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I found the best glide speed of cessna 152 is 60 kts. My question is why it is not more or less than 60 kts? What are the factors upon which the best gliding speed is determined?

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  • $\begingroup$ Because that gives the most distance per unit drop. A lower speed will have a steeper descent and a higher speed will also have a steeper descent. Remember when you are gliding, your only speed control is pitch, and pitch also affects your descent rate. Its also not that simple as you state, best glide is dependent on the weight of the aircraft, usually the POH states "best glide" at maximum weight. $\endgroup$ – Ron Beyer Jul 27 '16 at 13:22
  • $\begingroup$ I understand a higher speed will steeper may descent more than a lower speed. But how that controls the best gliding speed? Does it only depend on the weight? If my weight is less/more how that will control the best gliding speed? $\endgroup$ – Mirajul.Pias Jul 27 '16 at 13:32
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    $\begingroup$ do these posts answer your question? aviation.stackexchange.com/q/606/1467 aviation.stackexchange.com/q/3610/1467 $\endgroup$ – Federico Jul 27 '16 at 13:37
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    $\begingroup$ The weight doesn't "control" the best glide speed, it dictates it. Best glide speed means the most distance traveled per distance dropped. The lower the weight, the lower the speed. See this FAA document for some more information. $\endgroup$ – Ron Beyer Jul 27 '16 at 13:38
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Basically, the faster you go, the more lift and drag a wing will produce. These two values are not proportional, though. When speeding up, the amount of drag produced is higher than the extra amount of lift - which is why you need additional power to maintain level at higher speeds.

When you go slower, the amount of drag will reduce more than the amount of lift produced - at least for a while. That's why going slow is better in terms of gliding distance. The amount of "drag per lift" is very low. However, slowing down beyond a certain point, the wing will rapidly start producing less lift, because airflow separates from the wing. This is what's known as a stall. The best glide speed is the speed as which the drag is as low as possible while the wing is still producing a relatively large amount of lift.

This is illustrated on a speed polar, like this one:

enter image description here

The black line indicates the rate of sink for a given airspeed. The optimum glide speed is the speed corresponding to the point where the red line touches the black line (Vbg).

The red line is a straight line going from (0,0) and touching the speed polar at exactly one point.

A change in aircraft mass will shift the curve along the vertical axis, which is why a heavier aircraft has a higher best glide speed than a lighter one. The intersecting point between the black and red line would shift to the right as the black line is shifted down, and vice versa.

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  • $\begingroup$ Now I have one question, according to graph at Vmd the sink rate is less and by this I can remain up in the air for more time. And at Vbg sink rate is greater. Then why my best glide speed is not at the speed of Vmd? Why it is a higher speed at which sink rate is more? $\endgroup$ – Mirajul.Pias Jul 27 '16 at 16:47
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    $\begingroup$ At Vmd you are indeed sinking slower, but you are also moving forward slower. Your air time will be higher, but you will cover a shorter distance because your forward speed is slower. Vmd/Minimum sink is the speed that will keep you in the air for the longest time. The Vbg/best glide is the speed that will let you cover the greatest distance. $\endgroup$ – expeditedescent Jul 27 '16 at 19:30
  • $\begingroup$ This is kind of a fine point, but in general, slowing down does not actually cause a reduction in Lift. See my recent answer for a different view. $\endgroup$ – quiet flyer Jun 25 at 17:18
  • $\begingroup$ @quiet flyer that’s very much a valid point - unless accelerating upwards or downwards the lift must equal the weight of the aircraft. However, it does depend on which variables you change and which you don’t - if the airspeed is reduced and the attitude is unchanged then you’re right, but if the angle of incidence is maintained then the aircraft will accelerate downwards (initially at least). $\endgroup$ – Frog Jun 25 at 21:37
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The most important factors for the best glide speed are the aircraft's wing loading, the air density, the wing's aspect ratio, and the aerodynamic quality of the aircraft.

The aircraft must create lift equal to its own weight. The drag for doing so varies with airspeed, and to find the point where the glide ratio has its maximum, drag must be minimal. To find this speed, we describe drag mathematically as the sum of two components:

  1. Parasitic drag, which goes up with the square of airspeed. We express this as the zero-lift drag, a drag component which is independent of lift: $D_0 = \frac{\rho}{2}\cdot v^2\cdot S\cdot c_{D0}$
  2. Lift-dependent or induced drag which goes down with the inverse of the square of airspeed: $D_i = \frac{\rho}{2}\cdot v^2\cdot S\cdot \frac{c_L^2}{\pi\cdot AR\cdot\epsilon}$

Now it helps to find the lift coefficient to create the needed lift at a given speed: $$c_L = \frac{m\cdot g}{\frac{\rho}{2}\cdot v^2\cdot S}$$ Which, when inserted into the formula for induced drag, produces $$D_i = \frac{(m\cdot g)^2}{\frac{\rho}{2}\cdot v^2\cdot S\cdot \pi\cdot AR\cdot\epsilon}$$ Now it should be obvious that induced drag is indeed proportional to the inverse of flight speed squared. We can simplify this a little by inserting $AR = \frac{b^2}{S}$ and express the total drag as the sum of both components: $$D = \frac{\rho}{2}\cdot v^2\cdot S\cdot c_{D0} + \frac{(m\cdot g)^2}{\frac{\rho}{2}\cdot v^2\cdot \pi\cdot b^2\cdot\epsilon}$$ Next, we differentiate with respect to speed $v$ and need to set the result to zero to arrive at an equation for the speed of lowest drag: $$\frac{∂ D}{∂ v} = \rho\cdot v\cdot S\cdot c_{D0} - \frac{(2\cdot m\cdot g)^2}{\rho\cdot v^3\cdot \pi\cdot b^2\cdot\epsilon} = 0$$ $$\rho\cdot v^4\cdot S\cdot c_{D0} = \frac{(2\cdot m\cdot g)^2}{\rho\cdot \pi\cdot b^2\cdot\epsilon}$$ $$v = \sqrt[4]{\frac{(2\cdot m\cdot g)^2}{\rho^2\cdot\pi\cdot b^2\cdot\epsilon\cdot S\cdot c_{D0}}}$$ $$v = \sqrt{\frac{2\cdot m\cdot g}{\rho\cdot S\cdot\sqrt{\pi\cdot AR\cdot\epsilon\cdot c_{D0}}}}$$ There you have it: The best glide speed is proportional to the square root of both the wing loading $\frac{m\cdot g}{S}$ and the inverse of air density $\rho$, and the fourth root of the inverse of the aspect ratio $AR$, the Oswald factor $\epsilon$ and the zero-lift drag coefficient $c_{D0}$. The Oswald factor is a measure of the quality of lift production and is close to unity in most cases.

Thankfully, quiet flyer reminds me in the comments that I need to point out that I always talk about true air speed here. If you want to set your speed using the speed indicator, reduce the value by the applicable density correction.

Nomenclature:
$c_{D0} \:$ zero-lift drag coefficient
$c_L \:\:\:$ lift coefficient
$S \:\:\:\:\:$ reference area (wing area in most cases)
$v \:\:\:\:\:$ airspeed
$\rho \:\:\:\:\:$ air density
$\pi \:\:\:\:\:$ 3.14159$\dots$
$AR \:\:$ aspect ratio of the wing
$\epsilon \:\:\:\:\:$ the wing's Oswald factor
$m \:\:\:\:$ the aircraft's mass
$g \:\:\:\:\:$ gravitational acceleration
$b \:\:\:\:\:$ wingspan

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  • $\begingroup$ Is this the same as the L/D max speed (Vldmax)? $\endgroup$ – Max von Hippel Jul 28 '16 at 23:24
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    $\begingroup$ @MaxvonHippel: Yes. Minimum drag at constant lift means that L/D is at its maximum. $\endgroup$ – Peter Kämpf Jul 29 '16 at 6:27
  • $\begingroup$ Re "There you have it: The best glide speed is proportional to the square root of both the wing loading m⋅gS and the inverse of air density ρ..." -- since you have brought air density into it, it might be helpful to specify that you are speaking of TAS (if you are). $\endgroup$ – quiet flyer Jun 25 at 17:01
  • $\begingroup$ @quietflyer Yes, I am. Good point! $\endgroup$ – Peter Kämpf Jun 25 at 18:42
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(it's simpler than it may looks)

If you are at a certain height, you have a certain amount of potential energy (or height energy). The only thing you can do is to convert it to kinetic energy (or speed, which then creates lift). The problem: drag takes up energy too. So all the energy you loose due to drag means a loss in kinetic energy (=speed) and therefore a loss in lift.

The question actually is: how to reduce drag to a minimum?

It's actually quite simple: there are roughly two different kinds of drag:

Drag diagramm

  • induced drag, induced by the angle of attack of the airplane. The more your nose goes up (so the lower your airspeed is), the higher the induced drag. This is an exponential relation.

  • parasitic drag, comes from the air and is the "usual" drag you also feel with a car or bike. It depends exponential on the airspeed.

The total drag consists of the sum of both. The minimum is the best gliding speed.

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  • $\begingroup$ Wouldn't the best glide speed be a bit faster than the minimum drag speed (since, by definition, the airplane covers more distance per unit time at higher speeds?) $\endgroup$ – reirab Jul 27 '16 at 21:24
  • $\begingroup$ Sure. But your goal is not to fly the longest distance in the shortest time, which means speed is irrelevant, only the efficiency matters. If you loose, say, 500 feet, you better need 2 minutes for that with a speed of 50 knots instead of 1 minute with a speed of 70. We only look for the best height-loss versus distance covered ratio. We do not care about the time at all, it is completely irrelevant. $\endgroup$ – Mayou36 Jul 28 '16 at 12:53
  • $\begingroup$ @reirab --no. We can show that the best glide angle occurs at the highest ratio of L to D. We can also show that this occurs exactly at the point of minimum D. (By the way, L is nearly constant and nearly equal to Weight; more precisely, L = Weight * cosine glide angle.) $\endgroup$ – quiet flyer Jun 25 at 15:21
  • $\begingroup$ This is kind of a fine point, but in general, slowing down does not actually cause a reduction in Lift. $\endgroup$ – quiet flyer Jun 25 at 16:37
  • $\begingroup$ Also, if we were really trying to minimize the rate of energy loss, I think we'd end up at the airspeed for min sink rate, not max glide ratio. So, generally good answer but some room for improvement-- $\endgroup$ – quiet flyer Jun 25 at 16:37
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I've never heard of the term maximum gliding speed, there's no special limitation to how fast you can fly a c152 without an engine as opposed to with it working. I think what you are talking about is best glide speed, also known as Vbg, which is the speed that gives you the farthest horizontal distance travelled per unit of height lost. If I remember correctly 60kts is the best glide with flaps extended, 65kts was best glide with no flaps.

The best glide speed actually varies based on weight, as do most of the V-speeds. A heavier airplane would mean a faster Vbg and a lighter one a slower Vbg. On a c152 the difference is pretty small, maybe 2 kt either way, so giving a 1 speed answer makes sense as it's easy to remember. Best glide speed on a big airplane will vary much more and would need to be calculated based on weight estimate at that point in the flight.

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In this answer, we'll confine ourselves to steady linear flight with the wings level.

We can show that the highest glide ratio (i.e. smallest or flattest glide angle) occurs at the highest ratio of L (Lift) to D (Drag). In fact we can show that the glide ratio is exactly the same as the ratio of L to D. We can also show that the highest ratio of L to D occurs exactly at the point of minimum D. (L is nearly constant and nearly equal to Weight; more precisely, L = Weight * cosine glide angle.)

I found the best glide speed of cessna 152 is 60 kts. My question is why it is not more or less than 60 kts? What are the factors upon which the best gliding speed is determined?

Because that is where D (Drag) is at a minimum. Nothing else matters.

Keep in mind that Drag is proportional to the product of the drag coefficient-- which increases as we increase the angle-of-attack-- and the square of the airspeed.

Other answers give additional insight as to why the point of minimum Drag occurs where it does-- it is the point where Induced Drag and Parasite Drag are exactly equal.

By the way, the ratio of L/D is exactly equal to the ratio of Cl / Cd, where Cl and Cd are the coefficients of Lift and Drag. Unlike Lift, Cl is not nearly constant across a wide range of angles-of-attack or airspeeds. It is always much greater at high angles-of-attack (low airspeeds), at least until we get very close to the stall angle-of-attack. Cd is also always much greater at a high angle-of-attack (low airspeed) than at a low angle-of-attack (high airspeed). So we can also say that the maximum glide ratio occurs at the maximum ratio of Cl to Cd. Lest there be any confusion, we should also observe that this is the point of minimum Drag, but not the point of minimum drag coefficient. If we were flying at the angle-of-attack for the maximum glide ratio, and we then moved the stick or yoke forward to reduce the angle-of-attack (and increase the airspeed), this would reduce the drag coefficient (as well as the lift coefficient), but this would also increase Drag, and would decrease L/D and Cl / Cd.

In comments you indicate that you want to know how best glide speed varies with weight. A good approximation is that it scales up or down in proportion to the square root of any change in your wing loading. In gliders where we don't have drag from a windmilling prop to complicate things-- or in any airplane where we've stopped the prop-- this should be exactly true, except for effects due to changes in Reynolds number.

For completeness, we should say something about where the minimum power-off sink rate occurs. That turns out to happen at the angle-of-attack that gives the maximum value of (Cl cubed) / (Cd squared). This always occurs at a higher angle-of-attack (lower airspeed) than does the best glide ratio.

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