Shuttle Loom Sley Acceleration Ratio Calculation

This solution calculates the ratio of sley acceleration at the front and back centres for a shuttle loom, based on the given crank radius and connecting rod length.

Mechanism Parameters

Identify the key parameters provided:

• Crank Radius: $r = 10 \text{ cm}$
• Connecting Rod Length: $l = 40 \text{ cm}$

Sley Acceleration Formula

The acceleration ($a$) of the sley (reciprocating part) in a crank-driven mechanism can be approximated using the formula:

$a \approx \omega^2 r \left( \cos \theta + \frac{r}{l} \cos(2\theta) \right)$

Where $\omega$ is the constant angular velocity of the crank and $\theta$ is the crank angle.

Acceleration at Front Centre

The front centre position is typically when the crank angle $\theta = 0^\circ$.

• At $\theta = 0^\circ$, $\cos \theta = \cos(0^\circ) = 1$ and $\cos(2\theta) = \cos(0^\circ) = 1$.
• Substituting into the formula, the acceleration at the front centre ($a_{front}$) is: $a_{front} \approx \omega^2 r \left( 1 + \frac{r}{l} \times 1 \right) = \omega^2 r \left( 1 + \frac{r}{l} \right)$

Acceleration at Back Centre

The back centre position is typically when the crank angle $\theta = 180^\circ$.

• At $\theta = 180^\circ$, $\cos \theta = \cos(180^\circ) = -1$ and $\cos(2\theta) = \cos(360^\circ) = 1$.
• Substituting into the formula, the acceleration at the back centre ($a_{back}$) is: $a_{back} \approx \omega^2 r \left( -1 + \frac{r}{l} \times 1 \right) = \omega^2 r \left( -1 + \frac{r}{l} \right)$

Ratio Calculation

Calculate the ratio of the sley acceleration at the front centre to the back centre:

$\text{Ratio} = \frac{a_{front}}{a_{back}} \approx \frac{\omega^2 r \left( 1 + \frac{r}{l} \right)}{\omega^2 r \left( -1 + \frac{r}{l} \right)} = \frac{1 + \frac{r}{l}}{\frac{r}{l} - 1}$

Substitute Values and Compute Ratio

First, find the value of $\frac{r}{l}$:

$\frac{r}{l} = \frac{10 \text{ cm}}{40 \text{ cm}} = 0.25$

Now, substitute this value into the ratio formula:

$\text{Ratio} = \frac{1 + 0.25}{0.25 - 1} = \frac{1.25}{-0.75}$

Simplify the fraction:

$\text{Ratio} = -\frac{1.25}{0.75} = -\frac{125}{75} = -\frac{5}{3}$

Convert the fraction to a decimal:

$\text{Ratio} \approx -1.67$

Final Answer

The ratio of sley acceleration at the front centre and back centre is approximately -1.67.