Subject: [harryproa] Re: Schooner v. Unarig From: "Herb Desson" Date: 11/29/2006, 6:16 AM
 To: harryproa@yahoogroups.com.au Reply-to: harryproa@yahoogroups.com.au

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
>

__._,_.___
Yahoo!7 Groups

Start a group

in 3 easy steps.

Connect with others.

. __,_._,___