DDWFTTW - Directly Downwind Faster Than The Wind

[Rick Willoughby]

Originally Posted by InetRoadkill

We determined that the energy demands were 33 Watts. The air is moving past the prop at 1 m/s apparent. So ...

Power / velocity = thrust

33 / 1 = 33 N thrust.


edit: the reason it's not 9 is because your pushing against a moving airstream.

IRK
I am trying to determine if the plane in the attached diagram will accelerate or slow down. I am having trouble with the AST and ASD factors.

As you can see the prop is producing 2000N thrust in the airstream, the air drag is 1000N and the wheel drag is 200N. I do not have a problem with the 200N from the wheels as this is not in the airsteam but the prop thrust of 2000N is pushing on an airstream and the aircraft drag of 1000N is being applied by the airstream.

You will note that the plane is forced to take off in a slight tail wind of 3m/s and the ground speed of the aircraft is currently 10m/s. From my understanding of the IRK Law of Forces Pushing on Air (ILFPA for short) I believe these are essential data to work out the force derating factors.

What are the AST and the ASD in this case? If there is not enough thrust with 2000N how much would the prop need to produce to keep accelerating.

[I hope I am accurate in giving you credit for the ILFPA. I am not a pilot so it might be part of pilot training that I am completely ignorant about. Please advise a reference if you have one on this important part of flight training whereby forces are derated when acting on an airstream.]


Rick W.

[InetRoadkill]

Well clearly there is a problem. We can either balance the forces or we can balance energy. So how do we get both to balance?

[InetRoadkill]

Originally Posted by Rick Willoughby

IRK
I am trying to determine if the plane in the attached diagram will accelerate or slow down. I am having trouble with the AST and ASD factors.

As you can see the prop is producing 2000N thrust in the airstream, the air drag is 1000N and the wheel drag is 200N. I do not have a problem with the 200N from the wheels as this is not in the airsteam but the prop thrust of 2000N is pushing on an airstream and the aircraft drag of 1000N is being applied by the airstream.

You will note that the plane is forced to take off in a slight tail wind of 3m/s and the ground speed of the aircraft is currently 10m/s. From my understanding of the IRK Law of Forces Pushing on Air (ILFPA for short) I believe these are essential data to work out the force derating factors.

What are the AST and the ASD in this case? If there is not enough thrust with 2000N how much would the prop need to produce to keep accelerating.

[I hope I am accurate in giving you credit for the ILFPA. I am not a pilot so it might be part of pilot training that I am completely ignorant about. Please advise a reference if you have one on this important part of flight training whereby forces are derated when acting on an airstream.]


Rick W.

Oops. we cross posted.

Actually, I'm not familiar with those abbreviations. But that's ok.

Since the plane is taking off with a 3 m/s tailwind, the indicated airspeed would be 7 m/s.

The derating comes in from the need to accelerate to a higher kinetic energy before taking off. In other words, if we say that the plane can lift off at 10m/s, we have a certain amount of kinetic energy. But with the tailwind, we now have to accelerate to 13 m/s with a correspondingly higher KE. The power needed to spin the wheels has increased too since they are now traveling faster across the ground. The engine can only supply the KE so fast (limited by its horsepower), so it takes longer to get off the ground.

[Rick Willoughby]

Originally Posted by InetRoadkill

Oops. we cross posted.

Actually, I'm not familiar with those abbreviations. But that's ok.

Since the plane is taking off with a 3 m/s tailwind, the indicated airspeed would be 7 m/s.

The derating comes in from the need to accelerate to a higher kinetic energy before taking off. In other words, if we say that the plane can lift off at 10m/s, we have a certain amount of kinetic energy. But with the tailwind, we now have to accelerate to 13 m/s with a correspondingly higher KE. The power needed to spin the wheels has increased too since they are now traveling faster across the ground. The engine can only supply the KE so fast (limited by its horsepower), so it takes longer to get off the ground.

All I would like to know in the case with the airplane is what does the propeller thrust of 2000N reduce to because it is acting on an airstream and likewise what does the 1000N become because it is being applied by an airstream. Show how you use the KE and horsepower to derate the forces.

I have no pilot training so I am interested in how pilots calculate derating factors for forces acting on, or applied by, an airstream. You did it for my buggy model but I am having trouble applying the same principle to an airplane. I understand this is an area where pilots have a lot of training and knowledge.

Rick W

[Rick Willoughby]

Originally Posted by InetRoadkill

Well clearly there is a problem. We can either balance the forces or we can balance energy. So how do we get both to balance?

The problem is that you refuse to accept that the air can apply power to the system. Once you get over this point it all works out. Irrespective if it a sheet of nylon with the wind applying a force to it or the wind applying a force to a propeller the air is doing work on the system. The power developed by the wind is the force times the windspeed.

To be honest I have never heard anything so absurd as having to derate a force because it is acting on an airstream. A force is a force. It has nothing to do with what it is acting on. If a plane propeller is producing 2000N it is always 2000N. Exactly the same with my buggy 9N is 9N. It does not magically need to increase because it is acting on an airstream. Why did it not have to increase in the earlier examples where it is also acting on the airstream?

Also how can 1N reduce to 0.25N because it is being applied by air.

You are so convinced that it cannot work that you are coming up with stupid calculations. Go back and review what you have written. You will see it is utter stupidity.

[InetRoadkill]

Actually, I did go back and review. And I believe I have solved problem with the forces and energies not balancing. I did make an error with the 33N of thrust. But that doesn't get the cart out of trouble with energy.

Let me ask a couple questions:

* How much energy is the cart demanding? Are we still in agreement that it's 33W? (Actually, it's 36W. The windage should be multiplied by 4m/s like it originally was. Not 1m/s. Another minor mistake.)

* Were is the energy coming from?

[spork]

Originally Posted by Rick Willoughby

You are so convinced that it cannot work that you are coming up with stupid calculations. Go back and review what you have written. You will see it is utter stupidity.

Now you're approaching my level of frustration.

Inet. Let me ask you a serious question. Are you open minded enough to believe this cart may be doing exactly what it appears to be doing? If so, we have all the time in the world to explain how it can and does do that. If not, no worries. We won't burden you with facts.

Gratuitous 2nd question: do you believe that ice-boats can achieve and maintain a downwind velocity component of 3X to 4X the wind speed?

[InetRoadkill]

Originally Posted by spork

Now you're approaching my level of frustration.

Inet. Let me ask you a serious question. Are you open minded enough to believe this cart may be doing exactly what it appears to be doing? If so, we have all the time in the world to explain how it can and does do that. If not, no worries. We won't burden you with facts.

Gratuitous 2nd question: do you believe that ice-boats can achieve and maintain a downwind velocity component of 3X to 4X the wind speed?

According to the ice boat people I've talked too, they can only obtain that kind of speed on a reach with the downwind component less than windspeed. I'll take their word for it.

Regarding the problem at hand, as I've pointed out the numbers presented aren't adding up which means there's a flaw in the proposed model. Once I had a chance to get some sleep and revisit the problem, I was able for see what's going on.

As was pointed out earlier, the cart in the example requires 36W of power to meet the demands of driving the prop and overcoming the other drag factors. The wind is providing 27W under the given conditions which means there's an energy shortfall. That's a problem.

As it turns out, once you resolve the energy discrepancy, you'll find that you can theoretically exceed wind speed and meet the the demands of conservation of energy and balanced forces. (The maximum speed works out to be twice the wind speed.) However, it requires that the prop have 100% efficiency and that the cart have zero drag. That's not likely to happen.

The equations suggest that you can achieve a fraction over wind speed with some drag and prop losses. However, the drag budget is pretty small. It would be difficult to achieve in practice.

[spork]

Originally Posted by InetRoadkill

According to the ice boat people I've talked too, they can only obtain that kind of speed on a reach with the downwind component less than windspeed. I'll take their word for it.

So that's a "no" - you don't believe they can. However, there is plenty of physical evidence and very simple analysis that says otherwise. I'm not sure if you're one for facts or data, but you should consider the GPS data in the attached files.

As it turns out, once you resolve the energy discrepancy, you'll find that you can theoretically exceed wind speed and meet the the demands of conservation of energy and balanced forces.

As it turns out, I resolved any such discrepancy years ago. I've since built several vehicles that go directly downwind faster than the wind, steady state, by a pretty fair margin.

The maximum speed works out to be twice the wind speed.

Wrong. Given arbitrarily small losses, you can achieve an arbitrarily large multiple of the wind speed.

However, it requires that the prop have 100% efficiency and that the cart have zero drag. That's not likely to happen.

Wrong again.

The equations suggest that you can achieve a fraction over wind speed with some drag and prop losses. However, the drag budget is pretty small. It would be difficult to achieve in practice.

And still wrong - even you can do it if you follow my build videos closely; and you'll find you can do better than "a fraction over wind speed".

I'm really curious to know why you trust your flawed analyses over real world results.

[Rick Willoughby]

Originally Posted by spork

Now you're approaching my level of frustration.

Inet. Let me ask you a serious question. Are you open minded enough to believe this cart may be doing exactly what it appears to be doing? If so, we have all the time in the world to explain how it can and does do that. If not, no worries. We won't burden you with facts.

Gratuitous 2nd question: do you believe that ice-boats can achieve and maintain a downwind velocity component of 3X to 4X the wind speed?

Actually IRK amuses me tremendously. Have you seen his theory on force adjustment factors when the force acts on an airsteam. A force of 33N must be applied to create an "adjusted" force of 9N because the force is being applied to an airsteam. Someone has actually given this guy a pilot's licence if you believe what he says although I have my own view on that.

Rick W

[spork]

Originally Posted by Rick Willoughby

Actually IRK amuses me tremendously.

Admittedly, he amuses me as well. But it's in sort of an "oh my god" kind of way.

Someone has acrually given this guy a pilot's licence if you believe what he says although I have my own view on that.

I have little doubt he has a pilot's license. Having a pilot's license qualifies you to discuss aerodynamics and engineering like driving a bus qualifies you to design an F1 engine. Did you happen to see the pilot on Mythbusters that felt his plane could not take off from a moving treadmill!?

[Rick Willoughby]

Originally Posted by spork

... Did you happen to see the pilot on Mythbusters that felt his plane could not take off from a moving treadmill!?

I came into that story at the end of the model testing and wondered what they were on about. Then when I saw the test set up and heard the pilots comment I was flabbergasted. That guy probably went to the same pilot training course as IRK.

I have been trying to get IRK to explain the force derating factor that applies to a force that happens to be acting on an airstream. It seems to change with different conditions of velocity. Did you get to the force derating lecture during your pilots training course? If you have the lecture notes I would like to see them. I am just an ignorant electrical engineer not versed in all those key aspects of physics that enable an airplane to fly. Maybe if I could just grasp the force derating factor for a force acting on an airstream I would have a better understanding of how planes fly.

Rick W

[spork]

Originally Posted by Rick Willoughby

Did you get to the force derating lecture during your pilots training course?

I tried pretty hard not to listen to the aero descriptions in my flight training, and I must have been absent the day they covered that in my undergrad and/or grad aero programs.

[Rick Willoughby]

Originally Posted by InetRoadkill

...
* Were is the energy coming from?

The energy literally comes out of the air. The ground speed of the undisturbed airflow is 3m/s. After the vehicle passes there will be a cylindrical airstream immediately behind the cart prop that has been slowed down. Over time this will gradually be entrained into the general environment as happens with any propeller or wing.

Let me provide some actual numbers. The prop has an apparent wind of 1m/s. The airflow is being accelerated as it passes through the prop. In my particular example the airflow reaches 1.28m/s. The mass flow rate to achieve my nominated prop efficiency with thrust of 9N is calculated as 44kg/s.

So in this case the far field velocity for the prop is 1m/s and the near field is 1.28m/s. When the cart is doing 4m/s ground speed the far field velocity with respect to the ground is 3m/s and the near field velocity in the vicinity of the prop is 2.72m/s.

The kinetic energy reduction in the propeller airstream due to the cart passing can be determined as:
Energy/sec = 0.5 x 44 x (3^2 - 2.72^2) = 36W.

I have done some rounding here but it shows where the energy comes from. If you set up a smoke trail you would see the air speed over the ground is slowed down in the vicinity of the prop as the cart passes by. This energy goes into overcoming the losses in the cart.

Rick W

[InetRoadkill]

Present a general set of equations describing your device.