How good is the Rocna?

[Dockhead]

Originally Posted by Lodesman

You forget the heavier ball also has more inertia - which is why it doesn't fall faster than a lighter object (minus other factors). Here's a hammer dropping experiment, reminding you the hammer has more gravitational force. . .

In a vacuum, there is no buoyancy, and no aerodynamic or hydrodynamic drag to counteract the force of gravity, and all objects of whatever size, density, or shape fall at the same rate, because the only force acting is the force of gravity.

It gets interesting when something is falling in a fluid (air or water). Then other forces appear -- namely buoyancy, and drag -- which change the rate of acceleration.

[Lodesman]

Originally Posted by Dockhead

To observe clearly what balance does to an object in motion, you would really need to be dropping a steel-headed hammer. The balance of a rubber hammer will not be displaced enough, to make the effect easy to observe.

Have you done your own experiment?

But are we clear now on your two balls, and how buoyancy works?

Eh?

Your steel and tungsten balls wouldn't land at the same time, even if dropped off the Tower of Pisa. I think the root of all of this is that you've misinterpreted Galileo's experiment, which only works in a vacuum. Aerodynamic drag is a force opposing gravity, just like buoyancy, just like hydrodynamic drag. So dropped off the Tower of Pisa, your two balls would experience the same aerodynamic drag, but the tungsten one will be subject to 2.45x as much gravitational force. That means that aerodynamic drag will reduce acceleration of both balls proportionate to the amount of force opposing gravity (Newton's Second Law), so the steel ball will see a greater reduction of acceleration, than the tungsten one, and will land later. The tungsten ball will also have a higher terminal velocity, than the steel one.

This is the formula for terminal velocity, which should make the last point clear:

Attachment 132939

where:

Vt Terminal velocity (m/s)
g Standard gravity
d Diameter of the object (m)
Cd Drag coefficient (dimensionless)
ρs Density of the falling object (kg/m3)
ρ Density of the fluid through which the object is falling (kg/m3)


Note that density of the falling object is an operator, and buoyancy force is also considered, even in air. Buoyancy will affect the terminal velocity even of a tungsten ball falling through air, although the effect will be extremely small. But the point is that gravity fights buoyancy + drag, for any object falling through a fluid, that is, any falling object other than one falling in a vacuum. A falling object will stop accelerating (but keep moving at a constant speed) when gravity = buoyancy + drag (has reached terminal velocity). It will stop MOVING, when gravity = buoyancy (which is a state of neutral buoyancy).


Acceleration comes from force (Newton's second law!), and not just from the force of gravity. When some other force is acting on an object besides gravity, you have to do a vector sum of all the different forces, in order to know how it will accelerate and in what direction. If different forces are acting on different centers of the object, then you get angular acceleration (torque) as well as linear acceleration. And that's where Righting Moment comes from!

I assume you agree with Galileo's position that the two balls of different mass in a vacuum will fall at the same rate, despite the fact that one has a LARGE gravitational force and the other has a small gravitational force? Would you agree that if I cut a 1 kg chunk off each ball, they'll still fall at the same rate?
Instead we'll add air - drag is a function of cross-sectional area, coefficient of friction, density of the medium and speed. We can agree that these factors are equal for both balls at the outset (ie. speed is 0)? So both will accelerate at the same rate and at any given speed would feel the same force of resistance - as long as the force down exceeds that force of resistance, the items will continue to accelerate. When the force of resistance is equal to the gravitational force, the item will stop accelerating and maintain that speed - its terminal velocity, which we both agree will be lower for a lighter object. So when the steel ball tops out, the tungsten one will keep accelerating until it eventually hits its own terminal velocity. I said this some time back, but I still don't think we'd see that occur in shallow depths with wildly differing densities as between steel and tungsten, let alone the minor density difference between an Ultra anchor and a solid steel contemporary.

[Dockhead]

Originally Posted by Lodesman

I assume you agree with Galileo's position that the two balls of different mass in a vacuum will fall at the same rate, despite the fact that one has a LARGE gravitational force and the other has a small gravitational force? Would you agree that if I cut a 1 kg chunk off each ball, they'll still fall at the same rate?
Instead we'll add air - drag is a function of cross-sectional area, coefficient of friction, density of the medium and speed. We can agree that these factors are equal for both balls at the outset (ie. speed is 0)? So both will accelerate at the same rate and at any given speed would feel the same force of resistance - as long as the force down exceeds that force of resistance, the items will continue to accelerate. When the force of resistance is equal to the gravitational force, the item will stop accelerating and maintain that speed - its terminal velocity, which we both agree will be lower for a lighter object. So when the steel ball tops out, the tungsten one will keep accelerating until it eventually hits its own terminal velocity. I said this some time back, but I still don't think we'd see that occur in shallow depths with wildly differing densities as between steel and tungsten, let alone the minor density difference between an Ultra anchor and a solid steel contemporary.

YES -- now this is correct

In water, add the force of buoyancy to the force of drag, both counteracting gravity, and you've got it.

As to whether it's a big difference or not WELLLLLL -- that's not a difficult calculation, even for a non-physicist like (as my brother, who is a actual professor of physics, likes to point out )

Leaving aside drag (which as you correctly say, is not present at the moment of dropping the balls), the initial acceleration of the two balls dropped in water will be as follows:

Both balls displace 3.125 liters of water, weighing approximately (depending on temperature and salinity) 3.125kg.

The tungsten ball weighs 61.25kg, so the force of buoyancy reduces the initial acceleration from 9.80665 m/s2 by 5.102% to 9.30631 m/s2.

The steel ball weighs 25kg, so the force of buoyancy reduces the intial acceleration from 9.80665 m/s2 by 12.5% to 8.58802 m/s2.

It's a pretty significant difference! You would definitely be able to see the difference if you actually did this experiment. Hydrodynamic drag would widen the difference even more.

By the way, buoyancy is the reason why lead keels are so much better than iron ones. Iron is only 7x or 8x as dense as water, so buoyancy significantly reduces the ballast effect from a given mass of keel.

[Lodesman]

Originally Posted by noelex 77

Lodesman, if the toe was heavier and/or the shank lighter, do you think there would be no improvement?

While it's sitting on the seabed? Sure, why not.

[Dockhead]

Originally Posted by Lodesman

While it's sitting on the seabed? Sure, why not.

And why not while the anchor is dropping? I think we've now established that denser bits sink faster, no?

If one end of the anchor was full of lead, and the other had a hollow shank, this would make a really big difference between how fast each part sinks.

[Dave_S]

At the risk of being flamed.:-)

I don't think any of this is relevant. Any anchor can land any way up dependant on what it lands on.

What is important is what happens when you pull on the chain and the result will be determined by geometry. The few grams difference created by the air pocket is insignificant once the anchor has landed.

[Kenomac]

Originally Posted by Dockhead

And why not while the anchor is dropping? I think we've now established that denser bits sink faster, no?

If one end of the anchor was full of lead, and the other had a hollow shank, this would make a really big difference between how fast each part sinks.

But... the anchor falls as one solid connected object, not two. If dropped separately in water, the shank would sink slower than the lead tip. ????

Back to your hammer example, the handle is made from wood and the head from steel. When you drop the hammer via air, it doesn't always land on it's head.

But then again... if one were to drop the same hammer in water??? I can't do the experiment here because I'm landlocked. Can someone drop a wood handle hammer into water and get back to us on what actually happens. This would solve the mystery once and for all.

My hunch is, that the hammer will sink head first with the wood handle straight up all the way to the bottom, no matter how it's dropped into the water. The Ultra with the hollow shank would then be asumed to do the same.

Anyone have a hammer and swimming pool?

[Jim Cate]

Guys, all this theorizing about free falling bodies is good fun, but I must again point out that anchors in use are not free falling. They are attached at the end of their shanks to a chain, and the chain is routed out of a locker, through a chain pipe, around a gypsy, over a squeeky roller and thence into the water. All of those factors induce a large amount of drag, drag that is not distributed uniformly over the anchor, but right at the attachment point. I believe that this is the primary thing that influences the attitude of the anchor as it descends through the water and encounters the sea bed. Any motion of the chain away from vertical after the anchor hits bottom, as in the boat moving a bit, will influence the way the anchor then falls over from the vertical position it has while being lowered. Once it is lying on its side, geometry will quickly over ride any influence of buoyancy in how it engages the seabed and begins to set.

At least this is the way it seems to me, especially after watching some of Panope's videos.

Jim

[Lodesman]

Originally Posted by Dockhead

I'm not sure why you're not getting this -- do you not understand that gravity is just one force, among many? How do you think anything floats, or flies? Only because buoyancy, or lift, counteracts the force of gravity. A hovering helicopter is producing lift which equals the force of gravity, and will start to gradually accelerate upwards as the lift exceeds the force of gravity.

You're verging on becoming insulting.

Originally Posted by Dockhead

And why not while the anchor is dropping? I think we've now established that denser bits sink faster, no?

If one end of the anchor was full of lead, and the other had a hollow shank, this would make a really big difference between how fast each part sinks.

No I don't believe we've established that.

Originally Posted by Dockhead

Both balls displace 3.125 liters of water, weighing approximately (depending on temperature and salinity) 3.125kg.

The tungsten ball weighs 61.25kg, so the force of buoyancy reduces the initial acceleration from 9.80665 m/s2 by 5.102% to 9.30631 m/s2.

The steel ball weighs 25kg, so the force of buoyancy reduces the intial acceleration from 9.80665 m/s2 by 12.5% to 8.58802 m/s2.

It's been some time since my skool daze, but as I recall the reason we use 9.8 m/s/s is because we can generally ignore the gravitational pull of a ball, anchor or hammer, when it's compared to the relatively massive pull of the Earth. Yet you figure the relatively piddly gravitational pull of 3 litres of water reduces the Earth's pull on the steel ball by an 1/8. I'm not convinced. And btw, I dropped a steel hammer, and just as with the mallet, it landed in the same orientation it started in.

Frankly I'm with Jim. Drag of the anchor chain will play are far greater role.

[Dockhead]

Originally Posted by Lodesman

You're verging on becoming insulting.

No I don't believe we've established that.


It's been some time since my skool daze, but as I recall the reason we use 9.8 m/s/s is because we can generally ignore the gravitational pull of a ball, anchor or hammer, when it's compared to the relatively massive pull of the Earth. Yet you figure the relatively piddly gravitational pull of 3 litres of water reduces the Earth's pull on the steel ball by an 1/8. I'm not convinced. And btw, I dropped a steel hammer, and just as with the mallet, it landed in the same orientation it started in.

Frankly I'm with Jim. Drag of the anchor chain will play are far greater role.

No insult intended; I'm really trying to help you with this.

OK, so you've got now that gravity does NOT produce constant acceleration, in the presence of other forces, right? You've now written that you understand that your steel ball and tungsten ball will have different terminal velocities. Obviously acceleration has stopped altogether when terminal velocity has been reached and is being reduced constantly as drag builds up with velocity. Clear, right?

The force of buoyancy works the same way as drag -- it counteracts gravity. Therefore, acceleration in the presence of a force of buoyancy will NOT be constant, just like acceleration in the presence of a drag force is not constant.

How to calculate the force of buoyancy is very simple -- it's equal to the mass of water displaced. So the "piddly" 3.125 liters of water displaced produces 3.125kg of bouyancy (+/- a little depending on temperature and salinity) -- is that piddly? Well, it's 12.26% of the weight of the steel ball, and so will reduce initial rate of acceleration by exactly 12.26%. This is not controversial, Lodesman -- Newton's Second Law in black and white. It has nothing to do with the gravitational attraction of the displaced water -- it has to do with the mass of the displaced water -- because any motion down of the ball requires motion up of the same volume of water (and that's how buoyancy works).

The greater density of the tungsten ball will also make a big difference in the effect of hydrodynamic drag (but you already know that). So due to the combination of drag and buoyancy, the steel ball, being less than half as dense as the tungsten ball, will fall much more slowly than the steel one will, in air and even much more so in water. The physics of this are very simple and uncontroversial, and have been laid out for you.


Whether the chain has a bigger effect on the sinking attitude of the anchor, than the balance and distribution of denser and less dense materials, I don't know. But it is objectively true and cannot be doubted that having denser material in the toe of the anchor (lead) and lighter material in the shank (hollow steel with trapped air, or at least a lightweight space frame construction like the Spade) will produce quite a bit of righting moment. The chain would not fall faster than an anchor with lead ballast. That might mean the chain will pull the shank up -- I don't know. But the chain will probably not turn the anchor around its long axis, and this is the orientation which is critical for getting a good set -- and the lack of this orientation while falling is exactly what caused the fail posted by Noelex above. Balancing the anchor towards the point of the fluke, and creating righting moment, will certainly help the anchor land the right way, and not like the anchor in Noelex's post above. That's why Ultra and Spade have lead ballast.

[Dockhead]

Originally Posted by Kenomac

But... the anchor falls as one solid connected object, not two. If dropped separately in water, the shank would sink slower than the lead tip. ????

Back to your hammer example, the handle is made from wood and the head from steel. When you drop the hammer via air, it doesn't always land on it's head.

But then again... if one were to drop the same hammer in water??? I can't do the experiment here because I'm landlocked. Can someone drop a wood handle hammer into water and get back to us on what actually happens. This would solve the mystery once and for all.

My hunch is, that the hammer will sink head first with the wood handle straight up all the way to the bottom, no matter how it's dropped into the water. The Ultra with the hollow shank would then be asumed to do the same.

Anyone have a hammer and swimming pool?

Yes, you've got it.

The hammer will TEND to land on its head in air due to balance, but of course not always and not unless there is some at least slight force creating torque. Think of knife-throwers.

But in water, the much stronger buoyancy effect in water will create a strong righting moment which will reliably make the hammer fall head-down.

Same with anchors, and that is why the Ultra has the trapped air. That will work together with the lead ballast to provide even more righting moment than the Spade has. You can see from Noelex's photo, what happens if the anchor doesn't orient itself properly while falling.

By the way, a little drift from this fun physics discussion --

there was some discussion about whether the Ultra shank is actually sealed or not. I have been thinking about it, and I realized that it doesn't really matter. If there is a hole at the bottom which will let water drip out while the anchor is stored, the cavity will stay full of air anyway at least until the anchor is down, after which it doesn't matter. If it's designed like that, then that's really clever and my hat's off to the Ultra people. I thought this was just a cheap Spade copy and it looks like I was wrong.

[noelex 77]

I hope these photos help illustrate anchor behaviour and the importantance of a light shank.

This photo of my Mantus shows the typical setting position of a modern anchor. You can see how the toe of the fluke is aggressively forced into the substrate.




I don't have any photos of the Mantus in a shank down position because it is not stable in this orientation. I would have to hold it down, but imagine if the shank was heavier. As the shank is forced down the toe is forced up. In the worst case with a very heavy shank the anchor would be most stable with the shank at the bottom and fluke at the top sitting above the sustrate.





From this setting position for the anchor to work well it has to rotate level. This means forcing the shank upright. The anchor's geometry achieves this as it is pulled through the substrate, but once again this is more difficult to do if the relative weight of the shank becomes heavier.

This is the intermediate stage of the anchor starting to set:




This is the ideal finish position. The fluke is completely level, which means the shank has to be at the top. Not so easy if the shank is heavy. In this level position the anchor will bury all of its fluke and be in the best position to cope with a change in direction of pull.




This is why anchor manufacturers go to so much trouble and expense to make the shanks as light as possible. Many use high tensile steels, others use fabricated shanks consisting of numerous bits of steel that need to be welded together. They could save all this trouble, as the shank can be made just as strong out of slightly thicker simple mild steel.

Increasing the buoyancy of the shank is simply another method of reducing the effective weight of the shank.

[a64pilot]

All this talk about anchor buoyancy reminds me of a story.
UH-60 Blackhawk helicopter has a neat feature, the main spars of the helicopter blades are pressurized with nitrogen, this is so if a crack develops the nitrogen will leak out and the crack will be detected by the pressure drop. But like everything periodically the nitrogen pressure has to be topped off.
Lt. walks up to a crewchief that has a large steel bottle with the word NITROGEN clearly labeled in large block yellow letters, and is connecting it to a blade, Lt asks what he is doing, and without cracking a smile he says he is filling the blades with Helium as they have found that a helicopter with Helium filled blades will lift more weight and burn less fuel than one that is not.

I have a strong suspicion that if in fact there is trapped air in an anchors shank, it will have about as much effect as pressurized Helium would in a helicopters blades

[a64pilot]

Nolex, methinks the pivot point may have more effect on how the anchor drops tip down than the weight of the shank?

More than one way to skin a cat, and I do think that these anchors are logically designed and through experimentation, the designs refined.

The anchor that has carried the high strength steel / lightweight shank the farthest that I have seen is my Deep Set Danforth. That thing is I believe thin spring steel, but I think it's not done for weight but for other reasons.

[conachair]

Originally Posted by Dockhead

The force of buoyancy works the same way as drag -- it counteracts gravity. Therefore, acceleration in the presence of a force of buoyancy will NOT be constant, just like acceleration in the presence of a drag force is not constant.

Thats a very big "therefore"...., a force of buoyancy is a constant if the liquid stays the same, how can the acceleration not be constant if the force is constant?
(ignoring a tiny change for increase in density).
Drag is completely different and changes with velocity through the liquid.
A bit of buoyancy inside an object just means the object is a bit lighter.