Thrust / HP equivalence

Discussion in 'Boat Design' started by Guest1578132542, Jul 12, 2010.

  1. hi all.

    (this is to help me with my planing hulls)


    Question;
    if you are pushing a 1 ton weight up a frictionless surface, on an inclined plane, say a slope of 1 degree, how many Kg thrust is needed to hold the 1000 Kg steady?



    (and so what horse power outboard would hold the weight perfectly steady?)


    onya,
    mal.
     
  2. Paul Kotzebue

    Paul Kotzebue Previous Member

    Depends on the speed. Thrust = Power divided by Speed (in consistent units). The force required to hold a weight on a frictionless inclined plane = Weight times the sine of the incline angle, with the force directed along the incline angle.
     
  3. daiquiri
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    daiquiri Engineering and Design

    The component of 1t weight along the inclined plane is 1 ton x sin(1°) = 17.5 kg (or 38.6 lb), and that's the thrust necessary to keep it steady.

    The horsepower is a measurement of power, and the power is defined as force (or thrust) times velocity (speed). So, you need a velocity, in order to calculate a power.

    However, a boat is not just a brick staying still on a frictionless inclined plane... ;)
     
  4. lifting a hull up and over the bow wave height

    ignore time please, here's a picture


    [​IMG]




    so instead;


    17.5 kg?

    12 Kg thrust per Hp
    ( according to http://www.boatdesign.net/forums/outboards/how-many-lbs-thrust-equal-1-hp-24964-2.html#post263432




    so;



    you say an average 2 horse power outboard motor strapped on a 1 ton weight that is by the side of the jetty on a slope of 1 degree can easily hold the weight steady




    so if the hull of this 1 ton boat is 1/sin 1 degree or

    5.7298 m long

    then to lift the boat ten centimeters up it's bow wave takes 2 hp?


    (width of the hull = whatever gives 10 cm displacement on this perfectly rectangular hull)
     
  5. daiquiri
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    daiquiri Engineering and Design

    DB,

    From time to time "converting thrust to HP" quests appear on this forum, each one looking for some phantomatic single universal conversion multiplier.
    Evidently, it is often not so intuitive to grasp the difference between the thrust and the power.

    Guess I understand what you are looking for. You want to know what HP, as written in the engine name, is necessary to move your boat at some speed and at some trim angle.
    Please note - speed and angle. Until now you have only named the angle.

    But even if you named the speed, you just can't calculate the thrust by modeling the boat as just a brick on an inclined plane. By doing so, you completely ignore wave, friction, spray and aerodynamic drag, for example.

    There are much more consistent methods to perform your calculation, and you should use them. It is useless to re-invent something that has been already invented and well-developed. For example, you can use the power prediction method for planing hulls by D. Savitsky. Just make a little search in the forum and you'll find many resources about it, and also a few Excel spreadsheets for the calculation of hull drag according to Savitsky.

    Also, please take note that:
    1) 1/sin(1°) = 57.3 and not 5.73
    2) displacement is not measured in cm.

    If you are referring to the post #20, those guys have measured the bollard pull of their engine, which is valid only for zero speed. In that case the engine power is entirely used to keep the prop rotating against the moment (or couple) produced by the drag of the blades. No effective power is produced for the forward motion, because the hull is not moving.

    If your aim is to "lift the boat ten centimeters up it's bow wave" then you're talking about a boat which is moving past it's displacement speed, so the bollard pull figure wil be of little help. There is some nice info in the thread you've cited. You have chosen to take for good the only one which is of no use for you. ;)

    Cheers!
     
  6. gonzo
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    gonzo Senior Member

    The question is "to keep it steady". If I interpret that as being static, the answer is 1000Kg
     
  7. Jeremy Harris
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    Jeremy Harris Senior Member

    I think that the single thing that confuses people when they ask this seemingly straightforward question is that power (be it in watts, hp or any other unit) is a measure of force times distance per unit time.

    Thrust is expressed in units of force, so to derive power you cannot exclude time.

    I'm in SI-land, so my preferred units are Newtons for force, metres for distance and seconds for time. Power (in watts) = Newton metres per second.

    You can do the conversion for any other arbitrary system of units you choose, but you still have to include force, distance and time in the equation.
     
  8. daiquiri
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    daiquiri Engineering and Design

    Jeremy, that explanation is perfectly correct, but maybe it is better to simplify it for the people who are not too much into physics by explaining that:

    "distance per unit time" = speed

    So the power becomes simply "force times speed" (when force is kept constant). Or, imho even more simple, "thrust times speed" - because thrust and speed are concepts understood by every boater... ;)

    Cheers

    P.S.
    If even this stuff won't make people flee away from this thread, I don't know what else can we do... ;)
     
  9. jehardiman
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    jehardiman Senior Member

    The way I explain it is to point out that engine SHP is torque * RPM while propulsion EHP is speed*thrust (or drag if you prefer). What connects the two is the gearbox and propulsor; which is a black box function of speed, RPM and losses. Therefore there cannot be a direct conversion of engine HP to thrust; it will always depend on propeller, speed and shaft RPM.
     
  10. jonr
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    jonr Senior Member

    Or put it this way - a 1 HP engine can produce any thrust you want (with the right gears and prop). It can also produce any torque you want.
     
  11. Jeremy Harris
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    Jeremy Harris Senior Member

    The last point about an engine producing any thrust you want is a pretty good one. In my world (aviation) I usually manage to convince the 'thrust must be directly proportional to power' people by using a helicopter as an example. They can produce very high 'thrust' (disc lift) from a modest amount of power, by virtue of the very large diameter, low pitch, low rpm rotor. The same engine on a fixed wing aircraft driving a conventionally sized propeller will deliver the same power, but very much less thrust.

    I'm always intrigued by the way that the low power electric trolling motor companies rarely, if ever, give the power of their motors - they almost always talk about static thrust, a virtually meaningless number as it starts to reduce the moment that speed starts to increase from zero.

    Jeremy
     
  12. no - a 1 HP engine can not produce any thrust

    no - a 1 HP engine can not produce any thrust (with the right gears and prop). itcan produce a maximum thrust (measured in Kilograms) of;


    1 horse power = 750 watts = 750 joules = 750 newton / meters / second = 75 kilograms at gravity
    (note time is ignored - ie 1 unit of time on every single side of the equals signs)



    so the max thrust when no efficiency losses/friction losses/ thermal engine losses / energy transducing losses

    is 75 kg thrust


    in the real world, you loose 50% straight of the top for thermal engine losses, and looks like another 50 % for mechanical/efficiency losses

    giving as the other thread
    http://www.boatdesign.net/forums/outboards/how-many-lbs-thrust-equal-1-hp-24964-2.html

    mentions, 12 Kg thrust










    however, yes
    It can produce any torque


    there is a museum display online with 200 gears connected to each other and a little electric motor running continuously. the last gear is welded unmoveable, yet the motor runs continuously. its calculated not to break until 4 billion years from now.


    the torque produced by this tiny 12v electric motor over the next billion years will be astronomical










    "a modest amount of power"???????

    99% of helicopters have turbines, a single tiny 4cm helicopter turbine blade produces about 50 horse power - incredible power density

    these tiny little blades are sometimes single crytal titanium that take a year to grow


    to get higher energy density you have to go a nuclear weapon.

    is there any thing on this planet with a higher energy density? (than a single turbine blade)

    does anybody know of anything?





    m
     
  13. is pushing that one ton perfectly flat hull up a 10 cm hill

    the bit i'm interested in,

    is pushing that one ton perfectly flat hull up a 10 cm hill
    and then holding it there.

    and apparently i need 2 Horse Power to do it


    m
     
  14. Jeremy Harris
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    Jeremy Harris Senior Member

    I think this thread is rapidly denigrating into the equivalent of the Flat Earth Debating Society, but here are some observations, by way of a final post:

    1. The kilogramme is a unit of mass, not force.

    2. You cannot relate power and thrust without including time - it's a basic law of physics.

    3. Your statement that you need 2hp to move a mass of 1 ton up a 10cm hill is wholly fallacious. Here are some examples of the differing amounts of power needed to do exactly as you have asked, by way of illustration:

    a) Assume you want to move a mass of 1 ton (I'm assuming a short ton of 2240lbs, not a metric tonne) up a frictionless slope of 10cm in 100 hours. This means applying a force of 9967N over a distance of 0.1m for a time duration of 360,000S. The power required is therefore 9967 x 0.1 / 360000 = 0.00277 watts, or 0.00000371hp

    b) Assume you want to move a mass of 1 ton up a frictionless slope of 10cm in 1 second. This means applying a force of 9967N over a distance of 0.1m for 1S. The power needed is therefore 9967 x 0.1 / 1 = 996.7 watts, or 1.336hp

    c) Assume you want to move a mass of 1 ton up a frictionless slope of 10cm in 0.001 second. This means applying a force of 9967N over a distance of 0.1m in 0.001S. The power needed is therefore 996,700 watts, or 1336hp.

    So, for the same basic question I've proved by the above examples that the power could range anywhere from 0.00000371hp up to 1336hp, just by varying the time required from 100 hours to 0.001 second. Does this make sense to you now? Do you now see why you can't ignore time in this equation?

    BTW, there are very large numbers of helicopters flying with ordinary, non-turbine, petrol engines. In fact some of the most popular types ever made all have internal combustion engines and fly on relatively low hp (typically 120 to 150hp) engines. Look up Robinson, the world leading helicopter manufacturer, as an example; their best selling models, the R22 and R44, are internal combustion engine powered.
     
    Last edited: Jul 14, 2010
    1 person likes this.

  15. jonr
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    jonr Senior Member

    Dung's mistakes started at:

    750 watts = 750 joules
     
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