Taylor’s tool life relationship For cutting speed V (m/min) and tool life T (min), which form expresses the classical Taylor equation (n is the tool–work exponent, C is a constant)?
-
AV^n T = C
-
BV T^n = C
-
CV^n / T = C
-
DV / T^n = C
-
EV * T = C^n
Answer
Correct Answer: V T^n = C
Explanation
Introduction / Context:Taylor’s tool life equation is foundational in machining economics. It relates cutting speed to tool life and underpins choices for minimum cost, maximum production rate, and given-life cutting conditions.
Given Data / Assumptions:
- Speed V in m/min; life T in minutes.
- n (0.1–0.3 typical) depends on tool/work material system.
- C is an empirical constant for a given setup.
Concept / Approach:The canonical form is V T^n = C. Log-transforming gives log V = log C − n log T, a straight line often used for plotting speed–life data and extracting n and C from tests.
Step-by-Step Solution:Recall standard form: V T^n = C.Check dimensional and empirical consistency: V decreases as T increases for n > 0.Select the option matching this relationship.
Verification / Alternative check:Doubling tool life requires reducing speed by factor 2^(−1/n), consistent with experience.
Why Other Options Are Wrong:(a), (c), and (d) misplace the exponent on V or T. (e) is not the classical Taylor form.
Common Pitfalls:Confusing n placement or assuming linear V–T directly; the relationship is a power law, not linear.
Final Answer:V T^n = C