Endurance not at Best L/D [Archive]

Intruder

26th Mar 2005, 20:24

Since you already understand the drag curve, consider next that in unaccelerated flight lift = weight and thrust = drag. So, at a given weight, lift is constant in level flight, and thrust required is directly proportional to drag.

So, with constant lift, L/D max is where the D is smallest -- that same point on the bottom of the drag vs airspeed curve. Minimum thrust is required at that point, therefore minimum fuel flow and maximum endurance.

What you apparently misunderstand is that max endurance IS at L/D max. However, max range is at an airspeed and power setting above the max endurance airspeed. Max range is a function of power required, not thrust.

redblue

27th Mar 2005, 00:22

I am refering to single engine propeller aircrafts.
Intruder. I think you were meant "What you apparently misunderstand is that max range IS at L/D max" instead of "What you apparently misunderstand is that max endurance IS at L/D max".

The problem is that the Flight Test Notes p.78 says Best Glide is at the best L/D ratio (lowest point on the drag vs speed curve). And then max endurance is at a lower airspeed and higher drag point along the same drag vs speed curve. This I find doesn't quite make much sense.

Old Smokey

27th Mar 2005, 00:28

redblue,

I note from your profile that you fly C152 aircraft, and that Intruder, who just responded flys B747 aircraft. We still need to know whether your enquiry relates to propeller or Jet aircraft to respond appropriately.

What Intruder has said in the second sentence is absolutely correct for a jet aircraft, but certainly not for a propeller aircraft.

Both Max Endurance and Max Range are highly dependant upon Fuel Flow. There is a direct relationship between Fuel Flow and Power for a propeller aircraft, and between Fuel Flow and Thrust for a jet aircraft. Consider the following -

Power = Force X Velocity, and putting it into aircraft terms -

Power = Thrust X TAS

So, for the prop aircraft, Fuel Flow is on one side of the equation, and for the jet, on the other.

For the jet aircraft, thrust (and therefore Fuel Flow) is ALMOST (but not quite) constant over the entire speed range. Therefore, without even knowing the engine installed, I can extract the best holding speed from the conventional drag curve as the speed for best L/D ratio, i.e. VMD, the bottom of the Drag curve, and Intruder is absolutely correct.

For the propeller aircraft, thrust, for a given power setting declines constantly as speed increases, and the optimum holding speed can NOT be directly found from the conventional Drag curves. Instead, you need to refer to a POWER Required curve, a different beast to the Drag (Thrust required) curves. From this you can extract the speed for Minimum Power, which will be the speed for minimum Fuel Flow, that is, best holding sped. This speed will invariably be BELOW the optimum L/D speed. Just how far below will depend upon the characteristics of the propeller installed.

And we haven't talked of Maximum Range yet, that too, depends upon whether you're asking about a propeller or jet aircraft.

Regards,

Old Smokey

redblue

27th Mar 2005, 01:09

OldSmokey, thanks for pointing out the difference in propeller and jet engine aircrafts. I was referring to single engine propeller aircraft. I found a site here that offers a very very good explanation, like yours. It is a very good site:
http://www.eaa1000.av.org/technicl/perfspds/perfspds.htm

To quote:
To maximize the endurance, we want to maximize the amount of time that we can stay in the air. In order to do this, we must minimize the fuel flow. Since the fuel flow is proportional to the power required, the fuel flow will be minimized at the point where the power required is a minimum. The speed corresponding to the bottom of the power required curve is the speed for maximum endurance
and,
To minimize the pounds of fuel per nautical mile, we can minimize the ratio of power over velocity. Looking at the power required chart, a line from the origin to any point on the curve has the slope of power over velocity (P/V). As you trace a line from the origin to each point on the curve, the slope will be a minimum when the line is tangent to the power required curve. Therefore, the maximum range airspeed occurs where a line from the origin is tangent to the power required curve. This also corresponds to the minimum point on the thrust required curve (drag polar).
and for Carson's Speed,
Unfortunately the maximum range airspeed is generally a lot slower than most people wish to fly. After all, you built an airplane to get places fast. Since we are also interested in getting places fast, we must consider speed. So consider a parameter of fuel flow per knot (Fuel Flow/knot). This would tell us how much fuel per hour we are burning for each knot of velocity. The optimum speed would then be the speed where this parameter is a minimum. Mathematically, the derivative with respect to velocity would equal zero

Old Smokey

27th Mar 2005, 01:48

Hi redblue,

It seems that we were simultaneously earlier, don't you guys ever sleep in CYVR?

It seems that you've done a good job of finding the answers anyway. Knowledge is Power, and that will add more Thrust to your arguments.

Good Luck,

Old Smokey

Intruder

27th Mar 2005, 02:29

What Intruder has said in the second sentence is absolutely correct for a jet aircraft, but certainly not for a propeller aircraft.

So, at a given weight, lift is constant in level flight, and thrust required is directly proportional to drag.

AFAIK, the L/D max / max endurance equation works for any powered aircraft. I don't know how max endurance at a constant altitude can be at a lower airspeed.

OTOH, I am also aware of the minimum sink speed for a glider, which is at the maximum lift speed ( just above stall), and lower than the max endurance speed for a powered aircraft.

Maybe there are some technical nuances to the definintions used in the various publications...

Capt Pit Bull

27th Mar 2005, 06:54

OTOH, I am also aware of the minimum sink speed for a glider, which is at the maximum lift speed ( just above stall), and lower than the max endurance speed for a powered aircraft.

I don't think thats correct.

Obviously with a glider you remove any powerplant issues regarding thrust or power output versus speed, and it purely becomes a lift and drag versus airspeed problem.

I agree that the minimum sink speed is very low, but I don't think that is the point of maximum lift.

Given that Lift = W Cos (Angle of Descent) you will be at the point of maximum lift when you minimise the angle of descent, i.e. at best L/D ratio.

Where as the minimum sink speed is the point of minimum power required. as mentioned above, the point of minimum (Drag x TAS).

So in effect, by waffling along at low speed you put up with more drag (being below Vmd) for the bonus of a proportionally lower TAS. The downside is that this is not the optimum L/D ratio and therefore your descent angle is actually steeper, and hence lift is smaller.

So, in summary, minimum sink speed is not equal to maximum lift.

On the other hand, its been a while since I taught this so I could be talking out of my fundamental orifice. It wouldn't be the first time!

BTW since power=drag x TAS, and given that at any particular altitude TAS is proportional to RAS, you can find the minimum power required speed from a drag graph. Graphically, drag x TAS is proportional to the area of a rectangle formed by drawing lines from your chosen point on the graph to both axes. It is reasonably obvious (pictorially) that this point is way below Vmd.

CPB

bookworm

27th Mar 2005, 09:41

Both Max Endurance and Max Range are highly dependant upon Fuel Flow. There is a direct relationship between Fuel Flow and Power for a propeller aircraft, and between Fuel Flow and Thrust for a jet aircraft.

I don't know how much complexity it is appropriate to add, but the idea that power is dependent only on fuel consumption is an approximation, at least for a piston engine. Such engines are usually more efficient for high power delivery. Add in the efficiency of the propeller, and I believe there can be small but significant variations in brake specific fuel consumption with speed.

I say significant because we're finding the minimum of a curve, where the curve is by definition not very steep, so fairly large speed changes cause only small changes in drag. In the same way that the best rate of climb speed is usually substantially above the min sink speed because power available increases with speed, so the best range speed may be considerably higher than min drag, and best endurance may be higher than min power required.

(I presume that there is a similar effect with jets in that their TSFC is better at lower thrusts. For example this page (http://www.adl.gatech.edu/classes/dci/engines/dci05.html) claims to derive from Shevell a TSFC that increases substantially with Mach no. Thus best endurance might actually be below minimum drag speed.)

Intruder

27th Mar 2005, 14:52

Given that Lift = W Cos (Angle of Descent) you will be at the point of maximum lift when you minimise the angle of descent, i.e. at best L/D ratio.
Actually, not. Maximum lift will give you the minimum RATE of descent, without regard to angle of descent or distance covered in the glide.

In a glider there are several speeds normally considered (these are only quick descriptions, not technical diatribes):

Minimum sink: Just above stall speed, it gives you the least rate of descent, which translates to the greatest rate of climb when in a column of lift. It also gives you the smallest turn radius, helpful for staying inside narrow columns of lift.

Max glide: At L/D max, gives you the greatest still-air distance in the glide, without regard to the time it takes to get to the ground.

"Speed to fly": Calculated dynamically, based on the specific aircraft, winds, available lift/sink, distance to go, and time constraints. Used when racing, to get the farthest the fastest without hitting the ground.

Capt Pit Bull

27th Mar 2005, 15:28

Intruder,

hmm - I don't disagree with anything you say in your discriptions regarding minimum sink and max glide, where you find them in speed terms or the applications thereof.

I just disagree with your assertion that minimum sink = maximum lift.

1.) Best Glide Speed = Best L/D ratio.
2.) Best L/D ratio = smallest descent angle.
3.) Smallest descent angle yields biggest possible value for lift in the equation L=W Cos (Angle Of Descent.)

Could you clarify which of points 1, 2 or 3 you feel is invalid and maybe we can thrash it out from there.

Cheers,

CPB.

enicalyth

27th Mar 2005, 16:37

HI redblue!!!

The trick is to remember thrust is proportional to the square of speed whereas power is proportional to the cube of speed. Your best endurance speed conforms to the bottom of the bathtub curve of required power versus speed. Clearly you stay in the air longer if you use as little power as possible. (Provided that that speed is above the stall!). Looking at the same bathtub curve of required power versus speed or indeed of drag versus speed if you draw a line passing through the origin and tangent to the curve then you have the speed for maximum range which is also best glide slope speed. For aircraft where compressibility drag is negligible this occurs when parasite and induced drag are equal as you have discovered. But the speed for minimum power is 76% of the speed for maximum range because the two are in the ratio of 1/(3^0.25). I think your ground school can show you "Fuel Efficiency of Small Aircraft," which is AIAA Paper AIAA-80-1847 and was written by B H Carson.

Vancouver eh? How goes the Gastown Steam Clock and its steam whistle "chimes"?

Intruder

27th Mar 2005, 19:29

I just disagree with your assertion that minimum sink = maximum lift.

3.) Smallest descent angle yields biggest possible value for lift in the equation L=W Cos (Angle Of Descent.)

Could you clarify which of points 1, 2 or 3 you feel is invalid

Not invalid, just misinterpreted... You have to consider descent RATE as well as ANGLE.

Using your same equation, note that weight is constant, while lift and descent angle are varied with airspeed.

When airspeed is decreased from L/D max to max lift (stall), L increases. So, for a constant weight the ANGLE of descent will indeed increase. However, the sine of the descent angle is defined as the vertical speed divided by the airspeed. If airspeed is decreased sufficiently, even a slight increase in descent angle will result in a lesser vertical speed.

For example, assume a glider has an L/D max speed of 45 knots (4557 fpm), and a min sink speed of 300 fpm at 40 knots (4051 fpm). [Note: These are not "real" numbers, but are in the ballpark for a low-performance glider like an SGS 2-33.] At min sink speed the descent angle is 4.24 degrees -- arcsin 300/4051. Now assume that at 45 knots (L/D max) the sink speed is 325 fpm. The descent ANGLE will be only 4.09 degrees -- arcsin 325/4557 -- even though the descent RATE is greater!

Note that the descent rate at 45 knots will have to be 337 fpm to equal the descent angle at min sink speed. The situation satisfies your equation.

Capt Pit Bull

27th Mar 2005, 21:39

Hmm.

I agree with the idea that a steeper descent angle can be consistent with a lower descent rate (along with 99% of the rest of your post). You just have to fly at the minimum power required speed to make that happen, which is well below best range / best L/D ratio speed.

What I don't see is how, given that you accept L=w cos (descent angle), you can maintain that maximum lift happens anytime other than minimum descent angle.

(Take it to extremes. If you really dirtied up, (say by opening a huge parachute) you'd decrease your descent rate by virtue of massive drag, even though your descent angle would be 90 degrees and lift had disappeared.)

The only thing I can think is that perhaps when you say maximum lift, you mean maximum lift coefficient.

Hmm. Its late. I need more beer.

Regards,

CPB

Intruder

27th Mar 2005, 22:54

You just have to fly at the minimum power required speed to make that happen, which is well below best range / best L/D ratio speed.
...which is also referred to as stall speed! Continuing remembering your basic aerodynamics... An airplane will continue flying at some point below the power-off stall speed, but power must be increased to do so.

Since a glider has no internal power available, glider pilots make use of this principal when they fly at "minimum sink" speed while thermalling.

What I don't see is how, given that you accept L=w cos (descent angle), you can maintain that maximum lift happens anytime other than minimum descent angle.
The key is here:

However, the sine of the descent angle is defined as the vertical speed divided by the airspeed. If airspeed is decreased sufficiently, even a slight increase in descent angle will result in a lesser vertical speed.

Maximum lift represents minimum descent RATE, not ANGLE. The descent angle in the equation you cite is actually a result, not an input parameter. Maybe the equation would be better stated as

Descent angle = arccos (L/W)

The only thing I can think is that perhaps when you say maximum lift, you mean maximum lift coefficient.

That may well be... Aerodynamics equations always make my head hurt. :sad: I'm thinking of the top of the hump in the curve that represents stall speed, so if that's a Cl vs airspeed (or AOA) instead of a Lift vs AOA curve, that's what I mean.

Wizofoz

28th Mar 2005, 04:00

Intruder,

Continuing remembering your basic aerodynamics

OK, in an aircraft flying straight and level, LIFT = WEIGHT. If lift is any other value than weight, the aircraft will accelerate vertically either up or down as LOAD FACTOR = LIFT/WEIGHT.

When talking about "Maximum lift" you obviously mean "Maximum CO-EFFICIENT OF LIFT" which is indeed achievied at the stalling angle of attack. This has nothing to do with either max endurence in a powered aircraft (achieved at Max excess thrust) or minimum sink in a glider (achieved at minimum drag).

Both these speeds are well above stall,as Total Drag rapidly increases as stall is approached because of a massive increase in Induced Drag.

The term "Maximum lift" is meaningless. L= CL*1/2rho*Vsquared. As V can have any value, Lift can increase exponentially until the wings fall off!

bookworm

28th Mar 2005, 08:27

Look at it via the basic equations for flight performance for unaccelerated flight. T= thrust, D = drag, L = lift, W = weight.

T = D + W * sin(flight_path_angle)
L = W * cos(flight_path_angle)

Thus

(T - D)/L = tan(flight_path_angle)

The maximum flight_path_angle is at maximum (T - D)/L, i.e. at maximum excess thrust. For T = 0, the maximum flight_path_angle is best glide angle and reduces to maximum -D/L, which is equivalent to maximum L/D.

Multiply through by airspeed v to make the right hand side vertical velocity and get:

(T - D)*v/L = rate_of_climb

Thus best rate of climb occurs at maximum excess power and for the T = 0 case, best rate of climb is minimum rate of descent and reduces to maximum -D*v/L, which is equivalent minimum power required. This is unrelated to stall speed, though it's possible in some cases that it might occur very close to stall.

Maximum lift as an absolute quantity for unaccelerated (1G) flight will indeed occur at minimum flight path angle, i.e. level flight if you have the power to stay level, but it's not a very useful concept. Maximum lift coefficient does indeed occur at stall, but by the time you've reduced the speed to that point, it's not the same as maximum lift.

Dick Whittingham

28th Mar 2005, 08:56

I hope that this, concerned only with staying level or going down, may be of value.

There are three basic speeds defined in level flight.

The lowest is Vmp, where the minimum power is needed to keep the aircraft flying. This is not Vs. It is a little bit above Vs, where it is difficult to fly, as you are on the wrong side of the drag curve.

The next highest is Vmd, where the drag is minimum. Remember, this is in level flight, so the thrust required to keep the aircraft flying is also minimum

The third is the speed that gives the best Speed/drag ratio. This would be 1.32 times Vmd if the theory is working right, but in real life it could be rather more or less.

If you are gliding, and there is no power or thrust delivery to worry about, then these speeds, based only on the aircraft aerodynamics, apply to pistons, jets, and gliders. To get the minimum sink rate in the glide, fly at Vmp. This will allow you to stay up longer – gliding for endurance. If you are heavy you will sink faster and vice versa. If you want to glide as far as you can then fly at Vmd. This gives you the angle of attack for best lift/drag ratio, and your best glide angle is, in fact, exactly your lift/drag ratio and as long as you fly the correct speed is not affected by aircraft weight. This is gliding for range.

These speeds are defined in level flight, and, as lift required equals weight, speeds (not best glide angle) vary as the weight of the aircraft changes. They are faster for a heavy aircraft, slower for a light aircraft. When you are gliding lift no longer equals weight. It is now the lower value, W x Cosine of the glide angle. This makes no practical difference up to glide angles of about 15deg, but in theory, you should fly in the glide at slightly lower speeds than the level flight quoted values.

Now, if you are under power, holding level flight, it makes a difference what type of power delivery you have. In piston engines, approximately, fuel flow is proportional to the Engine Horsepower (EHP) delivered. So, to use minimum fuel you would fly at the minimum power required speed, Vmp. In practice it is so difficult to do this that your Flight Manual will probably suggest a higher speed. In jet engines, fuel flow is approximately proportional to thrust delivered. In this case, to stay up longest you would fly at the minimum thrust required speed, minimum drag speed, Vmd. Again, to make things easier for you your Flight Manual may recommend a higher value

A similar difference applies if you want maximum range in level flight. Best speed/power required works out as Vmd, so Vmd is range speed for piston aircraft. Best speed/thrust required works out at 1.32Vmd and is jet range speed. For various practical reasons, including increasing engine efficiency at higher speeds, your Flight Manual may quote higher speeds, and at altitude, transonic drag problems will define range speed at a different value.

Glider heads will recognise that at the beginning of the day, with lift all over the place, they will take off with full ballast tanks. Vmd will be high, but best glide angle is unchanged, so they transit rapidly from Cu to Cu. As the lift dies, they will dump the ballast to lighten ship and fly for endurance to stay up that last minute longer.

Dick W

Intruder

28th Mar 2005, 10:02

Wizofoz said:
for the T = 0 case, best rate of climb is minimum rate of descent and reduces to maximum -D*v/L, which is equivalent minimum power required. This is unrelated to stall speed, though it's possible in some cases that it might occur very close to stall.

Dick W said:
The lowest is Vmp, where the minimum power is needed to keep the aircraft flying. This is not Vs. It is a little bit above Vs, where it is difficult to fly, as you are on the wrong side of the drag curve.
Hmmm... I think that's what I said at first, before confusing myself with all the terms: mininimum sink speed in a glider is just above stall speed, and less than the speed for best glide angle.

I also found the graph I was looking for, and it is indeed the Cl vs AOA graph, which is often used to describe the stall AOA (or airspeed in unaccelerated flight), that I had in mind. So where I said "max lift" earlier, I did really mean max Cl.

Wizofoz

28th Mar 2005, 16:28

Dick,

The lowest is Vmp, where the minimum power is needed to keep the aircraft flying. This is not Vs. It is a little bit above Vs, where it is difficult to fly, as you are on the wrong side of the drag curve.

In unaccelerated flight, Thrust=Drag. therefore the point at which minimum thrust is required (different from power, I realise, but as has been pointed out earlier, min power actually occurs at a higher speed) will be at the point where drag is minimum, e.g what you refer to as Vmd.

If you want to glide as far as you can then fly at Vmd

No, this gives you minimum sink. Max range is, as you state, achieved (in nil wind) at best L/D, but this is a different, higher speed to minimun sink.

Dick Whittingham

28th Mar 2005, 19:07

Sorry, in level flight where lift equals weight and is constant, and drag is at a minimum, you have found your best lift/drag ratio. So Vmd is, in fact, the speed in level flight that gives the angle of attack for best L/D. We agree that best L/D is the speed to fly for best glide angle. The speed that requires minumum power, which is drag times speed, in level flight is lower than Vmd. Graphically, it is where the speed and drag values form a square that touches the drag curve. This is the speed to fly if you want to use up your potential energy, your height, at the lowest rateand that is minimum sink speed.

Dick W

Intruder

28th Mar 2005, 19:09

Wiz,

You're getting all confused all over again! :*

Max endurance for a powered airplane, and best glide angle (power-off max range) for powered or unpowered aircraft are both found at L/D max, which is the same as Vmd.

Max range speed for a powered airplane is greater than max endurance speed. Min sink speed for a glider is less than max glide range speed.

Wizofoz

28th Mar 2005, 20:11

Dick 'N' Int,

I hit the books in response to your answers and have found you are mainly correct.

Yes Max L/D is achievied at min drag. Min sink is a lower speed.

Intruder, read Dicks post re max endurence. It is achieved at maximun excess thrust which, whilst close to min drag, will be slightly different. You were also under the impression that min sink occurs at stall speed. Not so. I own an ASW 19. Its' min sink occurs at 47kts, its Vs is 34.

Dick Whittingham

29th Mar 2005, 07:53

The only small point that separates us is edurance speed for powered aircraft. If you had the same airframes, one powered by a jet and one by a prop, the jet would fly at Vmd for endurance, the prop at Vmp.

This is only because fuel flow in a prop setup is proportional to the power delivered by the engine, not the thrust delivered by the propeller, for prop efficiency has intervened. In the jet, fuel flow is proportional to thrust produced and delivered, so you try to get minimum thrust required, ie. min drag in level flight.

I used to own part of an Eagle, and always thought I knew what Howard Huges must have felt like in his monster wooden flying boat - all creaking and groaning around you!

Live long and prosper

Dick W

VRThomas

1st Apr 2005, 23:21

Hi folks,

I just got through reading a text from United. They stated that in the example of a 737 the max endurance speed is actually below Vmd, on the back side of the curve, just sightly because at a lower airspeed there is more thrust available from a jet engine therefore requiring a lower fuel flow to maintain level flight.

It makes a little sense but I still like the idea of flying at Vmd without going to back side of curve.

VRT

bookworm

2nd Apr 2005, 07:48

Not just more thrust available, but more thrust per unit fuel consumed at lower speed. That's the point I was making in my first post. Is the text online?

Old Smokey

2nd Apr 2005, 12:40

Consider the following –

Aircraft need Thrust, they always have, and probably always will.
Aircraft don’t need Power, they never have, and probably never will.

Jet engines DIRECTLY produce Thrust, and Fuel Flow is directly related to Thrust output.

Propeller engines DIRECTLY produce Power, and Fuel Flow is directly related to Power output. The propeller converts Power to Thrust as a function of speed (TAS).

Power = Force X Velocity, i.e. Power = Thrust X TAS, so, for the propeller aircraft –
Thrust = Power Divided by TAS, as TAS increases, Thrust decreases.

Speed Effect for propeller aircraft is that Thrust declines directly with speed, not quite linear due do propeller characteristics, but slightly kinked.

Speed Effect for Jet aircraft is that Thrust is fairly constant with speed, but declines slightly up to about M0.5, and then increases again.

If we plot Thrust available from a propeller on a Drag Vs TAS graph, a constant quite steep decrease is noted with increasing speed. If we plot Thrust available from a jet on the same Drag Vs TAS graph, it is almost ‘flat’, but dips slightly, descending slightly at typical speeds in the vicinity of VMD.

A picture is worth a thousand words –

http://www.gunboards.com/forums/uploaded/Y.%20Hunt/20054283815_Jet%20Vs%20Prop.jpg

For the Jet, we note a very shallow negative gradient for the Thrust line, leading to tangency with the Thrust Required curve very slightly below VMD. For practical purposes, it is VMD, and, in the interests of speed stability, a speed slightly above VMD will result in negligible fuel penalty. (For the purists, if the holding is at a considerably higher Altitude where VMD was above about M0.5, Best Holding Speed would be slightly above VMD).

For the Propeller however, we note a quite steep negative gradient for the Thrust line for a given Power, leading to tangency with the Thrust Required curve somewhat below VMD. Flight at VMD would require considerably more Power, and be fuel inefficient.

The relative constancy of jet thrust with increasing speed, compared to the rapid decline in thrust from the propeller with increasing speed, serves to illustrate why Sir Frank Whittle went to all the bother of inventing the jet engine.

Best Regards,

Old Smokey

bookworm

2nd Apr 2005, 16:11

What was the oil price when Sir Frank worked his magic? ;)

Jet engines DIRECTLY produce Thrust, and Fuel Flow is directly related to Thrust output.

It may be related but it's by no means dependent only on thrust. It varies significantly with speed.

The dotted line in your diagram is a good representation of the thrust available for a jet. Thus if you're looking at angles of climb to find maximum excess thrust, or if you're solving a ceiling problem where you want to know what speed permits you to stay level at the ceiling, then I think your diagram explains it well.

But the dotted line is not a good representation of the thrust per unit fuel consumed. For example the graphs that I'm looking at (from Anderson's Aircraft Peformance) for an RB211 show a Thrust Specific Fuel Consumption proportional to approx (1 + 2*M) at sea level, and proportional to (1 + 1.5*M) at 10,000 ft.

Let's take the second for holding. With the speed of sound at 660 kt (warm day at FL100), we can get an idea of the gradient:

000 kt TSFC = 1.00 unit
110 kt TSFC = 1.25 units
220 kt TSFC = 1.50 units
330 kt TSFC = 1.75 units

So for thrust available per unit fuel, it looks like:

000 kt -> 1.00 unit
110 kt -> 0.8 units
220 kt -> 0.67 units
330 kt -> 0.57 units

That's much steeper than your thrust-available line, and shifts the speed for max endurance down considerably further, though not down as far as the Vmp point.

On a rather more abstract level:

Aircraft need Thrust, they always have, and probably always will.
Aircraft don’t need Power, they never have, and probably never will.

As with many problems in physics, you can solve this equally well using force (thrust) or multiplying by speed and using power instead. The maths has to work in both regimes. Which one you choose depends on which is more convenient, and it's entirely appropriate to choose force when your thrust doesn't vary much with speed. So in that sense aircraft need both thrust and power.

What is less arbitrary is the nature of propulsion based on combustion. A ton of fuel contains a certain amount of chemical energy, which the engine converts at a certain rate to power used for propelling the aircraft by pushing air backwards.

Naively, one might expect the efficiency of this process to be independent of speed. For props, this is a reasonable assumption. For jets, the efficiency increases with speed, and leads to a power specific fuel consumption with a very strong negative speed dependence. This makes it almost worth dividing through by speed to consider the thrust specific fuel consumption, which will have a strong positive speed dependence. That simply reflects huge inefficiency at low speeds, not some magical feature of Whittle's invention that defies energy conservation.

Props are also less efficient at turning fuel into useful power at low speed. The difference between props and jets is only one of degree, in the slope of the speed dependence.

wondering

11th Apr 2005, 17:36

I had a look at 'Jet Airplane Performance' , chapter 5, by Lufthansa Consulting.

That´s what they are saying (at least for jet aircraft):

Max Endurance:

Lines of constant fuel flow (under consideration of altitude and temperature) can be added to the drag-diagram. These lines have a negative slope. The tangency point of each curve group locates the value for minimun fuel. At the same time, these points represent the speed at which the airplane will remain airborne the longest, referred to as 'maximum endurance' . Since these fuel lines have a negative slope max endurance speed is below Vmd. As has been pointed out before for speed stability reasons a higher speed, closer to Vmd, is used when holding.

Max Range:

Similar to fuel flow, specific range lines can be included in the drag-diagram (with consideration to altitude). These lines have a positive slope. The tangency points of both curve groups locate the speed schedule for maximum range under the appropriate conditions.

In summery max endurance is below Vmd amd max range is above Vmd.

Appologies for not being able to include any diagrams. I know a picture says more than 1000 words.

Alex Whittingham

11th Apr 2005, 20:03

Old Smokey,

I've a couple of problems with your diagram.

The first is that the thrust available curve shows a reduction in thrust with forward speed then an increase. This is the classic thrust line from aerodynamics textbooks produced in the 1960s which, of course, reflects a 1960s engine, typically an Avon. Modern, high bypass ratio, jets tend never to recover the thrust by ram effect.

The second problem is that you have shown the thrust available curve just touching the thrust required curve in such a way as to suggest that the point of contact is significant. If the way you have illustrated it was correct the thrust available would always be less than the thrust required and you'd never get off the deck. In fact, thrust available is (usually) greater than thrust required and what is important is not where the graphs touch when they are artificially brought together but the vertical gap between them, which indicates angle of climb. If you displace your jet thrust available curve upwards to where it belongs, well above the thrust required, the greatest difference between thrust available and thrust required occurs AT VMD not at the artificial point of contact you illustrate.