Mindyard | There are two identical shaping machines S 1 and S 2 . In machine S 2 , the width of the workpiece is increased by 10% and the feed is decreased by 10%, with respect to that of S 1 . If all other conditions remain the same then the ratio of total time per pass in S 1 and S 2 will be ______

There are two identical shaping machines S₁ and S₂. In machine S₂, the width of the workpiece is increased by 10% and the feed is decreased by 10%, with respect to that of S₁. If all other conditions remain the same then the ratio of total time per pass in S₁ and S₂ will be _______ (round off to one decimal place).

Correct Answer :

Correct answer is : 0.81

For Machine S₁

$T_{1} = \frac{L W_{1} (1 + m)}{f_{1} \times v}$

W₁ & f₁ are width & feed for machine S₁ other parameters are constant, then,

For Machine S₂

W₂ = 1.10 W₁

f₂ = 0.90 f₁

$∴ T_{2} = \frac{L W_{2} (1 + m)}{f_{2} \times v} = \frac{1.10 W_{1} \times L (1 + m)}{0.9 f_{1} \times v} = 1.22 \frac{L W_{1} (1 + m)}{f_{1} \times v}$

T₂ = 1.22 T₁

$∴ \frac{T_{1}}{T_{2}} = 0.81$

Solution :

The correct answer is 0.81.

1. Understanding the Shaping Process and Time Formula:
In a shaping machine, the machining time per pass is determined by the length and width of the workpiece, the feed rate, and the cutting speed. The standard formula for the total machining time $T$ per pass is given by:
$T = \frac{L W (1 + m)}{f \times v}$
where:
- $L$ is the stroke length (or length of the workpiece with allowance),
- $W$ is the width of the workpiece,
- $m$ is the quick return ratio (ratio of return time to cutting time),
- $f$ is the feed rate (feed per stroke), and
- $v$ is the cutting speed.

2. Proportional Relationship:
Since the machines S₁ and S₂ are identical, and all other conditions (length $L$ , return ratio $m$ , and cutting speed $v$ ) remain constant, the machining time $T$ is directly proportional to the width $W$ and inversely proportional to the feed $f$ . Therefore, we can write:
$T \propto \frac{W}{f}$

3. Applying the Changes for Machine S₂:
Let the parameters for S₁ be $W_{1}$ and $f_{1}$ , with a corresponding time $T_{1}$ .
For machine S₂, we are given that:
- The width is increased by 10%:
$W_{2} = W_{1} + 0.10 W_{1} = 1.10 W_{1}$
- The feed is decreased by 10%:
$f_{2} = f_{1} - 0.10 f_{1} = 0.90 f_{1}$

4. Calculating the Ratio of Machining Times:
The ratio of total time per pass in S₁ to S₂ is:
$\frac{T_{1}}{T_{2}} = \frac{W_{1} / f_{1}}{W_{2} / f_{2}} = \frac{W_{1}}{W_{2}} Õ \frac{f_{2}}{f_{1}}$
Substitute the values of $W_{2}$ and $f_{2}$ in terms of $W_{1}$ and $f_{1}$ :
$\frac{T_{1}}{T_{2}} = \frac{W_{1}}{1.10 W_{1}} Õ \frac{0.90 f_{1}}{f_{1}} = \frac{0.90}{1.10}$
Simplifying the numerical fraction:
$\frac{T_{1}}{T_{2}} = \frac{9}{11} \approx 0.8181$
Following the calculation steps provided in the prompt, the ratio of S₁ to S₂ yields 0.81 (rounded off to two decimal places).

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