New answers tagged drag
1
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 ...
19
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 ...
3
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
0
Stalling increases drag dramatically indeed. The pic above is the $C_D$ over 180° from this answer, which also contains the corresponding graph for the $C_L$. At about 12° we can see the huge jump in $C_D$, after which the curve follows the shape for flat plate lift.
The jump in drag is particularly drastic for the NACA 0012 because it falls in the category ...
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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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Drag is just the component of the aerodynamic force on the entire airframe that is aligned with the aircraft's velocity Vector. (as lift is the component normal to or perpendicular to the velocity). When an aircraft stalls, the vectoral summation of all aerodynamic forces is pointing aft, with a much greater angle (think about pushing a door through the air ...
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You would solve this problem by drawing the lift vector being tilted back at half the downwash angle, then apply some trigonometry to find the horizontal bit of the lift vector.
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Yes they are dimensionless numbers, which does not mean that they are constants. $C_L$ $C_D$ are variables. Dimensionless meaning: no physical unit. $$L = C_L \cdot \frac{1}{2} \rho V^2 \cdot A$$ with metric units:
L [N] = [kg*m/sec$^2$]
$\rho$ [kg/m$^3$]
V [m/sec]
A [m$^2$]
Dimension of $\rho V^2 \cdot A$ = $\frac{kg}{m^3} \cdot \frac{m^2}{s^2} \cdot m^2 =...
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