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59b023af44103_Propellers2.jpg.aa758914ecda5801e6e423e4f7bc111c.jpg

Often wrongly captioned as the propellers on the RMS Titanic these are in fact those of her sister ship RMS Olympic

 

One of the most compelling moments in Cameron's film Titanic for myself are the engine room scenes as the order is received to throw the two huge tripe-expansion steam engines, along with the Parson's central screw turbine, into reverse. And here it is:

Spoiler

 

But steam engines are for another topic, I am concerned with propellers here.

I had a general appreciation for the Archimedes screw in history, but that was about it, so I had a look into the subject of propellers and it is off the deep end in complexity, pardon the pun. Feel your eyes glaze over scanning this if you dare:

Spoiler

Propeller thrust

Single blade

Taking an arbitrary radial section of a blade at r, if revolutions are N then the rotational velocity is 2πNr{\displaystyle \scriptstyle 2\pi Nr}\scriptstyle 2\pi Nr. If the blade was a complete screw it would advance through a solid at the rate of NP, where P is the pitch of the blade. In water the advance speed is rather lower, Va{\displaystyle \scriptstyle V_{a}}\scriptstyle V_{a}, the difference, or slip ratio, is:

Slip=NP−VaNP=1−Jp{\displaystyle {\mbox{Slip}}={\frac {NP-V_{a}}{NP}}=1-{\frac {J}{p}}}{\mbox{Slip}}={\frac  {NP-V_{a}}{NP}}=1-{\frac  {J}{p}}

where J=VaND{\displaystyle \scriptstyle J={\frac {V_{a}}{ND}}}\scriptstyle J={\frac  {V_{a}}{ND}} is the advance coefficient, and p=PD{\displaystyle \scriptstyle p={\frac {P}{D}}}\scriptstyle p={\frac  {P}{D}} is the pitch ratio.

The forces of lift and drag on the blade, dA, where force normal to the surface is dL:

dL=12ρV12CLdA=12ρCL[Va2(1+a)2+4π2r2(1−a′)2]bdr{\displaystyle {\mbox{d}}L={\frac {1}{2}}\rho V_{1}^{2}C_{L}dA={\frac {1}{2}}\rho C_{L}[V_{a}^{2}(1+a)^{2}+4\pi ^{2}r^{2}(1-a')^{2}]b{\mbox{d}}r}{\mbox{d}}L={\frac  {1}{2}}\rho V_{1}^{2}C_{L}dA={\frac  {1}{2}}\rho C_{L}[V_{a}^{2}(1+a)^{2}+4\pi ^{2}r^{2}(1-a')^{2}]b{\mbox{d}}r

where:

V12=Va2(1+a)2+4π2r2(1−a′)2dD=12ρV12CDdA=12ρCD[Va2(1+a)2+4π2r2(1−a′)2]bdr{\displaystyle {\begin{aligned}V_{1}^{2}&=V_{a}^{2}(1+a)^{2}+4\pi ^{2}r^{2}(1-a')^{2}\\{\mbox{d}}D&={\frac {1}{2}}\rho V_{1}^{2}C_{D}{\mbox{d}}A={\frac {1}{2}}\rho C_{D}[V_{a}^{2}(1+a)^{2}+4\pi ^{2}r^{2}(1-a')^{2}]b{\mbox{d}}r\end{aligned}}}{\begin{aligned}V_{1}^{2}&=V_{a}^{2}(1+a)^{2}+4\pi ^{2}r^{2}(1-a')^{2}\\{\mbox{d}}D&={\frac  {1}{2}}\rho V_{1}^{2}C_{D}{\mbox{d}}A={\frac  {1}{2}}\rho C_{D}[V_{a}^{2}(1+a)^{2}+4\pi ^{2}r^{2}(1-a')^{2}]b{\mbox{d}}r\end{aligned}}

These forces contribute to thrust, T, on the blade:

dT=dLcos⁡φ−dDsin⁡φ=dL(cos⁡φ−dDdLsin⁡φ){\displaystyle {\mbox{d}}T={\mbox{d}}L\cos \varphi -{\mbox{d}}D\sin \varphi ={\mbox{d}}L(\cos \varphi -{\frac {{\mbox{d}}D}{{\mbox{d}}L}}\sin \varphi )}{\mbox{d}}T={\mbox{d}}L\cos \varphi -{\mbox{d}}D\sin \varphi ={\mbox{d}}L(\cos \varphi -{\frac  {{\mbox{d}}D}{{\mbox{d}}L}}\sin \varphi )

where:

tanβ=dDdL=CDCL=12ρV12CLcos⁡(φ+β)cos⁡βbdr{\displaystyle {\begin{aligned}tan\beta &={\frac {{\mbox{d}}D}{{\mbox{d}}L}}={\frac {C_{D}}{C_{L}}}\\&={\frac {1}{2}}\rho V_{1}^{2}C_{L}{\frac {\cos(\varphi +\beta )}{\cos \beta }}b{\mbox{d}}r\end{aligned}}}{\begin{aligned}tan\beta &={\frac  {{\mbox{d}}D}{{\mbox{d}}L}}={\frac  {C_{D}}{C_{L}}}\\&={\frac  {1}{2}}\rho V_{1}^{2}C_{L}{\frac  {\cos(\varphi +\beta )}{\cos \beta }}b{\mbox{d}}r\end{aligned}}

As V1=Va(1+a)sin⁡φ{\displaystyle \scriptstyle V_{1}={\frac {V_{a}(1+a)}{\sin \varphi }}}\scriptstyle V_{1}={\frac  {V_{a}(1+a)}{\sin \varphi }},

dT=12ρCLVa2(1+a)2cos⁡(φ+β)sin2⁡φcos⁡βbdr{\displaystyle {\mbox{d}}T={\frac {1}{2}}\rho C_{L}{\frac {V_{a}^{2}(1+a)^{2}\cos(\varphi +\beta )}{\sin ^{2}\varphi \cos \beta }}b{\mbox{d}}r}{\mbox{d}}T={\frac  {1}{2}}\rho C_{L}{\frac  {V_{a}^{2}(1+a)^{2}\cos(\varphi +\beta )}{\sin ^{2}\varphi \cos \beta }}b{\mbox{d}}r

From this total thrust can be obtained by integrating this expression along the blade. The transverse force is found in a similar manner:

dM=dLsin⁡φ+dDcos⁡φ=dL(sin⁡φ+dDdLcos⁡φ)=12ρV12CLsin⁡(φ+β)cos⁡φbdr{\displaystyle {\begin{aligned}{\mbox{d}}M&={\mbox{d}}L\sin \varphi +{\mbox{d}}D\cos \varphi \\&={\mbox{d}}L(\sin \varphi +{\frac {{\mbox{d}}D}{{\mbox{d}}L}}\cos \varphi )\\&={\frac {1}{2}}\rho V_{1}^{2}C_{L}{\frac {\sin(\varphi +\beta )}{\cos \varphi }}b{\mbox{d}}r\end{aligned}}}{\begin{aligned}{\mbox{d}}M&={\mbox{d}}L\sin \varphi +{\mbox{d}}D\cos \varphi \\&={\mbox{d}}L(\sin \varphi +{\frac  {{\mbox{d}}D}{{\mbox{d}}L}}\cos \varphi )\\&={\frac  {1}{2}}\rho V_{1}^{2}C_{L}{\frac  {\sin(\varphi +\beta )}{\cos \varphi }}b{\mbox{d}}r\end{aligned}}

Substituting for V1{\displaystyle \scriptstyle V_{1}}\scriptstyle V_{1} and multiplying by r, gives torque as:

dQ=rdM=12ρCLVa2(1+a)2sin⁡(φ+β)sin2⁡φcos⁡βbrdr{\displaystyle {\mbox{d}}Q=r{\mbox{d}}M={\frac {1}{2}}\rho C_{L}{\frac {V_{a}^{2}(1+a)^{2}\sin(\varphi +\beta )}{\sin ^{2}\varphi \cos \beta }}br{\mbox{d}}r}{\mbox{d}}Q=r{\mbox{d}}M={\frac  {1}{2}}\rho C_{L}{\frac  {V_{a}^{2}(1+a)^{2}\sin(\varphi +\beta )}{\sin ^{2}\varphi \cos \beta }}br{\mbox{d}}r

which can be integrated as before.

The total thrust power of the propeller is proportional to TVa{\displaystyle \scriptstyle TV_{a}}\scriptstyle TV_{a} and the shaft power to 2πNQ{\displaystyle \scriptstyle 2\pi NQ}\scriptstyle 2\pi NQ. So efficiency is TVa2πNQ{\displaystyle \scriptstyle {\frac {TV_{a}}{2\pi NQ}}}\scriptstyle {\frac  {TV_{a}}{2\pi NQ}}. The blade efficiency is in the ratio between thrust and torque:

blade element efficiency=Va2πNr⋅1tan⁡(φ+β){\displaystyle {\mbox{blade element efficiency}}={\frac {V_{a}}{2\pi Nr}}\cdot {\frac {1}{\tan(\varphi +\beta )}}}{\mbox{blade element efficiency}}={\frac  {V_{a}}{2\pi Nr}}\cdot {\frac  {1}{\tan(\varphi +\beta )}}

showing that the blade efficiency is determined by its momentum and its qualities in the form of angles φ{\displaystyle \scriptstyle \varphi }\scriptstyle \varphi and β{\displaystyle \scriptstyle \beta }\scriptstyle \beta, where β{\displaystyle \scriptstyle \beta }\scriptstyle \beta is the ratio of the drag and lift coefficients.

This analysis is simplified and ignores a number of significant factors including interference between the blades and the influence of tip vortices.

Thrust and torque

The thrust, T, and torque, Q, depend on the propeller's diameter, D, revolutions, N, and rate of advance, Va{\displaystyle V_{a}}V_{a}, together with the character of the fluid in which the propeller is operating and gravity. These factors create the following non-dimensional relationship:

T=ρV2D2[f1(NDVa),f2(vVaD),f3(gDVa2)]{\displaystyle T=\rho V^{2}D^{2}\left[f_{1}\left({\frac {ND}{V_{a}}}\right),f_{2}\left({\frac {v}{V_{a}D}}\right),f_{3}\left({\frac {gD}{V_{a}^{2}}}\right)\right]}T=\rho V^{2}D^{2}\left[f_{1}\left({\frac  {ND}{V_{a}}}\right),f_{2}\left({\frac  {v}{V_{a}D}}\right),f_{3}\left({\frac  {gD}{V_{a}^{2}}}\right)\right]

where f1{\displaystyle f_{1}}f_{1} is a function of the advance coefficient, f2{\displaystyle f_{2}}f_{2} is a function of the Reynolds' number, and f3{\displaystyle f_{3}}f_{3} is a function of the Froude number. Both f2{\displaystyle f_{2}}f_{2} and f3{\displaystyle f_{3}}f_{3} are likely to be small in comparison to f1{\displaystyle f_{1}}f_{1} under normal operating conditions, so the expression can be reduced to:

T=ρVa2D2×fr(NDVa){\displaystyle T=\rho V_{a}^{2}D^{2}\times f_{r}\left({\frac {ND}{V_{a}}}\right)}T=\rho V_{a}^{2}D^{2}\times f_{r}\left({\frac  {ND}{V_{a}}}\right)

For two identical propellers the expression for both will be the same. So with the propellers T1,T2{\displaystyle T_{1},T_{2}}T_{1},T_{2}, and using the same subscripts to indicate each propeller:

T1T2=ρ1ρ2×Va12Va22×D12D22{\displaystyle {\frac {T_{1}}{T_{2}}}={\frac {\rho _{1}}{\rho _{2}}}\times {\frac {V_{a1}^{2}}{V_{a2}^{2}}}\times {\frac {D_{1}^{2}}{D_{2}^{2}}}}{\frac  {T_{1}}{T_{2}}}={\frac  {\rho _{1}}{\rho _{2}}}\times {\frac  {V_{{a1}}^{2}}{V_{{a2}}^{2}}}\times {\frac  {D_{1}^{2}}{D_{2}^{2}}}

For both Froude number and advance coefficient:

T1T2=ρ1ρ2×D13D23=ρ1ρ2λ3{\displaystyle {\frac {T_{1}}{T_{2}}}={\frac {\rho _{1}}{\rho _{2}}}\times {\frac {D_{1}^{3}}{D_{2}^{3}}}={\frac {\rho _{1}}{\rho _{2}}}\lambda ^{3}}{\frac  {T_{1}}{T_{2}}}={\frac  {\rho _{1}}{\rho _{2}}}\times {\frac  {D_{1}^{3}}{D_{2}^{3}}}={\frac  {\rho _{1}}{\rho _{2}}}\lambda ^{3}

where λ{\displaystyle \lambda }\lambda is the ratio of the linear dimensions.

Thrust and velocity, at the same Froude number, give thrust power:

PT1PT2=ρ1ρ2λ3.5{\displaystyle {\frac {P_{T1}}{P_{T2}}}={\frac {\rho _{1}}{\rho _{2}}}\lambda ^{3.5}}{\frac  {P_{{T1}}}{P_{{T2}}}}={\frac  {\rho _{1}}{\rho _{2}}}\lambda ^{{3.5}}

For torque:

Q=ρVa2D3×fq(NDVa){\displaystyle Q=\rho V_{a}^{2}D^{3}\times f_{q}\left({\frac {ND}{V_{a}}}\right)}Q=\rho V_{a}^{2}D^{3}\times f_{q}\left({\frac  {ND}{V_{a}}}\right)

The math on reverse thrust is likely longer as you are pulling a ship instead of pushing it (?).

Up to the 1830s the British Admiralty maintained the view that screw propulsion would be ineffective in ocean-going service, believing that screw propelled ships could not be steered efficiently (as opposed to paddle steamers). The first ship to employ propellers (called screws) was the SS Archimedes, appropriately named after the inventor of the Archimedes screw in 1838, although fully sail outfitted as a backup given steam engine reliability at the time.

So much for a brief history, I came at this topic in starting to compare various ships of mine when sailing at full speed and then full reverse, with the rudder hard over presumably further increasing the drag and stopping distance (does it in fact?). It`s especially important for British cruisers given the small smoke area you will be setting--my recent acquisition of the Belfast really brought this home.

In addition to that I had noticed that reverse speed on some ships was far more effective (as in response time and final top speed in reverse) on some ships as opposed to others. I've seen other posts here that note that the Belfast's stop time seems longer than other cruisers.

Items:

--How detailed are the physics of thrust reversal modeled on ships in WoWs?

--Are players with Propulsion Mod 2 outfitted noticing a dramatic effect in reverse?

--Any insights into which ships stop faster/fastest and who wins the full reverse speed test would be interesting to hear about.

 

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16 minutes ago, AspiringCodger said:

I was under the impression going full reverse in game didn't slow you down any faster than going idle.

There isn't. The theory is that WG's intent is that the ships employ the fastest method of reaching 0 knots, IE going full reverse until the speed is 0, then cutting throttle.

Speaking of theories and the Titanic ... A current working theory is that if Titanic had either gone full reverse but remained at neutral rudder, or gone rudder hard over but remained at forward thrust, she would've survived.

- Rudder neutral + full reverse: The ship would've hit the iceberg nose on, the forward few compartments would've crumpled, but the ship would've lost two or three compartments, not 6, and thus remained afloat with a blunt nose.

- Rudder hardover + ahead thrust: The ship would've turned tighter because the rudder would've been more effective at deflecting the ship's course with actual thrust going over it, instead of the props trying to pull a vacuum behind them, where the rudder is.

But then, those are theories I've heard, and I'm not aware of any actual research into these.

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8 minutes ago, Lert said:

rudder hard over but remained at forward thrust, she would've survived.

Yes, the idea being she might have been able to "port around" the berg, something I often do in a DD when laying some smoke for a BB. Basically the Titanic swings it prow to the left of the berg and once having missed it, hard starboard to allow the stern to swing to the port and miss it as well.

Edited by Stauffenberg44

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Looking at some info on Titanic found this interesting little tidbit on the lifeboats.

Titanic carried a total of 20 lifeboats: 14 standard wooden Harland and Wolff lifeboats with a capacity of 65 people each and four Englehardt "collapsible" (wooden bottom, collapsible canvas sides) lifeboats (identified as A to D) with a capacity of 47 people each. In addition, she had two emergency cutters with a capacity of 40 people each. Olympic herself did not even carry the four collapsibles A–D during the 1911–12 season. All of the lifeboats were stowed securely on the boat deck and, except for collapsible lifeboats A and B, connected to davits by ropes. Those on the starboard side were odd-numbered 1–15 from bow to stern, while those on the port side were even-numbered 2–16 from bow to stern.

Both cutters were kept swung out, hanging from the davits, ready for immediate use, while collapsible lifeboats C and D were stowed on the boat deck (connected to davits) immediately inboard of boats 1 and 2 respectively. A and B were stored on the roof of the officers' quarters, on either side of number 1 funnel. There were no davits to lower them and their weight would make them difficult to launch by hand. Each boat carried (among other things) food, water, blankets, and a spare life belt. Lifeline ropes on the boats' sides enabled them to save additional people from the water if necessary.

Titanic had 16 sets of davits, each able to handle four lifeboats. This gave Titanic the ability to carry up to 64 wooden lifeboats which would have been enough for 4,000 people—considerably more than her actual capacity. However, the White Star Line decided that only 16 wooden lifeboats and four collapsibles would be carried, which could accommodate 1,178 people, only one-third of Titanic's total capacity. At the time, the Board of Trade's regulations required British vessels over 10,000 tons to only carry 16 lifeboats with a capacity of 990 occupants.

Therefore, the White Star Line actually provided more lifeboat accommodation than was legally required. At the time, lifeboats were intended to ferry survivors from a sinking ship to a rescuing ship—not keep afloat the whole population or power them to shore. Had the SS Californian responded to Titanic's distress calls, the lifeboats may have been adequate to ferry the passengers to safety as planned.

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Rather incredibly they actually photographed the iceberg that was believed to have been hit by Titanic. You can see why as there is scraps of Black paint on the iceberg particularly along the waterline.

280px-Titanic_iceberg.jpg

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Also suggest watching this. it's quite long but really worth it and a great watch.

 

Edited by Fog_Repair_Ship_Akashi
Better video found.

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From what I remember gameplay wise. When the ship is given the order from the player to slow down to full stop or full reverse, the engines are reversed and the props turn in full reverse to get to 0 knots as fast as possible. That is what I remember.

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59 minutes ago, Stauffenberg44 said:

Yes, the idea being she might have been able to "port around" the berg, something I often do in a DD when laying some smoke for a BB. Basically the Titanic swings it prow to the left of the berg and once having missed it, hard starboard to allow the stern to swing to the port and miss it as well.

I did that once in my boss's 300ZX at about 40 mph.  Of course, I wasn't missing an iceberg, but a Chevy pickup...

  • Cool 1

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How much power you have to go in reverse in part depends on the type of engine the ship has.  For the period of WoW the two main engine types are steam vertical expansion (usually triple expansion) and steam turbine.  VTE engines produce the same HP in reverse as forward.  Steam turbines require a set of reverse turbine wheels and are operated separately from the ahead turbine, even if the two are combined in one casing.  Usually the reverse turbine is only a fraction of the ahead turbine in terms of power.

So, the ahead turbine might have say, 8 rateau wheels and the reverse a curtiss wheel or 2 rateau's.  The alternative to a astern turbine is to have a clutch system to reverse the motion of the reduction gears.  This would be unusual as it would be a very, very big clutch to take that sort of power and would only see seldom use.  Easier to install a reverse turbine instead.

Turbo-electric ships like some US battleships, simply put the electric motors in reverse changing the direction of rotation.  This does take a couple of minutes to do as you have to coordinate generator output with the electric motor input, the US using AC motors rather than DC where you have finer speed control.

Diesels work like VTE engines.

One disadvantage of the reverse turbine is that it's inefficient and usually you can't maintain reverse operation for any long period of time, unlike the other systems mentioned here.

 

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