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Math Help - Transposition...

  1. #1
    Newbie
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    Transposition...

    Hi..

    I have the following problem that i cannot seem to solve..I have to rearrange & solve for "S" in the following equation...


    1*10^6=2{((1.8/S)^2*1/pi)^-0.256 - (20*10^-6)}/2*10^-6 * S^2.513 * pi^1.256 * ( -0.513)


    The answer for S is 2.13.. This answer is no good to me alone, i must see how this is done

    Regards....
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  2. #2
    Super Member Deadstar's Avatar
    Joined
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    Quote Originally Posted by Bailey View Post
    Hi..

    I have the following problem that i cannot seem to solve..I have to rearrange & solve for "S" in the following equation...


    1*10^6=2{((1.8/S)^2*1/pi)^-0.256 - (20*10^-6)}/2*10^-6 * S^2.513 * pi^1.256 * ( -0.513)


    The answer for S is 2.13.. This answer is no good to me alone, i must see how this is done

    Regards....
    Really unclear what that formula is. Gonna do this one step at a time so I don't get lost. Learn to use Latex and post up the real formula if this is wrong.

    10^6 = \frac{2((\frac{1.8}{S})^2 \cdot \frac{1}{\pi})^{0.256} - (20 \cdot 10^{-6})}{2 \cdot 10^{-6} \cdot S^{2.513} \cdot \pi^{1.256} \cdot (-0.513)} ?

    => 5 \cdot 10^5 = \frac{((\frac{1.8}{S})^2 \cdot \frac{1}{\pi})^{0.256} -  (20 \cdot 10^{-6})}{2 \cdot 10^{-6} \cdot S^{2.513} \cdot \pi^{1.256}  \cdot (-0.513)}

    => -2.565 \cdot 10^5 = \frac{((\frac{1.8}{S})^2 \cdot  \frac{1}{\pi})^{0.256} -  (20 \cdot 10^{-6})}{2 \cdot 10^{-6} \cdot S^{2.513} \cdot \pi^{1.256}}

    => -2.565 \cdot 10^5 \cdot \pi^{1.256}= \frac{((\frac{1.8}{S})^2 \cdot   \frac{1}{\pi})^{0.256} -  (20 \cdot 10^{-6})}{2 \cdot 10^{-6} \cdot  S^{2.513}}

    => -2.565 \cdot 10^5 \cdot \pi^{1.256} \cdot 2 \cdot 10^{-6}=  \frac{((\frac{1.8}{S})^2 \cdot   \frac{1}{\pi})^{0.256} -  (20 \cdot 10^{-6})}{S^{2.513}}

    => -0.513 \cdot \pi^{1.256} =  \frac{((\frac{1.8}{S})^2 \cdot   \frac{1}{\pi})^{0.256} -  (20 \cdot 10^{-6})}{S^{2.513}}

    => -0.513 \cdot \pi^{1.256} =  \frac{(\frac{3.24}{\pi S^2})^{0.256} -  (20 \cdot 10^{-6})}{S^{2.513}}

    => -0.513 \cdot \pi^{1.256} =  \frac{(\frac{3.24^{0.256}}{(\pi  S^2)^{0.256}}) -  (20 \cdot 10^{-6})}{S^{2.513}}

    => -0.513 \cdot \pi^{1.256} =  \frac{(\frac{3.24^{0.256}}{\pi^{0.256}   S^{0.512}}) -  (20 \cdot 10^{-6})}{S^{2.513}}

    => -0.513 \cdot \pi^{1.256} =   \frac{(\frac{3.24^{0.256}}{\pi^{0.256}   S^{0.512}}) -  (20 \cdot  10^{-6})}{S^{2.513}}

    => -0.513 \cdot \pi^{1.256} =    \frac{(\frac{3.24^{0.256}}{\pi^{0.256}   S^{0.512}})}{S^{2.513}} -  \frac{(20 \cdot   10^{-6})}{S^{2.513}}

    => -0.513 \cdot \pi^{1.256} =    \frac{(\frac{3.24^{0.256}}{\pi^{0.256}   S^{0.512}})}{S^{2.513}} -   \frac{(20 \cdot   10^{-6})}{S^{2.513}}

    => -0.513 \cdot \pi^{1.256} = \frac{3.24^{0.256}}{\pi^{0.256}   S^{3.025}} -   \frac{(20 \cdot   10^{-6})}{S^{2.513}}

    =>  S^{3.025} \cdot -0.513 \cdot \pi^{1.256} = \frac{3.24^{0.256}}{\pi^{0.256}} - S^{0.512} (20 \cdot   10^{-6})

    Now it's time to use some approximations...

    =>  -2.16041709548325134 \cdot S^{3.025} = 1.00792717675284593 - 0.00002 \cdot S^{0.512}

    =>  -2.16041709548325134 \cdot S^{3.025} + 0.00002 \cdot S^{0.512} - 1.00792717675284593 = 0

    To give a rough idea of the answer...

    2.16 S^3 = -1

    => S^3 = \frac{-1}{2.16}.

    => S = -0.7735981390...




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  3. #3
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    Hi

    I have attached a word doc containing the formula.. The formula that i am having trouble rearranging is used in fracture mechanics. S being sigma which is stress.
    I have an answer of S = 2.13...

    Thanks for the help....

    Bailey..
    Attached Files Attached Files
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  4. #4
    Super Member Deadstar's Avatar
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    10^6 = \frac{2((\frac{1.8}{\sigma})^2 \cdot \frac{1}{\pi})^{-0.256} - (20 \cdot 10^{-6})^{-0.256}}{2 \cdot 10^{-6} \cdot \sigma^{2.513} \cdot \pi^{1.256} \cdot (-0.513)}

    => 10^6 \cdot 2 \cdot 10^{-6}  \cdot \pi^{1.256} \cdot (-0.513) = \frac{2((\frac{1.8}{\sigma})^2 \cdot  \frac{1}{\pi})^{-0.256} - (20 \cdot 10^{-6})^{-0.256}}{\sigma^{2.513}}

    => -0.513 \pi^{1.256}=  \frac{((\frac{1.8}{\sigma})^2 \cdot  \frac{1}{\pi})^{-0.256} - (20  \cdot 10^{-6})^{-0.256}}{\sigma^{2.513}}

    => -0.513 \pi^{1.256}= \frac{1.8^{-0.512}}{\sigma^{-0.512} \pi^{-0.256}} - 0.00002^{-0.256}

    Taking approximations...

    -2.160417096\sigma^{2.513} - 0.9921351694 \sigma^{0.512} + 15.95645167 = 0

    => 2.160417096\sigma^{2.513} + 0.9921351694 \sigma^{0.512}  -15.95645167 = 0

    Solve for \sigma using Maple...

    \sigma = 2.132815069 to 9 decimal places...
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  5. #5
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    Thanks for the help Deadstar,, you're a good'un !!

    Really appreciate it...
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