Qustion About Circumfrence

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Qustion About Circumfrence

#1  Postby theropod » Jun 12, 2011 11:57 pm

All,

Since I'm not really all that great with math I have a request of those of you that are.

In this issue of Parade®, a little insert that is often found in the Sunday edition of American newspapers, there is a section called "Ask Myrilyn" wherein a the following is published;

Zan White, Elkins, WV writes:
Say I place a 25,0000-mile-long metal band snugly around the earth. (Assume a smooth planet.) The I cut the band and splice another 50 feet into it, thus loosening it all around. Can I get my finger between the new-length band and the earth? Can I crawl under it?


Marylin vos Savant replies;
Amazingly, even the tallest basketball player could walk under the band, which would float about eight feet off the ground around the planet.

The circumference of the object is irrelevant. Adding 50 feet to any sized band-one that wraps around a cantaloupe or the moon-will produce the same answer: The longer band will be about eight feet from the object it circles.


If this is so, which I find hard to accept, could one of you please educate me as to the formula used to calculate this?

Thanks,

RS
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Re: Qustion About Circumfrence

#2  Postby LucidFlight » Jun 13, 2011 12:18 am

C = 2πr gives (an extra) 7.958 feet radius for 50 feet of circumference.

No matter what circumference you start with, adding 50 feet more will increase the radius by about 8 feet.

Hmm, I can see why that would be hard to accept. Could it be that thinking about the proportion of the additional 50 feet to the overall starting circumference is the stumbling point in that particular logic? :think:

Edit: Incorrect rounding. :whistle:
Last edited by LucidFlight on Jun 13, 2011 1:13 am, edited 2 times in total.
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Re: Qustion About Circumfrence

#3  Postby Calilasseia » Jun 13, 2011 12:23 am

Oh, let's take a look at this.

If a circle has radius r, then the circumference of that circle is given by C = 2πr.

Now, (C+50) = 2π(r+x), where x is the additional radius to be determined.

Substitute the original relationship for C into the equation, and you get:

2πr + 50 = 2π(r+x)

2πr + 50 = 2πr + 2πx

50 = 2πx

Therefore x = 50/2π, which is approximately 7.95.

Problem solved.
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Re: Qustion About Circumfrence

#4  Postby Paul Almond » Jun 13, 2011 12:37 am

The maths you have been given above is correct, so I won't bother with that.

What I will try to help with is why we might have a tendency to think that a small change in the circumference shouldn't make much difference to the height above the planet.

Part of your intuition is correct. A small change in circumference will only make a small change to... something. And what that something is is where the problem starts. The small change is to the radius. And it is a small change! The circumference changes by only a small fraction and the radius also changes only by a small fraction - but it only has to change by a small fraction to lift the band eight feet into the air. It may look like a big change to someone seeing it move eight feet up into the air - but it is still only a small percentage change in the radius. Our intuition, though, is to think of that 8 feet change in height as "a big change" - forgetting that the band's radius is already thousands of km - and it has now gone to thousands of km + 8m - it has hardly changed at all.

Here is one idea - instead of a spherical planet, imagine that we live on a cube, and the band has to go around the cube. If you think in those terms, you'll easily see that a small change in the length is enough to life the band away from the cube.
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Re: Qustion About Circumfrence

#5  Postby theropod » Jun 13, 2011 12:40 am

Thanks all, I knew I was looking at the issue wrong.

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Re: Qustion About Circumfrence

#6  Postby LucidFlight » Jun 13, 2011 1:05 am

Earth's circumference is roughly 25,000 miles * 5280 = 132,000,000 feet.
Diameter for this circumference is 132,000,000 / π = 42,016,904.976 feet.
Radius is 132,000,000 / π / 2 = 21,008,452.488 feet. Let's call this radius E.

With an extra 50 feet of circumference, i.e., 132,000,050 feet:
Diameter is 132,000,050 / π = 42,016,920.892 feet.
Radius is 21,008,460.445 / π / 2 = 21,008,460.446 feet. Let's call this radius E1.

So, radius E1 - radius E
= 21,008,460.446 - 21,008,452.488
= 7.958 feet

That's an extra 7.958 feet "slack" all around the world.
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Re: Qustion About Circumfrence

#7  Postby LucidFlight » Jun 13, 2011 1:23 am

Paul Almond wrote:It may look like a big change to someone seeing it move eight feet up into the air...

And especially if you think "OMG! That eight-foot difference isn't just here, but goes right around the globe!"

It's an interesting little puzzler of sorts. Methinks I shall be trying it on a few unsuspecting folks this week. :)
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Re: Qustion About Circumfrence

#8  Postby Calilasseia » Jun 16, 2011 2:32 am

Jay, the point of my mathematical derivation above was to establish that this is true regardless of the circumference involved. Because the circumference drops out of the algebra, which means that the result is independent of the original choice of circumference. Which is what Marilyn Vos Savant was saying in a roundabout way.

EDIT: basically, it's a consequence of the fact that the circumference is a linear multiple of the radius (and vice versa as a corollary). Adding a fixed value to the circumference will always add the same fixed value to the radius.
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Re: Qustion About Circumfrence

#9  Postby Delvo » Jun 16, 2011 2:59 am

Another way to look at it is simply the distributive property of mathematics: a circle whose diameter equals the sum of the diameters of two or more smaller circles will have a circumference equal to the sum of their circumferences... or one whose circumference equals the sum of two or more smaller circumferences will have a diameter equal to the sum of their diameters...

aπ + bπ + cπ + dπ + eπ + fπ = (a+b+c+d+e+f)π

a/π + b/π +c/π +d/π + e/π + f/π = (a+b+c+d+e+f)/π

...because ANY single factor can be multiplied by a series of other numbers to yield products whose sum would equal the product of that factor multiplied by the sum of those same original other numbers; ANYthing can go where I put π in those above lines:

az+bz+cz+dz+ez+fz = (a+b+c+d+e+f)z

I'm sure everybody here knows that, but people often don't think anymore about stuff they've known for a long time, such as what the distributive property does in particular examples like this one with the circles where
z=π. I've seen the same thing also temporarily stump most of a class full of calculus students.
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Re: Qustion About Circumfrence

#10  Postby LucidFlight » Jun 16, 2011 8:24 am

Calilasseia wrote:Jay, the point of my mathematical derivation above was to establish that this is true regardless of the circumference involved.

Yes. Exactly.

JayWilson wrote:No matter what circumference you start with, adding 50 feet more will increase the radius by about 8 feet.


Calilasseia wrote:Because the circumference drops out of the algebra, which means that the result is independent of the original choice of circumference. Which is what Marilyn Vos Savant was saying in a roundabout way.

EDIT: basically, it's a consequence of the fact that the circumference is a linear multiple of the radius (and vice versa as a corollary). Adding a fixed value to the circumference will always add the same fixed value to the radius.

Indeed, this is precisely what I was thinking the other day. Thank you for clarifying.

Also... what Delvo said. :thumbup:
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