Wind speed vs angle of heel

Discussion in 'Stability' started by rader, Dec 7, 2012.

  1. Eric Sponberg
    Joined: Dec 2001
    Posts: 2,004
    Likes: 209, Points: 73, Legacy Rep: 2917
    Location: On board Corroboree

    Eric Sponberg Senior Member

    Check that again: Turning the equation around to solve for GM:

    GM = 57.3*486*21.5*1/(18*7005) = 4.75'

    I also checked with a fellow naval architect and good friend of mine that designs a lot of race boats and cruiser-racers. He had a design that had the following characteristics which were close to your boat:

    Loa = 34' 6"
    Lwl = 33' 10"
    B = 11' 9"
    Draft = 7' 0"
    SA = 815 ft^2
    Displ = 7,200 lbs.

    The GM for that boat was 5' 7"

    So, the GM for your boat should be in this realm. Have you contacted any of the other Tarten Ten owners to see if anyone has any information or data on GM to confirm the number more precisely?

    Alternatively, your initial inquiry was to determine what are the wind speeds that would cause various angles of heel. You can determine this experimentally. As you use the boat, use a good inclinimeter to record the angle of heel for the different wind conditions and plot them on a chart--wind speed vs. angle of heel.

    Eric
     
  2. rader
    Joined: Dec 2012
    Posts: 20
    Likes: 0, Points: 0, Legacy Rep: 10
    Location: Chicago

    rader Junior Member

    Metacentric height resolved

    This is in response to Eric's post (which I don't see yet on this forum).

    Metacentric height: 4.75 ft.
    Got the same result after re-calculating (the assumption is that the Tartan Ten has average stiffness). It did not seem to make sense that the fore-aft axis of rotation would be so low. The ORR certificate shows the CG about 5" below the water plane. As for actual sailing and observation and data collection, that would depend on my wife (and that's not going to happen) and other willing crew to hike out on the rail. Next I want to calculate the 'rail-meat' equivalent needed for additional righting moment to maintain a heel under 10 degrees as the wind builds.
     
  3. Eric Sponberg
    Joined: Dec 2001
    Posts: 2,004
    Likes: 209, Points: 73, Legacy Rep: 2917
    Location: On board Corroboree

    Eric Sponberg Senior Member

    Calculate the difference in heeling moment between 10° heel and whatever greater heel that you are considering (12°, 15°, etc), and divide that by the distance between the boat centerline and the area where the extra crew sits. That will give you an appoximate weight of crew necessary to maintain 10° heel.

    Eric
     
  4. rader
    Joined: Dec 2012
    Posts: 20
    Likes: 0, Points: 0, Legacy Rep: 10
    Location: Chicago

    rader Junior Member

    'Rail meat'

    Eric,
    My ORR certificate shows Arm and RM for every 2 degrees of heel. Is RM what you meant by heeling moment?

    For example

    10 degrees heel RM 1547 ft-lbs

    16 degrees heel RM 2342

    Difference: 795 ft-lbs/ 4.7 feet (distance from midline of boat to rail) = 169

    which seems to say that, whatever crew is already on the rail at 10 deg of heel, another 169 lbs is needed (if I did this right...)

    Rader
     
  5. Eric Sponberg
    Joined: Dec 2001
    Posts: 2,004
    Likes: 209, Points: 73, Legacy Rep: 2917
    Location: On board Corroboree

    Eric Sponberg Senior Member

    Correct. RM stands for Righting Moment, which is equal and opposite to Heeling Moment.

    Eric
     
  6. TANSL
    Joined: Sep 2011
    Posts: 6,317
    Likes: 318, Points: 93, Legacy Rep: 300
    Location: Spain

    TANSL Senior Member

    Not always both moments are equal. Only when equilibrium is reached.
     
  7. Eric Sponberg
    Joined: Dec 2001
    Posts: 2,004
    Likes: 209, Points: 73, Legacy Rep: 2917
    Location: On board Corroboree

    Eric Sponberg Senior Member

    It is generally understood that the boat is at equilibrium and in steady state motion when we do most analyses, such as reflected in an ORR certificate or doing investigations such as Rader is doing.

    Eric
     
  8. TANSL
    Joined: Sep 2011
    Posts: 6,317
    Likes: 318, Points: 93, Legacy Rep: 300
    Location: Spain

    TANSL Senior Member

    Eric. I understand what you mean but when Rader calculates RM values ​​at various angles of heel, he is not working at the equilibrium point. I suppose that, among other things, what he wants is to find that equilibrium point.
    If I'm wrong, then I apologize: I misunderstood your words.
     
  9. rader
    Joined: Dec 2012
    Posts: 20
    Likes: 0, Points: 0, Legacy Rep: 10
    Location: Chicago

    rader Junior Member

    From comments in this forum it seems that cosine squared of the heel angle is a way to account for a reduction of both the sail area and the heeling arm. But, does not the computation of wind heeling arm already take the angle of heel effect on effective sail area into account since the sail area is one factor in the calculation of wind heeling moment?
     
  10. Eric Sponberg
    Joined: Dec 2001
    Posts: 2,004
    Likes: 209, Points: 73, Legacy Rep: 2917
    Location: On board Corroboree

    Eric Sponberg Senior Member

    You use the equations with the full sail area and the upright distance for heeling arm (as if the boat is not heeled) because those are easiest to measure in design. The cosine factors reduce both sail area and heeling arm according to angle of heel.

    Eric
     
  11. Convenor 12217
    Joined: Jan 2013
    Posts: 9
    Likes: 1, Points: 0, Legacy Rep: 13
    Location: Southampton, UK

    Convenor 12217 Convenor 12217

    TANSL (Post #2) is right. The power of 1.3 applied to the cosine results from wind tunnel testing done by the Wolfson research unit in Southampton, UK.

    As the boat heels the resultant sail sideforce has a downward component as well as lateral, so one cannot just cosine the area and cosine the lever as doing so ignores this. The 1.3 index is empirical and takes account of these effects.
     
  12. rader
    Joined: Dec 2012
    Posts: 20
    Likes: 0, Points: 0, Legacy Rep: 10
    Location: Chicago

    rader Junior Member

    Heeling moment - units

    What are the units for heeling moment (depicted by Tansl in the diagram) in metric measurement units? How can I convert these to English units, foot-pounds?

    Thanks.
     
  13. rader
    Joined: Dec 2012
    Posts: 20
    Likes: 0, Points: 0, Legacy Rep: 10
    Location: Chicago

    rader Junior Member

    What are the measurement units of wind heeling moment in metric units, as depicted by TANSL in the diagram above. What is the conversion to English units, foot-pounds?

    Thanks.
     

  14. Convenor 12217
    Joined: Jan 2013
    Posts: 9
    Likes: 1, Points: 0, Legacy Rep: 13
    Location: Southampton, UK

    Convenor 12217 Convenor 12217

    Both wind heeling and righting moments in ISO 12217 are calculated in SI Units: Newton-metres or kiloNewton-metres, depending on the units used in the definitions associated with the formula in question.

    According to http://www.convertunits.com/from/newton meters/to/foot pounds, the answer is 1 Newton-metre = 0.73756 foot-pounds (force).
     
Loading...
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.