want some to help me in that question

Discussion in 'Stability' started by abdelrhman shah, Dec 15, 2014.

  1. abdelrhman shah
    Joined: Dec 2014
    Posts: 9
    Likes: 0, Points: 0, Legacy Rep: 10
    Location: egypt

    abdelrhman shah Junior Member

    I want know. Why is the Initial metacentric height (GM).is the point of intersection when a tangent drawn at the origin and the purpendicular at 57.3°

    why especially 57.3

    and why the tangent drawn ....
    I need explaination please ....
     

    Attached Files:

  2. Canracer
    Joined: Aug 2009
    Posts: 621
    Likes: 9, Points: 18, Legacy Rep: 47
    Location: Florida

    Canracer Senior Member

    When you say tangent, do you really mean a straight line? When I hear the word tangent, I think of a 90 deg angle.
     
  3. TANSL
    Joined: Sep 2011
    Posts: 5,861
    Likes: 218, Points: 73, Legacy Rep: 300
    Location: Spain

    TANSL Senior Member

    Refers to a straight line that is tangent to the curve of GZ at the origin.
    Canracer, I think you have to learn as much or more than he.
     
  4. Ad Hoc
    Joined: Oct 2008
    Posts: 6,208
    Likes: 416, Points: 83, Legacy Rep: 2488
    Location: Japan

    Ad Hoc Naval Architect

    The reason is because the 1st assumption is that the GZ curve is drawn for small angles of inclination. Thus the GZ = GM.Sin(theta). Where theta is small the angle is measured in radian.s Therefore 1 rad = 57.3 degrees.

    And since this is true at small angles only, the line is drawn tangentially from/to the origin at 57.3 degree. Because at larger angles of inclination this assumption is no longer valid.

    So that line passes through the origin, and the GZ curve for a few degrees is considered a straight line (since in rads). How straight of course depends upon the shape of the hull under investigation.
     
  5. abdelrhman shah
    Joined: Dec 2014
    Posts: 9
    Likes: 0, Points: 0, Legacy Rep: 10
    Location: egypt

    abdelrhman shah Junior Member

    Thanks alot.....but I still feeling confused. ..... or iam so stupid to under stand it
     
  6. Canracer
    Joined: Aug 2009
    Posts: 621
    Likes: 9, Points: 18, Legacy Rep: 47
    Location: Florida

    Canracer Senior Member

    I don't see that line to be drawn as a tangent to the curve. A tangent would be 90 degrees to the radius, right?
     
  7. TANSL
    Joined: Sep 2011
    Posts: 5,861
    Likes: 218, Points: 73, Legacy Rep: 300
    Location: Spain

    TANSL Senior Member

    "90 degrees to the radius" drawn from the center of curvature of the curve. Do you know where the center of curvature of the curve is?. I do not.
    The process is reverse, the straight line is drawn from the origin to the point of coordinates (1 radian, initial GMt) and GZ value curve must be tangent to this line.
     
  8. cmckesson
    Joined: Jun 2008
    Posts: 161
    Likes: 7, Points: 18, Legacy Rep: 55
    Location: Vancouver BC

    cmckesson Naval Architect

    Abdel

    Try this:

    What is the slope of the GZ curve? It is d GZ/ d (theta)

    Calculate the value of that derivative when GZ = GM sin(theta)

    At theta=0, what is the value? In other words, at theta=0, what is the slope of the GZ curve?
     
  9. abdelrhman shah
    Joined: Dec 2014
    Posts: 9
    Likes: 0, Points: 0, Legacy Rep: 10
    Location: egypt

    abdelrhman shah Junior Member

    I know that 57.3 = 1 radian....why we draw line from this point ....why especially 57.3
     
  10. Eric Sponberg
    Joined: Dec 2001
    Posts: 2,002
    Likes: 205, Points: 73, Legacy Rep: 2917
    Location: On board Corroboree

    Eric Sponberg Senior Member

    Abdelrhman Shah, Ad Hoc gave the explanation above. At small angles of heel, the righting arm, GZ = GMSin(theta). And also at small angles, Sin(theta) = theta when theta is expressed in radians. Therefore, GZ = GM x theta at small angles.

    So on your diagram in your first post, at small angles of heel, you can calculate the angle between the blue line or the red line, which we'll call alpha, with the equation Tan(alpha) = GZ/theta. You show the horizontal axis in degrees of heel, but in the equation and at small angles, theta is actually in radians. In the paragraph above, we see that GZ = GM x theta, and if we divide both sides of the equation by theta, we get GZ/theta = GM. Substituting GZ/theta for GM we get:

    Tan(alpha) = GM

    Another way to look at this is to realize that if theta equals zero, then the tangent equation becomes undefined. That is, Tan(alpha) = GZ/0 = infinity. But if theta = 1 (that is, if it equals one radian = 57.3 degrees) then we have Tan(alpha) = GZ/1 = GZ which is still equal to GM. That is why when you plot the tangent to the blue curve by drawing in the red line, and you are using degrees as the labels on the horizontal axis, you have to go across one radian = 57.3 degrees and the vertical erected at that point will cross the red line at the value of GM, which is really GZ/1.

    Does that make it any clearer?

    Eric
     

  11. abdelrhman shah
    Joined: Dec 2014
    Posts: 9
    Likes: 0, Points: 0, Legacy Rep: 10
    Location: egypt

    abdelrhman shah Junior Member

    Yeah I finally understood.....thanks very much
     
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.