Subject: Re: [harryproa] NIS cat boats |

From: "Rob Denney" <proa@iinet.net.au> |

Date: 12/3/2006, 7:57 AM |

To: |

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

G'day,

They have comparatively low righting moment, so
yours will need beefing up, but the idea is sound.

The strongest, lightest and easiest way is to
glass the tabernacle onto the deck. Don't need to build the extra
tube, don't need to cut the hole in the side of the boat. Just ensure the
load is spread over a reasonable area.

regards,

Rob

----- Original Message -----From:Doug HainesSent:Sunday, December 03, 2006 6:47 PMSubject:[harryproa] NIS cat boats

Looked at theses boats in Australian Amatuer Boat Builder, and realized just continue the tube up out of the hull and cut a 'side' out and put in a hinge .Would that mean you would not even build the lower part in the hull of the mast. Effectively it becomes deck stepped, but with the tabernacle/tube extending through the hull to the bottom for maximum strength.Dougwrote:Doug Haines <doha720@yahoo.co.uk> GDay,Can someone in the know give the data used in calculating the amount of carbon used?I remember that you get the strength of carbon as a constant, then you work out how much you are using and in what geometry, then you can give a value for the amount of working load.We (at uni), used to do the sums on steel, wood etc, but almost never on glass or carbon structures. When I tried to do sums I had a question as to how thick is the layer supposed to be?Say you put a 5mm Kiri timber strip down, then you put a layer of 450g/sqm uni glass over that. You work out the strength of this 'beam' by using the thicknesses, 5mm kiri and how thick is the glass/resin?If you look side on it looks like about 1mm or a bit less for each layer used. This however may be more if you use more resin. Is there a standard value for a single layer of so many grams per square metre? How do you calculate a composite cloth/resin material's strength?dOUGwrote:Herb Desson <squirebug@yahoo.com> Hi,

On consideration I think you are right about not designing masts to

break. It would take a lot of luck to work right - and luck is

usually in short supply in problem situations.

The idea of canting the masts to leeward is an interesting one. If it

doesn't cause any problems with the mast swinging to leeward because

of the tilt it might be just the solution.

With regard to the weight I have clearly been confused the last few

days. I am now back to considering my first calculation (which

implied a 29% reduction in weight for a schooner) to be more correct.

The reason is that for the angle of deflection to be the same, the

ratio L^2/R^2 must be constant. Which implies that if L is reduced by

2^.5 then R must be also.

I am not quite sure what to make of the smaller sail area, but I think

it is clear that in any given weather there will be less force on each

mast for the schooner than for the single mast of the sloop.

I look forward to seing the results of your calculations. I know FEA

costs money, but would it be possible to include an analysis of

exactly the same sail shape to get comparability? I am not sure how

comparable a jibless schooner is to a balestron sloop from a weight

point of view. My first thought is that it wouldn't make much

difference, but clearly my first thoughts are not very reliable in

these matters.

Best regards

Herb

--- 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

> >

> >

> >

> >

> >

> >

> >

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>

>

>

>

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