How to calculate the best glide speed if there is none on the POH?

Question

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.

Answer 1

According to the FAA, the short answer to your question is approximately "halfway between Vx and Vy."

While there is a lot of physics and calculus that can give a precise answer to this question, there is another answer that uses "pilot math". The FAA has a nice publication titled Best Glide Speed and Distance, which explains that there are two different types of "best" glide speeds.

1. The glide speed that gives you maximum distance covered for a given altitude (Best Glide Distance); and

2. The glide speed that gives you the maximum time in the air (Best Glide Time).

Per the FAA document, the glide speed that gives you the most distance covered is approximately halfway between Vx and Vy. For example, on a C172 with Vx at 53 and Vy at 73 Max Glide would be about 65. The speed provided in POHs is usually calculated at Max Gross Weight, so your actual Best Glide Distance speed will be a little lower if your weight is less than max gross.

Alternatively, if your goal is to stay in the air as long as possible, then you'll want a speed that minimizes the descent rate. This is typically slower than Best Glide Distance speed, and can be easily determined by pitching for an airspeed that gives you the lowest vertical descent speed on the VSI. There are times when you don't care how far you glide, and instead want to maximize the time in the air in order to troubleshoot. For example, if you are over an area with no suitable landing location (like over the ocean), you'll want to maximize your time to try to restart your engine. Or if you are over your landing location, but want extra time to finish your emergency checklists.

Answer 2

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... $