How good is the Rocna?

[chris95040]

Originally Posted by Lodesman

A number of you continue to labour under the illusion that anchors have a centre of buoyancy - there is none, as there is no part of an anchor that is buoyant.

Anything that occupies a volume of space has a "center of buoyancy". Center of buoyancy its the center of mass of a displaced volume of water.

[Dockhead]

Originally Posted by chris95040

Anything that occupies a volume of space has a "center of buoyancy". Center of buoyancy its the center of mass of a displaced volume of water.

Exactly!

The buoyant force is equal to the weight of the water displaced. Just because it's not enough to float the object, doesn't mean that there's no force being exerted!

Therefore, immersed bodies which do not float (because their weight is equal to or greater than the weight of displaced water and so the force of buoyancy), nevertheless can have stability, and righting moment, in case the centroid of the displaced water volume is not coextensive with the center of gravity of the object.

It works exactly like stability of a floating body (say, a ship). The centroid of the displaced water (so you ignore the part of the ship which is above the waterline) is the center of buoyancy. The ship is stable as long as the center of gravity is below that, and the ship will be more stable, the greater the lever arm between these centers.

If this were not true, then submarines would not have stability! But they do.

[Dockhead]

Originally Posted by Lodesman

Back to physics - if I take two identically-sized perfectly smooth metal balls - one of steel weighing 10 kg, and one of tungsten that weighs 25 kg and drop them from the same height above the seabed, they will land simultaneously. Do we agree?.

We do not agree!

The force of gravity is exerting 25kg/f on the steel ball and 61.25kg/f on the tungesten ball. However, the two balls would each displace 3.125 liters of water, producing 3.125 kg/f of buoyancy (at whatever temperature water has that density), which opposes the force of gravity. Therefore, the acceleration of gravity is reduced proportionately. For the tungsten ball, the acceleration would be reduced by the force of buoyancy by 5.102% to 9.306311 m/s2 from standard gravitational acceleration of 9.80665 m/s2, and for the steel ball by 12.5% to 8.53114 m/s2.

The acceleration of gravity will be further reduced by hydrodynamic drag, which I'm not going to try to calculate, but this force will be equal on both balls.

Net net, the sinking tungsten ball accelerates faster and gets to the seabed faster.


If you doubt this, then try this thought-experiment --

Take an object which is barely denser than water. A submarine with slightly negative buoyancy, say, or a solid piece of polyester.

Drop it in the water.

Will it sink like a lead weight? Noooo.

[Panope]

Originally Posted by noelex 77

..........It would be interesting to measure the volume of trapped air in a Mantus roll bar (and in a Rocna if you block the breather hole)..........

I no longer own my Mantus anchor as I have given it some dear friends who desperately needed an upgrade.

However, I believe that trapped air in a Mantus roll bar will provide no "righting" effect because the centroid of the arc is very near the vertical center of gravity of the anchor. This, combined with the fact that the Mantus roll bar is ventilated at both ends, means that whatever "bubble" of air exists will always be positioned directly above the C.G.

Now that I think about it, the roll bar arc centroid might even be BELOW the vertical C.G. . If so, trapped air might have a slight NEGATIVE righting effect.

Steve

[Jim Cate]

All of the discussion of physics is fine, and I believe that most of you are coming around to understanding a bit about buoyancy and COG/COB... BUT in the case of an anchor being lowered on the end of a chain, its attitude will not be much influenced by the minor effects of its buoyancy, whether it has a hollow shaft or not. That attitude will be determined by the relative positions of the anchors COG relative to the attachment point of the chain at the end of the shank. If the anchor is lowered in "free fall", and there is little friction (not the usual case IME) then some hydrodynamic forces may alter its attitude slightly, but by and large, it is just dangling there.

Once it reaches the sea bed, I suspect that the shape of the portion that first engages the bottom and any drag from the chain being pulled away from vertical by boat motion will determine how it falls over and begins the setting process. I find it very hard to believe that a kg or two of buoyancy (out of an anchor weight of tens of kg) will overcome those influences.

Once it has fallen over and some force applied to the chain, the anchors weight distribution and its shape (roll bar, etc) come i nto play, and it is there that any slight benefit from shank buoyancy might be realized. Personally, I believe that such an effect will be pretty damn small relative to all the other things going on.

Cheers,

Jim

[noelex 77]

Originally Posted by Panope

However, I believe that trapped air in a Mantus roll bar will provide no "righting" effect because the centroid of the arc is very near the vertical center of gravity of the anchor.

I agree buoyancy in the roll bar is different to buoyancy in the shank.

Buoyancy in the roll bar will not have much impact on the "righting effect", but will increase the effective tip force (therefore holding power). Buoyancy in the shank will not impact the tip force but will increase "righting effect" (therefore the setting and resetting ability).

I think extra buoyancy in either of these areas (roll bar or shank) is beneficial, but for different reasons.

The Hydrobubble anchor placed the buoyancy chamber very close to where the top of the roll bar would be located on an anchor like the Mantus. It should be noted the volume of buoyancy for the Hydrobubble anchor is much greater than the effect of the trapped air in either the roll bar or the shank of the anchors we are talking about.

Nevertheless, I think there is a (small) positive benefit of extra buoyancy in the roll bar or the shank.

This is not one of my photos (see this website Unique EasyStow BreakAway Anchor Assembles in Seconds), but it shows the location of the buoyancy on the Hydrobubble anchor.

[Bleemus]

No to take away from all the useful physics being discussed but I have to say I find talking about buoyancy and anchors to be hilarious.

[noelex 77]

Originally Posted by Bleemus

No to take away from all the useful physics being discussed but I have to say I find talking about buoyancy and anchors to be hilarious.

There was a good April fools joke in one of the yachting magazines (I think it was PBO) about a floating anchor.

Not as good as the new cruising yacht available as a flat pack from "IKEA" April fool spoof (Allen key included for quick assembly ).

However, it is not April 1 and while most of the anchor benefits from being as heavy as possible there are parts such as the fluke and rollbar which work better when made as light as possible.

I guess this is no different to yacht design. A heavy keel and light mast and deck is offen the recipe for the best performance.

[uncle stinkybob]

Originally Posted by noelex 77

I agree buoyancy in the roll bar is different to buoyancy in the shank.

Buoyancy in the roll bar will not have much impact on the "righting effect", but will increase the effective tip force (therefore holding power). Buoyancy in the shank will not impact the tip force but will increase "righting effect" (therefore the setting and resetting ability).

I think extra buoyancy in either of these areas (roll bar or shank) is beneficial, but for different reasons.

The Hydrobubble anchor placed the buoyancy chamber very close to where the top of the roll bar would be located on an anchor like the Mantus. It should be noted the volume of buoyancy for the Hydrobubble anchor is much greater than the effect of the trapped air in either the roll bar or the shank of the anchors we are talking about.

Nevertheless, I think there is a (small) positive benefit of extra buoyancy in the roll bar or the shank.

This is not one of my photos (see this website Unique EasyStow BreakAway Anchor Assembles in Seconds), but it shows the location of the buoyancy on the Hydrobubble anchor.

keep in mind Mantus does not call the roll bar anything but that. and when the anchor hits bottom, if it lands side wise it all escapes anyway.

[noelex 77]

Originally Posted by uncle stinkybob

keep in mind Mantus does not call the roll bar anything but that. and when the anchor hits bottom, if it lands side wise it all escapes anyway.

I think you are right some of the air escapes the rollbar during the setting process. I am also not convinced that the small volume of air that remains in the Mantus rollbar is enough to make any practical difference, but as we are moving to higher performing anchors it would be nice to see buoyancy given some greater consideration. I think Ultra have managed to achieve a practical difference.

You can see the difference between the location of the breather holes in the Rocna and the Mantus in these photos. (On both anchors the rollbar is also open under the fluke).

These refinements are a bit bit like the racing sailor that removes 20kg of deck weight on a 5000kg yacht. The difference is too small for anyone to measure a practical difference but in theory the change will result in better performance.

[uncle stinkybob]

[QUOTE=noelex 77;2235374]I think you are right some of the air escapes the rollbar during the setting process. I am also not convinced that the small volume of air that remains in the Mantus rollbar is enough to make any practical difference, but as we are moving to higher performing anchors it would be nice to see buoyancy given some greater consideration. I think Ultra have managed to achieve a practical difference.

You can see the difference between the location of the breather holes in the Rocna and the Mantus in these photos. (On both anchors the rollbar is also open under the fluke).

These refinements are a bit bit like the racing sailor that removes 20kg of deck weight on a 5000kg yacht. The difference is too small for anyone to measure a practical difference but in theory the change will result in better performance.



Right. I use a 25lb Mantus. The bottom holes at ends are approx 5/16, holes on roller bar approx 1/8. The only position any air would remain in roll bar is if it landed near perfect face down. So far the Mantus has been great, only slipping once in very soft oozy mud.

[Lodesman]

Originally Posted by Dockhead

We do not agree!

The force of gravity is exerting 25kg/f on the steel ball and 61.25kg/f on the tungesten ball. However, the two balls would each displace 3.125 liters of water, producing 3.125 kg/f of buoyancy (at whatever temperature water has that density), which opposes the force of gravity. Therefore, the acceleration of gravity is reduced proportionately. For the tungsten ball, the acceleration would be reduced by the force of buoyancy by 5.102% to 9.306311 m/s2 from standard gravitational acceleration of 9.80665 m/s2, and for the steel ball by 12.5% to 8.53114 m/s2.

The acceleration of gravity will be further reduced by hydrodynamic drag, which I'm not going to try to calculate, but this force will be equal on both balls.

Net net, the sinking tungsten ball accelerates faster and gets to the seabed faster.


If you doubt this, then try this thought-experiment --

Take an object which is barely denser than water. A submarine with slightly negative buoyancy, say, or a solid piece of polyester.

Drop it in the water.

Will it sink like a lead weight? Noooo.

With due respect you are talking nonsense. Gravitational acceleration is equal for the two objects, as they are roughly the same distance from the centre of the gravitational mass, ie. the Earth - 9.8 m/s/s. Acting against them is hydrodynamic drag, which you agree is the same for both balls as they are identical in size and frictional coefficient. Also acting against them is, what I will concede to, is called the force of buoyancy. Both balls displace the same volume, roughly 1.3 litres, which is not applied proportionately, but equally reduces the mass force of both balls - making the steel ball effectively 8.7 kg, and the tungsten ball 23.7 kg. Both balls will accelerate at the same rate, but increased speed equals increased drag (both form and frictional), and at a certain speed the force of that drag will match the force of mass and the object will cease to accelerate - it is said to have reached terminal velocity. Of course the lighter object will top out at a terminal velocity much sooner than the heavier object, so given enough depth the tungsten ball will eventually overtake the steel one. But I don't imagine that would be the case in the less than 10 m depths, in which most of us typically choose to anchor (I admit I don't know the extent of water-induced drag - it would take someone with a lot of impressive postnominals to break down the math on that or perhaps someone with a couple balls, enough water and time on his hands ). Naturally, a polyester ball with an SG barely above 1 will hit terminal velocity very soon after release, so will appear to waft slowly down.
That said, since the hollow shank and the lead-weighted tip are rigidly connected, the terminal velocities of the two won't be different - there will be one terminal velocity for combined unit. As I said previously, how the water drag acts on the anchor (and of course the drag of the anchor chain) will affect the orientation of the anchor. Regardless, I hazard to guess that the broad, flat fluke is going to produce a whole lot more drag than the streamlined shank.

[Kenomac]

Originally Posted by Lodesman

With due respect you are talking nonsense. Gravitational acceleration is equal for the two objects, as they are roughly the same distance from the centre of the gravitational mass, ie. the Earth - 9.8 m/s/s. Acting against them is hydrodynamic drag, which you agree is the same for both balls as they are identical in size and frictional coefficient. Also acting against them is, what I will concede to, is called the force of buoyancy. Both balls displace the same volume, roughly 1.3 litres, which is not applied proportionately, but equally reduces the mass force of both balls - making the steel ball effectively 8.7 kg, and the tungsten ball 23.7 kg. Both balls will accelerate at the same rate, but increased speed equals increased drag (both form and frictional), and at a certain speed the force of that drag will match the force of mass and the object will cease to accelerate - it is said to have reached terminal velocity. Of course the lighter object will top out at a terminal velocity much sooner than the heavier object, so given enough depth the tungsten ball will eventually overtake the steel one. But I don't imagine that would be the case in the less than 10 m depths, in which most of us typically choose to anchor (I admit I don't know the extent of water-induced drag - it would take someone with a lot of impressive postnominals to break down the math on that or perhaps someone with a couple balls, enough water and time on his hands ). Naturally, a polyester ball with an SG barely above 1 will hit terminal velocity very soon after release, so will appear to waft slowly down.
That said, since the hollow shank and the lead-weighted tip are rigidly connected, the terminal velocities of the two won't be different - there will be one terminal velocity for combined unit. As I said previously, how the water drag acts on the anchor (and of course the drag of the anchor chain) will affect the orientation of the anchor. Regardless, I hazard to guess that the broad, flat fluke is going to produce a whole lot more drag than the streamlined shank.

Me thinks this guy is making sense.

The key fact appears to be that they are connected and fall as one object. Dockhead's submarine example doesn't work, the sub rises and sinks as one solid object, the propeller and stabilizer flaps are what controls the motion and balance. At least... that's what I understand from watching "Voyage to the Bottom of the Sea" as a yute.

[Dockhead]

Originally Posted by Kenomac

Me thinks this guy is making sense.

The key fact appears to be that they are connected and fall as one object. Dockhead's submarine example doesn't work, the sub rises and sinks as one solid object, the propeller and stabilizer flaps are what controls the motion and balance. At least... that's what I understand from watching "Voyage to the Bottom of the Sea" as a yute.

No -- a submarine's propeller and "stabilizer flaps" (you mean the control surfaces) most certainly do not control buoyancy. Submarines have buoyancy and will rise and sink without making way at all. Buoyancy is controlled separately by changing the weight of the vessel by pumping water into and out of the buoyancy tanks.

[Dockhead]

Originally Posted by Lodesman

With due respect you are talking nonsense. Gravitational acceleration is equal for the two objects, as they are roughly the same distance from the centre of the gravitational mass, ie. the Earth - 9.8 m/s/s. Acting against them is hydrodynamic drag, which you agree is the same for both balls as they are identical in size and frictional coefficient. Also acting against them is, what I will concede to, is called the force of buoyancy. Both balls displace the same volume, roughly 1.3 litres, which is not applied proportionately, but equally reduces the mass force of both balls - making the steel ball effectively 8.7 kg, and the tungsten ball 23.7 kg. Both balls will accelerate at the same rate, but increased speed equals increased drag (both form and frictional), and at a certain speed the force of that drag will match the force of mass and the object will cease to accelerate - it is said to have reached terminal velocity. Of course the lighter object will top out at a terminal velocity much sooner than the heavier object, so given enough depth the tungsten ball will eventually overtake the steel one. But I don't imagine that would be the case in the less than 10 m depths, in which most of us typically choose to anchor (I admit I don't know the extent of water-induced drag - it would take someone with a lot of impressive postnominals to break down the math on that or perhaps someone with a couple balls, enough water and time on his hands ). Naturally, a polyester ball with an SG barely above 1 will hit terminal velocity very soon after release, so will appear to waft slowly down.
That said, since the hollow shank and the lead-weighted tip are rigidly connected, the terminal velocities of the two won't be different - there will be one terminal velocity for combined unit. As I said previously, how the water drag acts on the anchor (and of course the drag of the anchor chain) will affect the orientation of the anchor. Regardless, I hazard to guess that the broad, flat fluke is going to produce a whole lot more drag than the streamlined shank.

Back to the woodshed, Lodesman! Shoo!

You are confusing hydrodynamic forces with buoyancy, which are entirely different and entirely separate things.

"Terminal velocity" is reached when the force of a falling object's aerodynamic/hydrodynamic drag equals its weight minus any force of buoyancy (and all objects have buoyancy in air as well as water). At that point, acceleration stops, because the force of gravity has been cancelled out. The object will then continue to fall at a constant speed.

Buoyancy is a different and separate force from aerodynamic/hydrodynamic drag, that is, the fluid dynamics force. Buoyancy opposes the force of gravity and reduces gravitational acceleration. Newton's Second Law says that:

"The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object."

So you see, the acceleration is produced by the FORCE acting on the body, and varies according to the magnitude of the force. Change the force, change the acceleration -- that's the whole point of Newton's Second Law. Note that Newton says NET force -- that is, the SUM of forces acting on the object.

That's why a neutrally buoyant object has no acceleration (drag could not do that) and a positively buoyant object has acceleration in the other direction, when fully immersed.

The math in my post above was done correctly. The acceleration of an object immersed in water is derived from the sum of gravitational and buoyant forces.