Weight under water

[John Vigor]

I presume everybody knows that a 25-pound anchor weighs less under water than it does in air--but how much less? Can somebody give me the formula that I must have learned in a physics class a few aeons ago, but have now forgotten?

I seem to remember it has something to do with specific density, and volume of displaced water, but how it is applied is a mystery to me. How much does 1/4-inch chain weight per foot under water, for example? And (something that has always interested me) how much do those large aluminum anchors weigh under water? Can there possibly be any weight left to help them dig into the bottom?

I used to be able to weigh a 35-pound CQR on an all-chain 5/16th-inch rode without the aid of a windlass, but tempus has fugitted and I wouldn't like to try it now. I can remember raising it from 90-foot depths in the island of St. Helena, but I sometimes wonder what kind of weight I was actually lifting. Can anyone figure it out?

Cheers,

John V.

[Ed Haley]

The weight that a submerged object loses while under water is related to the volume of the object. The weight of the volume of water that the object displaces is the weight loss.

You can also do it with density if you know the weight of the anchor.

[Tom in Cambria]

Figuring the volume in an irregularly shaped item such as an anchor or chain is a challenge without calculus, so you can always do it the way Archimedes did it back when. Fill a container with water right to the brim and then submerge the anchor in it. The water will rise and overflow. The amount that overflows is the volume of the anchor. So you can put a bucket inside a tub and submerge the anchor so that the water overflows the bucket and collects in the tub. That's the volume in the anchor. Eureka! Now you could ladle it into quart bottles and since when it comes to water a pint is a pound that's 2 pounds per quart or 8 pounds per gallon. Of course you could also just weigh the water that overflows and not worry about how many quarts and pints it might be.

If you actually go to the trouble to do this how about putting it in a future book and posting it here on the BB so we'd all know without getting wet.

[Neil Gordon]

>>And (something that has always interested me) how much do those large aluminum anchors weigh under water? Can there possibly be any weight left to help them dig into the bottom?<<

They don't weigh anything. That's why they need those pointy things to hold them onto the bottom.

>>I used to be able to weigh a 35-pound CQR on an all-chain 5/16th-inch rode without the aid of a windlass, but tempus has fugitted and I wouldn't like to try it now.<<

Proving that the anchor and chain got heavier over time? What's the formula for that/

[Greg Ross]

John,
Start with this link for a good cross section of weights and densities;
http://www.reade.com/Particle_Briefings/spec_gra2.html
For a quick estimate, Iron/ steel will run approximately 500 lbs per cubic foot and a cubic foot of sea water is 64 lbs. Therefore your 35 Lb anchor representing 7% of 500 Lbs would displace 7% of the 64 lbs of water. That means the net weight submerged should be 35 minus 4 Lbs. equals 31 Lbs.
Your 5/16" chain should run approximately 1 Lb per foot, 90 Lbs being roughly 20% of of the 500 Lb/ Cu Ft of steel tranlates to 20% of the 64 Lb/ Cu Ft of sea water being displaced.
That's minus 12 Lbs, so all up 31 Lbs of anchor and 78 Lbs of chain, and you were hoisting 109 Lbs.
I think!

[tartansailor]

That's the way to calculate.

Dick

[darmoose]

The weight that a submerged object loses while under water is related to the volume of the object. The weight of the volume of water that the object displaces is the weight loss.

so let me understand this?

two golf ball sized balls, one made of lead and one made of aluminum will lose the same amount of weight based on their volume?

i would think the weight loss has more to do with the density of the water than the volume of the object. more weight loss in salt water than fresh water. you know?

darrell

[darmoose]

....or better question?

two objects, that weigh the same, but have different volumes (perhaps because one is hollow) will lose different amounts of weight when submerged?

not sure thats true?

darrell

[Chris Reinke]

The story goes something like this.....
Archimedes was a Greek philosopher. The story goes that the king of the day wanted a new crown made entirely of gold. After the goldsmith made it, the king was suspicious that the goldsmith made it of iron and coated it with gold. The king did not want to destroy the crown if in fact it were gold, so he asked Archimedes to figure out if it was pure gold or not without destroying the crown. Archimedes pondered this day in and day out until one day while
getting into his bath he discovered the principle that bears his name. The tub was full to the rim, and when he sat down in it, he noticed that the more he sank himself into the water, the more water that spilled over the side of the tub. He was supposedly so happy to make this discovery that he ran out into the streets naked shouting "I found it!" What he discovered is that the amount of water displaced by an object depends on the mass of that object (not the weight). If he knew the mass of that object, and the volume of fluid it displaces, he could determine its density. Since the densities of iron and gold are different, he did a test. He determined the density of the crown and compared it to the density of pure gold to see if they were the same. Legend says they were not the same, so the king was tricked. Shortly thereafter the gold smith was put to death....he was not a big fan of Archimedes.

[BillNH]

two objects, that weigh the same, but have different volumes (perhaps because one is hollow) will lose different amounts of weight when submerged?

not sure thats true?

darrell

Yes, it's totally true. That's why steel ships float, yet the same amount of steel packed into a solid block will sink like, you know, a steel block! Same masses but different volumes gives different average densities. If the average density of an object is less than that of water, it floats. If greater than water, it sinks...

Here's a puzzler to really think about density and buoyancy: A man is paddling a canoe in a swimming pool. He is holding a big iron cannon ball on his lap. He tosses the cannon ball out of the canoe and into the pool. What happens to the water level in the pool, and why? Answer to follow...

[Carl Thunberg]

Darrell,

I think you may be confusing buoyant force with weight. There's never any loss of weight, since the mass of the object doesn't change, and the acceleration by gravity at sea level is a constant. What does change is the magnitude of the buoyant force in the opposite direction. For solid objects, Greg's post provides the correct answer to John's original question.

A hollow object with the same mass as a solid object of the same material will displace more water as it gets submerged, and therefore has a higher buoyant force. There's never any loss of weight, just an offsetting buoyant force in the opposite direction that makes a submerged object easier to lift. I think that's where the confusion may be.

I'm a diver. There was a time when I occupied less volume than I do now. As my volume has increased, I have found I need to add more weights to my weight belt in order to get submerged. I'm sure Sea Hunt doesn't have this problem.

P.S. - The water level in the pool shouldn't change, since the canoe already displaced the weight of the cannon ball. Man, I hope I'm right.

[Ed Haley]

Sure there's a weight loss when an object is submerged. It loses an amount of weight exactly equal to the weight of the water it displaces.

You're confusing weight and mass. The mass remains the same but the weight changes.

To give you an example, there were members of our yacht club who used to grab a 5 gallon plastic pail and fill it with cement. They would use that as a mooring anchor for their $20,000 party barge. The problem being the pail of concrete lost about half it's weight when submerged and failed horribly as a mooring. Then they began adding rebar and chunks of iron to the pail (increase density) before pouring in the cement. It was better but not as good as a lead mushroom anchor.

An astronaut loses all his/her weight in about a minute as he/she is placed in orbit. Mass is still the same.

[Carl Thunberg]

The astronaut example works because the gravitational constant changes in space, which is specifically why I stated that the acceleration of gravity is the same at sea level. The weight does not change. It is constant in a downward direction. The buoyant force is equal in magnitude to the weight of water displaced by the volume of the bucket and it is in an upward direction. The weight of the object (directed downward) minus the buoyant force (upward, causing it to tend to want to float), is the net downward force.

Using the concrete-filled bucket example, let's pretend the volume is one cubic foot just to make the math easier. The weight of the concrete filled bucket is 150 pounds since the density of concrete is 150 pounds per cubic foot. The buoyant force in salt water is roughly 64 pounds. The net force in a downward direction is 150 minus 64, or 86 pounds. So, yes, the bucket is less effective as mooring than you would hope. For solid objects that sink, it's convenient to think of it as a submerged weight, as long as we realize there are actually two forces at work here and it's not 100% accurate. The only reason I brought this up was in response to Darrell's questions. It really is an esoteric point.

[M. R. Bober]

The story goes something like this.....
Archimedes was a Greek philosopher. The story goes that the king of the day wanted a new crown made entirely of gold. After the goldsmith made it, the king was suspicious that the goldsmith made it of iron and coated it with gold. The king did not want to destroy the crown if in fact it were gold, so he asked Archimedes to figure out if it was pure gold or not without destroying the crown. Archimedes pondered this day in and day out until one day while
getting into his bath he discovered the principle that bears his name. The tub was full to the rim, and when he sat down in it, he noticed that the more he sank himself into the water, the more water that spilled over the side of the tub. He was supposedly so happy to make this discovery that he ran out into the streets naked shouting "I found it!" What he discovered is that the amount of water displaced by an object depends on the mass of that object (not the weight). If he knew the mass of that object, and the volume of fluid it displaces, he could determine its density. Since the densities of iron and gold are different, he did a test. He determined the density of the crown and compared it to the density of pure gold to see if they were the same. Legend says they were not the same, so the king was tricked. Shortly thereafter the gold smith was put to death....he was not a big fan of Archimedes.

Eureka!! As they say.

Mitchell Bober
Sunny Lancaster (where the design of the anchor is more important that its weight), VA

[BillNH]

Here's a puzzler to really think about density and buoyancy: A man is paddling a canoe in a swimming pool. He is holding a big iron cannon ball on his lap. He tosses the cannon ball out of the canoe and into the pool. What happens to the water level in the pool, and why? Answer to follow...

P.S. - The water level in the pool shouldn't change, since the canoe already displaced the weight of the cannon ball. Man, I hope I'm right.

Thanks for being brave enough to take a stab at it! Actually the water level in the pool will go down. When the cannonball is in the floating canoe, the canoe sinks down in the water until it displaces an additional weight of water equal to the weight of the cannonball. This water displaced will have a greater volume than the cannonball since water is less dense than the cannonball. However, when the cannonball is thrown into the pool, it sinks and now only displaces a volume of water equal to its own volume. Between the canoe and the cannonball less water in total is now being displaced, so the water level in the pool will adctually drop... counter-intuitive, isn't it!