Determining the acceleration of an aircraft

[GeForce7]

Hello everyone. I'm currently doing a research on aircraft performance for my school project, which require me to design an airfoil to suit a specific mission. I'm not very familiar with the formulas used for calculating the aircraft performance and I would like to seek help from the forum.

Using the NASA FoilSim, I came out with an airfoil profile with a maximum coefficient of lift of 1.888. Using the lift equation, I calculated the minimum speed which I must fulfill in order to gain enough lift to counter the weight which is approximately 39.18 m/s.

V_stall=√(W/(CL_max×0.5×ρ×S))
V_stall=√(39240/(1.888×0.5×1.225×22.1))
V_stall=√(1535.425122)
V_stall=39.18 m∕s or 141.10 km/h
​

Assuming that the runway distance is 1800 m, how do I calculate the minimum runway distance in order for the aircraft to take-off? What can I do to reduce the distance needed to take-off while keeping the final velocity at 39.18 m/s? Is there a way to increase the acceleration of the aircraft while keeping the same final velocity?

Here are some requirements that my aircraft must fulfill:

1. Maximum take-off weight of 4000 kg (39240 N)​
2. Take-off must take place at 1.25 times the stall speed, but less than 200 km/h​
3. Landing must take place at 1.3 times the stall speed, but less than 180 km/h​
4. Aircraft must take-off and land at a runway with a distance of 1800 m or less​

 

[SteamKing]

Hello everyone. I'm currently doing a research on aircraft performance for my school project, which require me to design an airfoil to suit a specific mission. I'm not very familiar with the formulas used for calculating the aircraft performance and I would like to seek help from the forum.

Using the NASA FoilSim, I came out with an airfoil profile with a maximum coefficient of lift of 1.888. Using the lift equation, I calculated the minimum speed which I must fulfill in order to gain enough lift to counter the weight which is approximately 39.18 m/s.

V_stall=√(W/(CL_max×0.5×ρ×S))
V_stall=√(39240/(1.888×0.5×1.225×22.1))
V_stall=√(1535.425122)
V_stall=39.18 m∕s or 141.10 km/h
​

Assuming that the runway distance is 1800 m, how do I calculate the minimum runway distance in order for the aircraft to take-off? What can I do to reduce the distance needed to take-off while keeping the final velocity at 39.18 m/s? Is there a way to increase the acceleration of the aircraft while keeping the same final velocity?

Here are some requirements that my aircraft must fulfill:

1. Maximum take-off weight of 4000 kg (39240 N)​
2. Take-off must take place at 1.25 times the stall speed, but less than 200 km/h​
3. Landing must take place at 1.3 times the stall speed, but less than 180 km/h​
4. Aircraft must take-off and land at a runway with a distance of 1800 m or less​

You're missing some key components of aircraft performance to make an intelligent study.

Chiefly, what sort of power or thrust can your aircraft's engines produce? Are they recips w/propellers, turboprops, turbojets, ramjets, what?

You can make a barn door fly if you attach enough power to it.

Aircraft performance is dependent on much more than just picking an airfoil for the wing.

Here is a bizjet which has roughly your max. TO weight:

https://en.wikipedia.org/wiki/Cessna_Citation_Mustang

 

[GeForce7]

Thank you for the reply. Let's say my aircraft can produce a maximum take-off thrust of 14000 N, how can I work out the calculation? Can I use the F=ma formula to work out the acceleration? Do pilots always go for maximum engine thrust during take-off?

 

[SteamKing]

Thank you for the reply. Let's say my aircraft can produce a maximum take-off thrust of 14000 N, how can I work out the calculation? Can I use the F=ma formula to work out the acceleration?

You pretty much have to. Of course, F is not just the thrust, it's Fnet, which includes the rolling friction on the landing gear, and the drag on the plane as it increases speed during the takeoff roll, all acting against the thrust of the engines.

Do pilots always go for maximum engine thrust during take-off?

That's what it's there for.

The aircraft on take-off typically has a max load: full fuel, full passengers, full cargo, whatever.

Say the airfield is at a higher elevation; there's no wind blowing down the runway; the air temperature is high, so the air density at altitude is relatively low; you have a short runway with height obstructions just beyond the end of the field. This is the time to crank the throttles wide open, cross your fingers, and hope that your take-off calculations are right, and that you don't have an engine malfunction halfway down the runway.

Once the aircraft is off the ground and flying, the engines are required only to produce enough thrust to counter the drag of the aircraft as it is flying at the set cruising speed and altitude.

 

[David Lewis]

You can both increase acceleration and reduce take off distance by reducing weight. (I don't know if your assignment permits that.) Another way to reduce take off distance is increase maximum lift coefficient or wing area. But in any case you want to plug the total lift of the airplane into your formulae, which is normally wing lift minus stabilizer download, assuming a relatively conventional configuration.

Bear in mind that lift and weight are equal and opposite only in level flight. Climbing or descending, weight is equal and opposite to the vector sum of total aerodynamic force and thrust.

 

[GeForce7]

Thanks you guys for providing me the assistance. Right now, I need to calculate the thrust that can a turbofan engine can produce using F=ma equation. However, I'm not sure how I can approach it. Can I use a hypothetical thrust value and work out the calculation backward? For example, if my maximum take-off thrust is 14,000 N, how do I calculate the mass of the air and acceleration?

 

[SteamKing]

Thanks you guys for providing me the assistance. Right now, I need to calculate the thrust that can a turbofan engine can produce using F=ma equation. However, I'm not sure how I can approach it. Can I use a hypothetical thrust value and work out the calculation backward? For example, if my maximum take-off thrust is 14,000 N, how do I calculate the mass of the air and acceleration?

It's not clear what you want to do here.

Are you trying to figure out how fast an aircraft can move with a set max. thrust, or are you trying to design an engine to give a max. thrust value?

If the latter, it will take much more than F = ma to figure the performance of a turbofan engine.

https://www.grc.nasa.gov/www/k-12/airplane/turbfan.html

 

[GeForce7]

You are correct. I'm trying to figure out an engine that can give me a maximum thrust value of 14,000 N. The formula from NASA's website is too difficult for me to figure it out myself. The thrust value is taken from Pratt and Whitney PW615F which is capable of generating a maximum take-off thrust of 6000 N. Since I'm clueless on the engine thrust calculation, I can only use the thrust value of an engine and pluck it into an equation to calculate the acceleration of my aircraft which is 1.49 m/s^2. In theory, my aircraft can lift off easily and only require a runway distance of 850.05 m which is significantly lesser than requirement. My only issue now is that I need to show the lecturer that the engine thrust value can be calculated.

 

[SteamKing]

You are correct. I'm trying to figure out an engine that can give me a maximum thrust value of 14,000 N. The formula from NASA's website is too difficult for me to figure it out myself. The thrust value is taken from Pratt and Whitney PW615F which is capable of generating a maximum take-off thrust of 6000 N. Since I'm clueless on the engine thrust calculation, I can only use the thrust value of an engine and pluck it into an equation to calculate the acceleration of my aircraft which is 1.49 m/s^2. In theory, my aircraft can lift off easily and only require a runway distance of 850.05 m which is significantly lesser than requirement. My only issue now is that I need to show the lecturer that the engine thrust value can be calculated.

P & W make more than one model of jet engine. There are other engine manufacturers out there, like GE, Rolls-Royce, etc. A thrust of 14000 N is about 3100 lbf, so maybe one of the makers of smaller engines, like Garrett or Williams, would be more what you are looking for. A Williams FJ44-4 is rated at 16000 N.

http://www.williams-int.com/products/fj44-4

Most of these performance data, like static thrust, fuel consumption, etc., are taken not from calculations, but actual engine tests after the design is finished.

 

[GeForce7]

Ok. I will try to convince them that the engine thrust calculation is not feasible and I should just use the thrust value given by the manufacturer for better accuracy.

 

[cjl]

Do pilots always go for maximum engine thrust during take-off?

It's pretty rare for them to use max thrust at takeoff actually - they calculate thrust required based on runway length, air density, takeoff weight, and whether or not there are any obstacles to clear (like a large hill or building just past the end of the runway), and then base their takeoff thrust on that. Full thrust is rarely used because it increases engine wear.

 

[David Lewis]

You can calculate thrust using F=ma if you know engine power and the diameter of the engine's ideal (Rankine) actuator disk.

 

[cjl]

You can calculate thrust using F=ma if you know engine power and the diameter of the engine's ideal (Rankine) actuator disk.

Jet engines don't have a fixed power though (nor do they have a fixed thrust, though that's often a simplifying assumption that's made with them). It's rather complicated to actually get a decent estimate of the thrust of a jet engine.

 

[David Lewis]

Good point. Thrust varies depending on speed even at a fixed throttle setting. At the beginning of the take off run, you have static thrust, but as the airplane picks up speed, thrust changes.

I believe the relationship between thrust and power is:
power = sq rt [(thrust cubed) divided by (2 * air density * area of actuator disk)]

 

[cjl]

What kind of power are you trying to get? Power applied to the air? Power applied to the aircraft? The first is rather complicated for jet engines, as the core air has a pretty significant amount of power added in the form of heat, while the bypass air is pretty much just passing through a ducted fan. The second one is easy (but largely useless), as it will simply be the thrust multiplied by the aircraft velocity. I don't see how your formulation could be correct for any definition though - there should definitely be a velocity dependence somewhere in there..

 

[David Lewis]

The formula only tells you how much power is consumed in producing a given amount of thrust. It doesn't matter how fast the airplane is going. If the airplane is not moving, for example, the engine still puts out power and still produces thrust.

 

[cjl]

The formula only tells you how much power is consumed in producing a given amount of thrust. It doesn't matter how fast the airplane is going. If the airplane is not moving, for example, the engine still puts out power and still produces thrust.

It does matter how fast the airplane is going though - the same amount of power will produce different amounts of thrust both based on the speed of the air traveling through the engine and the massflow of the air traveling through the engine. You really can't separate the two.

 

[mfb]

V_stall=39.18 m∕s or 141.10 km/h

1. Landing must take place at 1.3 times the stall speed, but less than 180 km/h​

Those two don't work together.

 

[GeForce7]

Hi guys, I would like to ask a question with regards to TSFC (Thrust Specific Fuel Consumption). The formula for calculating TSFC is fuel flow (kg/hour) divided by the thrust being generated by the engine at that point of time. Is it correct to assume that the fuel flow changes as you increase or decrease the thrust? If not, is the fuel flow constant? For example, the engine will burn less fuel (lower fuel flow) during cruising and burn more fuel (high fuel flow) during take-off?

 

[mfb]

Is it correct to assume that the fuel flow changes as you increase or decrease the thrust?

Sure.

 

[jim hardy]

Airklners use movable extensions to change the airfoil shape
ever watch out the window on approach to landing? The wing becomes much wider and curved.

 

[cjl]

Airklners use movable extensions to change the airfoil shape
ever watch out the window on approach to landing? The wing becomes much wider and curved.

They're called flaps and slats, specifically, and they're important in allowing the airplane to have an airfoil shape that can provide low drag in cruise while still being able to generate enough lift to land at a fairly low speed.

 

[GeForce7]

Using an example taken from Wikipedia (https://en.wikipedia.org/wiki/Williams_FJ44#Specification), I took the values from FJ-44-1A and did some calculation.

Formula (https://www.grc.nasa.gov/www/k-12/airplane/sfc.html):

TSFC = Fuel Consumption in lb per hour/Thrust in lbf
0.456= Fuel Consumption in lb per hour/1900 lbf
Fuel Consumption in lb per hours=0.456*1900
Fuel Consumption in lb per hours=866.4 lb/h OR (392.99 kg/h)

Assuming my calculation is correct, the TSFC given by the manufacturer is for maximum thrust? If so, the fuel consumption at maximum thrust is 866 lb per hour (392.99 kg per hours)? Is there a way for me to calculate the fuel consumption at a certain thrust value (for example, during low speed cruise)?

 

[David Lewis]

The power (as a function of thrust) formula I gave you gives you induced power, if my understanding is correct. Therefore power must be divided by the efficiency of the propulsion system.

System efficiency is sometimes divided into two parts: How efficiently chemical energy in the fuel is converted to heat and mechanical energy, and then what portion of that energy is converted to thrust.

 

[cjl]

Using an example taken from Wikipedia (https://en.wikipedia.org/wiki/Williams_FJ44#Specification), I took the values from FJ-44-1A and did some calculation.

Formula (https://www.grc.nasa.gov/www/k-12/airplane/sfc.html):

TSFC = Fuel Consumption in lb per hour/Thrust in lbf
0.456= Fuel Consumption in lb per hour/1900 lbf
Fuel Consumption in lb per hours=0.456*1900
Fuel Consumption in lb per hours=866.4 lb/h OR (392.99 kg/h)

Assuming my calculation is correct, the TSFC given by the manufacturer is for maximum thrust? If so, the fuel consumption at maximum thrust is 866 lb per hour (392.99 kg per hours)? Is there a way for me to calculate the fuel consumption at a certain thrust value (for example, during low speed cruise)?

Yes, TSFC is usually specified at full (non-afterburning) thrust. Unfortunately, there isn't an easy relationship to calculate fuel consumption at lower speeds - there are some calculations that can be done, but they usually end up rather complex (I'll dig out my propulsion textbook in a bit and see what it says).

 

[cjl]

The power (as a function of thrust) formula I gave you gives you induced power, if my understanding is correct. Therefore power must be divided by the efficiency of the propulsion system.

System efficiency is sometimes divided into two parts: How efficiently chemical energy in the fuel is converted to heat and mechanical energy, and then what portion of that energy is converted to thrust.

I still don't believe your power formula is correct for any kind of power, but maybe I'm misunderstanding what you mean here by "induced power". Could you explain how you arrived at your formula?

 

[David Lewis]

My hypothesis is that if you specify the diameter of the propeller, fan, or nozzle of your engine (called the actuator disk), then you know how much air (or gas) the system is able to work on. And as soon as you know the mass of that air then you know how much that mass must be accelerated in order to generate any specified thrust.

This gives you the induced power, which is the smallest possible amount of power required to generate a given amount of thrust. To that you need to also add in some power lost to friction in the bearings, gearbox, boundary layer, and unequal pressure distribution across the actuator disk.

 

[cjl]

Yes, but without knowing the inflow velocity, you can't calculate any meaningful power number (you don't know how much acceleration you need to apply to the air, and you don't know the mass flow through your control volume), and your formula is missing the inflow velocity. Also, the smallest possible power to generate a given amount of thrust is zero, if you're allowed to fiddle with the parameters however you want.

 

[insightful]

I believe the relationship between thrust and power is:
power = sq rt [(thrust cubed) divided by (2 * air density * area of actuator disk)]

What are the units of this power? Please show a dimensional analysis of your equation.

 

[cjl]

That's an excellent point - I didn't even bother to check the units on that equation...

 

[David Lewis]

My basic assumption is that, even if the propulsion unit is 100% efficient, it will still consume some power -- an unavoidable consequence of adding momentum to a fluid.

 

[cjl]

But that power can be arbitrarily low by increasing the mass flow sufficiently. The power required for a given thrust depends on the energy required for a given change in momentum, which will depend both on inflow velocity and on the mass flow rate affected by the propulsion unit.

 

[insightful]

The power (as a function of thrust) formula I gave you gives you induced power...

Perhaps you could provide an example. For instance, the 1903 Wright flyer produced 90 lbf thrust with two 8.5 ft long propellers.

 

[David Lewis]

cjl wrote: "Yes, but without knowing the inflow velocity, you can't calculate any meaningful power number."

David Lewis wrote: Good catch. I forgot that kinetic energy is proportional to the square of velocity. It takes 3 times as much energy to increase the velocity of an air mass from 1.0 to 2.0 as it does from zero to 1.0. Yet delta v in both cases is the same. The power formula only applies to static thrust, if my reasoning is correct.

Acceleration seems to be greatest at the beginning, before rolling friction and air resistance come significantly into play, and initial acceleration is roughly equal to static thrust divided by gross takeoff mass. Acceleration gradually drops off to practically nothing before the airplane leaves the runway.

 

[cjl]

Acceleration gradually drops off to practically nothing before the airplane leaves the runway.

I wouldn't necessarily say that's accurate - at 150mph or so, modern jetliners still have a rather dramatic excess of thrust, and can still accelerate fairly hard. Once they leave the ground though, they will tend to throttle back a bit, and once they reach climb speed, they also climb at a gradient such that the airspeed remains constant (so all the excess thrust goes towards climbing, rather than increased speed).

 

[David Lewis]

I agree totally. My observations are not representative.

 

[insightful]

Perhaps you could provide an example. For instance, the 1903 Wright flyer produced 90 lbf thrust with two 8.5 ft long propellers.

The "induced power" formula actually is dimensionally correct. I don't know where the "2" factor comes from, so leaving that out and applying it to the Wright flyer data gives P = 2240 watts or about 3 hp. Their engine actually produced about 12 hp, so this formula (as pointed out) has little practical value.

 

[David Lewis]

This is my conception of acceleration versus time. Acceleration is zero until the brakes are released. Then acceleration quickly ramps up to the thrust divided by mass value and stays fairly constant until liftoff. After liftoff, the airplane noses up and the airplane sees gravity as an additional acceleration component.

The area under the curve (shaded yellow) is the speed at liftoff.

 

[cjl]

Why would acceleration increase after liftoff? It should drop, since the airplane is now climbing (thus some of the excess thrust goes into climb rate, rather than acceleration) and the drag rises dramatically after liftoff (especially induced drag).

 

[David Lewis]

I assume the maximum rate (or angle) of climb speed is higher than the lift off speed and, to reduce stress on the tires, the pilot postpones some acceleration until after liftoff. Additionally, when the airplane transitions from horizontal motion to an ascent, I counted the component of gravitational force parallel to the flight path the same as an acceleration to simplify performance calculations.

As a quick verification, when you are taking off in an airplane, the force with which the seat presses against your back is proportional to acceleration.

 

[cjl]

The pilot definitely will not postpone acceleration - that would increase runway length required. The pilot will simply rotate when the speed is appropriate, and the plane will take off. As for climb speed, yes it is higher than takeoff speed, but the acceleration will still slow at takeoff, since the thrust from the engines will be the same (or reduced) from the takeoff setting, and the drag will dramatically increase when the aircraft takes off.

 

[mfb]

I counted the component of gravitational force parallel to the flight path the same as an acceleration to simplify performance calculations.

Gravity acts downwards, the aircraft is flying upwards. Gravity is reducing the achievable acceleration.
Acceleration doesn't lead to much stress on the tires - they are passive anyway, unlike in cars where they transmit the accelerating force.

As a quick verification, when you are taking off in an airplane, the force with which the seat presses against your back is proportional to acceleration.

It is not, the orientation of the aircraft plays a role as well. Lift the nose and you get force between your back and the seat without any acceleration.

If aircraft would keep accelerating with the same magnitude after liftoff, they would reach their cruise speed within something like 2-3 minutes. That is not the case, they accelerate slowly while mainly gaining altitude.. The acceleration drops significantly after liftoff.

 

[cjl]

Gravity acts downwards, the aircraft is flying upwards. Gravity is reducing the achievable acceleration.
Acceleration doesn't lead to much stress on the tires - they are passive anyway, unlike in cars where they transmit the accelerating force.
It is not, the orientation of the aircraft plays a role as well. Lift the nose and you get force between your back and the seat without any acceleration.

If aircraft would keep accelerating with the same magnitude after liftoff, they would reach their cruise speed within something like 2-3 minutes. That is not the case, they accelerate slowly while mainly gaining altitude.. The acceleration drops significantly after liftoff.

To be fair, the acceleration is reduced because once best climb speed is reached, the aircraft is pitched up so all the excess thrust is used to climb, and the speed is held fairly constant for a good chunk of the climb (usually at something like 250-300mph - it's wherever the excess power is largest). If the aircraft were trying to accelerate rather than climb, it would reach cruise speed quite fast, but there are a lot of very good reasons why they do not accelerate to cruise speed at low altitude before climbing.

 

[mfb]

Sure, it could accelerate faster by staying lower. That would be very unpleasant for those living below the flight path, and it would increase drag way too fast in the dense atmosphere close to the ground. It wouldn't do that for long, however. Drag at sea level is more than twice the drag at typical cruise altitudes of larger airplanes.

 

[cjl]

Yes, but jet engine thrust at sea level is also more than twice the thrust at typical cruising altitude. The limit to safe jet aircraft speed at low altitude isn't whether it has the power to overcome the drag, it's the risk of structural damage (which, admittedly, is still drag-related).

 

[David Lewis]

mfb wrote: "It (force pressing against your back) is not (proportional to acceleration), the orientation of the aircraft plays a role as well. Lift the nose and you get force between your back and the seat without any acceleration."

David Lewis wrote: You're right. My graph attempts to depict what an accelerometer would show. During most of the climb regime, speed would be constant, and the accelerometer readout would display:
g * sin (angle of climb),
where g stands for acceleration of gravity.

 

[cjl]

Yep. That won't necessarily correlate well to engine excess thrust though, since the flight path angle won't be the same as the cabin deck angle.