The statue of Ino problem:

Dave C, 2014

Just before I begin:

This is a VERY HASTY summary of a lesson I gave to a class today and lacks the typesetting care and illustrations I normally give to an online lesson.

It means that I will replace this slideshow with a much more intelligible one someday. But for now just pretend that there are beautifully-crafted diagrams everywhere and the text is all nicely laid out around the illustrations.

ENJOY!

-DC

A life-size gold statue of Ino has been found on the seabed and is to be raised to the water surface using air-filled steel drums as floatation devices.

The question is ‘How many of these things will we need?’

First, some input numbers.

Ino can be approximated by a rectangular slab roughly 1.5m high by 0.4m wide by 0.2m thick.

Multiply these numbers together to get a very rough idea of Ino’s volume.

Yes, it’s rough but it will be accurate to maybe plus or minus twenty percent, and that will give me a fair idea of how heavy the statue will be.

Okay, so the dimensions produce 0.120 cubic metres.

The density of gold is 19 tonnes per cubic metre. That means that each full cubic metre of gold weighs 19 tonnes. If I had two of them I would have 2x19=38 tonnes. If I had three cubic metres the weight would grow to 3x19=57 tonnes. You get the idea that you multiply the volume (whatever it is) by 19 to get the weight.

0.120 x 19 = 2.28 tonnes.

That’s a very heavy statue, but it SHOULD be heavy because gold itself is heavy. Even one cubic metre of gold weighs as much as several cars put together.

Okay, so now I introduce the idea of buoyancy, which is that every submerged object has a lifting force associated with it, which is equal to the amount of water that would occupy the same volume.

The volume of the statue is 0.120 cubic metres, and since the density of water is ‘1’, the weight of water that would fill the shape of the statue is 0.120 tonnes. This is the uplift weight of the statue. It means that if a set of scales was underneath the statue, it would measure 2.28 - 0.120 tonnes.

We all know this from our own experience in the swimming pool. We feel lighter because water is trying to push us out of the pool. The amount to which we are lighter is equal to the weight of water our bodies would have if our bodies were made of water. (Hmmm... That would be an interesting class topic someday).

Okay so now I know how much weight I have to lift off the seabed.

2.28 – 0.12 = 2.16 tonnes.

Buoyancy isn’t helping much.

But look what buoyancy does to an air-filled steel drum.....

I’ll do the calculations for you in a wee while, but just to get through this story quickly, the uplift weight of the drum is equal to the volume of the drum times the density of water, which will be huge because the drum is quite big and it would weigh a lot if it was filled with water.

But the actual weight of the drum is much smaller because the only thing making weight in the drum is the steel wall. The rest of it is air, which we treat as zero weight.

So the uplift weight will be much bigger than the actual weight, which means the drum will rise, and if I tie the statue to it, I might be able to bring the statue to the surface.

Okay, so the uplift weight is the density of water (1) times the volume of the drum.

Volume = pi-r-squared x length for a cylinder.

If the drum is 40cm in diameter and 1.2m tall, you can do the math to get the volume, which is 0.15 cubic metres. The uplift weight therefore is 0.15 tonnes.

Notice that I’m not being very fussy with accuracy here, because I want to get the answer quickly and I don’t want to be bogged down with streams of decimal numbers. Remember that the value for the weight of the statue is very rough, so combining accurate numbers with rough numbers is a waste of effort.

Now the actual weight of the drum depends on the skin of the drum, which is made of steel, which has a density of about 8 t/m3.

Let’s say the wall of the drum is 5mm thick. This is a bit thick but it makes the calculations slightly nicer.

The volume of the two end disks = 2 x (pi-r-squared x thickness)

... Which is 2 x 3.14 x 0.2 x 0.2 x 0.005 m3 = 0.001256 m3

The volume of the side, straightened out to form a rectangle would be

1.2 x (2-pi-r) x thickness = 1.2 x 2 x 3.14 x 0.2 x 0.005 = 0.007536 m3

The total skin volume = the sum of these two numbers = 0.008792 m3

And the weight of the drum is eight times this number = 0.07 tonnes

Okay so the buoyant weight of the drum is

0.07 – 0.15 tonnes = -0.08 tonnes.

Oi! That’s a negative number! But that’s okay because it means the drum wants to float up to the surface. It means that it can lift as much as 0.08 tonnes off the seabed.

Is this good enough? Absolutely not! The statues weighs 2.28 tonnes and we can lift only 0.08 tonnes with one of these drums.

Solution? Use more drums.

How many? Well ‘how many’ x ( - 0.08 tonnes) balances 2.28 tonnes?

Mathematically, it means divide 2.28 by 0.08 to get the required number of drums.

I get 28 or 29.

28 or 29 steels drums to get a gold statue off the seabed?

Hmmm... Remember that my calculations are rough. But by doing this quick-and-dirty analysis I can see that the required number of drums will not be one or two but somewhere in the range 25-35. That very quickly helps me to determine whether the task is affordable or not, and what kind of practical constraints I am up against.

For example, how do you tie 30 steel drums to a submerged object? You probably need strong cables, and those cables will have their own weight, which means you will need a few more drums which means a few more cables and a few more drums....

The answer will spiral upwards for a while but you get the idea that we might need 30 or 40 drums in all to get the job done.

THAT’S REAL-WORLD MATHS! We don’t know all the input numbers so we make some good guesses and approximations, and refine our ideas as we go.

Now there’s a lot of number-crunching and a lot of new concepts in this problem. But there are key points which you should get from this even if you have trouble digesting it all at once:

•That objects in water will float if they weigh less that the amount of water that would fill their shapes.

• That objects that are heavier than an equal amount of water will sink to the bottom. But they will still weigh less than they would on land because water is still trying to lift them off the seabed. In other words, work out how heavy the object would be if its shape was filled with water, and subtract this from the real weight of the object. This is the buoyant weight.

•That hollow objects (like drums) will generally float because the uplift weight depends on the enclosed volume whereas the natural weight comes from the weight of the skin.

•That every object has an uplift weight as well as a natural weight, and whether the object sinks or floats depends on which of these two numbers is greater.

• That the weight of an object is found by multiplying its material density by its volume.

•That the density of water is ‘1’ and the density of steel is about ‘8’ and the density of gold is about ‘19’.

•That the VOLUME of a cylinder is the area of the circular end times the length.

•That the weight of an empty drum is found by peeling the drum part and working out the volume of each part. This would be the volume of two thin disks and the volume of a flattened-out rectangular plate.

•ALTERNATIVELY you could work out the volume of the skin of the drum by working out the volume of the air inside the drum and subtracting this from the volume of a solid steel cylinder.

[END]