Mechanical Engineering - Production Engineering - Discussion
Discussion Forum : Production Engineering - Section 4 (Q.No. 12)
12.
The ratio between two consecutive spindle speeds for a six-speed drilling machine using drills of diameter 6.25 to 25 mm size and at a cutting velocity of 18 m/min is
Discussion:
12 comments Page 2 of 2.
Pravin said:
9 years ago
@Vishwas.
Why are you taken min/ max dia= 3. 14* 6.25/25?
Why are you taken min/ max dia= 3. 14* 6.25/25?
Vishwas Patel said:
10 years ago
If n = no of spindle speeds.
The ratio between two consecutive spindle speeds is:
R = (Max rpm/min rpm)^1/(n-1).
Max rpm = (1000 * Cutting velocity)/Min diameter.
= (1000 * 18)/3.14 * 6. 25.
= 917.197.
Min rpm = (1000 * Cutting velocity)/Max diameter.
= (1000 * 18)/3.14 * 25 = 229.299.
Now ratio R = (917.197/229.299)^1/5.
= 1.32.
The ratio between two consecutive spindle speeds is:
R = (Max rpm/min rpm)^1/(n-1).
Max rpm = (1000 * Cutting velocity)/Min diameter.
= (1000 * 18)/3.14 * 6. 25.
= 917.197.
Min rpm = (1000 * Cutting velocity)/Max diameter.
= (1000 * 18)/3.14 * 25 = 229.299.
Now ratio R = (917.197/229.299)^1/5.
= 1.32.
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