# How to calculate optimal prop diameter, pitch, torque and power coefficient given required thrust?

this is my first post here, so bear with me if this isn't the right place. I'm coming here from Stack Overflow, and this seemed to be the most relevant Stack Exchange for my question.

So, I'm building a multicopter (quadcopter, preferably, due to it's relatively simple design) on a scale that's more-or-less never been done before. I've decided to funnel all the money I made over the summer into this project, and I've just begun the planning stage.

I've worked out that I need about 2500 Newtons of thrust: enough to move 180 kilograms (about 80kg for the person and 100kg for the electronics, maybe more) at a rate of ~4.5 m/s^2 more than standardized gravity—13.888 m/s^2 in total, to be specific. I'd need approximately 750 Newtons of maximum force on each rotor, as my experience with building and coding (from scratch) an Arduino flight controller has taught me that the upper motors often need to work much harder during x and z movement than they would during standard y elevation.

Some other notes:

1. I'm pretty sure that π * diameter-in-cm * rpm / 6000 must be <340.29 (Mach I). Maybe I should be using a turbofan? Especially considering the next note...
2. The diameter of the rotors shouldn't exceed 30cm for convenience purposes. I.O.W. it can't be too big or else lugging it around would be a hassle.
1. Fuel source will (hopefully) be batteries, most likely an array of LiPo cells in series and in parallel (probably something like 10x10), each cell containing about 10,000mAh.

I struggle to continue from this point. I haven't the slightest idea as to how to pick a prop design that is most efficient for this project. My question for Aviation is this:

Given a required maximum sustainable thrust of 750 Newtons, a mass of about 180kg, and the assumption that maximum velocity of the craft is slow enough that air resistance and other drag is pretty much irrelevant...

How would I be able to calculate the optimal prop diameter, pitch, Kp (power coefficient), Kt (torque coefficient), as well as optimal number of blades and optimal manufacturing material? Or, would a turbofan be more optimal?

I wouldn't be surprised if your first thought was something along the lines of "this is ridiculous." Some may even suspect me of being a troll. I can assure you, I am highly committed to this project. My flight controller is insanely stable, and I believe it would be worthwhile to test its abilities on a build that is more heavy, powerful, and top-heavy than most multirotors ever built.

Could you guys point me in the right direction?

• Have you looked at off the shelf props? It would probably be a lot easier. – Notts90 Oct 20 '16 at 6:27
• Have you seen this answer? – Koyovis Jun 14 '17 at 8:51

## 1 Answer

How would I be able to calculate the optimal prop diameter, pitch, Kp (power coefficient), Kt (torque coefficient), as well as optimal number of blades and optimal manufacturing material? Or, would a turbofan be more optimal?

• Optimal prop diameter is simple: the largest you can accommodate = 30 cm. Disk area A = 0.07 m$$^2$$
• Optimum number of blades: as few as possible, let's start with 2.
• Use ducted rotors for a more vertical alignment of the rotor tips.
• The quadcopter has horizontal rotors. Tip speed should not exceed 0.8M = 270 m/s which equates to $$\Omega$$ = 570 rad/s = 5,520 RPM
• We have a prop diameter, RPM and thrust, now we need to determine power P and torque Q to drive the prop. We'll do so using expanded momentum theory for the hover, like in this answer. For a rotor with solidity ratio $$\sigma$$ = 0.1, the thrust coefficient $$C_T$$ is maximum 0.01, let's use a safety margin of 20% and take 0.008. Then the maximum thrust for this rotor becomes:

$$T = C_T \cdot \rho A (\Omega R)^2 = 0.008 \cdot 1.225 \cdot 0.07 \cdot 270^2 = 50N$$

We need 15 times this thrust! In order to verify the value of $$C_T$$ there is another graph in Leishman that relates $$C_T$$ with blade-pitch angle. It turns out that $$\sigma$$ = 0.01 is at a blade-pitch angle of about 12 degrees, we cannot go much higher than that. Neither can we increase tip speed, our only choice is disk area. $$A = \frac{T}{ \rho \cdot C_T \cdot (\Omega R)^2} = \frac {750}{1.225 \cdot 0.008 \cdot 270^2} = 1.05 m^2$$

You need four rotors of 1.2 m in diameter, with a solidity ratio of 0.1. With the $$C_T$$ of 0.008, we find a corresponding $$C_P$$ of 0.0007, which results in

$$P = C_P \cdot \rho A (\Omega R)^3 = 0.0007 \cdot 1.225 \cdot 1.05 \cdot 270^3 = 17.7 kW$$