One of the most common forms of the constraint mechanisms is that it permits only relative motion of an oscillatory nature along a straight line. The mechanisms used for this purpose are called straight line mechanisms. These mechanisms are of the following two types:

*1. In which only turning pairs are used, and*

*2. In which one sliding pairs are used,*

**Approximate Straight Line Motion Mechanisms**
The approximate straight line motion mechanisms are the modifications of the four- bar chain mechanisms.

**Principle**

The principle of working of an accurate straight line mechanism is based upon the simple geometric property that the inverse of a circle with reference to a pole on the circle is a straight line. Thus, referring to Figure 2.1 , if straight line $OAB$ always passes through a fixed pole ** O** and the points

**and**

*A***move in such a manner that : $OA\times OB = constant$, then the end**

*B***is said to trace an inverse line to the locus of**

*B***moving on the circle of diameter $OC$. Stated otherwise if**

*A***be a point on the circumference of a circle diameter $OP$, $OA$ by any chord, and**

*O***is a point on $OA$ produced, such that $OA \times OB = \text{a constant}$, then the locus of a point**

*B***will be a straight line perpendicular to the diameter $OP$. All this is proves as follows :**

*B**Fig.- 2.1 Straight Line Motion Mechanism*

Draw a horizontal line from ** O**. From A draw a line perpendicular to $OA$ cutting the horizontal at

**. $OC$ is the diameter of the circle on which the point**

*C***will move about**

*A***such that $OA * OB$ remains constant. Now, $\Delta s \ OAC \ and\ ODB$ are similar**

*O*Therefore, $\frac{OA}{OC}= \frac{OD}{OB}$

$\Rightarrow OD =\frac{(OA\times OB)}{OC}$

But $OC$ is constant and so that if the product $OA \times OB$ is constant, $OD$ will be constant, or the position of the perpendicular from ** B** to $OC$ produced is fixed. This is possible only if the point

**moves along a straight path $BD$ which is perpendicular to $OC$ produced. A number of mechanisms have been innovated to connect**

*B***,**

*O***and**

*B***in such a way as to satisfy the above condition. Two of these are given as follows :**

*A***1. The Peaucellier Mechanism**
The mechanism consists of isosceles four bar chain $OKBM$ (Figure 2.2). Additional links $AK$ and $AM$ from, a rhombus $AKBM$. ** A** is constrained to move on a circular path by the radius bar $CA$ which is equal to the length of the fixed link $OC$.

*Fig.- 2.2 The Peaucellier Mechanism*

From the geometry of the figure, it follows that

$OA\times OB=(OL-AL)(OL+BL)=OL^2-AL^2[as\ AL=LB]$

$OA\times OB=(OK^2-KL^2)-(AK^2-KL^2)=OK^2-AK^2=constant$

Hence, $OA \times OB$ is constant for a given configuration and ** B** traces a straight path perpendicular to $OC$ produced.

**2. The Hart’s Mechanism**

The lengths of the links comply with the conditions: $AD=DC, AB=BC, AE=EF=EH\ and\ CG=GF$. Figure $ABCD$ is a rhomboid linkage. Link 1 turns about fixed axis ** A** and is connected by turning pairs

**and**

*B***to links 3 and 5. Link 4 turns about fixed axis**

*E***and is connected by turning pairs**

*D***and**

*C***to links 3 and 2. Links 2 and 5 are connected by turning pair**

*G***. When link 1 turns about axis**

*F***, point**

*A***describes straight line $Aa$ which is perpendicular to $AD$, and point**

*F***describes straight line $Ab$ which coincides with $AD$. Points of link 5 describe ellipses. Angle $∠ AEF$ is always equal to angle $∠ FGC$. Angle $∠ FCB$ equals 90°.**

*H*[https://youtu.be/g-4WCCT5G7c]

This is also known as crossed parallelogram mechanism. It is an application of four-bar chain. $PSQR$ is a four-bar chain in which $SP = QR\ and\ SQ = PR$.On three links $SP, SQ\ and\ PQ$, then it can be proved that for any configuration of the mechanism : $OA \times OB = Constant$

*Figure 2.3 The Hart’s Mechanism*

It is therefore, concluded if the mechanism is pivoted at ** Q** as a fixed point and the point

**is constrained to move on a circle through**

*A***the point**

*O***will trace a straight line perpendicular to the radius $OC$ produced.**

*B*The proof is given as follows,

let $SP=QR=a$

$SQ=PR=b$

$PQ=x$

and $SR=y$

then $OA\times OB=\frac{OS}{a}\times x\times \frac{OP}{a}\times y=xy\times constant$

But $x=SM-NM\ [QK \perp SR]$

$y=SM+NM\ [QN\parallel PS]$

$xy=(SM^2-NM^2)$

$xy=(b^2-QM^2)-(a^2-QM^2)=(b^2-a^2)$

Hence, $OA\times OB=xy\times constant=Constant$

It is therefore, concluded if the mechanism is pivoted at ** Q** as a fixed point and the point

**is constrained to move on a circle through**

*A***the point**

*O***will trace a straight line perpendicular to the radius $OC$ produced.**

*B***Approximate Straight-line Mechanisms**
With the four-bar chain a large number of mechanisms can be devised which give a path which is approximately straight line. These are given as follows:

*1. The Watt Mechanism*

$OABO$’is the mechanism used for Watt for obtaining approximate straight line motion (Figure 2.4). It consists of three links :$OA$ pivoted at ** O**. O’B pivoted at O’ and both connected by link AB. A point P can be found on the link AB which will have an approximate straight line motion over a limited range of the mechanism. Suppose in the mean position link $OA$ and $O’B$ are in the horizontal position an $OA’$ and $O’B’$ are the lower limits of movement of these two links such that the configuration is $OA’B’O’$. Let

**be the instantaneous centre of the coupler link $A’B’$, which is obtained by producing $OA’$ and $B’O’$ to meet at**

*I***. From**

*I***draw a horizontal line to meet $A’B’$ at $P’$. This point $P’$, at the instant, will move vertically.**

*I*

*Figure 2.4 The Watt Mechanism*

Considering angles $\Phi$ and $\Theta$ being exceedingly small, as an approximation,

$\frac{A'P'}{B'P'}=\frac{\Theta}{\Phi}=\frac{AA'}{a}+\frac{BB'}{b}=\frac{b}{a}$

Where a and b are the lengths of $OA$ and $O’B$ respectively. Since, both $OA$ and $OB$ are horizontal in the mean or mid-position ever point in the mechanism then moves vertically.

Hence, if ** P** divides $AB$ in the ratio

$AP :BP = b : a$

then ** P** will trace a straight line path for a small range of movement on either side of the mean position of $AB$.

**2. The Grass-hopper Mechanism**

It is shown in Figure 2.5. It is a modification of Scott-Russel mechanism. It consists of crank $OC$ pivoted at ** O**, link $O’P$ pivoted at

**O**

****’** and a link $PCR$ as shown. It is, in fact, a laid out four-bar mechanism. Line joining

**and**

*O***is horizontal in middle position of the mechanism. The lengths of the link are so fixed such that:**

*P*$OC =\frac{(CP)^2}{CR}$

If this condition is satisfied, it is found that for a small angular displacement of the link $O’P$, the point ** R** on the link $PCR$ will trace approximately a straight line path, perpendicular to line $PQ$.

*Figure 2.5 The Grass-hopper Mechanism*

In Figure 2.5, the positions both of ** P** and

**have been shown for three different configuration of links. It may be noted that the pin at**

*R***is slidable along with link $RP$ such that at each position the above equation is satisfied.**

*C*