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Line of best fit - for logarithmic-like dataset

  1. Aug 4, 2014 #1
    Hi all,

    Firstly, hope I'm posting in the right place!

    I'm working with a company that works with air filters. Without going into specifics etc, I have 4 data points:

    | x | y |
    |0mm|21%|
    |150mm|59.6%|
    |300mm|83%|
    |550mm|90%|

    where x is Filter Height and y is Filtration.
    (Apologies for bad formatting - I looked but couldn't find the syntax anywhere for posting in a table!)


    In the title, I've called it 'logarithmic-like' and what I mean by that is that it will not go above 100% (obviously) and looking at the data points, it's got a definite exponential fall off.

    21% at 0mm I know sounds wierd, but it's a characteristic of the system which, through laziness and some contractual NDA stuff, I won't go in to.

    I've used 3 different graph-drawing softwares (Excel, JMP 11 and graph) + WolframAlpha and have come to the conclusion that it isn't possible to alter the curve of a logarithmic plot enough to fit this data.

    I've also tried flipping the data and an exponential won't fit - x^2 does, but doesn't asymptote at 100%, so isn't really accurate enough. Increasing the power just makes any graphical software peak between the 1st and second points and then rise to the rest.

    TLDR;

    I have this data set (above) and I'm attempting to find a line of best fit so that we can design a unit and "know" what filtration it will give. Can anyone help me with a way of finding a formula that fits or a piece of software that will?

    Thanks in advance!
     
  2. jcsd
  3. Aug 4, 2014 #2

    mathman

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    One method that might work is to add one more data point: x=1 billion, y = 100%.
     
  4. Aug 5, 2014 #3
    You could try modeling it with [itex]y = 1 - (1 - a_1)e^{-a_2 x}[/itex] and perform a nonlinear fit to estimate the parameters. And if you know that [itex]f(0) = 0.21[/itex] then you know one of the parameters, [itex]y = 1 - 0.79e^{-a_2 x} , a_1 = 0.21[/itex].

    ExpNonLinFit1407220716.png
     
  5. Aug 10, 2014 #4

    haruspex

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    You could make it linear:
    ln(1-y) = ln(1-a1) - a2x
    Plot ln(1-y) against x.
     
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