I have just one problem with this: where does the force generated go?=20 One thing pops in my mind, communicating vessels.=20 As soon as the additional pressure appears it would just get equalized,=20 so I don't see how that could generate lift. Well... that much about my phy= sics skills :-) Now, pushing down on the air, should increase air pressure under the plane.= =20 So when planes fly close above our head there should be an increase of=20 atmospheric pressure for a short time, no?=A0=20 And if a 747 flies 20 meters above my head from the pressure=20 probably my eyes should pop out or skull get crushed? Well I bet that's not= gonna happen... maybe I'll go deaf because of the engine noise, otherwise = will be fine. --- On Fri, 1/28/11, Olin Lathrop wrote: From: Olin Lathrop Subject: Re: [TECH] How planes fly To: "Microcontroller discussion list - Public." Date: Friday, January 28, 2011, 9:27 PM Gerhard Fiedler wrote: >> When looking at the airplane as a whole, the physics says nothing >> about how wings or any other structure accomplishes the pushing down >> of air, only that there must be a certain net downward momentum per >> time imparted on the air. > > "Momentum per time" is what exactly, physically speaking? I don't > think you wrote this with your physicist's hat on. My hat is intact, but you might want to check where your duncecap is. Momentum per time is both correct and unambiguous.=A0 Momentum is mass time= s velocity.=A0 Momentum per time is therefore mass times velocity per time, o= r mass times accelleration, or force.=A0 Remember, F=3Dma. So "momentum per time" is a fancy way to say "force", but I chose to say it that way to help illuminate where that force is coming from. > Which one of Newton's rules says that air must move down to keep a > plane in steady height? I think he mumbled something about equal and opposite forces. > And please, don't put it in fuzzy and easy to > misunderstand (both in the writer's and in the reader's mind) words -- > put it in clear formulae, using proper terms used in physics. I think I've been plenty clear enough, especially if you read back thru thi= s thread.=A0 Your problem seems to be with basic physics.=A0 Surely there is = a lot of accessible stuff out there that explains forces, momentum, velocity, accelleration, and Newton's laws. However, let's do a example.=A0 Suppose a plane weighs 100 Newtons.=A0 Let'= s further stipulate we're talking about a normal airplane that uses power to push itself forward.=A0 In other words, the force applied by the propeller, jet engine, or whatever, is in the forward direction, which is also the direction of motion.=A0 As said before, the plane is in steady level flight flying into still air. Gravity is pulling down on the plane with 100N.=A0 Something else must be pushing up on the plane with 100N since it's not accellerating.=A0 Accordin= g to Newton's law (the equal and opposite one), this in turn means the plane is pushing down on something.=A0 That something is the air around it. So the plane is pushing down on the air with a force of 100N. =A0 Force =3D 100N =3D 100 Kg m / s**2 =3D (Kg m/s) / s Note that (Kg m/s) is momentum.=A0 So every second, the plane has to impart= a momentum of 100 Kg m/s on the air around it, and in the downward direction since the resulting force on the plane is up. One of the original points many posts ago is that there is a tradeoff here between Kg and m/s.=A0 The plane could push 100 Kg of air downward at 1 m/s= , or 10 Kg at 10 m/s, or 1 Kg at 100 m/s, etc, all resulting in the same net force.=A0 At a fixed speed and weight of the plane, the tradeoff actually chosen is dictated largely by the wing span.=A0 A wider wingspan gives acce= ss to more air to push over the same time. The point I made about this tradeoff earlier is that while the force is the same, the energy is not.=A0 The energy it takes to move the air is proportional to the square of the speed.=A0 To impart a momentum of 100 Kg = m/s to 100 Kg of air is: =A0 (1/2) 100Kg (1m/s)**2 =3D 50 Joules Since this is what happens each second, the power is 50 Watts.=A0 Now consi= der when only 10 Kg are moved.=A0 To keep the same momentum, those 10 Kg must b= e moved at 10 m/s.=A0 The energy required to do that is: =A0 (1/2) 10Kg (10m/s)**2 =3D 500 Joules which means a power of 500 Watts is required to keep doing this.=A0 This ba= sic physics explains why wide wingspans are more efficient.=A0 They move more a= ir, which therefore needs to be moved with a lower speed, which therefore requires less power to sustain the flight. ******************************************************************** Embed Inc, Littleton Massachusetts, http://www.embedinc.com/products (978) 742-9014.=A0 Gold level PIC consultants since 2000. --=20 http://www.piclist.com PIC/SX FAQ & list archive View/change your membership options at http://mailman.mit.edu/mailman/listinfo/piclist --=20 http://www.piclist.com PIC/SX FAQ & list archive View/change your membership options at http://mailman.mit.edu/mailman/listinfo/piclist .