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I've seen an equation related to stall and angle of attack in a few places like this patent that goes more or less:

$$ \frac{CAS_{alert}}{CAS_{current}} = \sqrt{\frac{C_{L,current}}{C_{L,alert}}} $$

Where CAS is your airspeed and C_l is the coefficient of lift at certain angle of attack.

It seems intuitive, for example if you are in unaccelerated flight and using 90% of your available lift then you are about 5% over your alert speed. However I'm confused about a few details like why there's a square root in the equation, or whether there are scenarios where this is just an approximation and it breaks down. Where does this speed ratio come from?

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Not sure the complete context of your question but much information can be found at https://www.ecfr.gov/current/title-14/chapter-I/subchapter-C/part-25/subpart-B/subject-group-ECFR1cd58fc7d4bfee0/section-25.207 for part 25 certified aircraft. With regards to your equation, it does correctly reflect the proportionality between CL and airspeed for 1-g level flight with the caveat that technically, it's for Vtrue, not Vcas. {CL = W/(rhoVtVt*Sref/2)}

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  • $\begingroup$ $\rho$ times $\text{TAS}^2$ is approximately $\text{CAS}^2$. (More accurately it's equivalent airspeed squared). $\endgroup$
    – Chris
    Commented May 17 at 21:28
  • $\begingroup$ Ve = Vt * sqrt(sigma) = Vt * sqrt(rho/rhoSL) $\endgroup$
    – AeroAndy
    Commented May 17 at 21:47
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It's a simple scaling law. The lift is proportional to $C_L$ times the square of the equivalent airspeed. In other words, for a given load factor, $C_L(\text{EAS})^2$ is a constant. So $\frac{\text{EAS}_1}{\text{EAS}_2}=\sqrt{\frac{C_{L,2}}{C_{L,1}}}$ is true for any two flight conditions at the same load factor.

At lower mach numbers equivalent and calibrated airspeeds are the same so they are often used interchangeably.

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