Introduction: there are four forces acting on an airplane. Two are thrust and drag that influence the forward motion. And two are lift and weight that influences the upward motion. I’m concerned with the upward motion because the lift is almost entirely generated by the wings redirecting air downwards (Newtons third law).
Problem: the weight of an airliner is immense, but takeoff speeds are low and air is light. Can redirected air actually move an airplane upward?
Analysis (please help because my physics knowledge is weak):
Let’s use a Boeing 747-400 for our example. Link.
And the takeoff speed is 290 kph from here.
So, let’s use the numbers 350,000 kg of weight, 300 km/h of takeoff speed and the wings redirect air at a 30 degree angle, the wing area is 525 meters^2. These are generous parameters.
300 km/hr = 83.33 m/s. So the wings are sending 525 x 83.33 = 43750 m^3/s volume of air downward at 30 degrees. The vertical component is sine(30) or .5 = 21875 m^3/s
The density of air is 1.225 kg/m^3. So the mass flow rate of the redirected air is 26797 kg/s. Multiply speed * rate: 83.33 m/s * 26797 kg/s = 2,233,073 kg*m/s^2 this is the lift force.
The weight of the plane is 350,000 9.8 = 3,430,000 newtons (also kgm/s^2)
So, the force of gravity is 1.5 times the lifting force of the wings.
I believe there is something called “ground effect,” not sure if that is what you mean, which is that when the airplane is on the ground or near the ground, yes, the air helps to cushion/redirect/do-something and help get the airplane into the air - it also can make it hard for an airplane to land properly.
One major problem is that your intuitive attempt to approximate the lift of an airfoil is badly flawed. Aside from other details, the most glaring issue is that lift goes with v^2 rather than v.
Lift is produced by the aggregate *pressure *difference from the bottom to the top of the wing, not by volume flow. Ground effect helps but is not required.
The OP is also ignoring the lift generated by the fuselage, which at any angle above zero degrees is net positive (I don’t know the percentages though). Additionally, with the 747, the “hump” right after the cockpit also generates some lift, from what I’ve read. Hopefully the pilots and engineers can clarify that.
The lift/drag ratio of an airliner fuselage is much lower than the wings, so I think efficiency considerations mean that they are designed for the fuselage to contribute minimal lift. Many fighter planes are quite different, however. The fuselage is obviously much sleeker. An Israeli pilot landed an F-15 with one wing!
The wing area, 525m[sup]2[/sup], is the wingspan times the chord length (the distance from the leading edge to the trailing edge). It’s like if you were to measure the area of the shadow the wings cast on the ground, you’d get 525 square meters.
So to get the volume, you’re taking a left-right measurement (the wingspan), a fore-aft measurement (the chord length), and another fore-aft measurement (the plane’s speed in meters/second). I don’t think that really defines a volume of air.
Consider instead that as the wing moves through the air, it doesn’t just affect the air that is in contact with the wing. Air below the wing is compressed and forced downwards, air above the wing is affected by low pressure and pulled downward. Suppose that the wing affects the air from 1 meter above to 1 meter below. Now, the volume of air deflected per second is the wingspan (left-right) x 2 meters (up-down) x the speed of the plane (fore-aft).
I have no idea if that 1 meter above and below is even in the ballpark, but I think that’s the approach to take for the way you’re trying to figure it.
The lift/drag ratio of an airliner fuselage is much lower than the wings, so I think efficiency considerations mean that they are designed for the fuselage to contribute minimal lift. Many fighter planes are quite different, however. The fuselage is obviously much sleeker. An Israeli pilot landed an F-15 with one wing!
Those are the same thing - the momentum of the air thrown downward is what causes the pressure difference. That momentum has to offset the weight of the plane, for flight to happen. I haven’t gone through K364’s figures, but if I were to do some back-of-the-napkin calculations and get an answer that’s off by a factor of 1.5x, I’d just say that’s pretty good and that I overlooked a detail somewhere.
…but in a different way than the OP envisaged. Per Bernoulli: when there is ordered directional movement of the air molecules, the resulting dynamic pressure relates to the kinetic energy of the molecules. So, like kinetic energy, dynamic pressure is proportional to velocity squared. And it’s dynamic pressure that generates lift, so velocity squared appears in the lift equation.
I would think that this part of the calculation needs to have different units. You need to say that the wing is throwing down some volume of air (m[sup]3[/sup]) by some speed (m/s). Then you can figure the mass of that air volume and you’d have the force.
In any event, the bottom line is that OP’s intuitive attempt to approximate lift is just so badly wrong that the result is meaningless. Even on its own terms the attempt to calculate “volume” is wrong, as noted by Robot Arm. And, well, that’s just not the way it works anyway. An approximation that ends up with proportionality to v rather than v^2 cannot possibly be valid.
The idea that it has nothing to do with downward air momentum is what’s been discredited - it has everything to do with downward air momentum. If there is no downward momentum, then there is no lift. The force required to throw the air downward is the lift.
Bernoulli is still correct, but applying it to a wing is fraught with issues, so the Bernoulli explanation of wing lift is deprecated.
It’s not that Bernoulli’s principle has been discredited – it’s just that the high school explanation is too simplistic and does not account for the effect of Newton’s third law. If Bernoulli’s principle is the only explanation behind lift, then planes should not be able to fly upside down (which is clearly not the case).
We don’t really notice the volume of air that gets displaced by aircraft wings as they move through the air, but it’s not insignificant. If you think of helicopter rotors and the incredible amount of air they push downwards, it’s obvious that downward air momentum plays a large part of lift created by wings.
Don’t get hung up on the (incorrect) idea that you get some lift from Bernoulli and some from the downward air momentum. They are two ways of looking at the same thing.
The force required to throw the air downwards IS the lift.
And the lift is also the pressure difference between the top and bottom sides of the wing.
i think the OP’s question could be answered by saying nothing more than, “Wings are big, and generate LOTS of lift.”
That aside, when I was a full-time flight instructor I did a deep dive on how to teach the four forces to students. Read a lot about the Bernoulli explanation, the alternative Newtonian explanation which focuses more on the deflection / redirection of air… Got really sick of it.
Ended up teaching it thusly:
Here’s the FAA way (Bernoulli)
Here’s another way of looking at it (Newtonian)
Go read these several articles and see what makes sense to you.
Pilots only need a working knowledge of how lift and drag work. Unless you’re designing a wing, the rest is superfluous IMHO.
