Froude and planing

Discussion in 'Hydrodynamics and Aerodynamics' started by sandhammaren05, Feb 26, 2017.

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  1. gonzo
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    gonzo Senior Member

    To make this clear: your position has changed to accept models as valid. In the previous posts you claimed that hanging out the transom of a boat and looking at the water was the only way to validate lift.
     
  2. sandhammaren05
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    sandhammaren05 Senior Member

    Not at all, this is your usual irritating confusion. I make models, I wrote that I do not trust simulations. You probably don't know the difference.
     
  3. CT249
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    CT249 Senior Member

    I show the deck because they are the available photos, and they show the relevant point which is the lack of a transom. The top board does not have a transom and yet it planes efficiently under low power therefore it is yet more proof that your claim "a hull with no transom cannot plane" is incorrect.
     
    Last edited: Apr 15, 2018
  4. gonzo
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    gonzo Senior Member

    Your claims read that leaning over the transom is a scientific method. However, simulation and models based on simplified systems are not. At the same time you claim that you do use models and simulations, but you have not changed your original position. No wonder I am confused.
     
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  5. sandhammaren05
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    sandhammaren05 Senior Member

    Very nice paper, also the historic Prandtl video. Thanks for the links. The paper is a 2D calculation, very pretty, and the Kutta condition is satisfied in the calculation when lift begins. The paper shows details that we (rightly) ignore in the inviscid fluid treatment. The inviscid fluid treatment is not wrong, it simply predicts that the trailing vortex (backflow) must start at the sharp trailing edge as soon as the speed is nonzero (it ignores the startup time for boundary layer separation). In the viscous treatment the boundary layer separates from the wing, that's the trailing vortex. The paper takes a large angle of attack greater than the stall angle, presumably in order to see the vortex formation early in the calculation. In the 2D picture the bound and trailing edge vortices must start at the same time to conserve circulation. The 2D picture misses an important part of the reality, namely, that in 3D the leading edge and trailing vortex are a single horseshoe vortex. This is because no vortex can end in mid-air or in mid-water. So the leading and trailing edge vortices form simultaneously and always conserve circulation. The trailing edge vortex ends on the fuselage, as does the bound vortex. The former experiences a downward-lift so that as the speed increases the end of that vortex slides downward (the sliding of the vortex end against the fuselage experiences drag, the vortex is circulating fluid sliding against a solid object). When the speed is large enough then the trailing vortex slides off the fuselage, connects with the vortex from the opposite wing to form a free horseshoe vortex that is washed downstream. At this pint the wing experiences a net lift, that due to the bound vortex. That the bound vortex of the wing and the free trailing vortex are a single vortex is shown in many flow visualization photos. So, the trailing vortex from the transom of a boat must in reality end on the sides of the boat, but once it's washed downstream there is only vortex stretching from the sides of the boat because no new trailing vortex/backflow forms at the transom (the Kutta condition is satisfied). That detail has not, I think, been simulated, but maybe someone will do it.
     
    Last edited: Apr 16, 2018
  6. Joakim
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    Joakim Senior Member

    In that paper the lift starts at the very same moment as the foil starts to move (0.5 s, not 0 s). If you stop the video at 0.5 s, you will see the starting vortex is beginning to develop and the velocity vector at the trailing edge points well upwards.

    But since you don't really believe in simulations, did you read the paper I linked you about measured lift and velocities. There was also lift all the way from the beginning and there was still backflow at the trailing edge. Here's the link again.
     
  7. sandhammaren05
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    sandhammaren05 Senior Member

    With the Wagner effect you're getting into insect flight. Neither a wing nor a boat bottom flaps up and down, never steady.

    I doubt simulations done elsewhere because I know the usual pitfalls, I'm not a believer.

    First sentence above. Velocity vector points upward at trailing edge and bound vortex begins to form. Bound vortex begins to form with backflow. No lift, the circulations cancel each other.

    Dumb question: are you identifying the existence of the bound vortex as lift? It produces no lift until the backflow ends because the two circulations cancel each other. The Kutta condition must be satisfied before there is lift.

    I will look at your insect flight link.
     
  8. Joakim
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    Joakim Senior Member

    No. The simulation paper shows the lift coefficient curve as a function of time and there is lift from 0.5 s when the foil starts to move. They show both the pressure and friction part of the lift and thus it has been defined as an integral of pressure and friction over the walls of the foil.

    Also the "insect" paper shows lift as soon as the flat plate starts to move. This lift is measured with a force sensor. Clearly there is lift while there is still backflow (starting vortex at the trailing edge). Note that the flat plate moves just like an airplane would, it is not flapping. This paper very clearly shows that your theory about onset of lift and Kutta condition is false.
     
  9. sandhammaren05
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    sandhammaren05 Senior Member

    You continue to rely on your reading of simulations, throwing insect wings into the argument. Unnecessary. There must be a net circulation in order to get a net lift. Circulation is conserved.
     
  10. Joakim
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    Joakim Senior Member

    Well, if you don't want to look at actual measurements of lift during start up, you can always try to understand it through the circulation. Look at the picture below. Circulation around the full area including the foil and the starting vortex must be zero before the foil moves and after that. So to get the correct lift from circulation one must exclude the starting vortex from the integral path (use the dotted line).

    You seem to acknowledge that the starting vortex starts to develop at the trailing edge. What happens, if you move the dotted line very close to the trailing edge? Now the dotted line goes through the starting vortex at the beginning. Thus part of the starting vortex is in the "starting vortex" integral and part in the "bound vortex" integral. This means there is already lift, but just not as much as when the starting vortex is further away from the foil.

    That is one of the problems with circulation theory. You get different results depending on where you put the integral path. If you include the starting vortex, you always get zero. Integrating the pressure over the foil surface has no such problem. Nor does measuring the lift with a force sensor.

    Most often starting vortex is just forgotten, since it is very far away in steady state simulation. Then you need to enforce the Kutta condition for an inviscid simulation in order to get the correct flow field and lift. No such need for a viscid simulation or real life.



    [​IMG]
     
  11. DCockey
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    DCockey Senior Member

    From Newman, Hydrodynamics p 231 after discussing the Wagner function "This problem is identical to that of a foil accelerating instantaneously from rest at t = 0 to a constant velocity U, with a constant angle of attack, since in both cases the perturbation vanishes for t < 0."
     
  12. sandhammaren05
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    sandhammaren05 Senior Member

    IMG_1294.jpg IMG_1294.jpg IMG_1294.jpg IMG_1295.jpg

    You have not answered my questions above. Instead, you continue throwing out red herring. What, e.g., is the point of the picture above that I obviously know? You also
    insist on relying simulations. Do you really think that I don't understand integration contours!? I insist on circulation conservation, not simulations. which may violate it via numerical error.

    I will try one last time to eliminate your confusion. As long as the backwash is there, both vortices are bound to the wing and their equal but oppositely directed lifts cancel. Once the trailing vortex is free, as in your drawing above, then to order -G (=circulation) it exerts no force on the wing. The backflow is zero, the Kutta condition is satisfied, and the wing experiences a lift due to the circulation G of the bound vortex. The free trailing vortex exerts a force on the wing to order G^2, that is the induced drag. It reduces the effective angle of attack of the inflow.

    I emphasize: the lift is and must be zero as long as there is backflow because the separate lifts of the two vortices of opposite sign of circulation cancel each other. This is all very elementary. See Newman. Ao long a both vortices are bound to the wing there is a torque (because their centers of pressure do not coincide) but no lift. If a simulation shows otherwise then the simulation is wrong.

    Here are two photos of the transom of the v153 in motion. Let e=epsilon. The first photo is taken at 8-e mph, there is backflow and hence no lift. The send photo is taken at 8+e mph, the Kutta condition is satisfied hence there is lift, although buoyancy dominates lift here. At 13 mph lift is not negligible compared with buoyancy, at 18 mph lift overwhelms buoyancy.
     
  13. gonzo
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    gonzo Senior Member

    According to Sandhammer this boat is not experiencing any dynamic lift, because the transom is still wet:
    upload_2018-4-17_17-50-47.png
     
  14. CT249
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    CT249 Senior Member

    If lift is zero when there is backflow, what is lifting this windsurfer?



    At around 7 seconds, the board is almost stationary and because its volume is low, the sailor is almost knee deep. Later, the board rises and yet there is no evidence at all that there is zero backflow over the stern.

    Here's some underwater pics of another sinker windsurfer which show the lack of a "dry" transom.

    Sinker Up-haul and Tack. https://thewavehobbit.wordpress.com/windsurfing/wave-riding-the-bottom-turn/sinker-up-haul-and-tack/

    The underwater photos clearly show that there IS backflow. The board is clearly NOT going to lift onto the surface without dynamic lift, because at 69 litres volume its buoyancy is clearly insufficient to float the sailor and rig when static, as demonstrated by the fact that it's clearly completely underwater. And yet this sinker, like others, WILL develop lift, rise to the surface, and plane at 30 knots or so.

    So planing lift can clearly develop without backflow and without a dry transom. If this is such a special case (as claimed earlier) then why is it so? How can one be sure that this is a special case, rather than something that proves the theory incorrect?
     

  15. sandhammaren05
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    sandhammaren05 Senior Member

    You need either glasses or more concentration. Look again. The water separates cleanly from the bottom at the transom.
     
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