Tendencies

reynolds_master · January 11, 2015, 02:44

Hi! How should the pitching moment (mca) AND the pitching moment coefficient (cmca) respect to the aerodynamic center vs the Angle of Attack and velocity behave?

Thanks!

t.teschner · January 11, 2015, 09:15

it depends on the aircraft and what sort of stability you require. for passenger / cargo airplanes you require static stability while military aircrafts (jets) may have different priorities such as maneuverability which in turn can have an unstable static nature.

for static stability (which i assume you are after?!) you would require that the derivative of the pitching moment coefficient

over the angle of attack $\alpha$ is less than zero or $\frac{d c_m}{d \alpha} < 0$ .
this follows from the fact that if you pitch the aircraft nose up (increase of $\alpha$ ) you want a restoring moment that brings the aircraft back into its original state, hence $\frac{d c_m}{d \alpha}$ needs to be negative.
now using a bit of math (chain rule) you can express the derivative in terms of the lift coefficient as

$\frac{d c_m}{d \alpha} = \frac{d c_m}{d c_l} \frac{d c_l}{d \alpha}$

we know that the second derivative $\frac{d c_l}{d \alpha}$ is positive (in the linear region as we assume that we are not operating the aircraft in stall conditions) and hence since $\frac{d c_m}{d \alpha}$ needs to be negative but $\frac{d c_l}{d \alpha}$ is positive, $\frac{d c_m}{d c_l}$ needs to be negative as well. the derivative of the pitching moment over the lift coefficient is just a simple linear function with a (positive) intersection of the

(pitching moment) axis. since the slope needs to be negative, it will intersect the

(lift coefficient) axis somewhere with a positive value. the intersection with the

axis is your trimmed condition and is a very important point. if you increase

(for example increase of $\alpha$ ),

gets negative (restoring moment, i.e. pitch down) but if you reduce

(for example reduction of $\alpha$ ) you get a positive

(restoring moment, i.e. pitch up).

you could also look at this from a velocity point of view. increasing the velocity will increase your lift (for the same $\alpha$ and

) since

$L = c_l (\frac{1}{2}\rho u^2 A)$

but since the lift needs to stay constant for trimmed flight, you have to lower your $\alpha$ (and therefore

). different $\alpha$ have different trim states so for each $\alpha$ there is a

for which the aircraft has static stability. in the same sense, lowering the velocity will increase your $\alpha$ and

.

now the pitching moment with respect to the aerodynamic center is constant (or its derivative over

is zero i.e. $\frac{d c_m}{d c_l} = 0 |_{ac}$ ). so changing $\alpha$ ,

or

does not change your pitching moment (at the aerodynamic center). so if this is what you was looking for then i think I explained it too much in detail but for your understanding, what i described above holds in general but there exist exactly one point on the airfoil for which all of that becomes redundant and that point is defined as the aerodynamic center.

January 11, 2015, 02:44	Tendencies	#1
reynolds_master New Member Join Date: Dec 2014 Posts: 5 Rep Power: 11	Hi! How should the pitching moment (mca) AND the pitching moment coefficient (cmca) respect to the aerodynamic center vs the Angle of Attack and velocity behave? Thanks!

January 11, 2015, 09:15		#2
t.teschner Senior Member Tom-Robin Teschner Join Date: Dec 2011 Location: Cranfield, UK Posts: 204 Rep Power: 16	it depends on the aircraft and what sort of stability you require. for passenger / cargo airplanes you require static stability while military aircrafts (jets) may have different priorities such as maneuverability which in turn can have an unstable static nature. for static stability (which i assume you are after?!) you would require that the derivative of the pitching moment coefficient over the angle of attack $\alpha$ is less than zero or $\frac{d c_m}{d \alpha} < 0$ . this follows from the fact that if you pitch the aircraft nose up (increase of $\alpha$ ) you want a restoring moment that brings the aircraft back into its original state, hence $\frac{d c_m}{d \alpha}$ needs to be negative. now using a bit of math (chain rule) you can express the derivative in terms of the lift coefficient as $\frac{d c_m}{d \alpha} = \frac{d c_m}{d c_l} \frac{d c_l}{d \alpha}$ we know that the second derivative $\frac{d c_l}{d \alpha}$ is positive (in the linear region as we assume that we are not operating the aircraft in stall conditions) and hence since $\frac{d c_m}{d \alpha}$ needs to be negative but $\frac{d c_l}{d \alpha}$ is positive, $\frac{d c_m}{d c_l}$ needs to be negative as well. the derivative of the pitching moment over the lift coefficient is just a simple linear function with a (positive) intersection of the (pitching moment) axis. since the slope needs to be negative, it will intersect the (lift coefficient) axis somewhere with a positive value. the intersection with the axis is your trimmed condition and is a very important point. if you increase (for example increase of $\alpha$ ), gets negative (restoring moment, i.e. pitch down) but if you reduce (for example reduction of $\alpha$ ) you get a positive (restoring moment, i.e. pitch up). you could also look at this from a velocity point of view. increasing the velocity will increase your lift (for the same $\alpha$ and ) since $L = c_l (\frac{1}{2}\rho u^2 A)$ but since the lift needs to stay constant for trimmed flight, you have to lower your $\alpha$ (and therefore ). different $\alpha$ have different trim states so for each $\alpha$ there is a for which the aircraft has static stability. in the same sense, lowering the velocity will increase your $\alpha$ and . now the pitching moment with respect to the aerodynamic center is constant (or its derivative over is zero i.e. $\frac{d c_m}{d c_l} = 0 \|_{ac}$ ). so changing $\alpha$ , or does not change your pitching moment (at the aerodynamic center). so if this is what you was looking for then i think I explained it too much in detail but for your understanding, what i described above holds in general but there exist exactly one point on the airfoil for which all of that becomes redundant and that point is defined as the aerodynamic center.