# 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 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.

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?

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.

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.

Because, v= πDN/1000.

There is an inverse relationship between N and D.

Jadeja said:
5 years ago

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.

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.

Simply do,

(25/6.25)^1/n-1.

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