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- From: jaf@nynexst.com (John Frieman)
- Subject: Re: How much force needed to propel boat ?
- Message-ID: <1992Dec21.150453.11225@nynexst.com>
- Sender: news@nynexst.com (For News purposes)
- Organization: Nynex Science and Technology
- References: <1h42csINN27c@leela.CS.ORST.EDU>
- Date: Mon, 21 Dec 92 15:04:53 GMT
- Lines: 30
-
- In article <1h42csINN27c@leela.CS.ORST.EDU> 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.
-
- From Dave Gerr, author of THE NATURE OF BOATS, although he did not
- include the formula in the book, the horse power required is relative
- to the displacement as follows:
-
- lbs displacement per HP = (10.665 / Speed to length ratio) ^3
- or required HP = displacement (lbs) / (10.665 / SL)^3
-
- That is the displacment divided by the quantity ((10.665 divided by
- the speed to length ratio) cubed).
-
- The speed to length ratio is the speed in knots divided by the
- square root of the water line length in feet.
-
- This is a good estimate for most small (ie. less than 100 ft.) boats.
- If you want to get technical, look up a copy of THE SPEED AND POWER
- OF SHIPS by Taylor. It is the classic text on the subject.
-
- By the way, the a reasonable maximum speed for a displacement boat
- can be found by dividing 8.26 by the displacement to length ratio
- raised to the 0.311 power and multiplying the result by the square
- root of the water line length. (8.26 / (DL ^ 0.311)) * sqrt(LWL)
-
- Enjoy,
-
- John
-
-