In this video, I want to
say something about making an equation dimensionless that's called
nondimensionalization. For an example, I'll
use this equation which governs the
oscillation of a pendulum. If you look at this equation, there's the mass of the pendulum, the length of the rod connecting the mass
to the pivot point. There's time, we're taking the second derivative of
the angle with respect to time. There is a friction
co-efficient C. There's gravity,
the gravitational acceleration. If you are on Earth, it's 9.8 meters per second squared and the units
of meters and seconds. But maybe you're running
an experiment on Jupiter or Mars so g could be
considered a parameter. Then you have an external force, so it has an amplitude f naught and it has a frequency omega. So if you look at this equation, you have a lot of parameters. You have m, l, c, g, f naught, and omega. You have six parameters. But those parameters
have dimensions, and if you then do what we say we nondimensionalize
this equation, it's possible to reduce those six parameters
into a smaller set. When you do that, then you can, if you were saying exploring the solutions
to this equation you have a fewer number of parameters that you
need to explore, and also another very
important point may be that the
non-dimensional equation here may be equivalent to other physical systems
provided you also nondimensionalize
those equations associated with the other physical systems. So how do you
nondimensionalize equation? You need to know the units
of the various things. So first of all you need to
have the fundamental units. Those are things like the length, the time, the mass. Here the fundamental units
are going to be a mass, and we're going to
denote the units of mass by m. M just means mass. It could be kilograms, it could be pounds. It's not really relevant
to what we're doing here. What is the precise unit of mass? The mass could be in units
of the weight of this pen. That's a valid unit for mass. So we just call mass, m. We have length, l. So we denote the unit
of length by l. Again, meters, feet, the length of this pen that's a valid unit
of length also. Then we have time, and we denote the units
of time by t. Seconds, minutes, hours, days, years, these are all valid
units of time. Then you can ask, what are the units of
the different terms here? For instance, what is the units of one of the terms
in this expression? So we use this bracket symbol. What are the units
of a mass m times the length l times the second derivative of
the angle with respect to time? The angle itself
doesn't have units. The angle is supposed
to be in radians. Radians are pure numbers. There are two Pi radians
in a circle. Radians doesn't have units. So we can do things like
take the sine of an angle. The angle is a pure number. So what are the units
of ml Theta double dot? So this symbol here
means that question. There's a mass,
there is a length, and then there's a d
square Theta dt squared. So there's a t squared
in the denominator. Every time you take the
derivative with respect to time, you introduce a time
in the denominator. So the units of this expression, this term in this equation is mass times length times
time to the minus 2 power, ml divided by t squared. That happens to have
units of force. A very fundamental understanding of physical equations is that a physical equation
is only valid if every term in that equation
has the same units. So ml Theta double dot
has to have the same units as cl Theta dot has to
have the same units as mg sine Theta and the same units
as f naught cosine Omega t. The Units across an equation always
have to be the same, otherwise it's not
a valid physical equation. So we know the units of mass, we know the units of length, we know that the units of time. We have this damping
parameter c here, so we can figure out
what are the units of c. So we have a length
divided by a time. So in order to match
the terms in this equation, we need a mass divided by a time. So the units of c is mass per unit time or mass times
t to the minus 1. Then, we have cl will be ml, and then d Theta dt is one over t. So it matches the terms. Similarly, g, that's
the gravitational acceleration, we know what the
units have to be. If we match the terms, then it would be a length
per unit time squared, so lt to the minus 2. You know g is 9.8 meters per
second squared on the Earth, so it's a length per time squared or length times
t to the minus 2. Then f naught and Omega, let me squeeze that in here, the units of f naught. This is our force has to match the units of
ml Theta double dot. So is mlt to the minus 2, and the units of Omega has to make the argument of
cosine dimensionless, so the units of Omega has
to be t to the minus 1. Now, that we're
familiar with what the units of everything
is in this equation, we can try to make a dimensionless
form of this equation. Let's divide through by ml. So then we have Theta double dot plus
divide through by ml c over m Theta dot plus
g over l sine Theta, and that's equal to f naught over ml times cosine Omega t. So how do we make
this equation dimensionless? The independent variable is t. The dependent variable is Theta. Theta is already
dimensionless in radians. So t is the only variable that we need to
make dimensionless. We have to choose
how to measure time. We make this equation
dimensionless by choosing how to measure time. There are actually several
choices on how to do that. But a common choice would be
to use this combination of g divided by l. So g divided by l has units of one
over time squared. G divided by l has units
of one over time squared. If you remember, if you know about a simple
harmonic oscillator, the square root of g over l would be the natural frequency of a simple harmonic oscillator provided that sine Theta
was approximately Theta. So that's something I will
talk about in the course. But here we just need to
know that g over l has units of t to the minus two. So we can define Omega naught. Remember Omega has units of
one over t. So we can define Omega naught to be
the square root of g over l, which also has units
of one over t. We can then use Omega naught to
make time non-dimensional. So we can define the time Tau, which will be
our dimensionless time, equal to Omega naught times
t. So t has units of time, Omega naught has units
of one over time. So Tau is dimensionless,
our dimensionless time. What that means is that when
Tau goes from zero to one, t goes from zero to
one over Omega naught. So we're using one
over Omega naught then as our unit of time. If we do that, we need to replace
the derivatives. So we can write
something like d Theta, dt, the first
derivative of Theta. We can write that as d Theta, d Tau, d Tau, dt using the chain rule. Chain rule, d Tau, dt is just Omega. So this is just Omega
naught, I should say. D Tau, dt is Omega naught. So this is just Omega naught
times d Theta, d Tau. You can go through
the chain rule. You can do a shortcut. You just replace t by Tau
divided by Omega naught. You treat this thing
like a fraction, you substitute t equals Tau over Omega naught and you pop
up an Omega naught there. Similarly, the second derivative, so Theta double dot,
the second derivative, is going to be
Omega naught squared times d squared
Theta d Tau squared. So we take our dimensional
differential equation here and we replace it by
a dimensionless equation. So we have the equation
then will be Omega naught squared times d squared
Theta d Tau squared, plus c Omega naught over m, times d Theta d Tau, plus Omega naught squared sine Theta equals
our right-hand sine. So it's equal to f naught
over ml times cosine, and then we replace t by
Tau over Omega naught. So that becomes Omega over
Omega naught times Tau. So Omega over Omega naught is
a dimensionless parameter. It's basically Omega in
units of Omega naught. What do we have left to do? We divide through by
Omega naught squared and we end up then with the dimensionless
differential equation. So that will look like d squared Theta d Tau squared plus, divide through by
Omega naught squared. So we have c over m Omega naught, d Theta d Tau plus sine Theta. The sine Theta term is cleared. That's because we used
the term in front of sine Theta to make
our time non-dimensional, the g over l. So we've cleared
that term and then that's equal to f naught over
ml Omega naught squared. Then we have our cosine Omega over Omega naught times Tau. That's our
dimensionless equation. Now, there are these groupings of dimensional parameters c over m Omega naught and this one, f naught over ml Omega naught
squared, and this one. We can write this equation
then as d squared Theta, d Tau squared plus, call this the Alpha d Theta
d Tau plus sine Theta. Then call this term Gamma, and then call
the inside Beta, Beta Tau. These are what are called
dimensionless parameters. We have our Alpha. We have our Gamma, and we have our Beta. What have we done? We started with an equation
that had six parameters, m, l, c, g, f naught, and Omega, and by non-dimensionalizing
the equation, we end up with
a final equation that has only three parameters:
Alpha, Gamma, and Beta. The reason we were
able to go from six to three is because there are three fundamental units
in this problem: mass, length, and time. So six minus three
gives us three. That has a funny name. It's called the
Buckingham Pi Theorem. It's interesting to
learn about that, but actually in
practice engineers don't really need to
use this theorem. All they need to do is know how to non-dimensionalize
an equation. Believe me, if you're doing
a numerical solution, you don't want to deal with six parameters when you
can deal with only three. Let me review. In this video, I'm trying to show
you how if you have a physical problem where
everything has units, you can redefine
your variables to be dimensionless so
that you can reduce that physical problem with
so many parameters to a lot fewer parameters and all of the parameters
are dimensionless. I'm Jeff Chasnov. Thanks for watching. I hope as engineers, you have a lot of use
with this new knowledge.