Interesting topic and great to see so many different techniques and opinions.
After a not so pleasant dragging experience in a storm I wanted to figure out why we were dragging and I got almost as many different opinions as there are colors in the rainbow. Being a pilot I knew that mathematics had to be involved, but I could not find any info until the moment I ran into a book called: "Dass ist Motorboot Fahren' written by Joachim Schult. It is in German, but for me that is no problem.
He explained the mathematical part behind anchoring and when you read that you realize that all the different scopes are just quick reference guides for people who don't want to calculate.
After some more searching I actually found a complete calculation program where you can fill in your boat specifics, the weather, the current and the waves. The program will then tell you what the forces involved are and thus you can figure out what you will need to set in anchor and chain.
The most important information I took away from his book are:
1. It is possible to calculate all the environmental forces on the boat (wind, current and waves).
2. It is also possible to calculate the holding power of the anchor and chain
3. As long as the holding power is greater than the total sum of the environmental forces you will not drag. That is of course if the anchor is set well.
This however also means that, if the holding power is less than the environmental forces you will drag, no matter how well you have set the anchor.
And how does the dragging start ?
The chain gets lifted of the seabed, which raises the shank of the anchor and once the shank of the anchor reaches 25 degrees the anchor has lost all holding power and you will drag.
In other words, it is imperative the keep the angle of the shank to the seabed preferably at 0 degrees and for that you need the chain.
Another thing to take away is that there is no magical anchor. What you need to find is the anchor that holds best is the least favorable seabed. In good seabed most anchors hold well, but what we should be looking at is the holding power in seabeds the anchor does not favor. And only then do you start looking at the types of seabed you will encounter most.
That way of reasoning did make a lot of sense to me, since the chances on dragging are greater when the seabed is less favorable to the anchor.
So what determines the holding power of the anchor ?
The total holding power is equal to the anchor weight x the seabed factor. You can find the tables with the seabed factor online for anyone who is interested.
Automatically you will understand that the heavier the anchor is the more holding power it has. So oversizing your anchor does not only look nice in the marina, but also makes sense when it comes to anchoring in adverse weather.
Then comes the question: 'how do you calculate the environmental forces ?'
When it comes to wind it is basically the forces the wind can excercise on the hull that is in contact with the wind. In other words, if the wind comes onto the bow you will need to calculate the frontal surface of your boat. That will give you a number of square meters (or square feet in the US) and that you then have to multiply by an amount of N which is given for each windforce.
E.g. at 4 Bft the force of the wind is around 15 N/m2, but at windforce 10 it is 204 N/m2.
When your frontal surface is 10 m2 you will find 150 N at 4 Bft, but 2040 N at 10 Bft.
Your anchor and chain therefore have to withstand those forces and usually at low windspeeds that is not a problem. But since the forces of the wind are squared when the wind increases you can understand the required holding power increases likewise.
Also the current excercises forces on the hull and those forces are quite high. In fact the density of water is 835 times that of air, so we can expect the forces to be very high quite quickly.
At 1 kt current and e.g. a surface area of 1.25 m2 it is only 70 N, but at 5 kts it is already 1650 N. And the surface area is of course the frontal part that is in the water.
Comparing wind to current we see that a current of 2 kts has the same effect as a windspeed of 6 Bft, so definitely something to take into consideration.
Lastly we have the wave factor. These forces are more difficult to calculate, but basically we have 3 different forces.
The first one is the up and down motion, the pulling on the chain when the boat goes up to the crest of the wave.
Second is the motion of the water in the wave, which acts as a current.
Third is the force of a breaking wave on the hull.
These 3 forces, wind, current and waves determine the total environmental forces and as has become clear the stronger the wind and the current are the environmental forces increase with dramatically. Not linear, but exponentially.
When it comes down to how much anchor and chain you need to throw out it must be clear that, since the environmental forces change drastically when wind or current increases, there cannot be such a thing as a fixed scope.
In fact, the 3 : 1, 5 : 1 and even 7 : 1 scope are all based on calm weather, calm seas and hardly any waves.
None of us starts dragging at calm weather, it happens only when either wind or current, or both increase drastically. The forces on the hull become such that the chain is lifted off the seabed which raises the shank of the anchor and with each degree the shank gets lifted the holding power of the anchor becomes less. At 20 degrees off the seabed the holding power is only 50 degrees and at 25 degrees there is no more holding power.
After reading the book I started calculating the figures for my boat and came to a frontal surface of about 18 m2. The frontal subsurface area is about 3 m2.
Where it becomes interesting is when current and wind don't come from the same direction. Then you have to recalculate everything to figure out how much surface is in the wind and how much surface is in the current.
On my boat I have a 50 kg (500 N) Sarca Excel and 140 mtr of 13 mm chain, which weighs 4 kg/mtr or 40 N.
So how do I decide how much chain to throw out ?
I want to have at least 10 mtr of chain lying on the seabed behind the anchor. After that the chain can be lifted off the seabed in bad weather, which means I calculate the expected wind forces (we have no current here) based on the forecasted weather. For safety purposes and based on experience I add at least 1 Bft and sometimes 2 Bft, depending on the location where we are.
That way I can calculate the expected environmental forces in N and since that is the force that will lift the chain off the ground I divide that number by 40 N. That will give me the length of the chain that I need as a minimum. After that I add 10 mtr, which is the part that always has to stay on the seabed. And I leave the holding power of the anchor completely out of the equasion. That is extra bonus for me.
As an example, let's say the expected wind force is 8 Bft, which is 37 kts.
I add 1 Bft for safety, which makes it 45 kts (takes care of the gusts).
At 45 kts the windforce is 157 N/mtr2 and in my case that comes down to 18 x 157 N = 2826 N
This 2826 N I have to divide by 40 N, which gives me 71 mtrs. To that I add 10 mtr behind the anchor and in total I will therefore throw out a minimum of 81 mtr. And all of this is based on a max angle of 25 degrees between the bow to the seabed.
In shallow water (up to 7 mtr) I don't add extra chain for the water depth, but over that I add the water depth to the bow as well.
Some people will call this overkill, some told me I have to buy a different anchor, but as long as nobody was able to give me a mathematical explanation of what exactly happens when you start dragging and why you start dragging............I basically left all those opinions for what they are.
Here in the Med we do stern to docking and anchoring most of the time, in fact I would say about 95 % of the time. The other 5 % is free anchoring, without a land line. And if there is one thing you cannot afford is to start dragging your anchor when you have a rocky shore 20 or 30 mtrs (or less) behind your stern. You need to know for sure that you won't drag.
Currently we are in the Aegean, which is known for very strong winds most of the time. Yesterday we came into an anchorage with steady 30 kts and gusts to 40 kts. We only wanted to go for a swim, nothing else. Waves were almost none, no current. Water depth was 8 mtrs.
So I dropped 100 mtr in total, stabilizers on and we were not moving anymore. 4 sailing vessels tried to anchor as well, all gave up and left the anchorage. There were 3 other boats anchored, but they were closer in to the shore, less exposure to the wind and shallower water. We stayed for 2 hours, hauled the anchor up and sailed to Kos.
So at the helm I have a page in my checklist that gives me an instant number of chain to put out at certain windspeeds. And if we have the chance we will drop even more. The number in the checklist is only the minimum, not the maximum.
I am glad I read the book, later I found the online calculator and an explanation for the commercial shipping
https://www.ocimf.org/document-liba...environmental-loads-on-anchoring-systems/file
https://www.ocimf.org/anchoring-calculator/