Balancing Of V-Engines

• Consider a symmetrical two cylinder V-engine as shown in figure. The common crank OC is driven by two connecting rods PC and OQ are inclined to the vertical OY at an angle $\alpha$ as shown in figure.

• Let m = Mass of reciprocating parts per cylinder,
  $\quad$ l = Length of connecting rod,
  $\quad$ r = Radius of crank,
  $\quad$ n = Ratio of length of connecting rod to crank radius = l / r
  $\quad$ $\theta$ = Inclination of crank to the vertical at any instant,
  $\quad$ $\omega$ = Angular velocity of crank.

• We know that inertia force due to reciprocating parts of cylinder 1, along the line of stroke
  = $m \omega^{2} r\left[\cos (\alpha-\theta)+\frac{\cos 2(\alpha-\theta)}{n}\right]$
  and the inertia force due to reciprocating parts of cylinder 2, along the line of stroke
  = $m \omega^{2} r\left[\cos (\alpha+\theta)+\frac{\cos 2(\alpha+\theta)}{n}\right]$

• The balancing of V-engines is only considered for primary and secondary forces as discussed below :
  Considering primary forces, we know that primary force acting along the line of stroke of cylinder 1,
  $\mathrm{F}_{\mathrm{P} 1}=\mathrm{m} \omega^{2} \mathrm{r} \cos (\alpha-\theta)$
  $\therefore$ Component of $F_{\mathrm{P} 1}$ along the vertical line OY,
  $=F_{\mathrm{P} 1} \cos \alpha=m \omega^{2} r \cos (\alpha-\theta) \cos \alpha$ ....(i)
  and component of $F_{P 1}$ along the horizontal line OX
  $=F_{P 1} \sin \alpha=m \omega^{2} r \cos (\alpha-\theta) \sin \alpha$ ....(ii)
  

• Similarly, primary force acting along the line of stroke of cylinder 2,
  $F_{\mathrm{P} 2}=m \omega^{2} r \cos (\alpha+\theta)$
  $\therefore$ Component of $\mathrm{F}_{\mathrm{P} 2}$ along the vertical line OY
  $=F_{P 2} \cos \alpha=m \omega^{2} r \cos (\alpha+\theta) \cos \alpha$ .....(iii)
  and component of $F_{P 2}$ along the horizontal line OX'
  $=F_{P 2} \sin \alpha=m \omega^{2} r \cos (\alpha+\theta) \sin \alpha$ .....(iv)
  

• Total component of primary force along the vertical line OY
  $F_{P V}=(i)+(i i)=m \omega^{2} r \ cos \ \alpha[\cos \ (\alpha-\theta) + cos( \alpha + \theta $
  $\quad \quad =m \omega^{2} r \cos \alpha \times 2 \cos \alpha \cos \theta$ $\quad \quad [\therefore \cos (\alpha-\theta)+\cos (\alpha + \theta)=2 \cos \alpha \cos \theta]$
  $\quad \quad =2 m \omega^{2} r \cos ^{2} \alpha \cos \theta$
  and total component of primary force along the horizontal line OX
  $F_{P H}=(i i)-(v)=m \omega^{2} r \ \sin \ \alpha[ cos \ (\alpha \ - \ \theta ) - \ cos(\alpha + \ \theta)$
  $\quad \quad = m \omega^{2} r \ cos \ \alpha \times 2 sin \ \alpha sin \ \theta$ $\quad \quad [\therefore cos(\alpha \ - \ \theta) - cos \ (\alpha \ + \ \theta) = 2 sin \ \alpha \ sin \ \theta]$
  $\quad \quad = 2 m \omega^{2} r \ sin^{2} \ \alpha \ sin \ \theta$
  
  $\therefore$ Resultant primary force, $F_{P} = \sqrt { (F_{P V})^{2} + (F_{P H})^{2}}$
  $\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =2 m \omega^{2} r \sqrt { (cos^{2} \alpha cos \theta)^{2} + ( sin^{2} \alpha sin \theta )^{2}}$ ....(iv)

• NOTE : The following results, derived from equation (v), depending upon the value of $\alpha$ may be noted when :
  

• When $2 a=60^{\circ}$ or $a=30^{\circ}$.
  $F_{P}=2 m \omega^{2} r \sqrt{\left(\cos ^{2} 30^{\circ} \cos \theta\right)^{2}+\left(\sin ^{2} 30^{\circ} \sin \theta\right)^{2}}$
  $\quad = 2 m \omega^{2} r \sqrt{(\frac {3}{4} cos \theta)^{2}+(\frac {1}{4} sin \theta )^{2}}$
  $\quad =\frac{m}{2} \omega^{2} r \sqrt{9 \cos ^{2} \theta+\sin ^{2} \theta}$ ....(vi)
  
• When $2 a=90^{\circ}$ or $a=45^{\circ}$.
  $F_{p} =2 m \omega^{2} r \sqrt{\left(\cos ^{2} 45^{\circ} \cos \theta\right)+\left(\sin ^{2} 45^{\circ} \sin \theta\right)^{2}}$
  $\quad =2 m \omega^{2} r \sqrt{\left(\frac{1}{2} \cos \theta\right)^{2}+\left(\frac{1}{2} \sin \theta\right)^{2}}=m \omega^{2} r$ ....(vii)
  

• When $2 a=120^{\circ}$ or $a=60^{\circ}$.
  $F_{P}=2 m \omega^{2} r \sqrt{\left(\cos ^{2} 60^{\circ} \cos \theta\right)^{2}+\left(\sin ^{2} 60^{\circ} \sin \theta\right)^{2}}$
  $\quad =2 m \omega^{2} r \sqrt{\left(\frac{1}{4} \cos \theta\right)^{2}+\left(\frac{3}{4} \sin \theta\right)^{2}}$
  $\quad =\frac{m}{2} \omega^{2} r \sqrt{\cos ^{2} \theta+9 \sin ^{2} \theta}$ .....(viii)

• Considering secondary forces, we known that secondary force acting along the line of stroke of cylinder 1,
  $F_{S 1}=m \omega^{2} r \times \frac{\cos 2(\alpha-\theta)}{n}$
  $\therefore$ Component of $\mathrm{F}_{\mathrm{S} 1}$ along the vertical line OY $=F_{S 1} \cos \alpha=m \omega^{2} r \times \frac{\cos 2(\alpha-\theta)}{n} \times \cos \alpha$ .....(ix)
  and component of $F_{\mathrm{SI}}$ along the horizontal line OX
  $=F_{S 1} \sin \alpha=m \omega^{2} r \times \frac{\cos 2(\alpha-\theta)}{n} \times \sin \alpha$ ....(x)

• Similarly, secondary force acting along the line of stroke of cylinder 2, $F_{\mathrm{S} 2}=m \omega^{2} r \times \frac{\cos 2(\alpha+\theta)}{n}$
  $\therefore$ Component of $\mathrm{F}_{\mathrm{S} 2}$ along the vertical line OY
  $=F_{S 2} \cos \alpha=m \omega^{2} r \times \frac{\cos 2(\alpha+\theta)}{n} \times \cos \alpha$ .....(xi)
  and component of $\mathrm{F}_{\mathrm{S} 2}$ along the horizontal line OX'
  $=F_{S 2} \sin \alpha=m \omega^{2} r \times \frac{\cos 2(\alpha+\theta)}{n} \times \sin \alpha$ .....(xiii)

• Total component of secondary force along the vertical line OY,
  $F_{S V} = (ix) + (xi) = \frac{m}{n} \times \omega^{2} r \cos \alpha[\cos 2(\alpha-\theta)+\cos 2(\alpha+\theta)]$
  $\quad \quad =\frac{m}{n} \times \omega^{2} r \cos \alpha \times 2 \cos 2 \alpha \cos 2 \theta \quad =\frac{2 m}{n} \times \omega^{2} r \cos a \cos 2 \theta$
  and total component of secondary force along the horizontal line OX,
  $F_{S H}=(x)+(x i i)=\frac{m}{n} \times \omega^{2} r \sin \alpha[\cos 2(\alpha-\theta)-\cos 2(\alpha+\theta)]$
  $\quad \quad =\frac{m}{n} \times \omega^{2} r \sin \alpha \times 2 \sin 2 a \sin 2 \theta =\frac{2 m}{n} \times \omega^{2} r \sin \alpha \sin 2 \alpha \sin 20$
  $\therefore$ Resultant secondary force $\mathrm{F}_{\mathrm{S}}=\sqrt{\left(\mathrm{F}_{\mathrm{SV}}\right)^{2}+\left(\mathrm{F}_{\mathrm{SH}}\right)^{2}}$
  $=\frac{2 m}{n} \times \omega^{2} r \sqrt{(\cos \alpha \cos 2 \alpha \cos 2 \theta)^{2}+(\sin \alpha \sin 2 \alpha \sin 2 \theta)^{2}}$ ....(xiii)

• Note : The following results, derived from equation (xiii), depending upon the value of $\alpha$ may be noted when :
  

• When $2 \alpha=60^{\circ}$ or $\alpha=30^{\circ}$,
  $F_{S}=\frac{2 m}{n} \times \omega^{2} r \sqrt{\left(\cos 30^{\circ} \cos 60^{\circ} \cos 2 \theta\right)^{2}+\left(\sin 30^{\circ} \sin 60^{\circ} \sin 2 \theta\right)^{2}}$
  $\quad =\frac{2 \mathrm{m}}{\mathrm{n}} \times \omega^{2} \mathrm{r} \sqrt{\left(\frac{\sqrt{3}}{2} \times \frac{1}{2} \cos 2 \theta\right)^{2}+\left(\frac{1}{2} \times \frac{\sqrt{3}}{2} \sin 2 \theta\right)^{2}}$
  $\quad =\frac{\sqrt{3}}{2} \times \frac{m}{n} \times \omega^{2} r$ ....(xiv)
  
• When $2 \alpha=90^{\circ}$ or $\alpha=45^{\circ}$,
  $F_{S}=\frac{2 m}{n} \times \omega^{2} r \sqrt{\left(\cos 45^{\circ} \cos 90^{\circ} \cos 2 \theta\right)^{2}+\left(\sin 45^{\circ} \sin 90^{\circ} \sin 2 \theta\right)^{2}}$
  $\quad =\frac{2 m}{n} \times \omega^{2} r \sqrt{0+\left(\frac{1}{2} \times 1 \times \sin 2 \theta\right)^{2}}$
  $=\frac{\sqrt{2} \mathrm{m}}{\mathrm{n}} \times \omega^{2} \mathrm{r} \sin 2 \theta$ .....(xv)
  
• When $2 \alpha=120^{\circ}$ or $\alpha=60^{\circ}$, $F_{S}=\frac{2 m}{n} \times \omega^{2} r \sqrt{\left(\cos 60^{\circ} \cos 120^{\circ} \cos 2 \theta\right)^{2}+\left(\sin 60^{\circ} \sin 120^{\circ} \sin 2 \theta\right)^{2}}$
  $\quad =\frac{2 m}{n} \times \omega^{2} r \sqrt{\left(\frac{1}{2} \times \frac{-1}{2} \cos 2 \theta\right)^{2}+\left(\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} \sin 20\right)^{2}}$
  $\quad =\frac{m}{2 n} \times \omega^{2} r \sqrt{\cos ^{2} 2 \theta+9 \sin ^{2} 2 \theta}$ .....(xvi)