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Cylinders in Spheres

datagenetics.com

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Re: Cylinders in Spheres

#21

Post by dotrob »

fleitz wrote: post #20
davtbaum wrote: post #4

It wasn't (initially) clear to me that the cylindrical hole must enter and exit the sphere. With that knowledge the solution seems pretty intuitive.

How can one drill a 6" long hole through a sphere of more than 6 inches in diameter? What I mean is if you drill 6 inches into the earth, you haven't passed through the other side... edit: I think they mean drill a 6 inch hole of maximum width, which of course would just leave a very thin ring of the earth 6 inches tall.

In the puzzle statement, "A six inch high cylindrical hole is drilled through the center of a sphere," through is the keyword, rather than into.

I made the same initial mistake of misreading through as into.

Re: Cylinders in Spheres

#22

Post by svantana »

Nice, I especially appreciate the "cheat answer" to the Gardner Puzzle at the bottom of the page. I have found that kind of meta-reasoning about questions quite useful, on exams and in games like Trivial Pursuit, for example.

Re: Cylinders in Spheres

#23

Post by phkahler »

chton wrote: post #2

The cheat is absolutely brilliant reasoning.

Agreed. Given that it is a puzzle and not a request for proof. Once you realize that the hole and sphere diameter are related and the situation has one degree of freedom not specified, it is perfectly valid to conclude that the answer must be independent of that and choose a case you can work out in your head.

Re: Cylinders in Spheres

#24

Post by delluminatus »

dotrob wrote: post #21
fleitz wrote: post #20

Earlier quoted context omitted.

How can one drill a 6" long hole through a sphere of more than 6 inches in diameter? What I mean is if you drill 6 inches into the earth, you haven't passed through the other side... edit: I think they mean drill a 6 inch hole of maximum width, which of course would just leave a very thin ring of the earth 6 inches tall.

In the puzzle statement, "A six inch high cylindrical hole is drilled through the center of a sphere," through is the keyword, rather than into . I made the same initial mistake of misreading through as into.

In that case, the statement from the article "We could have a sphere as large as a planet, bore a hole 6" in length through it..." seems inconsistent. Either the cylinder is not 6" long, or it does not go through the sphere.

Re: Cylinders in Spheres

#25

Post by eps »

There is a simpler puzzle that's similar to Gardner's, but as unintuitive.

Take an orange and wrap a string around its diameter. Now extend the string by 1 inch and redistribute it around the orange so that it floats an even distance from it. Do the same with the Earth, i.e. wrap, extend by 1 inch and even out into a circle. The gap betwen the string and the orange/Earth - which one is bigger?

Re: Cylinders in Spheres

#26

Post by eps »

throwaway283719 wrote: post #10

It reminds me of two other neat problems - 1. Imagine a band stretched taught around the diameter of the earth (which, for the purposes of this question, is a smooth sphere). Now imagine that the band is raised one metre from the ground at every single point along its length. How much longer is it? 2. Imagine perfectly parallel lines painted on the floor, exactly one foot apart, and a rigid needle of length one foot.…

#2 is 3/Pi.

It's a good puzzle and it helps to know the answer :)

Re: Cylinders in Spheres

#27

Post by Someone »

drcode wrote: post #6
Someone wrote: post #5

Earlier quoted context omitted.

Likely Kopfball ( http://de.wikipedia.org/wiki/Kopfball_(Show) ), although that is late eighties (and the precursor late seventies)

I found it through some extra searching starting from your link- "Kopf um Kopf" ... Is that the precursor you were thinking of? Here is a video on youtube that shows the awesomeness of this show (don't need to know German to appreciate it- On the start of the show, putting your finger under the device makes it spin in the opposite direction... why? The audience member with the right answer would get a reward.) https:…

I'm not sure. Wikipedia claims that he precursor also was called Kopfball, and has some overlap in the times when the series ran.

I just googled the "kopf" that I remembered in combination with "wissenschaft" (science) and "fernsehen" (television), and that popped up, and I thought "that must be it".

I should have been more cautious, though. German TV at the time had quite a few interesting programs about science (in a broad sense) I remember programs where they taught differentiation and integration, live chess (or at least, it seemed like that; if a player takes ten minutes to think about a move, the commenters would explain what the players might be thinking about) by such players as Karpov, Timman, and Anand (http://en.m.wikipedia.org/wiki/Chess_of_the_Grandmasters), Hobbythek (http://de.m.wikipedia.org/wiki/Hobbythek) about DIY, (with subjects such as "we build a hot air balloon", "book binding", and "stereo photography"), and that program about personal computing whose name I don't remember.

Re: Cylinders in Spheres

#28

Post by galvan »

delluminatus wrote: post #24
dotrob wrote: post #21

Earlier quoted context omitted.

In the puzzle statement, "A six inch high cylindrical hole is drilled through the center of a sphere," through is the keyword, rather than into . I made the same initial mistake of misreading through as into.

In that case, the statement from the article "We could have a sphere as large as a planet, bore a hole 6" in length through it..." seems inconsistent. Either the cylinder is not 6" long, or it does not go through the sphere.

I thought that at first, but the length of the hole is dependent on the width of the hole. The wider the hole, the shorter, because wider holes remove bigger caps.

With a sufficiently wide hole, you could indeed drill a 6" hole through a spherical Earth, it'd just look more like a thin ring the diameter of the Earth than a sphere.

Re: Cylinders in Spheres

#29

Post by carlob »

stephencanon wrote: post #19
carlob wrote: post #7

One of those cases where using integrals rather than geometry is much simpler. \pi \int_-3^3 (R^2 - x^2) dx = \pi (6 R^2 - 18) is the volume of the rotational solid without removing the cylinder. While the volume of the cylinder is given by: \pi \int_-3^3 (R^2 - 3^2) dx = \pi 6 (R^2 - 9) As you can see the difference between the two volumes is 36 \pi. You can actually show the solution does not depend on \pi without…

Nit: the solution doesn’t depend on “R” (as you note, this is clear from looking at the structure of the two integrals); it does depend on \pi.

typo! And I'm out of the edit window.

Re: Cylinders in Spheres

#30

Post by Chinjut »

eps wrote: post #26
throwaway283719 wrote: post #10

It reminds me of two other neat problems - 1. Imagine a band stretched taught around the diameter of the earth (which, for the purposes of this question, is a smooth sphere). Now imagine that the band is raised one metre from the ground at every single point along its length. How much longer is it? 2. Imagine perfectly parallel lines painted on the floor, exactly one foot apart, and a rigid needle of length one foot.…

#2 is 3/Pi. It's a good puzzle and it helps to know the answer :)

Rather, 2/pi.

The nice answer (referred to above as a "cheat") is to note that the sought probability is the mean number of crossings made by a 1 foot needle with the lines on the floor, where we are implicitly supposing our throw-distribution to be uniform with respect to both translation and rotation. This uniformity, along with "linearity of expectation", is such that the mean number of line-crossings from throwing any shape is simply proportional to the length of the shape (imagine breaking the shape up into many tiny straight lines, all identical except for location and orientation, the number of which is proportional to the length). Note that a circle of 1 foot diameter always makes exactly 2 crossings. Thus, the constant of proportion is 2/pi per foot, and accordingly the sought probability is 2/pi.

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