This article is a theoretical venture that aims to answer a series of practical questions, such as:

– given a certain electrical setup that can generate a certain power, what is the maximum thrust that we can achieve ?

– can a human-powered aircraft be built ?

– can I tell the expected thrust generated by a copter simply by knowing the power it consumes and vice-versa ?

– are larger propellers really more “efficient” ?

– which multicopter configuration is more efficient: tricopter , quadcopter , hexacopter or octacopter ?

– I heard that for every gram of mass I add to my multicopter, the flight time will decrease by 1 second, can you put any sense in this ?

When selecting motors for an RC model you will often come across the thrust tables that tell you how much thrust (or “pull”) a specific motor can generate. This usually depends on few factors: the propeller used and the voltage applied. Next to these two parameters you will usually see the current and power consumed and the resulting thrust generated, usually expressed in grams. If you want your model to hover, you must ensure that your motors can generate a thrust equal to or greater than the mass of the model.

The power that your motor consumes is calculated based on the simple formula:

**P = V * I** , where V is the voltage and I is the current.

Power represents the amount of energy we’re consuming per second, which is expressed as follows:

**P = E / t**, where E is the energy.

The efficiency of a specific setup consisting of a propeller and a motor is often calculated as:

**Eff = Thr / P** , where **Thr** is the thrust force generated expressed in grams and **P **is the power consumed expressed in Watts.

However this efficiency is not expressed in percentage as you would usually expect so it does not tell us if our motor-propeller setup is consuming energy exclusively to generate lift or wasting it somewhere else. To compute an efficiency value in percentage we would need to know the minimum or “ideal” power necessary to generate a specific thrust given our setup and environment.

We have to mention from the start that such an “ideal” model, taking into account all the possible factors would probably be too complex to use for practical purposes. So instead our goal is to first deduct and then define a “theoretical” or “reference” power formula. Finally this formula will be applied to a series of practical experiments and the resulting “theoretical” power will be compared to the real-world observed power. If the we can prove that our practical results tend to fluctuate around our theoretical value, we can propose that this formula be used for computing a “reference” power . To avoid confusion with the other efficiency let’s call this ratio “Theoretical Thrust Ratio” or **TTR**:

**TTR = P(real) / P(theoretical)**

So let’s dive into our calculations. But before we do that, for the impatient reader, here are the formulas that we’ll be deducting and using for computing our “theoretical” power (or thrust – if we flip the formula around):

(Formula 1.1)

(Formula 1.2)

Here **P **is the power expressed in **W**atts., **m** is the thrust expressed in **kg**, **r** is the radius of the propeller expressed in **meters **and **K** is a constant, with a nominal value of **22.35** assuming air pressure of **1atm** and air temperature of **20C**.

The formulas above are expressed in international physical units, however for those doing RC I am also providing the following convenient formula for computing thrust in grams **m(grams)**, based on a propeller diameter expressed in inches **d(inch)**:

(Formula 2.1)

(Formula 2.2)

(Formula 2.3)

Here **C **is a constant with a nominal value of **6.86** assuming air pressure of **1atm** and air temperature of **20C**.

Let’s get back to the main question of this article – *“how much power is needed to hover an object ?” ,* or in other words how much energy we need to consume per second to maintain this object “in the air”. Well, as it turns out “it all depends…”. For example we do not need to consume any energy to keep a book on a shelf or a tire suspended on a rope. Clearly we did not formulated our question correctly! So let’s try again and this time be more specific. “How much power does a craft powered by a spinning propeller needs in order to hover in the air ?” Here you go – much better ! Let’s see if we can answer this question.

You might recall from school physics a rule called “preservation of energy”, it simply states that the energy does not like to go away or disappear, all it does – it transforms from one form to another. We certainly know that our craft consumes energy in the form of electricity or burned fuel, but where does it go ? Didn’t we say that we don’t need energy to hover an object such as a book on a shelf or a tire hanging on a rope ? The reality is that not every object is as lucky as a book on a shelf or a tire suspended on a rope. In both these cases – the book and the tire have the benefit of a “reaction force” of the shelf and rope respectively. The book pushes on the shelf with its weight and the shelf will always push back with an equal force in the opposite direction. If you’re confused it’s probably because of a common misconception that an existence of a “force” implies that energy is consumed. The later is not true – a “force” can exist simply because of the setup of the nature, for example the gravitational force, the magnetic force and the reactive force they all simply exists without consuming energy. It’s their action that might or might not lead to the transfer of energy.

Back to our model of a propeller-powered craft, since our craft is suspended in the air, there must clearly exist an opposing force **F **that is directed in a opposite direction:

F = m * g , where **m** is the mass of our craft and **g** is the **acceleration of gravity (** approximately 9.8 m / s^2 ).

In fact this force is not much different from the “reaction force” that a shelf exerts on the book, except our medium “the air” is less dense than the shelf and as a result of this force, as we push “against” it with the propeller, the air starts moving. Here you go! We found were our energy goes.** The electrical or fuel energy consumed by an aircraft is transferred to the kinetic energy of the moving air. **Back to our school physics this energy can be expressed as follows:

Here, to avoid confusion with the mass of the craft **m **, we expressed the mass of the air as **Mair**.

Let’s examine our model for a period of time **t **. Assuming that the air moves at a constant speed **v**, during** ** this time period our propeller will generate a column of air with the height of **h = v * t**.

The volume of this cylinder can be calculated as follows:

Knowing the volume and the density of the air expressed as **Qair** we can now calculate the mass of the air:

Finally, putting this into the kinetic energy formula above we get:

Since power is nothing more than energy divided by time we immediately get the following formula for power:

This formula calculates power as a function of propeller radius and speed of air generated. It may be useful in some cases if you can measure the speed of air **v**, but it is not quite what we came here for. We need a formula as a function of the the thrust generated , which in case of a hovering aircraft is equal to the mass of the aircraft. To get that we’ll need to dive once again into our school physics book.

Let’s recall that the mechanical power of a system that exerts a force **F** on an object that is moved with the speed**v** is:

This formula results from the concept of “Work” that in a nutshell is the amount of energy transferred from one form to another. A better known formula in physics is **W = F * s** , where **W** is work, **s** is the distance, but since**P = E/t = W / t** and **v = s/t **it immediately results that **P = F * v**. We substituted **F** with **mg** since in our case of a hovering aircraft our reaction force is equal to gravitation force.

All we have to do now is solve the system of the previous two equations. To get rid of the **v **factor we raise second equation to power 3 and divide by the first equation.

Finally, we are deducting our main “theoretical” power formula:

Where K is an air-density dependent constant:

We can calculate a nominal value for K assuming an air density of **1.2 kg/m^3** , when air pressure is at **1atm**and air temperature is **20C. K =~ SQRT( 2 * 9.8^ 3 / 2.35 / 3.14) =~ 22.35.**

This concludes the deduction of our main formula for “theoretical” power required to hover an aircraft of mass**m,** equipped with a propeller of radius **r**.

The other related formulas presented in the beginning of this article can be easily deducted from this one.

To verify these theoretical results I conducted a series of thrust test using different motors and propellers commonly used in RC. I also used some thrust tests performed by third parties. The TTR (Theoretical Thrust / Real Thrust Ratio) was fluctuating around 80%-120% which is a good indicator that the formulas above work. Reader is invited to analyze these results by consulting this Excel spreadsheet: http://starlino.com/wp-content/uploads/data/power_thrust/Motor_Thrust_Tests.xlsx

A:A trained cyclist can generate 500W peak power, let’s a assume the cyclist weights 70Kg and the weight of the copter is 30Kg, thus the thrust needed for lift off is 100Kg. From this formula:

we can compute the radius of the propeller that this human-powered helicopter will need to have :

r = K * m ^ (3/2) / P = 22.35 * 100 ^ (3/2) / 500 = 44.7 (meters)

Thus this helicopter will need a propeller with a radius of 44 meters !!! This is not practical, however not impossible !In fact human powered machines have been built. Let’s analyze for example the Gamera II built by University of Maryland:

Weight: ~37kg + Pilot ~ 90kg

Rotors Radius:~ 7.2m ( x 4)Since this craft has 4 rotors , each rotor would need to produce at least 1/4 of the thrust or 90Kg / 4 = 22.5Kg . Using our formula we can calculate power needed for each rotor:

we get m = 22.35 * 25 ^ (3/2) / 7.2 =~ 331 W , thus the total power required is estimated at 331W*4 = 1324 W ! This is however much higher than the cyclist can produce using both legs and hands ( they can only peak at 750W). The reason this aircraft took off was due to “ground effect”. As air is pushed into the ground its density increases, as a result our K constant will increase. Thus , the discrepancy is due to the fact that we assumed that the K value of 22.35 would also apply to close-to-ground situation.

A:Simply apply these two formulas,

Eff= 6.86 * 10^(2/3) / 200 ^ (1/3) =~5.44 g / W

Thrust(grams)= 6.86 * (200 * 10) ^ (2/3) =~1089 g