conversion factor from HP to LBS pull

[catalac08]

in the process of making a hard top and I am aware that this will give me a bit more windage. I have the wind speed/PSI figures but I need also the conversion for HP to lbs pull - I know on occasion the mags do ouboard comparison tests and one of their comparators is a bollard pull test bbut I have not managed to find this. Specifically the connection between 30HP and lbs pull. I can relate the engine power to sail power to get a feel for the additional wind resistance particularly from the nose as that is where I will have max additional frontal area.

 

[peterb]

There is no direct connection between horsepower and pull. It depends on the propeller.

If you can imagine a "propeller" with the blades all aligned with the shaft (as if they were feathered), then turning the shaft would give no pull, irrespective of how fast it turned. But it would take a lot of power to turn it fast.

Now turn the blades to give them a slight twist, and you will get some pull. Turn them further, and the pull will increase. But as you continue to increase the twist the pull will reach a maximum, then start to decrease again until when the blades are roughly at rightangles to the shaft there will again be no pull.

But the angle giving the maximum pull will vary with the speed of the boat. A very fast boat will need a large pitch propeller, i.e. will need the blades at only a small angle to the shaft. A slow boat will need a low pitch propeller, one with the blades at a large angle to the shaft. Hence there is no direct connection between the engine horsepower and the bollard pull.

 

[colinmi]

1 hp equals 33,000 pound-feet/minute

 

[peterb]

[ QUOTE ]
1 hp equals 33,000 pound-feet/minute

[/ QUOTE ]

But that's nothing to do with bollard pull.

 

[lw395]

Bollard pull is probably not quite the right thing anyway, as the thrust may vary a lot at zero hull speed.
You could make a rough estimate from the pounds/feet per second number, then make a reduction for propellor effieciency. (50 to 70 per cent efficient?)
In non-redneck units
1hp = 748Watts= 748Nm/s
1m/s =3.6 km/h, 1nm=1.842km so 1knot=1.95m/s
so 1hp@1knot=748/1.95=383N x efficiency
383N=39.1kgf=~86LbF
Doesn't sound much does it? suggests horse pulling canal boat at 4kts with force of 21Lb? ???
Any gsce stoodents on here?

 

[Eygthene]

If you are looking for an estimate of the HP required to overcome your extra windage, lw395 has the right approach. I would do the calculation for your expected speed through the water when motoring - say 6 kts.

As lw395 says 1HP expended at 1kt requires a "force" of 86lbs (he points out that referring to force, in pounds is not strictly correct)
So 1HP expended at 6kts requires a force of 86/6 = 14 pounds.
That is every additional 14 lbs of extra windage will need another 1 HP delivered to the boat to keep you at 6 kts.
But if you want to know the extra engine power needed to provide this 1HP, you must take into account prop (and transmission) efficiency say 50% to 70%.

To maintain speed you would need 1/0.5 to 1/0.7 = between 2HP and 1.4HP from the engine for every extra 14lbs of windage.

 

[Talbot]

Alternatively, you can accept that a 30hp engine with a hi thrust prop, sited deep enough to avoid cavitation, will be fine with an 8m Catalac (empiracal test!)

The real problem is getting the prop deep enough. On my 9m Catalac, the skeg of the outboard was the deepest part of the boat. I also fitted wings on the cavitation plate - the sort that are designed to lift a power boat onto the plane. They have little real effect on raising the boat out of the water, but they do two things that make progress over deep seas possible. They move the pivot point of the boat further aft, thus reducing the tendency to hobbihorse, and they reduce cavitation.

In the 14 years with my outboard configured this way, I never had any cavitation from my prop.