The Question:
Someone asked me how fast a piston travels within the cylinder
at a given RPM. This question intrigued me. The following is my
solution. I also have worked the following scenario for the case of the
offset crankshaft.
I recommend reading this page before jumping to work with the offset example.
Disclaimer: I do not guarantee the accuracy of any of this. If you have any
questions or comments, I would love to hear from you. Email me at rollins@wfu.edu.
Equations:
To derive an answer, I first
thought I would represent the rotation of the crank as a circle where the
center of the circle would be the axis of the crankshaft and the
circumference represents the rotation of the center of the journal. The
piston rod is then represented as a straight line. Since the piston and
the top of the piston rod travel the same vertical velocity, the piston
itself is not a factor. Figure 1 shows this representation.
 |
| Figure 1 |
Next, I derived a displacement equation to relate the following:
- The length of the piston rod
- The radius of the crankshaft (This is my term for the distance from
the center of the journal to the center axis of the crankshaft.)
- The angle of the radius vector relative to the path of the piston
This relationship was straight forward by using the Law of Cosines.
Then I solved for
displacement (s) by the quadratic formula. The result can then be simplified
by a little Trigonometry.
 |
| Figure 2 |
I have written a Java Applet to
demonstrate the motion of above equation.
The displacement equation is then used as a function of time by substituting
a function of time for the angle. This relation is demonstrated in
the following table:
| RPM |
|
degrees/minute |
|
radians/minute |
| 1 revolution/minute |
=> |
360 degrees/minute |
=> |
2 π radians/minute |
| 10 revolutions/minute |
=> |
10 × 360 degrees/minute |
=> |
10 × 2 π radians/minute |
| 100 revolutions/minute |
=> |
100 × 360 degrees/minute |
=> |
100 × 2 π radians/minute |
| ω revolutions/minute |
=> |
ω × 360 degrees/minute |
=> |
ω × 2 π radians/minute |
| Table 1 |
From this data we can establish the relationship between revolutions
per minute and radians per minute. Making a leap forward,
we can calculate the number of radians traveled by the crankshaft
by multiplying the ω × 2 π radians/minute by the number of
minutes. Thus we have, θ(t) =2πωt. By substituting
this function as our angle
in the displacement equation, s(θ), we
then have the displacement as a function of time. This
is probably also a good time to reveal one of my biggest assumptions.
I am assuming that the rotational speed of the crank is constant
when the engine is held at a constant RPM.
(If you disagree with this assumption,
here is some of my reasoning.)
 |
| Figure 3 |
The next step is to take the first and second derivative of the displacement
equation to obtain an equation for velocity and acceleration respectively.
The derivatives of the displacement equation were derived using a
wonderful program called maple.
The graphs and equations shown here are derived from images produced with
maple.
 |
| Figure 4 |
Example:
Now let's use some numbers for these equations and generate some graphs.
I will spare you the actual equations for this example because they just
ain't perty.
I have not been able to actually measure the radius of a crankshaft nor
the length of a piston rod. A few people have given me rough estimates.
So, I am going to use the radius of 3.5 inches and the piston rod length
of 5.5 inches. For the RPMs, let's use 6000 revolutions per minute.
Graph 1 is our displacement equation, s(t).
If these dimensions are far off from
reality, then you can use an Excel spread sheet that
I created for customized numbers. The spread sheet should give a good
estimate of the maximum velocity and it will graph the equations.
r=3.5 inches, b=5.5 inches, ω=6000
 |
| Graph 1 |
Given that the revolutions per minute are 6000, then one revolution takes
1/6000th of a minute. This implies that one period of the displacement
graph is between t=0 and t=1/6000. Note that 1/6000 = 0.0001666. We can
also see that the displacement ranges from 9 inches down to 2 inches. The
9 comes from 3.5 + 5.5 (radius + rod length) at &theta=0 and 2π radians (360°). The 2 comes from 5.5-3.5 (rod length - radius) at &pi radians (180°)
The velocity graph is a little more interesting. I would have expected more
symmetry in this graph, but the calculus does not lie. We will come back to the
maximum velocity after the acceleration graph is examined.
 |
| Graph 2 |
The acceleration graph is by far the most interesting. I thought this was
going to be some simple problem with just some simple sin wave. But,
this graph shows that there is nothing simple about the piston's motion.
 |
| Graph 3 |
I tried to deduce the time for the intersection of the acceleration graph
and the x-axis. I came up with a time of 0.000029996 which would give
the acceleration of zero. This time is important because it is the time
at which the velocity is maximum (well actually, it is at its minimum
here...like, going very fast in reverse).
 |
| Graph 4 |
So what is the answer?
How Fast?
So, back to our original question, "How fast is the piston going?"
The velocity of the piston comes out to be
v(0.000029996) = -158954.7782 inches/minute
This number doesn't mean much to me, so let's look at it from
a different perspective. If we convert -158954.7782 inches/minute
to miles per hour what do we get?
-158954.7782 inches/minute
× (1 mile / 63360 inches)
× (60 minutes / 1 hour) = -150 miles / hour
Now, keep in mind that the piston went from 0 to 150 MPH in
0.00179976 seconds! That is FAST!
What is the displacement and angle?
I was also curious at where the piston was at its maximum
speed. The displacement tells us a little about this.
s(0.000029996) = 5.987615285 inches
This means that the piston came from 9 inches down to 5.9 to reach its
maximum velocity. But, I am also interested in the angle of the crankshaft at
this point. The angle can be found by converting 2πωt to degrees.
Example:
- t=0.000029996,
- θ = 2πωt = 2π(6000)(0.000029996)=1.130822 radians,
- 1.130822 radians × 360°/2π radians = 64.8°
To summarize the findings, here is a table that shows the interesting
points along the travel of a piston. I hope you now have an appreciation
for just how fast the pistons are traveling at 6000 RPMs. After this,
I sure do.
| Time (minutes) |
Height s(t) (inches) |
Velocity v(t) (MPH) |
Angle (degrees) |
| 0.000000000 |
9.00 |
0 |
0.0 |
| 0.000029996 |
5.99 |
-150 |
64.8 |
| 0.000083333 |
2.00 |
0 |
180.0 |
| 0.000136671 |
5.99 |
150 |
244.8 |
| 0.000166666 |
9.00 |
0 |
360.0 |
Table 2