1
$\begingroup$

For piston propeller aircraft, Vy calibrated airspeed[CAS] is equal to CAS for best Lift/drag at the plane's absolute ceiling, and at lower altitudes Vy increases above best L/D as excess power is available.

For example a Cessna 172m has a best glide(approximation of best L/D) of 80mph at a MTOW of 2300 pounds, but a Sea level Vy of 90mph. Now if you operate at 1700 pounds the L/D would be reduced approximately to $\sqrt{17/23}\times{80}= 68.8mph$ (Using mph because the 172M poh uses mph)

It would seem reasonable that sea level Vy would also be reduced by a similar factor if achieving some optimal AOA is the goal.$\sqrt{17/23}\times{90}= 77.4mph$ for the c172m example.

However the lighter weight increases the ceiling (where best L/D and Vy are equal), and because they start diverging higher the relative speed difference between best L/D and Vy will be greater at any given lower altitude.

So do the two sides fully cancel out leaving Vy unchanged? (90mph in the 172m example) What is the correct formula to make this adjustment? (exact or reliable approximation) Do you have a reference to empirical demonstrations in support of, or in lieu of a formula?(Something with reasonably accurate metrology, not eyeballing a typical VSI)


My primary question is only about Vy. For Vx everything implies lower speed at lower gross weight, though it would be somewhat interesting to know if a basic formula exists for a Vx adjustment. Setting aside possible complications like adequate cooling at continuous low speeds.

$\endgroup$

1 Answer 1

1
$\begingroup$

Vx and Vy are a combination of effects between the engine performance, the propeller performance, and the aerodynamics of the airplane.

Best L/D is just determined by the aerodynamics of the airplane.

If the prop is fixed pitch, it will behave much like the airplane. If it is constant speed, it can be a bit more complex, but it still behaves mostly like the airplane.

We generally ignore the performance of the engine.

So, that leaves us with the aerodynamics of the aircraft. This is the dominant factor.

Vx, Vy, and best L/D each happen at a certain lift coefficient.

For a simplified piston-prop,

Vy occurs at best $\frac{{C_L}^{3/2}}{C_D}$. Which, for a simple parabolic drag polar occurs when $C_L=\sqrt{\frac{3\,C_{D,0}}{K}}$.

Best L/D occurs at best $\frac{C_L}{C_D}$ (obviously). Which, for a simple parabolic drag polar occurs when $C_L=\sqrt{\frac{C_{D,0}}{K}}$.

Your error is that the point of absolute ceiling occurs at best $L/D$. For a simplified piston propeller aircraft, we assume that the power available is constant with airspeed. Therefore, ceiling will happen at the speed for minimum power required.

In a jet aircraft, we assume that the thrust available is constant with airspeed -- in that case, ceiling will happen at the speed for minimum drag (best L/D).

The rest of this should be familiar to you -- I typed it up before I re-read your question to focus on the issue of ceiling...

$C_L = \frac{W}{q\,S}$

So, we want to look at what happens when we change $W$ but keep $C_L$ constant.

Re-arranging for dynamic pressure

$q = \frac{W}{C_L\,S}$

We see that any increase in $W$ needs a proportional increase in $q$. The dynamic pressure is:

$q = 0.5\,\rho\,V^2$

Substituting gives

$0.5\,\rho\,V^2 = \frac{W}{C_L\,S}$

Solving for V gives

$V = \sqrt{\frac{2\,W}{\rho\,C_L\,S}}$

So, any velocity flown at the same density altitude ($\rho$) and a fixed lift coefficient (say that for Vx, Vy, or best range) will scale with the square root of weight -- because everything else in the equation is constant.

I.e. you could write this equation twice -- once for a current condition, and a second time for a reference condition. Then divide current by reference and cancel whatever is constant. You will be left with this....

$\frac{V}{V_\mathrm{ref}} = \sqrt{\frac{W}{W_\mathrm{ref}}}$

(at the same density altitude)

This was done in terms of true airspeed, but it works out the same in terms of equivalent airspeed. Just use the equation for dynamic pressure in terms of equivalent airspeed instead.

$q = 0.5\,\rho_0\,V_e^2$

This works out to

$\frac{V_e}{V_{e,\mathrm{ref}}} = \sqrt{\frac{W}{W_\mathrm{ref}}}$

Which, due to the magic of equivalent airspeed, is true for any density altitude.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .