Steel belt around a sphere

Because you are such quick and clever puzzle-solvers, I'm going to raise the (ahem) belt a bit. This one is hard. Acknowledgements to Martin Gardner.

Imagine that you are standing on a perfectly smooth sphere that is as large as the Earth (in circumference). A steel belt is wrapped snugly around one of the equators of the sphere.

Now, a length of steel, one meter long, is added to the belt (the belt is "opened", the new segment added, the belt fastened again). The segment adds to the length of the belt sufficiently to raise the belt off the surface of the sphere by the same distance all the way around. Will this segment lift the belt high enough so that you can:
- Slip a playing card under it?
- Slip your hand under it?
- Slip a baseball under it?

For bonus points:
- Is your answer different if the sphere is the size of the sun?

For extra do-decko bonus points:
- Explain your answer.


richard said...

I don't understand the question. Do you mean a cubic metre of steel, or just a metre of some other thickness? And how is it "added": what width does the new sub-belt come in?

Either way, my instinctive response is that the new sub-belt will only be about a single atom thick, which (let me look at my hand for a sec...) won't allow much to get under it.

fiona-h said...

revising to make more clear

Molly said...

How flexible is this belt? And does it have to stay exactly the same distance from the earth or can you deform it a bit?

fiona-h said...

Not flexible. Does not stretch, that is.

For the purposes of this puzzle, imagine that it is the same distance off the sphere all around. You're trying to figure out this: If it were the same distance off the sphere all 'round, about how far off would it be? (playing card, hand, or baseball distance)

Mark Hanington said...

You can get a card under it. The addition of a meter to the circumference adds .0159155 meters to the radius. Which is about 1.6 cms. Of that, .8 cm will be on either side of the globe, an amount that does not change with the increase of the initial circumference from 25,000 miles to 100,000 miles, the figures I used.

(I have no idea what the circumference of the sun actually is, but I see a trend.)

I will think why this is true - I just did the math. I don't claim to understand it yet.

fiona-h said...

Not quite right. You are correct that the answer is the same regardless of the size.

I'm going to paste in the answer that Brad sent me on this; he articulated it nicely:

First we have to define our variables:

Let d equal the diameter of the earth at the equator

Then d' will be the change in the diameter when the belt is lengthened. In this case d' =1m.

Let r equal the radius if the earth at the equator.

Then r' will be the change in the radius of the belt- this is what we are trying to solve for.

pi = 3.1417....

As I recall from the times I was actually awake in math class the formual for the diameter of a circle is: d = 2 pi r

and it follows that d / 2 pi = r (* remember this for later k?)

we can express the diameter of the belt after the 1m segment is added as follows:

d + d' = 2 pi (r + r')

solving for r' gives us:

r' = d/2pi + d'/2pi - r however recall*above: d/2pi = r

therefore r' = r+d'/2pi - r

r' = d'/2pi

r' = 1m / (2 x 3.1417) = 0.159m or 15.9 cm ... about the size of a baseball.

So, according to my calculations, I'm going to have to go with the baseball... which seems a bit much. Intuitively, I figured that the palying card would be the answer.

Presumably, because the answer only depends of the variable d', the answer for the sun would be the same (ie the baseball) as well.

fiona-h said...

what I like about this is that it is sooo non-intuitive!!

Mark Hanington said...

Good god! I got the right answer, but then failed to correctly go from meters to centimeters! I thuink I should get partial crredit, and a lot of it, for my answer.

fiona-h said...

How about 51 percent?

Mark Hanington said...

By the way I used an Exel spreadsheet using only 4 cells. I should have used a fifth for the unit conversion,evidently. How about 91%

fiona-h said...

oh all right.

Mark Hanington said...

God, you're a real push-over.

fiona-h said...

that is why it's lucky that I am not a teacher.

Anonymous said...

Thank you for solving this problem because it helped with my homework. I can't thank you enough for the math lesson