What is the difference between Vs and Vs1g?

Question

As the name suggests, what is the difference between Vs1g and Vs? If both of them are referring to the stall condition, why they aren't the same? Which one is usually greater and why? Why do we need a distinction between them?

Answer 1

User Sidestick_n_Rudder posted on PPRuNe (emphasis mine):

Vs1g is the lowest speed, at which the airplane can maintain 1g, i.e. level flight. It corresponds to the Cl max

Vs is the lowest speed attained during stall testing of the airplane. The pilots were able to reach this speed lower than Vs1g, but the plane was already losing altitude.

The way I understand it, the certification rules were changed at some point, and new airplanes are required to base their performance on Vs1g, whereas airplanes certified earlier were referencing to Vs.

However, this puts the new airplanes at a disadvantage, as Vs1g is higher than Vs. To counter this, Airbus managed to convince the certifying authorities, to allow for lower margins during performance calculations (eg. Vref is 1.23*vs1g instead of 1.3), because the airplane is speed protected anyway....

P.S. strangely enough, ATR-42/72 also use Vs1g and lower safety margins, even though they are a conventional plane, not speed-protected

I found another resource (AviationThrust) that mentions (emphasis mine):

Vs1g and Vs are two speeds majorly abbreviated for stalling conditions.

• Vs1g corresponds to maximum lift coefficient i.e. Clmax. Clmax is the condition when angle of attack is maximum and load factor is equal to one.
• Vs corresponds to conventional stall condition. This is the time when lift suddenly collapses and load factor is less than one.

Vs is less than Vs1g

We usethe definition of lift from here: $$ L = C_L \frac{1}{2} \rho V^2 S \tag{1}\label{eq1}$$ and the definition of the load factor: $$ n = \frac{L}{W} \tag{2}\label{eq2}$$

Plugging \eqref{eq1} in \eqref{eq2} gives:

$$ n \cdot W = C_L \frac{1}{2} \rho V^2 S \tag{3}\label{eq3}$$

Rewriting for $V$ gives:

$$ V = \sqrt{\frac{n \cdot W}{C_L \frac{1}{2} \rho S}} \tag{4}\label{eq4}$$

Equation \eqref{eq4} can be satisfied in different ways:

1. Initially, there was only Vs, with no requirement on n, so set n to any value (with n<1 meaning falling from the air) to reach $V_s$

$$ V_{s} = \sqrt{\frac{(n<1) \cdot W}{C_{L} \frac{1}{2} \rho S}} $$

2. The regulations were updated to define straight flight, so set n to 1 and $C_L$ to $C_{L_{max}}$ to calculate $V_{s_{1g}}$

$$ V_{s_{1g}} = \sqrt{\frac{1 \cdot W}{C_{L_{max}} \frac{1}{2} \rho S}} $$

For $V_{s_{1g}}$, we always require 1-g condition, meaning that lift and weight are balanced. For $V_s$, there is no such requirement. By setting the requirement for 1-g, it is possible to more rigidly define the stall speed.

When slowing down in level flight, you will first experience $V_{s_{1g}}$, then the aircraft will start to fall (n<1) and you will reach $V_s$. Therefore, $V_s$ is lower than $V_{s_{1g}}$, see the image from this Quora link

(In the image the CAS arrow is in the wrong direction, right is lower CAS)