Machining: Taylor's Tool Life Equation Analysis

The problem involves Taylor's tool life equation, which describes the relationship between cutting speed (V) and tool life (T) as:

$VT^n = C$

where 'n' is the exponent and 'C' is a constant.

Applying the Given Conditions

Let the initial cutting speed be $V_1$ and the initial tool life be $T_1$. The equation is:

$V_1 T_1^n = C \quad (1)$

According to the problem:

• The cutting speed is doubled: $V_2 = 2V_1$
• The tool life becomes one-eighth: $T_2 = \frac{1}{8}T_1$

The equation for the new conditions is:

$V_2 T_2^n = C \quad (2)$

Calculating the Exponent 'n'

Equating the two conditions (since C is constant):

$V_1 T_1^n = V_2 T_2^n$

Substitute the values of $V_2$ and $T_2$:

$V_1 T_1^n = (2V_1) \left(\frac{1}{8}T_1\right)^n$

Simplify the equation:

$V_1 T_1^n = 2V_1 \frac{1}{8^n} T_1^n$

Divide both sides by $V_1 T_1^n$ (assuming $V_1 \neq 0$ and $T_1 \neq 0$):

$1 = 2 \times \frac{1}{8^n}$

$1 = \frac{2}{8^n}$

Rearrange to solve for $8^n$:

$8^n = 2$

Express 8 as a power of 2 ($8 = 2^3$):

$(2^3)^n = 2^1$

$2^{3n} = 2^1$

Equate the exponents:

$3n = 1$

Solve for 'n':

$n = \frac{1}{3}$

Therefore, the exponent 'n' in Taylor's tool life equation is $\frac{1}{3}$.