UPDATED 25.08.2022

Now also available English digests for catenary calculations for various vessel sizes. See below

**Note: The content of this web page takes a while to read and digest. If you are one of those folks that have little patience and want answers within minutes – as I occasionally come across on some sailor’s fora – then just stop reading right here and now! You are not my intended audience! You may still want to install my AnchorChainCalculator app on iPhone or Android, though, it is easy to use… 😉**

On this web page I try to motivate my choice of the correct anchor chain and its length and how to deploy it based on a mathematical analysis of the main forces and energies involved when anchoring. I ignore friction and any form of energy dissipation, as this will only work in my favour and not lead to any increase in the chain length required. The models developed here will hopefully help you to have a better understanding of the physics of anchor chains and snubbers / bridles and to make the right choices.

The documents and sections further below work out diagrams for the **MINIMALLY** **required anchor chain length** as a function of the water depth, *Y*, at the position of the anchor, the weight per meter of the chain in water, *m*, the wind strength, and the swell, **tailored for different vessel sizes**. The minimally required chain length is defined by the chain still pulling horizontally at the anchor, but rising to the water surface right at the anchor shank. Obviously, one can and should pay out more chain than that, but it is important to know what the minimum is to maintain the maximum holding power of the anchor.

## Catenary, what is this?

Catenary refers to the shape a chain assumes when it is hanging down between two points. You will have seen examples in parking lots, where a chain was perhaps used to rope off a certain section, or one parking space in particular. The same shape you can see with telephone or power lines running from one pole to another along the road side.

The importance of a catenary for the anchor chain lies in the fact that it is the catenary that allows the chain to store more energy. Once the chain is pulled completely bare-taught, the catenary is gone and no more energy can be stored. This is when shock loads will not get dampened by the chain anymore and need to be dealt with differently (i.e. a good snubber / bridle).

Thus, folks who say that in strong wind the catenary is gone, anyway, implicitly also say you need to have a very good, i.e. elastic, snubber / bridle, or very elastic rope, if you are using a mix of rope and chain.

## Anchor Chain Length Calculator App:

I have published an app for iPhone, iPad as well as Android. This app calculates, among other things, the minimally required anchor chain length and the associated anchor load as a function of a couple of vessel parameters as well as sea / weather conditions:

A brief introduction to the App can be found in **English**, auf** Deutsch**, and en **Français**.

**iPhone and iPad version available in the Apple Store! Android version available in Google Play Store! A tutorial video can be found here.**

**And a free Lite version of this app with slightly reduced functionality you can try out online here.**

**In a nutshell, our main results are**:

- The guidance of using a fixed scope like use 3 times or 5 times the water depth as length of chain, etc., is not adequate. It is not enough in shallow water, and it may be too much in very deep water. And it does not depend on the wind strength at all, which is rather odd. We consider this guidance to be dangerous, at least for the novice. Our results do depend – among other parameters – also on the wind strength.
- Neglecting swell and waves, the result is very simple, based on the catenary equation: Minimum required chain length is
*L*= square_root(*Y*(*Y*+ 2*a*)), where the parameter*a*depends on the wind strength, on*A*_{eff}, which is the effective cross section of the vessel towards the wind – i.e., the windage area – as well as on the mass*m*of the chain per meter.*Y*is the water depth at the anchor (and not the vessel!). - This simple non-linear catenary result already shows that the scope approach generally
**underestimates**the chain length in shallow water, and**overestimates**it in deep water. See graphs further below. - A conventional scope approach generally leads to a higher anchor load than a fully developed catenary.
- If the swinging circle around the anchor is not an issue, it is better to use
**a thinner but longer chain**, perhaps of higher quality. It allows to**anchor in deeper water**when comparing chains of same total weight in the locker. But in crowded anchorages with small tides and not too deep water a thicker and shorter chain may be more appropriate in order to have a reduced swinging circle. Also, one needs to bear in mind that for the same anchor depth*Y*, a thinner chain has less elasticity than a thicker chain has – despite of there possibly being still unused chain lying on the seabed. Note: When talking about the „elasticity“ of a chain, we do not mean the stretching of the bare metal, but rather the stretching of the entire chain when it gets pulled harder and harder at one end. - If I
**half**the weight of the chain per metre and use**twice**as much chain to anchor at**twice**as deep a place – thereby deploying the same total weight of chain – I get**twice**the elasticity of the chain. However, at the**same**anchor depth, the thicker chain has more elasticity than the thinner chain, and in a swell it adds less load to the anchor. - The ability of a chain of full catenary shape (with the anchor being attached horizontally) to absorb energy is greatest when
**in addition**it also fulfills the ’scope‘ condition*L*=~ 1.4 … 4*Y*– the closer to 1.4 the better. This is a sweet spot where the chain is most elastic and works best. In order to achieve this, one may have to relocate to a deeper anchorage place. - The ability of a chain of full catenary shape to absorb energy is also increased when the
**slope between anchor and vessel**is negative all the way. This comes at the expense of a longer chain that is required, to compensate for the negative angle at the anchor shank. This might be a worth-while consideration, if shallow water cannot be avoided. Vice-versa, whilst a positive slope between anchor and vessel is beneficial for the ability of the anchor to dig in, and it also reduces the amount of chain required, it does have a very negative effect on the chain’s elasticity in severe weather. - In
**shallow water**the effect of heavy swell head-on requires more chain to be laid out than the square-root catenary formula based on wind only would suggest. Worst case, in such a scenario the force on the anchor will be more than it can bear, and the only solution is to relocate to deeper water and use a little more chain. Swell is worse in shallow water. Again, see graphs further below. This is where snubbers and bridles are needed most. **The more shallow**it gets at the position of the anchor, the more difficult it is for the then almost horizontal chain to absorb energy bursts (e.g., caused by swell) as additional potential energy. This – in a storm dangerous – inherent shortcoming of a chain has to be covered by good, long snubbers or bridles. They are most needed in shallow water, but of course good to have in any situation.- If a large scope and shallow water cannot be avoided, it may be better to use a good long elastic rope and a shortish lead chain (to avoid chafing on the seabed). The difference in shape between a chain and a rope will be marginal in such a case, but elasticity is so much better.
- Vessels using excellent snubbers or a mix of chain and rope are happy to anchor in shallow water, whilst vessels with chain only will want to prefer deeper water. This is particularly true for vessels that move a lot at anchor when exposed to gusts or swells. Heavy, compact displacement vessels that stay put no matter what gust or swell is thrown at them, will be less affected by these dynamic effects.

**Now let’s dive deeper into the analysis.** We start by outlining the underlying **physical model**, then analyse the results obtained for the **elasticity of the chain** as a function of scaled variables such as scope, *L*/*Y*, or the ratio *a*/*Y*. Note: When talking about the „elasticity“ of a chain, we do not mean the stretching of the bare metal, but rather the stretching of the entire chain when it gets pulled harder and harder at one end. Because of gravity, it acts very much like a spring. A lot of insight can be gained from this analysis – when does a chain work well, when does it not, and where is a chain’s sweet spot of operation. Surprisingly, the answers to this are rather universal. In the following sections we provide some **example graphs that illustrate the strong effect of swell in shallow water** – what we call dynamic anchoring. This section is supported by **dedicated short digests**, tailored for different vessel sizes. Non-horizontal seabeds, the effect of kellets, as well as **snubbers and bridles** are briefly discussed thereafter. We conclude with an analysis of the **Tuamotus anchoring technique** that helps preventing damage to corals, and we also briefly touch upon how this analysis applies to large vessels. For a brief discussion of the differences and similarities with **Alain Fraysse’s excel spreadsheets**, please have a look further below on this web page. We conclude by comparing our results with the **British Admiralty** guidance.

## Outline of physical model – the interplay between potential energy of the chain, the energy of the vessel in the wind field, and kinetic energy of the vessel**:**

I am not interested in knowing the precise load at the anchor as a function of time. Nor for the load at the bow. What I am interested in is knowing the maximal load at the anchor when the stress is largest. Obviously, the load at the anchor is largest when the vessel has pulled as much as it possibly can and the chain is as much stretched as possible. This point is reached when the vessel, after a gust or swell, is farthest away from the anchor and has just come to rest for a brief moment. Very much like a pendulum that has reached its farthest point. The obvious approach to tackle this issue is to use the law of conservation of energy…

The first diagram shows the basic approach how to take wind forces and swell into account by relating them to the potential energy gained when the chain is rising further off the ground. Obviously, the basic law of conservation of energy needs to hold. Now, what are the energies involved?

**Potential energy of the chain.**We calculate the increase of potential energy of the chain, from an initial position where part of the chain is still lying on the seabed – the situation just before a slowly increasing wind, a sudden gust or swell hits the vessel – to a fully developed catenary, where the very last chain link is just about to lift off the seabed as a response to the wind and or swell. Wind, also gusts, or currents, which are all forces acting on the vessel, all feed to 100% into this increase of the chain’s potential energy. So, this is like a ‚charging‘ of the chain’s potential energy with 100% efficiency.**Spring energy of the rope or the snubber.**Another place to store energy is the snubber / bridle when it gets stretched, which in essence can be viewed as a spring. When using a mix of chain and rope, it is the rope’s elasticity, which allows to store energy in the rope.**Wind energy due to displacement in the wind.**Now, this comes into play when the vessel is moving downwind (or against it). The energy is simply wind force integrated along the distance travelled, Δ*x*. As long as there are only forces pulling slowly at the vessel, this energy gets to 100% transferred into the chain’s potential energy. In fact, one can view this energy as the charging mechanism for the potential energy above. In physics notion, this is a slow, adiabatic process. If a mix of rope and chain is used, and the rope is very elastic, this contribution can become very large and will need to be compensated for by larger increases of the potential energy of the chain and the snubber / rope.**Kinetic energy of the vessel due to a velocity component parallel to the wind.**Now, this is a different beast! As long as the vessel possesses some additional kinetic energy – possibly caused by**swell**– it will move either downwind, or opposite to it. Worst case in terms of anchor gear is when the vessel moves a further Δ*x*in the direction of the wind, so let’s assume that. In doing so the vessel will pick up wind energy as per the previous item: wind force*F*_{w}times distance travelled, Δ*x*. As just discussed, this wind energy will feed into the increased chain’s potential energy as the chain gets pulled harder, leaving less ‚room‘ in this increase of potential energy for the kinetic energy to get stored there as well.

Two other diagrams way further below in the section **Example Graphs** show typical results for little swell and for strong swell. Note that we did **NOT** account for **snubber** or **bridle** here. It is the naked chain. This is ok, as the effect of a snubber can be added later, modelled as a spring with a spring constant. One only needs to match the forces at the bow – see the discussion towards the end of this web page, which shows some sample graphs that include the effect of a snubber. Also, one of the links given at the end of this page leads to an online calculator that also includes the effect of snubbers and bridles.

A note on the potential energy of the chain: *E*_{pot} = ½ *m g* (*L* *Y* + *a* (*L* – *X*)) = – ½ *m g* *L* *Y* – ½ *F* (*L* – *X*), as measured from the **anchor**. The first term corresponds to the potential energy of an ideal completely straight *L*:*Y* scope chain, whilst the second term accounts for the fact that a real catenary has a bit of slack and thus always has a potential energy smaller than that. This 2nd term is the anchor load *F* times half the distance between chain length *L* and swinging circle *X* (measured at the bow roller) – perhaps somebody can find a nice intuitive physical understanding of this?

## Chain elasticity – or its ability to store additional energy:

Starting with the expression for the potential energy of the chain, *E*_{pot} = ½ *m g* (*L* *Y* + *a* (*L* – *X*)), when measured using the anchor as reference point, a lot of insight can be gained by studying the elasticity of the chain as a function of various parameters. **The elasticity is defined as the derivative of the potential energy with regards to the anchor load F**. So, if I increase the anchor load

*F*by a tad, how much more energy can I then store in the chain? Obviously, the more the better.

When taking this derivative one needs to keep ’something‘ constant, as otherwise there would be no bound to it. An unbreakable chain deployed in as deep a water as one likes, and with as much chain as one likes, will store any amount of energy I like it to. So, something needs to be kept constant when doing this analysis. In the following, I have kept either the anchor depth *Y* constant, or the chain length *L*.

**Here is a real situation where this graph applies:**

Suppose you are anchoring in shallow water and there is only a light breeze. But forecast has it that a storm is expected for the night, and in anticipation of that you have already paid out a LOT of chain.

For now, in the light breeze, most of that chain lies unused on the seabed. The effective scope of your chain – so the ratio of the chain that is actually off the seabed to the anchor depth *Y* – is small. For the sake of the argument let’s assume this effective scope to be 1.4. The chain thus operates in the peak of the elasticity curve, and all is fine and smooth.

Now the wind starts to pick up and more and more chain is lifted off the seabed. The anchor depth *Y* stays the same, but the effective scope becomes larger, as more and more chain is being used.

With this, we move to the right in the graph, so away from the peak. At an effective scope of 3.7 we have only 50% of the chains optimal elasticity left. And to be clear – this optimal elasticity is the elasticity of the effectively much shorter chain when there was only a light breeze! So, despite of the fact that more chain is now being used, it has less elasticity. It is much less efficient at absorbing energy than the much shorter chain at scope 1.4 was when there was only a gentle breeze.

As the wind further develops, even more chain gets lifted off the seabed. At an effective scope of 20:1 only 9.5% of the chain’s original elasticity has remained – despite there possibly being still some unused chain lying on the seabed! – and the cleats at the bow start to moan… It is getting uncomfortable. Snubber to the rescue… Or – a relocation to deeper water well before the storm sets in.

**If I repeat the above, but now at twice the anchor depth, 2 Y**, I will initially be slightly before the peak at a scope of 1.22 (one can work this out using the equation

*L*/

*Y*= sqrt((Y+2

*a*)/

*Y*) and the parameter

*a*being the same as before). Eventually, when the storm has fully developed, we find us at a scope of only 14.2, or 13.5% of the peak elasticity. But since we doubled the anchor depth, in absolute terms, this peak is also twice of what it was before! So, we have 2 x 13.5% = 27% of the ‚old‘ optimal elasticity at anchor depth

*Y*. This is almost a factor 3 difference! And I needed only 41.6% more chain. Doing the same exercise with a relocation to three times the original anchor depth, 3

*Y*, one ends up at a scope of 11.6 and as much as 49.4% of the ‚old‘ peak elasticity at the original anchor depth

*Y*– a gain by a factor 5! The chain will have to be increased by 73.6% only.

So, if the new anchorage location is not exposed to too much more wind and swell, it may well be worth relocating to deeper water and have the chain work closer to its optimal performance – at very little cost in terms of additional chain length.

Now, how do **chains of different thickness** compare with each other? Remember that for fixed anchor depth *Y* the elasticity of the chain is the product of *Y* and a function depending on the scope *L*/*Y* only. That function was plotted in the graph above. Since a thinner chain requires more chain to be paid out than a thicker chain for a given anchor depth *Y*, it follows that its scope is larger than that of the thicker chain. This puts the thinner chain at a disadvantage at any fixed anchor depth *Y*. Consequently, a vessel with a thinner chain will need to anchor in even deeper water than the same vessel with thicker chain, in order to compensate for the difference in elasticity. But you can carry more chain on board, if it is thinner, so this strategy could still work out ok… So let’s study this graph then…

**Again, the chain’s elasticity is best at a certain sweet spot**, where *a*/*Y* = 0.474 holds, with *a* = *F*/(*m* *g*). As before, the height of the peak itself scales with the anchor depth *Y*. Now, let’s compare two chains, one chain twice as heavy as the other one per metre. Clearly, when going for the lighter chain to twice the anchor depth, 2*Y*, I have exactly the same ratio *a*/*Y* as with the heavier chain, and therefore exactly the same value in this graph. But, as the graph scales with *Y*, in real terms, the elasticity of the lighter chain is** twice as high** as that of the heavier chain. It requires precisely a twice as long chain (so in terms of scope, we end up at the very same point in the previous graph!), and the total weight of these two chains is thus the same. But as the elasticity is so much better for the thinner chain I could even save on the total weight of this chain, if I wanted to. So, in total, I gain a lot when using a lighter chain. But this does come at the expense of having to anchor in even deeper water. (And the chain’s breaking load being reduced…)

**Just double checking:** Potential energy of the chain as measured from the anchor: *E*_{pot} = ½ *m g* (*L* *Y* – *a* (*L* – *X*)). I half the chain mass *m*, thereby double the chain parameter *a*, I then double *L* and *Y*, which leads to a doubling of *X*, and thus I get exactly twice the potential energy! So, it all seems to make perfect sense! 🙂

Some might now argue that this is not taking full advantage of the chain’s ability to store energy. And indeed, I limit it by requiring the chain to pull still horizontally at the anchor shaft. My preference is not to meddle with the maximal holding power of the anchor by allowing the chain to pull at an angle.

Still, let us briefly study an extreme case: Initially, there is no wind or swell at all. The vessel is at rest. The potential energy of the chain is then simply given by *E = m g/2 Y*^{2}, where *Y* is again the anchor depth. Next we assume a massive gust, a massive swell, and the chain is fully stretched, becoming in the extreme a straight line between anchor and bow. Now we have as potential energy *E = m g/2 Y* *L*, with *L* being the chain length again. The energy that the chain can thus absorb is the difference: Δ*E = m g/2 Y* (*L* – *Y*), which has a maximum at scope *L*/*Y* = 2. For a fixed length of chain, the ability of the chain to absorb energy becomes poorer and poorer in very shallow water or when the scope becomes very large, again pointing to the fact that chains like deep water and fail in shallow water, where other means of energy absorption needs to be found. A scope of 2 is slightly different from the previous results, but this is due to the different boundary conditions chosen (and in one case we look at the energy, in the other case at a derivative of the energy). And of course, this result does not mean one should anchor at a scope of 2, as this would reduce the holding power at the anchor and, even worse, such a straight line of the chain can only be achieved with an excessive load to begin with, and any further shock load would reach the anchor unfiltered.

It just proves the point – again – that chains have a sweet spot where they perform best.

## E**xample graphs:**

The following two diagrams show the effect of dynamic anchoring, so when a large swell needs to be absorbed by the anchor gear. In particular, it shows that swell is more difficult to deal with in shallow water, it leads to a higher load at the anchor, and a possible resolution could be to relocate to deeper water, using only minimally more chain.

## Are we cutting corners by going for the MINIMALLY required chain length?

**No** 😉 These diagrams have to be used as follows: Take the top value of the gusts predicted for your area and add a margin as you see fit – possibly adjusting for the overshooting effect in gusts as Alain Fraysse has described it. Make a judgement call on how much swell to expect – expressed as energy. A good indicator are the SOG values at anchor at the worst conditions to expect. Together with the vessel’s displacement, you can work out the kinetic energy of the vessel at anchor. Assuming you are using good long snubbers or bridles, the largest part of this energy will be absorbed by the snubber / bridle but some of it will reach the chain. For now, as I have not added snubbers / bridles in the analysis yet, this is some guess work, but do see the notes at the end of this web page. Perhaps only as much as 10-20% will reach the chain. Use that value of Δ*E* to pick the closest diagram in the digest. Note the water depth at the position of the anchor (not the vessel), making sure to use as reference point the bow roller. With all these data, find the closest graph in the digest and read off the minimally required chain length. **NOTE**: By plugging in the worst case values for gust and swell, and by adding margins for both, you have already accounted for the anticipated worst case. So, this is not cutting any corner.

**Windage area ***A*_{eff}:

*A*

_{eff}:

An important parameter to know for all these calculations is the windage area or effective cross section *A*_{eff} of your vessel – we will come back to that in a minute. We have created a list of tailored digests for various vessel sizes having different *A*_{eff}. As the vessel gets larger, the chain will have to be thicker as well. For each *A*_{eff}, we have worked out the minimum required anchor chain length for at least three different chains.

The parameter *A*_{eff} is the windage area, so the effective cross section of your vessel facing the wind, augmented by a streamlining factor. So, for example, the nominal cross section *A* of your vessel when viewed from the front (including rigg) is, say, 13 m^{2}. You have worked this out by taking the beam of you vessel and multiplied it by the height of the deck over the water line. Then you have estimated the cross section of any superstructure like cabins, davids, etc. You added to this perhaps 1 m^{2} for the lazy jack, and you did not forget to include the rigg, so perhaps height of the mast times 30 cm or so. Since the vessel is tacking at anchor, it will often expose a surface somewhat larger than this to the wind, so, perhaps add 10-20% to the total value to account for this effect. Because of the shape of the vessel, the wind can slip around it more easily than for a simple ‚brick‘ surface. This effect can be accounted for by multiplying the result with a factor *p* smaller than one. The windage area, or effective cross section, therefore, is *A*_{eff} = *p* *A*. Values of *p* = 0.7 or so are reasonable (but far from accurate), which would give in the above case *A*_{eff} = 10 m^{2}.

**Other, more accurate, ways of determining the windage area ***A*_{eff}:

*A*

_{eff}:

- Not necessarily more accurate, but convenient: The late Robert Smith did some measurements on a variety of (monohull) vessels and for a range of wind strengths to determine the wind force acting on the vessel. There is a link to his table at the end of this page. From this table one can derive the windage area as 0.071
*L*_{m}^{2}, where*L*_{m}is the length of the monohull in metres. Heuristic correction factors need to be applied for multihulls and for vessels that are slimmer or more bulky than the average. It is at least a start and will give you a ball park figure. - In the absence of swell and current, measure the force at the anchor chain and the wind strength. Then using equation (4) plus either the cosine of the chain’s angle to the horizon, or equation (8) provided the chain is nowhere deeper than the anchor, work out the horizontal force acting on the anchor, and subsequently
*A*_{eff}. - As a variant of the above, use a mooring line to attach the vessel to a strong free-standing post in the water (such as a dolphin or a pile mooring) and measure the force. This might allow for an almost horizontal measurement and can be very accurate.
- Again without swell and current, anchor at various depths and adjust the chain such that it is not lying on the sea floor but still attaches horizontally at the anchor. Take a note of the required chain length and wind strength, as well as water depth at anchor position. From this determine the paper’s parameter
*a*for each wind strength, and as above, finally*A*_{eff}. This will require somebody diving to the anchor… 🙂 - Ask your shipyard! 🙂
- Place your vessel in a professional wind tunnel! 😀

Beware, though, that *A*_{eff} will slightly depend on the vessel being fully loaded or being empty. The latter is the safer measurement.

**List of available digests:**

Here is a list of digests that focus on the **effect of dynamic anchoring** for vessels of various size. As a point of reference, an Etap 39 has been quite accurately measured to have an *A*_{eff} = 7 m^{2}. For our Neel 51 trimaran I am currently using *A*_{eff} = 20 m^{2} as a rough estimate, but this is subject to change. As of version 3, these digests also include diagrams showing the effect of snubbers / bridles.

As a guidance, for a mono hull Robert Smith did a lot of measurements (see link below), resulting in this rough mapping as a function of the vessel’s length (LOA):

8 m LOA -> Aeff = 4.5 qm

9 m LOA -> Aeff = 5.7 qm

10 m LOA -> Aeff = 7.1 qm

11 m LOA -> Aeff = 8.6 qm

12 m LOA -> Aeff = 10.2 qm

13 m LOA -> Aeff = 12.0 qm

14 m LOA -> Aeff = 13.9 qm

15 m LOA -> Aeff = 16.0 qm

16 m LOA -> Aeff = 18.2 qm

17 m LOA -> Aeff = 20.5 qm

18 m LOA -> Aeff = 23.0 qm

19 m LOA -> Aeff = 25.6 qm

If you cannot find your scenario in this list, drop me a line and I can perhaps create it, time permitting. If you like mathematics, then perhaps you can still find the correct digest by applying scaling laws. As briefly discussed in a footnote in the digest, all formulas depend only on the ratios of *A*_{eff}/*m*, *F*/*m*, Δ*E*/*m*, where *m* is the mass of the chain in water per meter, *F* is the force at the anchor or at the bow, and Δ*E* is the energy burst induced by the swell. Hence, a digest for *A*_{eff} = 10 and *m* = 2 is the same one as, e.g., for *A*_{eff} = 20 and *m* = 4. You need to keep in mind that *F* and Δ*E* are scaled as well, of course.

Please note:** The formulas can also be used when it is current and not wind pulling at the vessel.** Or a combination of both. The *A*_{eff} would be different, though, and depend on the shape of the vessel underneath the waterline. Another measurement of *A*_{eff}… 😉

**Some words on the conventional scope approach:**

When using large scopes like 9:1 or more, it is certainly true that the pulling angle at the anchor is small, no matter how hard the wind blows, and so the anchor’s effectiveness is not much reduced, if the chain does not quite pull horizontally at the anchor. Further increasing the chain length would not change things a lot as far as the pulling angle is concerned. And this is the good part of the scope approach – it gives me peace of mind regarding the maximum pulling angle at the anchor.

There are, however, a couple of short-comings of a pure scope approach: **A)** Its indifference to vital parameters such as anchor depth and wind strength, resulting in it being either too conservative or, worse, actually dangerous. **B)** Its ignorance of the poor performance of a chain at excessively large scopes to absorb any further energy.

**To the first point:** The catenary results indicate that even in a storm, when anchoring in deep water, the catenary shape of the chain still exists, provided the chain is long enough. Perhaps surprisingly, the scope approach would actually result in an even longer chain, which is not needed and only creates problems as the swinging circle is larger than really needed. On the other hand, in shallow water, the scope approach often underestimates the required chain length. Of course, one can modify the scope guidance by saying in shallow water I need a larger scope than in deeper water, but this is a very heuristic approach and requires a lot of experience to get it right without being way too conservative all the time.

**To the second point:** In shallow water in a storm, when scope gets very large, the difference between the catenary and a straight line may be very small and there is little point in arguing whether the catenary is gone when you need it most, or it still exists, if only the chain is long enough. The point is that at such a large scope the chain – even a very long chain – is extremely poor at absorbing additional energy, and this lack of elasticity is what matters, not the existence or not of the catenary shape. It may help to remember that a horizontal chain laid out on the beach cannot absorb any potential energy at all, and so it should not come as a surprise, really, that a chain at a very large scope is not performing well. In conclusion, in particular when the scope is very large, it is good practice to use a long (!) snubber (monohull) or bridle (multihull) to help relieve excessive energies (or even use an elastic rope with only a short lead chain). This is the main reason why the same swell and gust can better be tolerated in deeper water, provided one has enough chain.

## What if the seabed is not horizontal?

All calculations here have been done assuming a horizontal seabed. If this is not the case, results will be different. For the dynamic anchoring part it will be difficult to give quantitative answers as to how much different the results will be, since the potential energy of the chain before the swell hits the vessel is strongly affected by any deviation of the seabed from the horizontal. If the chain lies in a trove when there is no swell, it has more room for gaining potential energy, but when it is sitting on a ‚mountain‘ under water before the swell arrives, the gain for further potential energy will be **much** smaller than what is calculated in all these calculations. In essence, a positive slope of the seabed between anchor and vessel – i.e. when the vessel is in shallower water than the anchor – will greatly decrease the chain’s ability to absorb energy, simply because the chain is already raised by the seabed and the difference to a chain under full load is smaller than it were the case with a horizontal seabed. So, to conclude, whilst from a ‚digging in‘ point of view for the anchor such a positive slope is favourable, it is absolutely bad for the elasticity of the chain and is not at all advisable from this point of view. In essence, such a situation has a lot of similarities with a ‚tilted‘ shallow-water scenario.

However, for the static anchoring the answers can be inferred from the graph in the previous section. The formula for *F*_{g} tells us by how much the chain will need to be longer (L_{1} + 2 L_{2}) or shorter (L_{1}), if the anchor shank is horizontal with respect to the local seabed to make a perfect catenary, but is at a certain angle alpha with respect to the water surface.

Below is a graph of a **chain’s elasticity**, normalised to unit water depth, as a function of *a*/*Y* and for a couple of **different slopes of the seabed**. In a storm, the elasticity of the chain is greatly improved for negative slopes, shown in green. This is simply because the chain at the anchor shank will first follow the seabed to deeper water depths, before eventually turning upwards towards the vessel. Clearly, the stronger the wind is, or the stronger the slope in the seabed, the deeper the lowest point of the catenary will be, resulting in great elasticity. But it comes at a price – a rather long chain is required. For instance, for the α = -5° curve at *a*/*Y* = 100 – which is a very severe storm in shallow water – a chain of length *L* = 25.4 *Y* is required, which at an anchor depth of only 4 metres amounts to more than 100 metres of chain. This has to be compared to *L* = sqrt(1 + 2 *a*/*Y*) *Y* = 14.2 *Y*, had the seabed been flat, or only *L* = 7.8 *Y* at α = +5° – but at a tiny fraction of the elasticity: These numbers are as follows: α = -5°: 77.3%, α = 0°: 13.5% and α = +5°: 2.3%, respectively. All numbers given are for an extreme scenario of *a*/*Y* = 100, they will be less extreme for smaller values of *a*/*Y*. For example, for *a*/*Y* = 20 we find 66.7%, 29.6% and 13.0%, respectively. **Conclusion, if enough chain is available, and provided the chain is not at increased risk of getting entangled with stuff at the seabed, it seems beneficial to anchor ’slightly on the green side’… ** 🙂

## The effect of chain lying on the seabed (dragging friction)

Here we want to analyse what contribution a chain lying on the seabed has with regards to holding power. A chain pulled at one end will require some resistance to be overcome before it starts moving along the seabed. This is the frictional force and it comes in two flavours: The force to be overcome before a chain at rest starts moving, and the force to keep it moving. Generally the latter is smaller than the former. The physics is described by the friction coefficient 𝜇 = *F*/*m*, where *F* is the pulling force and *m* the weight of the chain lying on the seabed. Little information can be found about values for 𝜇, which will also depend on the type of seabed. Taylor and Valent report values for what they call static friction coefficient between 0.98 for sand, 0.92 for mud with sand, and 0.90 for mud/clay. So, it is a little less than the weight of the chain, actually.

Based on this one finds that, e.g., 10 metres of a 10 mm chain – which has a weight of 2 kp per metres in water – requires a force of 10 m * 2 kp * 0.9 /m = 18 kp = 17.7 daN before it starts dragging in mud. Or, in other words, the anchor load will be reduced by this amount, when I have 10 metres of this chain still on the seabed. This is not a lot, though, compared to the typical holding powers of an anchor that normally matches such a chain, which are more like 500 daN or much more.

We thus conclude that although the frictional force of the chain on the seabed does help to reduce the anchor load, **it is typically a small effect**, unless excessive amounts of chain are lying on the seabed.

## The effect of snubbers and bridles

So far, we have not included the effect that a snubber or bridle has. In fact, all diagrams in the digests are for the naked chain with no snubber or bridle used at all. This is, of course, not our advise, not to use a snubber or bridle 😉 . To the contrary, but it is useful to know what happens when there is only chain, as even when using a snubber / bridle, the behaviour of the chain will be unchanged in principle. It is simply damped by the snubber / bridle. So, in this sense the graphs in the digests represent worst cases as far as snubber / bridles are concerned.

**A snubber / bridle can be modelled to a good accuracy as a spring with a spring constant c.** To elongate the snubber / bridle by a distance

*s*, a force

*F*=

*c*

*s*has to be applied. The energy thereby stored in this spring is given by

*E*

_{s/b}= ½

*c*

*s*

^{2}= 1/(2

*c*)

*F*

^{2}.

Now, this force *F* has to match the chain load at the bow, *F*_{bow} = *F*_{ws} + *m* *g* *Y*, where *F*_{ws} is the horizontal anchor load with contributions from wind and swell. On the other hand, we have *F*_{ws} = *a*_{ws} *m* *g*, where *a*_{ws} is the catenary parameter pertaining to that particular scenario.

Plugging this into the formula for the energy of the spring yields *E*_{s/b} = (*m* *g*)^{2}/(2 *c*) (*a*_{ws}+*Y*)^{2}. The energy that the spring can store additionally when a swell hits the vessel is then given by the difference of the spring energies at and before this event: Δ*E*_{s/b} = (*m* *g*)^{2}/(2 *c*) [(*a*_{ws}+*Y*)^{2}– (*a*_{w}+*Y*)^{2}] = (*m* *g*)^{2}/(2 *c*) [*a*_{ws}^{2}– *a*_{w}^{2}+2*Y*(*a*_{ws}–*a*_{w})]. Here *a*_{ws} describes a catenary when the swell has just been fully absorbed, and *a*_{w} the catenary before the swell.

Now, *a*_{w} is independent of water depth *Y* – it is only determined by the wind load. But *a*_{ws} does depend on the water depth at the anchor, *Y*, as the ability of the chain to absorb kinetic energy bursts as additional potential energy gets worse and worse the more horizontal the chain becomes. This is what happens in shallow water and, as a result, *a*_{ws} has to increase substantially to compensate for that. (This is also seen in the graphs in the various digests when studying the anchor load as a function of *Y* for large values of energy bursts – it increases steeply.) Luckily, we see with the above result for Δ*E*_{s/b} that the snubber / bridle behaves just the opposite way. It becomes more and more effective as the water gets more and more shallow and thus it can – at least to a good extent – compensate for the shortcoming of the chain in shallow water. **This is why we need to use snubbers / bridles!**

As the effectiveness of the snubber / bridle does depend on the water depth, it is unfortunately not as easy as simply using only a certain fraction of Δ*E* in the graphs in the digest to account for a snubber / bridle. The effective Δ*E* is also a function of *Y*. 🙁

Also, it must be noted that the snubber / bridle can never compensate the chain’s shortcomings to 100%, as it needs to get stretched more to be able to store more energy. So, some additional force will always have to exist to achieve that. And this is why it is good to know the shortcomings of the naked chain. These shortcomings will be reduced by the snubber / bridle, but they are never completely gone!

Snubbers and bridles are also needed to battle the dynamic overshooting effect described by Alain Fraysse, i.e., when a gust or swell suddenly hits the vessel and it starts oscillating back and fourth before settling down.

Below are some example graphs with snubbers of various quality. They all deal with the same scenario as shown above, where 1000 J of energy had to be absorbed by the anchor gear. Clearly, the snubber is only effective in the regime of dynamic anchoring, and it needs to be a really good snubber that can stretch by a couple of decimetre to have any impact for anchoring depths that are accessible in the first place. **Short Mickey Mouse snubbers, which are just 1 – 2 metres long in total, will not be able to provide this kind of elasticity!**

When using a long snubber or bridle, it is advantageous to have a fair amount of chain hanging slack between the bow and the point where the snubber / bridle is attached to the chain. Not too much – you to not want it to touch the seabed, but it should not be taut either. If it is adjusted such that only at the peak of swells / gusts this part of the chain will be pulled somewhat straight, then this is another means to store potential energy. Not a lot, but every little helps. In essence, this can also be analysed using the graphs for chain elasticity. If the chain between these two points just goes straight down to the bottom of the seabed, it is too much, the chain will not rise a lot in a gust or well. But there is an optimum as indicated in the graphs above where the gain in potential energy will be maximised. To me a good approach seems to be to make the chain about 25% longer than the snubber / bridle when not stretched. This way you know when they are reaching their limits.

Should you be comfortable with deploying (and – more importantly – quickly retrieving) a kellet, then placing a kellet midway on this stretch of chain allows for even more potential energy to be stored.

Finally, please note: Don’t be fooled by the diagrams that things are fine on the far left side, where the curves are all nicely tending to zero, eventually. The snubber will most likely be overloaded in this region! For most of us, there won’t be enough water left under the keel there, anyway… 😉

Also please note that snubbers / bridles are consumables. They have only a limited number of stretch cycles they can do before fatigue occurs, possibly resulting in them breaking when you least need it. They should be replaced at regular intervals. Their life span can be increased by oversizing them (without them going too soft).

At the end of this web page is a link to rather interesting articles on snubbers and bridles.

## The effect of kellets

In my view kellets are overrated. They are a pain to attach to the chain and to lower them to the point along the chain where they are most useful, and it is an even greater pain to haul them back in when in a storm and one needs to weigh the anchor in an emergency. Whether this is worth the slight reduction in swinging radius or the slight increase in elasticity of the chain is debatable for any reasonably sized kellet that can still be handled.

The most effective position of the kellet for reducing the minimally required chain length is to place it as close as possible to the anchor, but such that it is still lifted off the seabed. The characteristic parameter of the kellet is *k* = *F*_{k}/(*m* *g*), where *F*_{k} is its weight in water. At most, it can reduce the required chain length by *k* meters, but generally it is a good deal less than that. For example, a kellet of 12 kp in water and used with a 10 mm chain, which is 2 kp in water, will shorten the minimally required chain length by only less than 6 m.

If the kellet is more used to increase the elasticity of the chain, my hunch is it should be more positioned towards the middle of the chain. (More detailed analysis perhaps to follow.)

## The effect of multiple floating kellets – the art of Tuamotus anchoring

I got triggered by this most interesting anchoring technique that helps to reduce the damage done to corals when anchoring: https://www.spiritofargo.com/2019/08/14/floating-your-chain-what-is-that/

The idea is to use buoys to raise the chain above the seabed and thus avoid damage to the corals. Here is my attempt at an analysis of this technique:

As before, we work out the elasticity of the chain by calculating the potential energy as a line integral and then differentiate with respect to the anchor load *F*. In what follows we ignore the elasticity related to the final leg *L*_{bow}, between last buoy and bow, since this can be calculated independently and simply added later. So, our calculation stops at the last buoy.

Clearly, the buoyancy of the buoys has a very positive effect on the elasticity of this system, and the more buoys are being used, the better it is. However, all these curves have their peak at rather small values of *a*/*Y*, so well before a strong storm has developed. In fact, their peak is at even smaller values of *a*/*Y* than for the normal curve without any buoys (black). Secondly, all curves stop at a certain maximum value of *a*/*Y*. After this point, the pulling angle at the anchor is not horizontal anymore and the last buoy will dip under water. Again, the more buoys, the later this effect will happen, but even with 10 buoys spaced 15 *Y* apart, we barely reach a value of *a*/*Y* = 8, which corresponds to a no-buoy scope of *L*/*Y* = sqrt(1+2*8) = 4.1. And this requires a lot of chain, often well in excess of 100 m! The fact that the elasticity of this system eventually drops below that of a simple no-buoy approach is ok, since this graph does not include the elasticity of the last leg *L*_{bow} yet. It needs to be added. Finally, it is obvious that the larger *L*_{0} is, the better the system performs in a storm, but this first segment of the chain is limited by the perimeter around the anchor that is free of corals.

**It appears beneficial to keep the distance between the buoys as large as possible** (of course, without the chain segments in-between touching the corals when slack), and rather have two buoys bundled together and then have twice the chain to the next pair of buoys. With the same total length of chain, this moves the peaks to the right, even increases them slightly, and only marginally reduces the maximum value for *a*/*Y*.

So, this looks like a great approach to keep the chain tidy at all times and away from the corals. In a severe storm it is at a disadvantage compared to the normal way of anchoring, though, and sooner will depend on the anchor holding well even when it is being pulled at an angle. Also, the amount of chain needed in comparison to the normal anchoring method is very substantial. In the example above it is about 2…3 times that length. But then again, with that much chain out, the pulling angle at the anchor cannot get very large, at least when a couple of buoys have been deployed.

If we require the pulling angle at the anchor to be the same as when not using any buoys, then obviously, the **swinging radius** will be larger when using buoys. To see this, simply look at the first diagram above. If no buoys are used, the first leg from the anchor – extended to the water surface – will determine the swinging circle without buoys. Clearly smaller than when using buoys…

If we are still worried about the **pulling angle at the anchor in a severe storm**, we could opt to reduce the buoyancy of the first buoy, thereby accepting that the buoys will all sink slightly below the water surface even in calm weather. This seems to be done by some folks using this anchoring technique. Some further adjustments may then have to be made to make sure the chain is not interfering with the corals. In order to maintain the last buoy as a marker for when the pulling angle at the anchor starts to increase, any reduction of the buoyancy of the first buoy will need to be compensated for with the last buoy. But all this fiddling will not change the asymptotic value of the pulling angle, which is simply given by scope.

Alternatively, a **kellet** close to the anchor could be deployed. Although for ordinary anchoring kellets do not improve the situation a lot, for Tuamotus anchoring it could just be the additional few metres of imaginary chain needed in the first chain segment.

**So, to conclude this section:** It is generally beneficial to make the first chain segment *L*_{0} as large as possible to postpone the onset of the lifting of the anchor as much as possible, but there is a limit set by the proximity of the corals. Also, a kellet in this first chain segment can be very effective to this end. However, in storm conditions the anchor will inevitably be pulled upwards, but the pulling angle can still be small, if sufficient chain has been paid out. How much chain is needed to keep this angle below a given value is as usual simply derived from the scope of the total chain. Concentrating the buoys in fewer places and making the chain segments in-between correspondingly larger helps to move the peak of elasticity to larger values of *a*/*Y*, at a slight expense of the maximally achievable *a*/*Y*. **No matter how the buoys are placed, the sum of their buoyancy needs to match the total chain weight between anchor and the point half way between last buoy and bow.** If this condition holds, this last buoy will act as a marker for when the anchor starts to get pulled upwards: It will then dip below the water surface. In storm conditions, only the chain segment between last buoy and bow will contribute to the elasticity, but it seems prudent to use good snubbers / bridles, as this is by and large a shallow-water situation with insufficient chain when it really matters.

Note: Fenders tend to be poor buoys, as they cannot withstand the higher water pressure in, say, 5 metres water and collapse. Better to use hard-plastic buoys.

Finally a note of warning: If there is no wind at all and you keep drifting over your anchor position with your vessel, there is a serious risk of entangling the chain and creating a mess with the buoys. See this report.

## Using a mix of rope and chain

Most of my analysis has dealt with the case of a pure chain, possibly aided by a snubber or bridle. In that case it became very clear that a chain does not work well in shallow water and, because of this, if no elastic snubber / bridle is there to take over the work of absorbing shock loads, one may well have to relocate to deeper water in order to reduce the maximal anchor load. However, things are quite different when we use a mix of chain and rope (longer than the chain). In that case it turns out that the absorption of shock loads is almost entirely done by the rope, if it is elastic enough, whilst the chain’s only function is to create the initial bend off the anchor and away from the seabed. The chain does not absorb any energy at all. The table at the top of this page shows an overview of various scenarios. The anchor load is almost independent of the anchor depth. Why? Simply because the rope does not care at which angle it is pulling at the end of the chain, it will always work equally well. This is not the case for the chain, which works by gravity.

## How does all this compare with the work of Alain Fraysse?

A link at the end of this web page points to some seminal work that Alain did on anchoring, which is often referenced. Now, what is new or different in my work compared to what Alain did?

First of all, as far as the case of **static anchoring** is concerned, our model and formulas are identical. Alain accounts for a few more cases, though, like the use of a kellet. The latter I have only dealt with in the long German essay, but not in the digests, nor in the app.

When it comes to **dynamic anchoring**, I believe we are using different models, at least as far as I can tell looking at the models presented on Alain’s web page. I did not look under the hood of the excel spreadsheets that he provides. On his web page, to illustrate his approach, he compares the chain with a perfect spring and how that spring can store energy, but when I look at the non-sinoidal curves that his spreadsheet produces, I believe that under the hood he is using a better model for the chain. Perhaps it is also the same formula for the potential energy of the chain as I do, but I really do not know.

There is a definite **difference** in the models, though, when looking **how a gust or swell is modelled**. Alain does this in the time domain where he switches on an additional force that represents the gust or swell – but which never gets switched off again. In the absence of any dampening, this leads to an oscillation with a considerable overshooting in the force. And this is the force to reckon with as the maximal load.

In my model, on the other hand, I deal with **gusts** entirely within the framework of the static anchoring. This should work as long as a gust builds up gradually and does not lead to a massive overshooting. („adiabatically switching on“ would be the physics notion of this. 😉 ) Good snubbers or bridles will help a lot to dampen such an overshooting. If one is worried that there is some oscillation back and forth with respect to the anchor position, one can account for that by picking a wind value larger than the gust and then still use the formula for static anchoring. Alternatively, one can treat a gust in the same way as a swell by looking at the maximal velocity at anchor that is causes. In this case one would use the base wind as wind speed and not the gust wind speed, as otherwise one would count the gust twice.

Now, **swell** is modelled completely differently. Alain models swell as another force that is switched on, and stays on, whilst I model swell as a singular burst of energy that causes the vessel to pick up some speed. So, it is kinetic energy that needs to go somewhere, and not a force over a certain period of time. This kinetic energy needs to be stored in the form of additional potential energy in the chain as well as the snubber / bridle. And if there are two waves of swell one after the other before the first one had been dealt with, it is the sum of their kinetic energies that one has to deal with.

In the end, if the nonlinear spring properties of the chain that Alain refers to are also based on the potential energy of the chain, then these two approaches should give similar results. The results would agree better if Alain’s approach were modified such that the additional force gets switched on for a finite time only, rather than indefinitely, to simulate a burst of finite duration.

However, an energy-balance approach like I did is much simpler than tracking down forces in the time domain. After all, I am only interested in knowing the maximal loads. How the vessel got there, I do not care, really. Plus, I find it easier to estimate the energy of a swell, rather than the force it exerts on the vessel.

**British Admiralty guidance and commercial vessels:**

It is also interesting to compare this approach with the guidance found in the **British Admiralty’s Manual of Seamanship** for *commercial vessels*, which states *L* = 1.5 √*Y*, where *Y* is measured in metres and chain length *L* in shackles (90 ft). This guidance is **independent** of wind strength, windage area and chain thickness. It is – I believe – a guidance for up to 60 kn of wind strength. Consequently, for most scenarios, it turns out this guidance is a very conservative one, which means yes, one is on the safe side, but also, the swinging circle of the vessel is much larger than really needed. It is not a useful guidance for recreational vessels. You would not make any friends at most popular anchorages… 😉

For now I have created two explorative digests for **commercial vessels**. One for *A*_{eff} = 1000 m^{2}, and another one for 2000 m^{2}. The choice of chains has been such that they cover a wide range: 56 mm, 81 mm and 102 mm. Likely this means that not all combinations of *A*_{eff} and chain thickness make sense. However, it is meant to show the impact these choices have. For those interested in scaling laws: The 102 mm chain is about a factor 100 heavier than the 10 mm chain. Consequently, the curves for *A*_{eff} = 2000 m^{2} and a 102 mm chain are very close to those for *A*_{eff} = 20 m^{2} and a 10 mm chain – you’ll find that the parameter *a* is almost identical in both cases. 🙂

I have currently no real good feeling for how strong large commercial vessels move about at anchor when in a severe storm. Presumably much less than smaller vessels do. Waves that would get a small vessel into serious trouble would just bounce off a large vessel. For now I have just assumed some values and looked at the results, but with further insight this may require reassessment…

As expected, the guidance given by the British Admiralty is safe in the vast majority of cases. However, corner cases do seem to exist: Extremely large swell head–on, a very thin chain or an excessive windage area of the vessel may all result in the British Admiralty guidance being inadequate in a severe storm. But then again, professional ship designers will know that and fit their vessels with an appropriately thick chain! 🙂

In any case, though, the discussion above about the sweet spot when the chain is working best in terms of elasticity is relevant also for larger vessels – particularly so when they can only rely on the chain and cannot use any other means to absorb energy temporarily.

Please note that a variant of the BA guidance exists for heavy chains of certain type: *L* = √*Y*, where *Y* is again measured in metres and *L* in shackles (90 ft).

## Energy dissipating effects:

So far we have not really discussed effects that allow energy dissipation (as opposed to temporary energy storage) – and as such are useful for us, as they quench some of the energy coming in via swell or gusts. One such effect is related to the chain getting dragged through the water as the vessel pulls on it in all kinds of directions. This is a highly irregular movement and we will not attempt to model it. Just to hint at one complication: This dragging of the chain would have to be calculated relative to the water surrounding it, but in a swell, there is a lot of irregular motion of the water and this becomes a daunting task.

Still, to get some idea about what amount of energy dissipation we are talking about, let us look at a VERY simple model, where the chain is a straight line connecting the anchor with the vessel. Furthermore, we assume that the vessel is moving with a velocity *v* perpendicular to the chain (of length *L*), i.e., the chain is rotating as a stiff rod around the anchor, through motionless water. The energy dissipated per second by having the chain drag through the water, *P*, is then easily given by

*P* = ¼ ρ_{ch} *L* *v*^{3}

where ρ_{ch} is the chain’s dragging coefficient. For a 10 mm chain one finds experimentally ρ_{ch} =~ 1.2 N s^{2}/m^{3}. Thus, a 50 m chain that is moving with 1 m/s at its end is dissipating only about 15 J/s. This is not a lot! And worse, this energy dissipation only dampens the swinging of the vessel at anchor from one extreme side to the other, but it does not help to reduce the energy that needs to get temporarily stored in the chain and snubber in a shock load. Not directly, anyway.

This additional – and more relevant – energy dissipation happens when the chain gets straightened and lifted off the seabed. But again, this contribution will be at most in the same ball park as above, more likely less.

We thus conclude that this effect is minor and may well be ignored in the calculation. In any case, it works in favour for us… 🙂

Another place where energy can get dissipated is the anchor, when it drags through the soil without breaking free. This energy, which can be quite substantial, is simply given by anchor load times length of the drag.

## Reversing the engine when digging in the anchor – how much anchor load does this correspond to?

The general advice for anchoring is to set the anchor, then drift backwards due to wind or current, and slowly pay out more chain until the desired length is reached. Then slowly crank up the engine in reverse to get the anchor dug in. „Slowly“ is important here, as the anchor needs time to set, otherwise it would only get dragged across the sea bed. The interesting question then is how much load have I tested the anchor with in this approach? To get an approximate answer, let me do the following calculation: As physics dictates, work is force times distance, and so power is force times velocity. So I do the following measurement of my vessel: Whilst not at anchor and without any sails set, I put the engine in straight reverse and measure the velocity *v* at full engine power *p*. Ideally, this can be done with an apparent wind speed of zero, and no current or waves. The force *f* which is pushing the vessel through water is then simply given by *f* = *p* / *v*, and this should be the anchor load with which the anchor is dug in when using full reverse throttle when setting the anchor. The relevant power is not the nominal engine power, but the effective horse power after accounting for all losses in the system (within the transmission line from the engine to the propeller as well as the propeller itself). A simple way of dealing with this is to introduce an efficiency factor 𝜂. In our case, SV SAN, this all amounts to – very roughly – 60% * 75 HP / 8 kn = 8054 N = 831 kp as anchor load when digging in. This is quite reasonable when comparing with wind forces in a light gale. Some more food for thoughts can be found **here** and **here**.

In fact, when I use my **Anchor Chain Calculator **tool, it turns out that for our vessel this corresponds to 41 kn in a dead calm sea, or 35 kn in a choppy sea with a *Velocity at anchor* of 0.6 kn. When I remove the snubber, this number goes down to only 25 kn.

On the other hand, this little analysis should also be a warning to all those who believe that after having dug in their anchor by reversing their engine hard is enough and then nothing can possibly happen. The load the anchor did witness in this procedure is far less than in a serious gale storm!

## How much anchor load does a current / tide correspond to?

This is a difficult question to answer and I will only make an attempt for the special case of the tide coming straight head on and not at an angle to the vessel.

As in the previous section, I start by calculating the force on the vessel needed to push it ahead with a certain velocity *v*. The difference to the above case is that I now measure the velocity of the vessel under full forward thrust of the engine, again preferably with zero apparent wind, no current, no waves. The force is then given by *f* = 𝜂 *p* / *v*, where both 𝜂 and *v* may be slightly different to the previous case. Next, with Bernoulli’s law of a quadratic dependency of the resistance in a laminar flow of velocity *v*_{c} (the current / tide), I can work out the load the current / tide will generate on the vessel simply by scaling, *f*_{c} = 𝜂 *p* / *v* (*v*_{c}/*v*)^{2}.

So, for example, if for simplicity I take 𝜂 and *v* to be the same as in the above case, then a current of 2 kn would mean a load of 8054 N * (2/8)^{2} = 503 N.

But this is the special case of a current head on. When at anchor, normally wind and tide will mean that the vessel is neither with the wind nor with the current in alignment, and so the load generated by the current will be substantially larger.

**Useful links:**

**My Anchor Chain Calculator App: **The app is available for iPhone and iPad in the Apple Store! Android version available in Google Play Store! A brief introduction to the App can be found in English, auf Deutsch, and en Français. A tutorial video can be found here. A free online version of the app with reduced functionality can be found here. It is excellent for playing out what-if scenarios. The Apple and Android apps can also deal with a mix of rope and chain.

An online calculator for the chain / rode length that appears to be using the same catenary approach and same model for the energy burst to be absorbed as potential energy of the chain as I do. It can also include the effects of snubbers and bridles, and can also do a mix of chain and rope: http://svamanda.dk/anchor/intro

Another online calculator which also summarises a lot of the underlying maths: https://awelina.com/old_site/anchor_rode/rode_length_graph_only.html

A set of EXCEL sheets to calculate chain / rode lengths, anchor loads as a function of time, you name it. Gusts and swell are modelled differently than I have done it: http://alain.fraysse.free.fr/sail/rode/rode.htm

A simple catenary calculator (no swell) can also be found here (with an obvious mistake in the formula shown, but the program appears to be correct): http://abc-moorings.weebly.com/catenary-calculator.html

This one I find a bit difficult to use (plus it is operating with ft and lb, which is not my world 😉 ). Also, it seems not geared towards anchoring: https://www.spaceagecontrol.com/calccabl.htm

A very nice snubber calculator can be found here: https://www.snubberhead.com. It uses the same approach as I do, to take the vessel velocity at anchor as a measure for the kinetic energy the vessel has attained, which then needs to be absorbed in the anchor gear.

And here is a simple calculator for working out the wind force given wind speed and effective windage area: https://www.engineeringtoolbox.com/wind-load-d_1775.html

Some empirical data on anchor loads: https://www.practical-Asailor.com/sails-rigging-deckgear/the-load-on-your-rode

And the famous late Robert Smith’s Ground Tackle Load Tables interpolating real measurements of anchor loads for a variety of vessels: http://www.plaisance-pratique.com/IMG/pdf/30-17-Smith.pdf

This publication contains already a lot of the essential results presented by me as far as static and dynamic anchoring is concerned, or the advice to flee into deeper water when the anchor loads induced by swell are getting too large: http://northpacificresearch.com/downloads/anchor_load_revealed.pdf

Some very useful tips and interesting tests of snubbers / bridles can be found here: https://www.practical-sailor.com/sails-rigging-deckgear/anchor-snubber-shock-load-test and here: https://www.sailmagazine.com/cruising/anchor-snubber-tips

My favourite anchor testing source, SV Panope: https://www.youtube.com/user/flygoodwin

Discussion on Tuamotus‘ Anchoring technique: https://www.cruisersforum.com/forums/showthread.php?p=3202648&posted=1#post3202648

Whilst I disagree with Peter’s thoughts and statements about scope and the absence of catenary when it matters, this article on tandem anchors is very good: https://www.petersmith.net.nz/boat-anchors/tandem-anchoring.php

## Dieser Beitrag hat 24 Kommentare

Köstlich. Der Kettentrick klappt mit Physikern genauso wie der Fussball mit dem gemeinen Volk. Unwiderstehlich.

Der alte Werner von Siemens hat sich in seiner Autobiographie auch sehr über das Ausbringen von Kabeln auf hoher See (oder eher tiefer See) ausgelassen. Siehe

https://archive.org/details/lebenserinnerung00siemiala/page/n6

Ein paar Schmunzelaspekte:

– Schön, dem Leser zu erklären, dass << "sehr viel kleiner" heisst, aber gleichzeitig anzunehmen, er kenne den Sinus hyperbolicus.

– Eine Redewendung, die jeder versteht, die aber ausgerechnet hier fehl am Platz ist: "Die Kettenlänge von Bug zu Wasser fällt nicht ins Gewicht" oder so ähnlich.

Falsch: ins Gewicht fallen ist das Einzige, was sie tut, aber es hat keine Bedeutung.

– Die Präzision der Gravitations"feldstärke" (hat der Kurs in Quantenfeldtheorie doch was genutzt) wird dreistellig angegeben. Also musst Du noch Korrekturkurven für den Äquator mit geringerem g berechnen. Am Nordpol gibt's ja keine Segler.

– Vergiss nicht H.'s Ermahnung "Machen sie den linearen Ankerresponse nicht schlecht" Im Deliusverlag werden sie nicht gut auf Dich zu sprechen sein, wenn Du ihren Autoren auf die Füsse trittst.

Freue mich schon auf die Artikel, die Du auf der Reise schreiben wirst.

Bin’s nochmal. Warum die ganze Arbeit, wenn Anker sowieso nicht funktoinieren. Das steht hier: https://www.science20.com/chatter_box/why_anchors_dont_work

Wer die Logik der Renormierungstheorie der kritischen Phänomene in sich aufgesaugt hat wird damit auch keine Probleme haben. Dieser Post war auf der ersten Seite meine Google-Suche „anchor ship“ vor einer Woche. Mittlerweile findet man ihn gar nicht mehr (bei derselben oder ähnlicher Suche). Das versteh, wer will.

http://www.petersmith.net.nz

This guy really knows his anchors and catenaries….professional anchor designer 🙂

Ich vermisse bei der Betrachtung zum Ankern die Stromkomponente und den Tidenhub. Greetings the dentist

Stimmt! Strom kann man einfach per Kräftezerlegung mit hinzunehmen. Tide ist ein Problem. Da muss man gucken, was bei Hochwasser noch geht. Und dass man auch nicht aufsitzt. 🙂 Es kommt bald ein Update, wo ich auch noch dynamisches Ankern mit berücksichtige… Das ist ganz spannend in flachem Wasser…

The chain looks bright and shiny, is it still bright after being used a bit and being out on the high sea? The question in my mind is how long is that chain and what Anchor is about the end? Could it perhaps be a Bruce?

Must not be a Bruce or Mr. Murray would have commented on using „his“ Anchor which has been perfected since he engineered it back in the 70’s. I l always knew Bruce was a fount of knowledge and should not be surprised at his expoertise in anchoring. Perhaps it is a Rocna?

Regardless, as Bruce’s link illustrates the chain holds the key to a properly functioning anchoring system, much like life. The man may be an anchor but the woman turns the neck and therefore controls the „chain“ of sorts. Bon Voyage Mathias from your port in Portugal. Hope the chatter puts a smile on your face and family, until we meet again in person on the west coast of USA!

Hi Mark, the chain is still shiny now in Panama! 🙂 It is a duplex chain, so it should stay shiny for a while. The anchor is a 35 kg Spade anchor.

We will see how long it will take us to get to the west coast of the US, being stuck in Panama now due to Corona…

But, I have now discovered that I can answer in these blogs, so … 🙂

Der Link zum deutschen Artikel in den Szenarien führt wieder auf diesen englischen Artikel zurück.

Hallo Reinhold, ja, wenn man auf den Link in den Texten klickt, kommt man auf diese Seite wieder zuück und muss dann das Bild anklicken, um zum deutschen Text zu gelangen. Alles andere war aufwendig / hat langfristig Nachteile…

Für alle Interessierten, Deinen Artikel gibt es in deutsch auf:

https://www.blauwasser.de/ankern_kettenlaenge

Hallo Reinhold, ja, das ist richtig, aber nur den statischen Teil. Das dynamische Ankern ist erst später hinzugekommen. Einen langen Artikel auf Deutsch kann man aber auch hier auf unserer Webpage runterladen. Dazu einfach das Bild oben anklicken.

Hallo zusammen. Habe versucht das zu verstehen und die Kettenlängen für unser Boot zu bestimmen.. bei der Windangriffsfläche für unser Boot zu bestimmen bin ich dann gescheitert… ich mach wieder die Formel 4x die Wassertiefe. Hat auch bei 35Kn Wind gehalten in den Bahamas…

Viele Grüsse nach La Playita von Ecuador

Hallo Willi, welches Boot hast Du denn? Vielleicht kann ich da etwas helfen?

Wie ist es in Ecuador? Corona scheint da ja recht schlimm zu wüten… 🙁

Cheers, Mathias

Envirinmentally speaking, it would be nice if you included information of floating your chain. See:

https://www.spiritofargo.com/2019/08/14/floating-your-chain-what-is-that/

Oh, thank you so much for pointing out this technique! Absolutely fascinating! I do agree that we need to do anything to help the corals and not destroy them. In some sense, what you are doing is to attach kellets with negative weight. So that will have a negative effect on the length of the chain that is required, but I need to think about this a bit more. The farther away form the anchor traditional kellets are, the less beneficial they are, and so, reversing this argument, your buoyant kellets will also have lesser negative impact the closer they are to the vessel.

Cheers, Mathias

Tolles Engagement. Danke.

Für mich persönlich zuviel Text, zu kompliziert, zu unübersichtlich.

=> Praktische Anwendungswirkung verpufft leider.

OK, danke, dann muss ich daran noch arbeiten. Es ist zugegebenermaßen momentan immer noch ‚work in progress‘.

Zunächst: ich habe gerade nur den Beitragsauszug in der Yacht gelesen, und da vermisse die Komponennte „Wellen“.

Hier nur mal nach „Wave“ gesucht, gefunden, aber nicht gelesen. Sorry.

Mit dem Wind nimmt die Wellenhöhe i.d.R. zu, und die Kräfte, die wirken, wenn die Kette nicht mehr am Boden liegt sind gewaltig. Deutlich größer als die Windkräfte, da der gesamt Auftrieb am Bug an der Kette zerrt..

Wenn es wirklich ernst wird, etwa weil die zuvor sicher angenommene Bucht durch einen unerwarteten Winddreher zu Falle wird kann es eng werden.

Zudem: Wenn es wirklich kräftig bläst, wird zwar die Leine früher straff als die Kette, aber sie bleibt elastischer und kann harte Rucks vom Anker fern halten. Und da sie i.d.R. auch länger als die Kette gesteckt wird, bleibt der Winkel auch flacher.

Insofern halte ich für den schlimmsten Fall die ca. 7m fache Wasserteife (soweit möglich) als Länge nach wie vor besser, wenn den die Wind Und Wellenkräfte dazu führen, das Leine oder Kette nicht nach der Kettenlinie, sondern einfach eine Gerade bildet. Der Winkel bleibt unter 10°.

Für das gewöhnliche Ankern bei moderaten Verhältnissen mag ich dagegen Deinen Gedanken folgen.

Hallo Rolf,

Vielen Dank für das Feedback. Ich spreche nie von Wellen, sondern von Schwell, und dieser wird als Eintrag von kinetischer Energie betrachtet. Das ist eine Schätzung, ganz klar, aber es wird sonst leicht exponentiell schwierig.

Ich denke, mit der Leine meinst Du den Ankerstropp? Also Snubber auf Englisch? Ja, der bringt in flachem Wasser sehr viel, wenn er genügend lang ist. Er muß ordentlich Energie aufnehmen können, und das geht nur über Länge. Ohne Snubber ist die Belastung für Anker und Bugklampen einfach viel zu groß. Egal ob 7:1 oder nicht.

Es kann gut sein, dass effektiv ein 7:1 rauskommt. Meine Sichtweise ist jedoch, dass man nicht mit 7:1 starten sollte, sondern sich das physikalische Model und die Parameter genau anschauen sollte. Es kann z.B. gut sein, dass man gar nicht 7:1 braucht und mit weniger Kette auskommt, auch bei viel Wind. Gerade wenn es eng wird in der Bucht, ist es wichtig zu wissen, wo die Grenzen sind. Dass man hier noch einen Sicherheitszuschlag drauf legt, ist selbstverständlich.

Thank you for all your analysis! You might want to visit the catenary page at https://www.morganscloud.com/2018/12/10/anchor-chain-catenary-when-it-matters-and-when-it-doesnt/. This is a pay site, but it has many well-written articles on all aspects of cruising and is only about $20/yr. The comment threads on the articles are particularly good, and I think your comments on anchoring would be welcome.

Thank you Charles for this feedback and the suggestion. I am always hesitant about pay sites, but will think about it.

BTW – I am currently working on a simple app for iPhone / iPad to do my calculations, at least for one chain segment. A bit struggling with object oriented programming… 😉 But the number-crunching engine is working already… 🙂 So, perhaps worth checking back here occasionally.

Cheers, Mathias

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Eine kostenlose Möglichkeit, die App zu probieren, gibt es nicht. Irgendwo ist mal was von einer Web-App gestanden, finde ich aber nicht.

Gruß Edwin

Hallo Edwin, Ja, an der Web App arbeiten wir, aber wir haben enorme Schwierigkeiten, die online Version auf unserem Web Server laufen zu lassen / bzw. ist es nicht zugelassen, und so müssen wir einen anderen Weg finden. Mir ist klar, dass diese Web App wichtig ist, um mal damit rumzuspielen, aber momentan scheitern wir noch an den IT Problemen. Irgendwann werden wir das lösen, aber eigentlich wollte ich Weihnachten 2020 schon so weit sein… 🙁

Sorry, VLG, Mathias