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Lift and drag forces scale as the square of the relative flow speed, $v$. But speed is the magnitude of the velocity vector; that is, speed is the square root of the sum of the squared velocity vector components, by definition.

Why does NASA say that $v^2$ is velocity squared, rather than speed squared?

In aerodynamics, is the sum of squared velocity vector components referred to as velocity squared?

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    $\begingroup$ Where does NASA say that? $\endgroup$ Jul 9 at 3:31
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    $\begingroup$ Cf. the generalized use of airspeed instead of air velocity. Yes it is sometimes found on Nasa site, e.g. "The air velocity is the relative speed between the kite and the air. When the kite is held fixed by the control line, the relative air velocity is the wind speed". $\endgroup$
    – mins
    Jul 9 at 12:00
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    $\begingroup$ You really need to include a specific quote or a specific citation in the question itself. I'm sure that if you looked through the vast array of material NASA has published over the decades, you could find some instances of NASA "saying" that v represents "airspeed", or "speed". $\endgroup$ Jul 9 at 14:10
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    $\begingroup$ Also, instead of phrasing your question "why does NASA say" (who knows why anyone does anything), it might perhaps be better to phrase it more like "would it be technically more correct to say that v represents airspeed rather than velocity in this equation"-- or something like that-- but maybe that's not the question that you are most interested in having answered-- $\endgroup$ Jul 9 at 14:11
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    $\begingroup$ I don't entirely understand what your concern is. The square of a vector is generally taken to mean the dot product of the vector with itself, and for velocity and speed v dot v = |v|^2 = speed^2. They're the same thing, just different names $\endgroup$
    – llama
    Jul 9 at 17:35
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The proper technical term for the quantity is velocity, and it is a vector quantity. Speed is a colloquial name. If physicists use it, they only use it for the magnitude of velocity, but most of the time there is absolutely no reason to use it because the quantity is a vector one and acts as a vector one in all equations.

Now square of a vector quantity is the same as square of its magnitude, basically by definition, because magnitude is the square root of the square (in Euclidean vector spaces), so it does not really matter.

So given the magnitude is not actually use anywhere, why would they suddenly use the term speed if they don't use it anywhere else?

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  • $\begingroup$ re "given the magnitude is not actually use anywhere" -- I would suggest that the equation that appears in the link referenced in a comment by the asker of the question, grc.nasa.gov/www/k-12/airplane/lifteq.html , only works if you treat V and L as scalars, not vectors. If you treat V as a vector, don't you end up with L as a vector too, and pointing in the wrong direction, namely parallel to V? $\endgroup$ Jul 9 at 14:17
  • $\begingroup$ @quietflyer It also works if you treat CI as a non-scalar, say a rotation and scaling matrix. $\endgroup$
    – Yakk
    Jul 9 at 14:57
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    $\begingroup$ "Speed is a colloquial name." Pretty sure it's the technical name for a scalar value. $\endgroup$
    – nick012000
    Jul 9 at 16:26
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    $\begingroup$ @JanHudec "squaring" the velocity vector to get speed squared makes sense when thinking of squaring as taking the dot product of the velocity with itself. Llama's comment above clears it up for me. Thanks, $\endgroup$
    – user59327
    Jul 9 at 18:30
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    $\begingroup$ @user59327 Squaring vectors in Euclidean spaces (that physics uses to model space) is indeed defined as taking dot product with itself. And magnitude as square root of that dot product with itself. $\endgroup$
    – Jan Hudec
    Jul 9 at 18:36
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Because velocity (being a vector) is a little more precise than pure (scalar) speed - so the angle between the object (aircraft, airfoil, ...) and the incoming air is also taken into consideration.

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  • $\begingroup$ This doesn't sound right; you're talking about the angle of attack, which is accounted for in the lift and drag forces already, e.g. in the coefficient of lift / drag. But scaling as speed squared is different from scaling as velocity squared. $\endgroup$
    – user59327
    Jul 9 at 3:48
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    $\begingroup$ @user59327 speed squared and velocity squared is the same thing, because speed is a square root of velocity squared, by definition! $\endgroup$
    – Jan Hudec
    Jul 9 at 5:20
  • $\begingroup$ I would suggest that the equation that appears in the link referenced in a comment by the asker of the question, grc.nasa.gov/www/k-12/airplane/lifteq.html , only works if you treat V and L as scalars, not vectors. If you treat V as a vector, don't you end up with L as a vector too, and pointing in the wrong direction, namely parallel to V? $\endgroup$ Jul 9 at 14:20
  • $\begingroup$ @quietflyer no, it's a vector quantity: en.wikipedia.org/wiki/Lift_(force) $\endgroup$
    – eps
    Jul 9 at 15:22
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It’s just because it is derived from a larger more complete theory of aerodynamic forces where it’s necessary to make the distinction. In this equation the coefficient is simplified but a more complete calculation of the coefficient uses the directional components of velocity. https://www.grc.nasa.gov/www/k-12/airplane/eulereqs.html

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TL:DR

Relative flow over a lifting body has to be described as a vector because Lift and Drag are defined wrt the relative flow vector, and Angle of Attack is a parameter.

Is the sum of squared velocity vector components referred to as velocity squared, in aerodynamics?

Yes. Velocity is used to refer to the vector as well as to it's scalar magnitude in scalar formula. This isn't sloppy, it's compact, and more importantly, it maintains the connection to the underlying vector domain.

$$ =========== $$ The squaring-something operator actually represents a large number of different algorithms that are specific to the thing being squared.

Sometimes, there is more than one way to square a particular thing. For instance, a (scalar) distance squared might represent an area, or it might not. But the same symbol is used for the operation, and the associated units are often rendered identically. Sometimes it is important to keep track of the distinction and sometimes it isn't. An area has properties that the square of a distance does not have. Just because two things have the same units doesn't mean they have the same set of properties.

The same applies to velocities and speeds. Both speed and velocity have the same units, but they have different properties. Both can be squared, but the procedures are different.

So to get square of the speed of the relative flow, you take the magnitude of the relative flow, then square it. To get the square of the velocity of the relative flow, you square the velocity (a vector operation), and then take the magnitude if you want a scalar.

It's not luck that $ scalar square ( magnitude (vector))$ and $ magnitude (vector square ( vector)$ have the same numeric value and units, The operations were designed to work like that.

But do they have the same properties? Not really. When talking about velocity as a scalar, we are reminded that the domain is a vector domain and that different vectors can have the same magnitude.

Which brings us to a problem, different relative flow vectors with the same magnitude do not have the same Lift and Drag as each other. The only way the formula is valid is if the angle between the Lift and inflow remains fixed and the angle between the drag and the inflow remain fixed, which they do by definition. But all these have to remain fixed with respect to the lifting body as well. So the statement is only valid for constant angles of attack. To suddenly introduce a vector constraint on an entirely scalar chain of math is plain bad form.

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