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.


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... $

| improve this answer | |
  • 1
    $\begingroup$ Do you have some kind of authoritative source where that 1.32 number came from? $\endgroup$ – Terran Swett Nov 19 '18 at 17:36
  • 2
    $\begingroup$ @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
  • $\begingroup$ Please provide data from a real aircraft to support these calculations. $\endgroup$ – Robert DiGiovanni Apr 1 at 8:10
  • $\begingroup$ In actual practice in some lower-performance sailplanes like Schweizer 2-33 it seems the difference between min sink speed and still-air best glide speed is lower than this calculation would indicate. Not sure exactly why, or how this answer could best be improved to account for such variations. $\endgroup$ – quiet flyer Apr 1 at 12:43

One answer by a reputable website is "about half way between Vx and Vy".

Observation of glide polars shows Vbg to be slightly faster than "minimum sink rate".

Still an art in need of a little more science to explain, but not to apply.

Let's compare a flying squirrel to an albatross. Little doubt who can glide farther.

The flying squirrel uses mostly angled drag to move horizontally when "gliding", the albatross goes a little faster and uses its wing to move horizontally.

There is no set rule as to the magnitude of the lift vector when gliding. We simply use the Velocity of Best Gliding Distance, Vbg to glide as far as we can.

Vbg will vary from plane to plane, even before considering wind effects.

| improve this answer | |
  • 1
    $\begingroup$ "There is no set rule as to the magnitude of the lift vector when gliding." -- except that in a steady-state linear glide, lift is equal to weight * cosine glide angle. That's a significant rule I'd say. $\endgroup$ – quiet flyer Apr 1 at 15:27
  • $\begingroup$ @quiet flyer Roger that. I'm trying to picture it from an aircraft reference, which uses the weight component to offset the drag component (of the line of flight). The lift component would increase if V increased. From the original glide path this would "climb" the plane futher down range. $\endgroup$ – Robert DiGiovanni Apr 1 at 16:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.