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Thread: Linear interpolation

  1. #1
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    Linear interpolation



    I dont know what linear inteprolation is. Could someone explain how to solve this problem please. Cheers.
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  2. #2
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    ahh ... makes me reminisce about the old days when linear interpolation was used every day using tables in CRC manuals.

    linear interpolation uses a model of a linear rate of change (slope) between two successive values in the table to find an unknown value between those two values.

    \frac{0.8667 - 0.8593}{1.18 - 1.16} = \frac{0.8667 - x}{1.18 - 1.175}

    \frac{.0077}{.02} = \frac{0.8667 - x}{.005}

    0.8667 - x = \frac{(.0077) \cdot (.005)}{.02}

    x = 0.8667 - \frac{(.0077) \cdot (.005)}{.02}

    x \approx 0.8648
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