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- Path: sparky!uunet!dtix!oasys!hanners
- From: hanners@oasys.dt.navy.mil (Robert Hanners)
- Newsgroups: rec.boats
- Subject: Re: How much force needed to propel boat ?
- Message-ID: <28801@oasys.dt.navy.mil>
- Date: 21 Dec 92 14:02:01 GMT
- References: <1h42csINN27c@leela.CS.ORST.EDU>
- Reply-To: hanners@oasys.dt.navy.mil (Robert Hanners)
- Organization: Carderock Division, NSWC, Annapolis, MD
- Lines: 72
-
- 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
-