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Question 15.28: A mechanism for recording the distance covered by the bicycl...

A mechanism for recording the distance covered by the bicycle, as shown in Fig.15.32, is as follows: There is a fixed annular wheel A of 22 teeth and another annular wheel B of 23 teeth, which rotates loosely on the axis of A. An arm driven by the bicycle wheel through gearing not described, also revolves freely on the axis of A and carries on a pin at its extremity two wheels C and D, which are integral with one another. The wheel C has 19 teeth and meshes with A and the wheel D with 20 teeth meshes with B. The diameter of the bicycle wheel is 0.7 m. What must be the velocity ratio between the bicycle wheel and the arm, if B makes one revolution per 1.5 km covered?

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\text { Given: } z_{a}=22, z_{b}=23, z_{c}=19, z_{d}=20 .

Table 15.26 is used to find the speed of gears.

Table 15.26
Revolutions of Operation
Gear B, 23 Gear C, D, 19, 20 Gear A, 22 Arm, E

\frac{+Z_{a}}{Z_{c}} \times \frac{Z_{d}}{Z_{b}}

=\frac{+22}{19} \times \frac{20}{23}

=\frac{+440 x}{437}

 -\frac{Z_{a}}{Z_{c}}=\frac{+22}{19} +1 0 1. Arm E fixed, +1 revolutions given to gear A, ccw
\frac{+440 x}{437} \frac{+22 x}{19} +x 0 2. Multiply by x
y+\frac{440 x}{437} y+\frac{22 x}{19} y+x y 3. Add y

A fixed:      x + y = 0.

\frac{440 x}{437}+y=1 .

x=\frac{437}{3}=145.67 \text { revolutions for every } 1.5 km .

Revolutions made by bicycle wheel per 1.5 km    =1000 \times \frac{1.5}{(\pi \times 1.5)}=682 .

\text { Velocity ratio of gearing }=\frac{\text { Total velocity ratio }}{\text { Velocity ratio provided by mechanism }} .

=\frac{682}{145.67 \times 1}=4.682 .

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