Macabre indeed. Almost… Zombie like.
A Nimitz Class Nuclear Aircraft Carrier has two reactors and four main engines. Reactor No.1 powers main engines 1&4 and catapults 1&4 and Reactor No.2 powers main engines 2&3 and catapults 2&3. Each main engine has two steam turbines; one is an HP turbine (high pressure) and one is an LP turbine (low pressure turbine) for a total of eight steam turbines powering the main engines. The same flow path of steam goes through both turbines with a two story moisture separator in between. Together the two turbines produce 75,000 shaft hp per main engine. The shafts are 24 inches in diameter, hollow, have four inch thick walls and are made from high yield carbon steel. The shafts drive propellers that are 21 ft. in diameter and weigh 32 tons.
The most twist we measured during trials was 250 degrees (70%) at almost full power (at full power we had other priorities). While the main engines are incredibly powerful, the catapults when in operation are even more powerful as they take a 25 ton jet from 0 to 140 mph in 2.5 seconds and 340 ft.
Maybe even a little more Zombie like.
Still, over half a turn!
To help understand this, the shaft is very long – maybe 300 feet. Without considering this it can seem hard to imagine. E.g, this photo of a Ford-class propellor shaft only shows a short length:
https://www.navy.mil/ah_online/ussford/images/gallery/G-Photos/10-DCS12-1307-49.jpg
However this photo from a large container ship (not a carrier, but probably roughly similar) shows a better idea of total shaft length. It’s easier to visualize a 300-foot-long propeller shaft twisting 250 degrees: The Telegraph US - Breaking news, opinion & analysis
They say this carrier shaft is a bad mother…
(Shut your mouth)
Some decades ago when a new CV was commissioned (I forget which one) during a news conference after the shake-down cruise the skipper was asked how fast they’d gone. “In excess of 33 knots,” was the straight-faced answer. Seeing the disappointed looks on the reporter’s faces he added with a small smile, “More in excess than we’ve ever gone before.”
Perhaps they meant circumference? A 24-inch diameter shaft would have a circumference of 75 inches, or just over 6 feet.
As an aside, race car engines are designed to accommodate heavy twisting of the cam shafts at full racing RPM’s. The high RPM’s combined with heavy valve springs result in cam designs that offset the lobe positions more and more depending on how far they are from the timing chain. It’s one of the reasons, especially at oval tracks, that the engines idle so badly when in the pits. Computer controlled ignition and injection has mitigated this a lot along with improved material design, however at low RPM’s some of the pistons will still have more advanced valve timing than what’s optimal.
Considering ship’s drive shaft length and load it’s not too surprising they twist more, but regardless it’s shocking to me how much more.
Tales of super fast aircraft carriers have been floating (ha!) around for decades, probably since CVN-65. As usual, the truth is less impressive.
I just dug up this Quora post whose author claims first-hand knowledge of the performance of the USS John F Kennedy (last non-nuclear US carrier).
He says he witnessed 36.4 kts, with the #4 (longest) main shaft showing 1.5 turns of twist.
But I’m torquin’ about shaft.
Even the end shafts on container ships are thicker than that.
When you have a 100+ foot long propeller shafts they will need to be pretty large. Note this image of CVN 78 propellers. The intermediate shafts will be much larger.
What ? They literally measure this ??? How does he “personally witness” it if its not measured.
I think the constant answer is too much to believe,quora is not referenced,verifed or deletable in any way whatso ever. So no one can trump his answer. Can anyone say "I measured it and it was only 2 degrees max ? ".
I haven’t seen any CVN propeller shafts, but I have seen nuclear-powered cruiser and submarine shafts under torque.
There are red striped lines painted down the length of the shaft, so any twist is emphasized. The twist only appears when the shaft is under high torque and rotating (and straightens back out when the torque is removed as the shaft is slowed and/or stopped). Higher shaft rotation corresponds to more torque, and more torque corresponds to more twist.
Because the shaft is rotating, there may be an optical illusion going on that makes it difficult to eyeball exactly how much twist there is. IIRC, the twist is definitely noticeable, but I’d say it’s closer to a just few degrees than the >360 degrees that has been reported.
Caveat: I’m going by 20-year old memories and the difficulty of eyeballing the amount of twist for a rapidly rotating shaft.
First of all, I respect Una’s calculations and I repeated the same with some modifications and feel the one and a half rotation is possible. The calcs are based partially on jcrredsox numbers.
0> I used the ** metric system ** because I grew up using the metric system and believe it keeps units and dimensions clean.
1> RPM → Doing a google search, it seems that the large and fast ships use propellers that are high displacement (torque) and low speed. I used **40 RPM ** by trial and error to match numbers.
This works out to be 4.2 radians /per second
2> T, Torque (N.m) jcrredsox listed 75,000 horsepower which works out to 56 mega watts. So the torque works out to be 56 MW / 4.2 rad/s = 13,400,000 N.m
3> ** L, Length of Shaft, m ** The rule of thumb is to have 0.5 to 1 degrees of angle of twist per meter of shaft on normal shafts. Since jcrredsox listed a twist of about 300 degrees, I suspect the length of shaft is around 300m.
4> **J, The Polar moment of Inertia, m4 ** From jcrredsox the outer and inner radius works to be 0.3 m and 0.2 m respectively and J works out to be 0.011 m4
5> ** G, The modulus of elasticity N/m2 ** This is typically from 75 to 85 GPa. For this analysis it was assumed to be 75 GPa which is 75,000,000,000 N/m2
6> ** Theta, Angle of twist ** Theta = (T * L) / (J * G) = 4.9 radians which is around **280 degrees ** which comes close to the observed twist by jcrredsox
Feel free to criticize the above 
I concur, looks like Una Persson’s numbers are in order. Don’t look like you guys need me, but give me a holler if you get stumped. 
I’m glad this thread was resurrected. I haven’t the foggiest notion how to do the numbers on this, but gosh, it’s still fascinating to see how great minds work, and no wonder Cecil has Una Persson as a consultant.
Both yours and Una’s calculation in 2007 are correct, I believe, for the assumptions. For convenience here’s an online calculator for it:
https://www.amesweb.info/Torsion/TorsionalStressCalculator.aspx
The variation is result is due to different assumptions, where I’d comment as follows:
Hp: let’s take 75,000 though all published references AFAIK say it’s 70,000.
RPM: let’s use 171 as per an ostensible first hand account above. Full power RPM on a ship like this is something like that, nowhere near as low as 40, Una assumed 200.
Length of shaft: Una assumed 150’, it’s probably more like 250-300 ft, but no way 300 meters
Shaft outer diameter 24", thickness 4" as per another first hander above
Modulus of rigidity I assume 80 Gpa
So, in the online calculator:
torque=3.134mil N-m
rpm=171
shaft outer radius=12"=305mm
shaft inner radius=8"=203mm
shaft length=300’=91,440mm
shaft deflection=.33 radians=19 degrees
I think your calculation shows how extreme the assumptions have to be to get 280 degrees. That doesn’t seem right.
I’m 99.9% sure first hand accounts of very high speed of these ships are not right, so ‘first hand’ alone is not conclusive, no offence to anyone. The ships are capable of around 31 kts. despite many stories of sometimes entirely implausibly higher speeds from some people who’ve served on them. Here there’s still some uncertainty, need to nail down better the shaft length for example. And it would useful to know how exactly the deflection is measured.
Corry El - You are absolutely correct. The results have been bothering me equally. And then almost like a Christmas miracle - the answer came to me :D.
Walk with me please and I’ll show you what is happening :
So here is the Official Battleship Iowa Crew Handbook.
And I extracted the relevant information from the handbook :
1> 53,000 horsepower at the shaft after the reduction gear (See Page 56, - under section Reduction gear)
2> 202 Shaft RPM - same page and section as 1
3> Shaft is 23 inch diameter and 4.25 inch thick (Page 57, under section Shaft)
4> The longest shaft is 340 ft located Starboard Outboard (See table on Page 59)
Plugging the above numbers into my calculations gives me a shaft deflection angle of **15 degrees **. Its probably the same in your calculator too.
** But then I realize ** that the shaft is not one continuous piece as Una, you and I have assumed in the above calculations. It actually comes in sections that are flanged together and they are supported on bearings at two or more places. From page 57:
Now based on my experience with piping, the flanged joints are very capable of rotational deflection much more than the pipe. So my conclusion is that the " one and a half turns" of deflection reported above occurs predominantly at the joints and you will need the flange characteristics (bolt circle diameter, bolt diameter, etc. etc. ) to actually do the calculations. However, there is nothing strikingly wrong in the observation of this large twist.
Just another way to come to nearly the same result, based only on strain.
For high-strength carbon steels (yield strength between 60,000 and 80,000 PSI), a popular method of demonstrating the strength is to measure the load required to produce a total axial strain of 0.5%. For these steels, this produces a small amount of permanent set (or yielding) and the load required to produce this strain would be called the yield strength, or proof strength. Now, shear strain is half the axial strain, so for these materials, if axial yielding occurs at 0.5%, torsional yielding would occur at half that, or 0.25%
OK, for a shaft 300 feet long, 0.25% is 300120.0025 = 9 inches. That is, to produce a torsional strain of 0.25%, the OD of a 300 foot long shaft is going to rotate 9 inches. If the diameter of the shaft is 24", that is an angular displacement of 9/12 or .75 Radians (a tad less than 45 degrees). Now, proper design rules to prevent fatigue is going to limit operating stress (and strain) to about 1/3 the yield stress (or strain), which would be .25 Radians (about 14 degrees). My guess is that maximum allowable operating conditions would be less than half of that, since shaft failure at maximum torque due to fatigue is probably something the US Navy would want to avoid.
Now, the analysis based on strain is ignoring the actual torsional load, material strength, hollow or solid, as well as other variables. It is entirely based on a 300 foot round shaft 24 inches in diameter made from a material that has an axial yield strain under load of 0.5% designed for fatigue. While it doesn’t agree exactly with the other calculations in this thread, it is surprisingly close. What it tells me is that the claim of 1.5 revolutions is absurd.
Material scientists have a name for a material that could sustain an elastic torsional strain of 1.5 turns (or even half a turn) on a 2 foot diameter shaft 300 feet long. It’s called unobtainium.
The flanges are larger diameter than the shaft, and should be the stiffest part of the assembly. We are talking about flanged joints like this, right?