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Lenth of line around an object? 
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#1
May3004, 02:02 PM

P: 94

Lets say i had a cylinder x long with y radius with a string attached to the edge and I wrapped a string around it z number of time until the string reached the other end. How would I go about finding the length of this string. What if the object was something else like a cone or shere or a pyramid. thanks.



#2
May3004, 02:52 PM

P: 43

Well, finding the length of the string for one twist around the cylinder is easy using the formula for the circumference of a circle, pi times diametre. Then just multiply that number with the number of twists needed.



#3
May3004, 02:58 PM

P: 1,004

The area that a line of length L and width D is S = LD. The problem is you don't have the width of the line (its thickness), D. But you have Z, the number of times you wrap the line, so you can use it to find D = X/Z. Going back to the area S, that's also the area of the zone the line covers which is S = 2πYX. So finally you have:
[tex]S = LD = L\frac{X}{Z} = 2\pi YX[/tex] [tex]L = 2\pi YZ[/tex] Grizzlycomet's method is of course shorter and easier, but this is a more general description of how to go about solving such problems when dealing with objects more complicated than cylinders. 


#4
Jun104, 05:19 PM

P: 94

Lenth of line around an object?
I am afraid I am confused and I may have confused you with my question. So I have a cylinder and a piece of string. If I just wrap the string around the can the length of the string will be the circumference. I want to find the length of the string when it is spiraled around the cylinder from one end to the other. Picture a barber shop pole or the stripes on a candy cane.
I have tried doing this with a tin can and fishing line. The fishing line has basically no thickness so I am not worried with that. The circumference of the can I measured 32.5cm, the length 16.5cm. When I wrap the fishing line around the can once (like a barber shop pole) I measure 36cm, when I wrap it twice I get 66cm. Now this does match with the equation L = 2pi *Y*X where y is the radius and x is the length. The length of the string is 3cm longer when wraped around the cylinder end to end than just the circumference I can't see is any relation to the length. 


#5
Jun104, 05:40 PM

Sci Advisor
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P: 2,226

Well the surface of a Cylinder is Euclidan, so imagine drawing the line your string makes and unwrapping the surface of the cyclinder, what would it look like?
If you wnat the string to go from top to bottom (or bottom to top) and once around the cylinder it's lenght will be given by Pytahgoras's theorum where r is the radius and h the height: [tex]L^2 = 4\pi^2r^2 + h^2[/tex] For n twists it will be: [tex]L^2 = n^24\pi^2r^2 + {h^2}[/tex] 


#6
Jun104, 05:44 PM

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Imagine a label on the can. If you unroll it, it's a rectangle with the same height of the can and the width of one circumference. The string wrapped once like a barberpole on the can would be the diagonal of that rectangle.
sqrt(32.5^2+16.5^2)=36.44 If you wrap the string twice, that's the diagonal of a rectangle that is twice the width of label. sqrt( (2*32.5)^2+16.5^2)=67.06 for 3 turns, sqrt( (3*32.5)^2+16.5^2)=98.88 


#7
Jun104, 06:06 PM

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Clearly this problem is easier to solve (jcsd's post) than if the object were a cone or sphere or pyramid. Also, you would have to add more information to completely define the path of the string.
For a sphere you could require theta(phi,n+1)  theta(phi, n) = constant, where theta = polar angle (latitude), phi = azimuthal angle (longitude) and n = turn number 


#8
Jun104, 06:31 PM

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The only finite value allowed for n on a pyramid is 0, if you use the conditon that the line must be straight and have the same gradient on all faces, so other information must be specified.
I haven't worked out the equation yet but I'm pretty sure it's doable on a cone with the information specified (it is after all a conical helix), but it involves a little differentiation. 


#9
Jun104, 06:49 PM

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For the pyramid :
the length of the line will converge even if n diverges...so I don't think is so bad to assume a constant gradient. This requirement of a constant gradient was not stated in the original post. It seemed reasonable to assume it in the case of the cylinder, with the slope being dy/dx. In other cases you might want to use dy/d(phi) or d(theta)/d(phi) to quantify the slope. So you want specify how you calculate slope and whether or not it needs to be constant or some other function. 


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