1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Fluid force on flat object

  1. Aug 25, 2009 #1
    I'm trying to make a computer simulation of ship.
    Currently i'm working on a force of wind on the ship sail.

    This is the page which describes the basics:

    Basicaly, wind (W - vector) has some staring direction and then that direction is chaged because of sail (S - vector). So wind change vector (Fw) is: Fw = S - W.
    Because of Newton low, it means that force on boat is -Fw, or:

    Fb = W - S

    So far so good.
    But when the angle between the wind and sail becomes close to PI/2 my physics break apart.
    What happens then is that the resulting Fb is either down left or down right, both of
    which is wrong. As it is intuitively clear, when angle between wind and sail is PI/2, the resulting force on sail should be in the same direction as wind.

    Clearly, the problem is that wind current will split into two streams, left and right, and when their separate forces are summed, everything in fine again.
    But how to put it into simple equations?
  2. jcsd
  3. Aug 25, 2009 #2


    User Avatar

    I'm afraid you can't simply adapt your given equation to cover this. The problem is there are two different regimes the sail is operating in...

    1. As long as the angle between sail and wind is small enough, the wind hits the leading edge of the sail (nearest the mast, the part of the sail known as the 'luff') and is redirected along the length of the sail (to its 'leech') the resulting momentum change exerting a force on the boat.

    2. However, once the boat is pointing sufficiently downwind the sail 'stalls', and to a first approximation just traps the wind --- the momentum change now is just the original momentum of the wind before it hits the sail. A simple view of the wind splitting off to both sides of the sail as you suggest gives the same basic result (these two momentum will cancel), but in reality what tends to happen is that turbulence develops behind the sail.

    So you will have to write your computer program to treat each regime separately. But remember, you want to be working with the relative (or 'apparent') wind, not the absolute wind. It is possible for instance that on a heading that is downwind but not parallel to the wind (a very 'broad reach') a sail could start off stalled but as the boat speed builds begin to operate as a foil again due to changes in the relative wind vector.

    Precisely when you switch between these two regimes is also not entirely clear cut I'm afraid. For an entirely flat sail you could argue that it always has a leading edge until the angle becomes exactly PI/2 but this is not very realistic and turbulence would likely develop before this. An accurate answer could only be obtained by proper fluid dynamical modelling. For a toy model like this, you might get acceptable results by slowly increasing the percentage of 'stalled air' after the angle reaches some threshold. I'm afraid I don't know enough about detailed sail aerodynamics to offer anything better than that.

    Also be careful with your orientations. You said that:
    But if you meant by "down" the direction opposite to that in which the boat is pointing, then you are confused. Your formula (although wrong in this regime) should still predict a forward force on the boat when the boat is headed directly downwind ('on a run'). The only time a PI/2 angle between wind and sail would give a backward force on the boat is if you 'back' the sail --- i.e. the wind is hitting the sail from the wrong side.

    As an interesting aside, square rigged sales and spinnakers are sometimes trimmed so that instead of stalling, they operate by redirecting the wind downwards (i.e. their upper edges become the leading edges) allowing better downwind progress for the boat. Also there are a few small sailing dinghies (such as the Olympic Laser class) that allow their mainsail to be let out at such a large angle to the boat that the sail can channel the air in the reverse direction (the part of the sail furthest from the mast, known as the 'leech', becoming the leading edge and that past nearest the mast, the luff, becoming the trailing edge) --- known as "sailing by the lee" and resulting with the sail being on the windward rather than leeward side of the boat. But I don't suggest you try and model these complications!
  4. Aug 25, 2009 #3
    Actually, i started to programm this game as a simple drive-your-small-boat-left-right-up-down, so i want to keep things as simple as possible (read, no turbulence :)).

    I'm handeling this by switching direction of vector of the sail when angle between sail and relative wind is less then PI/2

    That's exactly what i need, predicting amount of this "stalled air"?

    I'm currently thinking about using sin[tex]^{(3, or higher)}[/tex](angle sail<-->rel. wind) to get amount of stalled air?

    Yes, as i read some materials on the net about sailing, it seems that LAZER is something sacred, right? :)

    And thnx for the answer!
  5. Aug 26, 2009 #4
    Solved it with approximation.

    I used the following code to approximate amout of water that would flow on the opposite side and it works satisfactory!

    double angle2 = Math.PI - relativeSpeed.angle(vKeel);
    double oppositeStream = (Math.pow(angle2, 4))/(2.0*Math.pow(Math.PI/2.0, 4));

    http://img233.imageshack.us/img233/8995/rjeseno.th.jpg [Broken]
    Last edited by a moderator: May 4, 2017
  6. Sep 25, 2009 #5


    User Avatar

    Sorry for replying so late.

    Glad to here it works. Although I can't say I know of any physical justification for the approximation you've used!

    If you ever finish the game send me a PM, since as you may have guessed I'm a bit of a sailing buff!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook