Power required and TAS

[balderax]

Hi guys,

I read a Flying magazine article some months ago talking about the movie "The Martian" and if it was scientifically "accurate".

It got me thinking about power required and I realised I never really understood why power required increases with TAS (I completed ATP theory).

Could you explain me why more power is required to fly 150 KIAS at FL300 than at 5000 feet ? I know it's not an engine perf consideration, so is it a physical law ?

Many thanks in advance

[keith williams]

The short answer to your question is “ The TAS at 150 KIAS at FL300 is greater than the TAS at 150 KIAS at 5000 ft, so the power required at FL 300 is greater.

If that explanation is insufficient for your purposes a rather longer (but be no means the longest possible) answer is:


For an aircraft to fly at any given airspeed, work must be done to overcome the drag force.

That work is equal to the drag multiplied by the distance flown through the air.

Power is the rate of doing work.

So the power required to fly at any given airspeed is equal to the drag multiplied by the distance flown through the air divided by the time taken.

Distance flown divided by time taken = TAS

So Power Required = Drag x TAS


Drag = CD ˝ Rho V squared where V is the TAS.

So drag is proportional to TAS squared.

So drag multiplied by TAS is proportional to TAS cubed (which is TAS squared x TAS )

So Power Required is proportional to TAS cubed.


The Indicated Airspeed (IAS) at any given altitude is determined by the TAS and the air density. As altitude increase the air density decreases, so more TAS is required to produce any given IAS.

At mean sea level in ISA conditions (ignoring instrument errors and pito system errors) the TAS is equal to IAS.

So at ISA MSL 150 KIAS = 150 Kts TAS.

But at 5000 ft in ISA conditions 150 KIAS = approximately 161 Kts TAS

And at FL300 in ISA conditions 150 KIAS = approximately 242 Kts TAS.

Looking at these last two figures, 242 kts is approximately 1.5 times greater than 161, So the power required at 150 KIAS at FL300 will be approximately 1.5 cubed = 3.375 times the power required at 150 KIAS at 5000 feet.

[Capn Bloggs]

So the power required at 150 KIAS at FL300 will be approximately 1.5 cubed = 3.375 times the power required at 150 KIAS at 5000 feet.

Dodgy? How do you actually measure "power"?

Blogg's jet science: Drag requires Energy to be used = fuel burnt. Drag, roughly speaking, is mostly created by IAS.

At 250KIAS at Sea Level (TAS around 250), I use about 2100kg/per hour. N1 is way down, around 60%

At 250KIAS at FL300, I will still use around 2100kg/hr, but TAS is around 400. N1 is way up, around 85%.

If 3.375 times the power is required at FL300, there would be no point in going high at all. You only get a 80% increase in speed. Far cheaper (but longer) to fly low.

[keith williams]

Capn Bloggs

You may have overlooked it, but In my post I included the phrase (But by no means the longest possible explanation).

Yes of course we can go into the added complications of Varying specific fuel consumption and also varying propulsive efficiency. But Balderax's post ruled out engine performance and the tone of the question suggested (to me at least) that he/she was looking for the most basic analysis.

By all means give us an explanation which encompasses all of the possible factors, but I suspect that this is not what he/she was looking for. But every day should be a day in school, so please go for it.

[Capn Bloggs]

So the power required to fly at any given airspeed is equal to the drag multiplied by the distance flown through the air divided by the time taken.

Keith, you've confused air miles and ground miles. Power required should be measured in this case as numbers of molecules of air being flown-through, not nautical miles flown over the ground.

My understanding is that the V in the drag formula is velocity with respect to the air the aircraft is in, ie IAS, not TAS.

The power required at Seal levl and FL300 for the same IAS is roughly the same.

[balderax]

Thank you for your answers.

There is this quote from the article which says :

"It takes more fuel to indicate 300 kts at 10000 feet than at 5000. The air particles hit the aircraft with the same force, but in a given amount of time, more of them hit it."

That's the point that confuses me and which seems contrary to Capn Bloggs statement (I understand this is only a part of the problem, and that other variables may offset the above statement) :

At 250KIAS at Sea Level (TAS around 250), I use about 2100kg/per hour. N1 is way down, around 60%

At 250KIAS at FL300, I will still use around 2100kg/hr, but TAS is around 400. N1 is way up, around 85%.

Do you agree on the fact that IAS (then parasite drag) is determined by the force with wich the air particles strike the aircraft, and not the amount of air particles ? Instinctively, I would think that if more particles hit the Pitot tube, the pressure felt would increase (and IAS increase as a consequence).

You are right Keith, I want to make sure I understand the basics of power required before we talk about propulsive efficiency, SFC etc...

Thanks again

[keith williams]

The air particles hit the aircraft with the same force, but in a given amount of time, more of them hit it.

The above statement is not strictly correct.

It would be more accurate to say that at the higher altitude the individual particles hit the aircraft with greater force (because the TAS at any given IAS is greater), but because the air is less dense, less particles hit the aircraft.

The Airspeed Indicator (ASI) is simply a pressure instrument which measures the dynamic pressure, 1/2 Rho V squared, Where Rho is air density and V is the TAS. Every time the ASI senses a given value of dynamic pressure it will give the same indication (IAS).

So flying at different altitudes at constant IAS means that we are flying with the same dynamic pressure. But because the air becomes less dense as our altitude increases, we require a greater value of TAS to achieve the same dynamic pressure.

If for example the value of Rho at the higher altitude is 1/4 of the value at the lower altitude, then the value of V squared at the higher altitude must be four times that at the lower altitude. To increase V squared by a factor of four we must increase V by a factor of 2. So although our IAS remained constant, our TAS at the higher altitude must be twice that at the lower altitude.

Do you agree on the fact that IAS (then parasite drag) is determined by the force with wich the air particles strike the aircraft, and not the amount of air particles ? Instinctively, I would think that if more particles hit the Pitot tube, the pressure felt would increase (and IAS increase as a consequence).

The force exerted by the airflow is determined by the force exerted by each individual particles and the number of particles hitting the aircraft in a given time. As I have explained above, if the IAS is constant then the dynamic pressure must also be constant.

It takes more fuel to indicate 300 kts at 10000 feet than at 5000.

We need to be careful here. We are talking about 300 Kts IAS, so the dynamic pressure and drag will be the same at both altitudes, so the thrust required will also be the same at both altitudes. But the amount of fuel flow required to produce this thrust will also be influenced by the type of engine, the varying specific fuel consumption and the varying propulsive efficiency.

In piston engines fuel flow is proportional to power, but in jet engines fuel flow is proportional to thrust. So in a jet aircraft fuel flow at the two altitudes would be about constant (as observed by Capt Bloggs). But in a piston aircraft more fuel flow would be needed to provided the increased power required to maintain the higher TAS.


Capn Bloggs

Blogg's jet science: Drag requires Energy to be used = fuel burnt. Drag, roughly speaking, ismostly created by IAS.

That is correct, but we need to remember that not all of the fuel burned is converted into thrust and power.

At 250KIAS at Sea Level (TAS around 250), I use about 2100kg/per hour. N1 is way down, around 60%

At 250KIAS at FL300, I will still use around 2100kg/hr, but TAS is around 400. N1 is way up, around 85%.

As I have explained previously, constant IAS means constant dynamic pressure and this means constant drag. So yes your fuel flow may have remained approximately constant. But this does not mean constant power output. The constant drag required constant thrust and in jet engines fuel flow is proportional to thrust. But it is not proportional to power. As you climbed at constant 250 kts IAS your TAS increased while drag remained constant. Propulsive power required is drag times TAS, so as you climbed the power output must have increased. So although your fuel flow was constant so power output increased.

If 3.375 times the power is required at FL300, there would be no point in going high at all. You only get a 80% increase in speed. Far cheaper (but longer) to fly low.

Let’s test your argument over a distance of 2000 miles, using the figures you have given.

MSL 5000 nm / 250 kts = 20 hours. 2 hours x 2100 kg/hr = 42000 kgs

FL300 5000 nm / 400 kts = 12.5 hours. 12.5 hours x 2100 kg/hr = 26250 kgs.

So flying at MSL used an extra 15750 kgs of fuel and took and extra 7.5 hours.

My understanding is that the V in the drag formula is velocity with respect to the air the aircraft is in, ie IAS, not TAS.

I think that you need to read that again. The velocity with respect to the air is the TAS. The IAS is just an indication.

The power required at Seal levl and FL300 for the same IAS is roughly the same.

You are just plain wrong her. The points that you are missing are:

1. Fuel flow in jet engines is proportional to thrust and is not proportional to power.
2. As altitude increases at constant IAS, the drag is constant, so the thrust required
is constant. This should mean that fuel flow is also constant, but see other
factors listed below.
3. As altitude increases the specific fuel consumption decreases, thereby reducing
the fuel flow required to maintain constant thrust.
4. As altitude increases the propulsive efficiency increases thereby reducing the
amount of energy that is being wasted in the exhaust gasses.
5. As altitude increases at constant IAS the TAS increases thereby reducing the time
taken to cover a given distance.
6. As altitude increases items 4 and 5 above enable the engines to produce the higher
power that is required, without increasing the fuel flow.

[Capn Bloggs]

Balderax, what sort of aeroplane/engine are you talking about? What do you actually mean by "power"? Throttle position? N1 gauge indication? Fuel Flow indication? It seems to me that Keith and I are arguing about two completely different things and that needs to be resolved before I go on.

[balderax]

Balderax, what sort of aeroplane/engine are you talking about? What do you actually mean by "power"? Throttle position? N1 gauge indication? Fuel Flow indication? It seems to me that Keith and I are arguing about two completely different things and that needs to be resolved before I go on.

The root of my question is the following part of the article :

"One of the schemes for exploring more of the Martian surface is to use drones. They would have to be very light and have a large wingspan to fly in the wispy Martian atmosphere, which is the equivalent of Earth's somewhere above 100,000 feet, but it helps that things on Mars weigh only 3/8 of what they do on Earth. More of a challenge would be achieving an indicated speed of 20 knots in the first place; the TAS would have to be around 200 kts. The problem is that at a given IAS, the power required is proportional to the TAS; that's why it takes more fuel to indicate 150 kts at 10,000 feet than at 5,000. Air particles are hitting the airplane with the same force, but in a given period of time, more of them hit it."

For his example, he says that 20 kias would be enough to make the drone fly. This speed seems easy to achieve. Then he says the problem is the 200 ktas needed, due to the fact that the air is very thin. But hey, these 200 ktas are needed precisely BECAUSE the air is thin, and if the air is that thin, drag must also be very low. So, if drag is so low, why is TAS a problem in the first place ?

In his article he doesn't mention the kind of engine the drone uses. But I'm personally more concerned by jet engines.

[Capn Bloggs]

OK, thanks. I profess to not know much about "power" and props. But what I do know is that at any level, in a jet, the fuel flow for the same indicated speed will be basically the same, notwithstanding minor variations caused by compressibility and other slight changes in the various drag types. That is, the thrust ie the stuff going out the back to counter the drag, will be the same. I leave the explanation of why to the rocket scientists.

So I do not agree that it will take more fuel to fly (a jet) at 150IAS at 10,000ft than 5000ft. The "power" required might be double (or 3.375 times at FL300) but the thrust/FF certainly isn't.

I simply cannot understand why the drag at FL300/250KIAS/440TAS is allegedly over three times that at 250KIAS at sea level. Drag is all about the aerofoil and the body pushing through the local airflow. At 250KIAS, that force much must be the same. The nil-wind speed over the ground ie the TAS, in my view, has no bearing on anything to do with drag.

Going to extremes, how about the ISS in low earth orbit? Massive TAS, but no IAS, no lift and no drag. No thrust required to keep it up there (apart from a few squirts occasionally to resist G ravity). Is there any power required? Apparently yes...

[balderax]

Going to extremes, how about the ISS in low earth orbit? Massive TAS, but no IAS, no lift and no drag. No thrust required to keep it up there (apart from a few squirts occasionally to resist G ravity). Is there any power required? Apparently yes...

That's EXACTLY what I was thinking. Two (minor?) differences, though : the ISS doesn't produce any induced drag and doesn't experience any compressibility effect.

I couldn't find the scene in the Martian where the drone flies, but I think it has a piston engine. If yes, the article would then make sense, right ? I understand trying to fly with a piston engine at FL100 might be difficult.

[keith williams]

Whoever wrote the article is using some very loose language and in at least one place is simply wrong.

The problem is that at a given IAS, the power required is proportional to the TAS; that's why it takes more fuel to indicate 150 kts at 10,000 feet than at 5,000.

This statement is right about the power required, but may be right or wrong about the fuel flow, dpeening on enine type. As I have explained previously, both drag and IAS are related to dynamic pressure (1/2 Rho V squared). (Drag = CD1/2 Rho V squared S) and IAS (is derived from ˝ Rho V squared). Every time the ASI senses any given value of dynamic pressure it will produce the same indicated airspeed, regardless of changes in altitude. And provide the values of CD a S (the surface area) do not change, that same dynamic pressure will produce the same value of drag. So the drag at any given Indicated airspeed will not change with changes in altitude.

But As I explained in my first post, Power required is proportional to TAS Cubed. Fuel flow in jet engines is proportional to thrust, and to maintain a constant IAS the thrust must equal the drag. So as Capn Bloggs has said, the fuel flow required to maintain any given IAS will not change with altitude. But in piston engines the fuel flow is proportional to power, so in piston/prop aircraft the fuel flow required to maintain constant IAS will increase with increasing altitude.

Air particles are hitting the airplane with the same force, but in a given period of time, more of them hit it."

This statement can be interpreted in (at least) two ways. If it is taken to mean that the overall force exerted by all of the particles hitting the airplane is constant, then it is true. Because the TAS is higher each particle hits harder, but because of the lower density there are less hits per second. If however the statement is taken to mean that each individual particle exerts the same force, then this is untrue. If I throw a small stone at you at low speed, it will exert a low impact force. If I then throw it at you at a much higher speed it will exert a much greater impact force.

For his example, he says that 20 kias would be enough to make the drone fly. This speed seems easy to achieve. Then he says the problem is the 200 ktas needed, due to the fact that the air is very thin. But hey, these 200 ktas are needed precisely BECAUSE the air is thin, and if the air is that thin, drag must also be very low. So, if drag is so low, why is TAS a problem in the first place ?

Again the problem probably stems from poor wording. While it will not require very much thrust to maintain an IAS of 20 kts, we must first accelerate the drone up to that speed. If we assume that it requires 20 kts IAS to support its own weight, we may well have a problem ensuring that it can accelerate up to this speed with falling to the ground. IAS is just an indication but the real speed required is the 200 kts TAS. We could perhaps use some kind of gun, but this would exert huge acceleration forces, so the drone will need to be very strong to survive the process. This will obviously have weight implications, which would in turn require greater acceleration forces from the gun.

Capn Bloggs

So I do not agree that it will take more fuel to fly (a jet) at 150IAS at 10,000ft than 5000ft. The "power" required might be double (or 3.375 times at FL300) but the thrust/FF certainly isn't.

I suggest that you might wish to read my post again. I did not say that 3.375 times as much fuel would be required. What I actually said was:

Looking at these last two figures, 242 kts is approximately 1.5 times greater than 161, So the power required at 150 KIAS at FL300 will be approximately 1.5 cubed = 3.375 times the power required at 150 KIAS at 5000 feet.

As I have said several times previously in this thread, Fuel flow in a jet engine is proportional to thrust, while fuel flow piston engines is proportional to power.

You do not appear to understand the difference between thrust and power. Thrust is a force, power is a rate of doing work.

Although the throttle levers in many jest are often referred to as the “Power Levers” they do not actually control the power. They control the fuel flow, which in turn influences the thrust. But the propulsive power being produces varies with TAS and altitude. As an example a jet aircraft has its engine running at full power prior to brake release, no propulsive power is being produced. We have lots of thrust, but the TAS is zero. Propulsive power = thrust x TAS which in this case is Lots x zero = zero.

If we now climb to some selected altiude (10000 ft for example), then push the throttles fully open the IAS And TAS will both increase. The increasing TAS will cause the propulsive power (TAS X thrust) to increase. We now have power increasing while our throttle lever angles remain unchanged.

[balderax]

Thank you again to both of you for your sound answers. This subject is a lot clearer now.

I would like to move on to another topic : the propulsive efficiency of turbojet engines. Here is a quote from Wikipedia, which basically says the same as my Oxford ATP books :

For all airbreathing jet engines the propulsive efficiency (essentially energy efficiency) is highest when the engine emits an exhaust jet at a speed that is as close as possible to the vehicle velocity. The exact formula for air-breathing engines as given in the literature,[2][3] is
η p = 2 1 + c v {\displaystyle \eta _{p}={\frac {2}{1+{\frac {c}{v}}}}} where c is the exhaust speed, and v is the speed of the aircraft.

We can derive that the propulsive efficiency will be 1 if the speed of the accelerated air that exits the engine equals the speed of the aircraft, right ?

But how can a jet engine create thrust if it doesn't accelerate the air faster that the speed at which it is flying ? Intaking air at 450 kts and exhausting it at 450 kts won't create any action/reaction effect...

[keith williams]

The statement that propulsive efficiency will be 100% when the exhaust velocity is equal to the aircraft velocity is theoretically correct. But is not possible to achieve this in reality.

To understand what this is all about we need to get behind the maths and understand what is really going on.

Let’s suppose that we are in our jet on the runway with the brakes on and the engines running at max rpm. We are burning fuel to release chemical energy, which is then converted into heat and then into mechanical energy in the form of pressure within our jet pipes. This pressure exerts a rearward force on the exhaust gas, and the exhaust gas in turn exerts an equal and opposite reactionary force as predicted by Newton’s Third Law. This reaction force is our thrust.

But the rearward force which we exerted on the exhaust gas causes it to be accelerated rearwards, thereby giving the exhaust gas an enormous amount of kinetic energy. In effect we have taken all of the useful energy which we extracted from the fuel and thrown it away in the exhaust gas. None of this energy is being used to propel our aircraft forward, so our propulsive efficiency is zero.

The important thing to note in all of this is the fact that the thrust was not actually produced by the acceleration of the exhaust gas. It was produced by the rearward force which we exerted on the exhaust gas. The acceleration of the gas and the resulting loss of energy was just a side effect of the process. It happens because the air is not sufficiently stiff to resist the rearward force which we exerted upon it.

If you have a problem with the idea that the acceleration of the gas does not actually produce the thrust let’s look at a slightly different scenario. You are driving down a motorway at 70 mph. The thrust required to maintain this speed is produced by your wheels exerting a rearward force on the surface of the road. This causes the road to exert an equal and opposite reaction on your wheels, thereby pushing the car forward.

But because the road is stiff enough to resist the forces involved, it does not experience a rearward acceleration. The road surface meets your wheels at 70 mph and leaves your wheels at 70 mph. In this case all of the energy that your engine is providing to the wheels is being used to propel you forward, so your propulsive efficiency is 100%. But this is only possible because the material on which you are exerting your rearward force (the road) is stiff enough to prevent it form being accelerated rearwards.

Now it might be argued that the entire Earth is actually being accelerated backwards relative to your car, and because the mass of the Earth is enormous, the resulting acceleration is to small to detect. To test this argument let’s imagine that we have 2 identical cars standing back to back and tied together with a strong rope. The drivers start their engines and put the cars into first gear then slowly let out the clutches. The cars move forward until the rope becomes stretched and are then brought to a standstill. The cars are pointing in opposite directions so they cannot both be accelerating the road rearwards. But both cars are producing thrust and it is this which is exerting tension on the rope.

For another example we could look at an inflated baloon. The gas in the boloon exerts an outward force which is balanced by an equal and opposite inward force exerted by the baloon envelope. There is no acceleration but there are equal and opposite forces.

Now getting back to our aircraft sitting on the runway, if we release the brakes the aircraft will start to accelerate forward. As it accelerates forward its velocity becomes closer and closer to the exhaust speed. This means that the amount of acceleration that we giving to the exhaust gas is becoming less and less. This in turn means that the amount of kinetic energy that are we throwing away in the exhaust gas is becoming less and less, so more of the energy derived from the fuel is being used to propel us forward. So our propulsive efficiency is gradually increasing.

This process will continue after lift-off. If we climb at constant IAS the increasing TAS will produce an increasing propulsive efficiency. But we will never get to 100% because the exhaust gas is not sufficiently stiff to resist being accelerated rearwards.

[balderax]

Thank you Keith for your reply. It now makes sense.

[oggers]

Keith Williams:

So Power Required is proportional to TAS cubed.


The Indicated Airspeed (IAS) at any given altitude is determined by the TAS and the air density. As altitude increase the air density decreases, so more TAS is required to produce any given IAS.

At mean sea level in ISA conditions (ignoring instrument errors and pito system errors) the TAS is equal to IAS.

So at ISA MSL 150 KIAS = 150 Kts TAS.

But at 5000 ft in ISA conditions 150 KIAS = approximately 161 Kts TAS

And at FL300 in ISA conditions 150 KIAS = approximately 242 Kts TAS.

Looking at these last two figures, 242 kts is approximately 1.5 times greater than 161, So the power required at 150 KIAS at FL300 will be approximately 1.5 cubed = 3.375 times the power required at 150 KIAS at 5000 feet.

The power required is only proportional to TAS cubed if ρ is constant, which of course it won't be at two different altitudes. So at constant IAS the power at FL300 will be 1.5 times that at FL50, not 3.375 times. Bloggs was right to call that error out.

Also:

The important thing to note in all of this is the fact that the thrust was not actually produced by the acceleration of the exhaust gas. It was produced by the rearward force which we exerted on the exhaust gas. The acceleration of the gas and the resulting loss of energy was just a side effect of the process. It happens because the air is not sufficiently stiff to resist the rearward force which we exerted upon it.

It is the mass of the jetwash/propwash/propellant that is the measure of its inertia, not “stiffness”.

You are driving down a motorway at 70 mph. The thrust required to maintain this speed is produced by your wheels exerting a rearward force on the surface of the road...because the road is stiff enough to resist the forces involved, it does not experience a rearward acceleration. The road surface meets your wheels at 70 mph and leaves your wheels at 70 mph. In this case all of the energy that your engine is providing to the wheels is being used to propel you forward, so your propulsive efficiency is 100%. But this is only possible because the material on which you are exerting your rearward force (the road) is stiff enough to prevent it form being accelerated rearwards.

In that case your propulsive efficiency is 100% because the chosen frame of reference is the same thing you are pushing against. Replace the car and Earth analogy with a hamster wheel and you find the wheel is spinning whilst the hamster is stationary. The hamster wheel is 'stiff' and yet it spins noticeably because it has a relatively small mass compared to the hamster. It is not about stiffness (except insofar as pushing against something stiff may transfer the force to something more massive).

Now it might be argued that the entire Earth is actually being accelerated backwards relative to your car, and because the mass of the Earth is enormous, the resulting acceleration is to small to detect.

Well, it would be very strange to argue that the Earth was accelerating relative to a car that was 'maintaining 70mph' relative to the Earth. But it would not be the least bit strange to point out that the force between tyre and Earth is a de facto torque on the Earth causing a – albeit infinitesimal – change of angular velocity. According to NASA:

"Any worldly event that involves the movement of mass affects the Earth's rotation, from seasonal weather down to driving a car"

To test this argument let’s imagine that we have 2 identical cars standing back to back and tied together with a strong rope. The drivers start their engines and put the cars into first gear then slowly let out the clutches. The cars move forward until the rope becomes stretched and are then brought to a standstill. The cars are pointing in opposite directions so they cannot both be accelerating the road rearwards. But both cars are producing thrust and it is this which is exerting tension on the rope.

That would not test the argument anyway. If the cars are pointing in opposite directions the net force is zero so there is no torque on the Earth only compression of the surface and tension in the rope.

[keith williams]

Oggers

If I may take your comments in turn:

The power required is only proportional to TAS cubed if ρ is constant, which of course it won't be at two different altitudes. So at constant IAS the power at FL300 will be 1.5 times that at FL50, not 3.375 times.

You are correct. I made the error of overlooking the fact that in the special case of a change in altitude at constant IAS, the reducing density and increasing TAS squared cancel each other out, so the increase in power required is proportional to the increase in TAS. So it would be more accurate for me to have said:

1. At any given altitude and IAS the power required is proportional to TAS cubed.
2. For any change in altitude at constant IAS the change in power required is
proportional to the change in TAS.

So in my example calculation the power required at FL300 should have been 1.5 times that at FL50.

Bloggs was right to call that error out.

Bloggs did not actually call me out for that error. What he/she actually said was:

Keith, you've confused air miles and ground miles. Power required should be measured in this case as numbers of molecules of air being flown-through, not nautical miles flown over the ground.

My understanding is that the V in the drag formula is velocity with respect to the air the aircraft is in, ie IAS, not TAS.

The power required at Seal levl and FL300 for the same IAS is roughly the same.

All of which is incorrect.

It is the mass of the jetwash/propwash/propellant that is the measure of its inertia, not “stiffness”.

You are correct in saying that the mass is the measure of its inertia. We only need to look at Newton’s Second Law equation F = MA to see that. But Newton assumed that the bodies in question were free to move. And if for example, one of the bodies were moving around a pivot such as your hamster wheel, then the stiffness of the bearing would also affect the outcome.

Let’s imagine that we took your hamster wheel and added an adjustable friction brake to the pivot. With the brake fully off the hamster would accelerate the wheel up to some given speed and then maintain this speed for as long as it kept running. But if we gradually tight the friction brake while the hamster continued to run, the wheel would gradually decrease. At some point the friction would be sufficient to stop the wheel, and if the hamster kept running it would follow a vertical looped path inside the wheel. So although we did not ever change the mass of the hamster or the mass of the wheel we have completely changed the outcomes. In order to determine which situation (wheel spinning / hamster stationary or Wheel stationary / Hamster spinning) gave the better propulsion efficiency, we would need to know what the hamster was trying to achieve. And as neither of us speak hamster we can never know that.


The problem with propellers and/or jet engine sin aircraft is that in order to generate thrust they must exert a rearward force on some object or material which is not itself part of the aircraft or propulsion system. The only obvious candidate is the air. But because the air is a gas, the exertion of any rearward force upon it will cause it to flow rearwards. This results in the propulsion system transferring kinetic energy to the air. Any energy which is transferred to the air cannot be used to do any useful work to propel the aircraft forward, so it represents a loss of energy. But if we were able to create a propulsion system which could exert the required rearward force on something which would not flow rearwards, we would have a more efficient propulsion system. I do not claim to be able to identify such a system, but that does not mean that it is impossible to do so. During my lifetime a great many products have been created which I could not possible have imagined beforehand.

In that case your propulsive efficiency is 100% because the chosen frame of reference is the same thing you are pushing against.

We can of course select any frame of reference we may wish, and the results will often appear to be very different. But in this case we are considering the motion of the car over the surface of the Earth, the Earth is the most appropriate reference frame.

Replace the car and Earth analogy with a hamster wheel and you find the wheel is spinning whilst the hamster is stationary. The hamster wheel is 'stiff' and yet it spins noticeably because it has a relatively small mass compared to the hamster. It is not about stiffness (except insofar as pushing against something stiff may transfer the force to something more massive).

Perhaps stiffness is too loose a term. The hamster wheel is certainly stiff in that it will resist deformation. But it is not stiff in terms of rotation. As I have said above if we increase the friction on the wheel pivot, the resulting motion will become very different. So the motions of the bodies concerned ( in this case hamster and wheel) are not determined entirely by their masses alone.

Well, it would be very strange to argue that the Earth was accelerating relative to a car that was 'maintaining 70mph' relative to the Earth.

I did not suggest that it was a good argument or even that it was a valid argument. I simply stated that some people might choose make it.

But it would not be the least bit strange to point out that the force between tyre and Earth is a de facto torque on the Earth causing a – albeit infinitesimal – change of angular velocity. According to NASA:

"Any worldly event that involves the movement of mass affects the Earth's rotation, from seasonal weather down to driving a car"

I really hope that we do not need NASA to tell us that. But that does not mean that every car moving over the surface of the Earth causes the Earth to accelerate. In many cases (including the example I gave above) there will be similar cars exerting similar forces in opposite directions. Where pairs of these individual forces cancel each other out there will be no acceleration exerted on the Earth but the cars will still be producing thrust.

That would not test the argument anyway. If the cars are pointing in opposite directions the net force is zero so there is no torque on the Earth only compression of the surface and tension in the rope.

But the driver of each car is not interested in the net force and the propulsive efficiency of the car does not depend upon it. He/she simply wants to move forward. The tension in the rope is entirely due to the equal and opposite thrust forces that are being produced by each car. The fact that the cars are unable to move does not mean that there is no thrust. Simply cut the rope and we will see both cars accelerate away. But there will be no acceleration or even net torque applied to the Earth.

The point which I am arguing is that the thrust in our car or aircraft or whatever, is not generated by the rearward acceleration of whatever we may be pushing against. It is simply the equal and opposite reaction predicted by Newton’s third Law. The rearward acceleration of the air caused by our propeller or jet engine is a wasteful by-product of this process. The process of generating thrust would be far more efficient if we could find a means of eliminating this waste product.

If I stand on the ground my feet exert a downward force equal to my weight. If the ground is stiff enough it exerts an equal and opposite upward force which prevents me from sinking into the ground. But this process does not cause the Earth to accelerate downwards.

[scifi]

Can we leave the motor car experiments, for a while, and get back to the original question....
Could you explain me why more power is required to fly 150 KIAS at FL300 than at 5000 feet ?


The real solution is that, to generate Lift equal to the Weight of the aircraft, at a higher altitude. The wing needs a higher Angle of Attack, as the air is thinner. This higher AoA creates a higher Drag... Simple as that.


Now what was all that about cars pushing the earth around....?

[keith williams]

Can we leave the motor car experiments, for a while, and get back to the original question....

Could you explain me why more power is required to fly 150 KIAS at FL300 than at 5000 feet ?

The real solution is that, to generate Lift equal to the Weight of the aircraft, at a higher altitude. The wing needs a higher Angle of Attack, as the air is thinner. This higher AoA creates a higher Drag... Simple as that.

Scifi your explanation is wonderfully simple, but just plain wrong.

The term “150 KIAS” means 150 knots Indicated Airspeed. The ASI produces its indication (The Indicated Airspeed) by measuring the dynamic pressure. The ASI has no means of knowing what the altitude is, so every time it senses a particular value of dynamic pressure it will give out the same Indicated Airspeed. This means that if we fly at the same IAS at any number of different altitudes, we will always be flying at a constant dynamic pressure.

Both lift and drag are proportional to dynamic pressure so the constant dynamic pressure at constant IAS produces constant values of lift and drag, without any requirement to change the angle of attack. But as our altitude increases, the TAS at which we must fly to achieve our constant IAS must increase. It is this increasing TAS which causes the increase in power required.

All of this is explained in my initial post in this thread. The only error in that post is the final number 3.375

[oggers]

Keith Williams, what I am taking issue with is this:

because the road is stiff enough to resist the forces involved, it does not experience a rearward acceleration....Now getting back to our aircraft ...we will never get to 100% because the exhaust gas is not sufficiently stiff to resist being accelerated rearwards.

The stiffness of the road is irrelevant except insofar as it transfers force to something with more inertia. Stiffness is for springs and structures. In terms of propulsive efficiency it is nothing more than a distraction. The important metric is mass - it is one of the 3 basic quantities of mechanics. Stiffness isn't. A kilo of spuds accelerated by a metre/sec˛ will give you the same reaction as a kilo of air.

To test this argument let’s imagine that we have 2 identical cars standing back to back and tied together with a strong rope. The cars move forward until the rope becomes stretched and are then brought to a standstill....If I stand on the ground my feet exert a downward force equal to my weight. If the ground is stiff enough it exerts an equal and opposite upward force which prevents me from sinking into the ground. But this process does not cause the Earth to accelerate downwards.

Both examples of static equilibrium - no net external force on either car, body or ground and no motion. Propulsion is different, there is a net external force causing motion of one body and therefore by conservation law there has to be an equal and opposite change of momentum on another. Stiffness cannot replace the momentum change. Something with mass has to be accelerated.

As I have said above if we increase the friction on the wheel pivot, the resulting motion will become very different. So the motions of the bodies concerned ( in this case hamster and wheel) are not determined entirely by their masses alone.

Well just assume the normal fricition for a hamster wheel spindle. The wheel spins. The resistance is mainly due to friction in the spindle and to a lesser extent on the very small inertia of the wheel. None of the resistance is "because the material on which you are exerting your rearward force is stiff" like the ground under a car.

The point which I am arguing is that the thrust in our car or aircraft or whatever, is not generated by the rearward acceleration of whatever we may be pushing against.

According to Aerodynamics for Naval Aviators: “the force of thrust results from the acceleration provided to the mass of working fluid”.