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Math Help - Linear Increment Across Fixed Distance

  1. #1
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    Question Linear Increment Across Fixed Distance

    Hi all,

    Could someone please give me a few pointers on how to solve the following:

    I need to divide a 9mm distance into linearly incrementing portions, the first of which must measure 0.01mm and the last must measure 1.2mm.

    As the size of each portion increments linearly across the entire 9mm distance, the rate increase must be constant from portion to portion, but how do I calculate this increment?

    Any help would be most appreciated, as this problem is driving me insane

    Thanks in advance, Carla.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by CarlaT View Post
    Hi all,

    Could someone please give me a few pointers on how to solve the following:

    I need to divide a 9mm distance into linearly incrementing portions, the first of which must measure 0.01mm and the last must measure 1.2mm.

    As the size of each portion increments linearly across the entire 9mm distance, the rate increase must be constant from portion to portion, but how do I calculate this increment?

    Any help would be most appreciated, as this problem is driving me insane

    Thanks in advance, Carla.
    This is an arithmetic series for the interval size.
    a_n = a_1 + (n - 1) \cdot \Delta r
    where a_1 is the first interval (0.01 mm), n is the number of intervals, a_n is the size of the last interval (1.2 mm), and \Delta r is how much each interval increases per step.

    We know that we need the sum of all these intervals to be 9 mm, so
    \sum_{i = 0}^n(a_1 + i \cdot \Delta r) = n/2 \cdot (a_1 + a_n) = 9~mm

    So we know that
    n = \frac{2 \cdot (9~mm)}{0.01~mm + 1.2~mm} = 14.876

    So we need 14 intervals. (If you can go over the 9 mm, you can use 15 intervals.)

    Now, what is \Delta r? Well,
    a_{14} = a_1 + 14 \cdot \Delta r

    \Delta r = \frac{a_{14} - a_1}{14} = 0.091538~mm

    So your first interval is
    a_1 = 0.01~mm

    Your second is
    a_2 = a_1 + (2 - 1) \cdot 0.091538 = 0.101538~mm

    a_3 = a_1 + (3 - 1) \cdot 0.091538 = 0.193077~mm

    etc. for a total span of 8.46996 mm. (If you do this using the 15 intervals \Delta r = 0.085~mm and the total span is 9.075 mm.)

    -Dan
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  3. #3
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    Quote Originally Posted by topsquark View Post
    We know that we need the sum of all these intervals to be 9 mm, so
    \sum_{i = 0}^n(a_1 + i \cdot \Delta r) = n/2 \cdot (a_1 + a_n) = 9~mm
    Oops. There are a couple of minor errors in Dan's solution. Summing from 0 to n yields n+1 intervals, so his first expression should be
    \sum_{i = 0}^{n-1}(a_1 + i \cdot \Delta r).

    And a_{14} = a_1 + 13\cdot\Delta r, not 14.

    But the key point here is there's no solution that meets all the constraints. You have to fill almost all of the 9 mm gap, or overfill it a bit, or accept that the last portion won't be exactly 1.2 mm wide. Compare this to the result of a similar problem, in which the first portion is 0.01 mm wide, the last portion is 0.81 mm wide, and the gap is 2.05 mm wide altogether. Applying the same equations yields an integer value (5) for n.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by OldFogie View Post
    Oops. There are a couple of minor errors in Dan's solution. Summing from 0 to n yields n+1 intervals, so his first expression should be
    \sum_{i = 0}^{n-1}(a_1 + i \cdot \Delta r).

    And a_{14} = a_1 + 13\cdot\Delta r, not 14.
    Whoops! Gotta get back to Kindergarten and learn how to count again. Thanks for the spot.

    -Dan
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  5. #5
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    Thumbs up Thanks Guys!!!

    Ok...

    Firstly, a very big thanks to topsquark for providing a solution to this most ellusive problem (well, maybe not for you gurus, but for me it appeared impossible).

    Secondly, an equally big thanks to OldFogie for finally silencing my biggest demon surrounding this type of problem (i.e. is there an exact solution for every set of parameters).

    Guys, thank you so much for taking the time to answer this question. Please don't stop helping us mathematical peons, as this type of thing isn't particularly easy for the average Joe.

    Once again, thanks guys; I could not have done this without you!

    Take care, Carla xxx
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