Quote:
Originally Posted by manyboats
I read this as saying that as long as there is chain on the bottom there's no pull on the anchor. I can't come close to agreeing with that ...

Eric,
you are right if you don’t agree, it was simplified too much and it is indeed not correct in every case. It is correct in case 1 (please refer the enclosed figure), it can be correct or not in case 2 and it is not correct in case 3. In a nutshell, valid for an all chain rode while anchoring on sand / mud ground and again very simplified: there is no force acting at our anchor as long as the weight of the chain laying directly on the ground is 1/3 higher than the force which is pulling our boat backward.
Let’s assume our boat at anchor (see figure), anchor at x = y = 0, boat at x = x0 and the bow at y = f above the ground. The rode has the length L and the weight per length q.
Wind or current or whatever is pulling at our boat with the horizontal force Hw. The result is that the rode becomes a catenary y(x) following the hyperbolic cosine between the boat position x0 and the position x1 given by the point where the rode loses contact to the ground. The length Lx of the rode which is not laying on the ground (formula for Lx given in the figure) may be equal or less the rode length L (case 1 in the figure: Lx = f < L, case 2: Lx < L, case 3: Lx = L).
The rode can only transmit a tensile force i.e. a force tangential to the catenary y(x). This tangential force is given by T = sqrt (H² + V²) where V is the vertical force component resulting from the rode weight and H is the horizontal force component. While H is constant along Lx the vertical component V is a function of the coordinate s(x) along the lifted rode, V(x) = q * s(x). Obviously V becomes a maximum at our bow with V = q * Lx i.e. the highest tensile load of the rode is located at our bow.
Back to the horizontal force Hw pulling our boat backwards and which has to be compensated by our ground tackle (anchor + rode) to keep our boat in place. To make it more handy I will give some figures which fit to our own boat as an example.
Let’s assume we are anchoring at a water depth of 4 m, our bow (1.7 m above water line) is then f = 5.7 m above the ground. Our 8 mm chain has a mass of 1.4 kg/m giving a weight per length of q = 13.7 N/m. We have put out all our chain, i.e. L = 55 m (180’), and since there is no wind and no current there is no horizontal force Hw acting at the boat, i.e. Hw = 0. The chain is hanging straight vertical, Lx = f = 5.7 m, and the position x0 of the bow is 49.3 m (distance from the anchor). This situation is case 1 in the figure.
Now a moderate breeze establishes. The breeze acts as a horizontal force Hw > 0 pulling on our boat. Hw may be estimated from the wind velocity, a drag coefficient and our boats area against the wind, e.g. I’m assuming Hw = 38 daN (85 lbf) for our AMS 40’ at a constant wind velocity of 15 knot. Hw will push our boat backwards. More and more chain will be lifted from the ground and the chain becomes a catenary with a certain slope angle alfa at our bow. Our boat will stop to move backwards when the horizontal component H = T * cos(alfa) of the rode force T(x0) pulling at our bow equals Hw. The corresponding data Lx and alfa of the catenary have to be derived by an iterative calculation from the formulas. For our example Hw = 38 daN it is Lx = 18.7 m (alfa = 33.9°). The wind has pushed us back 4.5 m (x0 = 53.8 m). This is situation 2 in the figure and there are still L – Lx = 36.3 m (or 50.8 kg!) of our chain laying directly on the ground. Since there is a certain friction between the chain and the ground the chain on the ground can develop a horizontal force Hc = (LLx) * q * fm, in our case Hc = 40 daN with a friction coefficient fm = 0.8 (text book data for a track vehicle on farmland).
In this example the chain laying on the ground alone is able to pull horizontally at our rode with the force needed to keep our boat in place, even without anchor!
The weakness of this is of course the friction coefficient. It will vary according to the ground material, vegetation, the chain geometry etc. It is uncertain to be 0.8, it might be less and I wouldn’t rely on it, but with fm = 0.75 it gives us at least a rule of thumb: there is no force acting at our anchor as long as the weight of the chain laying directly on the ground is 1/3 higher than the force which is pulling our boat backward.
However, we have an anchor at the end of our chain and we will need it definitely if the breeze develops to a 30 knots steady wind: the force is now Hw = 153 daN (335 lbf) and the corresponding catenary would be Lx = 36.1 m (x0 =54.4 m), alfa = 17.9°. There are only 18.9 m of chain left directly on the ground, even with our optimistic fm = 0.8 those 26.5 kg steel would only be able to pull with Hc = 20.7 daN, we need the anchor to hold the chain at the ground and us in place.
What is now the benefit of those 18.9 m chain laying on the ground? They have to be lifted if a gust or a high swell pushes our boat back. The kinetic energy of our boat is transformed into potential energy of the lifted chain and this is slowing down / damping the movement of our boat!
Wind develops further to a gale with 46 knots so that the complete chain has been lifted (case 3 in the figure). Corresponding data are Hw = 360 daN (810 lbf), Lx = L = 55 m, x0 = 54.6 m and alfa = 11.8°. Fine, but if we are facing a gale it will not be a steady load on the boat and our ground tackle due to swell and gusts. A small movement of our boat of just 0.1 m further backwards and our chain becomes complete stiff and all of a sudden there is a horizontal force of H = 725 daN (1’630 lbf) tearing brutally with a slope angle of 6° at our anchor. Now we have to rely severely on the holding power of our anchor, if might start to drag now.
And we wish to have an elastic element between the chain and our bow to reduce the impact loads on boat and ground tackle by damping effects, this elastic element can be the nylon rope of a combined chain / nylon rode or a snubber for an all chain rode.
Conclusions from our nice theoretic excursus?
If you want to anchor, use an anchor (! :)) and a chain at the anchor.
The weight of the rode chain is useful since its catenary will slow down movements of our boat + it helps to ensure a small slope angle at the anchor shaft.
And at ultimate load conditions we want to have an elastic element between chain and bow.
best regards / med venlig hilsen
wadden