Klien’s construction: 1 Klien’s velocity diagram,
2 Klien’s acceleration diagram

**KLIEN’S CONSTRUCTION**

Let *OC* be the crank and *PC* the connecting rod of a reciprocating steam engine, as shown in Fig. 15.2 (*a*). Let the crank makes an angle θ with the line of stroke *PO* and rotates with uniform angular velocity ω rad/s in a clockwise direction. The Klien’s velocity and acceleration diagrams are drawn as discussed below:

**Fig. 15.2. **Klien’s construction.

**1 Klien’s velocity diagram**

First of all, draw *OM* perpendicular to *OP*; such that it intersects the line *PC *produced at* M*. The triangle* OCM *is known as* Klien’s velocity diagram*. In this triangle

*OM *may be regarded as a line perpendicular to* PO*,

*CM *may be regarded as a line parallel to* PC*, and ...(Q It is the same* *line.)

*CO *may be regarded as a line parallel to* CO*.

We have already discussed that the velocity diagram for given configuration is a triangle *ocp* as shown in Fig. 15.2 (*b*). If this triangle is revolved through 90°, it will be a triangle *oc*1 *p*1, in which *oc*1 represents *v*CO (*i.e.* velocity of *C* with respect to *O* or velocity of crank pin *C*) and is paralel to *OC*,

*op*1* *represents* v*PO* *(*i.e. *velocity of* P *with respect to* O *or velocity of cross-head or piston *P*) and is perpendicular to *OP*, and *c*1*p*1* *represents* v*PC* *(*i.e. *velocity of* P *with respect to* C*) and is parallel to *CP*.

A little consideration will show, that the triangles *oc*1*p*1 and *OCM* are similar. Therefore,

Thus, we see that by drawing the Klien’s velocity diagram, the velocities of various points may be obtained without drawing a separate velocity diagram.

**2 Klien’s acceleration diagram**

The Klien’s acceleration dia-gram is drawn as discussed below:

**1. **First of all, draw a circle with *C* as centre and *CM* as radius.

**2. **Draw another circle with *PC* as diameter. Let this circle inter-

sect the previous circle at *K* and *L*.

**3. **Join** ***KL*** **and produce it to intersect** ***PO*** **at** ***N*. Let** ***KL*** **intersect** ***PC*** **at** ***Q*.

This forms the quadrilateral *CQNO*, which is known as *Klien’s acceleration **diagram***.**

We have already discussed that the acceleration diagram for the given configuration is as shown in Fig. 15. 2 (*c*). We know that

A little consideration will show that the quadrilateral *o'c'x p'* [Fig. 15.2 (*c*)] is similar to quadrilateral *CQNO* [Fig. 15.2 (*a*)]. Therefore,

**SOLVED PROBLEMS**

** **

1.
*The crank and connecting rod of a reciprocating
engine are 200 mm and 700 mm respectively. The crank is rotating in clockwise
direction at 120 rad/s. Find with the help of Klein’s construction: 1. Velocity and acceleration of the
piston, 2. Velocity and acceleration
of the mid point of the connecting rod, and 3. Angular velocity and angular acceleration of the connecting rod,
at the instant when the crank is at 30° to I.D.C. *(

**Solution. **Given:** ***OC*** **= 200 mm = 0.2 m ;** ***PC*** **= 700 mm = 0.7 m ;** **ω** **= 120 rad/s

The Klein’s velocity diagram *OCM*
and Klein’s acceleration diagram *CQNO*
as shown in Fig. 15.5 is drawn to some suitable scale, in the similar way as
discussed in Art. 15.5. By measurement, we find that

*OM *= 127 mm = 0.127 m* ; CM *= 173
mm = 0.173 m ;* QN *= 93 mm = 0.093 m ;* NO *= 200 mm = 0.2 m

**1. Velocity and acceleration of
the piston**

We know that the velocity of the piston *P,*

*v*P* *=* *ω* × OM
*= 120 × 0.127 = 15.24 m/s* ***Ans.*** *and*
*acceleration of the piston *P*,

*a*P* *=* *ω* *^{2}* × NO *= (120)^{2}* *× 0.2 =
2880 m/s* *^{2}* ***Ans.**

**2. Velocity and acceleration of
the mid-point of the connecting rod**

In order to find the velocity of the mid-point *D* of the connecting rod, divide *CM*
at *D*1 in the same ratio as *D* divides *CP*. Since *D* is the
mid-point of *CP*, therefore *D*^{1} is the mid-point of *CM*,*
i.e. CD*1* *=* D*1*M*. Join* OD*1. By measurement,

*OD*1* *=
140 mm = 0.14 m

Velocity of *D*,
*v*D = ω × *OD*1 = 120 × 0.14 = 16.8 m/s **Ans.**

In order
to find the acceleration of the mid-point of the connecting rod, draw a line *DD*2 parallel to the line of stroke *PO* which intersects *CN* at *D*2. By measurement,

*OD*2* *=
193 mm = 0.193 m* *∴ Acceleration of *D*,

*a*D* *=* *ω* *^{2}* *×* OD*2* *= (120)^{2}* *×
0.193 = 2779.2 m/s* *^{2}* ***Ans.*** ***3.
Angular velocity and angular acceleration
of the connecting rod**

We know that the velocity of the connecting rod *PC* (*i.e.*
velocity of *P* with respect to *C*),*
v*PC* *=* *ω* *×* CM *=
120 × 0.173 = 20.76 m/s

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