What are the factors required to estimate the minimum required power for an experimental aircraft of a specific mass?

I have already viewed these questions:

But is there a comprehensive list of factors, and a single formula for estimating the minimum required power, including the take-off power, for an aircraft of a certain mass?

EDIT: I'd like to be a bit more specific - I'm looking to estimate the minimum required power for an experimental micro-light (not ultra-light), for which I have a 3D model, and estimated MTOW in kg. The primary objective is to fly at a 'safe' speed (above stall speed - probably around 80 - 100 knots). So I have the following with me:

• 3D Model of Airframe: Composite + Foam (Burt Rutan style), closely resembles a Cirrus SR22
• MTOW estimate: approx. 340 kg
• Speed range: 60 - 100 knots

Can I calculate the minimum required engine power from these factors? Do I need more? How do I calculate the minimum required power (or is it thrust)?

• The short answer is no... The long answer is probably no as well. Estimating minimum required power requires a set of performance specifications for the aircraft. These will vary based on what the aircraft is being designed for; recreation, light passenger, transatlantic passenger and cargo, intercept, ground attack etc. Achieving these performance specifications and dealing with all the compromises associated with each is what aircraft design is about. We would all be out of a job if there was a single equation to do it all. If you want a single equation you would need to make it. – DJ319 Oct 30 '17 at 7:51
• @DJ319: Your comment has many valuable insights - I'm sorry I wasn't specific. Please refer to the question again, and see if you can answer it. Thanks! : ) – ClobberXD Nov 1 '17 at 7:33
• Does this answer help? – Koyovis Nov 1 '17 at 10:45
• @Koyovis: Unfortunately, I don't understand any of that - sorry! Can you help me understand? You can also answer the question, as I believe you know a great deal about this. Thanks! : ) – ClobberXD Nov 1 '17 at 11:23
• Ah yes sorry I forgot about that. well if you know your propeller information and you know your required thrust you can calculate hp required. Basically just use the propeller equations backwards. Your required thrust will either come from a time to climb requirement or a ground run requirement most likely. I doubt your critical design point is top speed in a micro lite. But if it is this might be your critical thrust requirement as well. – DJ319 Nov 2 '17 at 10:13

A good engineer will first check existing designs: How much power is installed in comparable designs? Use airplanes of similar speed and built quality, such as the Super Diamond Mk 1 which needs 50 to 60 hp. Cruise speed is 90 knots and the MTOW is 450 kg.

Next, try to estimate the minimum wing area. Starting from the minimum speed requirement of 35 kts ( = 18 m/s), and assuming a maximum lift coefficient with flaps down of 1.6, the area to support 340 kg at sea level is $$S = \frac{2\cdot m\cdot g}{\rho\cdot c_L\cdot v^2} = \frac{2\cdot 340\cdot 9.81}{1.225\cdot 1.6\cdot 18^2} = 10.5 m^2$$

Now calculate the drag coefficient in cruise, using the parabolic drag equation. First, establish the lift coefficient at which drag is minimized: $$c_{L_{opt}} = \sqrt{c_{D0}\cdot\pi\cdot\epsilon\cdot AR}$$ The total drag coefficient at this point is simply twice the zero-lift drag coefficient $c_{D0}$, so a low drag design is important. Still, with a fixed gear your zero-lift drag coefficient will hardly be lower than 0.035, so your cruise lift coefficient is 0.938 (assuming an aspect ratio $AR$ of 10 and an Oswald factor $\epsilon$ of 0.8), resulting in a flight speed of just 23.51 m/s = 45.7 kts. Total drag at this point is $$D_{min} = 2\cdot c_{D0}\cdot S\cdot\rho\cdot\frac{v^2}{2} = 249 N$$

To sustain flight at this point only requires $P = v\cdot D$ = 5.85 kW, and assuming a prop efficiency of 0.75, the installed power should be 7.8 kW. But you want to fly faster, so we need the drag at 100 kts ( = 51.4 m/s): $$D = \left(c_{D0} + \frac{c_L^2}{\pi\cdot\epsilon\cdot AR}\right)\cdot S\cdot\rho\cdot\frac{v^2}{2}$$ There, your lift coefficient is only 0.196 but the dynamic pressure rises to 1,621 N/mm². Since the Reynolds number is higher, your zero-lift drag could drop to 0.031, resulting in a drag force of 550 N. At that speed, the required power is 28.3 kW. Under the heroic assumption that your prop will still be 75% efficient at that speed, the installed power needs to be 37.8 kW or 50.65 hp.

If you "only" want to achieve a TAS of 100 kt at altitude, here is what you need to do in the case of cruise at 10,000 ft ( = 3048 m). First you need the density at that altitude, which is 0.9 kg/m³ or 74% of the value at sea level. This means that the dynamic pressure is 1,191 N/mm² and the lift coefficient 0.267, resulting in a drag force of 419 N. This needs a continuous power of 21.56 kW to overcome. Now I assume the 75% efficient propeller again and that you run the engine at 75% of max. power, so the installed power in 10,000 ft should be at least 38.3 kW or 51.4 hp. Assuming a normally aspirated engine, this would translate to a rated power of 70 hp at sea level.

Considering that similar designs require similar power, this looks about right. Normally, you now need to compute the climb speed with the excess power of 35.15 kW to check how usable this design is, but at 10.5 m/s I doubt this will be not enough.

If you manage to include a retractable gear with your limited mass budget, the zero-lift drag might be as low as 0.024. Now the drag force in cruise at 10,000 ft will be only 324.4 N and the installed rated power at sea level only 40.4 kw or 54 hp.

With piston aircraft, your power needs increase with the cube of airspeed. I leave it as an exercise to you to calculate how much more power the last 10 knots require: Repeat the calculation with only 90 kt cruise and your rated engine power can be as low as 51 hp with a fixed gear.

• Um - I made an error, which is corrected. Get 56 hp now! – Koyovis Nov 3 '17 at 15:45
• @Koyovis: Yes, that's better, only your minimum drag speed lift coefficient is actually that for minimum power. – Peter Kämpf Nov 3 '17 at 16:07
• In which equation have you given the cruise altitude / air density as an input? How do I calculate the minimum required power for a max. cruise altitude of 10000 ft. and what is the cruise altitude you have used as a parameter? – ClobberXD Nov 3 '17 at 16:56
• This is a great way to keep us busy :) – Koyovis Nov 4 '17 at 0:53
• @AnandS: The Diamond only cruises at 90 kt, not 100. This is a major difference. Altitude makes little difference - what you gain in lower drag, you lose in power with a normally aspirated engine as density drops. At high speed the larger weight does not hurt much, but a bigger wing does. Conversely, a bigger wing allows you to fly slower, so the minimum power goes down. But so does the speed regime you can fly in. – Peter Kämpf Nov 4 '17 at 5:58

Aircraft pre-design gives methods to compute this, based part on physics, part on statistical data of existing aircraft. For instance the method laid out in chapter 5 of Torenbeek, following this method, we would compute required power for several cases and take the maximum.

In the aircraft design phase there is not yet data such as wing area, gross weight, fuel etc that is normally used for performance calculation, we now have the opposite problem: determine combinations of design characteristics for power plant and wing to obtain desired performance. A very detailed method is given in Torenbeek, we'll shortcut as much as possible and take the SR22 wherever we can (from the wiki and from here).

1. Weight. You state an MTOW of 340 kg = 3,335 N.
2. Initial estimation of aeroplane drag. Most low-speed polars can be approximated by a parabola: $$C_D = C_{D_0} + \frac{{C_L}^2}{\pi \cdot A \cdot e}$$ with A being the aspect ratio $b^2 / S$. For now, let's take statistical data listed in Torenbeek for small single aircraft with fixed gear: $C_{D_0}$ = between 0.025 and 0.04 (take 0.035), e = 0.7. We take $C_{D_0}$ to be on the high side because of the small size, low speed, and associated low Reynolds number with a thick friction layer. A mid range value is chosen for e. For A, let's take the value of the SR22 which is 10.1.
3. Cruise: you specify 100 kts = 51.4 m/s. Horsepower $P_{CR}$ to fly at this speed and at 5,300 m altitude (ceiling of SR22): $$P_{CR} = \frac{1}{2} \cdot \rho \cdot V^3\cdot C_D \cdot S$$ With the wing area found under 4. we get $C_L$ = 0.52 and with drag parabola from 2. $C_D$ = 0.058. Substitute $\rho$ =0.73 for 5,300m and the $P_{CR}$ = 19 kW = 26 hp at this altitude. This is net power, typical prop efficiency is 0.78 and for unboosted engines power decreases with air density. Equivalent power at sea level = (26 / 0.78) * 1.225/0.73 = 56 hp
4. Stall: depending on the country there is a maximum stall speed imposed upon microlights. Your stated MTOW implies an FAA Light Sport Aircraft with a maximum stall speed of 45 knots = 23 m/s. Let's take a safety margin and take stall speed = 20 m/s. Stall speed of the SR22 = 58 kgs = 30 m/s => $$C_{L_{max}} = \frac {2W}{\rho \cdot V^2 \cdot S}$$ = 2.0 at sea level, let's take the same $C_{L_{max}}$ for the microlight. Substituting this value and a stall speed of 20 m/s, we then get a wing area of 6.7 m$^2$. With aspect ratio of 10.1, we get a wing span of 8.2 m
5. Climb, including airworthiness requirements. Let's take the same data as the SR22, ceiling = 5,300m, rate of climb C = 6.5 m/s @ sea level. For steady state climb power $P_{climb}$: $$\eta_p \cdot \frac{P_{climb}}{W} = C + \frac{C_D}{C_L}\cdot V$$ Minimum drag speed is minimum for $$C_L = \sqrt{3 \cdot C_{D_0} \cdot \pi \cdot A \cdot e}$$ = 1.53, equating to 23 m/s. Prop efficiency improves with airspeed and most favourable climbing speed is roughly 20% higher = 28 m/s. Typical $\eta_p$ = 0.78 for tractor piston engine in fuselage nose. $C_L$ = 1.0 and $C_D$ = 0.08 follow from lift equation and drag polar. Resulting in:

$$0.78 \cdot \frac {P}{3335} = 6.5 + \frac{0.08}{1.0}\cdot 28$$

$$P = 37.4 kW = 50 hp$$

1. Take-off performance. This one is quit lengthy and involves computation of the TO field length for a given engine power - which we found under 5. so we won't be doing this exercise for now.Procedure is given in Torenbeek 5.4.5

So the power required to cruise at 100 kts at 5,300m is higher than the power required to climb: $P_{cruise}$ must be applied = 56 hp. There are many improvements that can be made on above, for more detail I refer to the book.

• 8kW? Could you explain these calculations to me? I couldn't keep up and got lost somewhere in the Initial Estimation of Airplane Drag. : ) These are very difficult for me to understand, without a good deal of explanation. I'd be grateful if you could edit your answer to accommodate for these explanations, as it will also benefit those who come looking for your answer in the future. Thanks! – ClobberXD Nov 3 '17 at 0:37
• I'll edit the answer later in the day, and add take-off and altitude performance as well. A certain basic understanding of the physics behind aerodynamic principles can be found for instance in the FAA handbooks. – Koyovis Nov 3 '17 at 3:49
• Um - there was an error in the previous computation, is corrected. The values for $C_L$ and for climbing power make a lot more sense now. – Koyovis Nov 3 '17 at 15:06
• Now that's a better answer - for a moment there I was baffled by the huge margin of difference between your calculation and @PeterKampf's. Now both the estimates match up. So I see that you have estimated that I'd need around 56 hp to maintain 100 knots at around 5300 m / 17400 ft. How do I calculate the power required to maintain 100 knots at, say 10000 ft? – ClobberXD Nov 3 '17 at 15:46
• In the same way, first compute $C_L$ with the air density at 10,000 ft, then compute $C_D$ with the drag polar of 2., then compute cruise hp as in 3. – Koyovis Nov 3 '17 at 15:50

In order to calculate required thrust you need firm design requirements. These may come from user specifications and requirements or from the appropriate regulations.

You are likely going to find that your maximum required thrust is either to achieve a rate of climb requirement (likely to come from the regulations.), to achieve a specified service ceiling or a take off distance requirement.

The method would be to define your requirements for these three conditions. You then calculate required thrust for each condition. From required thrust you can use your propeller equations in "reverse" to work out what the required horse power will be. Which ever condition has the highest required horse power will be the critical case.

Remember naturally aspirated engines lose power with increasing altitude. This power drop off is approximately horse power = sigma * sea level horse power

where sigma is the relative density of air

so if you need 20 hp at 10 000ft you will need an engine that can put out about 27hp at sea level.

• I hope this helps you Anand. Let me know I will add more detail to this if you need more information – DJ319 Nov 2 '17 at 10:35
• I believe the maximum thrust is require to achieve take-off distance and climb rate - but I don't know how to calculate thrust with what I have - can you add the calculation of thrust to your answer? I don't have any other data, than those I mentioned in the question. If I need more parameters, can you please provide the formulas to derive them from existing parameters? I'm sorry I can't vote today, as I've reached my vote limit, but your answer was very helpful nonetheless, thank you! : ) – ClobberXD Nov 2 '17 at 10:45
• I see others were faster on the draw and got there examples and equations in before me. One thing that has been mentioned but I will say it again look and see if you can find a copy of Torebeek synthesis of sub sonic design this book will take you through everything you will need in order to do a basic aircraft design. Its pretty much the aircraft design bible. Another good one is Stinton the design of an airplane. – DJ319 Nov 3 '17 at 14:18
• @Koyovis provided a link to Torenbeek's book - very useful, but quite taxing on novices to aerodynamics... But still, great book! – ClobberXD Nov 3 '17 at 15:07