# The relation between the boat horse-power and the drag/thrust force

Discussion in 'Powerboats' started by pa_ad, Jul 12, 2009.

1. Joined: Jul 2009
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Location: USA

Dear listers,

I have a questions:

What is the relation between the boat horse-power and the drag/thrust force? and how the force is calculated in the following two scenarios:

1) A boat is tied from the back to a fixed bollard, slowly moving forward. The rope/chain became in full tension.Under the same gas rate to keep the pulling force on the bollard (not moving forward anymore). Assuming the horse power is known, what is the dragging force?

2) If you are moving a boat forward slowly until you touch a wall (or tugboat holding a ship). The boat is not moving but continue the pushing (by holding constant gas rate). What is the thrust/pushing force acting on wall in this case? again, the horse-power is known. Also, is the criteria for calculating the force in this case is similar to the above? (assuming the same horse-power in both scenarios)

I appreciate any help. Thanks

2. Joined: May 2004
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### daiquiriEngineering and Design

Hello,

The relations are (for a constant-speed navigation):
Drag = Thrust
Shaft power = (Thrust * Speed) / Prop efficiency
Engine power = Shaft power / Transmission (drive-train) efficiency.

The prop efficiency depends on propeller's geometrical characteristics, blade and hull fouling, rotational speed, boat speed, amount of ventilation, amount of cavitation, wake factor, shaft inclination etc. It can typically be anything from 0.35 to 0.65 for commercial small (boat) props and up to 0.8 for those optimized for propulsion of big ships.
A drive train efficiency depends, among other things, on wear and maintainance. so it can typically be anything from 0.92 (or even lower) to 0.97.

Bollard pull is a case of a prop operating at zero speed. Assuming that there is no significant difference in flow fields for push and pull configuration, there should also be no difference between the two situations you have described.
It can be estimated from the propeller experimental curves, reading the value of thrust coefficient at zero speed and correcting it for losses due to fouling, presence of hull - though the boat is not moving, the water still moves past the hull and is not having the same path as the experimental prop, etc.

Remember that, in order to avoid some very stupid errors, you should always use a consistant set of units. Means either all SI or all imperial, never a mixed one.
So, a "knot" is not a consistent imperial unit for speed, ft/s is.
And a horse-power is not a consistent imperial unit for power, (ft*lb)/s is.
Once you have obtained the numerical result, you can eventualy do the conversions between the two unit systems - or to knots and HP's.

If you need some quick-reference equations, you can find some in Dave Gerr's (pretty cheap and accessible) "Propeller Handbook", including the one for Bollard pull.

3. ### Submarine TomPrevious Member

Why are you asking this question, (twice)?

Tom

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### YellowjacketSenior Member

Static thrust is different

While the above relationships are correct, as the speed decreases to near zero, you are essentially generating static thrust and that is a different animal.

If you look at the above relationships as the speed goes to near zero the thrust becomes infinite. We all know that isn't true, but if the prop is large and the pitch is small, amount of thrust generated at low speed can be very high.

For this speed regime you need to know the prop size, pitch and speed and look at a correlation for static thrust. What is happening is that the prop is acting like a torque converter and the amount of slip comes into play.

5. ### apex1Guest

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### daiquiriEngineering and Design

Oooopssss, no no!
It only "apparently" goes to infinite, but you forget that as the speed goes to zero the efficiency does the same. So (don't know if you're good with the math of limits) you have a ( lim-->0 ) function divided by ( lim-->0 ) function, which is undetermined and can be a finite number. And in this case it is, because thrust is finite at V=0.

Now, the shaft power can also be written as
Shaft Power = Torque * Omega
(Omega = angular velocity = RPM/9.55)
and if I had written it in that form, there would have been no ambiguity of that kind. But then I wouldn't have replied to the original question, which was "what is the relationship between the power and the thrust force?"

All the rest you wrote is correct.
Cheers!

7. ### Submarine TomPrevious Member

But from his/her other post 0.3% of 2000 kw is from the engine so

about half that is going out of the propeller at best...

A 4 foot diameter (1.25 meter) diameter is claimed in his/her other post as

well. It's all very confusing and lacks sufficient information from the

8. Joined: May 2004
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Location: Italy (Garda Lake) and Croatia (Istria)

### daiquiriEngineering and Design

Yep, something is very wrong in the initial data. So much wrong that I think that maybe we are taking it too seriously.

9. ### Submarine TomPrevious Member

Perhaps.

I've asked pa_ad for clarification on both his/her posts and heard nothing.

Standing-by...

10. ### apex1Guest

Do´nt hold your breath Tom. I think daiquiri was right. Tooo much attention.

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