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.

Which way is the wind blowing and how hard?

That will be pretty hard to calc, but from experience, a boat sideways to current requires A LOT of force to hold steady.

no wind for this discussion

Let's leave the wind out of it for now. But major player in the real world.

Rocna or Delta? All-chain rode? Horizontal or vertical windlass?

No.

Sorry 8^)

Sorry, garbled message. See the next...

Try this:

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.

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.

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.

The angle of the dangle is equal to the mass of etc. Sorry. Second cocktail.

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.

Usually best to hang from a single anchor and thus minimize/compromise wind versus current forces.

dwhatty said:

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.

If the current increases to 2 knts

obthomas said:

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.

Click to expand...

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?

Read 4th of July Mandeville

markpierce said:

I'm more interested in your opinion after the fourth.

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Under gasp anchoring read 4rth of July mandeville.

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.

Scary said:

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.

Click to expand...

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.

Paul Swanson said:

...Several posters tried to change it into something they knew how to answer....

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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?

When you think about the loads you can generate

Hanging off a single anchor seems simple enough, When you very from the norm it appears things can ramp up very quickly. Say having a stern line to shore in a tidal bore with a bow anchor into the wind but tidal flow from the beam. I have a friend who in the early 60's father wanted to take his family down the Sacramento river from Red Bluff to the bay. He was a San Diego fireman who wanted run most of the major rivers in the US. He ended up running most including the great loop with a 16' outboard. This trip didn't end up as planned. The Coast Guard recommended rafting as the safest way to do the trip. My friend father decided to build a large raft of telephone poles banded together Huckfin style with a large turning oar at the stern. Fitted out with a redwood picnic table, green umbrella tent, coleman coolers, white gas lanterns and stove off they went down the river. The trip went well for several hours with the exception of the steering oar which just spun the raft in circles. About 1 pm the raft approached a large snag in the river. No problem the raft will just glance of the snag. This was not to be, the raft hit the snag head on, the stern of the log raft slumped down in the current and that raft flipped end for end like a paddle wheel all 60' or 70' All was lost including my friends transistor radio. Fortunately no one was hurt. That has always been a lesson to me about the force of water.

Smart Man

markpierce said:

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?

Click to expand...

plow, came with the boat, no recommendations from me on type. 3/8 all chain rode helps big time.

Scary said:

plow, came with the boat, no recommendations from me on type. 3/8 all chain rode helps big time.

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I'm also a 3/8 kind of guy.

markpierce said:

I'm also a 3/8 kind of guy.

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Me 3!

Can't answer this on a phone at an airport bar...

...but I actually did try to calculate this once. If anyone is familiar with the "Powell River Hulks", I re-moored them ~13 years ago.

I was relatively close. I hired a Coastal Engineer with a Masters in Engineering Physics model and do the real math.

IIRC, the most significant load wasn't current or wind, but wind generated waves. Modeled several different significant wave heights.

I'll dig out the file when I get home.

A chain rigged tight across 2 posts (anchors) with 1000lbs (boat) pulling from the center exerts 2000lbs pull on each post not 500lbs as most would assume.

If I wanted to know the force exerted on my rode, I'd connect a digital hanging scale to the rode and record the force. Any calculated values would be hypothetical and suspect.

I would agree with 1000# on each 'post'.

dimer2 said:

I would agree with 1000# on each 'post'.

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I think you will find 2000# is the correct number. Actually, if a chain could be placed straight between 2 posts, the force would be much much higher than 2000#, 2000# is the figure for 15 degree sag or 165 degree included angle. That is why anchoring in that manner is doomed to failure when current or wind picks up.

Dang! I had a 50/50 chance....or did I get that wrong also?