View Poll Results: chain length

3040ft


15 
8.72% 
4050ft


5 
2.91% 
6070ft


5 
2.91% 
7080ft


6 
3.49% 
8090ft


6 
3.49% 
90120ft


17 
9.88% 
120+ft


118 
68.60% 


10102007, 04:05

#76

Senior Cruiser
Join Date: Mar 2003
Location: Thunder Bay, Ontario  4829N x 8920W
Boat: (Cruiser Living On Dirt)
Posts: 34,024

I’ll admit that I’m seriously mathematics & physics/mechanics challenged.
Unlike Paul Simon whose “... lack of education hasn't hurt me none ...” (*1)  my ignorance leaves me essentially handicapped.
In principle, I see an anchor assembly as a class 2 lever (*2) , wherein the anchor shackle represents the fulcrum, the boat is the point effort (effort arm), and the rode & kellet represent the load (resistance arm).
Classically, the mechanical advantage is greater the longer the distance between fulcrum and effort, and the shorter the distance between fulcrum and load.
Since we wish to DECREASE the mechanical advantage (*3) (creating disadvantage), we would increase the distance between the fulcrum (anchor shackle) and the resistance (kellet), and increase the mass of the kellet resistance.
*1. “Kodachrome by Paul Simon
*2. The second order of levers has the fulcrum at one end and the effort at the other, with the resistance in the middle.
*3. Thus increasing the effort required to lift the resistance, and straighten the rode.
The Principle of Equilibrium of Rotation determines the force of effort by the ratio of the effective effort arm to the effective resistance arm.
Unfortunately, I don’t know how to quantify the lever principle, and perhaps I totally misunderstand the principle(s).
Can anyone help?
PS: When I first read Frasee’s ”Tuning an Anchor Rode” < Tuning an Anchor Rode >, Alain was kind enough to respond to my queries  but not articulate enough to break through my obtuse ignorance (*4).
*4. All of the mathematics, but in particular, where he says: “...the best improvement is obtained by putting the kellet near the anchor, where it equals the kellet weight K multiplied by the scope N ..."
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Gord May
"If you didn't have the time or money to do it right in the first place, when will you get the time/$ to fix it?"



10102007, 16:12

#77

Moderator Emeritus
Join Date: Feb 2004
Location: Pelican Bay, Great Sandy National Park
Boat: Steel Roberts Offshore 44
Posts: 5,173

Screw...
To me the closest theoretical approximation to an anchor is a screw.
Think of an anchor trying to screw itself into the bottom...
__________________



10102007, 16:56

#78

Senior Cruiser
Join Date: Nov 2005
Location: Tasmania
Boat: VandeStadt IOR 40'  Insatiable
Posts: 2,317

FWIW, you might want to specify the weight per foot of the chain... 300' of 1/8" chain isn't 3 times as good as 100' of 1/2" chain!



10102007, 22:38

#79

Senior Cruiser
Join Date: Dec 2006
Location: Seattle
Boat: Cal 40
Posts: 2,485

Quote:
Originally Posted by GordMay
I’ll admit that I’m seriously mathematics & physics/mechanics challenged.
Unlike Paul Simon whose “... lack of education hasn't hurt me none ...” (*1)  my ignorance leaves me essentially handicapped.
In principle, I see an anchor assembly as a class 2 lever (*2) , wherein the anchor shackle represents the fulcrum, the boat is the point effort (effort arm), and the rode & kellet represent the load (resistance arm).
Classically, the mechanical advantage is greater the longer the distance between fulcrum and effort, and the shorter the distance between fulcrum and load.
Since we wish to DECREASE the mechanical advantage (*3) (creating disadvantage), we would increase the distance between the fulcrum (anchor shackle) and the resistance (kellet), and increase the mass of the kellet resistance.
*1. “Kodachrome by Paul Simon
*2. The second order of levers has the fulcrum at one end and the effort at the other, with the resistance in the middle.
*3. Thus increasing the effort required to lift the resistance, and straighten the rode.
The Principle of Equilibrium of Rotation determines the force of effort by the ratio of the effective effort arm to the effective resistance arm.
Unfortunately, I don’t know how to quantify the lever principle, and perhaps I totally misunderstand the principle(s).
Can anyone help?
PS: When I first read Frasee’s ”Tuning an Anchor Rode” < Tuning an Anchor Rode >, Alain was kind enough to respond to my queries  but not articulate enough to break through my obtuse ignorance (*4).
*4. All of the mathematics, but in particular, where he says: “...the best improvement is obtained by putting the kellet near the anchor, where it equals the kellet weight K multiplied by the scope N ..."

I'll take a stab at it, but since this thread is what made me sit down and think about it, I could have messed up.
Let's take the end points. You're in 10 feet of water with a 100 foot nylon rode (to do away with catenary and make essentially straight lines).
Start out there is no load on the rode. You put a 10 lb. (weight in water) kellet out 10 feet down the rode such that the weight is entirely resting on the bottom. Now start to apply a load. Assuming that the angle change is not worth considering, how much force is required to lift the kellet off of the bottom? You're essentially pulling straight up, so 10 lbs of tension in the rode.
Now put it 90 feet out on the rode. The force in the up direction to lift the kellet off the bottom is still 10 lbs. How much tension in the rode is required to produce a 10 lb force in the Y (up) direction. The angle between the rode and the bottom is arc sine (10/90) = 6.4 degrees. The triangle we want to look at needs a 10 lb arrow pointing up, and has a the rode , hypotenuse, at a 906.4 =83.6 degree angle from the up direction. So the tension in the rode to apply a 10 lb up force is 10/(cos 83.6)= 90 lbs. After seeing this I realized that the triangles were complementary, so the hypotenuse is the same and the equation you gave scope N times the kellet weight applies. So far as the way I see it, scope applies to the kellet scope, I have only done this with ignoring the tension from the anchor direction because until the kellet lifts off of the bottom the tension is at right angles to the lift and doesn't affect the calculation. I need to work on it more to figure out what happens when the kellet is off the bottom.
John



10102007, 23:34

#80

Senior Cruiser
Join Date: Dec 2006
Location: Seattle
Boat: Cal 40
Posts: 2,485

Quote:
Originally Posted by GordMay
If the Alain Fraysse representations are accurate, re 1 & 4:
How would the kellet minimize swing radius & keep the rode "down"?
To do so, would require some significant horizontal deflection between kellet & anchor.
***
As I’ve previously indicated, I’ve repeatedly watched my Kellet / Rode work in 30 knot conditions, with considerable surge and current*.
The kellet sat near the bottom between gusts (< 28 kts), keeping the chain (twixt anchor & kellet) nearly horizontal  thus NO load on anchor.
As gusts build (> 28  32 kts), the kellet rose, and the chain begian to straighten.
The chain/rope rode only straightened completely, when gusts exceeded 30 knots, for more than several (30?) seconds.
* These observations were made, on several different occasions, at Normans Cay, Exuma (deep, soft sand, long fetch).
The boat was a 6800# C&C28, sitting to:
(1) 35# Delta on 75' 5/16" G4 Chain & 5/8" 3Strand Nyolon  with 20# kellet
(2) 15# Fortress (FX 23) on 30' 1/4" G4 Chain & 5/8" Nylon  20# kellet.

When you say there is no load on the anchor when the kellet is near the bottom, are you saying that there is no up angle load?
I think when Craig is talking about performance, he is talking about what it takes to drag the anchor. Assuming that the anchoring system you described is designed to hold your boat in winds higher than you described, then the kellet has no effect on the ultimate holding power of the anchor, as you stated that the rode is straight in sustained 30 knots, the kellet is doing nothing to change the angle of pull on your anchor at that wind speed and higher.
John



11102007, 02:13

#81

Senior Cruiser
Join Date: Mar 2003
Location: Thunder Bay, Ontario  4829N x 8920W
Boat: (Cruiser Living On Dirt)
Posts: 34,024

John:
Thanks for trying to explain the math’s. I’ll try to wrap my head around it.
When I claim no load on the anchor: I’m supposing that (practically) all the load is expended in raising the kellet, and straightening the rode. Until the rode is straightened, the anchor is essentially unloaded (excluding some “vectored” energy).
As to ultimate utility, when the rode is finally straightened: I expect the kellet (still exerting a downward force) would continue to reduce the pull on the anchor.
The following is offered less as an argument for my position (for I have little/no understanding), than as an account of my befuddled (perhaps “bumfuzzled”?) thinking.
Are we talking about “torque”, (being the product of the distance of the applied force from the axis of rotation and the component of the applied force that is perpendicular to the lever arm)?.
In order to maximize torque, you need to:
1. Maximize the magnitude of the force, F, that you apply to the lever (heavier kellet).
2. Maximize the distance, r, from the axis of rotation of the point on the lever to which you apply the force (increase the distance from the anchor to kellet).
3. Apply the force in a direction perpendicular to the lever (gravity exerts this force on the kellet).
Exert a force (F) of 20 Lbs (kellet) on a lever (rode), at a distance (r) 75 Ft from its axis of rotation (anchor). Assume that the pull is at an angle that is 60 degrees above the lever arm.
Torque (t) = Force (F) x Distance (r) x Sin ∅
t = 20 Lbs x 75 Ft x Sin 60 (= 0.866025404)
t = 1,291 Ft/Lbs
where:
Torque is the effort required to lift the kellet
Force is the upward component of the wind force
Distance is the Radius arm from the anchor (point of rotation)
Angle is the lifting angle between the kellet and the boat
This is killing me  I detest my own stupidity & ignorance.
__________________
Gord May
"If you didn't have the time or money to do it right in the first place, when will you get the time/$ to fix it?"



11102007, 02:48

#82

Marine Service Provider
Join Date: Jun 2005
Location: New Zealand
Posts: 406

Quote:
Originally Posted by GordMay
When I claim no load on the anchor: I’m supposing that (practically) all the load is expended in raising the kellet, and straightening the rode. Until the rode is straightened, the anchor is essentially unloaded (excluding some “vectored” energy).

You suppose wrong. All load is always transfered to the anchor, less any trivial reduction from the friction of the chain and kellet on the seafloor before the rode is lifted. All that is changed by the benefit of catenary from chain, or point loading of the kellet, is the angle of pull.
An anchor system consists chiefly of two points, the anchor and the boat, and a rode between them in tension if the boat is trying to go somewhere away from the anchor. That rode is affected by various forces including gravity and buoyancy, but unless there is a 2nd anchor point, the load from the boat is transferred entirely to the anchor. In the real world this is ALWAYS the case, with the exception of friction from the chain etc as mentioned above.
Changing the angle of pull is important, keeping it close to horizontal is desirable, but catenary from chain or kellets doesn't do a very good job.
Quote:
Originally Posted by GordMay
As to ultimate utility, when the rode is finally straightened: I expect the kellet (still exerting a downward force) would continue to reduce the pull on the anchor.

No, its benefit at that point is completely and totally lost.
Its downward force is irrelevant if the profile of the rode is unaffected. Tension equalizes and remains constant. See above.
Quote:
Originally Posted by GordMay
The following is offered less as an argument for my position (for I have little/no understanding), than as an account of my befuddled (perhaps “bumfuzzled”?) thinking.
Are we talking about “torque”, (being the product of the distance of the applied force from the axis of rotation and the component of the applied force that is perpendicular to the lever arm)?.

No, there is no lever arm because the rode is not rigid. There is no "torque".
To minimize angle of pull the kellet must be as close as possible to the anchor. This is another way of saying it must be as far away from the other end of the rode as possible. Why? To negate the kellet, a purely vertical force from its weight must be overcome. It is easiest to do this vertically with the same force that negates the kellet's weight. If the kellet is out on an angle, much more force must be exerted on that angle to effect the same vertical lift, as elemental trigonometry tells us.
To maximize the faux shock absorption of a kellet, it must be central, because this maximizes the distance the kellet is free to move (so maximizing the work to be done and the energy "absorbing" properties) while balancing the reduction of lifting force required.
To minimize swing radius, the kellet must be down the rode a length equal to the depth. This way the kellet may eliminate the greatest length of rode possible via its vertical pull. (I.e. it is trying to make vertical, and thus eliminate completely from the swing radius, a length of rode equal to the distance from the seabed to the bow roller. It can do no longer length).
Quote:
Originally Posted by GordMay
This is killing me  I detest my own stupidity & ignorance.

Don't worry, lots think the same thing... Heavy chain is good, all chain is important, plows are good anchors, and other myths...



11102007, 07:32

#83

Senior Cruiser
Join Date: Dec 2006
Location: Seattle
Boat: Cal 40
Posts: 2,485

Quote:
Originally Posted by GordMay
John:
Thanks for trying to explain the math’s. I’ll try to wrap my head around it.
When I claim no load on the anchor: I’m supposing that (practically) all the load is expended in raising the kellet, and straightening the rode. Until the rode is straightened, the anchor is essentially unloaded (excluding some “vectored” energy).

Break you anchor rode at any point and insert a spring scale. You will see that the tension in the rode is equal at every point with or without a kellet. Or try using a spring as an example. Your anchor rode is a spring stretched out a certain amount, now put a kellet on it. The stretch on the spring will be the same on either side of the kellet or for that matter everywhere on the spring. If the anchor were unloaded by the kellet, the stretch on the spring on the anchor side of the kellet would have to disappear.
The triangle I was describing is called resolving the forces. The hypotenuse is at the angle of the rode and the length is the tension on it. Draw a vertical line up from the bottom of the hypotenuse, and a horizontal line from there, completing the triangle. The length of the vertical line is the force exerted straight up, and the length of the horizontal line is the force exerted sideways.
Quote:
Originally Posted by GordMay
As to ultimate utility, when the rode is finally straightened: I expect the kellet (still exerting a downward force) would continue to reduce the pull on the anchor.

Actually since the weight of kellet is increasing the tension in the rode, you have increased the pull on the anchor, but by a trivial amount.
John
Quote:
Originally Posted by GordMay
The following is offered less as an argument for my position (for I have little/no understanding), than as an account of my befuddled (perhaps “bumfuzzled”?) thinking.
Are we talking about “torque”, (being the product of the distance of the applied force from the axis of rotation and the component of the applied force that is perpendicular to the lever arm)?.
In order to maximize torque, you need to:
1. Maximize the magnitude of the force, F, that you apply to the lever (heavier kellet).
2. Maximize the distance, r, from the axis of rotation of the point on the lever to which you apply the force (increase the distance from the anchor to kellet).
3. Apply the force in a direction perpendicular to the lever (gravity exerts this force on the kellet).
Exert a force (F) of 20 Lbs (kellet) on a lever (rode), at a distance (r) 75 Ft from its axis of rotation (anchor). Assume that the pull is at an angle that is 60 degrees above the lever arm.
Torque (t) = Force (F) x Distance (r) x Sin ∅
t = 20 Lbs x 75 Ft x Sin 60 (= 0.866025404)
t = 1,291 Ft/Lbs
where:
Torque is the effort required to lift the kellet
Force is the upward component of the wind force
Distance is the Radius arm from the anchor (point of rotation)
Angle is the lifting angle between the kellet and the boat
This is killing me  I detest my own stupidity & ignorance.




11102007, 08:06

#84

Marine Service Provider
Join Date: Jun 2005
Location: New Zealand
Posts: 406

Quote:
Originally Posted by cal40john
Actually since the weight of kellet is increasing the tension in the rode, you have increased the pull on the anchor, but by a trivial amount.

All right except that bit



12102007, 09:07

#85

Senior Cruiser
Join Date: May 2004
Location: annapolis
Boat: st francis 44 mk II catamaran
Posts: 1,180

OK, I'm leaping back in to clarify my own myths.
I understand that rode/chain combos are best for having the rode act as a shock absorber. Chain does nothing for releaving the stress on an anchor in high wind as at that point the chain is bar tight and in fact would increase the shock travelling to the anchor and be a bad thing. Here's my problem in answering the question on "how much anchor chain to carry"
Having a length of chain to prevent bottom chafe is important, very important. We all agree on that. Since you are leaving prepared to anchor in everything from 6 feet of water to 40 feet of water, how much rode you have to play out varies and once again if you want to ensure that coral heads don't cut your nylon rode, you would want only anchor chain on the bottom (unless someone can explain how nylon or polyester rode resting on the bottom with coral heads but further from the anchor would be less prone to chafing through). It seems that I would want my rode therefore to be no longer than the depth of the water at low tide in places with lots of coral. If I was anchoring in 30 feet of water at high tide, in a place with 18 foot tides, wanting a 5/1 length, 150 feet, I would want 132 feet of chain (150 feet  18) and in effect my desire for chafe resistance alone would cause me to have an anchoring configuration which would result in an all chain rode being used (most of the time I anchor in 5 to 10 feet of water). If I anchor in sand or mud, it's really not an issue. I've only been to cuba and bahamas and east coast for cruising. In those areas, finding spots to anchor in 5 to 10 feet of water isn't an issue and anchoring with a nylon rode really helped in absorbing the shocks of waves when riding out storms.



13102007, 00:00

#86

Senior Cruiser
Join Date: Dec 2006
Location: Seattle
Boat: Cal 40
Posts: 2,485

geek out
Quote:
Originally Posted by craigsmith
All right except that bit

Ok, let's see if I got this figured out. It reduces the load on the anchor, but at a fraction of the weight of the kellet dependent on the angle and load.
The resultant forces have to sum to zero. If the boat is applying a 141 lb tension to the rode at a 45 degree angle, the resultant forces are 100 lbs up and 100 lbs horizontal to the boat at the point where the kellet attaches to the rode. The kellet pulls down with 10 lbs (in my example), so the tension on the rode to the anchor is 100 lbs horizontal to the anchor, and 90 lbs down. Pythagoras gives 134.5 lbs of tension in the rode in the direction of the anchor, and the angle of the rode from the bottom is, arc sin (90/134.5) = 42 degrees. So, the kellet reduces the load to the anchor by a trivial amount in this unrealistic example.
I picked these numbers for me to see the problem better. A closer to real world example with the kellet near the bottom and a long scope means the kellets effect on load is almost nothing.
Somewhere it was stated that it would be better to invest that weight in the anchor rather than adding a kellet. I assume that is due to making the anchor have more weight to dig in better and probably increase fluke area which is what we are really depending on to hold the boat.
Earlier it was said that once the rode is straight the kellet has no effect, this was in reference to the reasoning for the kellet is to reduce the up angle of pull on the anchor.
Also earlier we said that the tension in the anchor rode has to be equal throughout. Well in order to bend my mind around the problem I firmly knotted the kellet to the rode. If we put the kellet on a snatch block, then the tension in the anchor rode has to be equal throughout. So I think that the tension in the rode is 134.5 lbs in this example. If the line that holds the kellet is kept at the same angle as the rode back to the boat, then the tension in the restraining line to the kellet is 141134.5 = 6.5 lbs.
I'm now trying to do it without numbers to get a real equation, but I think I'm doing it the hard way.
John



13102007, 01:43

#87

Marine Service Provider
Join Date: Jun 2005
Location: New Zealand
Posts: 406

You have all the basic ideas right. My reply above was in response to your statement that the load on the anchor is increased because the tension in the rode is also increased. The tension on the rode between the boat and the kellet is increased (trivially) but not on the anchor.
As you imply, the point at which the kellet is attached is affected by three loads  the force of the boat, the weight of the kellet, and the resistance of the anchor. For the situation to be static, all three must sum to zero. Because the weight of the kellet is downward and the rode is not rigid, the angles just naturally equalize in a different configuration to maintain the equilibrium.
Attached is my own drawing. Sorry for the poor handwriting.
Force is given by the horizontal load (L) * cosine of alpha (the angle of elevation of the rode relative to the seabed, essentially given by the scope if complicated catenary calculations are to be sidestepped). Nb. another reason to use decent scope  minimize that cosine. Buoyancy can only resist upward...
Without a kellet the force F and resistance force R are equal.
With a kellet, F is affected by the present of the kellet's weight force W. However R is unaffected. This holds true unless the anchor point is raised above that of the kellet attachment (maybe you've anchored on a dry beach with lots of deep water inbetween ).
Catenary from the chain complicates the real world scenario but this outlines the current topic of discussion.
In strong winds (high L) but with a constant and relatively low W, it should be no surprise that the gap between R and F 2 rapidly closes and the sum of the angles alpha plus beta 2 rapidly approaches 90 degrees (i.e. the rode is straight).



13102007, 11:14

#88

Registered User
Join Date: Apr 2007
Location: germany, Berlin, boat at Ft. Lauderdale
Boat: MANTA 42, before Morgan 41 Classic, GibSea 106
Posts: 91

Hi, Craig,
A question:
The Buegelanker manufacturer takes consideration for "Monohull" and "Multihull" sizes (weights) of the anchor.
Looking to your webpage, I see only length and displacement? No difference for Multihulls?



13102007, 11:32

#89

Marine Service Provider
Join Date: Jun 2005
Location: New Zealand
Posts: 406

It's not possible to take into account every different factor. Our chart is based on monohull LOAs but you can basically just go one size larger for multihulls. The difference is sort of constant.



13102007, 15:45

#90

Marine Service Provider
Join Date: Oct 2005
Posts: 1,659

Reading parts of this thread this springs to mind;
"In theory there is no difference between theory and practice. In practice there is."
said by Yogi Berra, for anyone interested.
Dealing as I do, day in day out with many many boats, many many users and in many many situations I can comfortably say that while the calculations appear possibly correct or as close as they can be when only using a tiny amount of the inputs that are actually in play when talking anchoring, the reality of it is they don't equate to real life very well at all.
Try replicating the calculations, some here are expounding, in real life. We have and you'll find many just don't stack up.
Didn't want to pop in but some people do take information from threads like this as gospel. All I would like to say is read and absorb, it is quite interesting, BUT don't stake your life on it without further investigation. Doing so may just kill you and yours.
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