Challenge Questions Blog

### Challenge Questions

So do you have a Challenge Question that could stump the community? Then submit the question with the "correct" answer and we'll post it. If it's really good, we may even roll it up to Specs & Techs. You'll be famous!

Answers to Challenge Questions appear by the last Tuesday of the month.

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# Metal Ring: Newsletter Challenge (05/24/05)

Posted May 24, 2005 7:00 AM

The question as it appears in the 05/24 edition of Specs & Techs from GlobalSpec:

Imagine a smooth metal ring running around the Earth's equator about the width of a sidewalk. Now imagine a metal cable wrapped around this first ring so tightly that nothing can squeeze between the cable and the ring. You may assume that the ring and cable are perfect circles. Now this cable, which does not stretch, has three extra feet added to it and is made to magnetically float up and off of the ring so that extra three feet of slack is then equally distributed around the Earth. This now gives you an equal distance between the ring and the cable anywhere you checked it around the planet. What is that distance? What about if you do this using a soccer ball instead of the Earth? Is there a way to make sense of this without resorting to math?

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The Engineer

Join Date: Feb 2005
Location: Albany, New York
Posts: 5170
#1

### I'm not sure....

05/24/2005 9:24 AM

Do we need to know the radius of the Earth for this question?

Anonymous Poster
#2

### Duh

05/24/2005 9:24 AM

It would be three feet longer than you started....

Participant

Join Date: May 2005
Posts: 1
#3

### Solution

05/24/2005 12:41 PM

Since the Circumference of a circle is approximately 3 times its diameter, pi= 3.14 (Approximately.) then you increase the circumference of a circle by three feet you should increase the diameter by around 1 foot regardless of the circle diameter so the cable will be floating 1/2 a foot apart from the ring.

Anonymous Poster
#4

### Re:Solution

05/24/2005 2:52 PM

Please explain how you can logicly add three feet of cable around the earth(24902 mi. circ) and get 1/2 foot of distance between the cable and the ring?

Member

Join Date: May 2005
Posts: 8
#5

### Metal ring

05/24/2005 4:26 PM

By math and theory we can show that the cable will be approximately 5.7 inches from the metal ring. By using the soccer ball that is about 1 ft in diameter we can tie a string around the ball to get the circumference. Remove the string and make another string three feet longer, and a third string six feet longer. Now we can place the strings on a flat surface in concentric circles and measure the distances between each string to show that the three foot addition to the circumference will create a 5.7 inch increase in the radius of the previous ring.

Member

Join Date: Apr 2005
Posts: 8
#6

### metal rings

05/24/2005 6:51 PM

As I sail in a 36ft sailboat across large waters. Like the Pacific and the Atlantic ponds, a 3 foot wave has little meaning. Thus, I believe a change of 3 feet in the ring is lost in the size of the numbers. I for one, cannot comprehend the number of 3 feets there are around the world, and if one of those 3 feet got lost, so what?

Commentator

Join Date: May 2005
Posts: 93
#7

### Challenge

05/25/2005 5:56 AM

I can't see any way to achieve this without mathematics. Assuming that all parameters remain constant (temperature, material properties, etc) the result will be a second diameter with a delta component resulting from the distances between the two radii. The answer seems to me to be the result of the equation R1-R2 (delta) regardless of initial dimensions. In this case I think the result would be so small as to be immeasureable

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Commentator

Join Date: May 2005
Posts: 93
#8

### further thoughts

05/25/2005 10:03 AM

On further thoughts......calculate the earth's circumference as C = ¦°d. If 3 is then added to C, the result becomes ¦°(d + ¦¤d) = C + 3. Thus, ¦°¦¤d = 3, ¦¤d ¡Ö 1 ¦¤r = 0.5' or approx' 6"

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Participant

Join Date: May 2005
Posts: 1
#9

### The hard way

05/25/2005 10:45 AM

Approx. 6", It took alot of string, an airplane and some very stupid freinds to figure this out

Active Contributor

Join Date: May 2005
Posts: 14
#11

### Re:The hard way

05/27/2005 12:48 AM

Must have had a skilled pilot to consistently fly 6" above sea level. I do look forward to you and your friends taking on further Challenges.

Active Contributor

Join Date: May 2005
Posts: 14
#10

### Metal Ring

05/26/2005 12:56 AM

The equatorial belt is indeed in need of a chastity belt but the cable eludes me - is it for pyjamas?

Friend of CR4

Join Date: Dec 2004
Posts: 1995
#12

05/31/2005 1:33 PM

As written in the 5/31 issue of Specs & Techs from GlobalSpec:

The circumference of a circle is always represented by the equation 2(pi)r. In the case of the challenge question, 3 feet has been added to the original circumference. Thus, the change in the radius is equal to 3 ft. divided by 2pi. It's a given that the value of pi is 3.14 (for some of us, this might have required reaching back a bit into our "stored knowledge"...). The answer is therefore 0.477 feet or 5.73 inches. And this would hold true for any circular object, regardless of whether it's the Earth or a soccer ball. Although most people are initially surprised by the answer, upon reflection it does make some intuitive sense: again, it is a property of a circle that the ratio of its circumference to its radius is a constant (2pi). Adding 3 feet to the circumference adds, as we have seen, 5.73 inches to the radius, no matter what the starting point. The surprise for most people is that for the case of the earth, the radius change is even measurable, but there it is: 5.73 inches. Of course, expressed as a fraction of the original radius, the change is 0.477/2.09E7 or a couple of millionths of a percent. For a soccer ball, on the other hand, the change is on the order of 100 percent.

__________________
Off to take on other challenges. Good luck everybody! See you around the Interwebs.
Participant

Join Date: Jul 2005
Posts: 1
#13

### Uses math, but not too much

07/06/2005 12:51 AM

If you change the legnth of the cable by X feet, the radius of that cable will change by "X/2pi" feet, since the length of a perimeter is 2*pi*radius. Since the radius here changes by 3/2pi feet, the distance between the cable and the ring must change the same distance.