[Alberg30] Alberg 30 center of gravity

[RABBIT649 at aol.com]

I guess the assumption of the method is that weight is already evenly 
distributed for her to float on her lines, ie if the waterline is the line on 
which she floats, weight has been distributed such as to make that happen, so 
that center of gravity = center of area.
Paul
#23 Ashwagh

In a message dated 4/22/02 1:01:27 PM Eastern Daylight Time, 
rjohns3 at optonline.net writes:


> Subj:Re: [Alberg30] Alberg 30 center of gravity
> Date:4/22/02 1:01:27 PM Eastern Daylight Time
> From:    rjohns3 at optonline.net (Bob Johns)
> Sender:    public-list-admin at alberg30.org
> Reply-to: <A HREF="mailto:public-list at alberg30.org">public-list at alberg30.org</A>
> To:    public-list at alberg30.org
> 
> 
> 
> 
> 
> On Monday, April 22, 2002, at 07:34  AM, chrishardy wrote:
> 
> > A very accurate way to find cg is to cut out the profile view of a 
> > picture of the a-30 enlarging the image may help. then fold the picture 
> > up like paper doll parallel to the waterline. then balance the folded 
> > paper on a pin mark it unfold it and there you have it. This info is 
> > right out of Chappells Yacht design that's how the pros used to do it 
> > before computers..  Chris
> 
> Chris,
> 
> I'm not sure what you mean by: "fold the picture up like paper doll 
> parallel to the waterline", but you are describing a method of locating a 
> "center of area", not a "center of gravity". If you cut out the profile of 
> the hull shape below the waterline and balance it on a pin you are finding 
> the "center of lateral area" (with the boat in an upright position).
> 
> You have to consider the locations of weights, not just area. Not sure if 
> I am saying this very clearly.
> 
> Bob Johns, Wind Call #397
> 


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