Question

A person cuts a piece of Bronze that is 5" long with a 2.5" diameter down to 2.100 Dia.

It is done in 2 roughing passes and 1 finishing pass.

First pass from 2.5" to 2.3"

Second Pass from 2.3" to 2.110"

Third pass from 2.110" to 2.100"

What is the machine cutting time (in seconds)?

CS= 110 fpm for roughing and 140 fpm for finishing

Feed = .02 "/rev for roughing and .01"/rev for finishing

Answer 1

Length of the rod (L)= 5 inches

Time taken for First roughing pass is

Diameter of the rod (D)= 2.5 inches

Cutter speed CS (v)= 110 feet per min = 1320 inch per min

Feed rate (S)= 0.2in/ rev

Cutting time (t₁) = ( pi * L* D ) / ( S * v )

                         = (3.14 * 5 * 2.5 ) / ( 0.2 * 1320 )

                           = 0.148 min.

Time taken for second roughing pass is

Diameter of the rod (D)= 2.3 inches

Cutter speed CS (v)= 110 feet per min = 1320 inch per min

Feed rate (S)= 0.2in/ rev

Cutting time (t₂) = ( pi * L* D ) / ( S * v )

                         = (3.14 * 5 * 2.3 ) / ( 0.2 * 1320 )

                           = 0.136 min.

Time taken for third pass (finishing pass) is

Diameter of the rod (D)= 2.110 inches

Cutter speed CS (v)= 140 feet per min = 1680 inch per min

Feed rate (S)= 0.1in/ rev

Cutting time (t₃) = ( pi * L* D ) / ( S * v )

                         = (3.14 * 5 * 2.11 ) / ( 0.1 * 1680 )

                           = 0.197 min.

Therefore, total cutting time (t) = t₁ + t₂ + t₃

                                                 = 0.148 + 0.136 + 0.197

                                                 = 0.481 min

                                              t = 28.86 secs.