Dimensional Analysis of a DC-Motor

Ralf Gerlich


For the construction of my first quadrocopter, I bought some brushless DC motors, some propellers, a set of drivers for brushless motors, a battery, built a geometrically matching frame and started working with that. This approach quite probably has not led to a good design: I might get more flight time with the same payload by some better design.

However, there’s so many parameters to tweak: Battery capacity, voltage, size and mass of the engines, size and pitch of the propellers, frame geometry, etc. All of these are interrelated by complicated relationships, most of which are not exactly known to us. Varying all the parameters to find some optimum may prove both time-consuming and costly: We’d have to order different parts with different parameters, measure them under a whole lot of different situations, and then find the optimal solution.

However, there’s a pretty clever tool to reduce the set of variables to consider: Dimensional Analysis. The very short version is: Any relationship between a set of physical quantities can be represented by a smaller set of physical quantities and their dimensionless relationship to the remaining quantities. This allows us to massively simplify experiment design.

In this article, we’ll have a look at the basics of dimensional analysis using the so-called Buckingham’s Pi Theorem ( Citation: , (). On physically similar systems; illustrations of the use of dimensional equations. Phys. Rev., 4. 345–376. https://doi.org/10.1103/PhysRev.4.345 ) , and try to use a combination of physical insight and dimensional analysis to characterise the performance of a DC motor in comparison to some of its basic parameters such as size or the strength of the magnetic material used.

Dimensional Analysis: An example

Schematic view of a mathematical pendulum

Let’s have a look at a mathematical pendulum. We have a mass [latex]m[/latex] fixed to an arm or rope of length [latex]l[/latex], and the strength of local gravity shall be given by the gravitational acceleration [latex]g[/latex]. We want to determine [latex]T[/latex], the period of the pendulum. Some basic physical insight tells us, that [latex]T[/latex] may depend on the mass, the length of the arm, local gravity and the angle of initial deflection [latex]\alpha_0[/latex]:

\begin{equation} \label{eq:mathematical-pendulum} T = f\left(m,l,g,\alpha_0\right) \end{equation}

Dimensional analysis allows us to get a general picture of these influences. Let’s try to express [latex]T[/latex] as a dimensionless multiple of some product of powers of [latex]m[/latex], [latex]l[/latex] and [latex]g[/latex]. Note that we do not consider [latex] \alpha_0[/latex] in this product, as it has no dimension itself and thus no power of [latex]\alpha_0[/latex] can contribute to making the product dimensionless:

\begin{equation} k_T = T m^{c_m} l^{c_l} g^{c_g} \end{equation}

Solving this equation so that [latex]k_T[/latex] is dimensionless leads us to

\begin{equation} k_T = T \sqrt{\frac{g}{l}} \end{equation}

Now, we can write our relationship from Eq. \ref{eq:mathematical-pendulum} in the following way:

\begin{equation} T \sqrt{\frac{g}{l}} = \bar{f}\left(m,l,g,\alpha_0\right) \end{equation}

However, on the left-hand side, we have a dimensionless number, so the function [latex]\bar{f}[/latex] must also map its parameters to dimensionless numbers. [latex]\bar{f}[/latex] cannot depend on any product of powers of [latex]m[/latex], [latex]l[/latex] and [latex] g[/latex], as such a product would not be dimensionless: No power of [latex]l[/latex] and [latex]g[/latex] can cancel the mass dimension present in any non-zero power of [latex]m[/latex], and the same is true for any other combination. There also cannot be a function that only depends on the value of the dimensioned parameter, as then the value of the function would change if we used feet as unit for the length instead of meters.

Thus, the function on the right can actually only depend on dimensionless quantities, which leaves [latex]\alpha_0[/latex]. So, we finally get

\begin{equation} T \sqrt{\frac{g}{l}} = T\left(\alpha_0\right) \end{equation}

So, dimensional analysis tells us, that the period of the pendulum only depends on [latex]\sqrt{\frac{g}{l}}[/latex] and the initial deflection angle [latex]\alpha_0[/latex]. We arrived at this conclusion only by analysing the dimensions of the quantities involved.

We might not know [latex]T\left(\alpha_0\right)[/latex] in general, but we could measure it for some set of values of [latex]g[/latex], [latex]l[/latex] and [latex]\alpha_0[/latex], and could then determine it for any other set of values of [latex]g[/latex] and [latex]l[/latex] if only we keep [latex]\alpha_0[/latex] constant. We say that [latex]\alpha_0[/latex] determines the similitude of two problems with different parameters.

If we were to look into the actual equations of motion, we’d get the same result — and we’d also get the form of . Depending on the complexity of the equations involved, this could prove very tedious.

The Buckingham Pi Theorem

What we just applied is known as the Buckingham Pi Theorem, named for Edgar Buckingham, an American physicist living from 1867 to 1940. Buckingham was not the first to notice and apply the principle, but still today it is named for him.

The theorem essentially states that, if there is a physically meaningful relationship between physical quantities, then there must be a function describing this relationship which only depends on dimensionless ratios of these quantities. He also described a procedure to find these ratios.

Let’s have a look at the proof idea: Assume that we have [latex]n[/latex] physical quantities, which we shall name [latex]q_1, \ldots, q_n[/latex], and there is some — possibly unknown — relationship between them, which we express in the form

\begin{equation} \label{eq:buckingham-relationship} F\left(q_1,\ldots,q_n\right) = 0 \end{equation}

Obviously, the value of [latex]F[/latex] does not have a dimension. If it did, we could simply divide by that dimension and still have a similar equation without dimension. Thus, [latex]F[/latex] also cannot depend on dimensioned quantities. If it did, its value would differ depending on the units used, even if the quantities were the same.

As a consequence, we can select [latex]k[/latex] of the quantities, and express the others as dimensionless multiples of these. Let’s say, that [latex]q_1, \ldots, q_k[/latex] were the selected quantities. Then we can write Eq. \ref{eq:buckingham-relationship} as follows:

\begin{equation} F\left(q_1,\ldots,q_k,\frac{q_{k+1}}{{q_1}^{r_{k+1,1}}\ldots{q_k}^{r_{k+1,k}}},\ldots\right) = 0 \end{equation}

Now, as argued before, [latex]F[/latex] cannot depend on dimensioned quantities, so [latex]F[/latex] must actually be independent of the [latex]q_1, \ldots, q_k[/latex] and can only depend on the dimensionless ratios:

\begin{equation} F\left(\frac{q_{k+1}}{{q_1}^{r_{k+1,1}}\ldots{q_k}^{r_{k+1,k}}},\ldots,\frac{q_{n}}{{q_1}^{r_{n,1}}\ldots{q_k}^{r_{n,k}}}\right) = 0 \end{equation}

Finding the Basis Quantities

Now, how do we find which quantities to use as our basis [latex]q_1,\ldots,q_k[/latex]? It may help to note that term “basis” is used intentionally here, as these quantities must form a basis for the vector space of dimensions used within [latex] q_1,\ldots,q_n[/latex]. In our pendulum example, we had quantities of time, length, mass and acceleration, which is length per time squared. Thus, we have three dimensions and four variables (or “vectors” in linear algebra parlance), and we can select three of these four to represent all the units.

There’s some useful rules for selecting them, but they do not uniquely determine the selection:

Specifically the third rule also may give us important insight into the problem, such as when our problem only contains a single variable referencing a specific dimension. In that case our relationship either does not actually depend on that quantity, or we are missing another quantity that influences the relationship. The missing dimension may guide us as to where to look.

How does a DC-Motor work?

Now, let’s use our new knowledge to analyse the performance of a DC motor and its relationship with some of its basic parameters. This will allow us to make better decisions on selecting motors in the future.

So, how does a DC motor work? Essentially, it turns electrical energy into mechanical energy via a magnetic field. For that, it uses two magnetic fields, leading to a magnetic force. That force results in a moment around the axis of the motor. When the motor turns around its axis, one of the fields needs to change its orientation so that the motor keeps turning. This is achieved either mechanically in brushed motors or electronically in brushless motors.

Configuration of a Brushless DC-Motor with coils in the center and permanent magnets on the shell

The figure above shows the basic mechanical configuration of a Brushless DC (BLDC) motor in outrunner configuration with permanent magnets for one part of the field, and the other field being generated by the coils in the middle. By switching the coils on and off, the inner field can be rotated. In an inrunner configuration, the fixed coils would be outside, and the static magnets would be on the axis in the middle.

For BLDC motors special brushless controllers are required that electronically determine the angular position of the magnets and switch the coils on and off as required. There are different setups for these controllers, some using hall sensors to determine the position of the magnets, others monitoring the induction voltage on the unconnected coil and using that to determine the proper timing for switching.

Schematic of the Electrical Configuration of a DC Motor

The figure to the right shows the basic electrical configuration of a DC motor, indicating the parasitic inductance and resistance of the coils in the motor. In addition, the turning motor also acts as a generator, inducing a counter-voltage, often called counter electromagnetic force (EMF) or back-EMF.

The torque acting on the axis is approximately proportional to the current flowing through the coils, and the reverse voltage generated is proportional to the rotational velocity. The proportionality factors are usually named [latex]k_t:=\frac{Q}{I}[/latex] for the torque coefficient and [latex]k_v:=\frac{U}{\omega}[/latex] for the voltage coefficient.

Efficiency Considerations for the DC-Motor

On the electrical side, we have losses on the resistor. On the magnetic-mechanical side, losses may come from constantly changing the magnetic field inside the metal of the stator. In addition, there may be aerodynamic losses from the turning motor.

The efficiency factor [latex]\eta[/latex] is found as the quotient [latex]\frac{P_{out}}{P_{in}}[/latex], where [latex]P_{in}[/latex] is the power injected into the system, and [latex]P_{out}[/latex] is the usable power provided by the system.

On the electrical side, we have

\begin{equation} \eta_{el} = \frac{k_v \omega I}{k_v \omega I + R I^2} = \frac{1}{1 + \frac{R I}{k_v \omega}} \end{equation}

On the mechanical side, we have

\begin{equation} \eta_{mech} = \frac{k_t I \omega}{k_v \omega I} = \frac{k_t}{k_v} \end{equation}

Dimensional Analysis of the DC-Motor

We are reaching the final goal of this article: Determining the influences of dimensions on the performance on a DC motor. Specifically, we want to check the influence of some relevant dimensional parameters on the voltage and torque coefficients.

From our previous insight, we assume that [latex]k_v[/latex] and [latex]k_t[/latex] depend on

Let’s have a look at the dimensions of our quantities. We denote the dimension of a quantity by square brackets. The dimension of [latex]D[/latex] and [latex]h[/latex] is a length: [latex]\left[D\right]=\left[h\right]=L[/latex]. The dimension of [latex]k_v[/latex] is voltage per rotational frequency, which is voltage times time. Voltage is energy per charge, or energy times time per current. So finally, we arrive at [latex]\left[k_v\right] = \frac{M L^2}{I T^2}[/latex], with [latex]M[/latex] denoting mass, [latex]I[/latex] denoting current and [latex]T[/latex] denoting time.

It is quite obvious that we have a dimension [latex]\frac{M}{I T^2}[/latex] which occurs in only one quantity, notably in [latex]k_v[/latex]. Obviously, the relationship must also include additional quantities. One reasonable quantity to consider is the strength of the of the magnets. The strength of the magnets is given by their remanence [latex]B_r[/latex], which happens to have the dimension [latex]\frac{M}{I T^2}[/latex], just what we need.

Now we have all the dimensions we need – [latex]L[/latex] and [latex]\frac{M}{I T^2}[/latex]. Let’s recapitulate the dimensions of all the quantities:

Quantity Symbol Dimension
Back-EMF Coefficient [latex]k_v[/latex] [latex]\frac{M L^2}{I T^2}[/latex]
Motor Size [latex]D[/latex] [latex]L[/latex]
Motor Height [latex]h[/latex] [latex]L[/latex]
Remanence [latex]B_r[/latex] [latex]\frac{M}{I T^2}[/latex]

The back-EMF coefficient [latex]k_v[/latex] is our dependent variable, so we cannot use that for expressing the dimensions. Thus, we use [latex]D[/latex] and [latex]B_r[/latex]:

Dimension Expression
[latex]L[/latex] [latex]D[/latex]
[latex]\frac{M}{I T^2}[/latex] [latex]B_r[/latex]

Now, we can express the unit of [latex]k_v[/latex] using these variables:

\begin{equation} \left[ B_r D^2 \right] = \frac{M L^2}{I T^2} \end{equation}

So, as a consequence [latex]\frac{k_v}{B_r D^2}[/latex] is dimensionless. Similarly, the units of [latex]D[/latex] and [latex]h[/latex] are the same, so that [latex]\lambda:=\frac{h}{D}[/latex] is a dimensionless quantity. We call [latex]\lambda[/latex] the aspect ratio of the motor.

Using the Buckingham Pi Theorem, the following relationship must hold:

\begin{equation} \frac{k_v}{B_r D^2} = f_v\left(n,p,\lambda\right) \end{equation}

We know that the dimensions of [latex]k_v[/latex] and [latex]k_t[/latex] must be the same – otherwise [latex]\eta = \frac{k_t}{k_v}[/latex] cannot be dimensionless. Also, it is reasonable to assume that [latex]k_t[/latex] also depends on the same parameters as [latex]k_v[/latex]. In consequence, the following relationship must also hold:

\begin{equation} \frac{k_t}{B_r D^2} = f_t\left(n,p,\lambda\right) \end{equation}

And as a final finding, we can thus also express the efficiency in a similar manner:

\begin{equation} \eta_{mech} = \frac{k_t}{k_v} = \frac{f_t\left(n,p,\lambda\right)}{f_v\left(n,p,\lambda\right)} =: f_\eta\left(n,p,\lambda\right) \end{equation}

These relationships enable us to compare motors with different sizes and magnet strengths, as long as the number of windings and the aspect ratio are the same. To determine the relationships expressed by [latex]f_t[/latex], [latex]f_v[/latex] and [latex]f_\eta[/latex], we only need to determine [latex]\frac{k_v}{B_r D^2}[/latex] and [latex]\frac{k_t}{B_r D^2}[/latex] for motors with different values of [latex]n[/latex], [latex]p[/latex] and [latex]\lambda[/latex]. We do not need to explicitly vary [latex]D[/latex] and [latex]B_r[/latex].


In this article, we learned about dimensional analysis, a very powerful tool which allows us to reduce the complexity of the analysis of physical relationships, just by looking at the dimensions. This method is widely used in aerodynamics and is the basis, for example, for the common formulae for lift and drag of wings and airfoils, or for thrust, torque and power for propellers.

Thus, to get a rough estimate on the performance of one system, we can scale up the performance data we have on another system, as long as the similarity parameters – which are the non-dimensional numbers to our coefficient parameters – are close enough. This strongly simplifies the dimensioning of a system, and allows us to get at least a rough ballpark setup for our design. We may later on refine the design and get some actual measurements, but we save a lot of up-front effort by getting to a simple estimate without much measurement.


Buckingham (1914)
(). On physically similar systems; illustrations of the use of dimensional equations. Phys. Rev., 4. 345–376. https://doi.org/10.1103/PhysRev.4.345