Multi Rotor Scalability

Multi Rotor Scalability With the rapid growth in popularity of quad copters and drones in general, there has been a small group of enthusiasts who propose full scale quad copter designs (usable payload of 100+ kg). This analysis will address the controllability of the quad copter as the size is scaled up from the small drone form factor to a full scale size with rotor disc area approaching that of a single seat helicopter such as the R22. The rapid growth in popularity, and rapid decrease in prices of the quad copter is partly due to the simplicity of its design. Electric motors are directly coupled to fixed pitch propellers. That s the heart of the quad copter. The rest is controls, navigation, frame, battery and payload (camera, etc) There is a defacto principle that this design has scale up limits, but as yet, I have not seen an attempt to quantify this limitation. That is the goal of this article. To provide the analysis of the dynamic response of a quad copter form factor as a function of some key design parameters the principal one being size. The first analysis will look at the most simplistic model of scale up which is a direct proportionality of linear dimensions. While not the most realistic approach, it will provide the mathematical relationships between key parameters which will lay the foundation for more refined models. Enough introduction. Let s get started! First let s describe our quad copter in very basic terms. Rotors: Injection molded solid plastic fixed pitch. Power: DC Electric motors. Direct drive to rotors Frame: + configuration (using a + frame vs X frame for ease of analysis). Sensors: 3 axis accelerometers. 3 axis gyro s. Control: PID closed loop control. Output of PID drives motor current. OK, before we get started, let me first give you an idea of what the journey looks like so you can put the pieces together as you go through this. My approach is to look at the chain reaction of events which occur within the control loop of the quad, to understand how each piece of that chain reaction is dependent on the size of the aircraft. In the most simplistic terms, you could describe a quads response to a wind gust as follows: A wind gust causes the quad to tilt. The gyros detect the tilt, and send a signal to the motors to counter act it, and re level the quad. But in actual fact it s WAY more complex. It s more like this: A wind gust causes lift dissymmetry in the rotors. One rotor begins to accelerate up, another down. The acceleration becomes a roll rate. After a fraction of a second that roll rate results in the quad tilting away from horizontal. As the roll begins, derivative control in the PID algorithm detects a change in the rate of roll, and sends more current to the motor. As the roll continues and becomes a noticeable offset from horizontal, the proportional control in the PID algorithm also sends more current to the motor. As this tilt is sustained for more and more time, the integral control in the PID algorithm also sends more current to the motor. These increases in motor current (and decreases in current to the opposite motor) increase the motor torque, which causes the rotor blade to begin to accelerate to ever increasing RPM s. As the RPM s increase, the lift of the rotor also increases, which applies a force to the quad

which counteracts the disturbance force from the wind gust. Eventually the increased rotor lift matches the wind gust disturbance, and the quad stops rolling, but is still tilted. Integral control continues to increase motor current, which increases rotor acceleration, and velocity, and lift, until the quad roll is reversed and heading back toward the desired horizontal orientation. So, you get the idea. It s a very complex dance of many variables which all interact with each other. The reason it s called a control loop is because a change in the aircraft position initiates a corrective action which results in a counteracting change in aircraft position back to where the loop started. If these corrective actions become too large, or occur too slowly, the loop can become unstable and start to cycle back and forth wildly until the aircraft crashes. The following analysis is my attempt to quantify how the model size affects each one of the steps in the chain reaction from both a magnitude and timing perspective. Motor The motor model will assume that motor torque is proportional to motor current. τ The motor model also assumes that the maximum torque output is limited by the supply voltage (less motor back emf) and the system resistance. // Where is the motor voltage constant in radians per second per volt. ((radian/sec)/volt) Rotor Rotational Inertia This is where a big challenge exists when scaling up the quad copter design. The polar inertia of a body is Let s assume we scale up a quad copter design and use the same materials for the rotor blade and scale it up exactly proportional by a factor of P in all dimensions. For any small particle of mass m in the original model, the corresponding part M in the scaled up model will have a mass of: Because the volume of the new mass M will increase by the cube of the scale up factor, and since the same material is being used, the density is the same. This new mass M will be at a different radius from the center of rotation also. The radius of the new mass M is:

Therefore, the polar inertia of the scaled up model relative to the original is: From this equation you can see that the polar inertia of the scaled up rotor increases with the 5 th power of the scale up factor. So doubling the size of the rotor increases the inertia by 2 32 times. Rotor Speed One of the big questions is how scaling up the quad copter model affects rotor speed. There are many options: Constant tip speed (under steady state hover) Constant rotor RPM (would likely go super sonic with full size scale) Constant Reynolds number Constant wing loading (Lift/rotor area) In any case, for this analysis we are interested in how the rotor speed changes as more or less power is applied to the motor. For a spinning mass the rotational speed is the integral of the rotational acceleration. ω Rotational acceleration is proportional the torque, and inversely proportional to the polar inertia I. τ It is important to note that the torque in the above equation is the sum of all torques or NET torque. The NET torque is the difference between the motor torque and the DRAG torque caused by aerodynamic drag of the rotor. Drag for an airfoil is: 1 2 It is also noteworthy that if you look at a step change in motor torque, that torque imbalance (motor torque drag torque) will eventually re balance itself as the rotor speeds up, and drag increases. So the response of speed to current isn t exactly an integral function, but rather a logarithmic approach to the new steady state conditions. We can determine the time constant of this first order lag response (blue line) by extrapolating the initial slope (yellow line) until it crosses the new steady state value (grey line). This occurs at time t=10, so the time constant of this first order lag is 10 seconds.

35 30 25 20 15 0 5 10 15 20 25 30 This is important because the dynamic frequency response of first order lag systems is well known and will be discussed later. Rotor Lift For an airfoil: For the scaled up model: 1 2 1 2 ω 1 2 ω The main point here is that lift is proportional to the square of velocity. This would apply for actual linear velocity, or for rotational speed ω since linear velocity is ωr. Ultimately, the actual scalar values will be needed to determine the gain of the system, but for now, we will just note the square relationship. Taking the derivative of this equation, you can see how the sensitivity of lift to velocity increases as velocity increases. ω ω A is the area of the wing or airfoil which is length * cord. is the coefficient of lift of the airfoil at a specific angle of attack (assume fixed) Note that for a rotor, the velocity changes along the length of the rotor, but as a whole, the relationship is valid. In order to compute actual numeric values, one must integrate the lift equation along the length of the airfoil to get the total scalar lift value. You might also note that lift is proportional to the area of the airfoil which of course is related to the scaling factor. If we assume that the same materials are used, and every detail of the design is scaled up proportionately (not necessarily a reasonable assumption) you know that the mass of the model will increase with. Which leads to the conclusion that the wing loading must increase with P to make up the difference. This might help us solve the lift equation to determine the nominal hover velocity of the rotor.

Now cancel a on each side (remember A is proportional to ) so we get: 1 2 Since we are only interested in proportionality, all scalar constants are irrelevant so we are left with: With our scale up factor, v, the velocity of the scaled up model is the product of angular velocity, original model rotor radius, and the scale up factor P, therefore: ωrp or: ω P In the above equation, r represents the original model radius, which is a constant so we eliminate it since we are only interested in how angular velocity, ω, (RPM) changes with the scale up factor P. ω or: ω So if we double the size of the quad, the RPM of the rotor should be the original RPM, or ~71% the original RPM. Now back to the lift derivative equation: Lets substitute in for v and A parameters based on the original quad combined with the scale up factor: = ωr ωr now in terms of original model: Combine all P terms: ω = ω. This shows that the Lift to RPM sensitivity of the scaled up model is proportional to the scale up factor raised to the exponent of 2.5. Airframe Inertia Now that we know how lift is affected by motor current through the chain of events, we need to understand how this change in lift affects the aircraft as a whole. Fortunately, we will re use some of the above equations since the same principles apply to the airframe, as they did to the rotor. Consider a quad with a + frame design. If we want to roll the quad to the left, we would increase the lift of the right rotor and decrease the lift of the left rotor. So the quad is going to pivot, or rotate around an axis lined up with the front and rear support arms. This rolling movement is the result of a torque applied to a mass which has rotational inertia. So just like the rotor, the inertia of the scaled up airframe will increase with P raised to the 5 th power.

Note: Although each rotor in a multi rotor has a significant of angular momentum which tends to resist pitching and rolling motions, it is important to note that typically half of the rotors are spinning clockwise, and half are spinning counter clockwise. By rigidly fixing the rotor axes in relationship to the aircraft frame, the angular momentum of the rotors cancel each other out (assuming they are spinning at roughly the same speed). So the angular momentum of the rotor, does NOT increase the polar inertia of the craft as a whole, because the rotor momentums cancel out. The polar inertial of the whole craft is computed as it would be if the rotors are not spinning. It is also significant to note that while the rotor angular momentums do cancel each other, they also produce significant stresses on the airframe when the aircraft pitches and rolls. The faster the roll rate, the larger the torques needed to make the rotor gyro s follow the airframe. You can see this in some drone videos. As the drone pitches, the two rotors in view will get closer and farther apart as the airframe flexes. The torque comes from rotor lift acting on the airframe through a lever arm l which is the length of the right and left motor support arms. Obviously, the scaled up support arms are P times longer than the original model. So the airframe will begin to roll with a rotational acceleration rate: τ Roll rate is: Roll angle is the integral of roll rate: ω θ ω RECAP Let s recap the relationships we ve analyzed, but in reverse order. Roll angle: θ ω Roll rate: ω Roll acceleration: τ Lift: Lift/RPM sensitivity: = ω. Rotor RPM or angular velocity: ω

Rotor hover speed: ω Rotor rotational acceleration: τ Motor torque: τ Dynamic response (or frequency response) As you can see from the above equations, the large increase in rotor inertia, and airframe inertia will necessitate larger control forces, however, the aircraft must still respond to these forces very rapidly in order keep it in the air. Now we will look at the dynamic response, or frequency response of the aircraft. This will tell us how quickly the aircraft responds to commands from the flight controller. First, visualize yourself driving straight down the highway, right on top of the centerline. Now wiggle the steering wheel very quickly back and forth a few inches. I have done this, and the car stays right over the centerline, but wobbles and rocks as you quickly shake the steering wheel. What happens now if you move the steering wheel back and forth the SAME AMOUNT, but MUCH SLOWER. Now the car curves like on a slalom course to one side of the centerline and back to the other. What would happen if you moved the steering even more slowly but by the same amount? Yes, you would end up in the ditch. This examples shows how a system can respond much differently to the same input when the only change is the frequency of the input. In the above equations, anywhere we had an integral symbol we will get that type of response. A large response for slow (low frequency) inputs and a small response for fast (high frequency) inputs. This is due to the integral formula for sinusoidal signals. And sinωt 1 ω cos ωt 1 ω sin ωt 2 cosωt 1 ω sinωt 1 ω cos ωt 2 Notice how the integral of a sinusoidal signal is equal to the original signal, multiplied by and lagging by radians (or 90 degrees). So each time we encounter an integral function in the chain of events, the amplitude of the function is dependent on frequency, and the output signal lags the input signal by 90 degrees. Here is where stability can be a problem. If the chain of events introduces enough lags, what is intended to be a corrective action, and actually be a harmful action because its effect occurs too late. Phase Margin / Gain Margin Analysis Just consider the response of the quad copter to (motor current) commands from the flight controller which are in a sinusoidal waveform. Now we need to start combining all of those equations together to get the total response of the quad copter. Let s ignore the hovering motor current and just look at the fluctuations of motor current above and below that nominal value, where Z is the amplitude of the motor current fluctuations.

Motor Torque: τ sin ω t Rotor Accel: τ Rotor Velocity: ω _ sin ω t Note: Rotor Velocity values reflect the change in velocity from the nominal hover velocity. From earlier, ω or ω (solve for K 0 using actual rotor speed/lift data) _ Rotor Lift: ω ω Roll acceleration: τ = Assume lift on opposite rotors is equal and opposite, so the torque of both combined is double that of a single rotor. Substitute L into equation. 2 _ 2 ω sin ω t 2 ω 2 ω sin ω t 2 _ ω _ Roll rate: ω Roll angle: θ ω Taa Daa!!!! WHAT DOES IT MEAN?!?!? The only thing of interest in the numerator of this equation is the bit. This is the phase shift of the response to a stimulus in radians. This corresponds to 270 degrees of phase shift. That s a red flag, because any system that has a phase lag of 180 degrees or more, has the potential to be unstable, or uncontrollable. There are two noteworthy items in the denominator:

First, the term in the denominator tells me that at any given frequency of control input, the scaled up model will respond with only a fraction of the movement of the original model. If the scale up factor is 2 (twice the size), the response will be 32 times smaller. Take a look at the table below, and you will see how this gets difficult very quickly. P factor Roll angle response 1 1 1.5 1/8 2 1/32 3 1/243 4 1/1,024 6 1/7,776 8 1/32,768 12 1/248,832 16 1/1,048,576 Secondly, the ω term in the denominator tells me that the roll angle response will be much greater for slow control inputs and very small for quick control inputs. The key to keeping the quad upright, is to detect a disturbance in orientation, and compensate for it very quickly with a control signal, before the disturbance has a chance to grow. Like balancing a yardstick on your hand. The key is to keep it as near vertical as possible, making small, but very quick movements. Once the stick leans past a certain angle, it is nearly impossible to recover, and that is what we re seeing here. The problem we re facing is that this dynamic analysis shows that scaling up the quad size has a drastic negative effect on its ability to respond quickly to control inputs. This is primarily due to the exponential increase in the polar inertia of the rotor blade.