767 Glide Ratio is better than C172

edit: see the A380 video as it punches lift/vortex pattern through clouds, further below

According to the link, the manufacturers’ published glide ratios for a 767 vs C172 is 20:1 vs 8:1

Is this common knowledge, that a commercial jet aircraft can glide 2.5 times further than a light private aircraft? Q&A - The New York Times

What about fighter jets I wonder? (I couldn’t find a reliable source)

I think fighter jets can’t glide far as they are built for speed. The aircraft would have to go down quite fast to avoid stall

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And to add to this, their wings are also tiny so that means less lift which is essential for gliding.

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I know from my flight training that it’s more like 10:1 for the C172 but I’d have to agree with that I mean with things such as ETOPS and RnD it’s almost assumed they can glide further. The 172 hasn’t been updated in many years and well ETOPS isn’t a thing for the 172.

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It is crazy to think that based on the ratio quoted, a 767 at 30,000 ft could glide approximately 113 miles. While it seems like a pretty long distance, I am sure it wouldn’t seem that to a pilot engine out on a north Atlantic track.

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Imagine the glide ratio for a 737-600. That thing could do go for days and still be in the air lmao

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That’s what I always assumed too. But I found this graph that seems to show military jets may not have inferior glide ratios unless they are flying over mach 1.

The L/D is the same as the glide ratio.

I think it’s kind of hard to believe (assuming it’s true).

graph 1

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That is an impressive distance. Especially when you consider that many people without much exposure to aviation seem to often believe an airliner will nose dive if all engine power is lost.

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Interesting find but not all that surprising. They have similar aspect ratios (8 vs 7.3), but note that the weight of the 767 is much heavier than a 172, so it’s inertia keeps it going. Same principal as why glider pilots load up with water ballast for races. Note that in @adit’s chart the B-52 has the best L/D. It’s aspect ratio is considerably larger than the others and it’s one heavy plane.

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I guess I was surprised that most fixed wing aircraft seem to fall in the range of about 10 to 20 L/D. I thought military jets might be much worse.

As far as load though is it not true that, as with water ballast added to gliders, it doesn’t affect the glide ratio?: You still get the same distance from the same height, but you just get there faster?

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This is a somewhat similar graph that shows more aircraft, a bit in the past but still relevant:
graph 2

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I guess that was a poor analogy to get my point across. I was trying to convey that just because the aircraft is larger and faster doesn’t mean it’s L/D would be any lower, as it might seem at face value. In other words speed doesn’t actually affect the maximum glide ratio, so it wouldn’t make sense that a light aircraft has a better L/D.

Are you that sure about that? Speed increases drag and would therefore presumably decrease glide distance for an aircraft with a similar drag coefficient, no? That is of course assuming the drag coefficient is constant which I know is a huge assumption, but still, I sort of doubt that speed having no effect is entirely accurate. I suspect L/D max has much more to do with the aircraft’s aerodynamics than weight or speed. Which would then make sense why planes of a general type with a common design goal are linked together in the charts shown above, not planes of a similar speed or weight.

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Correct. Speed will affect the L/D, but it will not affect the maximum L/D, that is a constant. Note that speed of max L/D will change as weight (represented below as wing loading) changes.
image

This is a derivation I was able to find in an old textbook. Maximum L/D can be thought of as a function of the zero-lift drag coefficient and the Oswald efficiency factor (k, but you may also see it written as e naught.)
image

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That seems to make sense, especially for the size, where you can just scale everything up. As you said previously:

Presumably you’d want a large aspect ratio to maximize L/D, except for fighter jets requiring shorter wings for high roll rate manoeuvres and “longer chord and thinner air foils involved in supersonic flight.”

So for fighter manoeuvrability and supersonic flight, the optimum subsonic L/D ratio is likely sacrificed to those other design needs. So inferior glide ratio might be expected?

L/D max is apparently associated with one particular AoA. So the reason the speed of max L/D changes with weight is presumably because the weight influences the speed you need to be able to hold that necessary AoA for max L/D.

So in a sense the concept of the efficiency of the design being constant is about the one particular aerodynamic form you hold the aircraft at with it’s fixed aerofoil profile at a fixed AoA to have the wind hitting it at that one most efficient direction(?).

And when you attain that, L/D is always the same, irrespective of weight, so the glide ratio is fixed.

edit: And as you said D has to increase with speed, but at L/D max, L increases at the same rate (for the weight dependence of speed) so that the ratio stays the same. And the far right term being a constant dependent on aerodynamic design suggests the constancy of “pay the same for what you get” (if that makes any sense??):

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Yeah, so I guess all that changes with weight is the amount of time you glide but not distance. So you will achieve the same distance, but have a longer time at a lower gross weight and the inverse. Since a plane’s best glide speed is at the point where induced drag and parasitic drag crossover. Since a higher weight would require a higher AOA for any given speed it would push that side of the graph up but the weight wouldn’t affect parasitic (at least not as much), so you would have the crossover point at a higher speed. The ratio of lift to drag is the same though as we already established, so you can go the same distance, you just do so at a higher speed and have a shorter time unless I’m thinking about this wrong. I’d love to just solve that equation, but I don’t have nay good numbers to sub in for the variables.

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I believe that is all true.

If I understand what you mean, the speed to get L/D max appears to be at a different point than the drag crossover point as shown below (crossover is after the peak and at a high AoA):

Another way I was thinking of it:

The fixed wing design “doesn’t care” in particular what load you put on it and what speed you fly it at (within reason). All it asks is that you fly it so that the relative wind hits it at the one (and only one) best angle. And that angle never changes, under any conditions (ignoring flaps etc.). That’s the AoA that always gives you L/D max (so for the figure example above, just over 6 degrees).

That best AoA angle always gives you the same L/D max value, irrespective of how you balance load vs speed. So that best AoA always gives you the best glide distance.

But because more load requires more lift, It’s up to the pilot to increase speed so that the AoA is held at the favoured value.

If you are too slow for the load you carry, your AoA will be pulled higher than the optimum value as you resist a sink rate (to generate more lift for the extra weight). This messes up your L/D max, and so also your best glide distance.

If too fast, then you have the opposite problem with AoA being too small (because you’re getting too much lift, and resisting the climb squeezes AoA smaller).

Either too slow or two fast for the weight and you miss the relative wind angle sweet spot.

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Also about the L/D relationship:

One might ask: lift doesn’t require energy?

I mean obviously, drag consumes energy, that’s why you need engine power (or during a glide, gravity’s assist)

Two forces are created, but only one requires energy while the other does not?

Both L and D come from change in momentum of the relative wind.

But why does D’s change in momentum require engine power while the large L which holds up the weight of the aircraft seems to come for free?

Momentum = mv,
So Force = change in momentum = m(dv/dt)

But because v is a vector (it has direction as well as size), it will cause a force from either a change in size OR direction (or both)

for D = m(dv/dt) the size of v changes so air loses kinetic energy according to
K.E. = (1/2)*mv^2

Kinetic energy goes down because the velocity of air has been resisted (so thrust is needed to compensate).

But for L = m(dv/dt) only the DIRECTION of v changes so the passing relative wind loses no kinetic energy to it’s path bending.

So the aircraft’s weight is supported by L without any expenditure of energy beyond overcoming D.

An analogy is using the brakes or accelerator in you car vs the steering wheel. All of these are associated with forces due to changing your momentum. But only brakes and accelerator change your kinetic energy; the steering wheel changes your direction (you feel the force), but doesn’t change the kinetic energy.

The L/D ratio tells you the cost of L in terms of D.

It tells you the cost of the force derived from the change in direction of relative wind in terms of the loss of energy in making that change in direction.

In contrast, if you used directed thrust instead of the fixed wing’s “bargain L” it becomes very expensive in terms of energy, because all of the momentum change to produce L comes from increasing the size of v rather than bending it’s direction.

On glide ratios (L/D), the Smithsonian has some interesting histories on dead stick landing (including helicopter), and the comparative glide ratios shown below.

Modern glider: 70
Hang glider: 15
Boeing 767: 12
Space shuttle orbiter during approach: 4.5
Human body: 1

It’s fascinating that a human body has a L/D ratio of 1! You can “glide” your body, generating lift, at a descent angle of 45 degrees! Watch out for the landing flare!

But I was curious about fighter jets. The F16, without a backup engine, has had some notable dead stick landings. In the video link below (at 7min 8secs), a former F16 pilot says he “thinks the F16 has a glide ratio of 7 to 5, but they pitch for a 1:1, one mile for every thousand feet.”

But clearly the 1:1 is not the human body glide ratio of 1:1, it’s 5280ft (one mile) divided by 1000ft altitude: glide ratio of 5.28