Thoughts on anchor rode wind loading

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ksanders

ksanders

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Vessel Name
DOS PECES
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BAYLINER 4788
I was thinking the other day, that with my all chain rode, just how much wind would it take to even lift the rode off the sea bed so that it is a relativly straight line to the boat.

I know this is a difficult to calculate subject, but I have suspected based on the mud clinging to my anchor rode that it just sits on the sea bed most of the time.

So I thought, If I was anchor'd in 100' of water, and deployed a 4:1 scope of 400' of chain, just how much force would need to be exerted on my boat to lift that chain from the sea bed.

Math gurus, please help me out here, but my basic thought is if the chain weighs 1.17 lbs per foot, and I have a water depth of 100', and have 400' or chain deployed then.

Just sitting there, no wind there is 300' of chain on the sea bed to lift so to me that requires a force of 300' * 1.17 lbs per foot or approx 350 pounds.

Then I tried to calculate wind force. That is impossible for me so I tried to guesstimate the equlivent square footage of the frontal area of my boat. My guess is 150-200 square feet.

So the handy dandy wind loading calculator I found online shows a 30 mph wind producing approx 370 pounds of force at 30 knots on a 200 sqft frontal area.

So... It looks to me that in a 30 knot wind my anchor chain might be just getting lifted off of the sea bed.

Now, the math geniuses, please fix my simplistic calculations and tell me how much wind is REALLY required to lift my anchor rode off the sea bed. :blush:
 
Kevin:
I am thinking your math is correct enough, but it really does not matter that much. I think the issue is not so much how much wind to lift the chain, but rather the angle of the chain to the anchor if and when it is lifted. If it is a shallow angle, then the angle should dig in further and hold better. Your example used a 4:1 scope in100ft, presumably because you often anchor in deep water. We usually anchored in much shallower water and when possible set a 7:1 scope, specifically to put more chain on the bottom and to minimize the scope angle.
 
I am hesitant to use a lot of numbers thrown around with anchoring.

My Fortress anchor guide says for a 40 foot boat, in 15 knots of wind there is 300 pounds of force on the anchor. Yet in 15 knots of wind, I can still push or pull my boat around in the slip.

Am I applying 300 pounds of force to overcome the wind? I wish I were that strong.

Now, is that 300 pounds just wind? Or does it include the tiny wave action that comes with it?

Even their scope calculations barely agree with the Compass Marine tests.

In true storm conditions, I contend wave action and yawing does more to pull an anchor than just pure wind force.

So basically, other than Steve anchoring videos...I classify most anchoring facts right up there with most boating BS that many have been fed for centuries.....

And somewhat passed along in TF by some.
 
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Don't forget the buoyancy of the chain itself; i.e. it doesn't have the same weight in water that it does on land.

And the force calculation is a vector in that it has horizontal and vertical components. So, when the wind load is applied to the chain, it tries to lift the chain vertically and pull it horizontally. Some of the horizontal load would be taken by the anchor (I suppose).

That's all I've got without pulling out a textbook. Engineering school was many moons ago.

Come to think of it, this would be a great problem for an engineering statics or dynamics class.
 
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I'm w Chrisjs,
What would yo do w this information if you obtained it Kevin?
There are people that just like to calculate. Are you one of them?
 
Nomad Willy said:
I'm w Chrisjs,
What would yo do w this information if you obtained it Kevin?
There are people that just like to calculate. Are you one of them?
Click to expand...

Thats too funny, and too close to the truth. :blush:

I was just curious. Simple curiosity driven by mud on my anchor rode.

Ihe information/answer won't be used for anything. :)
 
ABfish said:
Don't forget the buoyancy of the chain itself; i.e. it doesn't have the same weight in water that it does on land.

And the force calculation is a vector in that it has horizontal and vertical components. So, when the wind load is applied to the chain, it tries to lift the chain vertically and pull it horizontally. Some of the horizontal load would be taken by the anchor (I suppose).

That's all I've got without pulling out a textbook. Engineering school was many moons ago.

Come to think of it, this would be a great problem for an engineering statics or dynamics class.
Click to expand...

The answer would involve way too much calculus for me. I forgot that decades ago. ABfish is right, the weight of the chain is much less in water than in air (the mass is constant however). The total energy required to lift a 300' pile of chain off the bottom is different however than the force required to keep 300' of chain off the bottom connecting a boat and and anchor. Again as ABfish noted you would need to calculate the vectors involved. Add to that you would need to estimate the flat plate area of your boat, and I would have no way to guess that.

There are some very smart engineering types here for whom this type of thing would be child's play to figure out. To be honest, I do share your curiosity.

What would you do with the answer? Well, for me, knowing how much force is applied to the end of the road under what wind conditions would be helpful to determine if my 1/2" three strand bridle lines are sized large enough.
 
ksanders said:
So I thought, If I was anchor'd in 100' of water, and deployed a 4:1 scope of 400' of chain, just how much force would need to be exerted on my boat to lift that chain from the sea bed.

Math gurus, please help me out here, but my basic thought is if the chain weighs 1.17 lbs per foot, and I have a water depth of 100', and have 400' or chain deployed then.
Click to expand...

I'll take a cut at it:

400 ft in 100 ft of water will come taut at at about 15 degrees. You'll need to lift 400 ft times 1 lb/ft (guessimate for buoyancy), which is 400 lbs. Divide that by the arcsin(15 degrees) and you get about 1500 lbs of force needed.

That'd be about 35 knots of wind on a 40 ft boat.

(Ok, now where have I screwed this up? Been long enough I've probably got something backwards.)
 
Sounds good to me danderer but I question the legitimacy of calculating the drag of wind on a boat. If one were in a harbor fairly open to the wind and there was a fairly strong measurable wind one could cut the boat loose and then laying in the wind measure the pull or tension on the line holding the boat.

Calculating every little thing exposed to the wind would be nearly impossible. Even that would be fairly inaccurate as wind 10' up is considerably stronger than only 2' up. And many trawlwes are more than 10' up. Ask any pilot about wind gradiant. So drag of a stanchion or railing will be way different depending on way too many variables.

Square footage isn't even a good stab in the dark. Willard Boat Owners group on yahoo really got into that about 8 - 10 years ago. But boaters grab onto that as it seems the only way to produce numbers. And many or most have no faith in anything w/o numbers. And some of those will belive any level of falshoods if numbers are provided.
 
Nomad Willy said:
Sounds good to me danderer but I question the legitimacy of calculating the drag of wind on a boat.
Click to expand...

No argument on how good or appropriate that number is. I stole the wind number from Calder who credits ABYC as the original source.
 
All of these esoteric calculations has me smiling and I am an engineer!!

Forces on the anchor have a lot more to do with factors other than frontal area and wind velocity. How about yawing. That will easily double the frontal area for a few seconds.

And how about sea state surge? What is the sea condition in your 30 kts of wind? It might be negligible if you are in a tight hurricane hole, but I can testify that it will be very significant with a 1/4 mile or more of fetch. A three foot sea will easily double or triple the simple wind forces for a few seconds.

So my conclusion is to forget math. Rules of thumb can be ok if they are grounded in real world experience.

David
 
Don't forget to take into account any slope of the sea bottom, and also the influence of current. Perhaps also the bottom type will have an effect, as the chain may sink into mud or silt.

30 -35 knots sounds about right to me.
 
djmarchand said:
All of these esoteric calculations has me smiling and I am an engineer!!

Forces on the anchor have a lot more to do with factors other than frontal area and wind velocity. How about yawing. That will easily double the frontal area for a few seconds.

And how about sea state surge? What is the sea condition in your 30 kts of wind? It might be negligible if you are in a tight hurricane hole, but I can testify that it will be very significant with a 1/4 mile or more of fetch. A three foot sea will easily double or triple the simple wind forces for a few seconds.

So my conclusion is to forget math. Rules of thumb can be ok if they are grounded in real world experience.

David
Click to expand...

All very good points. I sometimes get nervous about "rules of thumb" because so often they seem to be close to WAGs.
 
Ok, So here are some numbers that I have calculated and actually use myself for those who asked.

Ksanders yes your chain will lift off the bottom in 30 knots. For a bayliner 47 with a bow roller 6' off the water, anchoring in 100' of water 400' and using 5/16" chain. The chain will give you 695 pounds of pull against the wind due to chain's weight and leverage alone. A 3/8" would give you 973 ponds of pull and the chain would remain on the sea bed. The wind force on your vessel at 30 knots is probably a little over 800 pounds.

Psneeld a 40' boat in 15 knots is by my calculations around 150 pounds of force. I posted a windload chart here many moons ago which I have found to be accurate enough to use.

For those doing the math themselves, frontal area may seem to be important in calculating wind loads, but if you watch and time a boat's movement you'll see most boat's spend 80% of their time not facing strait into the wind which also imposes the minimal load. Some boat's even sail forward turn broadside and then drift back at speed going sideways until the anchor yanks the bow back into the wind. Snatch loads like this can be six times the wind load and is why all chain rodes need to be used with some kind of snubber at the extremes.

Not sure how to post a video but I was so amazed at the way a catamaran was behaving in cape may last fall I had to video it. As approached to anchor I thought he was actually under way and leaving.
 
Put engine into reverse and dog throttle hard. If anchor does not slip, you are good to go.

That's my calculus.
 
Does one need to bark to dog the throttle?
 
Thanks cafesport....I have always been skeptical of wind load calculations.

I also always look for hidey hole anchorages or in the lee of a deep, treeline beach to avoid strong yawing and wave loading.
 
I don't have any numbers, but I have a great piece of anecdotal evidence that leads me to believe that our anchors are massively oversized. We have been halibut fishing at anchor for the last 8 years out in the middle of the straits of Juan de Fuca. The current often runs 2kn+, and the wind often picks up suddenly and a 3-4' chop picks up before we give up and head home. The conditions are way worse than I would ever consider anchoring in for the night. I usually anchor in about 110 feet of water. I use 600' of 1/4 yellow nylon line, about 12' of 1/4" chain, and a cheap steel danforth anchor that is probably around 12 pounds. I purposely use super light gear since I have heard horror stories about huge logs floating into anchor rodes of small boats in the strong current, and dragging the whole boat under water almost immediately. Even with the supper undersized gear, I have never had the anchor let go, or any sign that the rode was about to break once it was set. The anchor seems to bury itself very deeply in the mud, and can be a challenge to get out.

A coupe additional thoughts:
-The 600' of 1/4" nylon rode stretches a lot, minimizing shock loads on the anchor.
-I have tried this with small Bruce anchors and had a hard time getting them to set. Once set they seemed to work fine, but it is such a pain to pull in 600' of rode and try again, that I will never use another bruce anchor for anything, ever.
-For the areas where I live in Washington, where I am usually anchoring in mud I think the danforth anchor is the best, but it doesn't store well on the bow so I don't use one for my primary anchor. I have a Rocanna, which I don't think much of. It doesn't seem to set that well, and it brings up a huge clump of mud with it every time you pull it, that is a pain to get off due to the roll bar holding it on.
- I think typical chain and line sizes for anchors are totally excessive in terms of breaking strength, but probably make more sense for chafe and corrosion margin of safety.
 
One factor that seems to have been overlooked is momentum. The calculations are based on static loading. If you visualize much of the chain lying on the seabed, it takes less force to start picking it up. As the boat moves backward, you have the force of the wind pushing the boat combined with the moment force of the moving mass (the boat). Have no idea how to calculate this, but my experience is that it takes much less wind than you would expect to straighten the chain out, because of the boat's mass and momentum.

Ted
 
Snapdragon, very interesting. I have 300' of chain. If I had the room to put out 600' of rode at most anchorages, I would be happy to have lighter rode. Not sure I would go with 1/4" but certainly would be happy with 1/2" three strand.

I readily admits that my anchor and rode are definitely overkill for most of the anchoring that I do. Most of the time I could just toss a 100' of chain overboard and be fine. However, I like having a robust anchor, chain, and bridle to use just-in-case.
 
Frequently have 35 kts of wind when anchored. The chain appears bar tight. In these conditions however I am unwilling to dive on the anchor to check if the rode is off the bottom the whole way.
 
Snapdragon III said:
I don't have any numbers, but I have a great piece of anecdotal evidence that leads me to believe that our anchors are massively oversized. We have been halibut fishing at anchor for the last 8 years out in the middle of the straits of Juan de Fuca. The current often runs 2kn+, and the wind often picks up suddenly and a 3-4' chop picks up before we give up and head home. The conditions are way worse than I would ever consider anchoring in for the night. I usually anchor in about 110 feet of water. I use 600' of 1/4 yellow nylon line, about 12' of 1/4" chain, and a cheap steel danforth anchor that is probably around 12 pounds. I purposely use super light gear since I have heard horror stories about huge logs floating into anchor rodes of small boats in the strong current, and dragging the whole boat under water almost immediately. Even with the supper undersized gear, I have never had the anchor let go, or any sign that the rode was about to break once it was set. The anchor seems to bury itself very deeply in the mud, and can be a challenge to get out.

A coupe additional thoughts:
-The 600' of 1/4" nylon rode stretches a lot, minimizing shock loads on the anchor.
-I have tried this with small Bruce anchors and had a hard time getting them to set. Once set they seemed to work fine, but it is such a pain to pull in 600' of rode and try again, that I will never use another bruce anchor for anything, ever.
-For the areas where I live in Washington, where I am usually anchoring in mud I think the danforth anchor is the best, but it doesn't store well on the bow so I don't use one for my primary anchor. I have a Rocanna, which I don't think much of. It doesn't seem to set that well, and it brings up a huge clump of mud with it every time you pull it, that is a pain to get off due to the roll bar holding it on.
- I think typical chain and line sizes for anchors are totally excessive in terms of breaking strength, but probably make more sense for chafe and corrosion margin of safety.
Click to expand...

A very very excellent post that was a long time comming.
99.9% of trawler formers worship "bigger is better".
All chain man .. all chain.
Your post is breath of fresh air to be sure.
Cheers ....
 
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Rudy's book is good! Hope this helps!

 

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Come on guys. The math really isn't that difficult.

First of all, Kevin has 100 ft hanging from his bow and a further 300 ft laying on the mud (maybe IN the mud), so he wants to know how much wind it will take to lift that 300 ft and straighten out his 400 ft of chain.

One will need to use Differential Equations to get at the force as the chain comes up out of the mud, so taking the force required to pull one link out of the mud, at a sharp angle, as before that force is applied, the chain is hanging vertically, decreasing the angle as each subsequent link rises out of the mud, until the whole 300 ft has lifted, then adding a factor based upon the amount of fluke area has been buried in that same mud, you can easily calculate the wind speed required.

Sorry guys, I took Diff Eq from Prof McSkasey (sp?) at UBC in 1969 and got only a bare pass, and what I did learn correctly I have forgotten completely.
 
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The hassle is not the chain lifting , it is the tight chain has basically no stretch.

Once the chain is tight , it is a bar , and will pass on every increase in loading , directly to the anchor.

A good time to rig a heavy weight on the chain to regain some shock absorber.
 
Calculating the square area of the boat is not all the easy to translate to load/area as its somewhat aerodynamic in addition wind load is exponential not linear.

But back of the napkin I'd say 35 knots! ;-)
 
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