Thread: Linear interpolation

1. Linear interpolation

I dont know what linear inteprolation is. Could someone explain how to solve this problem please. Cheers.

2. 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.

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

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

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

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

$x \approx 0.8648$