View Full Version : Anchor rode length calcs

Can someone please point me to a calculation method for chain rode length/ depth of water. I'm not looking for 3:1, 4:1, 5:1 etc, but one that I have seen on this forum a few times that I think depends on the square (square root?) of the depth, rather than depth itself, to increase proportionately for shallower water.

Links to various spreadsheet and graphical methods not required, thanks, I reckon I have them all.

Thanks for all replies.

I work on three times the depth of water, but I have no means of measuring, so I look for a 45% angle off the bow. Then watch the plotter to see if I am moving. If you only have 100 feet in the locker then do not try anchoring in 100 feet. IT WILL NOT WORK.

I work on three times the depth of water, but I have no means of measuring, so I look for a 45% angle off the bow. Then watch the plotter to see if I am moving. If you only have 100 feet in the locker then do not try anchoring in 100 feet. IT WILL NOT WORK.

Small plastic cable ties are good at 10 m intervals.

We never use less than 15m of chain however shallow the water, however light the wind and however little current. I would rather do that than get out a calculator and then allow for the bit between the roller and the water and remember if the cable tie was meant to be in the roller or on the surface.

We have no marks at all in the chain until 15m. If it is that tight with other boats or a steep side I would be nervous about anchoring there at all.

Can someone please point me to a calculation method for chain rode length/ depth of water. I'm not looking for 3:1, 4:1, 5:1 etc, but one that I have seen on this forum a few times that I think depends on the square (square root?) of the depth, rather than depth itself, to increase proportionately for shallower water.

Links to various spreadsheet and graphical methods not required, thanks, I reckon I have them all.

Thanks for all replies.

IIRC it's 12x the square root of the depth for all chain rode, and 20x for a chain/rope combination.

This gives more scope than a linear formula in shallow water, and correspondingly less in deep water.

Can someone please point me to a calculation method for chain rode length/ depth of water. I'm not looking for 3:1, 4:1, 5:1 etc, but one that I have seen on this forum a few times that I think depends on the square (square root?) of the depth, rather than depth itself, to increase proportionately for shallower water.

Links to various spreadsheet and graphical methods not required, thanks, I reckon I have them all.

Thanks for all replies.

some rationale about where the square root of depth comes from can be found here, together with one version of the formula

http://www.anchorwatch.co.uk/index-page04.htm

fwiw, I go depth+freeboard times beaufort strength :smile:

Quote: fwiw, I go depth+freeboard times beaufort strength :smile:

What a brilliantly simple idea. I almost always put out more, but that is neat! Thanks!

Thanks to the last 2 posters for their formulae. I had been 'lurking' on this thread with interest. That helps.

Do the formulae change with either a.) a bower, genuine 25lb cqr that does work, plus 30m chain and miles of anchorplait, c60m, - versus b.)a kedge that is a genuine bruce c.15lb, plus miles of anchorplait, c40m. I try to anchor no deeper than 10m usually less, and I have a high windage Heavenly Twin.?

Thanks. Mike

Ubergeekian

06-10-09, 18:50

IIRC it's 12x the square root of the depth for all chain rode, and 20x for a chain/rope combination.

This gives more scope than a linear formula in shallow water, and correspondingly less in deep water.

I avoid these formulae because they are not scale-independent. Anchored in ten metres? That's 38 metres of chain. Anchored in five fathoms? That's 27 fathoms of chain - 40% more.

My personal rule is simple: three times depth for chain, unless it's blowing up in which case every last foot of chain goes out.

My personal rule is simple: three times depth for chain, unless it's blowing up in which case every last foot of chain goes out.

3 times depth AT HIGH WATER taking into account any echo sounder offset plus the height of the bow roller from the water. If it isn't going to blow.

Fair enough, but why keep any chain in the locker if you have swinging room?

- W

Very simple - always aim to have 2 X length of boat lying on the bottom (regardless of depth) plus (obviously) length equal to max depth at high tide. I have 11m boat and therefore typically use 35 metres in almost all conditions. Remembering that of course chain left in the anchor locker does no good at all!

Footnote: the 3x or 6x formula is basically flawed as it does not take account of shallow or deep water. For an anchor to work it needs a good long horizontal pull and a decent catenary - difficult in shallower water unless you use some kind of 'angel' (or disproportionately more chain)

Fair enough, but why keep any chain in the locker if you have swinging room

All the chain + rope for depth at HW generally works for me ;-)

(That's 10m chain, and I normally anchor from 1.5m to 5m)

Ubergeekian

06-10-09, 23:13

3 times depth AT HIGH WATER taking into account any echo sounder offset plus the height of the bow roller from the water. If it isn't going to blow.

Fair enough, but why keep any chain in the locker if you have swinging room?

Why not?

Can someone please point me to a calculation method for chain rode length/ depth of water. I'm not looking for 3:1, 4:1, 5:1 etc, but one that I have seen on this forum a few times that I think depends on the square (square root?) of the depth, rather than depth itself, to increase proportionately for shallower water.

Links to various spreadsheet and graphical methods not required, thanks, I reckon I have them all.

Thanks for all replies.

.................................................. ..................................

Think the answer is, there is not one, think you have to be a nerd to ask for one. There is no equation that can answer the question, except more chain is better than less. I never have a problem anchoring, but then I never asked stupid questions. I normally anchor in about 5/6 meters with 30 meters out. I dont try to put out the critical minimum, what ever that might be. If over night, I might well let out all 60 meters. It's like how much hand brake pressure to be applied to a car.

It is all about the angle of pull on the anchor when the boat is pulling as hard as possible.ie rode straight.

So if you draw a 2 dimensional diagram you will see that at rode twice as long as depth you get a 45 degree angle (very bad). If you have the rode 3 times the depth the angle changes dramatically to something less than 30 degrees and again if you go for 4 or 5 times the depth the angle decreases but at less reduction of angle as you increase rode length. So the real improvement value is in the 3 to 5 times range.

In practice it is the chain weight that also improves the angle at the anchor. If the boat is not tugging so hard then the chain lays on the bottom and the angle becomes zero. The more chain on the bottom the harder the boat can tug without lifting all the chain off the bottom.

In the end it is a question of how hard the boat tugs and how worried you are about drift.

(NB not a single mention of anchor type) olewill

Can someone please point me to a calculation method for chain rode length/ depth of water. I'm not looking for 3:1, 4:1, 5:1 etc, but one that I have seen on this forum a few times that I think depends on the square (square root?) of the depth, rather than depth itself, to increase proportionately for shallower water.

Links to various spreadsheet and graphical methods not required, thanks, I reckon I have them all.

Thanks for all replies.

There are several simple square root of depth formulae around that ignore snatching, but this printed in YM and PBO several times might just be what you are looking for:

http://www.anchorwatch.co.uk/index-page04.htm

Thanks for all replies, both helpful and arrogant ones. Contrary to the latter, I am neither a nerd nor inexperienced, anchoring for most of six months every year. The purpose of the question was research for a mag article, trying to find a method that takes into account the increased tendency for chain to lift from the bottom in lower wind speeds when using a linear method.

Thanks for all replies, both helpful and arrogant ones. Contrary to the latter, I am neither a nerd nor inexperienced, anchoring for most of six months every year. The purpose of the question was research for a mag article, trying to find a method that takes into account the increased tendency for chain to lift from the bottom in lower wind speeds when using a linear method.

FWIW, I tried applying the "anchorwatch" method I posted above to my boat, while it gives a theoretically satisfying scope decreasing with increasing depth (that's where the square root of depth comes into play), it also says that I should use a scope of only 2 while anchoring in 14m of water with 15kt wind (30m of chain needed for borderline catenary)... which appears very low indeed, maybe too low?

More funnily, the formula states for example that with a 5knot wind I should use 8m of chain with 10m depth...

Though the "anchorwatch" approach seems very sensible to me, I guess there might be some flaws ?

jimbaerselman

08-10-09, 14:49

Vyv,

I'm sure you've done your research and found Alain Fraisses (SP?) anchoring sums. Paraphrased very heavily, his point is simple. The biggest rode stresses occur when there's enough wind to cause a vessel to swing from side to side. These tacking stresses on the rode are then so high that a chain will temporarily lie with no significant catenary, and you can assume a straight line from stem to anchor shackle. If that angle is more than a certain critical angle (which varies between anchor types from 10 to 15 degrees) the anchor will work out of the ground.

Those considerations bypass wind strengths, catenary sums and square roots . . . and shift the emphasis to nice long springy nylon snubbers . . . but I bet you knew that already.

I subscribe to the view that a standard length should be laid on the sea bed and then add the depth of water. That length will depend on conditions and the nature of the bottom. If there is plenty of swinging room why not use your chain. It's no good in the locker!

Have just produced calculations for my boat using a spreadsheet using the Anchorwatch formula for a number of depths and wind speeds. Didn't get Roberto's problem at shallow depth and have double checked both my spreadsheet and the Anchorwatch algebra both of which are fine.

Flaws I can see are:

What about the force of the a tidal currant?

What about a sustained yaw on the wind caused by a tidal current working against the wind and thereby increasing the windage of the boat?

What about the holding?

All represent potential risks which can only be obviated by veering more cable.

The advantages of putting out the minimum required amount of scope are that your boat will swing less and you will have less cable to recover, either chain or rope. I have an angel and will factor it into the calculations once I have weighed it in order to understand its utility.

Whatever, if in doubt veer more cable and take some bearings or note some transits.

Really useful and interesting thread, thank you vyv.

Vyv,

I'm sure you've done your research and found Alain Fraisses (SP?) anchoring sums.

Here (http://alain.fraysse.free.fr/sail/rode/rode_b.htm), if anyone is looking for it.

The formula for chain, for the chain to be just horizontal at the anchor, is:

L = sqrt(D^2 + 2 x Lambda x D);

Where L = length in m

D = depth in m

Lambda = horizontal pull in kg you want to sustain / weight of chain per m in water.

That this has the correct characteristics is clear: for low force and very deep water all formulae must approach L = D (more precisely, dL/dD ==1 for large D). It is also dimensionally correct. It can be derived it from first principles, allowing only that the chain be flexible, using integral calculus and using the identity sinh(x) = (e^x - e^-x)/2.

Some measured numbers for Lambda:

On my boat chain weight is approx 2kg/m in water, and I estimated horizontal force at 13kts wind (F4 mean) as 64kg (I count in wenches, 64kg = 1 wench). So Lambda = 32. This is eerily close to L = 12 sqrt(D) and not too far from L = 3xD + 10, which is why these other formulas are reasonable approximations.

Don't forget that the force increases with the square of wind velocity. Graphs available if interested.

saltwater_gypsy

11-10-09, 20:16

An alternative approach is to find out the maximum angle to the sea bed that your particular anchor will tolerate before breaking out. As I recall this angle for a Bruce anchor is about 8 to 10 degrees.

If you are in shallow water (say typically 3metres in the Med.) there is a good likelyhood, in a strong wind, that the chain will be pulled taut so that there is no appreciable catenery.

If you do the trigonometry, this means that the scope of the chain will be about eight times the depth to keep within the 10degrees limit. I would be quite happy , if space allows to put out 30metres.

In deeper water, the calculation becomes more complicated as the chain will take up a catenery curve before terminating at the anchor at the required angle and this is where the three to five times depth rule comes from.

I previously had a spreadsheet which allowed these calculationsto be made with a mixture of chain and rope and also with a "chum". I can dig this out on another computer given a little time.

PS found it! http://alain.fraysse.free.fr

A suitable amount of scope to put out is dependent on many factors such as:

How much scope have other boats got out?

What is the expected wind strength?

What is the holding like?

What Fixed hazards will I swing into with more scope?

If there are no boats and hazards close why not let out lots, chain in a locker is not going to do any good. The point of diminishing returns for the maximum holding power of an anchor is not reached in until you have something like 14/1 out.

In practice other boats and hazards often limit the amount of scope that is desirable.

3X max depth (to the bow roller)+ 10m is a good formula (and avoids mental square root calculations), but let out more if you can.

jimbaerselman

13-10-09, 11:09

The formula for chain, for the chain to be just horizontal at the anchor, is:

L = sqrt(D^2 + 2 x Lambda x D);

Where L = length in m

D = depth in m

Lambda = horizontal pull in kg you want to sustain / weight of chain per m in water.

Your formula is fine. However, it very rarely applies - only in light and steady wind conditions. The normal situation is an oscillation. Initially a boat at anchor will set up a small fore and aft surging oscillation, which, in stronger winds, becomes a side to side 'tacking' oscillation.

The forces in the rode when movement is reversed (reversing the boat's momentum) are many times higher than any static forces, making static forces irrelevant. And they will routinely remove all significant catenary from the chain. Bar taught, you might say. Hic.

I agree! Except not necessarily that it applies to light conditions only. The issue I think is more that gusts and snatching make a huge difference. especially in shallow water. However where I do think it at least qualitively useful is that it shows that in deep water (say 25m - a depth I do anchor in from time to time, mostly in Scotland) one needs less chain than most rules of thumb would predict. Hence I encourage people to be less worried about anchoring in deep water - 3 to 4x depth is over pessimistic, you don't actually need as much chain as you think.

However Vyv asked for the formula, and I'm a physicist, and it was lunch time...

PS:

...The forces in the rode when movement is reversed (reversing the boat's momentum) are many times higher than any static forces, making static forces irrelevant. And they will routinely remove all significant catenary from the chain. Bar taught, you might say. Hic.

I've been wondering about this for some time. I have even designed a strain guage and inclinometer for my chain to measure Horizontal and Vertical components of the force, but as yet am too lazy to actually make it and record the data. Do people think it worthwile doing this experiment? The modelling done by Alain (surname escapes me) is fantastic, but is modelling only, whereas I want to know what it is in practice for my boat.

Instead what I have done is put a very long and light line out into shallow water, so the rode is weightless and horizontal, and use this to estimate horizontal component of anchoring force. I was surprised that I could hold it, the force being only about 60kg peak I estimate in a wind gusting to 20kts (this was in the Isle of Muck: great place). So in summary while the force is clearly not uniform while shearing around, peak force may not be as much as several times the static force; that must be a function of the period and amplitude of the oscillation, which are in turn dependant on the Q or damping, which is probably affected by the hull form and boat windage. Thus techniques to increase damping may be more useful than longer rode; in the Isle of Muck anchorage mentioned above I let chain from my bower lie on the bottom while lying to the kedge, and the damping effect was astounding with shearing all but stopped.

jimbaerselman

13-10-09, 16:23

The forces exerted when tacking are very much dependent on the elasticity of the rode, and the characteristics of that elasticity.

All chain rode offers trivial horizontal force as it first lifts off the ground, insufficient to stop the boat paying off from the wind. Then a first significant force comes in by the time it's angled at 30 degrees to the horizontal, shown by the yaw rate slowing. From then, force increases asymptotically over a mere further metre of movement as the rode lifts from 30 degrees to 15 degrees from the horizontal (simple geometry). The jerk detected standing at the foredeck (yes, that's me) is sufficient to unbalance most people in 20 to 25 kts of wind in fin keel yachts.

You're right, damping helps enormously. Yaw rate damping is high in long keel boats, so it takes a lot more wind for them to build up an oscillation in the first place. I guess hanging a long loop of chain over the bow would have some damping effect too. But both just postpone the wind speed at which unacceptable yawing sets in. Better is a forked moor.

Exerting that yaw force over a longer period of time also helps - 20m of nylon give you a linear increase of force over a 4m stretch, a much easier ride than applying the same momentum change over 1m distance!

It's all simple mechanics and geometry.

I've never looked at the deep anchoring situation (>20m) and why the ratio of scope to depth is commonly reduced. But I'm sure the explanation will lie in the fact that there's a greater roaming area, so a longer time to apply smaller forces, and lots of water damping hauling all that chain backwards and forwards. I still suspect, however, that a critical windspeed will be reached when that all counts for nothing.

timbartlett

13-10-09, 16:37

One of the funny things (I think) about these kinds of calculation is that factors such as the size and material of the chain or the buoyancy of the anchor seldom get a look in.

Buoyancy of an anchor?... imagine two identical anchors, one of steel and one of aluminium. If the steel one weighs 15kg, the ally one is probably about 5kg. But when immersed in water, they both experience a reduction in weight due to buoyancy. It's probably about 2kg each, so the steel one reduces to 13kg, while the ally one drops to just 3kg

The maths for different sizes or densities of chain is more complicated -- but it is sufficiently significant that the Admiralty Manual of Seamanship quotes different formlae for wrought and forged cables (forged cable needs 33% more than wrought cable)

What complicates things, at least to my mind, is where the bottom slopes - eg last week in Turkey (SO 36i) I dropped my anchor in 20m, but by the time I had let out 50m of chain, I was in 5m, with a line ashore about 20m long. I felt secure, but would be interested in any rules of thumb for dealing with sloping bottoms.

Regards

Neil

Artic Warrior

13-10-09, 20:11

3 times depth AT HIGH WATER taking into account any echo sounder offset plus the height of the bow roller from the water. If it isn't going to blow.

Fair enough, but why keep any chain in the locker if you have swinging room?

- W

right with ya there....let it all out, if there's space..can do no wrong

What complicates things, at least to my mind, is where the bottom slopes - eg last week in Turkey (SO 36i) I dropped my anchor in 20m, but by the time I had let out 50m of chain, I was in 5m, with a line ashore about 20m long. I felt secure, but would be interested in any rules of thumb for dealing with sloping bottoms.

Regards

Neil

It is the depth of water where the anchor is that is important. So if you anchor was dropped in 20m (from anchor roller to the bottom) you only had a ratio of 50/20 :1 (2.5 : 1) despite ending up in 5m water.

Anchoring into a slope (as you did) does however significantly help the anchors holding, conversely anchoring down the slope hurts holding. The effect of the slope is difficult to quantify and is very dependent on the local bottom topography directly at and behind the anchor.

I would not have felt secure in your situation, unless I was confident that the wind was going to stay from behind, or remain reasonably light.

right with ya there....let it all out, if there's space..can do no wrong

apart from having to lug 100m of 10mm chain back in again? You guys must have forearms like gorillas unless you are softies with electric winches :)

Vyv, excellent post. It's not something that had occured to me.

FWIW, I use 200-300 foot of rope plus 30 foot of chain when anchoring in 100+ foot when I'm fishing. The 20 x sq root depth makes sense.

Can someone please point me to a calculation method for chain rode length/ depth of water. I'm not looking for 3:1, 4:1, 5:1 etc, but one that I have seen on this forum a few times that I think depends on the square (square root?) of the depth, rather than depth itself, to increase proportionately for shallower water.

Links to various spreadsheet and graphical methods not required, thanks, I reckon I have them all.

Thanks for all replies.

There are many formulas and rules of thumb. In truth none are truly scientifically derived and many based on poor data. The correct amount of rode is dependent on the weight of anchor chain, weight and displacement of the vessel, type of bottom, type of anchor, wind and tide strengths and wave action. There is no single answer. In calm conditions and relatively shallow water the minimum is usually considered to be 3 times depth. This should be increased, if space allows, when conditions worsen and there is no theoretical maximum.

In practice the maximum is determined by how much chain you can reasonably carry. If this is not enough the effectiveness can be increased by using and angel or chummy as discussed in a recent thread.

Thanks again to everyone who responded. Sorry for delay, have been travelling back from Greece for the past week and was not able to get online.

The question relates to an upcoming question/answer in YM. I have used the Alain Fraysse, John Knox and Racnor pages on this subject many times, for answering other YM questions and for my own benefit. However, the calculation method I recall has not come up this time. It was possibly something the RYA came up with originally, and has been posted several times to my recollection. I have enough info to answer the question but would still like to see the method I recall, unless everyone has now had enough of the topic.

On the post about comparisons between calculated and measured load values, I answered a query on this topic in YM a year or so ago. The ABYC theoretical method seems to have several safety factors built in and it consequently predicts high loads. YM printed an article a couple of years ago, written by a man who runs a sailing school on W coast of Scotland. He had monitored actual loads using Anchorwatch permanently monitored by an onboard computer, in parallel with directly measured wind speeds. The article included a graphs of measurements over two years. Measured loads were surprisingly low, less than 200 kg in wind speeds to force 9 for a boat between 35 - 40 ft. I tend to agree with these figures, as I have used the same nylon snubber for nearly 20 years, a length of 11 mm nylon doublebraid with a breaking force of about 2.5 tonnes. In this time I have anchored several times in winds of 50 knots but the line has never given any concern. Most theoretical predictions would have snapped the snubber and probably pulled the cleat out of the deck.

tudorsailor

14-10-09, 19:46

My biggest problem in anchoring around Mallorca is that the anchorages are quite crowded. I let out 5:1 scope but doubt that my neighbours do, so worry that when we swing with the changing wind that we may meet. No real answer to this other an watch as others anchor and try and calculate how much they have let out and estimate the distance from them to me! No formula for this!

Generally hate anchoring as a result.

Tudorsailor

My biggest problem in anchoring around Mallorca is that the anchorages are quite crowded. I let out 5:1 scope but doubt that my neighbours do, so worry that when we swing with the changing wind that we may meet. No real answer to this other an watch as others anchor and try and calculate how much they have let out and estimate the distance from them to me! No formula for this!

Generally hate anchoring as a result.

Tudorsailor

Yes that bothers me too. Only way round it that I can think of is to ask people nearby how much chain they have out [assuming they are onboard to ask] and if I don't like it, I move somewhere else.

I don't hate anchoring - only anchoring near anybody else! :D

silver-fox

14-10-09, 20:24

Vyv

Not central to your current research on rode but it has been mentioned that yawing/sailing at anchor can be a big problem that increases loads on the anchor significantly. This is a much worse problem with the modern short keel fin and skeg designs

Some recommend a riding sail as a cure.

I would love to know what the trade off is when you fly a riding sail. Clearly that will keep the bow to the wind which reduces forces but also increases windage which will increase them again!

Does anyone know of any information on this. Are riding sails a good idea or is the cure worse than the problem?

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