Date: 5/28/2009, 7:13 am
Hi Jeff,
Thanks a lot for that detailed explanation... it makes perfect sense to me.
Regards,
Phil.
: No problem. Just make a donation to my fund.
: First, use the cross curve function in Freeship with a 250 lb displacement at
: 15 degrees. You will get a number something like .247. That is the short
: leg of a triangle and you have to calculate the different legs of the
: triangle. I use this site.
: http://www.carbidedepot.com/formulas-trigright.asp
: You know Angle A (15 degerees) and side A from Freeship. What you need to
: know is side B. In my case it is .9504. That is the height of the
: Metacenter above the waterline.
: Determine your waterline at 250 lbs. In my case it is .370
: The subtract the water line height from the Metacenter height.
: .950 - .370 =.580 That is the metacenter above the waterline in decimal feet.
: So convert that to inches. .580 x 12 = 6.966" That is Metacenter above
: the waterline in inches. That was the hard part to figure out. Getting to
: that number. Took me two days to figure out what everything was. Now that
: you have that the formula is simple.
: (Metacenter/7.5) x 100 = stabilty factor.
: (6.966/7.5)*100_92.88 stability factor.
: The definition was found on the Bear Boat web site.
: There is thread over on BoatDesign.net where I finally grasped this.
:
: http://www.boatdesign.net/forums/stability/freship-stability-again-27396.html
: Yea, I know this is confusing. Email me if you get lost and we work through
: one.
Messages In This Thread
- Other: "Stabilty Factor' conquered (finally)
Kudzu -- 5/23/2009, 12:32 pm- Re: Other: "Stabilty Factor' conquered (finally)
Phil Nelson -- 5/26/2009, 8:21 pm- Re: Other: "Stabilty Factor' conquered (finally) *Pic*
Kudzu -- 5/26/2009, 10:45 pm- Re: Other: "Stabilty Factor' conquered (finally)
Phil Nelson -- 5/28/2009, 7:13 am
- Re: Other: "Stabilty Factor' conquered (finally)
- Re: Other: "Stabilty Factor' conquered (finally) *Pic*
- Re: Other: "Stabilty Factor' conquered (finally)