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I am studing the POH of Seneca II and could not find the Best Glide Speed. Where can I get this information from? I thought to multiply Vs clean for 1.4 to get an approximate result but I am not sure if this is the correct way to do so.

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Gliding clean, with the engine at idle, you may find the IAS for maximum endurance, precisely where the variometer shows the minimum descent speed. Then multiply that IAS by 1.32 That's the best glide speed. The derivation of 1.32 follows. It was written by DeltaLima for an answer elsewhere on aviation.stackexchange.com.

The ratio is the same for all aeroplanes if you accept a number of assumptions:

  • The propulsion efficiency is constant, regardless of speed or power setting
  • Aerodynamic drag is the sum of parasite drag and induced drag
  • Parasite drag is proportional to the square of airspeed:$ D_p = k_p \cdot V^2$
  • Induced drag is inversely proportional to the square of airspeed: $D_i = \frac{k_i }{V^2}$
  • There is no wind

Since we assume efficiency is constant, the fuel consumption rate is directly proportional to the power. Power required is drag times airspeed: $P = D\cdot V = D_p\cdot V + D_i\cdot V = > k_p\cdot V^3 + \frac{k_i}{V}$

For the maximum endurance we need to minimise the fuel consumption and thus we need to find the speed that minimises the power.

$\frac{dP}{dV} = \frac{1}{3} k_p V^2 - > \frac{k_i}{V^2} = 0$ Solving for $V$ results in $V_{endurance} = > \sqrt[\uproot{1}4]{ 3\frac{k_i}{k_p}} $

For the maximum range we need to find the speed that minimises the fuel consumption per distance travelled, which is found when the ratio of power to speed over ground is minimal. As we assume there is no wind, the ground speed and the airspeed are equal. Since the ratio of power to airspeed is drag, we have to find the speed for minimum drag:

$\frac{dD}{dV} = 2 k_p V - 2\frac{k_i}{V^3} = 0$

Solving for $V$ results in $V_{range} = \sqrt[\uproot{1}4]{ \frac{k_i}{k_p}} $

We can now show that the ratio of maximum endurance speed to maximum range speed is: $\frac{V_{endurance}}{V_{range}} = \left. > \sqrt[\uproot{1}4]{3 \frac{k_i}{k_p}} \middle/ \sqrt[\uproot{1}4]{ > \frac{k_i}{k_p}} \right. = \sqrt[\uproot{1}4]{ 3} = 1.316... $

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    $\begingroup$ Do you have some kind of authoritative source where that 1.32 number came from? $\endgroup$ – Terran Swett Nov 19 '18 at 17:36
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    $\begingroup$ @TannerSwett: Look here for the answer. $\endgroup$ – Peter Kämpf Nov 19 '18 at 17:55
  • $\begingroup$ Is it possible to calculate it from a speed given by the POH? $\endgroup$ – Andrea Ghilardi Nov 20 '18 at 8:06
  • $\begingroup$ Links are great, but generally a good idea to include the most relevant part(s) in the answer itself. Links rot. Answers do not. Also, some reference for the 1.32 number would be great. $\endgroup$ – Jamiec yesterday
  • $\begingroup$ Please provide an an example of this "math" as compared with data from a real aircraft. $\endgroup$ – Robert DiGiovanni 8 hours ago
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Velocity of Best Gliding Distance Vbg is generally in a clean configuration and will be dependent on weight and headwinds.

Manufacturers of larger planes with multiple engines and a wider weight range (depending on cargo, passengers, and fuel load) seem to avoid publishing "Vbg", as it will vary far greater than a single engine trainer.

Even though best gliding distance is similar heavy or light, best gliding speed is variable. Headwinds demand greater speed as well. While a listed range would be desirable, practical experience is best.

For example, if the Piper PA-34 Seneca II stall clean is around 60 knots, Vx may be around 78 knots, Vy 91 knots. These values vary by weight as well, but should be in the POH.

Although a double engine out is rare, one may wish to start at around 85 - 90 knots and trim it from there. Just like landing, pick out a stationary point of reference, and try a slight change in glide speed. If it moves up, you've gone farther away from Vbg. If it moves down, the new glide speed is closer to Vbg.

Mathematicly, the effect of weight on Vbg can be derived from the Lift equation:

$Lift$ = Coefficient of Lift × Rho × Area × Velocity$^2$.

We derive the % change in Vbg airspeed as the the square root of the % change from the listed weight of the aircraft for the original Vbg.

Vbg values can also be derived $in situ$ (while flying) due to wind effects, which will determine ground speed. There for, the landing site downwind may be easier to reach by way of emergency glide.

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  • $\begingroup$ I wonder why the downvotes? The answer is correct about weight and wind effects, though published best glide airspeed numbers undoubtedly always presume still air, as the asker of the question likely did as well. $\endgroup$ – quiet flyer 5 hours ago
  • $\begingroup$ Seem to have hit a raw nerve here. When a plane loses power, it must pitch forward to maintain airspeed. The amount of vertical lift to maintain steady state flight is the same. This will be around the $cosine$ of the pitch angle × glide speed = 1 × minimum power airspeed. $\endgroup$ – Robert DiGiovanni 3 hours ago

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