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Old 10-13-2009, 09:25 PM
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lewisboats lewisboats is offline
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Cross Curves in Freeship

Ok...so could someone explain how to interpret cross curves? I understand that the left side is the distance from the keel to the center of buoyancy and the bottom is the displacement. What would an almost flat line for a curve indicate? What do the higher heel angle lines indicate when they curve the opposite direction of the lower ones (humped up instead of sloped down or flat-ish) and why when I put in a single displacement does the graph change to straight lines angling up and to the right...stopping at the displacement set (or as close as the increments will allow).

Steve

PS: Yes I read the manual and from the page to my brain it becomes confusing because KN sin(ø) is a meaningless statement to this non math person. From another thread I found out that this MIGHT be the Transverse Center of Buoyancy at angles of heel but I may not be reading that correctly at all so.....HELP!

Quote:
15.4 Cross curves.
Stability calculations are provided in the form of cross curves. For a number of heeling angles and
displacements KN sin(ø) is calculated and presented in a graph and table. If only one displacement
is provided the KN sin(ø) curve is displayed. If multiple displacements are provided the graph shows
the standard cross curves. The calculated data can be printed or saved to a textfile.
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Old 10-13-2009, 11:16 PM
Ad Hoc Ad Hoc is offline
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The "KN" is on the left hand side of the graph. The Displacement, is on the base of the graph.
You then have a series of curves all plotted at various 'angles'.

Select the displacement of the vessel, then draw a straight perpendicular from the displacement. Where this vertical line crosses the curves of angles, note the angle, and then note the value of KN.

First thought, you need to be sure that KN, is from the baseline/keel of the boat to the metacentre. (not all produced like this).

Then once you ahve the KN data you do the following:

GZ = KN - KG x sin (angle)
where
KN = cross curve ordinate
KG = centre of gravity above datum (Same datum as KN, corrected for free surface effects)
angle = heel angle, from cross curves

So, you draw up a table of the items you have
Angle, going down, and going across you have columns of
Angle, KN, sin (angle), KG x sin(angle), GZ.
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Old 11-25-2016, 12:12 AM
terrnz terrnz is offline
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Freeship cross curves and righting moments.

Excuse the resurrection of this old thread.

I have been working with FreeShip and in particular the curves of stability.

My project is a sailing yacht.

Is the formula GZ=KN - KG x sin (angle) valid if KG is negative?

my datum is the lowest point of the canoe body (Tc) Then I have a deep fin keel with a lead bulb and a ballast ratio of 55% so the centre of gravity is actually around 0.5m below the bottom of the hull.

I am also working with just one displacement (light ship mode)

also up to what angles is formula valid.


My purpose is to get a meaningful figure for the 30 deg heeling moment to use in mast inertia calculations.
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Old 11-25-2016, 12:16 AM
Ad Hoc Ad Hoc is offline
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Quote:
Originally Posted by terrnz View Post
Is the formula GZ=KN - KG x sin (angle) valid if KG is negative?


Check your datums are consistent!
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Old 11-25-2016, 04:21 PM
jehardiman jehardiman is offline
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Quote:
Originally Posted by terrnz View Post
My project is a sailing yacht.
KN curves really aren't useful for a sailing yacht. They are designed to quickly allow the calculation of GZ for vessels that greatly change their KG and draft, like cargo ships. Better to calculate moment to heeling moment directly, especially due to the lack of wall sided-ness and buoyancy distribution on most yachts.
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Old 11-25-2016, 05:07 PM
TANSL TANSL is offline
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The formula GZ = KN - KG x sin (angle) is valid for all heel angles.
You need to place your baseline through the bottom of the keel bulb and take all measurements with respect to that line. This way your KG will be positive.
In order to calculate the mast, in my opinion, you should use the ship's displacement at full load, which will normally be the worst situation that can occur.
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Old 11-25-2016, 08:15 PM
terrnz terrnz is offline
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1) thank you for your reply about the formula. That's very helpful
2) I am already reworking the baseline, I had been treating the canoe body and appendages separately.
3) for rig inertia calculations Previously I had been using the formulae from Eliasson and Larsson and calculated the righting moment by their method then multiplying the figure by max displ/light ship. I will revise my spreadsheet and input the new figures from the new baseline based on heavy ship RM30 which should be more accurate than the approximation.
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