What is the propulsive efficiency of oars?

Discussion in 'Hydrodynamics and Aerodynamics' started by daiquiri, Jun 27, 2013.

  1. Leo Lazauskas
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    Leo Lazauskas Senior Member

    30 strokes per minute (spm) is a bit low. Olympic class scullers start at > 40
    spm, sometimes higher. I would try 35 spm.
    Kleshnev's site is a treasure trove for biomechanical aspects of rowing, but
    he is very weak on the fluid dynamics.

    The attached short report shows most of what I use as inputs to my model.

    I haven't shown the acceleration from rest at the start of the race, but I
    can also model that. Rowers don't start with full length strokes, so you need
    to know their starting pattern. Also, as a race progresses, some rowers will
    fatigue at different rates. That makes multiple crews quite difficult to model:
    some athletes can maintain an almost constant output for the last 1500m of
    a race, others tire more quickly.

    As you can see, there is a lot of data needed to model an entire race!

    After all that "sophistication", modelling something as difficult as the air drag
    of silly hairstyles is done very crudely. Maybe CFD can be used to good effect here :)

    I have another few hundred papers, but it's time for you all to do a little of
    your own research now :p

    Have fun!
     

    Attached Files:

  2. El_Guero

    El_Guero Previous Member

    I think some folks misunderstand your paragraph .... shown below.

    The assumptions on drag versus lift will be over simplified, but nonetheless, if you have paddled, you have noticed the swirling of the water around the edges of the paddle.

    These vortices (swirls) are where the paddle is losing 'efficiency.'

    The term 'lift' is probably confusing for most. Rather than look at it as 'lift,' look at it as reducing the vortices (swirls). So, by changing the angle of attack for the paddle, you reduce the vortices, it feels as if the paddle is moving faster, it is, and it feels 'intuitively' as if that is less efficient.

    That is not quite true. The efficiency at this point would require very complex comparisons of the speed of the boat. I am not going to do that.

    But, I will point out that by reducing the 'vortices' around the edges of the paddle, the fluid (water) will 'stick' better to the surface of the paddle, or oar.

    The increased stickiness means there is a greatly increased amount of energy transferred between the person and the water, and there will be an increased velocity both in the water, and to the forward motion of the boat.

    WHEW!

    My short note became a paper ....

    I should note also, if you propel the paddle at 90% to the boat, you reduce vortices, but do not gain forward movement of the boat. The angle will probably be no more than a 3-5 degree (slight) angle of attack.

     
  3. tspeer
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    tspeer Senior Member

    The original question concerned the propulsive efficiency, but the answers seem to be aiming at the energy efficiency. Propulsive efficiency is quite different and specific.

    I think the propulsive efficiency for oars can be determined in a similar manner to the propulsive efficiency of a prop or jet. It's true that the ideal oar would act as though the blade were practically anchored in place and all the force applied to the oar resulted in the boat moving forward. But the oar does move fluid back, and this means its propulsive efficiency can be defined in the same way as for a prop.

    The propulsive efficiency (not to be confused with the total energy efficiency) is

    eta_p = 2/(1 + Ve/v)
    eta_p = propulsive efficiency
    Ve = exit velocity of the fluid propelled by the oar
    V = speed of the boat

    The blade of the oar is similar to the disk of a prop - it takes in fluid and imparts a change of velocity to it. Since the plane of the blade is approximately 90 deg to the velocity of the boat and there's no flow through the blade, Ve can be taken to be the speed of the blade relative to the boat. This would be the angular velocity of the oar times the distance from the oarlock to the center of effort of the blade.

    If the blade were truly stationary in the water, then Ve = V, because the water coming back from the blade (relative to the boat) would be equal to the speed of the boat. In that case the propulsive efficiency would be 100%.

    The more slippage of the blade through the water, the higher Ve and the lower the propulsive efficiency will be. So, just as with a prop, a small blade moved at a high rate will have a lower propulsive efficiency than a large blade moved at a low rate to produce the same force.

    Note also that for the case of a bollard pull, the propulsive efficiency is zero no matter what the oars do, because the boat isn't moving.
     
  4. philSweet
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    philSweet Senior Member


    I think we should set that definition aside since it assumes a quasi-steady-state operation. Once that goes away, the simplification is lost and it becomes an energy problem whether you want it to be or not.

    The various efficiency components (etas) that were conveniently multiplicative under steady-state prop/shaft/tranny/motor systems are no longer so well behaved. You now have to go instant by instant and particle by particle and consider the entire system at once. Cyclic integrals of component etas don't help much because the applied force vector is changing direction and magnitude during the cycle, and thus the propulsive efficiency (cyclical integral) is no longer multiplicative with respect to other efficiencies even if you did calculate it. (At least you can't assume it will be. It needs to be looked at.) And if you can't multiply them all together to get overall efficiency, they would be of limited interest.

    In short, unless steady-state can be assumed, the integral operator (in time) lacks the distributive property necessary to use the etas in the conventional way.
     
  5. DCockey
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    DCockey Senior Member

    Principles of Naval Architecture, Second Revision, Volume II 1988 defines propulsive efficiency as the Effective Power / Power Delivered with the Effective Power defined as the power used in overcoming the resistance to motion of the ship. (Section 1.4, p130).

    This is an approximate relationship for a propeller based on momentum theory.
     
  6. tspeer
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    tspeer Senior Member

    Yes, it is. My point is to propose one can treat the oar using momentum theory in much the same way. The question had to do with propulsive efficiency, not energy efficiency. Propulsive efficiency is a momentum theory concept.

    philSweet has a good point that rowing is an intermittent propulsion, while other forms are continuous. Would intermittent operation of a propeller be any different from the standpoint of propulsive efficiency than steady state operation?

    It might be possible to average the velocity of the boat and the velocity of the oar over the cycle period to arrive at an equivalent propulsive efficiency.

    Or it may be that propulsive efficiency is simply not an applicable concept to rowing, due to the cyclical nature of rowing, and energy efficiency is the only meaningful efficiency.
     
  7. DCockey
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    DCockey Senior Member

    I checked four references and they agree that propulsive efficiency is Effective Power / Delivered Power.
    Principles of Naval Architecture, Second Revision, Volume II, 1988, p130
    Basic Ship Theory by Rawson and Tupper, fourth edition, 1994, p403
    Ship Resistance and Propulsion by Molland, Turnock and Hudson, 2011, p3
    Hydrodynamics in Ship Design, Volume One by Saunders, 1957, p516​

    ___________________________________

    "eta_p" above is the ideal efficiency for a propeller which can never be achieved in reality. It can be derived starting with momentum thoery applied to a stream tube containing the propeller. Various assumptions and approximations are made including uniform flow across the stream tube, radial and swirling flow components are negligible, and no viscous losses. Basic Ship Theory has a derivation in Chapter 10.
     
  8. Petros
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    Petros Senior Member

    the only thing that matters is total energy out, vs. total average drag to overcome. And with oars or paddles where the speed varies, you have to average it over long periods of time. As an engineer, where the practical application of the physic is the most important aspect of designing any system, I would generalize this even further; efficiency should be measured as the total energy output over a give distance at a give average speed. when paddling to get somewhere, there are only two things that matter: the distance traveled, and the time it takes to get there. The least amount of effort it takes to accomplish that, the more efficient the overall paddle/hull system.

    taking the instantaneous efficiency at any given point in the paddle stroke is only relevant if trying to improve the component efficiency at that part of the stoke. This has no practical application for either long distance paddling or even for short sprints in a race, since the total energy output vs. the maximum speed achieved is the object of the race. The total energy output over a given distance is what is relevant in paddling trip. Optimizing some tiny portion of the stroke could result in overall efficiency being reduced. IOW, total power output (area under the curve) for either a given distance, or a given average speed, is what is relevant and the only thing worth optimizing. Otherwise it just becomes another meaningless intellectual physics exercise. The object after all is to design real paddles or oars that perform better in a real environment.

    You used that word "slip" again in describing the action of the paddle. I know many in the industry use this term, but you would not describe the lift of a wing on a sail plane as being generated by "slip", so why would you use it to describe the lift (or in this case the forward thrust) coming off a paddle or oar blade?

    I want to emphasize that to get a force out of a fluid, you must accelerate the mass of the fluid. F=ma. that is how wings on an airplane, blades of propeller, and paddle and oar blades all work. the physic is the same, to apply an obsolete term like slip to it only confuses the issue. You can not measure blade efficiency by comparing "slip", it can only be measured by energy in vs. useful thrust (or lift) out. The perceived slip does not enter into the equation.

    For example, using really giant oar blades would have very low perceived slip, vs smaller blades. But this does not say anything about the energy input it takes to create the same amount of forward thrust. Poorly designed blades will take more energy input to get the same amount of thrust output no matter their size.

    This is why using our perception as the unit of measure is not a reliable indication of the mechanics involved. And I think this is why studying the efficiency of paddles or oars has been so problematic, what is happening in terms of the fluid mechanics is not intuitively obvious.
     
  9. daiquiri
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    daiquiri Engineering and Design

    Since we are all boating-types here, it is pretty much logical that propeller theory is the first thing that came into our mind when talking about paddling efficiency.
    Few things that I have to note in this late evening hour, after a long working day (so pardon me if you read some nonsense here):
    1. "Paddling efficiency" vs. "paddle efficiency" (or "rowing efficiency" vs. "oars efficiency"). Which one is more appropriate to define? Because, as previously noted, rowing or paddling is a cyclic activity, and comprises a wide range of oars or paddle's efficiencies, since the speed relative to the water varies both in function of time and in function of the boat speed relative to the water. Messy stuff indeed, which implies that in this particular case, the efficiency is perhaps better defined via the energy output/input than via the instantaneous power output/input.
      ---
    2. If we wanted to treat this via the actuator disc theory, which model is then more appropriate - the propeller or the wind-turbine one? Oars work by creating a drag force. From the reference frame moving with the row blades, the water is flowing from the transom and is being slowed down by the blade. So in this reference frame the blades act like a wind-turbine, and hence one could expect a limit to exist even for an ideal efficiency. A Betz limit applied to oars, perhaps?
      ---
    3. I don't like the measuring techniques which involves measuring oxygen consumption by the paddler, because they don't allow a reliable definition of the pure mechanical efficiency. There's always the human-body metabolism in between the oxygen intake and the boat speed. And the metabolism varies on an hourly basis - besides having a memory effect. Imo, the efficiency should be measured via purely electro-mechanical means.
    Cheers
     
  10. Petros
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    Petros Senior Member

    daiquiri,

    you make some good points, however one point I think is an error.

    if this was true than the most efficient oars or paddles would be drag devices like a tennis racket, or you could increase paddle efficiency by wrapping the blade with carpet. A simple test would show this is incorrect: warp one blade of a kayak paddle, or one oar blade, with carpet, and the leave the other "clean" and go and try and use it. the clean blade, the one with less drag, will produce the most thrust.

    This means that the most efficient the lift creation off the blade, the more thrust than can be generated from the blade. And the most forward movement for a given amount of power input.

    This is what I mean that the mechanics of the action of a blade in the water is not obvious. When you heave against a set of oars or a paddle, the more resistance you feel seems like the more forward motion you are able to generate. This feels like drag in your hands, but I think you are actually pulling against the thrust (or lift in the forward direction) that the blades generate, not against the drag. "slip" or drag have little to no part in it, the more thrust or lift the blades generate, the more you can pull against them, and the more forward speed you can obtain.
     
  11. daiquiri
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    daiquiri Engineering and Design

    Agreed, the lift component contributes to the total hydrodynamic force R created by the blade. But it is the drag component of that total force that counts in terms of propulsion, because it is the drag component which acts in the direction of the water flow (by the definition of the drag force).

    See the attached picture:

    Oar.gif

    It shows how a more draggy blade will produce a bigger horizontal force (i.e. the drag, i.e. the thrust), when the lift force is kept unchanged. However it is true, as you correctly said, that the lift component does play a role in this process, because at zero lift (90° blade AoA) the drag component would be smaller.

    Besides that, all the other previous considerations remain unaffected by this clarification.

    Cheers
     
  12. DCockey
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    DCockey Senior Member

    Is the direction of the water flow being defined the direction relative to the hull or the direction relative to the blade?

    The blade is moving relative to the hull. The only time the direction of the water flow relative to the blade is parallel with the direction of the water flow relative to the hull is when the oar is perpendicular to the centerplane of the hull. (Assuming the hull is moving straight ahead.) During the rest the stroke the direction of the water flow is in a different direction relative to the blade than it is to the hull.

    If the "lift" and "drag" components of the hydrodynamic force on the blade are defined by the direction of the water flow relative to the hull, then only the "drag" force contributes to propulsive force.

    But if the "lift" and "drag" components of the hydrodynamic force on the blade are defined by the direction of the water flow relative to the blade, then both the "lift" and "drag" contribute to the propulsive force, with the relative amounts of contribution varying during the stoke.
     
    Last edited: Jul 24, 2013
  13. daiquiri
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    daiquiri Engineering and Design

    Addendum to my previous post (#41):

    The previous graph was sketched in a qualitative (non quantitative!) way, based on my general knowledge about airfoils at common (moderate) angles of attack.

    However, I was curious to see what happens at AoA's up to 0°, and have found this interesting page: http://www.aerospaceweb.org/question/airfoils/q0150b.shtml
    There are two graphs, showing Cl-AoA and Cd-AoA curves of an airfoil for angles up to 180°. I have taken those values, inserted them in the Excel spreadsheet and plotted them in a standard polar form Cl-Cd, up to 90°. Here is the result:

    Cl-Cd.gif

    The interesting thing is that the maximum R force, and also the maximum Drag force, occurs at nearly 90° AoA. The lift force is almost negligible at that angle.

    So, it likely that the oar blade (though it is a 3-D object) will behave in a qualitatively similar manner, just with a more rounded curve (due to the dampening effect of low aspect ratio on the stall characteristics). And hence an oar will be essentially a drag-generating body when set at it's optimum angle of attack for the generation of thrust. The lift component will then be very limited, if not even zero. If that is true, it would imply that an optimum oar motion should be as much as possible (and for as long as possible) parallel to the boat speed vector, with a blade angle set at nearly 90°.

    Cheers
     
  14. daiquiri
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    daiquiri Engineering and Design

    IMO, that will depend on the rowing kinematics - which is an input variable to be optimized, together with the oar shape. A wide range of near-parallel motion (in the vertical longitudinal plane) should imo be possible by a careful control of the oars by the crew, which comes with good technical training.
    Cheers
     

  15. DCockey
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    DCockey Senior Member

    The motion in the horizontal plane will not be "near-parallel" unless very short strokes are used. The fore-aft sweeping of the oar results in the blade moving away from the boat from the start of the stroke until the oar is perpendicular to the boat's centerline. Then the blade starts moving towards the boat until the end of the stroke.
     
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