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The engine was turned off at 1000m above sea level. If you ignore the landing pattern, how long can the runway be within a few kilometers of the landing to safely land? However, the wind is blowing at 72 km / h from behind.

  • Empty weight = 90kg, fuel = 30kg
  • zero lift drag coefficient = 0.05
  • Oswald = 0.75
  • Span 4.5 m
  • Wing w Area 3.7m^2
  • Cl =0.1(alpha + 1)

I am really confused if the wind is blew from head or tail how it influence on range.

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  • $\begingroup$ The influence of the wind is very important in what concerns range. For example, if you're gliding with a headwind (measured, as all winds, with respect to the ground) that has the same speed as the airspeed measured by your anemometer on board the aircraft, the descent trajectory will be vertical, and the range will be zero... $\endgroup$ – xxavier Dec 19 '18 at 12:59
  • $\begingroup$ The distance of the glide is computed, from the L/D ratio, related to the air mass. If the air mass moves related to the ground, the ground distance is changed according to the air mass speed. If you walk 10 meters in 10 seconds in a moving train, the actual ground distance depends on the train move during this time. $\endgroup$ – mins Dec 19 '18 at 16:20
  • $\begingroup$ I'm not sure if you're asking a) how to calculate gliding range, or b) how wind affects gliding range. The second question is probably answered here. $\endgroup$ – Pondlife Dec 19 '18 at 17:22
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    $\begingroup$ I am appreciate to your answers. Then, what I really want to know is first question. Even though, now l start to understand gliding little bit thank you $\endgroup$ – 설승우 Dec 19 '18 at 19:56
  • $\begingroup$ You do have everything needed to calculate the aircraft's L/D and best glide speed (assuming you $c_{d_0}$ is referenced to wing area too, because you don't have frontal area), but it's somewhat more math. $\endgroup$ – Jan Hudec Dec 20 '18 at 21:26
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I'm going to assume that when you say 1000 meters above sea level, what you really mean is 1000 meters above ground level. 1000 m AMSL can be underground in many locations.

First, you need to understand that the airplane, once in the air, always moves together with the surrounding air. The surrounding air can be moving relative to the ground – what's usually referred to as wind – but assuming sufficient altitude above local ground, the airplane is largely unconcerned with what the ground is doing under it.

Second, every airplane has some glide ratio. The glide ratio tells you, for each amount of distance covered, how much altitude you will lose. Typical values for light motor fixed-wing airplanes might be around 10:1 to 12:1, while commercial airliners might get 15:1 to 20:1, and high-performance gliders can exceed 50:1. For simplicity, I'll assume a glide ratio of 10:1, which is pessimistic, but which should be achievable by most light motor airplanes. (User XRF pointed out in a comment that these values might actually be on the optimistic side, but I'm sticking with them because the point is to illustrate the principle, not to give an exact calculation.)

Typically, the glide ratio is quoted at speed and configuration of best glide, so your first course of action when losing the engine should be to configure the airplane for best glide, and re-trim to match.

What the 10:1 tells us is that, for each 1*X meters (or feet) of altitude that you lose in a glide, you will cover 10*X meters (or feet) of distance. Since you start out at 1000 meters above ground, this means that you can cover 10*1000 meters or 10 km of distance before your flight path intersects ground.

However, again, that's distance covered through the surrounding air mass.

To figure out how far you can travel along your ground track, you need to figure out the amount of time for which you can remain in the air. Let's say that the speed of best glide in your airplane is 100 km/h indicated airspeed. Since you can travel 10 km through the surrounding air mass, this means that your travel time after losing the engine is 10/100 of an hour, or six minutes.

In six minutes, the 72 km/h wind will push your airplane 7.2 km along the ground, because six minutes is 1/10 of an hour, and 1/10 * 72 = 7.2.

Therefore, if you maintain that tailwind, with this highly simplified example, the point where you reach ground level will be 7.2 + 6 = 13.2 km from where you lost engine power.

If you instead make a 180° turn to head straight into the wind, you now have a 72 km/h headwind and a 100 km/h indicated airspeed, for a ground speed of 100-72 = 28 km/h. You can still travel 10 km through the airmass before you reach the ground, and this still takes 6 minutes, but in those 6 minutes, your ground track distance is only 2.8 km.

The major advantage of turning into the wind is that when you reach ground, you will do so at a relative speed of 28 km/h, rather than 172 km/h, assuming an unchanged 100 km/h indicated airspeed. (This is why you always land into the wind.)

Therefore, with that wind, depending on which direction you turn into, your point of contact with ground, with these assumptions, will be somewhere between 2.8 km from where you lose engine power, and 13.2 km from where you lose engine power, expressed in terms of ground track distance.

In practice, however, if you lose the engine while in the air, this is all awfully academic. Rather, you'd pick out a nearby spot on the ground that looks reasonably even, level, and large enough to set down on; and head that way. There are virtually always ways to lose altitude, but without engines in an airplane not designed as a glider, very few ways to gain altitude. With experience, you might be able to quickly judge if you'll make it to the runway or not; but any controlled landing, even an outfield landing, is always better than a crash. You can always worry about getting the airplane back to the airport once you're safely on the ground.

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  • $\begingroup$ Of course, covering 10Km of ground (as in your example) requires that the ground cooperates and remains level. If it should chose to rise up in front of you, you'll meet it much sooner. $\endgroup$ – FreeMan Dec 20 '18 at 14:41
  • $\begingroup$ @FreeMan For the sake of simplicity, in true physics textbook fashion, I assumed a perfectly flat ground of infinite extent. I also assumed some other things, including the pilot immediately attaining speed of best glide, that all turns are absolutely perfectly coordinated, that the pilot remains calm throughout the entire exercise, and so on. $\endgroup$ – a CVn Dec 20 '18 at 16:20
  • $\begingroup$ I believe the data given in the OP‘s question is sufficient to calculate best glide speed and achievable glide ratio for the given scenario when assuming quadratic drag polar (I am not in a position to try myself for the next couple of hours hence just very lazily commenting...). I also believe, looking at weights in original question, there won’t be a (human) pilot on board. $\endgroup$ – Cpt Reynolds Dec 20 '18 at 18:02
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    $\begingroup$ I think your glide ratio's are a little optimistic. A Cessna 172 does about 7:1, a DA-20 that was designed from a motor glider does 14:1. The "Gimli Glider", a Boeing 767 that ran out of fuel managed 12:1. While the best open class sailplanes can manage over 60:1, most gliders are in the 30:1 or 40:1 range. $\endgroup$ – XRF Dec 20 '18 at 20:36

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