# 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
1.02
1.32
1.66
1.82
Explanation:
No answer description is available. Let's discuss.
Discussion:
12 comments Page 1 of 2.

Vishwas Patel said:   8 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.

Pravin said:   8 years ago
@Vishwas.

Why are you taken min/ max dia= 3. 14* 6.25/25?

Akash said:   8 years ago
n(max) = (v x 1000)/(3.14 x min dia).

Parth said:   8 years ago
@Pravin.

Because it's the formula for conversation.
(1)

BANDAN BAURI said:   6 years ago
Spindle Speed = (3.14* D*N)/(V*1000).

Ravki said:   6 years ago
(max speed/min speed)^( 1/ (no of speeds-1) ) = (25/6.25)^(1/5) = 1.319.

Sumit kumar said:   6 years ago
You are correct @Vishwas Patel.

Bhawani Singh said:   5 years ago
@Pravin.

Because, v= πDN/1000.

There is an inverse relationship between N and D.

V=π*D*N/1000.
So, N1D1=N2D2.
N1/N2=D1/D2.
=25/6.25.
=4.

NOW,
(Nmax/Nmin)^1/(6-1).
=4^(1/5),
=1.319,
~1.32.
(1)

Samarth nawale said:   3 years ago
If 1000*18 and 3.14 is common then why uh putting values,
Simply do,
(25/6.25)^1/n-1.