Velocity and Acceleration of a Piston

An analysis of the velocity and acceleration of a piston.

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Contents

Introduction
Velocity And Acceleration Of A Piston
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Introduction

Key facts

For a piston with the crank radius $r$ , the rod length $l$ , the crank angle $\theta$ , and the crank angular velocity $\omega$ , its velocity can be written as:

$v=-r\omega \left( \sin \theta + \frac{\sin 2\theta}{2n} \right)$

and its acceleration:

$a=-r\omega^2 \left( \cos \theta + \frac{\cos 2\theta}{n} \right)$

where $\displaystyle n = \frac{l}{r}$ .

A reciprocating engine derives its power from the vertical motion of a piston within individual cylinders of the engine. The up-down motion of the piston, also known as reciprocating motion, is converted into rotary motion by use of a crankshaft. Such a mechanism in motion can be seen in Figure 2, where the crankshaft is depicted with red, and the pistons with gray (to see the animation click on the thumbnail). Power transmitted such, requires constant motion of the piston within the cylinder from one extreme to the other. During this motion the velocity of the piston changes constantly; becoming zero at each extreme, and accelerating to a maximum value proportional to the engine RPM (rotations per minute).

Velocity And Acceleration Of A Piston

In order to define the velocity and acceleration of a piston, consider the mechanism in Figure 1, where the crank $OC$ is driven with the uniform angular velocity $\omega$ .
Also, let $r$ be the crank radius, $l$ the rod length, $x$ the position of the piston pin from the crank center, $\phi$ the $\angle {CPO}$ angle, and $\theta$ the $\angle {COP}$ angle (the crank angle).

From Figure 1 it can be seen that:

$x = r \cos \theta + l \cos \phi$

(1)

It can also be noted from Figure 1 that:

$r \sin \theta = l \sin \phi$

(2)

Squaring equation (2) gives:

$r^2 \sin^2 \theta = l^2 \sin^2 \phi$

which can also be written as:

$r^2 \sin^2 \theta = l^2 (1- \cos^2 \phi)$

This eventually leads to:

$\cos \phi = \sqrt{1-\frac{\sin^2 \theta}{n^2}}$

where $\displaystyle n = \frac{l}{r}$ .

By using this expression of $\cos \phi$ in equation (1), we obtain:

$x = r \cos \theta + l \sqrt{1-\frac{\sin^2 \theta}{n^2}}$

which can also be written as:

$x = r \left( \cos \theta + n \sqrt{1-\frac{\sin^2 \theta}{n^2}} \right)$

and furthermore, as:

$x = r \left( \cos \theta + \sqrt{n^2 - \sin^2 \theta \right)$

(3)

In order to calculate the piston velocity we differentiate (3) with respect to time, when we get:

$v = \frac{dx}{dt} ( = \dot x ) = -r \omega \left( \sin \theta + \frac{\sin 2\theta}{2\sqrt{n^2 - \sin^2 \theta}} \right)$

The piston acceleration can then be calculated by differentiating again with respect to time, when we obtain:

$a = \frac{dv}{dt} ( = \ddot x ) = -r \omega^2 \left[ \cos \theta + \frac{n^2 \cos 2\theta + \sin^4 \theta}{(n^2 - \sin^2 \theta)^\frac{3}{2}} \right]$

Note that in these equations the positive direction of velocity and acceleration is away from the crankshaft.

In normal situations $\sin^2 \theta$ and $\sin^4 \theta$ can be neglected in comparison with $n^2$ , and thus, the above equations can be reduced to:

$v = -r \omega \left( \sin \theta + \frac{\sin 2\theta}{2n} \right)$

$a = -r \omega^2 \left( \cos \theta + \frac{\cos 2\theta}{n} \right)$

Example:

Example - Example 1

Problem

In the crank and slotted lever mechanism diagramed in Figure E1, the crank $OP$ is driven at a uniform speed of $\omega$ radians per second.

If $OL$ is the perpendicular from $O$ on $XQ$ , $XQ$ being the center-line of the slotted lever, prove that the angular acceleration of the slotted lever is given by:

$\ddot \phi = \omega^2 \frac{OL}{XP} \left( \frac{2PL}{XP} - 1 \right)$

Also, find the magnitude and direction of the acceleration of the point $Q$ when the crank angle is $\theta = 60^o$ , given that $OP = 3\; inch$ , $OX = 10\; inch$ , and $XQ = 18\; inch$ .

Figure E1

Workings

From the diagram we can write that:

$OL = OX \sin \phi$

$OL = OP \sin (\theta - \phi)$

From which:

$OX \sin \phi = OP \sin (\theta - \phi)$

(1)

By differentiating (1) with respect to time, we get:

$OX \dot \phi \cos \phi = OP (\omega - \dot \phi) \cos (\theta - \phi)$

Which leads to:

$\dot \phi [ OX \cos \phi + OP \cos (\theta - \phi) ] = OP \omega \cos (\theta - \phi)$

(2)

From Figure E1 we can write that:

$XL = OX \cos \phi$

$PL = OP \cos (\theta - \phi)$

Thus, equation (2) becomes:

$XP \dot \phi = PL \omega$

Or:

$\dot \phi = \frac{PL}{XP} \omega$

(3)

By differentiating (2) with respect to time, we obtain:

$\ddot \phi [ OX \cos \phi + OP \cos(\theta - \phi)] - \dot \phi^2 OX \sin \phi - \dot \phi (\omega - \phi) OP \sin (\theta - \phi) = -\omega (\omega - \dot \phi) OP \sin (\theta - \phi)$

(4)

By substituting $\dot \phi$ in equation (4) with the expression from (3), we can write:

$XP \ddot \phi - OL \frac{PL^2}{XP^2} \omega^2 - OL \frac{PL}{XP} \left( 1 - \frac{PL}{XP} \right) \omega^2 = -OL \left( 1 - \frac{PL}{XP} \right) \omega^2$

Or:

$\ddot \phi = \omega^2 \frac{OL}{XP} \left( \frac{2PL}{XP} - 1 \right)$

(5)

By replacing the numerical values in equation (1), we can write that:

$10 \sin \phi = 3 \sin (60 - \phi)$

(6)

As:

$3 \sin (60 - \phi) = 3(\sin 60 \cos \phi - \sin \phi \cos 60)$

Equation (6) becomes:

$10 \sin \phi = 3\frac{\sqrt{3}}{2} \cos \phi - \frac{3}{2} \sin \phi$

Which gives:

$\tan \phi = \frac{2.59}{11.5}$

Thus, we get that:

$\phi = 12^\circ 32'$

And, by using this value, we also obtain that:

$OL = 10 \sin \phi = 2.20 \; inch$

$PL = 3 \cos (60-\phi) = 2.035 \; inch$

$XP = 10 \cos \phi + PL = 11.79 \; inch$

Furthermore, by replacing numerical values in (3) and (5), we respectively get that:

$\dot \phi = \frac{2.035}{11.79} \cdot 10 = 1.75 \; \frac{rad}{s}$

$\ddot \phi = 100 \cdot \frac{2.2}{11.79} \cdot \left( \frac{2 \cdot 2.035}{11.79} - 1 \right) = -12.2 \; \frac{rad}{s^2}$

In order to calculate the acceleration of $Q$ , $a$ , we are going to split it in a tangential component, $a_T$ , perpendicular to $XQ$ , and a radial component, $a_r$ , in the direction of $XQ$ (see Figure E1). Also, let $\alpha$ be the angle which defines the direction of the acceleration. We can write that:

$a_r = XQ \dot \phi^2 = 18 \cdot 1.73^2 = 53.7 \; \frac{inch}{s^2}$

$a_T = XQ |\ddot \phi| = 18 \cdot 12.2 = 219.5 \; \frac{inch}{s^2}$

From which we get the magnitude of acceleration:

$a = \sqrt{a_T^2 + a_r^2} = \sqrt{219.5^2 + 53.7^2} = 226 \; \frac{inch}{s^2}$

Also, as:

$\tan \alpha = \frac{a_T}{a_r}$

We obtain the direction of acceleration:

$\alpha = \tan^-1 \frac{a_T}{a_r} = \tan^-1 \frac{219.5}{53.7} = 76^\circ 16'$

Solution

$\ddot \phi = \omega^2 \frac{OL}{XP} \left( \frac{2PL}{XP} - 1 \right)$

$a = 226 \; \frac{inch}{s^2}$

$\alpha = 76^\circ 16'$$