Hurricane wind effect on rigging on the hard

[Quadrille in JB]

If you don't believe how much wind load the rigging represents, look at this video of a Leopard 39 "Sailing the Southern Ocean."
Pretty impressive speeds with bare poles and only 65 kts of wind. And they've got a drogue out.
Remember, the waves never look as big and impressive (or scary) on a video as they do in real life. I'd say they look about 50-60% smaller on the video.

https://www.youtube.com/watch?v=hQ-svmgOxqw

[a64pilot]

Find the answer to this, how much psi does 100 mph of wind generate.
Then get a swag of how many square inches there are on your mast, I predict the force to be far less than you suspect.
I suck at math,so I won’t try, but was surprised at the number I got years ago from an Engineer. It had to do with pressure of a fuel tank assuming a vent placed into 150 mph wind.

[Jim Cate]

Originally Posted by Seas the Day

I'm moving from Sydney to Vanuatu over the next year where they get a cyclone (NH Hurricane) every year or so. My preparation choices are 1) to leave the boat on a 'cyclone rated mooring' in my deep harbour or hard stand in town. If I go the hard stand way then I would remove the mast and standing rigging and strap everything down hopefully away from other boats that leave their masts on.

I assume you are talking about Pt Vila... the history has shown that local boats are often poorly secured and go careening about the harbour, making the storm mooring somewhat of a crapshoot ( with you being the target!) no matter how secure the mooring. Being on the hard removes that particular worry, but substitutes damage from flying debris and, as you note, falling masts from other boats.

Damned if you do, damned if you don't! A cf member, "No ties", lives in Vila and might be able to advise better than I (only an occasional visitor). You may have already contacted him, but if not...

Jim

[DotDun]

Originally Posted by Quadrille in JB

If you don't believe how much wind load the rigging represents, look at this video of a Leopard 39 "Sailing the Southern Ocean."
Pretty impressive speeds with bare poles and only 65 kts of wind. And they've got a drogue out.
Remember, the waves never look as big and impressive (or scary) on a video as they do in real life. I'd say they look about 50-60% smaller on the video.

The boat has a lot more windage than the rigging. The bare poles wasn't pushing that boat.

[Quadrille in JB]

The bare poles are sure helping. But all of the windage works against you in a storm.

[NevisDog]

Originally Posted by DotDun

The boat has a lot more windage than the rigging. The bare poles wasn't pushing that boat.

Try going side-on to the wind and see how much that bare pole is being pushed. Half the time the hull is completely blanketed by waves but still storming along. More obvious in a mono p'haps.

[a64pilot]

Originally Posted by NevisDog

Try going side-on to the wind and see how much that bare pole is being pushed. Half the time the hull is completely blanketed by waves but still storming along. More obvious in a mono p'haps.

You get the same in a powerboat, to be honest being beam on in bad weather has always scared me, it’s when I Have the throttle pushed hard on a powerboat, cause I want to get through that fast.
Course running down wind worries me too. I feel best with the pointy end into the wind and waves.

However I think 100 mph of wind is something like .17 PSI?
If you have a 50’ mast that is .5’ wide, that is 25 sq ft?
25 Sg ft is 3,600 sq in?
If 100 mph is actually .17psi then force on mast is like 600 lbs?

Now, I really do suck at equations, so please check me, cause that sounds like an impossibly low number, point is that it can be calculated.

On edit I found with Googles help of course that it seems that 100 mph is 26 lbs per sq ft. So 26 times 25 is 650 lbs of wind force on the mast.
Unlikely to take a mast down with that kind of force, however wrap a piece of roofing tin around it and that number can get really big I’m sure. Maybe debris is what is taking down masts on the hard?

[zippy]

Here is a simplified calculation of the force generated by the wind on an object.

F=A x P x Cd

F= force generated
A= area exposed to the wind
P= pressure generated in psi = .00256 x V^2
V= wind velocity in mph
Cd= drag coefficient with depends on the shape of the object

Assume a 60' by 1' mast area for an A= 60 ft^2
Assume wind speed V = 150mph

P = .00256 x 150^2
=57.6 psi

The drag coefficient for a tall cylinder Cd=1.2, Cd = 2 for a flat plate. I will use 1.4 for a rounded rectangular mast.

F = 60 x 57.6 x 1.4
=4,838 lbs of force on the mast

The force would be half that if we used the narrow profile of the mast using half the width.

Increasing the wind speed to 200mph almost doubles the pressure to 102 psi, making the force increase to 8568 lbs of force on the wide cross section of our mast.

[CARGO]

MY FRIENDS NEAR ST. KITTS DID WHAT THEIR INSURANCE COMPANY RECOMENDED OTHERWISE THEY WOULD NOT HAVE BEEN COVERED.
THEY PUT THEIR BOAT IN A PIT UP ON DRY LAND

[Jim Cate]

Originally Posted by CARGO

MY FRIENDS NEAR ST. KITTS DID WHAT THEIR INSURANCE COMPANY RECOMENDED OTHERWISE THEY WOULD NOT HAVE BEEN COVERED.
THEY PUT THEIR BOAT IN A PIT UP ON DRY LAND

So, did the boat survive?

Jim

[a64pilot]

Originally Posted by zippy

Here is a simplified calculation of the force generated by the wind on an object.

F=A x P x Cd

F= force generated
A= area exposed to the wind
P= pressure generated in psi = .00256 x V^2
V= wind velocity in mph
Cd= drag coefficient with depends on the shape of the object

Assume a 60' by 1' mast area for an A= 60 ft^2
Assume wind speed V = 150mph

P = .00256 x 150^2
=57.6 psi

The drag coefficient for a tall cylinder Cd=1.2, Cd = 2 for a flat plate. I will use 1.4 for a rounded rectangular mast.

F = 60 x 57.6 x 1.4
=4,838 lbs of force on the mast

The force would be half that if we used the narrow profile of the mast using half the width.

Increasing the wind speed to 200mph almost doubles the pressure to 102 psi, making the force increase to 8568 lbs of force on the wide cross section of our mast.

While I suck at math, I do believe there is a mistake somewhere.
15O mph is no where near 57 psi, if it were my approx 2 sq ft of torso area (288 sq in ) on a motorcycle would have have 16,000 lbs of force acting on it? Be hard to hold on
288 x 57 = 16,416.

Maybe .57 ?

[Gary Mc]

Living on the Great Lakes so not exactly Cane knowledgeable my instinct would have been to put the boat in a sand pit but more importantly take the mast down!

Were there no facilities to do this? I read about many boats being prepped and secured, but the pictures showed the mast still up and the boat subsequently fell over.

Was there not time or facilities to take the mast down?

[mitiempo]

Originally Posted by sailon46

......Especially with the deck stepped masts all the new boats are required to have, so they can break away without sinking the boat.
FYI
Ernie on the Mary Jane

No such requirement as far as I know. Boats are built with both deck and keel stepped masts.

[Jim Cate]

Originally Posted by Gary Mc

Living on the Great Lakes so not exactly Cane knowledgeable my instinct would have been to put the boat in a sand pit but more importantly take the mast down!

Were there no facilities to do this? I read about many boats being prepped and secured, but the pictures showed the mast still up and the boat subsequently fell over.

Was there not time or facilities to take the mast down?

Seems that taking halyards out to the sides to anchors buried in the ground would give great support to the mast on a "pitted" boat.

Lots of places that don't have mast cranes available, and there might well be a lot of folks needing that service just before a cyclone strikes if there WAS one available! But agreed that pitting the keel has worked pretty well in Vunda Point, Fiji, where they've done it for years. Safer than cradles, "cause ya can't fall out of a pit!

Jim

[NevisDog]

Originally Posted by a64pilot

15O mph is no where near 57 psi, ...

Yep, 57 psi seems way out for 100 mph wind.
Keep it simple: two websites give wind pressure of 0.1 psi at the bottom end of Beaufort 12 (over 64 kts, or exactly 34 m/s) so about ten times the force you first calculated, roughly speaking. Not sure that helps much though.

Edit: or maybe exactly what you first came up with!