HP requirement?

[surekandoo]

The rule of thumb for engine size is 4bhp/ton.

How conservative/optimistic is this figure?

Also how important is the torque figure for a marine engine?

Thanks

 

[earlybird]

Presumably for a cruising sailing yacht?
4 hp per ton displacement is nowadays on the low side IMHO. 6 might be a better figure.
Torque figure is really irrelevant for a boat. Far more important is matching the propeller to the maximum power output and rpm.

 

[LittleSister]

Ours works out at 5.5hp per design ton, and is more than adequate, and that's with a long keel (lots of wetted area=friction, and prop partly masked by the thickness of the keel and turbulence behind it). I imagine the next size down of engine (comes out at 4.2hp/ton) would be perfectly adequate.

I suspect that a more sophisticated calculation would first take out a figure (around 1 hp?) for driving the alternator and water pump - presumably that doesn't change very much as you go up in size - and add it back in at the end (i.e. not multiply that portion by the displacement tonnage). In the case of our current engine that would work out at 5.2hp/ton plus 1 hp for the alternator.

Possibly more informative would be asking (in a poll?) what people have in their boats of similar displacement//loa/shape/purpose, and whether they consider it adequate.

Our 'new' project boat has just 2.9hp/ton, which I worry is probably a bit feeble for modern tastes and the amount of motoring we tend to do, but try to console myself with the thought it has apparently proved perfectly adequate for the last 36 years the boat has been in use and no-one got round to re-engining.

 

[bedouin]

The rule of thumb for engine size is 4bhp/ton.

How conservative/optimistic is this figure?

Also how important is the torque figure for a marine engine?

Thanks

The absolute torque doesn't matter much, but the shape of the torque curve does. As boats don't usually have gears you need a much broader torque curve than you would in a car.

 

[AntarcticPilot]

Moody 31: displacement 4.4 tons; Volvo 2003 - 28 BHP. So we've got about 6 BHP per ton nominally; it seems to be plenty.

 

[Tranona]

The rule of thumb for engine size is 4bhp/ton.

How conservative/optimistic is this figure?

Also how important is the torque figure for a marine engine?

Thanks

It is the propeller that drives the boat, not the engine. So if you want the best motoring performance then fit the largest diameter prop you can. You can work out what that is by feeding your boat's details into the Propcalc programme on www.castlemarine.co.uk using your chosen engine. The maximum propsize is one that allows clearance of ideally 15% diameter. So a 13 " prop would need nearly 7.5" from prop centre to hull.

In reality on your boat you will probably be constrained by the clearance between the prop and the hull so the maximum diameter will then determine the shaft speed, which in turn determines the gearbox reduction ratio so that your engine can achieve its maximum revs to give you hull speed.

On your boat any of the nominal 9 or 10hp motors will have enough power to achieve hull speed with the correct ratio and prop. For example I used a 1GM to power an old wooden boat that weighs well over 3 tons, but I had a 3.1:1 ratio reduction and a 15" prop. You will probably only be able to swing a 12 or 13" so will need a higher shaft speed therefore a 2:1 or 2.6:1 reduction.

Get a recommendation from the engine supplier for the appropriate ratio and prop size for your boat. Work it out first from Propcalc so that you know whether the recommendation makes sense.

 

[Rivers & creeks]

13.5 tonnes and 75hp so 5.5hp per tonne. But we've never come close to needing it all despite being in some big weather. It's rare we get above 50% revs and have to plan a high rev run to keep the engine healthy. In reality 60hp would be a better engine.

 

[alahol2]

3.5 Tonnes and 13HP is ample for hull speed (3.7HP per tonne). Might ultimately struggle against high winds and waves but cruising speed at about 3/4 throttle.

 

[puddock]

Westerly Storm is @ 5 tons+ and has 18 hp so works out as 3.6 hp/ ton.
Must admit, I'd feel better with a little more ooomph but never really had any problems due to lack of power.

 

[bigwow]

I opened this thread thinking it was about sauce

 

[doug748]

On an easily driven hull I was always quite happy with 4T - 12hp, 3hp per ton. This is a low windage boat that goes well to windward and I prefer, and have the time, to sail upwind. The advantages are less weight and a smaller prop for the sails to drag around
If your circumstances and inclinations are different you will want more power to forge ahead in waves and strong headwinds. This is particularly so if your boat boasts high windage accommodation on the motor sailor scale, many of the latest cruising boats do.
Going larger can get you a more sophisticated, multi cylinder, engine; often at very little extra cost. Soaking up the extra hp with a larger alternator is often useful.

 

[Salty John]

I think 1HP per foot of boat length is about right for a moderate to heavy displacement cruising boat.

 

[jdc]

What power law?

The assumptions seem to be that power requirement is, to a first approximation, linearly related to something: displacement or, in a recent post, length. These differ by a power of 3! It can't be both (and may be neither).

What should one expect? Considering first the power required to push the boat through the water:

Drag (R) is given by R = 1/2 x rho x C x S x U^2, where

rho is density of water and so is constant
C is a coefficient dependent on Froude or Reynolds number so only weakly dependent on LWL
S is wetted surface area so approximately proportional to length squared
U is speed, so approximately proportional to the square root of length, assuming you want to go at 'hull speed'

Power used = resistance x speed = R * U

So Power is proportional to L^2 x sqrt(L) x sqrt(L) x sqrt(L) = L^3.5 where
L = length.

Hence I'd think power should increase slightly faster than proportional to displacement (all other things being equal, displacement should be roughly proportional to L^3).

Now consider fighting against a gale. The boat speed is less relevant, being only a small fraction of the total wind speed, and can be disregarded. Again

R = 1/2 x rho x C x S x U^2, where this time

rho is density of air so is constant
C is a coefficient dependent on shape (how close is your boat to the windage of a brick, not size)
S is cross sectional area of hull and rigging above the water, so approximately proportional to length squared
U is wind speed, so independent of boat size

So Power is proportional to L^2.

Which dominates? The answer depends on the size of boat. For small boats the requirement to drive a boat against the wind dominates, whereas for larger craft the requirement to drive the boat through the water does.

Hence neither the proportional to displacement (P = k x L^3) nor the recent poster's proportional to length (P = k x L) are right!

I bet the function would be something like max(alpha x L^2, beta x L^3.5) where
alpha and beta are constants. If we have enough data points from peoples boats we can derive alpha and beta. This is an initial guess

About right?

 

[geem]

We have 14.5 tonne design (17tonne in cruising trim) 86hp. works out at 6hp/tonne at design weight.

 

[alahol2]

About right?

I'd say still a bit high at the bottom end. Continue the top curve down beyond where it 'kinks' and it would be about right.

 

[surekandoo]

The assumptions seem to be that power requirement is, to a first approximation, linearly related to something: displacement or, in a recent post, length. These differ by a power of 3! It can't be both (and may be neither).

What should one expect? Considering first the power required to push the boat through the water:

Drag (R) is given by R = 1/2 x rho x C x S x U^2, where

rho is density of water and so is constant
C is a coefficient dependent on Froude or Reynolds number so only weakly dependent on LWL
S is wetted surface area so approximately proportional to length squared
U is speed, so approximately proportional to the square root of length, assuming you want to go at 'hull speed'

Power used = resistance x speed = R * U

So Power is proportional to L^2 x sqrt(L) x sqrt(L) x sqrt(L) = L^3.5 where
L = length.

Hence I'd think power should increase slightly faster than proportional to displacement (all other things being equal, displacement should be roughly proportional to L^3).

Now consider fighting against a gale. The boat speed is less relevant, being only a small fraction of the total wind speed, and can be disregarded. Again

R = 1/2 x rho x C x S x U^2, where this time

rho is density of air so is constant
C is a coefficient dependent on shape (how close is your boat to the windage of a brick, not size)
S is cross sectional area of hull and rigging above the water, so approximately proportional to length squared
U is wind speed, so independent of boat size

So Power is proportional to L^2.

Which dominates? The answer depends on the size of boat. For small boats the requirement to drive a boat against the wind dominates, whereas for larger craft the requirement to drive the boat through the water does.

Hence neither the proportional to displacement (P = k x L^3) nor the recent poster's proportional to length (P = k x L) are right!

I bet the function would be something like max(alpha x L^2, beta x L^3.5) where
alpha and beta are constants. If we have enough data points from peoples boats we can derive alpha and beta. This is an initial guess

About right?

Sounds a bit ott to me! Graph shows a 25` boat needing 20hp? Or have I misunderstood?

 

[puddock]

Sounds a bit ott to me! Graph shows a 25` boat needing 20hp? Or have I misunderstood?

My thoughts also
Certainly doesn't seem to relate with a "slippery yacht", perhaps good for a motorboat/ fishing boat?

 

[Ex-SolentBoy]

I think the best way to do this is to follow Nigel Calders approach. Chapter five in his book.

His approach is to size an engine and fit it with a propeller that will get the boat to hull speed at maximum rpm.

The formulas are there and are easy to follow and they correctly use the relationship between displacement and waterline length. They also allow for the fact that the transmission and alternator take some power.

Using these my Rustler 36 wanted 34hp and my Rustler 44 wanted 85hp. Thats what we had and I would say they are dead right. For "normal" sailing boats windage is not a significant factor for the calculation.

I ran the calculation for several boats and got ranges from 3.9 to 4.9 hp per ton fully loaded weight. If I were to spec another boat I would look for between 4.5 and 5.

Of course, if you don't want or need to be able to get to hull speed, then you can use a lot less.

 

[surekandoo]

I think the best way to do this is to follow Nigel Calders approach. Chapter five in his book.

His approach is to size an engine and fit it with a propeller that will get the boat to hull speed at maximum rpm.

The formulas are there and are easy to follow and they correctly use the relationship between displacement and waterline length. They also allow for the fact that the transmission and alternator take some power.

Using these my Rustler 36 wanted 34hp and my Rustler 44 wanted 85hp. Thats what we had and I would say they are dead right. For "normal" sailing boats windage is not a significant factor for the calculation.

I ran the calculation for several boats and got ranges from 3.9 to 4.9 hp per ton fully loaded weight. If I were to spec another boat I would look for between 4.5 and 5.

Of course, if you don't want or need to be able to get to hull speed, then you can use a lot less.

I have an 1800kg light displacement Feeling 720. 7.2m overall length. How do I work out what max hull speed would be for this boat please?

 

[Tranona]

I have an 1800kg light displacement Feeling 720. 7.2m overall length. How do I work out what max hull speed would be for this boat please?

It is the waterline length that determines the maximum hull speed. Approx 1.35* sq rt LWL in feet. So your boat has a waterline length of approx 20 ft and max speed is 1.35*sq rt 20 = 6 knots approx.

As I suggested earlier either a Yanmar 1GM or a Nanni/Beta 10 have more than enough power to achieve that speed with the correct prop. It is pointless overpowering the boat as you will not be able to use the extra power. Maximum speed is limited by the waterline length.

Using Propcalc a 10hp Nanni/Beta 10 with a 2:1 box and a 13*10 3 blade or 14*10 2 blade will give you a maximum speed of 6.2 knots. A very good setup. Alternatively a Yanmar 1GM with 2.1:1 box and 11*9 will give 6knots.

Note these are my estimates and you should check with the engine supplier, but they won't be far out.