metacentric height

[BennyWailer]

hi,
I'm using Freeship to model a canoe. I want to find the metacenter at 15 deg of heel at a load. The program outputs stability cross-curves as KN*sin(theta)(ft).

Is there a formula to I can use to find the metacenter.

Cheers

[Guillermo]

I recommend you the reading of: http://web.nps.navy.mil/~me/tsse/NavArchWeb/1/toc.htm
Cheers.

[BillyDoc]

That's a very good link, Guillermo, thanks!

BillyDoc

[RANCHI OTTO]

Very usefull link! Thank you G.

[BillyDoc]

And another one I just stumbled onto, a low-level Navy course with pdfs of the chapters. Click on "Contents.pdf" and then ch01.pdf, etc. This is very basic stuff well explained. Just what I need most of the time!

http://www.usna.edu/naoe/courses/en200/

[knowme]

If I am right at 15 deg heel your metacenter M will coincide with your N. Hence you can find your metacenter from cross curves of stability. If I am wrong somebody correct me please.

[Guillermo]

Originally Posted by knowme

If I am right at 15 deg heel your metacenter M will coincide with your N.

Nope. Have a look at the attached figure.
Just a clue: KN = KM * sin θ

[EAP]

Mr G,
Something Dosent Jive Between The Diagram You Posted And The Kn*sin(theta) Given In The Cross Curves Cals Of Freeship. Actually The Diagram Makes Sense To Me, What I Dont Understand Is What The Vertical Axis Of The Freeship Diagram Is Telling Us. It Seems That The Freeship Diagram Should Either Say Kn -or- Km*sin(theta) As Giving The Righing Lever.
Any Thoughts???

[Guillermo]

Could you post the Freeship diagram, please? I do not use that program. Maybe it's a typo or a diferent meaning of notation there. Have you gone through the manual?
Righting lever GZ is equal to KN-KG sin(theta) in the notation I'm used to.
Cheers.