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Math Help - is this complete ?

  1. #1
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    Question is this complete ?

    using the equation VT(power n) = C

    cuttin speed tool life
    180m/min 30
    250m/min 12

    v1=180
    t1=30
    v2=250
    t2=12

    logv1+logt1=log c
    logv2+logt2=log c

    log180+nlog30
    log250+nlog12

    2.255+n1.477
    -2.397+n1.079
    =-0.142+n-0.602=0

    n=0.602 = 0.142

    0.142
    0.602
    =0.235

    180+0.235x1.477=180.3470

    is this the end of the equation?
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  2. #2
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    sorry missed out a bit of infomation n=0.13
    Last edited by jenko; December 21st 2007 at 11:22 AM.
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by jenko View Post
    sorry missed out a bit of infomation n=0.13
    But as I showed in my last post, n = 0.13 isn't going to work in your equation. If VT^n = C is correct for both sets of data then your problem is inconsistent.

    Are you giving us the full problem? If so, then what are you trying to find?

    -Dan
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  4. #4
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    full equation:

    a test was carried out on the centre lathe and the following results obtained:

    Cutting speed (V) Tool life (T)
    180m/min 30min
    250m/min 12min

    using the equation VT(small n at the top of T) = C
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by jenko View Post
    full equation:

    a test was carried out on the centre lathe and the following results obtained:

    Cutting speed (V) Tool life (T)
    180m/min 30min
    250m/min 12min

    using the equation VT(small n at the top of T) = C
    So you are trying to find n and C?

    -Dan
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  6. #6
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    i think thats what its asking :S i am realy unsure i do not have any other infomation on this question
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by jenko View Post
    full equation:

    a test was carried out on the centre lathe and the following results obtained:

    Cutting speed (V) Tool life (T)
    180m/min 30min
    250m/min 12min

    using the equation VT(small n at the top of T) = C
    Okay, start to finish, finding n and C. Here's how I would do this:
    First
    VT^n = C

    So
    V_1T_1^n = V_2T_2^n

    180 \cdot 30^n = 250 \cdot 12^n

    \frac{30^n}{12^n} = \frac{250}{180}

    \left ( \frac{30}{12} \right )^n = \frac{25}{18}

    \left ( \frac{5}{2} \right )^n = \frac{25}{18}

    log \left [ \left ( \frac{5}{2} \right )^n \right ] = log \left ( \frac{25}{18} \right )

    n~log \left ( \frac{5}{2} \right ) = log \left ( \frac{25}{18} \right )

    n = \frac{log \left ( \frac{25}{18} \right )}{log \left ( \frac{5}{2} \right )} = 0.358515
    (which means I screwed up in my earlier post. )

    Now we may take either set of data to find C:
    180m/min 30min
    180 \cdot 30^{0.358515} = 609.316

    -Dan
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