Subject: RE: [harryproa] Re: Schooner v. Unarig |

From: "Jim Baltaxe" <jim.baltaxe@vuw.ac.nz> |

Date: 11/30/2006, 4:03 PM |

To: |

Reply-to: harryproa@yahoogroups.com.au |

Jim Baltaxe

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From:harryproa@yahoogroups.com.au [mailto:harryproa@ yahoogroups. com.au] On Behalf OfaudeojudeSent:Friday, 1 December 2006 1:31 a.m.To:harryproa@yahoogroups.com.au Subject:[harryproa] Re: Schooner v. UnarigI would think that canting the masts to leward would be fraught with

problems. I can just see the boat hitting bridge pilings, and other

sailboats as it goes by. How far out to the side of the boat would the

top of your mast end up being? You now have effectively increased the

beam of the boat at the masthead by some ammount.

--- In harryproa@yahoogroups.com.au , "Rob Denney" <proa@...> wrote:

>

> G'day,

>

> You are right, the angle is more relevant than the amount. Same

weight is closer to the truth, but still low, I think. We will be

doing some numbers on a schooner rigged 22m Visionarry which we are

starting early next year. I will let you know the results.

>

> Designing a mast to break before the boat capsizes is fraught. It

implies no safety factors and some hard decisions about waves and

payloads. Then you are sailing along with full sail up, a big gust

hits and you fly a hull. Ease one sail and the other mast breaks. I

would prefer to cant both masts to leeward 10 degrees, keep the weight

low in the windward hull and have a chance of self righting from 80

degrees, or more realistically, not be able to capsize this far.

>

>

> regards,

>

> Rob

>

>

> Oops.

>

> There is an error in my calculation.

>

> The weight of each mast is not 1/((2^.5) * (2^.5)) as shown below. It

> should be 1/((2^.25) * (2^.25)). This is because the luff is being

> reduced by 2^.5, which means that the reduction in weight is square

> root (2^.5) = 2^.25.

>

> So with the schooner sail the weight of the luff portion of the

> mast is now 1 / ((2^.25) * (2^.25)) = 1/2^.5 = .707. And for two

> masts it would be 2*.707 = 1.415.

>

> However the force on each mast is still 1/2 of the force on a unarig,

> so we can still reduce (I think) the diameter of each mast by 1/2^.5.

> This gives the total weight as 2/(2^.5)(2^.5) = 1.

>

> Hopefully this is a bit closer to the truth.

>

> Also, I was looking at the same angle of deflection - not the distance

> of deflection, on the assumption that what was relevant was the angle

> at the top of the mast. The link quoted gives that the angle is

> proportional to the square of length and radius, whereas distance of

> deflection is proportional to the cube. Should we be concerned here

> with angle or distance of deflection?

>

> One other point, on a cruising boat after a drama with 1 sail up and

> one down, do we really want to have 1 mast pointing up or 2 masts

> pointing down? If the former, perhaps the breaking strenght of one

> mast should be less than the weight of the boat.

>

> Best regards

> Herb

>

> --- In harryproa@yahoogroups.com.au , "Rob Denney" <proa@> wrote:

> >

> > G'day,

> >

> >

> >

> >

> > Rob I don't mind the schooner rig, in fact on a big boat where a

> mainsheet winch is required, and if the fore boom of an Easy rig is

> too high to easily reach, they have a lot going for them. Horses for

> courses.

> >

> > Herb The angle of deflection of a circular thin wall beam is

> proportional

> > to the square of the length

> > ( http://en.wikipedia.org/wiki/ )Deflection

> > and the square of the radius

> > ( http://en.wikipedia.org/wiki/ ).List_of_moments_ of_inertia

> >

> > R

> >

> > The formula for cantilever deflection is

> >

> > (Load*length cubed)/8*E*I Therefore it is a cube function of

> length, not a square. Halve the length, one eight the deflection.

> >

> > E is the material properties, I is the 2nd moment of area about

> the neutral axis. I= pi*radius cubed *wall thickness. The radius is

> also a cube function, so half the radius, 8 times the deflection.

> >

> >

> > H

> >

> > However, each mast now has only 1/2 as much sail area, so the

force on

> > each mast is 1/2 of what it would be for the single mast.

> >

> >

> >

> > R

> >

> > Yes and no. Each mast has to be strong enough to capsize the

> boat, as it is possible that only one sail would be fully powered up

> in a capsize scenario. This does not make each mast as heavy as a

> single one as to be stiff enough they are already stronger than

> required. However, depending on bury and other variables, it does

> make each mast much more than half the weight of a single one. On a

> harry, the bury will be very similar, although although a higher

> percentage of the overall length, which reduces the sheer loads

> somewhat.

> >

> >

> > H

> >

> > So I am confused. Is a schooner rig really about 2/3 the weight and

> > cost of a unarig (ignoring sails), or have I misunderstood

something?

> >

> > R

> >

> > I would say 3/2 is closer to reality for the weight and cost.

> >

> > regards,

> >

> > Rob

> >

> >

> >

> > Best regards

> > Herb

> >

> >

> >

> >

> >

> >

> >

> --------------------- --------- --------- --------- --------- -

> >

> >

> > Internal Virus Database is out-of-date.

> > Checked by AVG Free Edition.

> > Version: 7.1.394 / Virus Database: 268.14.6/536 - Release Date:

> 11/16/2006

> >

>

>

>

>

>

>

>

--------------------- --------- --------- --------- --------- -

>

>

> Internal Virus Database is out-of-date.

> Checked by AVG Free Edition.

> Version: 7.1.394 / Virus Database: 268.14.6/536 - Release Date:

11/16/2006

>

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