On the physics behind speed vs. lift vs. kinetic energy
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On the physics behind speed vs. lift vs. kinetic energy
I'm making a presentation for new students on the subject "aircraft"... and was wondering about the equation we all use, or at least know..
If you double the speed, you'll quadruple the lift and drag.
It's also well established that if you double the speed, you'll quadruple the kinetic energy.
I simply don't remember, can we agree that lift/drag and kinetic energy is directly related in these equations? Meaning lift and drag are equal kinetic energy, so if you double the kinetic energy, you'll double the lift/drag.... guess I just answered my own question...
I was thinking as presenting it as a formula for speed vs. kinetic energy (the one we all leaned many moons ago in high school) and then relate that to lift/drag, and hopefully help the youngsters (who are closer to physics lessons in high school than I) understand why.
If you double the speed, you'll quadruple the lift and drag.
It's also well established that if you double the speed, you'll quadruple the kinetic energy.
I simply don't remember, can we agree that lift/drag and kinetic energy is directly related in these equations? Meaning lift and drag are equal kinetic energy, so if you double the kinetic energy, you'll double the lift/drag.... guess I just answered my own question...
I was thinking as presenting it as a formula for speed vs. kinetic energy (the one we all leaned many moons ago in high school) and then relate that to lift/drag, and hopefully help the youngsters (who are closer to physics lessons in high school than I) understand why.
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You may want to steer clear. Principles of flight involves a fair bit of hand waving and approximations of basic physics. Often the students don’t have their basic physics understood anyway.
however, for you own self confideence, it sounds like you are asking how to equate force units to energy units.
Try this:
Use work = force x distance
FWIW I often find it easier to visualise moving aerofoil and initially stationary air (rather than aerofoil in a wind tunnel).
imagine a flat plate area of 1 square metre moving at velocity V metres per seond, that has to accelerate the previously stationary air up to its velocity. (This is the same as a flat plate in a wind tunnel bringing the air to rest.)
Kinetic energy of 1 cubic metre of air at velicity V = 1/2 rho V^2
each second, the volume of the parcel of air needing to be accelerated by the 1 square meter plate = V.
V cubic metres of air, with a Kinetic Energy of 1/2 rho V^ 2 per cubic meter, is a K E of 1/2 rho V ^ 3
now use work = force x distance.
therefore force = work / distance
the flat plate has to do work = the KE change of the air it is working on. In this case 1/2 rho V ^ 3
but it is doing that work over a distance of V metres. I.e.
Force = work / distance = 1/2 rho V ^ 3 / V = 1/2 rho V ^ 2
Does that help at all?
however, for you own self confideence, it sounds like you are asking how to equate force units to energy units.
Try this:
Use work = force x distance
FWIW I often find it easier to visualise moving aerofoil and initially stationary air (rather than aerofoil in a wind tunnel).
imagine a flat plate area of 1 square metre moving at velocity V metres per seond, that has to accelerate the previously stationary air up to its velocity. (This is the same as a flat plate in a wind tunnel bringing the air to rest.)
Kinetic energy of 1 cubic metre of air at velicity V = 1/2 rho V^2
each second, the volume of the parcel of air needing to be accelerated by the 1 square meter plate = V.
V cubic metres of air, with a Kinetic Energy of 1/2 rho V^ 2 per cubic meter, is a K E of 1/2 rho V ^ 3
now use work = force x distance.
therefore force = work / distance
the flat plate has to do work = the KE change of the air it is working on. In this case 1/2 rho V ^ 3
but it is doing that work over a distance of V metres. I.e.
Force = work / distance = 1/2 rho V ^ 3 / V = 1/2 rho V ^ 2
Does that help at all?
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If you give a presentation, please please get at least the orders in your formulas right.
In laminar flow drag is proportional to velocity. Lift is proportional to angle (of attack) times velocity for small angles. That's for lower mach numbers; at jet-cruise speeds compressibility of air becomes an important factor too.
Most airfoils are designed to bend the air and only accelerate it temporarily to get a higher pressure under the wing and a lower above. The global deceleration (before vs after the wing) is kept small, because of the design goal to keep drag low. (Action=Reaction: the drag force on the plane is the acceleration force on the air)
In laminar flow drag is proportional to velocity. Lift is proportional to angle (of attack) times velocity for small angles. That's for lower mach numbers; at jet-cruise speeds compressibility of air becomes an important factor too.
Most airfoils are designed to bend the air and only accelerate it temporarily to get a higher pressure under the wing and a lower above. The global deceleration (before vs after the wing) is kept small, because of the design goal to keep drag low. (Action=Reaction: the drag force on the plane is the acceleration force on the air)
Plenty of stuff on the internet, eg
https://www.google.com/url?sa=t&sour...OOS4OYltUWdjdB
There have been threads on here before questioning eg the source of lift and it got a bit confusing
https://www.google.com/url?sa=t&sour...OOS4OYltUWdjdB
There have been threads on here before questioning eg the source of lift and it got a bit confusing
