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

From: "Herb Desson" <squirebug@yahoo.com> |

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^.

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@yahoogrou

>

> 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

> and the square of the radius

> ( http://en.wikipedia

>

> 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

>

Change settings via the Web (Yahoo! ID required)

Change settings via email: Switch delivery to Daily Digest | Switch format to Traditional

Visit Your Group | Yahoo! Groups Terms of Use | Unsubscribe

.

__,_._,___