Terminal velocity of a dropped anchor.....

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DaveNTL

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..... that isn't attached to a rode? /forums/images/graemlins/frown.gif

My new second anchor fell off by bouncing out of it's retaining pin when pitching around in about 2800 metres deep water. I decided not to dive for it.

Interestingly I think the depth guage picked it up as it went past. I had a 'last' depth of about 299 feet which changed to 8 in the really rough patch.

Anyway, as you do on longish trecks, I got to thinking how long it would take it to sink to the bottom assuming it didn't hit a submarine on the way down? I guessed at a terminal velocity of about 8m/s which would be about 6 minutes.

I was pretty sure it was Bernoulli's equation I needed to do it accurately or is it to do with developing a Reynolds number for a 20kG Delta? I'm sure someone will enjoy that kind of thing but I'm still in mourning.

RIP Delta. Much loved. Never used in anchor. /forums/images/graemlins/blush.gif
 
Could have been worse..........you could have dropped it on your foot! /forums/images/graemlins/smile.gif
 
you already know that a retarding force that is proportional to the first (linear) power of the velocity is typical in cases where the viscosity of the medium is significant. For objects falling through less viscous mediums, it is more typical that the retarding force is proportional to the second power (quadratic) of the velocity. In this case the retarding force is a result of the momentum exchanged when the object collides with the particles in the medium. In many circumstances the velocity dependent force does not necessarily obey the simple linear or quadratic dependence on velocity and the value of p is not even necessarily an integer.

You can model the movement of an object moving through a viscous medium with the following differential equation:

dv
p
m =mg - k v
dt

where m is the mass of the object,
k is a scaling or proportionality factor that accounts for
the area experiencing the viscous force,
v is the velocity, and
p will be determined from in vitro data.

I'll ask Bernoulli, when I get into work on Monday. /forums/images/graemlins/wink.gif
 
I wonder what the equation is for the changing pressure density of the water, if any?
Also, being a Delta, and being unfettered by chain, would it invert and the blades provide lift, thus inducing a glide path, or would it drop vertically with the shank providing a tail?

In addition, there would be currents affecting the path, so therefore the total distance travelled will be greater than the 2800m depth.

So there must be some extra vectors to add to the 2800m path.
 
[ QUOTE ]
Interestingly I think the depth guage picked it up as it went past. I had a 'last' depth of about 299 feet which changed to 8 in the really rough patch.

[/ QUOTE ]

Nice to think it was the dying echo of an anchor on its way to anchor nirvana, but probably a spurious reading caused by air bubbles in the rough stuff.
 
I did ask Bernoulli and he said it was more complicated than that because he wasn't sure that water could still be thought of as incompressible under those conditions - i.e. great depths and pressures (well, maybe Bernoulli knows, but I don't!)

At that stage I gave up and got back to work.
 
Personally I would not worry, the terminal velocity is zero! Now if you mean the velocity just before it hits the seabed, that is a different question altogether /forums/images/graemlins/smile.gif
 
Ah but we haven't considered the possibility of the anchor not exhibiting an oscillating or tumbling action thereby boundary layer effects would have to be modelled and calculate its Reynolds number and have a look at the works of Stokes etc.

In addition Bernoulli's law only holds for a non -compressible fluid of uniform density so as the density changes (temperature /salinity etc) could not be applied in a simple form.

EEEEEEEKKKKKK now going to read (if that's the word) a copy of the Daily Sport or simerlar /forums/images/graemlins/blush.gif and relax my mind -oh how I thought I'd forgotten all that stuff - used to write software that modelled the flow of "stuff" though pipes
 
[ QUOTE ]
had it been a spade, or Rocna, of course, it would have dug in to the water, and not fallen at all.

[/ QUOTE ] That's nothing - a CQR would have been so offended by the lack of chain that it would have absolutely refused to move from the foredeck in the first place.
 
As luck would have it, terminal velocities are my specialist subject, though usually of smaller things like crystals and such.
There is a problem in this case - the shape. Terminal velocity for spheres is straightforward. For non-spheres we normally calculate a Shape Factor, the ratio of Surface Area to Volume of the particle. I can easily calculate the volume of a 20kg Delta (about 0.0026m^3), but the total surface area is more than I can do in a tea break.
Simplifying the problem to a 0.17m diameter sphere, I find that the flow regime is turbulent so the drag coefficient is 0.1 giving a terminal velocity of 12m/s with a corresponding Renolds number of 2 million, confirming the turbulent flow regime.
Incidentally, my correlations for non-spheres are not applicable in turbulent flow so don't bother calculating the surface area, I can't use it.
So it took 230 seconds to sink. Does that help?
Derek
 
Anyway, as you do on longish trecks, I got to thinking how long it would take it to sink to the bottom

Yep, Iv'e been on those trips................ worst one was when somebody on the boat said "did you know that there are no pulling forces (or was that pushing)( who G A S) in nature" ......

Honest I didn't know the Mull was so far from Largs

I have a rule now that we can only talk Sex Drugs and Rock and Roll! /forums/images/graemlins/smile.gif

Tom
 
[ QUOTE ]
So it took 230 seconds to sink. Does that help?

[/ QUOTE ]

Very impressive responses one and all. I'm pleased with my 6 minutes guess. If you calculated about 4, but not taking into account that it might zig zag its way down as Fullcircle says I reckon I could have been in the right ballpark..... er.... anchorpark.

The lesson learned of course is to belay the retaining pin in future.
 
Did you know that the world record for anchoring is actually 2799 metres and that a species of fish has evolved that live just below that level???

MURDERER /forums/images/graemlins/smile.gif
 
[ QUOTE ]
Did you know that the world record for anchoring is actually 2799 metres and that a species of fish has evolved that live just below that level???

[/ QUOTE ]

Is that below chart datum or actual depth?
 
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