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Offset Slider Crank Mechanisms

Pantograph

Pantograph is a geometrical instrument used in drawing offices for reproducing given geometrical figures or plane areas of any shape, on an enlarged or reduced scale. It is also used for guiding cutting tools. Its mechanism is utilised as an indicator rig for reproducing the displacement of cross-head of a reciprocating engine which, in effect, gives the position of displacement. There could be a number of forms of a pantograph. One such form is shown in Figure 2.6. It comprises of four links : AB, BC, CD, DA, pin-jointed at A, B, C and D. Link BA is extended to a fixed pin O. Suppose Q is a point on the link AD of which the motion is to be enlarged, then the link BC is extended to P such that O, Q, P are in a straight line. It may be pointed out that link BC is parallel to link AD and that AB is parallel to CD as shown. Thus, ABCD is a parallelogram. Suppose a point Q on the link AD moves to position Q1 by rotating the link OAB downward. Now all the links and the joints will move to the new positions : A to A1, B to B1, C to C1, D to D1 and P to P1 and the new configuration of the mechanism will be as shown by dotted lines. The movement of Q (QQ1) will stand enlarged to PP1 in a definite ratio and in the same form as proved below : Triangles OAQ and OBP are similar.

Therefore, $OA/OB = OQ/OP$

In the dotted position of the mechanism when Q has moved to position $Q_{1}$ and correspondingly P to $P_{1}$, triangles $OA_{1}Q_{1}$ and $OB_{1}P_{1}$ are also similar since length of the links remain unchanged.

$\frac{OA_{1}}{B_{1}} = \frac{OQ_{1}}{OP_{1}}$

$But, \; OB_{1} = OB$

$OA_{1} = OA$

$\therefore \; \frac{OA}{OB} = \frac{OQ_{1}}{OP_{1}}$

$\frac{OQ}{OP} = \frac{QQ_{1}}{OP_{1}}$

As such triangle OQQ Simple Mechanisms1 and $OPP_{1}$ are similar, and $PP_{1}$ and $QQ_{1}$ are parallel and further,

$\frac{PP_{1}}{OP} = \frac{QQ_{1}}{OQ}$

$PP_{1} = OQ_{1} \times \frac{QQ_{1}}{OQ}$

$\; = QQ_{1} \times \frac{OB}{OA}$

Therefore, $PP_{1}$ is a copied curve at enlarged scale.

Hooke’s Joint (Universal Coupling)

This joint is used to connect two non-parallel intersecting shafts. It also used for shafts with angular misalignment where flexible coupling does not serve the purpose. Thus Hooke‘s Joint connecting two rotating shafts whose axes lies in one plane. There are two types of Hooke’s joints in use,

(a) Single Hooke‘s Joint

(b) Double Hooke‘s Joint

One disadvantage of single Hooke‘s joint is that the velocity ratio is not constant during rotation. But this can be overcome by using double Hooke‘s joint.

$Let \;\theta =$input angular displacement

$\phi =$output angular displacement

$\alpha =$shaft angle

$\;\;\boxed{\frac{tan\theta}{tan\phi} = cos\alpha}$

$tan\theta = cos\alpha\,tan\phi$

$Let\,\omega_{1} =$ angular velocity of driving shaft

$\;\;\;\; \omega_{2} =$ angular velocity of driven shaft

$\therefore\; \omega_{1} = \frac{d\theta}{dt}$

$\omega_{2} = \frac{d\phi}{dt}$

Now different above equation

$sec^2\theta\,\frac{d\theta}{dt} = cos\alpha\,sec^2\phi\,\frac{d\phi}{dt}$

$sec^2\theta\,\omega_{1} = cos\alpha\,sec^2\phi\,\omega_{2}$

$\therefore\;\frac{\omega_{2}}{\omega_{1}} = \frac{sec^2\theta}{cos\alpha\,sec^2 \phi} \;\;\;\;\; ...(a)$

$now\; sec^2\phi = 1 + tan^2\phi$

$\;\; = 1+ \frac{tan^2\theta}{cos^2\phi} \;\;\;\;[\because tan\phi = \frac{tan\theta}{cos\alpha}$

$\;\; = 1+ \frac{sin^2\theta}{cos^2\theta\,cos^2 \alpha}$

$\;\; = \frac{cos^2\theta\,cos^2 \alpha + sin^2\theta}{cos^2\theta\,cos^2\alpha}$

$\therefore\;$ Equation (a) will be

$\frac{\omega_{2}}{\omega_{1}} = \frac{sec^2\theta}{cos\alpha[\frac{cos^2\theta\,cos^2 \alpha + sin^2\theta}{cos^2\theta\,cos^2\alpha}]}$

$\;\; = \frac{cos\alpha}{cos^2\theta\,cos^2 \alpha + sin^2\theta}$

$\;\; = \frac{cos\alpha}{cos^2\theta\,[1 - sin^2 \alpha] + sin^2\theta} = \frac{cos\alpha}{cos^2\theta\,[1 - cos^2 \theta\,sin^2\alpha]}$

$\therefore\;\;\; \boxed{\omega_{2} = \frac{\omega_{1}cos\alpha}{1 - cos^2\theta\,sin^2\alpha}}$

Special Cases:

For a given shaft angle $\alpha$, the expresses given above is maximum when cos$\theta\pm$ 1

i.e., when $\theta$ = 0,$\pi$,etc.

And will be minimum when cos$\theta$ = 0

i.e., when $\theta$ = $\frac{\pi}{2},\frac{3\pi}{2}$,etc.

(i) Max. Velocity Ratio:

$\frac{\omega_{2}}{\omega_{1}}$ is maximum when $\theta$ = 0,$\pi$.

$\boxed{ (\frac{\omega_{2}}{\omega_{1}})_{max} = \frac{cos\alpha}{1 - sin^2\alpha} = \frac{1}{cos\alpha} }$

(ii) Minimum V.R:

$\boxed{ (\frac{\omega_{2}}{\omega_{1}})_{min} =cos\alpha}$

$\frac{\omega_{2}}{\omega_{1}}$ is minimum at $\theta$ = $90^{\circ}$, $270^{\circ}$.i.e. two times in one revolution.

(iii) For equal speed:

$\frac{\omega_{2}}{\omega_{1}} = 1 = \frac{cos\alpha}{1 - cos^2\theta\,sin^2\alpha}$

$\;\;cos\theta = \pm\,\sqrt{\frac{1}{1+cos\alpha}}$

Thus $\frac{\omega_{2}}{\omega_{1}}$ is unity at $\theta$ given by above equation i.e.,four times in one revolution of driving shaft.

Now

$\;\; cos^2\theta = \frac{1}{1+cos\alpha}$

$\;\; sin^2\theta = 1 - \frac{1}{1+cos\alpha} = \frac{cos\alpha}{1+cos\alpha}$

$\therefore \;\;tan^2\theta = cos\alpha$

$\boxed{tan^2\theta = \pm\,\sqrt{cos\alpha}}$

(iv) Maximum variation of the velocity of driven shaft

Variation of velocity of driven shaft

= $\frac{\omega_{2_{max}} - \omega_{2_{min}}}{\omega_{2_{mean}}}$

But $\omega_{2_{mean}} = \omega_{1}$, because both shafts complete one revolution during the same interval of time

maximum variation of velocity of $w_{2}$ = $\frac{\frac{\omega_{1}}{cos\alpha} - \omega_{1}\,cos\alpha}{\omega_{1}}$

$\boxed{maximum\,variation\,of\,\omega_{2} = sin\alpha\,tan \alpha}$

$\boxed{\omega_{2_{max}} - \omega_{2_{min}} = \omega_{1}\,tan\alpha\,sin\alpha}$

Maximum fluctuation of speed of driven shaft.

$\omega_{2_{max}} - \omega_{2_{min}} = \omega_{1}\,(\frac{1}{cos\alpha} - cos\alpha)$

$\;\;\omega_{1}\alpha^2 = w_{1}tan\alpha\,sin\alpha$

Since $\alpha$ is a small angle

$\therefore\,sin\alpha = tan\alpha = \alpha$

Average equation of driven shaft:

$\frac{d\omega_{2}}{dt} = \frac{d\omega_{2}}{d\theta}.\frac{d\theta}{dt} = \omega_{1}\,\frac{d\omega_{2}}{d\theta} = \frac{-\omega_{1}^2\,cos\alpha\,\times\,sin2\theta\,sin^2\alpha}{(1 - cos^2\theta\,sin^2\alpha)^2}$

Condition for maximum acceleration

$\frac{d\alpha_{2}}{d\theta} = 0 \;\;\implies\;\;cos2\theta\;\;\approx\;\; \frac{2sin^2\alpha}{2 - sin^2\alpha}$

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