Saturday, October 21, 2006

Water weighs how much?

Yesterday, I bumped into one of the dive master candidates who recently completed his dive master classes. He was having all kinds of problems with the physics portion of the course and was lamenting his travails in trying to pass the exams. We will be getting together next week sometime to work on his physics and try to get him over the hurdle.

In the interim, I am going to review some of the most salient points in the physics for dive masters. What I would like to do is boil the issues down to the most critical ones. My intention is not to cover the entire physics portion, but to go over the parts that most students seem to have problems with. If anyone is interested in learning more about dive physics, physiology, equipment, I would highly recommend the PADI encyclopedia of recreational diving. There is a printed version and a version that you can install on your computer that includes videos etc. Again, the ERD is highly recommended.

So, why do we need to know how much water weighs? After all we are not going to haul water around.

Just about everyone knows that things seem to weigh less when submerged in water. So what is the reason and why is this important.

Water has the odd effect that it "pushes things up" to the extent that you are "pushing" water away for an object. For example if you put a SCUBA tank weighing approximately 40 pounds in the water it may feel like it only weighs a fraction of the actual weight. The reason is that the SCUBA tank displaces (pushes water away) to make place for the object.

So to determine how much the SCUBA tank weighs (or seems to weigh) in water, you have to know how much water has been displaced (pushed away). The amount is called the displacement volume. For example if the object we are putting in the water is one foot by one foot by one foot (a cube) then we know that the displacement volume is 1 cubic foot.

Next we need to know how much a cubic foot of water weighs. For example one cubic foot of sea water weighs 64 lbs and one cubic foot of fresh water weighs 62.4 lbs.

Now that we know how much the water "pushes back" we can determine what the effect of the push back is on the object we are putting in the water. Lets say the 1 cubic foot object weighs 100 lbs on the surface, we can find out how much the object will appear to weigh in the water. So the apparent weight is 100 lbs - 64 lbs = 36lbs.

This means that the effect of the sea water "pushing back" on the object will make it seem to only weigh 36 lbs.

Does this make sense?

Say for example you were asked by a friend to recover this 1 cubic foot object from the local quarry that you dive in. To do this, you will need to know how much lift you need to use to lift the object (how large a lift bag to use).

Right now the object is negatively buoyant by 36 lbs. Therefore we need to add extra lift to the object of at least 36 lbs to make it neutrally buoyant. To lift the object we will have to use a little more lift to make it positively buoyant to make it rise to the surface.

So we must again ask ourselves how much does water weigh to find out how much water we must displace (push away) to cause lift of 36 lbs. We know that 1 cubic foot of sea water weighs 64 lbs, so to displace 36 lbs we have to displace 36 divided by 64 which is approximately 0.56 cubic feet. So to make the object neutrally buoyant, we need a lift bag that can be inflated with at least 0.56 cubic feet of air. In other words we need to displace at least 0.56 cubic feet of sea water to provide enough lift to make the object neutrally buoyant.

I hope that this is fairly straight forward. I tried to boil the issues down to the most essential in calculating displacement and how much lift is needed.

Lets do another example for practice.

Lets say you need to recover a motorboat engine that weighs 400 lbs. The engine displaces 3 cubic foot of water.

So lets see what we know. The engine displaces 3 cubic feet of water and the weight of the displaced water is: 3 cubic feet x 64 lbs = 192 lbs. To find out what the apparent weight is we subtract the weight of the displaced water from the weight of the engine: 400 lbs - 192 lbs = 208 lbs. In other words the object appears to only weigh 208 in the water (or: it is negatively buoyant by 208 lbs).

Next we need to determine what size lift bag we need to lift the engine. Since we need to provide at least 208 lbs lift we need to displace 208 lbs of sea water: 208 lbs divided by 64 lbs = 3.25 cubic feet.


Tonia said...

I think that the self study aproach is a reasonable starting point but when it comes to the physics and physiology in the divemaster class we really need to assist our students.

Your physic's post is most helpful and I'm looking forward to a physiology label on your page soon.

As you mentioned, although we are all trained by the same agency, the minimum ability and knowledge of scuba, there are many wonderful approaches to teaching.

BWRAF said...

Thanks for the kind words Tonia. I will post some entries on physiology in the coming weeks.