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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 utilized 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.

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Figure 2.6 Pantograph Mechanism

Suppose a point Q on the link $AD$ moves to position $Q-1$ by rotating the link $OAB$ downward. Now all the links and the joints will move to the new positions : $A\ to\ A_1, B\ to\ B_1, C\ to\ C_1, D\ to\ D_1 and\ P to\ P_1$ and the new configuration of the mechanism will be as shown by dotted lines. The movement of $Q (QQ_1)$ will stand enlarged to $PP_1$ in a definite ratio and in the same form as proved below : Triangles $OAQ$ and $OBP$ are similar.

Therefore,$\frac{OA}{OB}=\frac{OQ}{OP}$

In the dotted position of the mechanism when Q has moved to position $Q_1$ and correspondingly P to $P_1$, triangles $OA_1Q_1$ and $OB_1P_1$ are also similar since length of the links remain unchanged.

But $\frac{OA_1}{OB_1}=\frac{OQ_1}{OP_1}$

$OB_1=OB$

$OA_1=OA$

$\frac{OA}{OB}=\frac{OQ_1}{OP_1}$

$\frac{OQ}{OP}=\frac{QQ_1}{OP_1}$

As such triangle $OQQ$ Simple Mechanisms 1 and $OPP_1$ are similar, and $PP_1$ and $QQ_1$ are parallel and further,

$\frac{PP_1}{OP}=\frac{QQ_1}{OQ}$

$PP_1=OQ_1*\frac{QQ_1}{OQ}$

$PP_1=QQ_1*\frac{OB}{OA}$

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.

Figure 2.6 Universal Coupling

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

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

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

let $w_1$=angular velocity of driving shaft

$w_2$=angular velocity of driven shaft

$w_1=\frac{d\theta}{dt}$

$w_2=\frac{d\phi}{dt}$

Now different above equation

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

$sec^2\theta w_1=cos\alpha sec^2\phi w_2$

$\frac{w_2}{w_1}=\frac{sec^2\theta}{cos\alpha sec^2\phi}$..........(a)

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

$sec^2\phi=1+\frac{tan^2\theta}{cos^2\alpha}$ $\quad$ [as $tan\phi=\frac{tan\theta}{cos\alpha}$]

$sec^2\phi=1+\frac{sin^2\theta}{cos^2\theta cos^2\alpha}$

$sec^2\phi=\frac{cos^2\theta cos^2\alpha+sin^2\theta}{cos^2\theta cos^2\alpha}$

so equation (a) will be-

$\frac{w_2}{w_1}=\frac{sec^2\theta}{cos\alpha [\frac{cos^2\theta cos^2\alpha+sin^2\theta}{cos^2\theta cos^2\alpha}]}$

$\frac{w_2}{w_1}=\frac{cos\alpha}{cos^2\theta cos^2\alpha+sin^2\theta}$

$\frac{w_2}{w_1}=\frac{cos\alpha}{cos^2\theta[1-sin^2\alpha]+sin^2\theta}$

$\frac{w_2}{w_1}=\frac{cos\alpha}{1-cos^2\theta sin^2\alpha}$

$w_2=\frac{w_1cos\alpha}{1-cos^2\theta sin^2\alpha}$

Special Cases

For a give shaft angle $\alpha$ the expression 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.

(1)Max Velocity Ratio:

$\frac{w_2}{w_1}$ is maximum when $\theta$=0,$\pi$

$[\frac{w_2}{w_1}]_{max}=\frac{cos\alpha}{1- sin^2\alpha}$=$\frac{1}{cos\alpha}$

Thus $\frac{w_2}{w_1}$ is maximum when $\theta$=0,$\pi$ i.e two times in one revolution of the driving shaft

(2)Minimum V.R:

$[\frac{w_2}{w_1}]_{min}=cos\alpha$

Thus $\frac{w_2}{w_1}$ is minimum when $\theta=90,270$ i.e two times in one revolution

(3) For Equal Speed:

$\frac{w_2}{w_1}=1=\frac{cos\alpha}{1-cos^2\theta sin^2\alpha}$

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

Thus $\frac{w_2}{w_1}$is unity at $\theta$ given by above equation i.e. four times in one revolution of the driving shaft.

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

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

$tan^2\theta=cos\alpha$

$tan\theta=\pm\sqrt{cos\alpha}$

(4)Maximum variation of the Velocity of the driven shaft:

variation of velocity of driven shaft

$=\frac{w_{2max}-w_{2min}}{w_{2max}}$

But $w_{2max}=w_1$,because both shaft complete one revolution during the same interval of time

Maximum variation of velocity $w_2=\frac{\frac{w_1}{cos\alpha}-w_1 cos\alpha}{w_1}$

Max variation of $w_2=sin\alpha tan\alpha$

$w_{2max}-w_{2min}=w_1tan\alpha sin\alpha$

Maximum Fluctuation of Speed of Driven shaft

$w_{2max}-w_{2min}=w_1(\frac{1}{cos\alpha}-cos\alpha)$

$w_1\alpha^2=w_1tan\alpha sin\alpha$

since $\alpha$ is small angle ,$sin\alpha=tan\alpha=\alpha$

Average equation of Driven Shaft

$\frac{dw_2}{dt}=\frac{dw_2}{d\theta}\frac{d\theta}{dt}=w_1\frac{dw_2}{d\theta}=\frac{-w_1^2cos\alpha*sin2\theta sin^2\alpha}{(1-cos^2\theta sin^2\alpha)^2}$($\alpha_2)$

Condition for Maximum Acceleration

$\frac{d\alpha_2}{d\theta}=0 \Rightarrow cos2\theta \approx \frac{2sin^2\alpha}{2-sin^2\alpha}$