup] * &frac12; 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) * &frac12; 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] * &frac12; 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] * &frac12; A * rho * c[sub]w[/sub] / M

(1/T) = V[sub]0[/sub] * &frac12; A * rho * c[sub]w[/sub] / M

T = M / ( V[sub]0[/sub] * &frac12; 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] * &frac12; 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] * &frac12; A * rho * c[sub]w[/sub]

c[sub]w[/sub] = F / ( v[sup]2[/sup] * &frac12; A * rho )

c[sub]w[/sub] = 2143 10[sup]3[/sup] Newton / ( (10.8 m/s)[sup]2[/sup] * &frac12; * 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] * &frac12; A * rho * c[sub]w[/sub] )

T = 66,000 10[sup]3[/sup] kg / ( 10.8 m/s * &frac12; 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 * &frac12; 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 * &frac12; ( 280 m[sup]2[/sup] * 0.13 + 38 m[sup]2[/sup] * 0.8 )

F = v[sup]2[/sup] * rho * &frac12; ( 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 * &frac12; A * c[sub]w[/sub] )

T = 66,000 10[sup]3[/sup] kg / ( 1025 kg/m[sup]3[/sup] * 10.8 m/s * &frac12; 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.
 
Status
Not open for further replies.

Similar threads

Similar threads