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Old 07-13-2015, 04:02 PM   #1
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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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Old 07-13-2015, 04:06 PM   #2
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Which way is the wind blowing and how hard?
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Old 07-13-2015, 04:19 PM   #3
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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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Old 07-13-2015, 04:20 PM   #4
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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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Old 07-13-2015, 04:42 PM   #5
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Rocna or Delta? All-chain rode? Horizontal or vertical windlass?

No.

Sorry 8^)
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Old 07-13-2015, 04:46 PM   #6
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Sorry, garbled message. See the next...
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Old 07-13-2015, 04:52 PM   #7
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Try this:
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Old 07-13-2015, 05:03 PM   #8
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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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Old 07-13-2015, 05:09 PM   #9
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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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Old 07-13-2015, 05:15 PM   #10
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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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Old 07-13-2015, 05:57 PM   #11
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The angle of the dangle is equal to the mass of etc. Sorry. Second cocktail.
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Old 07-13-2015, 06:03 PM   #12
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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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Old 07-13-2015, 06:25 PM   #13
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Usually best to hang from a single anchor and thus minimize/compromise wind versus current forces.
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Old 07-13-2015, 06:26 PM   #14
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The angle of the dangle is equal to the mass of etc. Sorry. Second cocktail.
I'm more interested in your opinion after the fourth.
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Old 07-13-2015, 06:51 PM   #15
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If the current increases to 2 knts

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Originally Posted by obthomas View Post
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.
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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Old 07-13-2015, 06:54 PM   #16
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Read 4th of July Mandeville

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I'm more interested in your opinion after the fourth.
Under gasp anchoring read 4rth of July mandeville.
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Old 07-13-2015, 06:55 PM   #17
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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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Old 07-13-2015, 06:57 PM   #18
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Quote:
Originally Posted by Scary View Post
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.
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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Old 07-13-2015, 07:13 PM   #19
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...Several posters tried to change it into something they knew how to answer....
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Old 07-13-2015, 07:16 PM   #20
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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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