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Friday Puzzler -- String Theory (2 of 2)

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Friday Puzzler -- String Theory (2 of 2)

#1

Friday Puzzler -- String Theory (2 of 2)

Alex Y

For this one, you have the same two strings with the same burn characteristics and the same zippo lighter, but your challenge is to measure 20 seconds.

Re: Friday Puzzler -- String Theory (2 of 2)

#2

Kent B

Re: Friday Puzzler -- String Theory (2 of 2)

Kent B

Uncle

Re: Friday Puzzler -- String Theory (2 of 2)

#3

Hint

Alex Y

I was going to wait, but since you cried "Uncle"...

Giving you the version to get to 15 seconds first was a bit of a dirty trick, probably creating a mindset that will not work for the 20-second version.

Think outside the box. (No double entendre there.)

Re: Friday Puzzler -- String Theory (2 of 2)

#4

Kent B

Re: Hint

Kent B

I am a linear thinker. Stickley/Ellis repros, and designs that those two would certainly have built had they only thought to ask me first.

This 20-second thing, though - that does not fit. I suspect it may be a ritual of the tribe colloquially known as "Turners" who worship at "Lathe Idols" as they propagate what I have been told is called "Round Stuff".

Not a language I have the capability of deciphering. I stand by my original reply: Uncle.

[Dagnabbitall...... >( ]

Re: Friday Puzzler -- String Theory (2 of 2)

#5

Hint 2

Alex Y

Kent mentioned that the 20 seconds was creating difficulty in finding an answer. Would it help if I told you that the same method could be used to time 12 seconds, or with somewhat more difficulty 23 seconds?

Re: Friday Puzzler -- String Theory (2 of 2)

#6

Re: Hint 2

Larry Barrett

Maybe construct a pendulum with a string and the Zippo lighter as a weight. I'll have to do a little research to see what the period of a 4 foot string might be.

Re: Friday Puzzler -- String Theory (2 of 2)

#7

Close!


Re: Friday Puzzler -- String Theory (2 of 2)

#8

Hint 3

Alex Y

Well, you are 90% of the way there. The period of a pendulum is sqrt(l/g) where l is the length and g is acceleration due to gravity.

Using imperial units, that would be sqrt(4'/(32'/sec^2)) = sqrt(1/8) sec = approx .35 sec.

Of course, we don't know how much of the string is used to tie the ends, or where the center of mass of the lighter is. But really, I messed up the statement of the problem by stating the length of the string, which you really don't need. You have the tools to find the period of the pendulum without knowing the string length or the pendulum formula.

See the last piece of the puzzle?

Re: Friday Puzzler -- String Theory (2 of 2)

#9

Woops!

Alex Y

Forgot the 2 * pi in the formula. Knew that period didn't sound right.

2*pi*sqrt(4/32)sec

=approx 2.2 sec

But, the formula isn't needed.

Re: Friday Puzzler -- String Theory (2 of 2)

#10

Re: Hint 3

Larry Barrett

How about:

Construct a pendulum from one string and the Zippo.

Use the Zippo to light one end of the other string, which will burn for 1 minute.

Start the pendulum at the same instant.

Count the number of Tick Tocks, divide into 60 to find period, P.

If you want to measure N seconds, divide N by P to determine the (approx) number of Tick Tocks you need count.

Re: Friday Puzzler -- String Theory (2 of 2)

#11

Correct!


Re: Friday Puzzler -- String Theory (2 of 2)

#12

Kent B

Re: Hint 3

Kent B

Well done, Larry. Well done indeed.

Tom and Ray were good at misdirection, and I bit on the head fake, where you stood firm.

Reverse the order of the 2 puzzles, drop the 4', and the thought process is reset. But then, it would lose its essence.

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