It is worth looking at the relationship between the two displacements here, as I think that should give a reasonable guide as to the power relationship. My reasoning is based on the principle that the power required is directly proportional to the wetted surface area, at least at displacement speeds. WSA is proportional to displacement, hence the common use of the rough rule of thumb methods for power requirement prediction, as already discussed here a few days ago.
I would expect a 16 ton boat to require a little over 3 times the power of a 5 ton boat for any given speed. Using your example of 6kts needing 5kW on a 5 ton boat, it would be reasonable to estimate that the power required to do 6kts in a 16 ton boat would be around 16kW, rather than 7.6 to 8kW.
Without knowing the LWL, beam or draft of either boat it is hard to be more precise, but if we had those figures, and made some broad assumptions about hull shape, we could use one of the well-known formulae to derive the wetted surface area for each and refine the above estimate. In practice I don't think it would make much difference, as I suspect that the commonly used "1000W per ton" method gives a close enough figure for this size and type of boat.
Jeremy
--- In electricboats@yahoogroups.com, "Eric" <ewdysar@...> wrote:
>
> Hey Mike,
>
> No worries. For my calcs, I did the math and thought that the answer sounded familiar, so I looked up my last answer.
>
> Any time that you're doing this kind of work, you can stop and evaluate if your answer sounds reasonable. My 30' 5 ton sailboat uses 5kW to motor at 6kts. So 7.6-8kW seems reasonable for a 37' 16 ton sailboat.
>
> Alternatively, 43hp is 32kW. If 32kW is 6kts, that would mean that you would need a 65kW drive to motor your boat at 90% of hull speed. That does not sound reasonable.
>
> Honestly, this isn't rocket surgery.
>
> Eric
>
> --- In electricboats@yahoogroups.com, Michael Mccomb <mccomb.michael@> wrote:
> >
> > i understand, i just like getting the same answer from multiple approaches whenever i can find such alternatives.... i stumbled upon these new data while researching props... thank you for the input
> >
> >
> >
> > ________________________________
> > From: Eric <ewdysar@>
> > To: electricboats@yahoogroups.com
> > Sent: Monday, September 19, 2011 10:36 PM
> > Subject: [Electric Boats] Re: please take a stab at sizing a motor....
> >
> >
> > Â
> > Hi Michael,
> >
> > For gasoline, the conversion is GPH = 0.0226 * HP / TE. Diesel contains 115% the BTUs of gasoline.
> >
> > If we assume a 20% Thermal Efficiency for your engine at low RPM (just a guess), we get 1 * 0.20(TE) / 0.0226 = 8.8hp * 1.15 for diesel = 10.1hp
> >
> > That's 7.6kW. If you look back at post #19718, my answer was that you could expect a 16kW drive to push your boat at 7kts. Half the power and slow down 1 knot means 6kts = 8kW. Two completely unrelated calculations that come up with the same answer +/- 5%. Weird, isn't it.
> >
> > You can keep coming up with different ways to ask the question, but the answer is going to keep coming out the same.
> >
> > At least this answer is better than the 43hp that you guessed on your next post...
> >
> > Fair winds,
> > Eric
> > Marina del Rey, CA
> >
>
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