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Old Jan 03, 2013, 09:20 AM
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Perhaps you should consider taking a physic
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Old Jan 03, 2013, 09:28 AM
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You have asserted that air's rate of downward momentum change must be equal and opposite the lift acting on a helicopter rotor in ground effect.

-If the force exerted by the air on the ground accurately reflects the rate at which the rotor is transferring downward momentum to the air, then by placing a scale under a large plate, you can measure the air's rate of downward momentum change.

-You can directly measure the amount of lift generated by a helicopter rotor in ground effect.

The simple experiment described in post #952 offers a setup for performing simultaneous measurement of the force exerted by the air on the ground and the lift acting on the rotor. If introducing a plate above the rotor changes the force exerted by the air on the ground by one amount, and the lift on the rotor by a different amount, then one of two things must be true:

-The force exerted by the air on the ground does not accurately reflect the rate at which the rotor is transferring downward momentum ot the air (in which case measuring the force exerted by the air on the ground under helicopter in ground effect does not illustrate the momentum balance assertion).

-or-

-The assertion that the air's rate of downward momentum change must be equal and opposite the lift acting on a helicopter rotor in ground effect is not correct.

I have performed the experiment, and my results clearly indicate that introducing the upper plate can cause a substantial change in the force exerted by the air on the ground (a reduction of more than 80%) while changing the rotor lift by less than 10%.

This is not weird physics, it is a straightforward test of an assertion. The results do not in any way suggest Newton is wrong, they simply show that your application of Newton's Second Law (equating an object's rate of momentum change to only one of the forces acting on it) is incorrect.
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Old Jan 03, 2013, 09:58 AM
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Originally Posted by ShoeDLG View Post
You have asserted that air's rate of downward momentum change must be equal and opposite the lift acting on a helicopter rotor in ground effect.
I think you are the one who keeps the copter discussion in ground effect. Clearly in an effort to find fault with my statement that "copters react with lift to the air they accelerate downward".

But I'm with the scientists, who say that ground effect is still not thoroughly understood. So, trying to use something like ground effect, to understand lift, is kind of kooky, to me.

Can we stay away from the ground for a while?
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Old Jan 03, 2013, 10:27 AM
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You've made the general assertion that a lifting wing or rotor always results in the air's downward momentum increasing at a rate equal to the lift force.

I have suggested that this is not always true, that instead the air's rate of vertical momentum change depends not only on the wing/rotor lift, but also on how the air is bounded.

There's nothing mysterious about the experiment I have described. Introducing the upper plate either changes the force exerted by the air on the ground by the same amount that it changes the lift acting on the rotor or it doesn't

For your assertion to be true, it must hold under all conditions, including ground effect.
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Old Jan 03, 2013, 10:47 AM
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If'n the air recirculates- lift decreases
or if'n the air supply to the rotors is starved - net movement and pressure difference decreases
pretty straightforward.
Next?
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Old Jan 03, 2013, 01:30 PM
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Originally Posted by ShoeDLG View Post
You've made the general assertion that a lifting wing or rotor always results in the air's downward momentum increasing at a rate equal to the lift force.

I have suggested that this is not always true, that instead the air's rate of vertical momentum change depends not only on the wing/rotor lift, but also on how the air is bounded.

There's nothing mysterious about the experiment I have described. Introducing the upper plate either changes the force exerted by the air on the ground by the same amount that it changes the lift acting on the rotor or it doesn't

For your assertion to be true, it must hold under all conditions, including ground effect.
That's true - and it does, but in ground effect, it's not easy to separate it from the rest of what's going on. Real scientists have stated they don't understand ground effect - Again, I accuse you of keeping the discussion in ground effect because it gives you leeway to obfuscate the discussion.

Keep it simple. Understand it without the ground first - then you can more confidently try understanding what happens to the lift when close to the ground.

And you ain't understanding it one little bit, away from the ground. Here, I'll prove it by asking you one simple question ...

What object will react with an equal and opposing force to the air a chopper accelerates downward?
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Old Jan 03, 2013, 02:10 PM
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Seriously?
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Old Jan 03, 2013, 02:36 PM
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If a rotor or wing is the only object exerting an external force on the air, then the air's rate of momentum change has to be equal and opposite the aerodynamic force acting on the rotor or wing. This is a straightforward application of Newton's Second and Third Laws. I have never suggested otherwise.

However, as soon as another object (such as a wall, ceiling or the ground) exerts a force on the air, Newton's Laws no longer require the air's rate of momentum change to be equal and opposite the aerodynamic force acting on the rotor/wing. Newton's Second Law very clearly says that the air's rate of momentum change is equal to the sum of the forces acting on it.

Sum(F_air) = d(p_air)/dt

For a wing in level flight over the ground, Newton's third Law can be used to write the vertical component of this equation as:

Lift - F_air_ground = air's rate of downward momentum change

Only in the case where the net vertical force exerted by the ground on the air (F_ground_air) is zero is the air's rate of downward momentum change equal to the lift on the wing.

You have stated that the air's rate of downward momentum change is equal to the the lift on the wing under all conditions. You can't make this statement and then retreat to the position that the realtionship between lift and momentum is too hard to understand in the presence of the ground.

Interaction between the ground (or other rigid surface) and a lifting wing is not some poorly-understood mystery. Engineers have been accurately compensating for wind tunnel wall interactions for more than a century. As I stated in post #962, the results of the "upper plate" experiment clearly show that either:

-The force exerted by the air on the ground does not accurately reflect the rate at which the rotor is transferring downward momentum ot the air.

-or-

-The assertion that the air's rate of downward momentum change must be equal and opposite the lift acting on a helicopter rotor in ground effect is not correct.

There's no obfuscation here, just very simple logic.
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Old Jan 03, 2013, 04:24 PM
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Originally Posted by ShoeDLG View Post
If a rotor or wing is the only object exerting an external force on the air, then the air's rate of momentum change has to be equal and opposite the aerodynamic force acting on the rotor or wing. This is a straightforward application of Newton's Second and Third Laws. I have never suggested otherwise.

However, as soon as another object (such as a wall, ceiling or the ground) exerts a force on the air, Newton's Laws no longer require the air's rate of momentum change to be equal and opposite the aerodynamic force acting on the rotor/wing. Newton's Second Law very clearly says that the air's rate of momentum change is equal to the sum of the forces acting on it.

Sum(F_air) = d(p_air)/dt

For a wing in level flight over the ground, Newton's third Law can be used to write the vertical component of this equation as:

Lift - F_air_ground = air's rate of downward momentum change

Only in the case where the net vertical force exerted by the ground on the air (F_ground_air) is zero is the air's rate of downward momentum change equal to the lift on the wing.

You have stated that the air's rate of downward momentum change is equal to the the lift on the wing under all conditions. You can't make this statement and then retreat to the position that the realtionship between lift and momentum is too hard to understand in the presence of the ground.

Interaction between the ground (or other rigid surface) and a lifting wing is not some poorly-understood mystery. Engineers have been accurately compensating for wind tunnel wall interactions for more than a century. As I stated in post #962, the results of the "upper plate" experiment clearly show that either:

-The force exerted by the air on the ground does not accurately reflect the rate at which the rotor is transferring downward momentum ot the air.

-or-

-The assertion that the air's rate of downward momentum change must be equal and opposite the lift actingg on a helicopter rotor in ground effect is not correct.

There's no obfuscation here, just very simple logic.
Ok, in your own wierd way, you are agreeing with my one statement - a wing or rotors lift is a Newtonian reaction to the action of accelerating air downward.

Cool.

Wasn't easy - just look at the size of this thread.
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Old Jan 04, 2013, 05:52 AM
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I agree that, under a very specific condition, wing or rotor lift must be a pure Newtonian reaction to accelerating air downward.

That specific condition is: when the the wing or rotor is the only object exerting a force on the air. This circumstance is rare for wings and rotors that operate on planets.

You have declared that the air's rate of downward momentum change is always equal to the lift. If this is always the case, then it must be the case in ground effect.

I have offered a simple experiment demonstrating that placing a rigid surface near a rotor can significantly alter the balance between the air's rate of downward momentum change and the lift on a rotor. Your response has been to suggest the conditions of the experiment are too difficult to understand (post #963).

If you do not understand the balance between the air's rate of downward momentum change and the lift on a wing or rotor under all conditions, then you cannot (with any credibility) assert that a particular balance exists under all conditions. Are you agreeing that the air's rate of downward momentum change may not be equal to the lift in ground effect?
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Old Jan 04, 2013, 11:15 AM
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Originally Posted by ShoeDLG View Post
I agree that, under a very specific condition, wing or rotor lift must be a pure Newtonian reaction to accelerating air downward.

That specific condition is: when the the wing or rotor is the only object exerting a force on the air. This circumstance is rare for wings and rotors that operate on planets.
That so called specific condition sounds a lot like normal flight (no ground involved), non-specific in other words.

Where on earth did you come up with the idea that this was rare?
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Old Jan 04, 2013, 11:47 AM
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Do you agree that in ground effect, the air's rate of downward momentum change may not be equal and opposite the lift?
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Old Jan 04, 2013, 02:07 PM
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Do you agree that in ground effect, the air's rate of downward momentum change may not be equal and opposite the lift?
Yes - we all know that for a given air speed and AOA, a wing's lift increases near the ground.

Meaning there's something in ground effect, that's not there in normal flight, which makes the plane more efficient at lifting, when close to the ground.

Some say a cushioning effect of the air funneled between ground and wing at high AOA. Which is my fav, cause you can take this to the extreme, a low wing plane, with wheels embedded in wings such that the wing sat within inches of the ground - at this extreme, we all know that there'd be a cushion of air, and that it would have an effect. So, take away the extreme, put the normal landing gear back on the plane, and we still have the same air cushion effect, but much diminished due to greater wing to ground distance.

Some say that the upwash's negative effect on lift is decreased near the ground, making the wing more efficient.

In either case above, it would mean less downwash required for given lift at given AOA and air speed.

Some say no one's really nailed this decisively yet.
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Old Jan 04, 2013, 02:43 PM
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We agree that the air's rate of downward momentum change does not have to be equal and opposite the lift. For a lifting wing in ground effect there are two objects pushing on the air: the wing and the ground. Newton's Second Law says that the air's rate of downward momentum change is equal to the net downward force acting on it. Do we also agree that:

air's rate of downward momentum change = lift - F_air_ground

where F_air_ground is the net upward force exerted by the ground on the air?
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Old Jan 05, 2013, 02:53 PM
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Originally Posted by ShoeDLG View Post
We agree that the air's rate of downward momentum change does not have to be equal and opposite the lift. For a lifting wing in ground effect there are two objects pushing on the air: the wing and the ground. Newton's Second Law says that the air's rate of downward momentum change is equal to the net downward force acting on it. Do we also agree that:

air's rate of downward momentum change = lift - F_air_ground

where F_air_ground is the net upward force exerted by the ground on the air?
Nope - cause you are putting a stake in the ground - that ground effect is due to the air pushing back - the cushion effect. And as I said before, I'm not sure what causes ground effect yet.
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