From: hanners@oasys.dt.navy.mil (Robert Hanners)
Newsgroups: rec.boats
Subject: Re: How much force needed to propel boat ?
Date: 21 Dec 92 14:02:01 GMT
Reply-To: hanners@oasys.dt.navy.mil (Robert Hanners)
Organization: Carderock Division, NSWC, Annapolis, MD

In rec.boats, youngqud@ucs.cs.orst.edu writes:
>I need a ball park figure for the force needed to propel a 45 foot
>power boat, 10 foot beam, at at hull speed, 7 - 8 knots I think.
>This boat weighs about 8 to 10 tons.  How much forward force is exerted
>by the propellor to move the boat at 7 - 8 knots ?  If anyone has a
>guess or estimate please send me email.  BTW it takes I think it takes
>about 60 of the 83 horse power available to push it at 7 - 8 knots, if that'll

This is very much dependant upon many boat design factors that vary
too much to give you a hard number based on those factors alone.  If
you're confident of the horsepower value, we can help.  By the very
nature of your question, however, I suspect you don't really know that
either.

I would determine the rough answer to your question as follows:
    [I'll provide numbers in brackets purely for example.]

(1) Measure the maximum RPM your engine delivers for this boat. [3200]

(2) Use the published max torque or max horsepower curve for your engine
    to find either value at this max RPM.
	      (HP = Torque (lb-in) x RPM / 63,025)
			   -or-
    If you don't have this curve, you might assume that the max torque
    is approximately constant and use the advertized rated values as
    follows:
    (your max HP) = (Max Rated Engine Power) x (your Max RPM) / (RPM
						 of the max Rated Power)
      [your max HP = 83 x 3200 / 3600 =  73.8 hp]

(3) Next measure your RPM at the speed you're interested in.
      [2400 @ 7.5 knots]

(4) Now calculate the horsepower at this RPM using "The Propeller Law":
	HP = (your max HP) x (RPM / your max RPM)**3
					  (Note: "**3" = third power)

       [HP = 73.8 x (2400 / 3200)**3 = 73.8 x 0.422 = 31.1 hp!]

	     (This may be a lot less than you thought!)

(5) Now we get to the propeller efficiency.  (We've disregarded the
relatively small horsepower losses through the transmission, cutless
bearing and so forth.  This reasonably offset by the conservative
estimate that "torque is ... constant" in (2) above.)  Propeller
efficiency can vary quite a bit.  Generally, larger diameters with
more blades are less affected by the deadwater behind the shaft, strut
and keel.  Also there's an theoretically greater propulsive efficiency
for larger diameter, lower pitch, propellers.  Well spare you the formula.
I'm not too familiar with smaller props (I'm more conversant with props
that handle over 25,000 hp :-), so I'll guesstimate a propulsive
efficiency for your boat of 55%.  So now we can calculate (guesstimate)
the horsepower actually devoted to propelling the boat forward:
	   hp = .55 x HP     [= .55 x 31.1 = 17.1 hp]

(6) Finally, the propelling force is:

	   Force = hp x 550 / velocity       where velocity is the boat
					     speed in feet per second
    So for this example,

	   Force = 17.1 x 550 / (7.5 x 6076 / 3600)

		 = 744 pounds

Give or take a thousand. :-)

-- Bob

------------------------------------------------------------------------------
Robert J. Hanners         "My comments; only mine."          *DIASTOLE* 676673
CD, NSWC, Annapolis, MD                              hanners@oasys.dt.navy.mil
Newsgroups: rec.boats