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Topic: Glass spirals (Read 1633 times) |
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Noke Lieu
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pen... paper... let's go! (and bit of plastic)
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Glass spirals
« on: Apr 18th, 2011, 7:35pm » |
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I stuck a 30m long piece of fishing line to a 5cm diameter circular glass. I then attached a 1 cm wide pencil to the free end of the line. Keeping the line taught and running tangentally to the glass, I started to draw a spiral-inspired arc, stopping when the pencil touched the glass... How long was it?
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Grimbal
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Re: Glass spirals
« Reply #1 on: Apr 19th, 2011, 7:46am » |
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How much of the line have you lost in the knots? How elastic is the line? Don't tell me it is not. All fishing lines are somewhat elastic. When you start, is the line already tangential to the glass? Does the pencil keep its orientation, therefore also winding up some of the rope, or does the pencil turn as you run around, always facing the same side to the glass? In the first case, do you start with the rope tangential to the pencil. Are the glass and the pencil perfectly vertical and the line horizontal? (just searching excuses for not doing the calculations)
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Noke Lieu
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Re: Glass spirals
« Reply #3 on: Apr 19th, 2011, 3:28pm » |
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ah, whoops! I meant 30 cm...! But, yes.
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towr
wu::riddles Moderator Uberpuzzler
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Re: Glass spirals
« Reply #4 on: Apr 19th, 2011, 10:30pm » |
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In that case I get about 158.1824 cm length of wire: x(t) = 30 - 3t radius: r(t) = sqrt( (30 - 3t)2 + 32) -- Assuming wire is tangential to both pencil and glass at all times. derivative: r'(t) = (3t-30))/sqrt(t2-20 t+101) curve length: integral over t=0..10 of sqrt[ r(t)2 + r'(t)2 ] dt ~= 158.1824
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« Last Edit: Apr 19th, 2011, 10:32pm by towr » |
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Immanuel_Bonfils
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Re: Glass spirals
« Reply #5 on: May 24th, 2011, 2:40pm » |
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Wasn't it 30 cm long? And what means 3t ?
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Immanuel_Bonfils
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Re: Glass spirals
« Reply #7 on: May 25th, 2011, 12:52pm » |
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To a silly question a "good answer"? Obviously the question is what makes 3 overthere? Shouldn't be something X ? And what about the 30 cm ?
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towr
wu::riddles Moderator Uberpuzzler
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Re: Glass spirals
« Reply #8 on: May 25th, 2011, 1:17pm » |
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I can't remember why I chose to include a 3 there. It doesn't really matter, you could pick pi or e or phi, but choosing something other than a divider of 30 just gives ugly integration bounds. What's important to note is that speeding up or slowing down will not change the length of the curve. So make choices that simplify the calculation.
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Ant_Man
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Re: Glass spirals
« Reply #9 on: Sep 2nd, 2011, 8:41pm » |
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I think in polar coordinates ds=sqrt{ r'(t)2+[r(t) q'(t)]2 }dt, where q(t) is the angle of the (imaginary) line connecting the pencil to the center of the glass (NOT the angle of the string). That's not what towr has written so I'm confused about his method. I get 90 cm by sticking with a Cartesian parameterization parameterized by the angle, \theta, between the center of the glass and the point where the string leaves the glass. This has a neat form for a string of length L and glass of radius R, namely, s=0.5 L2/R.
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alien2
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Re: Glass spirals
« Reply #10 on: Sep 5th, 2011, 12:12pm » |
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on Apr 18th, 2011, 7:35pm, Noke Lieu wrote: Size doesn’t matter.
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towr
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Re: Glass spirals
« Reply #11 on: Sep 5th, 2011, 1:00pm » |
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on Sep 5th, 2011, 12:12pm, alien2 wrote:You wouldn't say that after walking a mile in shoes that are a size too small.
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