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Math Help - Practical problem

  1. #1
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    Practical problem

    Hi, I hope you can help, it's a long time since school and I find my self with a practical problem where I need to isolate a variable in a formula. The formula is

    S = ((DH/(H-D+F))-((DH/(H+D-F))

    Is it possible to derive a formula or solve for D i.e isolate D given I have the values of all the other variables?
    Many thanks for your help, this will help me a great deal
    Steve
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  2. #2
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    Re: Practical problem

    Quote Originally Posted by Stevekuk View Post
    Hi, I hope you can help, it's a long time since school and I find my self with a practical problem where I need to isolate a variable in a formula. The formula is

    S = ((DH/(H-D+F))-((DH/(H+D-F))

    Is it possible to derive a formula or solve for D i.e isolate D given I have the values of all the other variables?
    Many thanks for your help, this will help me a great deal
    Steve

    I can't even read this!

    First, I'll try to LaTex it up... then we'll see about an answer.

    S = \frac{DH}{H - D + F} - \frac{DH}{H + D - F}

    S = \frac{DH(H + D - F) - DH(H - D + F)}{(H - D + F)(H + D - F)}

    S = \frac{2DH(D - F)}{H^2 - (D - F)^2}

    Hmm...
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  3. #3
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    Re: Practical problem

    Quote Originally Posted by Stevekuk View Post
    Hi, I hope you can help, it's a long time since school and I find my self with a practical problem where I need to isolate a variable in a formula. The formula is

    S = ((DH/(H-D+F))-((DH/(H+D-F))

    Is it possible to derive a formula or solve for D i.e isolate D given I have the values of all the other variables?
    Many thanks for your help, this will help me a great deal
    Steve
    Get common denominators for your fraction by multiplying numerator and denominator of the first fraction by H+ D- F and multiplying the numerator and denominator of the second fraction by H- D+ F:
    S= \frac{DH(H+ D- F)}{(H-D+ F)(H+D-F)}- \frac{DH(H-D+ F)}{(H- D+ F)(H+ D- F)}
    S= \frac{DH^2+ D^2H- DFH- DH^2+ D^2F- DFH}{(H+ D- F)(H- D+ F)}= \frac{2D^2H- DFH}{(H+ D- F)(H- D+ F)}

    Multiply both sides by (D+ H- F)(D- H+ F) to get rid of the fraction
    S(D+ H- F)(D- H-+ F)= 2D^2H- DFH

    Note that D- H+ F= D- (H+ F) so that (D+ H- F)(D- H+ F) is a :"product of sum and difference"- (D+ (H-F))(D- (H- F))= D^2- (H- F)^2 so we have
    S(D^2- (H- F)^2)= SD^2- DFH
    SD^2- S(H- F)^2= 2D^2- DFH
    Subtract 2D^2- DFH from both sides to get
    SD^2- 2D^2- S(H-F)+ DFH= (S- 2)D^2+ (FH)D- S(H-F)= 0

    That's a quadratic equation and you can solve for D using the quadratic formula:
    "If ax^2+ bx+ c= 0 then x= (-b\pm\sqrt{b^2- 4ac})/(2a)
    with a= S- 2, b= FH, and c= -S(H- F).
    x= \frac{-FH\pm\sqrt{F^2H^2- 4(S-2)(-S(H-F)}}{2(S- 2)}

    (If S= 2, you cannot use that formula because you cannot divide by 0, then this is not a quadratic at all but is FHD- 2(H- F)= 0 which has solution
    D= 2(H- F)/FH= \frac{2}{F}- \frac{2}{H}.)
    Last edited by HallsofIvy; September 4th 2011 at 08:46 AM.
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  4. #4
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    Re: Practical problem

    Sorry I didn't know how to format it properly but your first line is a true reflection of the sart formula. Is it possible to isolate D from here?
    Many thanks
    Steve
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  5. #5
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    Re: Practical problem

    Wow, thanks, your reply came through as I was typing my reply to the first. I'll just digest this
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  6. #6
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    Re: Practical problem

    I have corrected my many formatting errors!
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  7. #7
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    Re: Practical problem

    Sorry but I'm clueless when it comes to this which is why I'm glad you're around. Are you saying that

    D = 2/F - 2/H

    I was expecting to see S still there?
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  8. #8
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    Re: Practical problem

    To explain what I'm trying to do: I'm trying to manipulate the formula used in photography to calculate the depth of field (DoF) within a photograph i.e the amount of image in acceptable focus. This is dependent on a few things, your camera's sensor, the focal length and aperature of your lens and the distance you are from the subject. It is a common formula that it used widely but I have only ever seen it to calculate the DoF which I have called S. I was trying to solve it for the distance required D, so I can work out exactly how far away I need to be for a certain amount of the image to be in focus without the trial and error involved using the more traditional formula. So, after a bit of basic formula engineering I got it to the point within my original post but got stuck trying to solve it for S (Sharp focus). F is the focal length of the lens and H is the hyperfocal point - the distance that maximises the DoF. So hopefully you can see this is a practical problem that would help me out greatly.

    As a practical example to work backwards to get D, i use

    D = 1000
    F = 50
    H = 44759.2311
    Which gives S = 42.468478, this is in millimetres so 4.246 cm should be in focus if my subject is 100cm away using a 50mm lens. But say I know that I want about 4cm to be in focus so what distance do I need to be at? hence solving it for D? Plugging the other known values back into the solved formula I hope to get D = 1000 but it doesn't seem to be there just yet?

    Once again many thanks for your help
    Steve
    Last edited by Stevekuk; September 5th 2011 at 03:06 AM.
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  9. #9
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    Re: Practical problem

    Quote Originally Posted by Stevekuk View Post
    Sorry but I'm clueless when it comes to this which is why I'm glad you're around. Are you saying that

    D = 2/F - 2/H

    I was expecting to see S still there?
    That was "If S= -2".
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  10. #10
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    Re: Practical problem

    Thank you for clarifying, as S is a distance it will always be positive so is what I'm trying to achieve possible? I haven't got the maths knowledge to solve it myself so I understand if it's not possible
    Many thanks
    Steve
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  11. #11
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    Re: Practical problem

    Note that D- H+ F= D- (H+ F) so that (D+ H- F)(D- H+ F) is a :"product of sum and difference"-
    Did you trip up here HallsofIvy?

    The original was (h-d+f)(h+d-f)

    for which I had

    h-d+f=h-(d-f)
    h+d-f=h+(d-f)

    From there I have...

    s=\dfrac{((dh(h+d-f) )-(dh(h-d+f) ))}{(h-(d-f))(h+(d-f))}\ \ =\ \ \dfrac{dh(2d-2f)}{(h-(d-f))(h+(d-f))}

    This reduces down to

    (2h+s)d^2-2f(h+s)d+(fs^2-2h^2 )=0

    and the best I could get for d was



    d=\dfrac{f(h+s)\pm\sqrt{f^2(h+s)^2-fs^2+2h^2}}{2h+s}
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  12. #12
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    Re: Practical problem

    Huge thank you for this. Just to prove how clueless I am with this, if I substitute the values in and solve for d using + and - I get 2 values as is the nature of the formula I suppose. The aim is to get one formula for d that will give a single answer, is this possible?
    Thanks again
    Steve
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  13. #13
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    Re: Practical problem

    Not really since the square root part is positive

    f^2 (h+s)^2-fs^2+2h^2=[f^2(h+s)^2+2h^2]-fs^2

    if the square root of this is less than f(h+s)
    then there are 2 solutions

    if it is greater than f(h+s) then you can disregard the -ve root

    hope this helps

    btw, I know nothing about photography. If any of your other variables can have a negative value, let me know which


    Pro
    Last edited by procyon; September 6th 2011 at 05:38 AM. Reason: added question
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  14. #14
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    Re: Practical problem

    None of the original variables will be negative. Using the values I gave as an example, the square is more than f(h+s) so ignoring the minus then I make d=50.03369 which is spookly close to f but I was expecting d to equal 1000?

    Thanks again
    Steve
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  15. #15
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    Re: Practical problem

    The values you gave in your example actually give S as

    S=42.468476

    not the 42.468478 you said

    That could be where you're getting the slight deviation

    Pro
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