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Re: Propulsion power: was [Nav-L] Barrels
From: Bill B
Date: 2005 Jan 19, 19:59 -0500

```> This is a response to a posting by Bill.
>
> Who wrote-
>
>> 1.  I do not know if fuel consumption is in direct proportion to rpm.

Thank you George, I good explanation.  If it were drag alone, the difference
between and increase from 14 to 18 kt would be 101 barrel/day x 1.65 (18/14
squared) or 167 barrels/day at 18 kt, not 215 barrels/day at 18 kt as stated
in Doug's problem.

Regarding:

> These are abstractions that are only approximately met by real ships, which
> don't always read the theory books.

LOL.  So true. Limiting the discussion to displacement hulls, a
nuclear-powered aircraft carrier that is 800 ft at the waterline certainly
beats the old "hull speed in kt = 1.34 square root of LAW," or 37.9 kt.
Hobie 16's (approx. 15' LAW would have a theoretical hull speed of 5.2 kt,
but have been clocked close to 25 mph.  To the best of my knowledge, neither
of the above craft have climbed their bow wave and gone on a plane.

Regarding the following statement, it raises several questions for me:

> (For a submarine, travelling deep, that wavemaking limit wouldn't apply).

I recall an article decades ago about an instructor teaching sail trim on a
racing yacht off San Francisco. With her guidance and participation they got
all of the speed out of her there was to get.  The students then matched the
speed without her guidance. Later she challenged them to match the earlier
speed under wind and wave conditions similar to the earlier runs.  They
could not get within 10% of the earlier speeds. The lesson she taught was
that besides bow waves, a displacement hull also radiates a pressure wave
toward the bottom.  Their failure to meet the earlier speed had to do with
relatively shallow water depth and resistance from that pressure wave.

My questions:

1. Do you know the formula/ratio involving LAW, speed and depth when this
limiting factor/resistance comes into play?

2.  Would a submarine making speed very close to the bottom be slowed down,
and or forced upward (perhaps bow up, stern down) because of the above
dynamic?

Bill

>
> It can't be as simple as that. For power (and hence fuel consumption) to be
> proportional to rpm, that could only be tha case if the driving torque
> remained constant. But to drive a vessel faster requires more torque on the
> screw as well as more speed. So the power has to increase more than
> proportionally with the speed.
>
> Indeed, the drag on any vessel increases according to the square of the
> speed, roughly speaking, up to some limiting speed, depending on length, at
> which wave-making becomes excessive. (For a submarine, travelling deep,
> that wavemaking limit wouldn't apply).
>
> In which case the power required to overcome that drag, being drag x speed,
> will depend on the cube of the speed
>
> So, if the engine efficiency and screw efficiency remained constant,
> halving the vessel's speed would cut the rpm by 2, the shaft torque by 4,
> and the power required and rate of fuel consumption by 8. And a voyage
> would take twice as long so the engine would be running for double the
> time, so the total fuel used for the voyage would be reduced by 4 compared
> with full speed.
>
> These are abstractions that are only approximately met by real ships, which
> don't always read the theory books. And it's only very roughly true to
> assume that engine and screw efficiencies and slip remain constant over
> such a wide speed range. For example, a diesel running at a certain rpm,
> even under no-load conditions on the shaft, requires to develop a certain
> minimum power just to keep itself turning, because it has to chuff through
> itself a constant volume of air per second, compress it and spit it out. So
> that results in a diesel's efficiency being near maximum at close to its
> designed running speed, and falling at lower speeds and torques.
>
> It would, then, be a very poor approximation to imagine that power would be
> proportional to ship's speed. In answer to the exam question, Doug appears
> to be presuming a cube-law, or something very close to it, rather as was
> guessed-at above.
>
> It was interesting that Doug's question presumed a propellor slip value of
> only 2%. I had always though that ship's propellors slipped much more than
> that, but it seems I was wrong, or very out-of-date. Is 2% slip a realistic
> value for the prop of a modern merchant vessel? Doug probably knows.
>
> George
>
>
>
>
> ================================================================
> contact George Huxtable by email at george@huxtable.u-net.com, by phone at
> 01865 820222 (from outside UK, +44 1865 820222), or by mail at 1 Sandy
> Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.
> ================================================================

```
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