# Calculating power requirements for full displacement hull

Discussion in 'Boat Design' started by Annode, Sep 2, 2019.

1. ### AnnodePrevious Member

It seems to take very little power to move even large and heavy full displacement hulls up until about 6-8kts. Then the power curve vs speed seems to go up exponentially after that, and nearly vertical at about 13kts.
Assuming that 8kts is sufficient, both in terms of efficiency and speed, what is the nominal (flat water) calculation for power vs weight.

After that is known, what is the calculation for additional power for manoevering a heavy steel hull with a lot of inertia, and for wind, sea. Since its not trawling, dont need monster trawler engine, but something extra I imagine.

How is this calculation typically made?

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### DCockeySenior Member

The speed at which the resistance begins to significantly increase varies with square root of the waterline length. The shape of the curve depends significantly on the displacement divided by the waterline length cubed (or equivalent ratio).

What is the reason for your questions?

3. ### AnnodePrevious Member

to calculate engine power... which for some strange reason is measured in horsepower in boat engines - that only rev to 1500 or 2000 rpm thus disguising the the torque that the engine generates... a more relevant number since a turbo diesel 500hp can generate 2000 ft/lbs of torque

Last edited: Sep 2, 2019
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### DCockeySenior Member

Where did you get the idea that "boat engines that only rev to 15 or 200rpm"? I assume you know that boats usually have "reduction gears" between the engine and propeller which reduce rotational speed while increase the torque. For instance 3:1 reduction gears reduce the output rotational speed to 1/3 of the input rotational speed while increasing the output torque to 3 times the input torque.

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### JSLSenior Member

(* wl length, displacement, power, etc etc.
Power, speed, & * etc. also governs propeller size.

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### Mr EfficiencySenior Member

For practical purposes, rather than strictly theoretical ones, it is the wave making property of the typical displacement hull that causes the rapid rise in resistance, and is related to boat length, the wave system set up that is parallel to the direction of travel, propagates predictably according to speed, in that a given speed gives a given wave length, it is when the crest of the second wave falls behind the stern, that the resistance goes off the chart. Waves on the ocean behave according to similar rules, long wave length equates to speed, short wave lengths slow. Simply, physics.

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As already noted by JSL, there are far too many variables to consider before attempting to provide any kind of reply.

This suggests you're already getting confused.
All engines are measured in power output, in either metric units, kW or imperial units, Hp.
The amount of power = resistance x speed - that's it.
Toque plays no part in this calculation - why are you fixated on torque?

8. ### AnnodePrevious Member

> boats usually have "reduction gears" between the engine and propeller which reduce rotational speed while increase the torque.

yeeeees ... just talking about engine right now. 800 - 2000 rpm for larger boats is normal I am told (with a 3:1 reduction in the gearbox)

Specifics... no hab
This is a THEORETICAL rough approximation discussion. I am going backwards from the power requirements for reasons that are not relevant right now.
answers that are basically "why is that the question?" are not helpful. It is the question.

so.. back to the topic...

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### Chuck LosnessSenior Member

Dave Gerr's "The Nature of Boats" discusses power requirements. It might be too basic and not give you what you are looking for. It would be a place to get started. I am sure that this topic has been discussed in the past. Try doing a search for calculating the power for displacement boats.

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### fredrosseUSACE Steam

A rough estimate of horsepower for displacement type hulls is fairly simple:

1. Maximum "hull speed", in knots, is equal to about 1.3 X the square root of the waterline length of the hull in feet. For example, say you have a displacement hull with a waterline length of 25 feet, then the hull speed is 5 x 1.3 = 7 knots.
Trying to propel the boat faster than this will result in enormous increasing power requirements.

2. For displacement hulls of reasonable shape (not a square box, which would require more power, and not a rowing shell used for competition, using less power), about 1 horsepower per long ton of displacement will get the boat to hull speed in calm water. This value is based on a propeller of good efficiency, generally a large prop at relatively low RPM. Some additional margin is usually prudent to cope with wind and waves, or non-optimum propeller conditions, but anything over about 2 horsepower per ton of displacement is not required.

3. Much more detail is provided in the FAQ section of thesteamboatingforum.net, as virtually all of the steamboats use displacement type hulls, a general exception to the pleasure boat industry which has much faster boats.

11. ### AnnodePrevious Member

Fred. Thank you. thank you. That was a great answer.
These numbers are for flat water and a reasonable shape hull obviously. OK. So the next part of the question is what facto do you use to give you some get out of trouble power.

I have been reading on this forum for a while and one thread about a backup 9.9hp outboard and stories of string winds and currents got me wondering how you would calcualte a reasonable margin of "get out of trouble" power on top of the requirement to move at hull speed in clam seas.

Obviously you can never have too much power, but these engines go up significantly in price with each step up in power. (for the sake of discussion lets assume a good size boat 25-30m with a steel hull and a weight of 100 - 150 tons. this is out of the category of small fibreglass hulls so things like wind, waves and inertia become significant.

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### KeithOSenior Member

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