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I've read Carson's paper on V(L/D)max, and while theoretically interesting, it does not seem to me to be useful in actually finding the V(L/D)max of an arbitrary aircraft, as it assumes knowledge of total parasite area (f) which can only really be calculated experimentally (and is very hard to calculate at that).

Another paper, by Russ Erb, claims that optimal glide speed, in the emergency section of the Pilot's Handbook for an aircraft, is equal to V(L/D)max for that aircraft. This sounds wrong intuitively as it does not seem to take into account weight, air pressure, height, or other such variables. As a math major, I am certainly suspicious of people who say "I'll spare you the derivations."

For an aircraft with an engine, such as a Cessna or a Bonanza, is best glide speed roughly equal to V(L/D)max?

I was told by an experienced pilot that

Vldmax at weight G = Vldmax at weight L * ( weight G / weight L )^0.5

This would seem to contradict the claim that V(L/D)max = best glide speed, as best glide speed (according to the Bonanza Operating Handbook, as an example), is constant.

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  • $\begingroup$ Would that answer your question? By the way, the experienced pilot is right. $\endgroup$ – Peter Kämpf Jul 28 '16 at 10:09
  • $\begingroup$ @PeterKämpf this is the best written synopsis I've seen yet, but it leaves me at the same impasse of not knowing how to discern form factor, roughness, or resulting total parasite area in order to calculate Cdo. $\endgroup$ – Max von Hippel Jul 28 '16 at 15:20
  • $\begingroup$ Calculation of $c_{D0}$ is an art in itself. You need to sum up all details to get close to a realistic value, or you get one from flight test which produces a wide scatter, so you need to measure multiple times. Sorry, there is no shortcut! $\endgroup$ – Peter Kämpf Jul 28 '16 at 16:44
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    $\begingroup$ Is there an easy way to estimate using flight tests that recreational pilots could do? I'm trying to create an easy to use recourse for recreational pilots. $\endgroup$ – Max von Hippel Jul 28 '16 at 17:16
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    $\begingroup$ @MaxvonHippel if your engine is on, you're not gliding. Because the engine is producing thrust (either positive or negative), it will change the range from a pure glide. $\endgroup$ – Kenn Sebesta Oct 19 '20 at 11:41
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A few questions here, starting with parasite drag. Here, experimental measurements in a wind tunnel would be helpful, as the old standby graph of induced lift going down and parasite lift going up with speed is pretty generic. However, after looking at the PBY-3 Catalina and where they mounted the props (and flew it as a long distance patrol aircraft), one may conclude the lions share of drag is induced, particularly at lower GA speeds.

Potential energy (altitude, glider fuel) = m×gravity×height. One can see this increases linearly with weight. Speed increase needed to lift (at constant optimal AOA) a given weight increase is the SQUARE ROOT of the weight increase. Drag produced by this speed increase is speed SQUARED. This produces a linear relationship between gliding distance gained from increased weight compared with gliding distance lost from increased INDUCED drag.

Increased speed will produce a minor increase in parasite drag. So, efforts to quantify this would focus on the drag coefficient of the prop, fuselage and tail. Here you're in luck as all studies are done at optimal WING AOA.

Keep in mind Vbg is based on Indicated Airspeed, so one need not worry about pressure or altitude. So, for the practical GA pilot, Vbg will be fairly constant within the given safe weight range of your aircraft (but could be calculated while you do your gross weight and CG location work pre-flight), cargo carriers need consult the POH as Vbg will vary greater with a wider weight range.

Now, the other factor for Vbg is wind speed. Here greater weight is advantageous for wind penetration. Hence nature's inexorable millions of year process of trial and error has produced the heavy, fast high aspect ratio albatross to easily glide 1000s of miles on the wind. Glider pilots add ballast under windy conditions, while the lighter wing loaded, slower vulture delights in riding the thermals. Glider pilots may choose to lighten ballast under those conditions.

However, if wind is a factor, best glide slope (from best glide speed) can be determined just like in a landing, by observing how a fixed object in the distance moves in your windscreen. At Vbg it will rise the slowest. And landing is what you will be doing soon.

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  • $\begingroup$ "Potential energy (altitude, glider fuel) = m×gravity" I'm quite sure you did not intend to write $E_{pot} h = m \cdot g$, but rather $E_{pot} = m \cdot g \cdot h$ $\endgroup$ – AEhere supports Monica Sep 16 '19 at 12:29
  • $\begingroup$ Noted and edited, thanks. $\endgroup$ – Robert DiGiovanni Sep 16 '19 at 15:12
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According to the FAA's "Best Glide Speed and Distance" handout provided by the General Aviation Joint Steering Committee ("GAJSC"), on most airplanes, the best glide speed is halfway between Vx and Vy. See Exhibit 1

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    $\begingroup$ Answers like these are very approximate, though. Where I got stuck was trying to find a way to calculate the exact correct speed, not an estimate. But getting the exact speed required surface roughness and wetted area, and I don't think I have any justifiable chance of really calculating either .... $\endgroup$ – Max von Hippel Jul 4 '17 at 15:55
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    $\begingroup$ … and yet this makes absolutely no sense. Note that there are two best gliding speeds—speed for best gliding distance and speed for best gliding time. The best gliding time speed is $V_y$. The best gliding distance speed, is always higher (and how much higher depends on weight). $\endgroup$ – Jan Hudec Jul 4 '17 at 18:46
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    $\begingroup$ @MaxvonHippel, the answer is not just approximate, it is clearly wrong. $\endgroup$ – Jan Hudec Jul 4 '17 at 19:00
  • $\begingroup$ What is wrong about it? The FAA document seems pretty straight forward. I just looked at a few POHs and the recommended engine failure speed by the manufacturer is approximately half-way between Vx and Vy. $\endgroup$ – Devil07 Jul 4 '17 at 21:05
  • $\begingroup$ @Devil07 you may likely be right in your interpretation of the FAA document, but there's a good chance (I think) that the FAA document in this case is wrong. The math involved in finding this speed is really complex, and I have a hard time believing any "rule of thumb" like that could ever be any good. That said, I'm a mathematician, not a pilot. $\endgroup$ – Max von Hippel Jul 5 '17 at 18:14

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