SOLUTION: i have a 20.5" cylinder (core) and I need to put 1800' of material on the outside of it. the material is .022 thick. what would the diameter be of the material after it has all o
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Question 982924: i have a 20.5" cylinder (core) and I need to put 1800' of material on the outside of it. the material is .022 thick. what would the diameter be of the material after it has all of the material on it?
better yet, how can I do this calculation with different size cylinders, different gauges and different lengths?
thank you in advance for your efforts.
-michael Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! i have a 20.5" cylinder (core) and I need to put 1800' of material on the outside of it. the material is .022 thick. what would the diameter be of the material after it has all of the material on it?
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If the 20.5" is the diameter:
The 1st turn adds 0.044 to the diameter.
The length of the 1st turn = pi*d = 20.5*pi =~ 64.4 inches
Each turn adds 0.044 to the ID --> adds 0.13823 to the circumference.
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For n turns,
Length = 20.5*pi + pi*(20.5 + n*0.044)*n
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1800' = 21600 inches
21600 = 20.5pi + 20.5*n*pi + 0.044n^2
0.044n^2 + 20.5*pi*n + (20.5pi - 21600) = 0
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--> 0.044n^2 + 64.4n - 21535.6 = 0
Quadratic equation (in our case ) has the following solutons:
For these solutions to exist, the discriminant should not be a negative number.
First, we need to compute the discriminant : .
Discriminant d=7937.6256 is greater than zero. That means that there are two solutions: .
Quadratic expression can be factored:
Again, the answer is: 280.606287357235, -1744.2426509936.
Here's your graph:
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Ignore the negative solution
n =~ 281 turns
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dia with 281 turns = 20.5 + 281*0.044
dia = 32.82 inches
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I'll do the general case later.