Straight and Level Flight

One important trim state is "straight and level flight". Straight means the vehicle isn't turning. Level means it's altitude isn't changing. Since it's a trim state, the speed is constant as well.

When a plane is cruising, we often want to fly straight-and-level at a particular speed. The questions we'll answer here are:

For a particular speed…

… how much thrust does the propulsion system need to provide?
… what angle of attack do we need to fly at?
… what elevator deflection is required?

We can use the equations we've developed in this lesson to answer these questions.

1. How much thrust is required?

We're going to make the simplifying assumption that thrust and drag point directly opposite each other, so we can say that thrust and drag are equal.

T = D

If we plug in our equation for D, this becomes

T = C_D \bar{q} S

And since \bar{q} = \rho V^2 / 2, we can write

T = \frac{1}{2} C_D \rho V^2 S

This gives us the thrust in terms of the drag coefficient C_D, the air density \rho, the airspeed V, and the surface area of the wings S.

2. What angle of attack?

At trim, we know that the weight of the vehicle (W) must be supported by the lift (L). We can use this to solve for the angle of attack at trim (\alpha_{\text{trim}}).

\begin{aligned} L &= W \\ C_L \bar{q} S &= W \\ \left[ C_{L_0} + C_{L_\alpha}\alpha_{\text{trim}} + C_{L_{\delta E}} \delta E\right] \bar{q} S &= W \\ C_{L_0} + C_{L_\alpha} \alpha_{\text{trim}} + C_{L_{\delta E}} \delta E &= \frac{W}{\bar{q} S} \end{aligned}

The last term on the left hand side of this equation will be very small compared to the other two terms, so we can safely ignore it and then solve for \alpha_{\text{trim}}.

\begin{aligned} C_{L_0} + C_{L_\alpha} \alpha_{\text{trim}} &= \frac{W}{\bar{q} S} \\ \\ \alpha_{\text{trim}} &= \frac{ \frac{W}{\bar{q} S} - C_{L_0}}{C_{L_\alpha}} \end{aligned}

What elevator deflection?

This positive angle of attack will cause a pitching moment. We need to use the elevator to balance that out. We can find \delta E_{\text{trim}} by setting the pitching moment to zero.

\begin{aligned} M &= 0 \\ C_M \bar{q} S \bar{c} &= 0 \\ C_M &= 0 \\ C_{M_0} + C_{M_\alpha} \alpha_{\text{trim}} + C_{M_{\delta E}} \delta E_{\text{trim}} &= 0 \\ \delta E_{\text{trim}} & = \frac{-C_{M_0} - C_{M_\alpha} \alpha_{\text{trim}}}{C_{M_{\delta E}}} \end{aligned}

Implications for Control

We just calculated T, \delta E, and \alpha at trim. We made some simplifying assumptions, so these values won't be exact, but they should be very close.

Eventually, you will write a controller that commands the thrust and elevator deflection of a fixed wing vehicle. You will be able to greatly improve the performance of this controller by passing in these trim state values for thrust and elevator deflection as "feed forward" parameters when you want the vehicle to be in straight and level flight.

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