[MHml] Engine power

[roy mills]

----- Original Message ----- 
From: "Martin Schöön" <martin.schoon at gmail.com>
To: "Informed discussion of multihull issues" <multihulls at steamradio.com>
Sent: Sunday, February 04, 2007 5:34 AM
Subject: Re: [MHml] Engine power


I have done some (w)reckless number crunching. The results are
summarised in a pdf-dcoument that can be downloaded at:
http://hem.bredband.net/mschoon/Modelling_Afterburner.pdf

        Very interesting Malcolm, as well as pretty to look at. Thoughts 
that occurred to me are that The Michelet drag graph, interpolated as best I 
could, indicates that the hull drag at 20 knots is only 7 times the drag at 
7 knots. The square formula for air drag would have that increase by 8.16, 
so the hull drag increases with speed at a lower rate than the air drag, 
which I did not think was the case. An anecdotal observation - when I sail 
dead down under my moderate sized spinnaker alone ( no nasty cracks about 
gybing downwind please, I have valid reasons) at speeds below about 6 knots 
the apparent wind equals the boat speed, so its below about 12 knots true. 
As the wind builds up and the boat speed increases, 8 knots of boat speed 
needs about 9 knots of apparent wind, so the hull drag is increasing more 
than the air drag. At 15 knots ddw the wind is running about 17-18 knots 
apparent, not easy to be precise here because in that sort of wind one 
accelerates and slows as one goes down the front of one wave then climbs up 
the back of the wave in front. None the less my experience is that 
hydrodynamic hull drag increases at a rate higher than aerodynamic air drag. 
Agreed I am not AB, just a generally similar, slightly fatter, somewhat 
shorter,probably somewhat heavier, boat from the same designer.

    Hp required to overcome that drag at various speeds. Curve shape is of 
course very similar to drag curve, and as you point out the graph does not 
factor in prop or transmission efficiency. I believe that for the past many 
years the advertised h.p. is that delivered at the prop and so has 
'absorbed" the transmission losses. Now according to that graph the power 
required to drive AB at 7 knots is about 2.5 kilowatts, which at 1.36 KW per 
h.p. comes to 3.4 h.p.  Bill needs pretty close to 20 in the real world, so 
the prop is only converting 17% of the horsepower into useful thrust. I knew 
it was not a high percentage but I did not think it was that low. Do you 
have a guesstimate of the additional drag of rudders,boards,and the airdrag 
of everything above water which are all being overcome by that 20h.p. Honda 
and which will increase the apparent efficiency, making it look better? Bill 
has to get a much better prop or he will need about 400h.p. to hit close to 
20 knots. Do we have any figures for the efficiency that can expected from a 
prop if the known speed is 20 knots but diameter and prop revs are 
negotiable as in this situation.?


    Anyway, thanks for going to all that trouble. Do we now all await with 
bated breath for a posting from Tom Speer?  I certainly do.   Roy Mills

-- 
Martin Schöön              "Problems worthy of attack
                                         prove their worth by hitting back"
                                         Piet Hein
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