Aspect ratio for surface piercing foil

Discussion in 'Hydrodynamics and Aerodynamics' started by revintage, Feb 22, 2020.

  1. revintage
    Joined: Nov 2016
    Posts: 240
    Likes: 45, Points: 28, Legacy Rep: 10
    Location: Sweden

    revintage Senior Member

    This is off course obvious to everyone schooled in aero- or hydrodynamics.

    In aerodynamics AR seems to be calculated from projected area.

    I want to use AR to calculate 3D CL. Projected area is off course used to find lift.

    AR=length^2/area, question is if length should be measured along the foil or projected and if one should use foil surface or projected area?

    In a semi scientific reference about hydrofoils(Grogono 1987), Alan Alexander have used length along the foil together with foil surface area, for the type of foil to the right.

    Should the left one be calculated the same way as the right, or should it be looked upon as two separate foils with projected area=500 respectively?

    surforproj.png
     
  2. DCockey
    Joined: Oct 2009
    Posts: 4,496
    Likes: 237, Points: 63, Legacy Rep: 1485
    Location: Midcoast Maine

    DCockey Senior Member

    What method are you using for the calculations?
     
  3. revintage
    Joined: Nov 2016
    Posts: 240
    Likes: 45, Points: 28, Legacy Rep: 10
    Location: Sweden

    revintage Senior Member

    AR calculation as in my post above. Just a question about projected or total surface area. In the ballpark lift calculations I have made, the difference in liftoff speed is within 20% of each other, depending of which of the two area alternatives you use.
    What method are you using?
     
  4. DCockey
    Joined: Oct 2009
    Posts: 4,496
    Likes: 237, Points: 63, Legacy Rep: 1485
    Location: Midcoast Maine

    DCockey Senior Member

    I assume by "AR calculation" you mean the formulas which can be derived from lifting line theory. Such theories use the plan view area and aspect ratio based as viewed normal to the span direction which would be tilted 60 degrees from vertical in your example. The lift for those equations in normal to the span direction. The vertical component of lift will be the total lift multipled by the cosine of the dihedral angle. For your example this would be vertical lift = 0.5 * total lift.

    For the V-shaped hydrofoil the interaction between the two sides will be different than for a straight foil, and that will affect lift. I assume there are tables/formulas available somewhere for the effect of large dihedral angles on lift.
     
  5. revintage
    Joined: Nov 2016
    Posts: 240
    Likes: 45, Points: 28, Legacy Rep: 10
    Location: Sweden

    revintage Senior Member

    Thanks for your concern to help up my lack of knowledge. The lift calculation is not the main issue here and is no problem. It is what area to be used when calculating AR in order to calculate 3D CL, surface 2000cm2 or projected 1000cm2 in the sketch above? In my example AR differs by 2:1, but thankfully lift differs less.

    Unfortunately "Such theories use the plan view area and aspect ratio based as viewed normal to the span direction which would be tilted 60 degrees from vertical in your example." is beyond my English knowledge:oops:.

    I also understand the foils of the V should interact, but haven´t found any info/tables/formulas. Have studied the NASA paper about dihedral foils, but unfortunately no clue about the definiton of AR..https://conservancy.umn.edu/bitstream/handle/11299/108301/pr041.pdf?sequence=1

    As I mentioned in my first post AR seems to be calculated from projected area in aerodynamics, in contradiction to the hydrofoil calculations of Alexander in my reference. Puzzled!

    To vizualise the differences in lift, I copied the results from a spreadsheet example I made, with chord 20cm and speed 10knots.
     

    Attached Files:

  6. DCockey
    Joined: Oct 2009
    Posts: 4,496
    Likes: 237, Points: 63, Legacy Rep: 1485
    Location: Midcoast Maine

    DCockey Senior Member

    Ignore my comments above. I assume that you are using the equations from the 1954 report. Your question is about how the authors of the report define aspect ratio. I am not certain but I believe the authors used the total area, not the projected area.
     
  7. revintage
    Joined: Nov 2016
    Posts: 240
    Likes: 45, Points: 28, Legacy Rep: 10
    Location: Sweden

    revintage Senior Member

    Thanks, I also initially guessed it is about surface area, but I am not sure. It is a mystery that that there are no explanations to be found.

    Attaching the spreadsheet "grog" where I use metric system. Please correct me if you find any faults in the formulas used, or have any better equations to get these ballpark numbers.

    Also added a spreadsheet with IRL data from the V-foils of Doug Halsey´s Broomstick V-foils, where the differences between projected and surface area shows to be small. Testing this one makes me wonder which area to useo_O?
     

    Attached Files:

    Last edited: Feb 23, 2020
  8. Heimfried
    Joined: Apr 2015
    Posts: 337
    Likes: 37, Points: 28, Legacy Rep: 37
    Location: Berlin, Germany

    Heimfried Senior Member

    Hi Lars,
    If you compare the equations 2 and 4 (at pages 15 and 16 of the linked 1954 paper), you will find (case f(t)/c = 0) that f(m)/c = a/(2 * c) * tan(gamma) = 1/4 * AR * tan(gamma)
    Multiplied by 4 and divided by tan(gamma) remains AR = 2 * a / c . According to the sketch at page 15 that means aspect ratio is given as the projected full span divided by the chord.
    I'm quite sure, this could be different in other papers. So you have to check each source to find out which way it is used.
     
  9. Doug Halsey
    Joined: Feb 2007
    Posts: 403
    Likes: 106, Points: 53, Legacy Rep: 160
    Location: California, USA

    Doug Halsey Senior Member

    If you only have a slanted planar foil, it's probably most sensible to use the projected area normal to that plane, and to calculate the induced drag for the total normal force on the foil. This force has both vertical & horizontal components (i.e. - lift & side force), but you can get away with having a single efficiency factor for the combination.

    If you have a more complicated geometry (like a V foil), it makes more sense to use either vertical or horizontal projections, depending on whether you are calculating the induced drag due to vertical lift or horizontal side force. (To get the total induced drag, you need two efficiency factors).

    I don't think it ever makes sense to use the wetted area.

    When dealing with surface-piercing foils, the height that the hull flies above the water varies considerably with speed, so the span, area, and aspect ratio are not convenient constants. In my VPP codes, I find it easier to forego the aspect ratio and force coefficient calculations altogether & calculate lift, side force, and drag directly in dimensional terms.
     
  10. revintage
    Joined: Nov 2016
    Posts: 240
    Likes: 45, Points: 28, Legacy Rep: 10
    Location: Sweden

    revintage Senior Member

    Hi Heimfried,
    Thanks for clarifying. I actally found out another way by comparing the third V-foil for test in Fig.2 page 6. Span 12", chord 2", AR=6. Then checked fig 13 page 18 and took the 2D CL value at 4 degrees and used equation (5) and got the same 3D CL. This also shows that they use projected area.

    Hi Doug,
    Great, have learnt a lot here, even if much is above my level. As I mentioned when coping with IRL foils, using Broomstick as example, the difference is very small. Still wanted something to get ballpark figures from in this discussion.
     

    Attached Files:

    Last edited: Feb 23, 2020
  11. Heimfried
    Joined: Apr 2015
    Posts: 337
    Likes: 37, Points: 28, Legacy Rep: 37
    Location: Berlin, Germany

    Heimfried Senior Member

    You are using a dihedral angle of 60° in your calculations and the empirical formula given by equation 5 (page 17, 1954 Paper). It is said in this paper that these formula is for surface piercing foils and delivers good results at dihedral angles of 25° up to 45°. Angles above 45° are not mentioned. So a result for 60° should be considered with caution.
     
    revintage likes this.

  12. revintage
    Joined: Nov 2016
    Posts: 240
    Likes: 45, Points: 28, Legacy Rep: 10
    Location: Sweden

    revintage Senior Member

    Yep, there is a reason to be a little skeptical, as when you insert 90 degrees you get the highest CL3 number. So one might expect that even at 60 degrees the numbers can be a little to high. In my last example speed does not differ with more than 0,5 knot when going from 60 to 45 degrees.
    EDIT: Compared Vellingas submergence factor with NASA 1954 and they where quite close up to just below 60 degrees. Both equations are determined empirically.

    3dcl.PNG
     
    Last edited: Feb 24, 2020
Loading...
Similar Threads
  1. Autodafe
    Replies:
    29
    Views:
    5,865
  2. Will Fraser
    Replies:
    150
    Views:
    22,495
  3. amjams
    Replies:
    12
    Views:
    4,379
  4. sailingdaniel
    Replies:
    95
    Views:
    22,968
  5. philSweet
    Replies:
    4
    Views:
    3,308
  6. DCockey
    Replies:
    153
    Views:
    34,829
  7. JohanH
    Replies:
    11
    Views:
    1,779
  8. CocoonCruisers
    Replies:
    5
    Views:
    841
  9. motorbike
    Replies:
    19
    Views:
    3,216
  10. surfbus
    Replies:
    44
    Views:
    7,424
Forum posts represent the experience, opinion, and view of individual users. Boat Design Net does not necessarily endorse nor share the view of each individual post.
When making potentially dangerous or financial decisions, always employ and consult appropriate professionals. Your circumstances or experience may be different.