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Math Help - Reel Fill Calculation - Need Help Please

  1. #1
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    Reel Fill Calculation - Need Help Please

    Hello,

    I have a real world application that I would like some help with.

    It has been a while since college, any help would be appreciated. I believe that this is an integral calculus problem. The reel spool diameter changes as line is wound on the reel.

    I have a fishing reel that has a line counter on it. As you reel in or let line out, it counts the number of revolutions of the spool and mechanically translates it into feet. The line counter is designed such that it is only accurate when the spool is full of line.

    When the reel is empty, one foot on the line counter equals four inches of line on the spool. The spool is 0.5 inches in diameter with no line.

    The reel is capable of holding 310 yards of line that is 0.019" in diameter, or 3.164 cubic inches of line.

    The line that I intend to use is 0.0157 inches in diameter.

    I have three reels and I want to fill each with 1000 feet line from a 1000 yard spool.

    The problem:

    Need to calculate the ending line counter reading when 1000 feet of line has been spooled up on the reel from empty.


    Thank you in advance for your help.
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  2. #2
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    Re: Reel Fill Calculation - Need Help Please

    do you know the length of the spool?
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    Re: Reel Fill Calculation - Need Help Please

    Thanks for your reply.

    I'm not sure that I understand what you are asking, but the spool itself is 1.75 inches wide, so the area that line can wrap onto the spool is 1.75 inches in width.

    Stated another way from the original post, the reel can hold a piece of line that is 310 yards in length and 0.019 inches in diameter.
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    Re: Reel Fill Calculation - Need Help Please

    Bummer, I thought someone would be able to help!
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    Re: Reel Fill Calculation - Need Help Please

    Quote Originally Posted by carpedium View Post
    Bummer, I thought someone would be able to help!
    Odd, I could have sworn I answered this. I'll redo it a bit later this morning
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  6. #6
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    Re: Reel Fill Calculation - Need Help Please

    This can be closely approximated by having $n$ layers of $k$ loops.

    The radius of the loops of the $n$th layer is given by $r_n=R+n d$ where

    $R$ is the radius of the spool
    $d$ is the thickness of the line

    There are $k$ loops per layer where $k=\dfrac L d$ where

    $L$ is the length of the spool.

    The length of a given loop in layer n is $2 \pi r_n$ so the total length of line is given by

    $2 \pi r_n n k = 2 \pi (R + n d) n k$

    your numbers are

    $R=0.5$

    $L=1.75$

    $d = 0.0157$

    plugging these in and solving for length=1000 we find

    $k\approx 111.465$

    $n \approx 2.64$

    or roughly 2 layers and then 71 loops in the 3rd layer. This corresponds to a count of

    $count=111.5 \times 2.64 \approx 294$
    Last edited by romsek; April 18th 2014 at 08:24 AM.
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    Re: Reel Fill Calculation - Need Help Please

    The only thing is that I would expect the count to be greater than 1000, since when reeling line when the spool is less than full, one count equals less than one foot?
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  8. #8
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    Re: Reel Fill Calculation - Need Help Please

    Quote Originally Posted by romsek View Post
    This can be closely approximated by having $n$ layers of $k$ loops.

    The radius of the loops of the $n$th layer is given by $r_n=R+n d$ where

    $R$ is the radius of the spool
    $d$ is the thickness of the line

    There are $k$ loops per layer where $k=\dfrac L d$ where

    $L$ is the length of the spool.

    The length of a given loop in layer n is $2 \pi r_n$ so the total length of line is given by

    $2 \pi r_n n k = 2 \pi (R + n d) n k$

    your numbers are

    $R=0.5$

    $L=1.75$

    $d = 0.0157$

    plugging these in and solving for length=1000 we find

    $k\approx 111.465$

    $n \approx 2.64$

    or roughly 2 layers and then 71 loops in the 3rd layer. This corresponds to a count of

    $count=111.5 \times 2.64 \approx 294$
    bloody hell I'm an idiot. I solved for 1000 inches.

    solving for 12000 inches $n\approx 20.75$ or

    20 layers and about 79 loops in the 21st layer

    This is a count of about 2313
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