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Old Nov 16, 2012, 06:13 AM
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greg
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somerset, nj
Joined Feb 2005
374 Posts
Quote:
Originally Posted by ShoeDLG View Post
I don't understand what you're saying. If you have a string of boxcars that form a closed loop, there is no way for the translational momentum of that string to change (as long as it stays on the track).
... on what scale?

Quote:
Originally Posted by ShoeDLG View Post
When Newton says that an object's rate of momentum change is equal to the unbalanced force acting on it, he is referring to the object's linear (or translational) momentum, not it's angular momentum.
i've been interpreting "rate of momentum change" as m dV/dT = ma = F. And when you say there is little or no rate of momentum change as being that ma = F = ~0. but certainly if you put something into motion, even if it's just angular motion, a Force is involved.

Some may be arguing about what happens to the air several wing chord lengths away from the wing and how ground effect does have an influence on lift, while others are simply concerned with what happens near the surface of the wing which affects the wing and is involved in the creation of lift.

So while you and others are correct to a large degree that there little translation of the air in the larger vicinity of the aircraft, the more local effects near the surface of the wing need to be considered differently.

I believe that near the surface, let's say within a box 1/4 chord length on edge, the pressure regions create near vertical forces resulting in lift and the vertical acceleration and translation of the air. What happens outside of the box has less impact on the creation of lift and on the aircraft, especially since an aircraft is moving away from the air affected by the process. (This of course may not be true of a helicopter).

greg
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