1.1 Unbalancing
Unbalance is most common source of vibration in machine with rotating parts. It is a very important factor to be considered in modern machine design, especially where high speed and reliability are significant considerations. Balancing of rotors prevents excessive loading of bearings and avoids fatigue failure, thus increasing the useful life of machinery.

1.2 Types Of Unbalance

1.2.1 Static Unbalance –
Static Unbalance is defined as the eccentricity of the center of gravity of a rotor, caused by a point mass at a certain radius from the center of rotation.
Static Unbalance

1.2.2 Couple (Moment) Unbalance –
In the case of a cylinder, shown in Fig. 13.2, it is possible to have two equal masses placed symmetrically about the centre of gravity, but positioned at 180° from each other. The rotor is in static balance, i.e. there is no eccentricity of the centre of gravity, but when the rotor turns , the two masses cause a shift in the inertia axis, so that it is no longer aligned with the rotation axis, leading to strong vibrations in the bearings. The unbalance can only be corrected by taking vibration measurements with the rotor turning and adding correction masses in two planes. The difference between static balance and couple balance is illustrated in Fig. It can be seen that when the rotor is stationary, the end masses balance each other. However, when it rotates, strong unbalance is experienced.
Couple Unbalance

1.2.3 Dynamic Unbalance 
Dynamic Unbalance is a combination of static and couple unbalance and is the most common type of unbalance found in rotors. To correct dynamic unbalance, it is necessary to make vibration measurements while the machine is running and to add balancing masses in two planes.
Dynamic Unbalance

1.3 Balancing
Balancing is the technique of eliminating unwanted inertia forces and moments in rotating and reciprocating masses and is achieved by changing the location of the mass centers. In rotor or reciprocating machines with the moving masses, if the parts are not properly balanced, the dynamic forces are set up. These forces not only increase the loads on the bearings and stresses in the various components but also it may cause unpleasant noise, vibration and cause wear and tear of the system.
When a mass moves along a circular path, it experiences a centripetal acceleration which generates a force and acts inwards. An equal and opposite force called centrifugal force acts radially outwards and it is a disturbing force on the axis of rotation. The magnitude of this remains constant but the direction changes with the rotation of mass.
The centrifugal disturbing force is given by Fc=mω2r−−−−−−−Eqn13.1${F}_{c}=m{\omega}^{2}rEqn13.1$
Where ,
m = mass of the rotating mass
ω = angular velocity of the component, rad/sec
r = distance of the centre of gravity of mass from the axis of rotation, m

1.3.1 Types of Balancing
(i) Static Balancing: A rotating mass is said to be statically balanced if the rotating mass can rest without turning at any angular position in its bearings. The condition is attained when the sum of the centrifugal forces on the rotating mass due to unbalanced masses is zero at any radial direction. Static balancing is valid for a rotor with only one disc or balancing is required in single plane only.
(ii) Dynamic Balancing: A rotating mass is said to be dynamically balanced when it does not vibrate in its running state. In dynamic balancing, resultant moments or couple which involved in the acceleration of different moving parts is equal to zero. To make a rotating mass dynamically balanced, it must first be statically balanced.

1.3.2 Balancing Of Single Revolving Mass By A Single Mass In The Same Plane :
Let,
m = disturbing mass, kg B = balancing mass
mω2r=Bω2b=>mr=Bb−−−−−−eqn13.2$m{\omega}^{2}r=B{\omega}^{2}b=>mr=Bbeqn13.2$
Balancing Of Single Revolving Mass By A Single Mass In The Same Plane (Fig. 13.4)
Dynamic reactions at the bearings will be zero. Thus the shaft will be free from any dynamic bending stress.

1.3.3 Balancing Of A Single Mass By Two Masses Rotating In Different Planes:
By this type of arrangement for balancing gives rise to a couple which tends to rock the shaft in its bearings. So two balancing masses are placed in two different planes parallel to the plane of rotation of the disturbing mass.
Condition 1. Net force acting on the shaft must be equal to zero
therefor, mω2r=B1b1ω2+Bmbmω2ormr=B1b1+Bmbm−−−−−−−eqn13.3$m{\omega}^{2}r={B}_{1}{b}_{1}{\omega}^{2}+{B}_{m}{b}_{m}{\omega}^{2}\phantom{\rule{1em}{0ex}}or\phantom{\rule{1em}{0ex}}mr={B}_{1}{b}_{1}+{B}_{m}{b}_{m}eqn13.3$
Balancing Of A Single Mass By Two Masses Rotating In Different Planes (Fig. 13.5)
Condition 2. The algebraic sum of the moments about any point in the plane must be zero. Taking moment about L and M axis
mω2rl=Bmω2bmd−−−−−−−eqn13.4$m{\omega}^{2}rl={B}_{m}{\omega}^{2}{b}_{m}deqn13.4$ , and
mω2rx=B1ω2b1d−−−−−−−−−−−eqn13.5$m{\omega}^{2}rx={B}_{1}{\omega}^{2}{b}_{1}deqn13.5$
Two possibilities:
(a) The distributing mass may be in between the planes of the balancing masses
(b) The distributing mass be on the left or right.

1.3.4 Balancing Of Several Masses Rotating In The Same Plane
Balancing Of Several Masses Rotating In The Same Plane (Fig. 13.6)
Consider any number of masses of magnitude m1, m2,m3 and m4 at distances of r1, r2, r3 and r4 from the axis of the rotating shaft. Let α , β and γ be the angles between the masses. Let these masses rotate about an axis through o and perpendicular to the plane of the paper with constant angular velocity, ω rad/\sec. The magnitude and position of the balancing mass may be found out analytically or graphically.

1.3.5 Analytical Method:
Fh=m1ω2r1Cos0+m2ω2r2Cosα+m3ω2r3Cos(α+β)+m4ω2r4Cos(α+β+Y)${F}_{h}={m}_{1}{\omega}^{2}{r}_{1}Cos0+{m}_{2}{\omega}^{2}{r}_{2}Cos\alpha +{m}_{3}{\omega}^{2}{r}_{3}Cos(\alpha +\beta )+{m}_{4}{\omega}^{2}{r}_{4}Cos(\alpha +\beta +Y)$
Fv=m1ω2r1Sin0+m2ω2r2Sinα+m3ω2r3Sin(α+β)+m4ω2r4Sin(α+β+Y)${F}_{v}={m}_{1}{\omega}^{2}{r}_{1}Sin0+{m}_{2}{\omega}^{2}{r}_{2}Sin\alpha +{m}_{3}{\omega}^{2}{r}_{3}Sin(\alpha +\beta )+{m}_{4}{\omega}^{2}{r}_{4}Sin(\alpha +\beta +Y)$
Resultant Bb=(√F2h+F2v)$Bb=\sqrt{(}{F}_{h}^{2}+{F}_{v}^{2})$ and Tanθ=FvFh−−−−−−−eqn13.6$Tan\theta =\frac{{F}_{v}}{{F}_{h}}eqn13.6$
Balancing force should be in opposite direction

1.3.6 Graphical Method:
Graphical Method (Fig. 13.7)
The effect of transferring the rotating mass in the refrence plane will be to produce a centrifugal force F=mω2r$F=m{\omega}^{2}r$ and ac couple C=Fl=mω2rl$C=Fl=m{\omega}^{2}rl$.L in the refrence plane.
Couple is represented by a vector perpendicular to the plane in which.

1.4 Balancing Of Several Masses In Different Planes
Balancing of several masses in different planes (Fig. 13.8)
Plane 
Mass (m) 
Radius (r) 
cent for a  ω2
(mr) 
Dist from pl
(l) 
Couple  ω2
(mrl) 
A 
ma${m}_{a}$

ra${r}_{a}$

mara${m}_{a}{r}_{a}$

−la${l}_{a}$

−marala${m}_{a}{r}_{a}{l}_{a}$ 
L 
Bl${B}_{l}$

bl${b}_{l}$

Blbl${B}_{l}{b}_{l}$

0

0

B 
mb${m}_{b}$

rb${r}_{b}$

mbrb${m}_{b}{r}_{b}$

lb${l}_{b}$

mbrblb${m}_{b}{r}_{b}{l}_{b}$

C 
mc${m}_{c}$ 
rc${r}_{c}$ 
mcrc${m}_{c}{r}_{c}$ 
lc${l}_{c}$ 
mcrclc${m}_{c}{r}_{c}{l}_{c}$ 
M 
Bm${B}_{m}$

bm${b}_{m}$

Bmbm${B}_{m}{b}_{m}$

d

Bmbmd${B}_{m}{b}_{m}d$

D 
md${m}_{d}$

rd${r}_{d}$

mdrd${m}_{d}{r}_{d}$

ld${l}_{d}$

mdrdld${m}_{d}{r}_{d}{l}_{d}$

Bl${B}_{l}$ and Bm${B}_{m}$ are obtained from force polygon and couple polygon

1.5 Balancing Of Reciprocating Masses

1.5.1 Analytical Determination Of Acceleration Of The Piston Of A Reciprocating Engine:
Analytical Determination Of Acceleration Of The Piston Of A Reciprocating Engine (Fig. 13.9)
Let;n=Lr$n=\frac{L}{r}$
Now by geometry we see that x=PPr=PrO−PO=L+r−(Lcosϕ+rcosθ)−−−−−−−−eqn13.7$x=P{P}_{r}={P}_{r}OPO=L+r(Lcos\varphi +rcos\theta )eqn13.7$,
Our interest is to present the equation 13.7 in terms of one variable i.e either θorϕ$\theta \phantom{\rule{1em}{0ex}}or\phantom{\rule{1em}{0ex}}\varphi $. Let us present it in terms of θ$\theta $ only.
the common side CR=Lsinϕ=rsinθorsinϕ=sinθn$CR=Lsin\varphi =rsin\theta \phantom{\rule{1em}{0ex}}or\phantom{\rule{1em}{0ex}}sin\varphi =\frac{sin\theta}{n}$, thus
Cosϕ=(√1−Sin2ϕ)=1n(√n2−Sin2θ)−−−−−eqn13.8$Cos\varphi =\sqrt{(}1Si{n}^{2}\varphi )=\frac{1}{n}\sqrt{(}{n}^{2}Si{n}^{2}\theta )eqn13.8$
Applying (Eqn.13.8) in (Eqn.13.7)
x=r(1−Cosθ)+L(1−1n(√n2−Sin2θ))$x=r(1Cos\theta )+L(1\frac{1}{n}\sqrt{(}{n}^{2}Si{n}^{2}\theta ))$
or, x=r(1−Cosθ)+r(n−(√n2−Sin2θ))orx=r(1−Cosθ+n−(√n2−Sin2θ))$x=r(1Cos\theta )+r(n\sqrt{(}{n}^{2}Si{n}^{2}\theta ))\phantom{\rule{1em}{0ex}}or\phantom{\rule{1em}{0ex}}x=r(1Cos\theta +n\sqrt{(}{n}^{2}Si{n}^{2}\theta ))$
Thus any instantaneous position of the piston can be found out byx=r(1−Cosθ+n−(√n2−Sin2θ))$x=r(1Cos\theta +n\sqrt{(}{n}^{2}Si{n}^{2}\theta ))$
the instantaneous velocity of piston can be found by differentiating the above equation with time.
V=dxdt=dxdθ.dθdt$V=\frac{dx}{dt}=\frac{dx}{d\theta}.\frac{d\theta}{dt}$, but dxdθ=ω,angularvelocity.$\frac{dx}{d\theta}=\omega ,\phantom{\rule{1em}{0ex}}angular\phantom{\rule{1em}{0ex}}velocity.$
Thus velocity of piston at any point will be V=−ωr(Sinθ+Sin2θ2n2−Sin2θ√)$V=\omega r(Sin\theta +\frac{Sin2\theta}{2\sqrt{{n}^{2}Si{n}^{2}\theta}})$
Since Sin2θ<<n2,$Si{n}^{2}\theta <<{n}^{2},$ we neglect Sin2θ$Si{n}^{2}\theta $, so the expresion for velocity change to
V=−ωr(Sinθ+Sin2θ2n)$V=\omega r(Sin\theta +\frac{Sin2\theta}{2n})$
Similarly for the knowing the instantaneous acceleration of piston we differentiate the velocity once again w.r.t time.
dVdt=dθdt.dVdθ=ω.dVdθ$\frac{dV}{dt}=\frac{d\theta}{dt}.\frac{dV}{d\theta}=\omega .\frac{dV}{d\theta}$
therefore instantaneous acceleration of the piton is ,f=−ω2r(cosθ+cos2θn)$f={\omega}^{2}r(cos\theta +\frac{cos2\theta}{n})$

1.5.2 The Effect Of The Inertia Force Of The Reciprocating Mass On The Engine Frame :
We now know that acceleration of the piston is given by
f=−ω2r(cosθ+cos2θn)$f={\omega}^{2}r(cos\theta +\frac{cos2\theta}{n})$
where f$f$ is the acceleration of the piston , θ is the inclination of the crank with IDC., ω is the annular speed of the crank , r is the radius of the crank., n is the ratio of length of connecting rod L to the crank radius r i.e. n=Lr$n=\frac{L}{r}$
The Effect Of The Inertia Force Of The Reciprocating Mass On The Engine Frame (Fig. 13.12)
Force to accelerate the reciprocating mass, F=mf=−mω2r(cosθ+cos2θn)$F=mf=m{\omega}^{2}r(cos\theta +\frac{cos2\theta}{n})$
where m = mass of the reciprocating parts.
The force applied to the frame is equal and opposite to the force required to accelerate the reciprocating mass. It is equal to the reversed effective force or the inertia force.
F=mω2rcosθ+mω2rcos2θn$F=m{\omega}^{2}rcos\theta +m{\omega}^{2}r\frac{cos2\theta}{n}$, the first part of the equation is the primary unbalanced force and the second part is the secondary unbalanced force. Thus F=Fp+Fs$F={F}_{p}+{F}_{s}$
Fp→ primary unbalanced force = mω2rcosθ$m{\omega}^{2}rcos\theta $
FS→ secondary unbalanced force = mω2rcos2θn$m{\omega}^{2}r\frac{cos2\theta}{n}$
Secondary unbalanced force is 1/n times the primary unbalanced force and may be neglected. The primary unbalanced force is maximum when θ = 0 or 180. The primary force is maximum force in one revolution of the crank.The secondary unbalanced force is maximum when θ = 0 , 90, 180 and 270, 360. Hence it is maximum four times in one revolution of the crank. Thus secondary force has twice the frequency of the primary force.The unbalanced force due to reciprocating masses varies in magnitude but constant in direction. (i.e. it acts along OP).The unbalanced force due to revolving masses is constant in magnitude but varies in direction.

1.6 Partial Primary Balance
F=mω2rcosθ$F=m{\omega}^{2}rcos\theta $
The primary unbalance force acts from O to P which may be considered as the component of the centrifugal force produced by a rotating mass 'm' placed at the crank pin . This is balanced by having a mass B at a radius b placed diametrically opposite to the crank pin C. It has two component.
Partial Primary Balance (Fig. 13.11)
Bω2bcosθ$B{\omega}^{2}bcos\theta $ along the line of stroke and Bω2bsinθ$B{\omega}^{2}bsin\theta $ perpendicular to the line of stroke.The primary force is balanced if
mω2rcosθ=Bω2bcosθ$m{\omega}^{2}rcos\theta =B{\omega}^{2}bcos\theta $ or mr=Bb$mr=Bb$
But in doing so another unbalanced vertical force of magnitude Bω2rsinθ$B{\omega}^{2}rsin\theta $ has been introduced . The maximum value of this force is equal to Bω2b$B{\omega}^{2}b$ when θ$\theta $ = 90 and 270. Thus balancing the force along the line of stroke, another unbalanced vertical force of the same magnitude has been introduced. As a compromise let a fraction 'e' of the reciprocating masses is balanced . i.e. Bb=emr$Bb=emr$.
Thus the new unbalanced force along the line of stroke, =mω2rcosθ−Bω2bcosθ$m{\omega}^{2}rcos\theta B{\omega}^{2}bcos\theta $
Unbalanced force = mω2rcosθ−emω2rcosθ=(1−e)mω2rcosθ$m{\omega}^{2}rcos\theta em{\omega}^{2}rcos\theta =(1e)m{\omega}^{2}rcos\theta $
The unbalanced force along the perpendicular to the line of stroke, =Bω2bsinθ=emω2rsinθ$B{\omega}^{2}bsin\theta =em{\omega}^{2}rsin\theta $
∴ Resultant unbalanced force at any instant
=mω2r(√(1−e)2cos2θ+e2sin2θ)$m{\omega}^{2}r\sqrt{(}(1e{)}^{2}co{s}^{2}\theta +{e}^{2}si{n}^{2}\theta )$
If the balancing mass is requested to balance the revolving masses as well as reciprocating masses then
B.b=m1r+emr=r(m1+em)$B.b={m}_{1}r+emr=r({m}_{1}+em)$
m1${m}_{1}$= magnitude of the revolving masses
m$m$= magnitude of the reciprocating masses

1.7 Partial Balancing Of Locomotives
The locomotives, usually it have two cylinder with cranks placed at right angles to each oher in order to have uniformity in turning moment diagram. The two Cylinder locomotives may be,
(a) inside cylinder locomotives
(b) outside cylinder locomotives
In inside cylinder locomotives, the two cylinder are placed inside the driving wheels, where as the outside cylinder locomotives , the two cylinder are placed outside the driving wheels.
Partial Balancing Of Locomotives (Fig. 13.12)
The locomotives may be single or uncoupled locomotive if the efforts is transmitted to one pair of the wheels.

1.7.1 Coupled locomotive :
The driving wheels are connected to the loading and troiling wheels by an outside coupling rod.
Since the cranks are placed at right angle to each other. which facilitates starting of the locomotive in any position. Because one crank will always be away from IDC. Position or the position of no torque on the crank. Further, the secondary forces for one set of reciprocating parts are equal and opposite to those for the other.

1.7.2 Effect Of Partial Balancing Of Locomotive
Due to this partial balancing of the reciprocating parts, there is an unbalanced primary force along the line of stroke which produces.
(a) Variation of tractive forces
(b) Hammer Blow
(c) Swaying couple and an unbalanced primary force perpendicular to the line of stroke which produce.
(a) Variation Of Tractive Forces
The variation of tractive force is caused by unbalanced part of the primary disturbing force due to reciprocating masses . the resultant unbalanced force due to the two cylinders. along the line of stroke known as tractive force.
Let,
m = mass of the reciprocating parts/cylinder
e = fraction of the reciprocating parts to be balanced
θ = inclination of the crank for 1st cylinder
(90+ θ) = inclination of the crank for 2nd cylinder
∴ For cylinder 1 . the unbalanced force along the line of stroke = (1−e)mω2rcosθ$(1e)m{\omega}^{2}rcos\theta $
∴ For cylinder 2 . the unbalanced force along the line of stroke=(1−e)mω2rcos(90+θ)$(1e)m{\omega}^{2}rcos(90+\theta )$, IF 'a' is the wheel to wheel distance then the swayin couple,C =(1−e)mω2r(cosθ+sinθ).a2$(1e)m{\omega}^{2}r(cos\theta +sin\theta ).\frac{a}{2}$
Maximum value of swaying
dCdθ=0$\frac{dC}{d\theta}=0$ or d[(1−e)mω2r(cosθ+sinθ)]dθ=0=>(1−e)mω2rd(cosθ+sinθ)dθ=0$\frac{d[(1e)m{\omega}^{2}r(cos\theta +sin\theta )]}{d\theta}=0=>(1e)m{\omega}^{2}r\frac{d(cos\theta +sin\theta )}{d\theta}=0$, which implies that sinθ+cosθ=0$sin\theta +cos\theta =0$, this is possibble at θ=45225degrees$\theta =45\phantom{\rule{1em}{0ex}}225\phantom{\rule{1em}{0ex}}degrees$
thus, Cmax=(1−e)mω2ra2[Cos45+Sin45]${C}_{max}=(1e)m{\omega}^{2}r\frac{a}{2}[Cos45+Sin45]$, and Cmax=(1−e)mω2ra2[Cos225+Sin225]${C}_{max}=(1e)m{\omega}^{2}r\frac{a}{2}[Cos225+Sin225]$
therefore Cmax,min=±(1−e)mω2ra2√${C}_{max,min}=\pm (1e)m{\omega}^{2}r\frac{a}{\sqrt{2}}$
(b) Hammer Blow
When vertical unbalanced force is considered in the plane of the wheel, it gives rise to the lifting of the wheels from the rails. The maximum value occurs when the center if gravity of the balance. Mass is either directly above the wheel or directly blow the wheel. The maximum value of the unbalanced force in hammer blow.
Hammer Blow
Hammer Blow =Bω2b=emω2r$B{\omega}^{2}b=em{\omega}^{2}r$
The effect of hammer blow is to cause the variation in pressure between the wheel and the rail and cause oscillation of the locomotive in a vertical plane about horizontal axis.
Let P$P$ = static wheel load. This will act in downward direction.
Net pressure between the wheel and the rail = P±Bω2b$P\pm B{\omega}^{2}b$
If (P−Bω2b$PB{\omega}^{2}b$)is negative then the wheel will be lifted from the rail. In order to prevent this lifting of the wheel , the limiting speed is given by
P=Bω2b$P=B{\omega}^{2}b$ => ωlimiting=(√P(Bb)rad/sec${\omega}_{limiting}=\sqrt{(}\frac{P}{(Bb)}\phantom{\rule{1em}{0ex}}rad/sec$

1.7.3 Balancing Of Multi Cylinder In Line Engine
The multicylinder engines with the cylinder center lines in the same plane and on the same side of the center line of the crank shaft, are known as Inline engine
Balancing of Multi Cylinder Inline Engines
For primary balance the following two conditions must be satisfied in order to given the primary balance or the reciprocating points
1.The algebraic sum of the primary forces must be zero i.e Σmiω2ricosθ=0$\mathrm{\Sigma}{m}_{i}{\omega}^{2}{r}_{i}cos\theta =0$
2.The algebraic sum of the couples about any point in the plane of the primary forces must be zero i.e Σmiω2rixicosθ=0$\mathrm{\Sigma}{m}_{i}{\omega}^{2}{r}_{i}{x}_{i}cos\theta =0$
Here line PQ is parallel to the line of stroke ab+bc = cd+da
The algebraic sum of these four forces is zero, the engine is balanced for primary forces. When the crank shaft are in the position shown. But suppose the crank shaft turns clockwise through an angle α$\alpha $ in the new position the algebraic sum is not zero. Therefore the primary forces can only balance for this new position, if the force polygon is a closed one. In a similar way it would be shown that the primary couples can only balanced if the couple polygon for the corresponding centrifugal force is closed.
For the primary force balance of the reciprocating parts of a multicylinder in line engine , it is convenient to imagine the reciprocating masses to be transferred & their respective crankpin and to treat the problem as one of revolving masses.
For Secondary Balance
Fs=mω2rcos2θn=m(2ω)2rcos2θ4n${F}_{s}=m{\omega}^{2}r\frac{cos2\theta}{n}=m(2\omega {)}^{2}r\frac{cos2\theta}{4n}$
Secondary Balance (Fig. 13.15)
Second force Fs is equivalent to the component parallel to the line of stroke , of the centrifugal crank force produced by an equal mass placed at an imaginary crank length r/4n and revolving at twice the speed of the actual crank. Force secondary balance
Σm(2ω)2r4ncos2θ=0$\mathrm{\Sigma}m(2\omega {)}^{2}\frac{r}{4n}cos2\theta =0$ (secondary force polygon must close) , and
Σm(2ω)2r4nxcos2θ=0$\mathrm{\Sigma}m(2\omega {)}^{2}\frac{r}{4n}xcos2\theta =0$ (secondary couple polygon must close)

1.7.4 Direct And Reverse Crank Method
This method is used in balancing radial or Vengines in which the connecting rods are connected to a common crank, The plane of rotation of the various cranks is same, therefore there is no unbalanced primary or secondary couple.
Direct And Reverse Crank Method (Fig. 13.16)
The direct crank revolves in a clockwise direction . The reversed crank will revolve in the anticlockwise direction.
Primary Force
Fp=mω2rcosθ${F}_{p}=m{\omega}^{2}rcos\theta $
It is assumed that a mass m/2 is fixed at the direct crankpin C and m/2 at the reverse crankpin C'.
centrifugal force acting on the direct reverse crank=mω2r2$\frac{m{\omega}^{2}r}{2}$
∴ Total component of the centrifugal force along the stroke =2mω2rcosθ2=mω2rcosθ=Fp$\frac{2m{\omega}^{2}rcos\theta}{2}=m{\omega}^{2}rcos\theta ={F}_{p}$
Hence the primary effect the mass 'm' of the reciprocating parts at P may be replaced by two masses and C and C' each of magnitude m/2.
Secondary Force
Fs=m(2ω)2r4ncos2θ${F}_{s}=m(2\omega {)}^{2}\frac{r}{4n}cos2\theta $
Secondary Force (Fig. 13.17)

1.8 V Engine And Redial Engine
