## Wednesday, February 08, 2006

### Lift Does Not Act Up

Comments on my last post revealed some confusion regarding the role and direction of the lift force on an airplane in flight. I now attempt to dispel and elucidate. The following is basic theory of flight, not anything dramatic or unique to jets.

Lift is the component of aerodynamic force that acts perpendicular to the direction the airplane is actually travelling. (The movement of the airplane through the air, relative to the air, is the same thing as a wind blowing towards a stationary airplane. Leonardo da Vinci figures that out). The direction and strength of the airflow caused by the motion of the aircraft, but before the airplane actually disturbs the air, is the relative wind. To repeat: lift acts perpendicular to the relative wind.

When the airplane is in level flight, as depicted in that ubiquitous four forces diagram, the flight path is parallel to the ground, so the lift does act straight up and thus is in direct opposition to the weight. Thrust balances drag, (and all the moments I haven't mentioned balance out), and thus there are no unresolved forces. Disturb that equilibrium and the airplane will accelerate (speed up, slow down or turn) in the direction of the unbalanced force. The change in motion will result in a change in the forces, until either the forces are balanced, or the airplane disintegrates or hits something.

It doesn't matter which way it is going or if it's an eight engine bomber or a glider, the lift vector is perpendicular to the relative wind. Really. In the glider, relative to still air, the unaccelerated flight path is always slightly down. (The glider can go up if the air it's in is moving up (a thermal or updraft), or if the pilot raises the nose to zoom (that's actually a technical term) the aircraft, trading airspeed for altitude, but I'm talking about the stable, sustainable case.) It applies equally well to a glider or an aircraft with power at idle.

So your flight path is inclined below the horizontal at an angle I'll call gamma (γ). There is no thrust. Drag acts directly backwards along the flight path. Lift acts perpendicular to the flight path, so up and a little forward. Weight acts straight down. We wave our wand of trigonometry and resolve weight into two vectors. One, of strength weight x cos(γ), acts perpendicular to the relative wind, directly opposite lift. The other, of strength weight x sin(γ) acts along the flight path. (I'm using x for multiplication because * is nerdy, and I'm pretending that non-nerds are going to be following my trigonometry). Inventory the forces according to the direction they pull, and you discover that the only force opposing lift is weight x cos(γ). A cosine is always less than one, so lift is less than weight.

You will also see that weight is the only forward force. That is why the flight path must be angled down: if gamma is not greater than zero then there is no force to offset the drag.

Shove the nose of your glider down, accomplished by lowering the elevator such that the airflow hauls the tail up, and everything changes. It's complex, more complex than the next few sentences indicate. The angle at which the wings meet the air decreases, decreasing lift. The unbalanced weight accelerates the airplane downwards. The change in direction changes the relative wind, changing the directions of the components of lift, drag and weight. The higher airspeed changes drag and lift. The airplane continues to accelerate until the change in drag can offset the change in the forward component of weight, and the new lift offsets the new perpendicular component of weight. Or the glider is still accelerating as it lawn-darts into the ground.

If instead, you haul the nose of the glider up, by shoving the tail down, what happens? There is the same complex dance of changing forces. Lift momentarily increases, because of the increased angle of attack, and the glider accelerates upward. Weight doesn't change, but with an upward flight path, now there is a component of weight that acts backwards along the flight path, and what is there to act forward? Nothing. Lift acts perpendicular to the flight path, opposite weight x cos(gamma;) but as weight and drag both slow the glider down, lift decreases. You can keep hauling the nose up to increase the angle of attack, and that will work right up to the critical angle, when the glider stalls, and if you don't get it flying again the only forces acting will be weight straight down and drag straight up. The weight will accelerate the glider downwards until the drag increases to equal the weight. Or you hit the ground. That's a good way to land a small airplane, by the way. Just arrange for the stall to occur just above the ground, so that you hit the ground before the plumetting starts. If you're going to be plummetting, you want to do it in something really draggy, like a parachute. With that, the drag equals the weight at a speed that is low enough to survive the impact.

If we want to go up for long, we need another force along the flight path. That's thrust. So here's our aircraft with a flight path inclined upwards by that same angle gamma (I hope html does &gamma; on everyone's computers.) As always, drag is opposite the flight path and lift acts perpendicular to it. Weight (W) is still W(cos(γ)) perpendicular to the flight path and W(sin(γ)) backwards along the flight path, just as in the previous paragraph. Thrust we'll consider aligned with the flight path. For the aircraft to be in equilibrium, forces along the flight path must balance and forces perpendicular to the flight path must balance. Along: Thrust = Drag + W(sin(γ)). Perpendicular: Lift = W(cos(γ)). Once again the cosine is always less than one, so the lift is less than the weight, and the steeper the angle of climb, the less lift is required.

Another way of explaining what is happening is that in a climb, the thrust is helping to hold the airplane up, and in a descent the drag is helping to hold the airplane up, so not as much lift is required. That explains why with the same power setting, you can't go as fast as level flight in a climb, but can go faster in a descent. The climb thrust has to help out lift and overcome downward drag, while the descent thrust is being helped out by downward drag. That's the handwaving, non trigonometry way of describing it, but I know some of you went "greek-greek-geek" to yourselves, as you skipped over the trig.

To answer anoynmous, yes, the thrust often is acting other than directly along the flight path, giving a component equal to the sine of that angle times the total thrust, that acts in the same direction of lift, and reducing the forward thrust to total thrust times the cosine of that angle. For small angles, the sine is small enough, and the cosine is close enough to one, that people don't worry too much about it. As the angle of thrust theta above the flight path increases, as it does when the airplane flies more slowly, then the upward forces in level flight must be counted as Lift + Thrust(sinθ)). The authors of my Air Canada-recommended textbook are not especially concerned about the off-flight path component of thrust at normal speeds, so I'm leaving it out. Directed thrust like on the Harrier jet (did you see the movie True Lies?) steers us towards the realm of rocket science. I once saw a t-shirt that had equations all over it, and said "As a matter of fact I am a rocket scientist." I wish I had bought it.

#### 1 comment:

Anonymous said...

Very interesting post. It sounds very much like the way that I approach an example problem that I work with my students in intro physics.