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    Re: Barrels was DR thread from Nov-Dec '04
    From: Bill B
    Date: 2005 Jan 19, 15:41 -0500

    Since I see only the results of your computations, not the input and
    calculations, I may well be talking out of school here.
    1.  I do not know if fuel consumption is in direct proportion to rpm.
    Potential variables, load or no load. Perhaps you or the list can help me
    here.  Common sense says my car will consume more fuel maintaining a
    velocity of 50 mph up a grade than it will maintaining 50 mph on a level
    road (ignoring wind velocity and direction).
    2.  Even if fuel consumption vs. rpm was a straight-line graph, Chapman
    indicates screw turns vs. speed is *not* directly proportional.  It is quite
    possible that when the speed was reduced to 14 kt, if fell on a different
    part of the speed-to-screw-turns curve.
    If memory and Hobie racing (also displacement hulls) experience serves me,
    when speed increased by 2X, hull/water drag is increases by 4X (proportional
    speed increase squared). Light wind to weather, skipper and crew as far
    forward and to leeward as possible to keep the transom(s) from dragging.
    Heavy air, skipper and crew as far aft and windward as possible to keep the
    bows from submarining.  Triple the wind velocity equals approx., 9X force on
    the sail.  Triple the speed means approx. 9X increase in hull drag.  The
    boat simply tries to trip over itself, stuffing a lee bow and sending a crew
    on the wire for a spin around the forestay. Painful. 
    As fuel consumption is part of the passgage/navigation equation for power
    craft, I would love to hear more about the methods you used to come up with
    your solutions.
    > As for Bill's question about turns and fuel consumption the easiest way for
    > me to show how much fuel these larger,laden vessels consume would be to take
    > a question out of the C.G. exam for 3rd mate,unlimited as an example.
    > Q: The speed necessary for reaching a point at the designated time is 18.0
    > kt.The pitch of the prop is 23.9 ft.How many rpm will the shaft have to turn
    > with a - 2 % slip to reach the point in the given time?
    > My calculated rpm = 77 rpm
    > If the fuel consumption at the above shaft rpm is 215 barrels/day what will
    > be the fuel consumption if the vessel's speed is reduced to 14.0 kt.?
    > My calculated consumption = 102.3 BBLS/day
    > The correct answer = 101 BBLS/day
    > Close enough to pick the correct answer out of the supplied answers but my
    > calculated rpm is apperantly high as shown by the 102.3 barrels instead of
    > 101 barrels.So,in real life, you can see that a fair amount of money would
    > be lost if calculated incorrectly.

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