General > General Technical Chat

Newton's third law problem.

<< < (23/55) > >>

PlainName:

--- Quote ---If by pushing you mean accelerating then you need to apply power to the propeller.
--- End quote ---

I stopped reading there since it's obvious you're already predetermined to go off on a distractions again. Recall:

"Forget where the power comes from for the moment and look at how it's actually working"

You even quoted that, yet the first second thing you say is "you need to apply power to the propeller".

And, to round it off, if you're pushing along at 5m/s or 10m/s or <any>m/s you are NOT accelerating. If it were accelerating I would say "it's accelerating".

electrodacus:

--- Quote from: Nominal Animal on November 24, 2022, 07:43:56 pm ---Well, it is better than unphysical nonsense and equations unrelated to the physical phenomena at hand.  You seem to believe that if an equation applies in some specific situation, it must apply always, and that just isn't so.

Like I explained, if a vehicle uses the fixed ground instead of surrounding air in its propulsion mechanism, the amount of energy it can obtain from wind is not relative to the vehicle speed, it is relative to the ground.  So no subtracting vehicle speed from the wind speed relative to ground.
Extracting the energy is only an engineering problem.

--- End quote ---


There is no such thing as ground energy. There is only wind energy.
But more importantly we are only talking about Power

You can get as much energy from the wind as long as direct downwind vehicle is below wind speed.
You can do anything you want with that energy like:
Wasting it as heat (maybe apply brakes).
Increase vehicle kinetic energy and thus speed.
Storing it for later use
Or a combination of the above.



--- Quote from: Nominal Animal on November 24, 2022, 07:43:56 pm ---You insist on using static surfaces, ignoring all pressure-related effects, as if the vehicle is and has to be a simple rectangular box, with no internal mechanisms and only interacting with its environment via elastic collisions.  It is silly, and quite annoying.

--- End quote ---

Surfaces can be static or dynamic like a propeller they will act as a sail and vehicle can gain kinetic energy if air particle hit the static or moving part of the vehicle and will lose kinetic energy if say propeller hits a air particle that has the same speed as the vehicle or it moves in the opposite direction before being hit by the propeller blade.   


--- Quote from: Nominal Animal on November 24, 2022, 07:43:56 pm ---You need to move past your preconceptions.  For example, you could start gently, and consider a vehicle with a very aerodynamic shape, say saucer-shape with a sharp edge, with upper and lower curvatures different (to balance out any vertical forces, including lift, due to the pressure differential), and with a Gorlov helical turbine poking up on top.  It has minimal drag, and the turbine efficiency is about 35%.  It typically operates at tip speed ratio over 1, which means that the induced flow rate due to rotation of the turbine is greater than the flow rate with respect to the axis.  Go read Gorlov's 2001 paper on it, but do recall that it only considers the properties of a stationary turbine.

And before you assert it, no, a moving turbine is not the same thing as stationary turbine in zero-wind situation.  It only matches if we assume the turbine is not moving in either case, and that's not going to happen with a moving turbine.  Because of the tip speed ratio being over 1, even when the wind speed approaches zero relative to the turbine axis, the turbine isn't going to stop (unless it has bad bearings or other significant losses).  Because it can keep accelerating the vehicle all the way, it will be less efficient at that particular point compared to speeds immediately above and below, but it isn't zero even there.

What makes that seemingly simple construction interesting, is the fact that airflow speed varies as a distance from static surfaces like the ground, below the boundary layer.  This means that along the vertical turbine axis, there is always a range of wind speeds relative to the axis, and because the wind direction does not affect the turbine, it can extract energy from the wind even when the vehicle is traveling downwind at the nominal wind speed.  This applies even to wind tunnels (and in those, it is most noticeable, because they have laminar flow patterns by design; in nature, wind tends to be a bit turbulent).  That sort of a vehicle does not even need wheels, it can even use a propeller for the propulsion, as long as it keeps close to a static surface (ground), and still achieve faster than nominal wind speed downwind or upwind.

When you wrap your mind around that, go look at aerodynamics, propulsion, and so on, and expand your understanding from there.  Reconsider the models and practical examples I've shown earlier in this thread. Don't get stuck in your preconceived notions just because you're emotionally heavily invested in them.  Learn to entertain ideas and theories and models that you do not believe are true, and learn how to rationally examine why your beliefs and the idea/theory/model do not agree.

--- End quote ---


OK I will give you a concrete example with numbers that you can verify in a real world test.

35% efficient wind turbine installed on top of a vehicle. Wind turbine swept area say for simplicity 1m2


a) vehicle drives at +10m/s in a day with no wind so wind speed 0m/s
b) vehicle is stationary 0m/s and head wind speed is -10m/s
c) vehicle speed is +10m/s and head wind speed is -20m/s
d) vehicle speed is +10m/s and tail wind is +10m/s
e) vehicle speed is +10m/s and tail wind is +6m/s

For all this cases you use this equation to find what the wind turbine power output will be.

Pw = 0.5 * 1.2 * 1 * 0.35 * (wind speed - vehicle speed)3

If you can prove that this equation will not provide the correct result for any of the a) to e) cases or any other case you can think of then you can say you proved me wrong.

The mistake you make is to think there is not always an equal and opposite reaction to any action basically the title of this thread.

electrodacus:

--- Quote from: PlainName on November 24, 2022, 07:50:19 pm ---
And, to round it off, if you're pushing along at 5m/s or 10m/s or <any>m/s you are NOT accelerating. If it were accelerating I would say "it's accelerating".

--- End quote ---

You can not get there (above wind speed directly downwind) unless you push the vehicle or stored energy pushes vehicle.

So from that point you mentioned 50% above wind speed without any stored energy your vehicle can only decelerate due to frictional losses and there is no way to accelerate as wind power available there is zero and there is no such thing as ground power.

fourfathom:
Nominal, what did I tell you???

Nominal Animal:

--- Quote from: fourfathom on November 24, 2022, 09:27:03 pm ---Nominal, what did I tell you???

--- End quote ---
I'm an easily socially manipulated sucker, I know.  :-[

But the demand to "prove them wrong" just goes over the top.  As if the physical Lego model that behaves as I described and contrary to their description was not proof enough.

Time to use the Ignore list, I guess.

Navigation

[0] Message Index

[#] Next page

[*] Previous page

There was an error while thanking
Thanking...
Go to full version
Powered by SMFPacks Advanced Attachments Uploader Mod