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 ½ 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... :confused:

[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 ½ 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 (https://en.wikipedia.org/wiki/Airbreathing_jet_engine) the propulsive efficiency (essentially energy efficiency (https://en.wikipedia.org/wiki/Energy_conservation)) 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] (https://en.wikipedia.org/wiki/Propulsive_efficiency#cite_note-2)[3] (https://en.wikipedia.org/wiki/Propulsive_efficiency#cite_note-spt-3) is
η p = 2 1 + c v {\displaystyle \eta _{p}={\frac {2}{1+{\frac {c}{v}}}}} https://wikimedia.org/api/rest_v1/media/math/render/svg/fdc028d9a1268aa45e6b66b31d8ad6e9dc5d7aca 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 (https://www.nasa.gov/home/hqnews/2005/jan/HQ_05011_earthquake.html):
"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/sec² 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”.

[keith williams]

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.


You appear to be arguing that if a body is in a state of equilibrium there can be no propulsive force acting upon it. Might I suggest that you look at an aircraft in straight and level flight at constant speed.

Lift = Weight That is why the flight is level.
Forces to the left = force to the right That is why it flies straight.
Thrust = drag. That is why the speed is constant.

The aircraft is in equilibrium but it has a propulsive force (thrust) to maintain this condition.


Stiffness cannot replace the momentum change. Something with mass has to be accelerated.

I have given you several examples where a Newton 3 force is being generated without anything being accelerated. Some of these examples were static, but at least one was not. Two identical cars driving at the same speed, in opposite directions on the same stretch of road. Because they are exerting equal force on the road, but in opposite directions they cannot both be accelerating the road backwards, in fact neither of them is doing so. But both cars are generating a propulsive force. This is only possible because the road is stiff enough to resist the forces being applied to it. In this example the mass of the road is irrelevant.
From these examples you should by now have recognised the fact that a propulsive force can sometimes be generated without accelerating anything.

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

Just how much thought do you think the author gave to the question of whether the acceleration was the cause of the thrust or a consequence of it?

[Tinstaafl]

Oh, bad luck, Keith! I hear that woman is so good even presidents favour her.

[keith williams]

Ah that would be the fourth law of motion (where there is a Bill there is a way.)

[oggers]

Keith Williams,

You appear to be arguing that if a body is in a state of equilibrium there can be no propulsive force acting upon it. Might I suggest that you look at an aircraft in straight and level flight at constant speed.

An aircraft in unaccelerated flight is not an example of static equilibrium.. The forces on the airframe may be in equilibrium but the thrust results from a momentum change of the jetwash and the drag results from a momentum change of the slipstream. Your example of standing on the ground is static equilibrium – no net force on either body or ground, nothing is moving and there is no momentum change.

I have given you several examples where a Newton 3 force is being generated without anything being accelerated. Some of these examples were static, but at least one was not. Two identical cars driving at the same speed, in opposite directions on the same stretch of road. Because they are exerting equal force on the road, but in opposite directions they cannot both be accelerating the road backwards, in fact neither of them is doing so. But both cars are generating a propulsive force. This is only possible because the road is stiff enough to resist the forces being applied to it. In this example the mass of the road is irrelevant.
From these examples you should by now have recognised the fact that a propulsive force can sometimes be generated without accelerating anything.
You can have a force and counter force without any change of momentum. That would be static equilibrium but it is not propulsion. There is no such thing as a propulsive force that does not involve a change of momentum. It is not always immediately apparent where the change of momentum is taking place but it is always there.

The conservation laws are the most fundamental principle in mechanics and there is no example of them ever having been broken. Your own pet hypotheses are not consistent with them and are therefore incorrect.

[keith williams]

You can have a force and counter force without any change of momentum. That would be static equilibrium but it is not propulsion. There is no such thing as a propulsive force that does not involve a change of momentum.

The conservation laws are the most fundamental principle in mechanics and there is no example of them ever having been broken. Your own pet hypotheses are not consistent with them and are therefore incorrect.


Let’s go back to the situation of the two cars.

Phase 1
They are tied together by a rope, standing back to back with their engines running, gears engaged and clutch pedal released. Their wheels are exerting a rearward force on the road, which in turn is exerting a forward force on the cars. These forward forces are balanced by equal and opposite forces exerted by the tension in the rope. Both cars and the road are stationary, so the total momentum of the system (2 cars plus road) is zero.

Phase 2
We now cut the rope and the cars accelerate away in opposite directions. The rearward forces exerted on the road by the two cars are equal and opposite so the road is not being accelerated and is not moving, so its momentum is zero. Although the velocity of the two cars is increasing, they are moving in opposite directions at the same speed, so their total momentum is zero. So the total momentum of the system is still zero. The cars are both accelerating so they must be experiencing propulsive forces, but the total momentum is not changing.

Phase 3
After a few seconds the increasing aerodynamic drag and rolling resistance acting on each car will equal the propulsive force being exerted by the road, so their speed will become constant. As before, the road is not moving or accelerating and not moving so its total momentum is still zero. The two cars are not accelerating, but are moving in opposite directions at a common speed, so their total momentum zero. But the fact that the cars are maintaining their speed while pushing against drag and rolling resistance means that they must be experiencing propulsive forces.

The total momentum of the system has remained constant throughout the entire sequence, so the law of conservation of momentum has not been violated.

The only objects which experience accelerations are the cars (during the second phase), But nothing at all is accelerating during the third phase. But in the second and third phases both cars are experiencing propulsive forces.

It is entirely possible that I have overlooked something in this analysis, and if you can find it and point it out to me I would be grateful


The fundamental argument in this thread is whether the thrust for an aircraft is produced by the rearward force we exert on the air passing through our propulsion systems, or by the rearward acceleration of that air.

If we were asked how our propulsion systems create thrust we could say:

1. By exerting a rearward force on the air passing through them, thereby causing this air to exert a forward acting force in accordance with Newton’s Third law.

Or:

2. We accelerate air rearwards, thereby causing this air to exert a forward acting force in accordance with Newton’s Third law.

Both answers sound reasonable, so we need to ask a second question to decide which is the more valid.

One such second question might be “How do we accelerate the air rearwards?” The answer to this question is “By exerting a rearward force upon it”, which is of course our first answer. We can say that we accelerate the air rearwards by exerting a rearward force on it, but we cannot say that we exert a rearward force on the air by accelerating it rearwards.

The rearward acceleration of air in aircraft propulsion systems is an inevitable, but wasteful consequence of the process by which we generate thrust. Over the years increasingly high by-pass ratios and ever larger fans have been introduced in jet engines in order to reduce the acceleration of the air passing through them. If the rearward acceleration were the hero in all of this we would not by trying to reduce it.


Your own pet hypotheses are not consistent with them and are therefore incorrect.

I do not have any pet hypotheses and I really don’t think that the use of such pejorative terms adds anything useful to the debate.

[oggers]

The only objects which experience accelerations are the cars (during the second phase), But nothing at all is accelerating during the third phase. But in the second and third phases both cars are experiencing propulsive forces.

It is entirely possible that I have overlooked something in this analysis, and if you can find it and point it out to me I would be grateful

What you have overlooked is the momentum change of the slipstream of each car (ie the source of the drag). The one is opposed to the other, hence you have two masses (or m-dot in this case) experiencing opposite momentum change and overall conservation of momentum of the entire 'system'. Whereas, with a single vehicle there is momentum change of the single slipstream and opposing momentum change of the Earth resulting from the net torque on the road surface. Where there is propulsive force there are equal and opposite momentum changes and conservation of momentum.

F = m.a = mΔv/t = rate of change of momentum

[keith williams]

No, I do not think that is a valid explanation.

All you have really done is to select a smaller system ( 1 car, the air around it and the road/earth beneath it). But you have ignored the rest of the wider system.

In reality there will be no net torque on the Earth because the torque applied by car 1 is equal and opposite to that applied by car 2. One of the conditions required when applying the law of conservation of momentum is that the system must be an isolated system. In your explanation you have ignored the fact that your smaller system is connected to the other car via the road. And these exert a force on your smaller system.

Perhaps a more valid explanation of the conundrum is that if we consider the whole system (2 cars, the road and the air around them), no overall propulsive force will be produced, because the propulsive force exerted on car 1 will be equal and opposite to that on car 2. This does not mean that the individual cars are not generating propulsive forces, they are doing so. But (and this is a very important but) they are not accelerating anything rearwards. And because they are not accelerating anything rearwards, they are not throwing away energy, so their propulsive efficiency is much higher than that for an aircraft in flight with a conventional propulsion system (prop or jet engine).

[oggers]

Keith Williams,

In reality there will be no net torque on the Earth because the torque applied by car 1 is equal and opposite to that applied by car 2
Well, it was I who pointed out in my first reply to your 'cars pointing in opposite direction' scenario that there was no net torque - and therefore it would not, as you put it, test the argument.

nothing at all is accelerating during the third phase...But the fact that the cars are maintaining their speed while pushing against drag and rolling resistance means that they must be experiencing propulsive forces...It is entirely possible that I have overlooked something in this analysis, and if you can find it and point it out to me I would be grateful

And you did overlook the acceleration of the air in the slipstreams that results in drag and is where most of the engine power is going. But the point you are avoiding is this; if there doesn't just happen to be a second car producing precisely equal and opposite torque to the first car there is a net torque on the Earth. Hence this:

You are driving down a motorway at 70 mph...because the road is stiff enough to resist the forces involved, it does not experience a rearward acceleration.

...is wrong.

I am not being obtuse here, I see the argument you are trying to make but it is a fallacy. Stiffness cannot resist the change in angular momentum, there has to be an equal and opposite change of momentum. You keep inventing scenarios with multiple cars and ropes where the net torque on the Earth is zero but the fact remains that in your original example of a single car there is a net torque, and the only thing that resists is the Earth's huge moment of inertia. The equation is simply:

τ = Ι α .....or the linear equivalent, F = m a

If you have an equation for 'resistance to acceleration' as a function of “stiffness” please go ahead and share it.

[keith williams]

But the point you are avoiding is this; if there doesn't just happen to be a second car producing precisely equal and opposite torque to the first car there is a net torque on the Earth.


And the point which you are avoiding is the fact that in the real world there just happens to be millions of cars moving over the surface of the Earth at any one point in time. It is true to say that each of these cars exerts a torque on the Earth, but it is not true to say that each of these torques produces an acceleration. There is only one Earth and it is sufficiently stiff to resist these forces, so it cannot be experiencing millions of individual accelerations simultaneously. If the Earth experiences any acceleration at all it will be the result of the vector sum of all of these individual torques. And it is highly unlikely that this net torque will produce an acceleration that happens to be proportional to the torque exerted by any individual car.

The millions of individual cars are separate bodies, so we must consider their propulsive forces individually. So in reality we have millions of individual cars generating millions of individual propulsive forces, with only a single net torque and acceleration of the Earth. And that net torque and acceleration are not what the laws of motion would predict for any individual car.

You have repeatedly argued that the stiffness of the material is irrelevant. Well if the Earth were insufficiently stiff to resist the millions of competing torques being applied by the millions of cars, the surface would be a swirling mass of debris moving in all directions. It is not. Why Not? Because it is stiff enough to resist these forces.

You have repeatedly stated various equations of motion, in what appears to be an attempt to convince me that they are true. You are wasting your time because I already know that they are true. I studied them in secondary school and again in various career courses, and spent several years teaching them to university students. But what you are unable or unwilling to do is to look beyond these equations and see what actually happens in the real world.

To see what I mean let’s go back to your hamster wheel. This time we will insert two identical hamsters and have them run in opposite directions. The torque reaction exerted on the wheel by hamster A is equal and opposite to that exerted by hamster B. This means that the two torques cancel out to zero and the wheel remains stationary. Instead, the two hamster run in vertical loops, one clockwise and the other anticlockwise, within the cage. Each of the hamsters is generating a propulsive force but there is no acceleration of the cage. This does not mean that the laws of motion are incorrect. It simply means that by putting two sets of the equations back-to-back, the predicted accelerations cancel to zero.


You keep inventing scenarios with multiple cars and ropes where the net torque on the Earth is zero but the fact remains that in your original example of a single car there is a net torque, and the only thing that resists is the Earth's huge moment of inertia. The equation is simply:




τ = Ι α .....or the linear equivalent, F = m a

As I have said previously:

1. In reality there are millions of cars moving over the Earth, so the actual results are
not what is predicted for a single car.
2. I already know and understand the equations, so you will achieve nothing by
repeatedly stating them

If you have an equation for 'resistance to acceleration' as a function of “stiffness” please go ahead and share it.


I have never argued that there is a direct relationship between the stiffness of a material and the acceleration produced. What I have done is to cite several examples of situations in which the stiffness of the intervening material enables the forces exerted by one body to be felt by another, in a way which alters the overall outcome. But to understand this you really must look beyond the equations and see what actually happens in the real world.

Life is not an equation, it is much messier than that.

[oggers]

Keith Williams

I have never argued that there is a direct relationship between the stiffness of a material and the acceleration produced.


This is what you argued in your own words:

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

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...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

Now getting back to our aircraft sitting on the runway, if we release the brakes the aircraft will start to accelerate forward. So our propulsive efficiency is gradually increasing. This process will continue after lift-off. But we will never get to 100% because the exhaust gas is not sufficiently stiff to resist being accelerated rearwards.

Perhaps a more valid explanation of the conundrum...

There is no conundrum. F = m.a covers it succinctly. You have confused yourself with this stiffness idea. Stiffness is irrelevant, it is for structures and springs. Mass is the measure of inertia.

Life is not an equation, it is much messier than that.

As anticipated, you don't have an equation for your "road is stiff enough to prevent it from being accelerated" concept because it is not part of any scientific theory. The job of scientists and engineers is not to marvel at the messiness of life but to unravel its mysteries and reduce them to their component parts.

[keith williams]

The words which you have quoted do not argue that there is a direct relationship between stiffness and acceleration.

Our aircraft propulsion systems exert rearward forces on the air in their immediate vicinity, but not in the wider atmosphere. If we were to gradually increase the stiffness of the air, this would cause a gradual increase in the volume and mass of the air affected. This in turn would reduce the rearward acceleration, but would not reduce the propulsive force. If we continued to increase the stiffness we would eventually reach a point at which the entire atmosphere would be affected. The rearward acceleration of the air would be vanishing small, but again the propulsive force would not have changed. If we now make the air even stiffer, the entire atmosphere and Earth would be affected. Once again the acceleration would be smaller but the propulsive force would be unchanged. If we now increase the stiffness even more, there would be no change in acceleration or propulsive force. So although we can influence the acceleration by changing the stiffness of the air, there is not a direct quantifiable ratio between stiffness and acceleration.

In an earlier post you said:

There is no such thing as a propulsive force that does not involve a change of momentum.

But my examples of the two cars and the two hamsters illustrate the fact that it is entirely possible to create a propulsive force without any change in momentum.

The job of scientists and engineers is not to marvel at the messiness of life but to unravel its mysteries and reduce them to their component parts.


You are correct, but it also not the job of scientists and engineers to carry out experiments then simply dismiss unwanted results with comments such as “If there doesn't just happen to be a second car”, which you used earlier.

The job of scientists is to compose hypotheses, carry out experiments to prove or disprove them, analyse the results of these experiments then amend their hypotheses accordingly. Your approach in this thread has been to ignore results which you do not like and then simply restate the laws of motion.

The job of engineers is to look at how the world really works then use these observations to devise solutions to problems. An example of this is contra-rotating propellers as used in the Fairey Gannet aircraft. The Gannet was powered by a Twin Mamba turbine engine, which produced a great deal of power. If this power were to be fed into a single propeller the torque reaction would cause the fuselage to rotate very rapidly in the opposite direction to the propeller. This would make the aircraft uncontrollable and would probably tear it apart.

To prevent this the designers used two propellers spinning in opposite directions, but on the same axis of rotation. The front propeller was driven by a drive shaft which passed through the centre of rear propeller drive shaft. In this way the torque reactions of the two propeller were set against each other within the gearbox, causing them to be cancel each other out, so no torque was applied to the fuselage.

For all of this to be possible it was necessary for the gearbox structure to be sufficiently stiff to resist the two opposing torque reactions without being destroyed. Had the designers simply looked at the equations of motion and not beyond them, they would simply have said “we’re stuffed”, and gone home.

[oggers]

Keith Williams,

Hmm. Having started from the position:

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 happens because the air is not sufficiently stiff...
...the road is stiff enough to prevent it form being accelerated [whilst] the exhaust gas is not sufficiently stiff to resist being accelerated

of late you have said:

I have never argued that there is a direct relationship between the stiffness of a material and the acceleration produced


And now:

[blah blah Fairey Gannet blah]... For all of this to be possible it was necessary for the gearbox structure to be sufficiently stiff to resist the two opposing torque reactions without being destroyed. Had the designers simply looked at the equations of motion and not beyond them, they would simply have said “we’re stuffed”, and gone home.

Now I said many posts back; "stiffness of the road is irrelevant except insofar as it transfers force to something with more inertia. Stiffness is for structures. In terms of propulsive efficiency it is nothing more than a distraction." Having spent literally thousands of words arguing you are clearly coming round to what I managed to say with about 30 words in my second post.

[keith williams]

Oggers.

In Post 16 you said:

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


In post 17 I replied saying:


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.

In post 20 you then continued to attempt to convince me of something to which I had already agreed by saying:

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/sec² will give you the same reaction as a kilo of air.

This response typifies the approach which you have taken throughout this thread. When faced with comments with which you disagree you ignore them and simply restate the laws of motion.

Your characterization of my comments as [blah blah Fairey Gannet blah] clearly illustrates the fact that you have no intention whatsoever of considering any opinion other than your own. This makes further discussion pointless, so I will make no further comment.

[oggers]

Keith Williams

The key point which you have never conceded is this is wrong:

we will never get to 100% because the exhaust gas is not sufficiently stiff to resist being accelerated rearwards.

F = m.a. For any given thrust, the acceleration of the exhaust gas varies directly as the inverse of its mass. Stiffness does not come into it.

In post 17 I replied saying:


Quote:
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. In post 20 you then continued to attempt to convince me of something to which I had already agreed


...”agreed” except the very next word was "But...":

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 point of the hamster wheel analogy is that you can see it spin because the moment of inertia is very small compared to the Earth. Your response is proof (in the colloquial sense) that you didn't take the point, even though you now claim to have agreed – in which case the thousands of words you have written seem unnecessary.

This response typifies the approach which you have taken throughout this thread. When faced with comments with which you disagree you ignore them and simply restate the laws of motion.


I disagreed and made counter points. But it's true there was quite a lot to ignore too, some rambling strawmen plus a few errors I didn't bother with. If there's something in particular you want me to answer feel free. I have some more points I want to make anyway, your withdrawal won't change that. After all, as you wrote: "every day should be a day in school, so please go for it". :ok:

[oggers]

Keith Williams,

Your characterization of my comments as [blah blah Fairey Gannet blah] clearly illustrates the fact that you have no intention whatsoever of considering any opinion other than your own. This makes further discussion pointless, so I will make no further comment.

Subjective. Maybe you could explain what happened to the flippant list of silly quotes from post 21, that only disappeared after you decided to get all sniffy about tone?