Different parts of a reciprocating engine is shown in Fig. 4.4 (a)

Let us consider this figure,

Point A will have linear motion towards point C along the line AC

Point B will have circular motion with an angular velocity

Thus,

Velocity at point A, V_{A} = _{0} AO

Similarly,

Velocity at point B, V_{B} = C_{0} BO

Or,

r = _{0} BO. _{0}r/ BO

V_{A} = _{0} AO

= (r) AO/ BO

Hence,

We can easily get the velocity at point A by knowing the values of all elements.

**Analytical Method to find V _{A}**

Fig. 4.4 (c) shows the graphical representation of different parts of an engine.

To find the velocity at point A, we can use different analytical methods.

**Method 1:**

Draw line CD ⏊ AC and line BE ⏊ AC

Now,

In AOB and BCD

⦟AOB = 90^{0} – = ⦟BCD

⦟OAB = 90^{0} – = ⦟BDC (we know that ⦟OBA = ⦟LCB)

Hence,

AO/ BO = CD/ BC = CD/ r

And,

tan = CD/ AC

CD = AC. tan

= (AE + EC) tan

= (L cos + r cos ) tan

= L sin + r cos tan

Since,

V_{A} = (r) AO/ BO

= (r) CD/ r

= (L sin + r cos tan)

**Method 2:**

Suppose the crank is at IDC (Inner Death Centre). In this condition,

A_{1}B_{1} = AB = L

B_{1}C = r

Hence,

AO/ BO = CD/ BC = CD/ r

tan = CD/ AC

we know that

CD = AC. tan

= (AE + EC) tan

= (L cos + r cos ) tan

= L sin + r cos tan

Thus,

V_{A} = (r) AO/ BO

= (r) CD/ r

= (L sin + r cos tan)

**Method 3:**

Considering the Fig.4.4 (d),

Let, A_{1}A be x

It can be written as,

A_{1}C – AC = x

(A_{1}B_{1} + B_{1}C) – (AE + EC) = x

(L + r) – (L cos + r cos ) = x

Differentiating the above equation with respect to time, we get-

=

= L sin+ r sin

Hence,

V_{A} = L sin+ r sin

Since,

BE = L sin = r sin

sin = r/ L sin

Differentiating the above equation with respect to time, we get-

cos =r/ L

= r/ L cos

= rcos / Lcos

Now, substitute this value to get V_{A},

Therefore velocity at point A,

V_{A} = L sin(rcos / Lcos ) + r sin

= r cos sincos + r sin

= (L sin + r cos tan)