Math puzzle

If the Earth was a sphere (I know, lol), and I mean a perfect sphere, and you wrapped a rope around the great circle of the sphere, how much length would you need to add to the rope in order for it to be able to be 1 meter off the ground?

Edit

Good job.  I am going to edit your post so that others have a chance.

Good job.  I am going to edit your post so that others have a chance.

Oh no, oppressive mods

I will abuse my powerz every chance I get. 

Edit

When are the next mod elections anyway? 

When are the next mod elections anyway? 

You quit, so it does not matter to you.

Edit

Edit

I can't hear you.

7.283r = Circumference? I suck at math.

2pi(R+1) - 2piR

R = Radius of the Earth Sphere. (not sure what number you wanna use)

I think...

2pi(R+1) - 2piR

R = Radius of the Earth Sphere. (not sure what number you wanna use)

I think...

You'll know when you get it right, because he'll delete your comment.

0.006283 km or 6.283 m or roughly 2 pi m

The size of the sphere doesn't matter.

It's 2 pi meters.

2pi

0.006283 km or 6.283 m or roughly 2 pi m

Incorrect. Your first two answers are roughly correct and your last is precise.

Quote from: sokarul on July 14, 2013, 09:41:20 AM

0.006283 km or 6.283 m or roughly 2 pi m

Incorrect. Your first two answers are roughly correct and your last is precise.

6.283 m is roughly equal to 2 pi.

Quote from: spoon on July 14, 2013, 10:11:32 AM

Quote from: sokarul on July 14, 2013, 09:41:20 AM

0.006283 km or 6.283 m or roughly 2 pi m

Incorrect. Your first two answers are roughly correct and your last is precise.

6.283 m is roughly equal to 2 pi.

Yes, the only problem is the unfortunate positioning of the word "roughly" in your original response.

To get an exact result of 1m, you need 2pi metres of extra rope, which is roughly 6.283m. 6.283m will get you an increase in radius of approximately 0.9999705074m, which is clearly less than 1m.

2pi(R+1) - 2piR

R = Radius of the Earth Sphere. (not sure what number you wanna use)

I think...

Uh... yes, but:

2pi(R+1) - 2piR =
2piR + 2pi*1 - 2piR =
2pi*1

You don't need to mention R at all in your final answer.

Quote from: sokarul on July 14, 2013, 11:43:53 AM

Quote from: spoon on July 14, 2013, 10:11:32 AM

Quote from: sokarul on July 14, 2013, 09:41:20 AM

0.006283 km or 6.283 m or roughly 2 pi m

Incorrect. Your first two answers are roughly correct and your last is precise.

6.283 m is roughly equal to 2 pi.

Yes, the only problem is the unfortunate positioning of the word "roughly" in your original response.

To get an exact result of 1m, you need 2pi metres of extra rope, which is roughly 6.283m. 6.283m will get you an increase in radius of approximately 0.9999705074m, which is clearly less than 1m.

I wasn't aware that there was a 2 pi exact answer. I just did it two different ways and got the same result. Then I noticed it was equal to 2 pi.
So as I said, I was in the right to say roughly.

Quote from: sokarul on July 14, 2013, 11:43:53 AM

Quote from: spoon on July 14, 2013, 10:11:32 AM

Quote from: sokarul on July 14, 2013, 09:41:20 AM

0.006283 km or 6.283 m or roughly 2 pi m

Incorrect. Your first two answers are roughly correct and your last is precise.

6.283 m is roughly equal to 2 pi.

Yes, the only problem is the unfortunate positioning of the word "roughly" in your original response.

To get an exact result of 1m, you need 2pi metres of extra rope, which is roughly 6.283m. 6.283m will get you an increase in radius of approximately 0.9999705074m, which is clearly less than 1m.

Quote from: Lorddave on July 14, 2013, 08:34:24 AM

2pi(R+1) - 2piR

R = Radius of the Earth Sphere. (not sure what number you wanna use)

I think...

Uh... yes, but:

2pi(R+1) - 2piR =
2piR + 2pi*1 - 2piR =
2pi*1

You don't need to mention R at all in your final answer.

Yeah.  I really should learn to simplify my equations.

Why would you simply the one part but not remove the *1? I'm so confused.

Anyway, tau.

I wasn't aware that there was a 2 pi exact answer. I just did it two different ways and got the same result. Then I noticed it was equal to 2 pi.

Out of curiosity, how did you solve it? The only way that comes to my mind arrives at 2pi with no approximation involved.

Why would you simply the one part but not remove the *1? I'm so confused.

I could have added one more step to remove the "*1", but I felt like that would be wasting space. What I was trying to demonstrate was clear at that point.

Alternatively, I could have dropped it arbitrarily between steps, but I wanted to keep things readable and visually consistent. In this case it might have been overkill, but it often helps a lot.

Anyway, tau.

Maybe in 20 years. Then again, probably not.

Out of curiosity, how did you solve it? The only way that comes to my mind arrives at 2pi with no approximation involved.

First I simply followed the formula already posted. You can't get 2 pi exactly because of sig figs.(not that I followed them).
Second, to check because I thought I was missing something, I used ratios to get the same result.
2 pi( r+1)*(1-(r/r+1))
I didn't check but it must simplify to 2 pi. I'm sure you will.

2 pi( r+1)*(1-(r/r+1))

That would work if you didn't mess up the brackets. What you want is 2π(r+1)*(1-r/(r+1))

I'm still surprised you somehow only reached an approximate result, but I suppose it's fair enough if you didn't follow standard procedures and filled in the variables too early.

Quote from: sokarul on July 14, 2013, 10:31:43 PM

2 pi( r+1)*(1-(r/r+1))

That would work if you didn't mess up the brackets. What you want is 2π(r+1)*(1-r/(r+1))

I'm still surprised you somehow only reached an approximate result, but I suppose it's fair enough if you didn't follow standard procedures and filled in the variables too early.

Yes, there is one missing. What I used in google calculator was  2 pi( r+1)*(1-(r/(r+1)). Which is equal to what you posted. That is how I got the same answer as the other formula.

If you have a problem with significant figures, I suggest you contact the proper controlling parties. You cannot get 2 pi if you enter valves for r. 

ITT: Chemists consider algebra to be beneath them.

Here's another puzzle which gives a counter-intuitive answer:

Imagine a wooden sphere which has a hole drilled from one side to the other through its centre so that the hole is one inch long.
What volume of wood is left after the drilling?

Here's another puzzle which gives a counter-intuitive answer:

Imagine a wooden sphere which has a hole drilled from one side to the other through its centre so that the hole is one inch long.
What volume of wood is left after the drilling?

There isn't enough information in that puzzle to solve it.

Here's another puzzle which gives a counter-intuitive answer:

Imagine a wooden sphere which has a hole drilled from one side to the other through its centre so that the hole is one inch long.
What volume of wood is left after the drilling?

Depends on what you mean? Is it the volume of wood left in the holey sphere, or the volume of wood total (which would be the same, since drilling only removes wood--just some of the volume would be sawdust).

If you mean the volume of the sphere, what's the radius of the sphere and the radius of the newly-formed hole?