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Endurance not at Best L/D

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Endurance not at Best L/D

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Old 28th Mar 2005, 16:28
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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.
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Old 28th Mar 2005, 19:07
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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
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Old 28th Mar 2005, 19:09
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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.
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Old 28th Mar 2005, 20:11
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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.
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Old 29th Mar 2005, 07:53
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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
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Old 1st Apr 2005, 23:21
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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
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Old 2nd Apr 2005, 07:48
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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?
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Old 2nd Apr 2005, 12:40
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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 –



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
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Old 2nd Apr 2005, 16:11
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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.
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Old 11th Apr 2005, 17:36
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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.
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Old 11th Apr 2005, 20:03
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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.
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