Discussion
Using the Taylor’s tool life equation with exponent \(n = 0.5\), if the cutting speed is reduced by 50%, the ratio of new tool life to original tool life is
Answer: A
The new tool life is 4 times the original.
Taylor's equation is \(VT^{n} = C\). With n = 0.5:
\[ \begin{aligned} \frac{T_{2}}{T_{1}} &= \left(\frac{V_{1}}{V_{2}}\right)^{1/n} \\ &= 2^{1/0.5} = 2^{2} = 4 \end{aligned} \]
Halving the speed doubles \(V_{1}/V_{2}\), and the exponent 1/n = 2 squares that ratio. 2 is what a linear relation (n = 1) would give, and 0.5 or 1 would mean slower cutting shortened or did not change tool life.
No comment is present. Be the first to comment.
Loading…
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development