Coordinate systems and Newton’s laws of motion

The right-handed Cartesian coordinate system OXYZ is fixed to the Earth, origin at point O, and it is assumed that this is an inertial system. That is, Newton’s laws of motion are valid in the Earth axis system. The unit base vectors in the OXYZ system are denoted by \((\hat{I}, \hat{J}, \hat{K})\). The right-handed Cartesian coordinate system Gxyz is fixed in the body of the aircraft, with its origin at the center of gravity, which is labeled G. The unit base vectors in the Gxyz system are denoted by \((\hat{i}, \hat{j}, \hat{k})\). Consider planar motion of the aircraft – that is, symmetrical maneuvers of the aircraft, and where the aircraft is symmetrical about its vertical fore and aft plane. Body axis Gx is directed forward, axis Gy is directed toward the starboard wing, and body axis Gz is in the normal direction. For symmetrical maneuvers there is no roll or yaw of the airplane, so symmetrical maneuvers imply \(\hat{j}=\hat{J}\) for all time t. Let \(\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{R}}}_{G}\) denote the position vector of the center of gravity G with respect to fixed point O. The flight path is a plane curve in the X-Z plane with the arc-length along the curve denoted by s. The unit tangent vector to the flight path at s is denoted by \(\hat{t}\), the unit normal vector at s by \(\hat{n}\), and the angle between the flight path and the unit tangent vector, or the x-axis, by \(\theta\). Note that \(\hat{t}=\hat{i}\). Angle \(\theta\) represents the clockwise rotation of the aircraft in pitch. See figure [fig2.1].

The unit tangent vector and its derivative along the path are \[\begin{aligned} \label{eq2.1} \hat{t}=\frac{d \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{R}}}_{G}}{d s}\mbox{, and }\frac{d \hat{t}}{d s}=\frac{d \theta}{d s} \hat{n}.\end{aligned}\] Let \(\theta^{\prime}=d \theta/d s\). The change in slope of the flight path with respect to arc-length \(\theta^{\prime}\) defines the curvature of the path, and the radius of curvature is \(R=1/\theta^{\prime}\). The velocity of the center of gravity G along the flight path is \[\begin{aligned} \label{eq2.2} \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{v_{G}}}}=\frac{d \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{R}}}_{G}}{d t}=\frac{d \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{R}}}_{G}}{d s} \frac{d s}{d t}=\hat{t} v_{G},\end{aligned}\] where the speed of the center of gravity of the aircraft along the flight path is \(v_{G}=d s/d t\). The acceleration of the center of gravity is \[\begin{aligned} \label{eq2.3} \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{a_{G}}}}=\frac{d \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{v_{G}}}}}{d t}=\frac{d}{d t}(v_{G} \hat{t})=\dot{v}_{G} \hat{t}+v_{G} \frac{d \hat{t}}{d s} \frac{d s}{d t}=\dot{v}_{G} \hat{t}+v_{G}^{2} \theta^{\prime} \hat{n}.\end{aligned}\] where \(\dot{v}_{G}=d v_{G}/d t\) is the acceleration component tangent to the path. The acceleration component normal to the path \(v_{G}^{2} \theta^{\prime}=v_{G}^{2}/R\), or centripetal acceleration, is directed toward the concave side of the path.

A free body diagram of the aircraft at time t and its time rate of change of momenta are shown in figure [fig2.2].

Derivatives with respect to time of the pitch angle are written as \[\begin{aligned} \label{eq2.4} \frac{d \theta}{d t}=\dot{\theta}\mbox{, and }\frac{d^{2} \theta}{d t^{2}}=\ddot{\theta}.\end{aligned}\] The mass of the aircraft is denoted by m, the moment of inertia about the center of gravity by \(I_{y}\), the local acceleration due to gravity by \(g\), and the weight of the aircraft by W where \(W=m g\). Equations for Newton’s second law at time t are \[\begin{aligned} \label{eq2.5} \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{F}}}=m\big(\dot{v}_{G} \hat{t}+v_{G}^{2} \theta^{\prime} \hat{n}\big) \quad \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{M}}}_{G}=I_{y} \ddot{\theta} \hat{j} \quad \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{M}}}_{\mathrm{O}}=\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{R}}}_{G} \times m\big(\dot{v_{G}} \hat{t}+v_{G}^{2} \theta^{\prime} \hat{n}\big)+I_{y} \ddot{\theta} \hat{j},\end{aligned}\] where the resultant force is denoted by \(\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{F}}}\), the moment about the center of gravity by \(\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{M}}}_{G}\), and the moment about the fixed point by \(\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{M}}}_{\mathrm{O}}\). These force and moment vectors are determined from \[\begin{aligned} \label{eq2.6} \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{F}}}=\sum_{m=1}^n \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{F}}}_{m} \quad \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{M}}}_{G}=\sum_{m=1}^n \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{r}}}_{m/G} \times \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{F}}}_{m} \quad \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{M}}}_{\mathrm{O}}=\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{M}}}_{G}+\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{R}}}_{G} \times \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{F}}},\end{aligned}\] where \(\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{r}}}_{m/G}\) is the position vector of the point of application of force \(\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{F}}}_{m}\) with respect to the center of gravity.

Principle of D’Alembert

D’Alembert in 1743 proposed a principle that would reduce a problem in dynamics to an equivalent one in statics by introducing so-called inertial forces. The inertial force acting at the airplane’s center of gravity is defined as \(-m \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{a}}}_{G}\), and the inertial moment about the center of gravity is defined as \(-I_{y} \ddot{\theta} \hat{j}\). These inertial forces are drawn on the free body diagram of the airplane in addition to all the applied forces. D’Alembert’s principle states that the applied forces together with the inertial forces form a system in equilibrium. Thus we write Newton’s second law as \[\begin{aligned} \label{eq2.7} \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{F}}}+\big(-m \dot{v}_{G} \hat{t}-m v_{G}^{2} \theta^{\prime} \hat{n}\big)=0 \quad \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{M}}}_{G}+\big(-I_{y} \ddot{\theta} \hat{j}\big)=0.\end{aligned}\] The free body diagram is modified accordingly as shown in figure [fig2.3]. In the free body diagram the inertial forces and moment are indicated by dashed lines. From the free body diagram we proceed as in statics to write the conditions of (dynamic) equilibrium.

The curvature of the flight path \(\theta^{\prime}\) can change sign. As shown in figure [fig2.4], the curvature is positive for a pull-up maneuver from a dive, and the curvature is negative for a push-down maneuver from a climb. Consequently, the inertia force normal to the flight path is directed toward the convex side of the path.

Relative velocity and acceleration

Often it is necessary to determine the inertial forces at locations within the airplane not coincident with the center of gravity. For these computations we need the relative velocity and acceleration formulas from rigid body dynamics. Consider two points A and G fixed in the body. The position of point A relative to G is taken as \(x_{A/G} \hat{i}\), as shown in figure [fig2.5].

The position vectors of points A and G are related by \[\begin{aligned} \label{eq2.8} \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{R}}}_{A}=\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{R}}}_{G}+x_{A/G} \hat{i}.\end{aligned}\] The velocity vectors of points A and G are then \[\begin{aligned} \label{eq2.9} \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{v}}}_{A}=\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{v}}}_{G}+\frac{d}{d t}(x_{A/G} \hat{t}).\end{aligned}\] Since vector \(x_{A/G} \hat{i}\) is embedded in the rigid body for all time, its rate of change is determined from its cross product with the vector of the time rate of change of pitch rotation. That is, \[\begin{aligned} \label{eq2.10} \frac{d}{d t}(x_{A/G} \hat{t})=\dot{\theta} \hat{j} \times x_{A/G} \hat{t}=x_{A/G} \dot{\theta} \hat{n}.\end{aligned}\] Hence, \[\begin{aligned} \label{eq2.11} \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{\nu}}}_{A}=\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{\nu}}}_{G}+ x_{A/G} \dot{\theta} \hat{n}.\end{aligned}\] The time rate of change of this velocity expression (2.11) relates the acceleration of A relative to G by \[\begin{aligned} \label{eq2.12} \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{a}}}_{A}=\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{a}}}_{G}+\frac{d}{d t}[x_{A/G} \dot{\theta} \hat{n}]=\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{a}}}_{A}+x_{A/G} \ddot{\theta} \hat{n}+\dot{\theta} \frac{d}{d t}[x_{A/G} \hat{n}]=\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{a}}}_{G}+x_{A/G} \ddot{\theta} \hat{n}+\dot{\theta}[\dot{\theta} \hat{j} \times x_{A/G} \hat{n}].\end{aligned}\] Perform the vector cross product in eq. (2.12) to find \[\begin{aligned} \label{eq2.13} \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{a}}}_{A}=\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{a}}}_{G}+x_{A/G} \ddot{\theta} \hat{n}-\dot{\theta}^{2} x_{A/G} \hat{t}.\end{aligned}\]

Uniform linear and angular accelerations

In some inertial load problems it is reasonable to assume that the acceleration of a particle and/or the angular acceleration of a rigid body are constant over a time interval. Let s denote the distance of a particle along a straight line, v its speed along the line, and a its constant acceleration. Then, we have the following formulas \[\begin{aligned} \label{eq2.14} v=a t+v_{0} \quad s=\frac{1}{2} a t^{2}+v_{0} t+s_{0} \quad 2 a\left(s-s_{0}\right)=v^{2}-v_{0}^{2},\end{aligned}\] where \(s=s_{0}\) and \(\nu=v_{0}\) at time \(t=0\). Similarly if the angular acceleration \(\ddot{\theta}\) is constant over some time interval, then \[\begin{aligned} \label{eq2.15} \dot{\theta}=\ddot{\theta} t+\dot{\theta}_{0} \quad \theta=\frac{1}{2} \ddot{\theta} t^{2}+\dot{\theta}_{0} t+\theta_{0} \quad 2 \ddot{\theta}\left(\theta-\theta_{0}\right)=\dot{\theta}^{2}-\dot{\theta}_{0}^{2},\end{aligned}\] where \(\dot{\theta}=\dot{\theta}_{0}\) and \(\theta=\theta_{0}\) at time \(t=0\).

Airplane design requirements

1. All parts of the airplane are designed so they are not stressed beyond the yield point at the limit load factor.

2. The airplane structure must carry the ultimate loads for at least 3 seconds without collapsing, even though the members may acquire permanent deformation.

Regulations

Limit load factors are specified by regulations, which depend on the type of aircraft (e.g., transport, aerobatic, etc.). Criteria for civil aerospace vehicles in the United States.

Code of Federal Regulations

Title 14, Aeronautics and Space

Parts 1–59

Federal Aviation Administration (Department of Transportation is the regulatory agency.)

Military requirements in the United States. issued in MIL–Specs covering specific topics of structural designof US Air Force, Navy, and Marine aircraft.

Specified data

Specified maximum positive load factor; \(n_{\max }>1\).
Specified maximum negative load factor; \(n_{\min }<0\).
Specified design airspeeds: \[\begin{gathered} V_{C}=\text{maximum level flight cruise speed }\\ V_{D}=\text { maximum dive speed } \sim 1.2 \text { to } 1.5 V_{C}\end{gathered}\]

Basic maneuver V-n diagram

This is predicated on pilot-controlled, symmetrical maneuvers in flight through calm air (i.e., no gust). Assumptions made for analytical purposes are that the pitching acceleration is assumed zero or negligible, the airspeed is constant during the maneuver, and there is no rolling or yawing of the aircraft, although rolling or yawing maneuvers may be considered in design as well. For no pitching acceleration the pitch rate \(\dot{\theta}\) is constant with respect to time. Use the chain rule to write the pitch rate as \[\begin{aligned} \label{eq2.23} \dot{\theta}=\frac{d \theta}{d t}=\frac{d \theta}{d s} \frac{d s}{d t}=\theta^{\prime} v_{G}.\end{aligned}\] Hence, in a steady state maneuver \(\theta^{\prime} v_{G}\) is constant with respect to time. The load factors ([eq2.21]) for a steady state maneuver are \[\begin{aligned} \label{eq2.24} n_{x}=\sin \theta \quad n_{z}=\left(\cos \theta+\frac{v_{G}^{2} \theta^{\prime}}{g}\right).\end{aligned}\] A pull-up from a dive, and a push-down from a climb are examples of steady state symmetrical maneuvers and are depicted in figure [fig2.4]. Also, a level flight, coordinated turn is considered a symmetrical maneuver even though the airplane does have a lateral acceleration in the turn. (Refer to practice exercise 2.) In general, the steady state symmetrical maneuvers will produce the maximum design wing loads. See figure [fig2.9].

Aerodynamic data

When a two-dimensional airfoil is subject to a relative wind there is a net pressure distribution over the airfoil that depends on the angle of attack, which is denoted by \(\alpha\). The angle of attack is the angle between the relative wind and the chord of the airfoil.The chord is the width of the airfoil and its length is denoted by c. The resultant action of the pressure distribution is a force R and no moment at the center of pressure, which is labeled C.P. in figure [fig2.10](a). The center of pressure location varies with the angle of attack. The resultant action of the pressure distribution is a force and a moment at any other location. The standard reference point for aerodynamic data is the aerodynamic center, which is labeled A.C. in figure [fig2.10](b). The aerodynamic center is the point where the pitching moment is independent of the angle of attack. For most subsonic wing sections the A.C. is around 25 percent of the chord.The net force of the pressure distribution is resolved at the aerodynamic center into a lift force perpendicular to the relative wind and a drag force parallel to the relative wind and the pitching moment.

The lift increases as the angle of attack increases. At some point, however, the flow can no longer stay attached to the upper surface and detaches. This results in a decrease in lift, which is called aerodynamic stall as shown in figure [fig2.10](c). The sharpness of the decrease in lift is dependent on the type of airfoil.

Airplanes are three-dimensional vehicles with three-dimensional aerodynamic surfaces, so the aerodynamic loads are spread over these surfaces. This distribution in the spanwise direction of the wing results in a force and moment at the root of the wing. The spanwise distribution of the airload is a function of the wing planform shape, the airfoil sections, and the geometric twist. The basic aerodynamic reference for three-dimensional wings is the mean aerodynamic chord (MAC). The thickness, chord length, and angle of attack of the MAC airfoil section is used as a reference for all aerodynamic data. For a rectangular wing planform, the MAC is equal to the wing chord, and for a trapezoidal planform of the semiwing the MAC is equal to the chord at the centroid of the trapezoid.

Maneuver V-n diagram including aerodynamic stall

The maneuver V-n diagram including aerodynamic stall is shown in figure [fig2.14].

Note:

1. \(C_{N}\) varies with compressibility, and varies with the C.G. location as shown in eq. ([eq2.41]). Generally we must consider different altitudes and weight configurations.

2. For flight in incompressible air, the dynamic pressure \(q=\frac{1}{2} \rho V^{2}\), where \(V\) is the airspeed and \(\rho\) is the air density, both at altitude. Define equivalent airspeed \(V_{\mathrm{EAS}}\) at sea level by \[\begin{aligned} \label{eq2.46} q=\frac{1}{2} \rho V^{2} \equiv \frac{1}{2} \rho_{\text {s.l. }} V_{\text {EAS }}^{2}.\end{aligned}\] Then the equivalent airspeed is given by \(V_{\mathrm{EAS}}=\sqrt{\frac{\rho}{\rho_{\mathrm{s}.1}}} V\). Use \(V_{\mathrm{EAS}}\) on V-n diagram to cover all altitudes. Some typical values of the load factors are shown in table 2.1

Gust alleviation factor

A more realistic, semiempirical treatment of gust effects, based on experience and analysis, is to replace the sharp-edge gust speed \(U\) by \(K_{g} U\), where \(K_{g}\) is the gust alleviation factor. In reality there is no such thing as a sharp-edged gust so we account empirically for gust build-up and airplane response. For transport airplanes, NACA specified \[\begin{aligned} \label{eq2.50} K_{g}=\frac{0.88 \mu_{g}}{5.3+\mu_{g}}<0.88,\end{aligned}\] where \(\mu_{g}\) is the airplane mass ratio defined by \[\begin{aligned} \label{eq2.51} \mu_{g} \equiv \frac{W/S}{(\rho \bar{c} g m)/2},\end{aligned}\] and where \(\bar{c}\) is the mean aerodynamic chord; \(\bar{c}=S/b\), and \(b\) is the wing span. See figure [fig2.19].

Gust load factor

For steady level flight, the gust load factor is \[\begin{aligned} \label{eq2.52} \boxed{n=1+\left(\frac{\rho m/2}{W/S}\right) K_{g} U V.}\end{aligned}\] Note that a lightly loaded airplane is more susceptible than when heavily loaded. This is because the increment in lift is independent of the weight. A heavily loaded airplane has more inertia with which to smooth out gusts than a lightly loaded airplane, all other things being equal.

NACA discrete gust conditions

Discrete gusts refer to sudden changes, or alleviated sharp-edged gusts, as opposed to continuous turbulence aircraft gust analysis. In continuous turbulence gusts are represented as a stationary Gaussian random process leading to specification of a power spectral density (Hoblit, 1988). For civil transport airplanes 0–20,000 ft., three discrete gusts are specified:

1. Rough air gust: \(U=66\,\mathrm{ft}/\mathrm{s}\) at \(V=V_{B}\) a speed related to the stall speed.

2. High speed gust: \(U=50\,\mathrm{ft}/\mathrm{s}\) at cruise speed \(V_{C}\).

3. Dive speed gust: \(U=25\,\mathrm{ft}/\mathrm{s}\) at dive speed \(V_{D}\).

The gust V-n diagram is shown in figure [fig2.20], and it is almost symmetrical with n = 1 line since gusts are as likely to act down as up.

Design V-n diagram

The extreme load factors of both maneuver and gust diagrams must be met. Superimpose the two and take the outer boundaries as shown in figure [fig2.21]. Generally, large airplanes are designed primarily by gust load factors. Small military and aerobatic airplanes are designed by maneuver load factors.

Problem statement:

Determine the design V-n diagram at sea level using the NACA formulas for gust loads.

First, draw the maneuver V-n diagram. The stall boundary is given by \[\begin{aligned} \label{ex2.6a} n_{\text {stall }}=\left(C_{N}\right)_{\max } \frac{\rho_{\mathrm{s}.1.}/2}{W/S} V^{2}.\end{aligned}\] The density of air at sea level is \[\begin{aligned} \label{ex2.6b} \rho_{\text {s.l. }}=0.002378 \frac{\text { slugs }}{\mathrm{ft.}^{3}}\mbox{, where }1\,\mathrm{lb}.=(1 \mathrm{slug})(1\,\mathrm{ft}./\mathrm{s}^{2}).\end{aligned}\] The wing loading is \[\begin{aligned} \label{ex2.6c} W/S=(10,000 \text {lb.})/(100 \mathrm{ft.}^{2})=100\,\mathrm{lb}./\mathrm{ft}.^{2}.\end{aligned}\] Substitute eqs. ([ex2.6b]) and ([ex2.6c]) into eq. ([ex2.6a]) to get \[\begin{aligned} \label{ex2.6d} n_{\text {stall }}=(2.07) \frac{\left(0.002378 \frac{\mathrm{lb} \cdot \mathrm{s}^{2}}{\mathrm{ft}.^{4}}/ 2\right)}{100\,\mathrm{lb}./\mathrm{ft}.^{2}} V^{2},\mbox{ or}\\ n_{\text {stall }}=2.46 \times 10^{-5}\,{V}^{2} \quad V~\text{in}\,\mathrm{ft}./\mathrm{s}.\end{aligned}\] The inverted stall boundary is \[\begin{aligned} \label{ex2.6e} n_{\text {istall }}=\left(C_{N}\right)_{\min } \frac{\rho_{\mathrm{s}.1.}/2}{W/S} V^{2}=\frac{-1.2(0.002378/2)}{100} V^{2},\mbox{ or}\\ n_{\text {istall }}=-1.43 \times 10^{-5} V^{2} \quad V~\text{in}\,\mathrm{ft}./\mathrm{s}.\label{ex2.6f}\end{aligned}\] The airspeeds at stall and structural limit factors are \[\begin{aligned} \label{ex2.6g} 2.46 \times 10^{-5} V_{T}^{2}=7.5 \quad V_{T}=552\,\mathrm{ft}./\mathrm{s}\left(\frac{60\,\mathrm{mph}}{88\,\mathrm{ft}./ \mathrm{s}}\right)=376\,\mathrm{mph},\mbox{ and}\\ -1.43 \times 10^{-5}\left(V_{T}^{\prime}\right)^{2}=-3.0 \quad V_{T}^{\prime}=459\,\mathrm{ft}./ \mathrm{s}=313\,\mathrm{mph}.\label{ex2.6h}\end{aligned}\] The maneuver V-n diagram is shown in figure [fig2.23].

Second, draw the gust V-n diagram using NACA formulas. Find the value of the gust alleviation factor as given in eq. ([eq2.50]). The airplane mass ratio is \[\begin{aligned} \label{ex2.6i} \mu_{g} \equiv \frac{W/S}{(\rho \bar{c} g m)/2}&=\frac{100\,\mathrm{lb}./\mathrm{ft}^{2}}{(0.002378\,\mathrm{lb} \mathrm{s}^{2}/\mathrm{ft}^{4})(4\,\mathrm{ft}.)(32.2\,\mathrm{ft}/\mathrm{s}^{2})(4.37)/2},\mbox{ where}\\ \mu_{g}&=149\quad\mbox{dimensionless.}\nonumber\end{aligned}\] Hence, the gust alleviation factor is \(K_{g}=0.85\). The change in the load factor due to the gust is \[\begin{aligned} \Delta n_{\text {gust }}&=\left(\frac{\rho m/2}{W/S}\right) K_{g} U V=\frac{0.002378(4.37)/2}{100}(0.85) U V,\mbox{ or}\label{ex2.6k}\\ \Delta n_{\text {gust }}&=4.42 \times 10^{-5} U V \quad U, V~\text{in}\,\mathrm{ft}./\mathrm{s}.\label{ex2.6l}\end{aligned}\] For gust (1) \(U=66\,\mathrm{ft}/\mathrm{s}\) at \(V=V_{B}\). To find the airplane speed \(V_{B}\) on the stall boundary we set the load factor in eq. (2.52) equal to its relationship to the airplane speed on the stall boundary; i.e., \[\begin{aligned} \label{ex2.6m} \underbrace{2.46 \times 10^{-5} V_{B}^{2}}_{\text {stall }}=1+\underbrace{4.42 \times 10^{-5}(66)}_{2.91 \times 10^{-3}} V_{B}.\end{aligned}\] Solve eq. (m) for speed \(V_{B}\) as follows: \[\begin{aligned} \label{ex2.6n} V_{B} &=\frac{2.91 \times 10^{-3} \pm \sqrt{\left(2.91 \times 10^{-3}\right)^{2}+4\left(2.46 \times 10^{-5}\right)(1)}}{2\left(2.46 \times 10^{-5}\right)},\nonumber\\ V_{B}&=\frac{2.91 \times 10^{-3} \pm 1.03 \times 10^{-2}}{2\left(2.46 \times 10^{-5}\right)}\quad\mbox{choose} +.\end{aligned}\] Hence, \(V_{B}=269\,\mathrm{ft}/\mathrm{s}=184\,\mathrm{mph}\), and the change in load factor for gust (1) is \[\begin{aligned} \label{ex2.6o} \Delta n_{1}=4.42 \times 10^{-5}(66)(269)=0.78.\end{aligned}\] For gust (2) \(U=50\,\mathrm{ft}/\mathrm{s}\) at \(V_{C}=500\,\mathrm{mph}\). Hence, the change in load factor is \[\begin{aligned} \label{ex2.6p} \Delta n_{2}=4.42 \times 10^{-5}(50)\left[500\,\mathrm{mph}\left(\frac{88\,\mathrm{ft}/ \mathrm{s}}{60\,\mathrm{mph}}\right)\right]=1.62.\end{aligned}\] For gust (3) \(U=25\,\mathrm{ft}/\mathrm{s}\) at 650 mph, and the change in load factor is \[\begin{aligned} \label{ex2.6q} \Delta n_{3}=4.42 \times 10^{-5}(25)\left[650\,\mathrm{mph}\left(\frac{88\,\mathrm{ft}/ \mathrm{s}}{60\,\mathrm{mph}}\right)\right]=1.05.\end{aligned}\] Thus, the six points on the gust V-n diagram are \[\begin{aligned} \label{ex2.6r} \begin{array}{ll} 1 \pm \Delta n_{1}=1.78,0.22 & V_{B}=184\,\mathrm{mph} \\ 1 \pm \Delta n_{2}=2.62,-0.62 & V_{C}=500\,\mathrm{mph} \\ 1 \pm \Delta n_{3}=2.05,-0.05 & V_{D}=650\,\mathrm{mph} \end{array}.\end{aligned}\] A sketch of the gust V-n diagram is shown in figure [fig2.23]. Here the design V-n diagram is the maneuver V-n diagram since the gust V-n diagram is contained inside the maneuver V-n diagram.

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[Peery] Peery, D. J., 2011, Aircraft Structures. Dover Publications, Inc., 2011. (Unabridged republication of the work originally published in 1950 by the McGraw-Hill Book Company, New York.) Chapter 3.