easy way to pull up well set anchor

[valhalla360]

Originally Posted by Putt-Putt

Think about it, if 100# horizontal resulted in 100# both vertical and horizontal at the anchor, where did the addition 100# of force come from?
If that worked and you could commercialize it, your riches would make Musk look like a pauper.

In your jeep example, you added a 150-200lb force downward on the winch line and pulled a 3500-5000lb truck forward thru thick mud...where did that force come from (certainly far more than the 150-200lb applied to the line)? Hint, the tree didn't move.

Look up mechanical advantage. It's already been commercialized in many thousands of ways. Ever use pulleys to lift a heavy weight. Apply 100lb of force and you can lift 400lb with a 4-1 pulley system...where did the extra 300b of force come from...mechanical advantage and the fixed point supporting the pulley system.

If you are thinking F=ma, that applies to an object that is free to move (a stuck anchor is not free to move...at least until it breaks free).

This is just a specialized version that has limited uses. Yes, the less elasticity in the system, the better it works but it's still works very well with nylon anchor lines (chain is better as it has less elasticity).

The vertical force comes from the buoyancy of the boat against the weight of the earth. Eventually assuming it breaks free, it exceeds the vertical holding power of the anchor buried in the earth as you apply a relatively small horizontal force at the top of the anchor line. In theory, you could pull the bow under if the anchor is lodged under a large rock or other debris but most people would stop before that happens.

[valhalla360]

Originally Posted by Dieseldude

Thanks Valhalla, not trying to knock your theory, but I would have to see a vector diagram to see how a lateral force applied to the upper end of a cable would result in strong vertical force on it. Is this because tension is tension, and independent of applied direction? I do know that if I want to lift a weight on a rope, I get better results for my efforts by pulling straight up than I would by pulling sideways. This can be shown by simple vector analysis.

If we assume a line is completely flexible and the line can only apply force when in tension and along the direction of the line. Not technically true but close enough as to make no difference in our use case.

Also assuming the line is near vertical, there will be no catenary effect of significance...ie we don't have to worry about the angle at the bow being noticeably different from the angle at the anchor.

In this case, the force vector in the line must follow the direction of the line.
- If you have infinite scope (ie. the pull is perfectly parallel to the bottom), there is no vertical component. (this is why it's good to use plenty of scope and the catenary in the chain is good for holding as it minimizes the vertical component at the anchor shank that might pull an anchor out)
- At 45 deg, it's a 1 to 1 ratio (100lb forward generates 100lb up). Not good for anchoring and with a poor set, might pull free of the bottom.
- At 90 deg, it creates infinite vertical force as the force vector of the line at the anchor has no horizontal component to resist the horizontal force at the bow. Of course, this is theoretical as the anchor would break free or something else would break first. In practice, it' simply not possible as elasticity (even in heavy steel cables) would result in some angle below 90deg but the closer to 90deg the higher the vertical component is.

If you take the tangent of the angle and multiply by the horizontal force (see my prior post), that corresponds to the vertical component of the force vector.

Keep in mind the forces discussed generated by the engine assumes no momentum (ie: you slowly increase power from the motor so there is almost no movement of the boat until the anchor breaks free). If you have the boat moving when you snub off the anchor line, the horizontal force from stopping even a slow moving 10,000-30,000lb boat can be far higher.

Setting out a 2nd anchor with a large scope works though less efficiently as there is a large horizontal component to the 2nd anchor lines force vector. Though still less than the force along the line, so you may need to apply 150lb to the line to get 100lb of horizontal force. Not convenient but ultimately it can work.

[Dieseldude]

Originally Posted by Putt-Putt

AH yes, grasping the Physics can be a problem.

Often using an analogy proves useful. Let's try.

Here's the set up.

You and your girlfriend are out in the 4x4 for an early rutting season frolic in the backwoods. On the way home you get stuck, big time. You're now a mud slinger.

Not to worry, you grab the 15' chain, spool out the 100' of cable on your new (cheap) winch, sling a large fir tree 100' away with the chain, shackle on the steel cable and signal the girlfriend to tighten the cable. The cable is drawn bar tight, and the winch stalls. The truck goes nowhere.

Half way back to the truck, feeling frustrated you do your best WWE impression of a bounce off of the ropes. Dusting yourself off from the obligatory face plant, you notice that the cable has now drooped. Something must have moved! It wasn't the tree, is was the truck, about an inch and a half.

Retighten, lunge, another couple. Rinse and repeat a dozen times and you're free.

The important parts here are you must use inelastic components, (almost nothing is truly inelastic but steel is way better than nylon) the 100' (the longer the better) bar tight to start with and 2 fixed points. (the tree and truck)

Two paycheques later you replace the winch with a real one, a month later the girl replaced me with a new guy who had a bigger........errr, truck.

Yep, I was the guy with the welts! That was 45 years ago.

From my read of your system you are lacking at least the steel and the 2 fixed points. Feeling that it works may be its only virtue.

My rudimentary engineering skills reveal that if somehow you could apply a 100# force horizontally to your 45º rode, the forces at the anchor shank are 25# both vertically and horizontal.

Think about it, if 100# horizontal resulted in 100# both vertical and horizontal at the anchor, where did the addition 100# of force come from?
If that worked and you could commercialize it, your riches would make Musk look like a pauper.

Then again, except for the girl part, I could be all wrong.

You slowly applied tension to the wire according to the limit of the winch. The power developed by the winch was stored in the wire. Once at maximum tension, the wire very much behaved like a solid rod, so it could be used as a lever. A good strong sideways lunge applied force to the seemingly "solid rod" wire. You have a total length of 115 feet of combined chain and wire. If you kicked it in the middle, you have 57.5 feet of leverage multiplied by tension force. Plus the distance that the wire deflected side ways comes into play. This gives quite a high magnitude of mechanical advantage. The force was delivered by impulse, so the vehicle lunged forward further than the elastic stretch of the wire. The tension was was relieved, so the wire went slack.


Power delivered impulsively can deliver impressive force. The power, being measured in joules per second appears to be greatly multiplied, as it is delivered near instantly. Impulse actually decreases the time which is in the denominator of joules/second, so the quotient (resulting force) is quite high. Of course this is why a sudden kick is so effective. You essentially kicked your vehicle out of the mud. Ok so much for the engineering analysis. To winch out a a stuck vehicle, giving the winch wire a good kick can get ya moving again. The old expression, "Ya can't push a string." is not necessarily so. If the string is under tension you sort of can. Beware of the safety issue here though. If the wire were to suddenly fail, it would not be good to get be in its path as all those joules per second snap your way.



Getting back to the stuck anchor though, how would the kick be applied to the wire?

[Dieseldude]

Originally Posted by valhalla360

If we assume a line is completely flexible and the line can only apply force when in tension and along the direction of the line. Not technically true but close enough as to make no difference in our use case.

Also assuming the line is near vertical, there will be no catenary effect of significance...ie we don't have to worry about the angle at the bow being noticeably different from the angle at the anchor.

In this case, the force vector in the line must follow the direction of the line.
- If you have infinite scope (ie. the pull is perfectly parallel to the bottom), there is no vertical component. (this is why it's good to use plenty of scope and the catenary in the chain is good for holding as it minimizes the vertical component at the anchor shank that might pull an anchor out)
- At 45 deg, it's a 1 to 1 ratio (100lb forward generates 100lb up). Not good for anchoring and with a poor set, might pull free of the bottom.
- At 90 deg, it creates infinite vertical force as the force vector of the line at the anchor has no horizontal component to resist the horizontal force at the bow. Of course, this is theoretical as the anchor would break free or something else would break first. In practice, it' simply not possible as elasticity (even in heavy steel cables) would result in some angle below 90deg but the closer to 90deg the higher the vertical component is.

If you take the tangent of the angle and multiply by the horizontal force (see my prior post), that corresponds to the vertical component of the force vector.

Keep in mind the forces discussed generated by the engine assumes no momentum (ie: you slowly increase power from the motor so there is almost no movement of the boat until the anchor breaks free). If you have the boat moving when you snub off the anchor line, the horizontal force from stopping even a slow moving 10,000-30,000lb boat can be far higher.

Setting out a 2nd anchor with a large scope works though less efficiently as there is a large horizontal component to the 2nd anchor lines force vector. Though still less than the force along the line, so you may need to apply 150lb to the line to get 100lb of horizontal force. Not convenient but ultimately it can work.

Thanks Valhalla for the great description. Essentially, tension on the cable is independent of the direction of applied force. It acts along the length. regardless of orientation in space. All the while, I had been considering vectors that would apply to rigid materials.

[valhalla360]

Originally Posted by Dieseldude

Getting back to the stuck anchor though, how would the kick be applied to the wire?

It's almost exactly the same except flipped sideways and the side force was applied to the middle rather than the end of the line.

If there was no resistance to the jeep sliding sideways and you pushed the jeep a foot or two to the side or up or down, it would increase the length of the winch line (Pythagoras came up with the formula for how much it lengthens). Within the working load limits, the greater the elongation percent, the greater the tension in the line, most of which is directed pulling the jeep towards the tree.

Now obviously pushing a stuck jeep or the tree over by a foot or two is not a good option. But pushing in the center of the winch line when straight offers no resistance initially as there is no portion of the force vector from the line to resist your push. Eventually the combination of angle and increased tension are enough to resist the side force you apply but by then there is a large increase in the force pulling the jeep toward the tree.

With a boat, moving the bow over by a foot or two in order to stretch the line is quite viable, so there is no need to try and apply the force in the middle of the anchor line.

Yes, the winch pretensions the line which is helpful but you can achieve the same thing on a boat by snubbing as the bow dips in a wave. This will get you closer to the vertical lift that you want and as an added bonus when the next wave hits, it will try to lift the bow increasing the tension even more. Often this is enough by itself to break an anchor free. But if it's not enough a little forward throttle will increase the vertical force dramatically.

[rgleason]

I like this technique. Thanks for posting it.

[jo_sail]

Originally Posted by valhalla360

It's almost exactly the same except flipped sideways and the side force was applied to the middle rather than the end of the line.



If there was no resistance to the jeep sliding sideways and you pushed the jeep a foot or two to the side or up or down, it would increase the length of the winch line (Pythagoras came up with the formula for how much it lengthens). Within the working load limits, the greater the elongation percent, the greater the tension in the line, most of which is directed pulling the jeep towards the tree.



… But if it's not enough a little forward throttle will increase the vertical force dramatically.

I can confirm the physics from a recent calculation and drawing of force paralelograms for a material ropeway made of dyneema. There the vertical force of the weight easily is leveraged by a factor of 3 to 5 on a long cable.



For lifting the anchor imagine only one side of the ropeway problem and a 90 degree flip of the image, so that you result with a thin vertically oriented right-angeled triangle.

The rope or chain under tension works like a strong lever! Force ratios can be estimated by Pythagoras. But be careful with the maximum working load of the cable / chain / rope and all equipment involved, rope for securing the chain, cleats, etc. avoid including the bow roller and windlass! Forces in the chain can become very high, especially if a little bit of speed of the boat, wave vertical motion and their huge mass add significant forces from stopping the boat on the chain.

I wish anyone who happens to need this for lifting a fouled anchor a lot of success and no accidents. Take care!

Jo

[jo_sail]

30:6 m = 5:1 lever built of a cable in tension. 400 kg resulting in 4 kN vertical force (horizontal after flipping by 90 degree for the boat-chain-anchor example) are leveraged to Q=20 kN horizontal (resp. vertical) force.

[jeanathon]

Originally Posted by valhalla360

99.99% of cruisers do have an engine, so very relevant.

99.99% of cruisers have an engine that has or will at some point not start when it is supposed to. Can we stop trying to inject an engine into the thread.
How you do it is not relevant unless you do it without an engine.

[rslifkin]

Originally Posted by jeanathon

99.99% of cruisers have an engine that has or will at some point not start when it is supposed to. Can we stop trying to inject an engine into the thread.
How you do it is not relevant unless you do it without an engine.

In my mind, if I'm on a sailboat with an engine that isn't working and there's no wind to sail, I'm going to be working on fixing the engine rather than pulling up the anchor.

[Dieseldude]

Originally Posted by rslifkin

In my mind, if I'm on a sailboat with an engine that isn't working and there's no wind to sail, I'm going to be working on fixing the engine rather than pulling up the anchor.

There is always the argument that a sailing vessel does not need an engine to sail. And this can be true when there is plenty of sea room. But raising anchor in confined or congested waters without the engine available seems like a needless risk.

[Exile]

Originally Posted by Dieseldude

There is always the argument that a sailing vessel does not need an engine to sail. And this can be true when there is plenty of sea room. But raising anchor in confined or congested waters without the engine available seems like a needless risk.

IMHO, it requires a higher level of seamanship that I respect & admire. Also not a bad idea for the majority of us with engines to perhaps practice from time to time, albeit with the engine running as a precaution.

https://www.youtube.com/channel/UCTm...3IL7Bvtf_7nTLw

[ColIn Ferguson]

Hi

Not wanting to stray into physics too much but to break out an anchor is a bit like breaking the seal on a stuck nut. You hit it hard and quickly rather than applying a progressive force over time.
It is therefore about impulse. Or change in momentum.
Force = change in momentum/ change in time.
F=dp/dt

Momentum = mass x velocity and thus has a direction as well as a time component

So the ancients knew that sailing your heavy boat hard at the problem worked. You use its mass and velocity to apply a considerable force over a short period of time to the static object .As well as using the technique of a float as in the Alderney ring previously described to convert the horizontal force into a vertical one.

This will be a much greater force than your windlass can apply.

[Dieseldude]

Originally Posted by Exile

IMHO, it requires a higher level of seamanship that I respect & admire. Also not a bad idea for the majority of us with engines to perhaps practice from time to time, albeit with the engine running as a precaution.

https://www.youtube.com/channel/UCTm...3IL7Bvtf_7nTLw

Nothing wrong with sailing totally by sail power. Fine sailing skills are fantastic. But if one must appear in court over an incident, and the skipper is questioned, "Why was the engine not made ready?" It is not a good position to be in. Unfortunately, potential liability issues can interfere with use of traditional skills.

[valhalla360]

Originally Posted by jeanathon

99.99% of cruisers have an engine that has or will at some point not start when it is supposed to. Can we stop trying to inject an engine into the thread.
How you do it is not relevant unless you do it without an engine.

You can use wind if the engine isn't working. Same basic principal.

If the engine isn't working and there is no wind...probably best to stay anchored if you have no means of propulsion.

So again, very relevant.