# Calculating Forces on Anchor Rode

Discussion in 'Boat Design' started by bristol27, Oct 20, 2012.

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### bristol27Junior Member

I am trying to determine the forces (in pounds) at work on an anchor rode. My specific questions are in bold, though the extra text may be useful for better understand the source of my questions.

The strength of the anchor rode should be sized based on the rode's ability to resist the forces of a 'worst-case load scenario'. This scenario could could be described as: a fully loaded cruising yacht, anchored in a soft, moderate holding bottom, with 50 - 64 knot winds and associated surge.

With the environment conditions in mind, the forces acting on an anchor rode include:

Vessel Displacement - The maximum load that could pull on the anchor rode would be the full displacement (cruising weight) of the vessel. In my case, that weight is ~9000 lb. To put this in perspective, this would be like trying to hoist the vessel by it's anchor rode without it snapping. After speaking with a Yale Cordage technician, they mentioned the loads on an anchor rode would never be the full displacement of the vessel. They thought the anchor rode well below the full weight of a vessel's displacement.

Wave & Wind Action - The force of winds and waves create a total load on an anchor rode. Research indicated that a 30', 5 ton Catalina was estimated to have 443 lb. of force being applied to the entire boat in a 40 knot wind with 3 feet waves (this figure includes topsides, hardware, rigging etc.). Since a 3/8" line is suggested to have a 249 lb. safe working load, this road size would be unsafe. A better choice would be 5/8" line which has a safe working load of 691 lb.

It's not clear how wave nor wind forces were calculated above, but an idea for calculating windage might be:

Area Exposed to Wind (inches) * Strength of Wind (pounds per square inch) = Total Windage Force (pounds)

It should be possible to calculate the total force (pounds) on an anchor rode during a worst-case scenario loading by using the vessel's mass and wave and wind action described above. The idea is that by using these forces, it should be possible to calculate the minimum WLL of an anchor rode component. As long as an anchor component can resist the forces during the worst case scenario loading described above, it should (hypothetically) never fail.

Below is the equation I believe could be used to determine minimum WLL of an anchor rode component:

Force of Wave & Wind on Vessel (lbs.) + Force of Vessel Displacement (lbs.) = Minimum WLL of Anchor Rode Component(s) (lbs.)

Below is an estimated calculation for my 27' cruising sailboat.

500 lbs. Force of Wave & Wind on Vessel (50 knot winds & swell) + 9000 lbs. displacement = 9500 lbs. Minimum WLL for Rode Component

There is a problem with the above result, however. The 9500 lb WLL is a higher strength than the WLL of any component I've researched thus far. For example, 5/16" G43 High Test (HT) chain has a WLL of 3900 lb. The chain is the highest strength component of the entire anchor rode, but yet it isn't even near the 9500 lb WLL that my calculation requires.

I think where I've gone wrong is in the calculation of the 'Force of Vessel Displacement'. I say 9000 lb. Perhaps, this should be a safety factor of total displacement. By taking my vessel displacement, and reduce it by a factor that better matches the WLL of the anchor rode components available. This ratio would be 1/5. So instead of 9000 lb, it would be 1800 lb of force placed on the anchor rode, by the vessel's mass.

If it's not the loaded displacement of a boat, how can the vessel's displacement be calculated in terms of force (pounds) acting on an anchor rode? In other words, what is the actual Force of the Vessel's Displacement acting on the anchor rode?

Perhaps the displacement of the vessel doesn't have such a large impact as I assume above? Maybe the water helps to absorb a lot of the load forces?

Below are some other references for sizing an anchor rode's strength:

• A poster on cruiser forum said that the USCG suggests that the rode WLL is 1/8 of displacement.
• Yale Cordage documentation mentioned that both wind and wave forces impart measurable loads (energy) on the hull and superstructure of a vessel. This energy must be absorbed either by movement of the boat through the water, or by the anchor system’s ability to absorb this energy.
• A Yale Cordage technician mentioned that water adds some reduction in the force of the full displacement of the boat.

I should note that I am aware that chafe and the anchor letting loose are the primary reasons for a boat to break free at anchorage. These calculations are simply meant to give a more measurable figure on the forces at work on an anchor rode.

Also, I do not want to oversize the components, as oversizing adds unnecessary weight and cost to an anchor rode. This said, an oversized component will have a greater safety margin against breaking.

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### dougfrolichSenior Member

The load put on your anchor rode is due to friction from wind and water passing by your vessel the displacement would only relate in terms of momentum of the vessel jogging back and forth on the mooring. Your assumption that anchoring a vessel is like hanging It from the anchor is not correct.
A basic formula for drag is a better place to start i.e.
1/2 density * viscosity * area * Cd * velocity^2 = force

But why not just look in the west marine catalog and read the adviser on ground tackle

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### jonrSenior Member

You are never trying to lift the boat out of the water, so I wouldn't use total displacement. On the other hand, the boat may be trying to slide backwards down the face of a large breaking wave while a strong wind is also pushing it backwards. Or you could get slack that could be taken up suddenly (shock loading + heat issues).

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### bristol27Junior Member

Thank you for your input Doug.

So, the force of a boat at anchor is actually the force of drag experienced by the vessel due to it's movement through water. I will use this formula to try to calculate the actual force at work.

Agreed, and I've definitely done that for rope, chain and shackles. That said, I want to better understand why and how vendors determine their suggestions, so I can be an informed designer.

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### dougfrolichSenior Member

Check the GL rules they provide a very good methodology for Yachts

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### JohnTTJunior Member

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### bristol27Junior Member

Doug, thank you for suggesting this source. I did review Germanischer Lloyd's (GL) Rules & Guidelines. Specifically, I looked at 'I - Ship Technology - Part III Special Craft - Chapter 3 - Yachts < 24 m', pages 1-85 -> 1-86.

They do provide a framework for choosing an anchor rode, by finding my boat's 'Equipment Numeral' and then using this figure to compare to a table table (F.1, p. F-3), which helps determine the rode nominal thickness. Here is an image of the formula:

Below is Table F.1 - Anchors, anchor cables and lines of sailing craft and motorsailers:

By completing this formula, I gain another suggestion for the sizing of my anchor rode chain, rope and shackles (similar to West System or other vendors). However, their input still doesn't get down the root of the forces at work on a boat at anchor.

John, thank you for the link. This source goes into more depth about the wind and wave forces. For example, they mention that a 40 foot sloop-rigged yacht in a hurricane rated at 100 mph will 3 - 400 lb drag from the hull and 7 - 800 lb from the mast and rigging; a conservative measurement was 2,000 lb total.

So, the wind and wave forces are a little more clear now, though still a little difficult to fully estimate.

I guess my question may be difficult to answer. Anchoring, as John mentions, has a set of complex and varying forces. Even so, I would still like to narrow down the forces at play, then accurately calculate the forces, allowing me to insure that the anchor rode is designed to hold when conditions get bad.

To move forward, does anyone have further input/formulas for the forces at play on a vessel at anchor?

Specifically, I would be interested in understanding how the mass of a boat at anchor translates to force pounds.

8. ### FrostyPrevious Member

A good healthy anchoring rig will be many times bigger than necessary.

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### bristol27Junior Member

The conditions, as alluded to above, are hurricane force. This scenario could could be described as: a fully loaded cruising yacht, anchored in a soft, moderate holding bottom, with 50 - 64 knot winds and associated surge.

You're also correct that oversizing the anchor rode may be a good route for addressing an unknown and complex set of forces. That said, oversizing isn't ideal, and high accuracy is always beneficial.

10. ### FrostyPrevious Member

Why should that be included --are you thinking the anchor will give a little reducing shock loads ---depends on the chain weight and scope etc

How soft how much give are you getting. How deep is this set.

Sorry to sound negative but you would not ever anchor in a hurricane, and if one was coming you would not be there or move out to deep water.

My point is that if your calcs come to a rode of 1.467 inches there isn't one you would be buying a 2.5 inch.

You can land a 50lb fish on a 10 lb line,---if you know what you are doing.

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### bristol27Junior Member

Hm, I guess I was describing the anchors holding power more than it's ability to reduce shock loads. Perhaps this isn't really a factor..

Oh I don't think you sound negative at all. This is just a discourse about at topic I am not sure on, so all critique is openly accepted.

I see. Yes, I've heard that it's better to be out in the blue water rather than at anchor. That said, I've also read that it is possible to anchor in 'Hurricane Holes', which is somewhat I'm envisioning.

Using a hurricane hole is partially described here - http://boatsafe.com/nauticalknowhow/61798tip2.htm

So, while my goal should be to always get out to sea when facing a hurricane, perhaps I get stuck and don't have a way out, then I will need my anchor rode to hold in these conditions. I mention getting stuck, because I will not be using an engine for my boat, just a Yuloh oar, so if the winds aren't right, I won't have much choice but to hunker down.

I don't quite understand. Are you saying something like "oversize no matter what"?

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### MikeJohnsSenior Member

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### daiquiriEngineering and Design

Mike, that's an excellently written site about anchors and rodes. Thanks.

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### FAST FREDSenior Member

"40 knot wind with 3 feet waves"

Not realistic if there is any fetch , and the wind blows for very long.

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### MikeJohnsSenior Member

Thanks Slavi !

People should thank the author directly too his email is on the site.

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