# 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.

Shravan said:   2 years ago
Cutting velocity= πDN ,

Where D=Diameter, N=speed of drill in r.p.m,
N1/N2=D2/D1=25/6.25=4.
The Ratio of spindle speeds =(4)1/6-1=(4)0.2. = 1.32Ans.
(3)

Mir said:   2 years ago
The speed ration = (Nmax/Nmin)^(1/n-1) = (Dmax/Dmin)^(1/n-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.

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)

Bhawani Singh said:   5 years ago
@Pravin.

Because, v= πDN/1000.

There is an inverse relationship between N and D.

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

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

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

Parth said:   7 years ago
@Pravin.

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

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