Thread: Discussion A non-aerodynamic proof that a lifting wing pushes down on the earth View Single Post
Dec 30, 2012, 03:15 PM
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Quote:
 Originally Posted by ciurpita Wheli = Fair = d(m.v)/dt = ma causing the air to accelerate downward and increase its velocity Fscale = Fair = d(m.v)/dt = ma causing the air to de-accelerate and decrease its downward velocity
This may seem like semantics, but neither of the above two expressions for the air's rate of momentum change are correct. Newton's Second Law says that an object's rate of momentum change is equal to the unbalanced force acting on it.

The only correct way to write Newton's Second Law is:

Sum (F_air) = W_helicopter - F_scale = d(mv)/dt = 0

By separating the forces and applying Newton's Second Law twice, you are implying that the air gains downward momentum at a rate equal to the helicopter's weight and then loses downward momentum at a rate equal to the reading on the scale.

The fact that it is incorrect to apply Newton's Second Law in "pieces" becomes readily apparent if the helicopter is supported not by the air but by a foam block. If you were to write two separate equations for this case you would get:

W_helicopter = F_block = d(mv)/dt = ma_block

suggesting the weight of the helicopter increases the block's downward velocity

F_scale = F_block = d(mv)/dt = ma_block

suggesting the force from the scale decreases the block's downward velocity

Obviously the downward velocity of the block never changes.

Newton's Second Law does not say that the air must acquire downward momentum at a rate equal to the weight of the helicopter any more than it says the foam block must acquire downward momentum at a rate equal to the weight of the helicopter. It is POSSIBLE that the air does gain momentum at a rate equal to the helicopter's weight, only to have it later removed by the scale, but Newton's Second Law certainly doesn't require this to be the case.

Imagine that the helicopter isn't pushing down on the air, but is instead pushing down on a stream of independent BB's. In this case, you could draw a box around the helicopter, and for the helicopter to be in a steady hover, the BB's would have to transport downward momentum out of the box at a rate equal to the helicopter's weight. In other words, the helicopter would be increasing the downward momentum of the stream of BB's at a rate equal to its weight. The scale would obviously remove the downward momentum at the same rate.

This is only true for the stream of BB's because they are independent. The path (and velocity) of a BB leaving the helicopter is not influenced in any way by the presence of the ground. Each BB is first influenced only by the helicopter and then influenced only by the ground. The performance of a helicopter using independent BB's to hover would not depend in any way on its height above the scale.

In the case of a helicopter using the air to hover, the acceleration experienced by an element of the air is at every instant influenced by both the rotor and the ground. This is borne out by the well established fact that the hover performance of a helicopter depends very strongly on the height of the rotor above the ground.

Quote:
 Originally Posted by ciurpita ok, the net result is no change in momentum, but what happens to the air between rotor and scale compared to the same mass of air without a helicopter? what is the velocity of the air above the rotor, halfway between the rotor and scale, and at the scale?
The answers to those questions depend on the height of the rotor above the scale!
Last edited by ShoeDLG; Dec 30, 2012 at 04:50 PM.