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

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

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

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.

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?

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.

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?

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

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"

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

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.

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

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.

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.