DC Motors

This section of the report gives an overview of DC motors and the importance of DC motor performance in relation to a quadrotor’s ability to hover. DC brushless motors are the most commonly used motors in RC hobbies due to availability, price, and performance. Any hobby website offers dozens of different DC motors with different specifications, and a method of selecting the appropriate motor is outlined below. Figure 1 shows a generalized torque vs. speed curve of a DC motor with a constant applied voltage.

Maximum torque can be achieved when the motor is fully braked and no torque is experienced when the motor is at its maximum rotational speed. DC motors perform the best when they are producing the most mechanical power. Power is the product of torque and rotational speed which is given in Equation 7.

In order to find the maximum power of a motor, all that is needed is to find the rotational speed at which the maximum power occurs. It is shown in Figure 4 that maximum power is achieved at a torque and rotational speed that is half of the motor’s maximum capabilities given a constant voltage. Maximum power output is important because it is desirable to match a motor’s best range of performance with the most common flight situations of the aircraft. In this case, the quadrotor helicopter’s most common mode of flight is to hover. Hovering occurs when a propeller/motor set produces thrust (mass) that equals the weight of the aircraft. Hovering should occur at 50% of the motor’s maximum capabilities which directly relates to the equivalent of half the battery’s voltage.

Since DC motors are rated in Kv (rpm/v), the rotational speed at which maximum power is achieved is found by multiplying the motors Kv by half the battery voltage and dividing the results by 2. This process is illustrated in Equation 8 below.

The resulting desired rpm occurs at 1/4th the maximum rpm of the motor at full voltage. This desired rpm will be used in the Propeller and Motor Selection section of this report. But before a proper propeller and motor can be selected, the mass of the aircraft must be estimated.

5 Responses to “DC Motors”
  1. Ganeshyam says:

    A perfect and handy compilation…. for the beginners.:) thank you for the info..

  2. Prototype says:

    Hi Jeremiah,
    First of all, I want to say that your website is very helpful for beginners in quadcopter design. I myself referenced your site when just starting my studies in quadcopter design last year. I have since studied DC motors and propellers more thoroughly and I find my knowledge in DC motors conflicts with your claim in Equation 8, and I can prove it mathematically, assuming I am not mistaking your intentions.
    Lets start off with the definition of rotational power,
    P=w*t (1)
    (w=rotational speed, t=torque)
    Now, lets refer to a motors torque speed curve, as you have shown in Figure 4. The equation for the line can be shown to be
    t=-(t_max/w_max)*w + t_max (2)
    Combining equations 1 and 2 results in the following
    P=-(t_max/w_max)*w^2+t_max*w (3)
    Taking the derivative of equation 3 with respect to rotational speed yields the following
    dP/dw = -2(t_max/w_max)*w+t_max (4)
    In calculus, optimization is used to determine the location of local maximums in minimums. This is done by setting an equations derivative to zero and solving for the independent variable. In equation 4, we have the derivative of power as a function of rotational speed. Using optimization, we can determine the rotational speed at which maximum power occurs, as seen below
    -t_max=-2(t_max/w_max)*w (subtract t_max from both sides)
    1=2(1/w_max)*w (multiply -1/t_max on both sides)
    0.5*w_max=w (multiply 0.5*w_max on both sides)
    w_max_p = 0.5*w_max (5)
    This shows that max power occurs at half the maximum rotational speed. The maximum rotational speed of a motor can be found by multiplying its Kv by the applied voltage.
    w_max = Kv * Applied_voltage (6)

    Combining equations 5 and 6 gives the following
    w_max_p = 0.5*Kv*Applied_voltage (7)
    which is close to what you have in equation 8, except you don’t divide by 2. Again, I am not sure what your intention is with your equation 8, so I want to apologize ahead of time if I missed something. Below is a link to a picture that shows an arbitrary motor performance graph that shows that max power indeed occurs at half the rotational speed.

    • jedickey says:

      Hi Prototype,

      Thank you for the kind words. I’m happy to hear that my blog was helpful. I’m also happy that you put so much thought into you question. I’ll start off with an explanation followed by an example.

      The concept here is to obtain congruence between equations 8 and 9 (my equations). Both equations approach the problem from different sides; equation 8 concerns motor and battery selection while equation 9 concerns propeller selection. In equation 9, we want to determine the rotational speed that produces enough thrust to counteract gravity (hover) given a certain mass and propeller. In equation 8, we want to determine which motor will give us the most effective rpm to hover given a certain battery voltage. Now to the point of your question, I chose to define hovering as an event that occurs at 50% throttle. For PWM signals or Electronic Speed Controllers, 1/2 throttle is essentially equivalent to cutting the battery voltage in half. Basically we want to find a motor that meets our needs with only half the battery voltage. Otherwise we define a system that hovers at 100% throttle, which is not a good system.

      Now I’d like to give you an example of this problem. I forgot to divide equation 8 by 2 on the first iteration of this project. This error lead me to establishing design variables that did not work with my design; I paired motors and props that were too small for the weight, and I made the arms long enough to just fit my props. I tried to fly this design, but it barely got off the ground because, much to my surprise and disappointment, I set hover to 100% throttle. In addition, I had to redesign and cut another frame because I made the arms too short to add larger props. This hurt my pride and my wallet because I made my frame out of carbon fiber sheet (I chose carbon fiber because I had other, larger ideas for this project in mind). As a check, follow my process using equation 8 with and without the divisor. You will find that the addition of the divisor will require you to choose a larger prop and more powerful motor.

      I hope I have convinced you that you need to divide equation 8 by 2. The method that I defined will not work unless you do. Please let me know if you agree or disagree. Either way, I have really enjoyed this correspondence! You have obviously put a lot of thought into this problem. Please send me a link to some of your project videos so I can post them here.

  3. Anonymous says:

    Hello Jeremiah,
    After doing my own research regarding the selection of props & motors, i have found that your page is very informative and unique. In fact, you are the only one that shows the math involved in choosing the right combination of propellers + motors + battery. I am an engineering student entering my junior year and unlike many people who just want to read recommendations and build their aircraft, I am interested in the math involved as well as making graphs and spreadsheets that will help me and other people better understand the process.

    I think you misunderstood prototype, and I understand what he is saying. After doing the math myself and following your steps and prototype’s, i believe i can point you to the area of confusion. I understand that you are approaching the same answers (ideal rpms that will match props and motors by following two different paths [eq. (8) for motors, and (9) for props] as there is no “true” answer for the combo that will lift you off the ground and get you in the air. For example, you can choose a prop with a really small diameter + a high pitch to lift a big aircraft, however you are going to have to spin your motors really fast, resulting in a very unstable aircraft, besides the point that you are going to have to choose a motor with a max rpm that equals 4X the ideal rpm for the set of props. I believe that what prototype is saying and i agree with him is that in eq. (8), there is a grammar error and not a math error.

    I should note that ———> RPM ideal/hovering = [KV[rpm/v]*Volts(v) / 4] and ***not*** RPM max = [KV[rpm/v]*Volts(v) / 4].

    Thank you for your time and i know this is was in 2014, but if you see this would you please verify for me and future visitors.

    • jedickey says:

      Hi Anonymous,

      Thanks for the comment. It always brings me great joy to know that people are critically reviewing my work. You are correct about Equation 8; this rotational speed is the “ideal” hovering rpm. However, I intended it to be the rpm at maximum power given 50% throttle. You’re statement is more accurate though. I made an assumption that torque would also be 1/4 stall torque, which is incorrect. I am working on a new analysis that I’m hoping will take the hobby world by storm. I’ve already proven this new model with empirical data and written a GUI. I’m just finishing the write-up now. I’m really excited. Until I refresh my blog, you can still follow my method here. Just choose the most powerful motor you can find that meets the ideal rpm criteria. So stay posted!!

      Best Regards,
      Jeremiah Dickey

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