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07-13-2015, 03:02 PM
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#1
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Guru
City: Walnut Grove Ca
Vessel Name: Cary'D Away
Vessel Model: Hatteras 48 LRC
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Posts: 887
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Anybody know how to figure this out
A boat is anchored across the flow of current, right angle to the current flow bow and stern. For this example the boat is square to the flow and not on an angle. How much water force is exerted on the anchors in 1 knt of current , and what would that be at 2 knts. In the case of my boat, it displaces 66,000lb and has about 220 sqft of exposed hull side.
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07-13-2015, 03:06 PM
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#2
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Guru
City: Sitka
Vessel Model: Transpacific Marine Eagle 32
Join Date: Dec 2014
Posts: 519
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Which way is the wind blowing and how hard?
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07-13-2015, 03:19 PM
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#3
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Technical Guru
City: Wilmington, NC
Vessel Name: Louisa
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Posts: 6,194
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That will be pretty hard to calc, but from experience, a boat sideways to current requires A LOT of force to hold steady.
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07-13-2015, 03:20 PM
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#4
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Guru
City: Walnut Grove Ca
Vessel Name: Cary'D Away
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Posts: 887
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no wind for this discussion
Let's leave the wind out of it for now. But major player in the real world.
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07-13-2015, 03:42 PM
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#5
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Guru
City: Pender Harbour, BC
Vessel Name: Gwaii Haanas
Vessel Model: Custom Aluminum 52
Join Date: Sep 2013
Posts: 3,791
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Rocna or Delta? All-chain rode? Horizontal or vertical windlass?
No.
Sorry 8^)
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Don't believe everything that you think.
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07-13-2015, 03:46 PM
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#6
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Guru
City: Pender Harbour, BC
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Sorry, garbled message. See the next...
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Don't believe everything that you think.
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07-13-2015, 03:52 PM
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#7
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Guru
City: Pender Harbour, BC
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Try this:
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Don't believe everything that you think.
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07-13-2015, 04:03 PM
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#8
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Guru
City: Seabrook Texas
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Or...... you could use velocity pressure which is calculated using the speed of the water in feet per second pushing against the side of the boat. There would need to be an adjustment for the dead rise that is not perpendicular to the water flow. The calculation would go something like this. 1/2(V^2)x 32ft/sec/sec x .433 One knot is about 1.69 ft/sec so the above says that there is about 2.78 pounds per square foot against the side of the boat if I done the math right.
O.B. Thomas, P.E.
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07-13-2015, 04:09 PM
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#9
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Guru
City: Pender Harbour, BC
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Sorry, but you're on your own to figure out the Reynolds Number. I don't know how to measure the parameters.
So once again, no.
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Don't believe everything that you think.
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07-13-2015, 04:15 PM
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#10
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Guru
City: Ft Pierce
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cant calculate without the angles of the rodes from the centerline.
but the greater the angle the less the pull on each anchor.
you can calculate the force of the water on the hull but without the angle of the anchors...that cant be calculated.
pick an angle or several then just apply trig to the force on the boat.
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07-13-2015, 04:57 PM
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#11
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Moderator Emeritus
City: Home Port: Buck's Harbor, Maine
Vessel Name: "Emily Anne"
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The angle of the dangle is equal to the mass of etc. Sorry. Second cocktail.
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David Hawkins
Deer Isle, Maine
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07-13-2015, 05:03 PM
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#12
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Guru
City: North Carolina for now
Join Date: Aug 2011
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Lacking wind, the boat cannot be perpendicular to the current. And it can take, depending on boat profile, quite a bit of wind to offset the effect of current, especially if you have a deep keel. In a wind against current situation, that's when things get fairly uncomfortable, and deploying some sort of kedge to keep the bow pointed into the resulting waves may be in order, as it does in some ocean coves where the inherent swell is counter to prevailing wind.
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George
"There's the Right Way, the Wrong Way, and what some guy says he's gotten away with"
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07-13-2015, 05:25 PM
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#13
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Master and Commander
City: Vallejo CA
Vessel Name: Carquinez Coot
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Posts: 12,559
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Usually best to hang from a single anchor and thus minimize/compromise wind versus current forces.
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Kar-KEEN-ez Koot
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07-13-2015, 05:26 PM
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#14
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Master and Commander
City: Vallejo CA
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Quote:
Originally Posted by dwhatty
The angle of the dangle is equal to the mass of etc. Sorry. Second cocktail.
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I'm more interested in your opinion after the fourth.
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Kar-KEEN-ez Koot
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07-13-2015, 05:51 PM
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#15
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Guru
City: Walnut Grove Ca
Vessel Name: Cary'D Away
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Join Date: Apr 2012
Posts: 887
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If the current increases to 2 knts
Quote:
Originally Posted by obthomas
Or...... you could use velocity pressure which is calculated using the speed of the water in feet per second pushing against the side of the boat. There would need to be an adjustment for the dead rise that is not perpendicular to the water flow. The calculation would go something like this. 1/2(V^2)x 32ft/sec/sec x .433 One knot is about 1.69 ft/sec so the above says that there is about 2.78 pounds per square foot against the side of the boat if I done the math right.
O.B. Thomas, P.E.
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Some of you jumped the gun, I'm getting around to wind and anchor angles. What I'm thinking about is what kinds of loads are possible to exert on an anchor while anchored cross current. The reason for anchoring in this manner would be do to prevailing winds and possible chop and ventilation advantages. This occurs often at the Mandeville cut fourth of July celebration. Usually there a number of anchor dragging and raft break ups in the wind and current. In fact I've bent an Anchor shank trying to anchor cross current here. So if we up the water speed to say 2 knts and Thomas is right in his math I believe the force cubes to 21.48 lb per square ft. Would that be right?
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07-13-2015, 05:54 PM
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#16
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Guru
City: Walnut Grove Ca
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Read 4th of July Mandeville
Quote:
Originally Posted by markpierce
I'm more interested in your opinion after the fourth.
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Under gasp anchoring read 4rth of July mandeville.
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07-13-2015, 05:55 PM
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#17
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Guru
City: Concrete Washington State
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Your anchor winch probably couldn't hold the rode's straight but if they could the tension on the lines would probably be so much that only a 500lb Navy anchor could hold it.
Give us a real world 1' of drift (or an angle of 10 degrees) and numbers could be generated.
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Eric
North Western Washington State USA
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07-13-2015, 05:57 PM
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#18
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Veteran Member
City: Ventura CA
Vessel Name: Proud Mary
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Join Date: Apr 2015
Posts: 92
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Quote:
Originally Posted by Scary
A boat is anchored across the flow of current, right angle to the current flow bow and stern. For this example the boat is square to the flow and not on an angle. How much water force is exerted on the anchors in 1 knt of current , and what would that be at 2 knts. In the case of my boat, it displaces 66,000lb and has about 220 sqft of exposed hull side.
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This is a well stated problem. Several posters tried to change it into something they knew how to answer. Engineer Thomas has the force on the boat right. For the 220 sq. ft area the force would be approximately 600 lbs. However the rode tension would be 600 lb. /sin(theta) where theta is the angle the rode makes with the centerline of the boat. If theta is zero, the force is infinite, forcing the anchors to drag until the force is low enough for the anchors to hold. This is, of course, almost impossible to calculate since it involves the anchor, bottom, scope and probably a bunch of other things. If the anchor can hold 1200 lbs, the angle would end up being 30 deg. 600 lb. /sin (30 deg) = 600/.5=1200 lbs.
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07-13-2015, 06:13 PM
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#19
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Senior Member
City: Sydney
Join Date: Apr 2012
Posts: 261
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Quote:
Originally Posted by Paul Swanson
...Several posters tried to change it into something they knew how to answer....
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John
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07-13-2015, 06:16 PM
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#20
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Master and Commander
City: Vallejo CA
Vessel Name: Carquinez Coot
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Join Date: Sep 2010
Posts: 12,559
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I have no desire to anchor fore-and-aft, cross-current to face 20-knot winds, let alone accompanying several hundred other boats attending a Fourth of July fireworks. That's why you wouldn't see us anywhere near there until the Fifth.
Single-anchored in the Delta on the Fifth:
Regardless, how did it all work out for you? Anchor held? What weight, what type?
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Kar-KEEN-ez Koot
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