# Trying to find the vertical acceleration by subtracting weight from lift

I'm trying to determine the vertical acceleration component of a plane with the following specifications:

• Density of air: 1.29kg/m3
• Speed: 33 m/s
• Wing area: 13.97 m2
• Lift Coefficient: 1.5
• Mass:560 kg

Lift formula : 1/2*1.29*33^2*13.97*1.5 = 14718 N.

(Lift-Weight)/Mass = (14718-(560*9.8))/560 = 16.48

From this I deduce that the plane has a vertical acceleration of 16.48 m/s2 which I find way too much.(I was expecting to get somewhere from 3 to 5 m/s2)

• your mass seems a little low and your speed and wing area a little high. Your lift coefficient is generous. So what combination of pilot, fuel, airframe, and engine gave you 560 kg? But you are close enough to a Cessna 152 to compare real data to your calculations. Also, remember increasing Lift coefficient to 1.5 on a wing that big is going to have a fairly big drag penalty, requiring more thrust, requiring a larger engine, which weighs more ... etc. Check that 152 out for starters. – Robert DiGiovanni Oct 11 at 16:02
• PS crop dusters have lots of power for just that type of performance, you may want to check them too. – Robert DiGiovanni Oct 11 at 16:04
• The lift grows with square of velocity and note that quadratic curve gets steep rather quickly. At typical approach speed of 1.3 times the stall speed, you can have vertical acceleration of 6.56 m/s² already and at twice the stall speed you can have almost 30 m/s². 3 to 5 m/s² would only be dangerously close to the stall speed. It's only available vertical acceleration of course—most of the time you simply don't have any, flying at lower angle of attack to just balance your weight. – Jan Hudec Oct 11 at 23:10

Your math looks right to me. If an airplane has the statistics that you listed there, then it will accelerate upwards at $$16.48\ \mathrm{m}/\mathrm{s}^2$$, as you calculated. The occupants will experience about 3 g's of proper acceleration (1 g from gravity, and about 2 g's from the acceleration).

But here's what would actually happen, in context.

To start, assume that the plane is in straight and level flight at 33 m/s. (Here's why I'm making this assumption: You're subtracting weight from lift, which only makes sense if the lift vector is pointing straight up. And if the lift vector is pointing straight up, then the plane is probably in straight and level flight.)

While the plane is in straight and level flight, the lift must equal the weight, so the coefficient of lift must be much less than the 1.5 you have here.

Then, the pilot suddenly pulls back on the yoke, causing the plane to pitch up. This causes the angle of attack to quickly increase, which causes the coefficient of lift to increase to 1.5. The plane starts accelerating upwards at $$16.48\ \mathrm{m}/\mathrm{s}^2$$.

At this point, let's assume that the pilot maintains a constant pitch after this initial pull-up.

As the plane's vertical speed increases, its climb angle also increases, which causes the angle of attack to decrease. This, in turn, causes the coefficient of lift to decrease again, until the lift and weight approximately balance out again. At this point, the plane is in a steady climb. The vertical acceleration is now $$0\ \mathrm{m}/\mathrm{s}^2$$, and the occupants are feeling just 1 g again.

(I'm making a couple of small simplifying assumptions in the above: I'm assuming that there is no updraft or downdraft, and that plane's engine is not producing a significant amount of upward force.)

• I would say what the original question describes is something like the start of a loop. The pilot's controI input that creates the 1.5 CL condition also initiates a pitch rotation. I would not say that the resulting upward acceleration automatically decreases the plane's angle-of-attack, because the plane is free to rotate in pitch. However two effects that do tend to decrease the angle-of-attack associated with any given elevator position are pitch rotational inertia (as the maneuver is initiated) and aerodynamic damping of pitch rotation rate (once a significant pitch rotation rate is – quiet flyer Oct 11 at 16:00
• However two effects that do tend to decrease the angle-of-attack associated with any given elevator position are pitch rotational inertia (as the maneuver is initiated) and aerodynamic damping of pitch rotation rate (once a significant pitch rotation rate is achieved). Despite these effects, to transition to a steady-state climb after initiating a 1.5 CL pullup, I think the pilot would probably have to relax his elevator input, at least until some of the airspeed bleeds off. – quiet flyer Oct 11 at 16:01
• @quietflyer That's a good point. I've edited this to point out that I'm assuming that the pilot maintains a constant pitch after initially pulling up. – Terran Swett Oct 11 at 16:26
• Comments now developed into an answer... – quiet flyer Oct 11 at 16:27

I would say what the original question describes is something like the start of a loop. The pilot's elevator controI input that creates the 1.5 CL (about 3G in this case) condition also initiates a pitch rotation. To a first approximation, I would not say that the resulting upward acceleration automatically decreases the plane's angle-of-attack, because the plane is essentially free to rotate in pitch to maintain the angle-of-attack that is commanded by the elevator position. However, looking in more detail, two effects that do tend to decrease the angle-of-attack associated with any given elevator position (or any given pull force exerted on the stick or yoke) are pitch rotational inertia (as the maneuver is initiated) and aerodynamic damping of pitch rotation rate (once a significant pitch rotation rate is achieved). Despite these effects, to quickly transition to a steady-state climb after initiating a 1.5 CL (about 3G in this case) pullup, the pilot would have to relax his elevator input, at least until some of the airspeed bleeds off.

Keep in mind that in a steady-state climb, lift must be LESS than weight.