mjcoon
Well-known member
And you are ignoring the figures given for rock and steel density in post #17 and assuming that rock is denser? Or perhaps failed O level physics?
Mike.
How much does my 40Kg anchor weigh when in sea water?
- Thread starter tudorsailor
- Start date
And you are ignoring the figures given for rock and steel density in post #17 and assuming that rock is denser? Or perhaps failed O level physics?
Mike.
Buck Turgidson
Well-known member
And you are ignoring the figures given for rock and steel density in post #17 and assuming that rock is denser? Or perhaps failed O level physics?
Mike.
hehe
my bad.
savageseadog
Well-known member
The anchor's mass is unchanged in or out of the water. The weight does decrease in water as calculated previously.
Neeves
Well-known member
As long as Tudor Sailor sticks to the edges of the Atlantic and does not hang around too long in the middle then I suspect he will not deploy all of his chain at once. Most of the chain deployed sits on the bottom and is retrieved bit by bit and its just the last length of chain, say 10m his anchor, type XXXX, weight 40kg, the ball of mud and the tension needed to break it out that are critical.
I suspect these will be well below the WLL of the windlass though model YYYY might carry less mud and weed than model XXXX.
I think the greatest load occurs when the anchor is turned over the bow roller - so advise must be get rid of the ball of mud/weed when it still hangs vertically. I might also suggest that if this causing angst - get a new anchor.
Jonathan
I suspect these will be well below the WLL of the windlass though model YYYY might carry less mud and weed than model XXXX.
I think the greatest load occurs when the anchor is turned over the bow roller - so advise must be get rid of the ball of mud/weed when it still hangs vertically. I might also suggest that if this causing angst - get a new anchor.
Jonathan
pathfinderstu
Member
If you really have to know how much your anchor weighs in water why not tie it to a piece of rope lower it into the water and weight it with appropriate calibre spring balance or like.
I personally think the specs the makers of the windlass concerning usage is enough let them worry about suitable loads, besides if it becomes overloaded the breaker should cut in allowing you to adjust the situation.
I personally think the specs the makers of the windlass concerning usage is enough let them worry about suitable loads, besides if it becomes overloaded the breaker should cut in allowing you to adjust the situation.
Neeves
Well-known member
What I love on some of these threads is that if collect the learned knowledge (and remember it) you could pass 'O' level physics
Jonathan
Jonathan
pyrojames
Well-known member
Well the equation is correct, but the arithmetic hasn't made O level!
Beaten to it by mistroma...
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noelex
Well-known member
If you want an accurate figure I think this is the only practical method.
The first problem is that anchors rarely weigh their nominated weight. Variations of +/- 10% and more are quite common. The second problem is we need to know the mix of materials. In water, aluminium will lose much more weight in water than steel which in turn will loose more than lead. So we need to know the relative mix of the different materials.
Then there is the issue of buoyancy. Vent holes placed in chambers for galvanising are not always at the top when when the anchor is in the setting or set position. It is quite common to see bubbles escaping as the anchor is retrieved. This buoyancy means the weight is not what would be calculated from a simple material calculation. Even worse, the weight is changing as the anchor is retrieved, or depending how the anchor falls to the seabed and sets.
We need to determine if it is fresh or saltwater and even if it is in the Med or Bay of Bengal. Finally, there is even the effect of magnetic attraction and variations in the earths gravitational field to consider (now we are being silly
).Well you did ask
.Of course, this is only a bit of fun, of academic interest only. It is difficult to know why the exact weight of the anchor would be important.
For most cruising boats with an electric windlass the majority of the weight will be in the chain. If you drag into deep water you will need a windlass that can lift this total weight (and the windlass rating is at stall speed, with no friction so a factor of x3 is often used). The anchor weight will typically only be a small fraction of the windlass pulling power that is required.
If you work on -5%, you are likely to be close for a steel anchor, but weigh the anchor first. It may not weigh in air what you expect.
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Cariadco
Active member
Yeah, Yeah......I still think my Lewmar Delta's the best anchor......
mjcoon
Well-known member
You have to remember to recognise and filter out the wrong answers...What I love on some of these threads is that if collect the learned knowledge (and remember it) you could pass 'O' level physics
Jonathan
Bouba
Well-known member
If something is submersed in water (as opposed to floating) then it’s weight is the weight of the water above it...so the deeper you anchor the more it must weigh
dombuckley
Well-known member
No, the weight of the water above increases the surface pressure experienced by an object, but this applies equally on all sides, so does not alter the weight of the object. Likewise, the density of seawater remains virtually unaltered with depth, so the apparent weight remains unchanged. Otherwise, every submarine would go into an uncontrollable terminal dive as it submerged.
Wansworth
Well-known member
Old car engine…..did art 
Bouba
Well-known member
The pressure on an object is its weight...if you carry a sack of potatoes on your back...then step on a scale...that is your weight...if you subtract the spuds from your weight..then all you get is your weight on the surface...and the question is what is your weight at depth...and apart from a little lighter due to being closer to the earths center, it has to be the water...or why the question
Neeves
Well-known member
I think some seawaters have very high concentrations of dissolved solids (some so high the solids precipitate out) - do we need to factor that into the equation....I think we might start to move from 'O' level to 'A' level physics soon.
This thread is 7 years old ..... is Cariadco bored or simply trying to provoke (if so - unsuccessfully)
Jonathan
Bouba
Well-known member
Here is all everyone’s problem..everyone is trying to calculate the mass…when the question is weight..ie the gravitational effect of an object by the earth…discounting a small variation due to the depth below sea level and where you are on the globe..the weight will be the anchor and everything that is resting on the anchor..shackle, chain and water
Neeves
Well-known member
Much depends on why the OP asked the question. After all - if its embedded in the seabed, in terms of the yacht (or windlass) or OP if he retrieves by hand - it weighs nothing. It only weighs, such that impacts the OP, when he tries to retreive.
Now its not the weight (whatever that is) but how deeply it is in the seabed - viz what is its hold and then he needs to wonder about the tension to break out (weight is not relevant)
So we need to know what sort of anchor, how much chain, ,,,,,,
once it breaks out - is the OP up to date and following anchor threads - has he upgraded to smaller high tensile chain......?
Too many unknowns - as usual not enough information to offer a sensible reply.
But its an anchor thread.....
Jonathan
Now its not the weight (whatever that is) but how deeply it is in the seabed - viz what is its hold and then he needs to wonder about the tension to break out (weight is not relevant)
So we need to know what sort of anchor, how much chain, ,,,,,,
once it breaks out - is the OP up to date and following anchor threads - has he upgraded to smaller high tensile chain......?
Too many unknowns - as usual not enough information to offer a sensible reply.
But its an anchor thread.....
Jonathan
dombuckley
Well-known member
OK, let's say you are right: let's calculate this "increase in weight" due to the water column. Take a small dinghy anchor, surface area of say 100cm2. That's 1/100 of a square metre. For every meter it is lowered, the water column above will weigh 1/100th of a ton. So the anchor would increase in weight by 10kg for every meter below the surface. By the time it's 10metres down, nobody would be able to recover it without a substantial mechanical assistance. My 25lb CQR, with a surface area of approx 900cm2, would be torn out of my hand before I even lowered it a meter below the surface. That is clearly not the case. As I stated before: water pressure has no effect on the weight of an object, as it applies equally all round (unlike your sack of spuds analogy, where the pressure is downward only).
Oh, and gravity increases as you get closer to the centre of the earth, not decreases.
Bouba
Well-known member
No..because when you recover it, you change it’s shape..that’s how it is designed to be retrieved..
rogerthebodger
Well-known member
The original question is "What is the weight of a 40kg anchor in seawater."
Any iten when submerged in any fluid will be subject to 2 forces.
1) Gravity
2 Buoyancy
the weight of a 40 kb anchor under gravity will be err 40 Kg
Buoyancy based on Archimedes principle is F = fluid density * acceleration due to gravity * fluid volume.
You can work out the formula or determine the reduction in weight by dunking your anchor in a tub of seawater and measure the displacement volume the increase in depth of the sea water.and multply this volume by the density of the sea water displaced.
You could also collect the increase in sea water volume then weigh it.
deduct the weight of sea water displaced and from the 40 Kg will give you the weight of anchor submerges in water
Don't forget the density of sea water various with the location around the world as in the plimsoll line on the sides of ships
Density is lowest at the surface, where the water is the warmest. As depth increases, there is a region of rapidly increasing density with increasing depth, which is called the pycnocline . The pycnocline coincides with the thermocline , as it is the sudden decrease in temperature that leads to the increase in density.
Any iten when submerged in any fluid will be subject to 2 forces.
1) Gravity
2 Buoyancy
the weight of a 40 kb anchor under gravity will be err 40 Kg
Buoyancy based on Archimedes principle is F = fluid density * acceleration due to gravity * fluid volume.
You can work out the formula or determine the reduction in weight by dunking your anchor in a tub of seawater and measure the displacement volume the increase in depth of the sea water.and multply this volume by the density of the sea water displaced.
You could also collect the increase in sea water volume then weigh it.
deduct the weight of sea water displaced and from the 40 Kg will give you the weight of anchor submerges in water
Don't forget the density of sea water various with the location around the world as in the plimsoll line on the sides of ships
Density is lowest at the surface, where the water is the warmest. As depth increases, there is a region of rapidly increasing density with increasing depth, which is called the pycnocline . The pycnocline coincides with the thermocline , as it is the sudden decrease in temperature that leads to the increase in density.