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Newton's third law problem.

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electrodacus:

--- Quote from: Nominal Animal on November 28, 2022, 11:31:37 pm ---
--- Quote from: electrodacus on November 28, 2022, 11:15:28 pm ---He is getting confused by the two equations and how they apply.
--- End quote ---
Nope.  They are perfectly in agreement with the experiments that you claim are misconstructed or exhibit some special slip-stick hysteresis or magical energy storage pixies.

They are also perfectly in alignment with my old University Physics textbooks on Classical Mechanics.  I have done the work and passed those courses without problems.  (Modern/quantum mechanics too, but that's not relevant here.)

The only thing there that can be discussed – among physicists – is how balancing static forces against the ground means the relative wind speed is irrelevant.  To laymen, the analog to keels and hydrofoils in sailing vessels provides the concept and the practical example.

Nothing can affect the misconceptions your kind of people have.  You refuse to acknowledge or even perceive anything that is contrary to your preconceptions, and instead disregard them as being "wrong" somehow.  That really reminds me of the people who insist that despite hundreds of millions of murders, communism is still a viable political system; it's just that because they themselves have not been in the lead, nobody has implemented it truly correctly yet.  Their argument, too, is that until you can prove it would not work when they are in charge, it is proven to work and everybody else is wrong.  Invalid logic, irrational thinking, and evasive random argumentation, just to prop up your ego and misplaced beliefs of your own 'understanding'.

It is horrible, and yet interesting, in the pathological sense.  There are those who interview monsters, so why not engage with delusional lunatics pushing irrational concepts?

--- End quote ---

You know why engineers exist ?  It is so that physicist have someone to look-up to :)

Is sad but you will be only convinced by doing the test and seeing the measurement results.  They will match what this calculator provides https://www.electromotive.eu/?page_id=12&lang=en

You do not have a good understanding of what power and energy is and always use just forces that clearly are not properly understood also.
Forces come in pairs always equal and opposite unless they accelerate an object.

Your first equation is a general case for a stationary object relative to ground like a wind turbine fixed to ground.
Second equation is the same equation but for the case where objects moves relative to ground so the speed is the relative speed between wind and the object  and it is the same equation I linked quite a few times https://scienceworld.wolfram.com/physics/DragPower.html  and all of you disagreed with.
The above linked calculator implements this equation correctly.

I need to remind you that all of you not even just on this forum used this equation that is completely wrong meaning it has no applications.
Same speed will be used for calculating the Force to overcome drag as it to calculate power to overcome drag.

Nominal Animal:

--- Quote from: electrodacus on November 28, 2022, 11:48:40 pm ---Second equation is the same equation but for the case where objects moves relative to ground so the speed is the relative speed between wind and the object  and it is the same equation I linked quite a few times https://scienceworld.wolfram.com/physics/DragPower.html  and all of you disagreed with.
--- End quote ---
That equation describes the power needed to overcome drag.  It has nothing to do with harvesting power from the wind, because harvesting power from the wind is not based on drag; the most efficient ones are based on aerodynamic lift.  Lift and drag are not collinear or opposite, they are typically perpendicular forces, and their magnitudes are only very loosely coupled together, based on the shape of the airflow, its orientation with respect to the airflow, and the relative speed of the airflow.  In most useful airfoils, lift is much, much greater than drag.

Scoop-type turbines, like the Savonius vertical-axis wind turbine, are based on drag, but they have a very low efficiency.  There are other "drag" type airfoils that rely on the difference in drag between the two airfoil surfaces.  All those are less efficient than "lift" type airfoils, and their tip velocity – the radial velocity at the blades – is slower than the relative wind speed.  For "lift" type airfoils, the radial velocity is much higher than the relative wind speed; they rotate faster than the wind.

Thus, the drag power equation has no relevance at all to limits of harvesting power from the wind.  None.  It only describes aerodynamic losses, and certain very poor turbine models, nothing more.

As an example, consider the Gorlov helical turbine I described some time back.  It has three airfoils, narrow wing-like elements, that "coil" around the common axis.  In wind tunnel tests, it has reached \$\kappa = 0.35\$, with drag coefficient on the order of \$C_D \approx 0.05\$).  More interestingly, even the drag force vectors are not parallel to the wind, so a vehicle using one could possibly use the small drag as kind of a keel.  Moreover, it typically rotates much faster than the wind speed relative to its axis.

(In water, according to Bachant and Wosnik's experiments in 2011, Gorlov helical turbine lift to drag ratio can reach 70, i.e. \$\kappa \approx 70 C_D\$.)

I don't know how more clearly that can be stated.  You claim I don't know this or that, only because you live in your own private little world with your private little custom physics.  I do not know those, that's for sure; nobody can.

electrodacus:

--- Quote from: Nominal Animal on November 29, 2022, 02:28:54 am ---
--- Quote from: electrodacus on November 28, 2022, 11:48:40 pm ---Second equation is the same equation but for the case where objects moves relative to ground so the speed is the relative speed between wind and the object  and it is the same equation I linked quite a few times https://scienceworld.wolfram.com/physics/DragPower.html  and all of you disagreed with.
--- End quote ---
That equation describes the power needed to overcome drag.  It has nothing to do with harvesting power from the wind, because harvesting power from the wind is not based on drag; the most efficient ones are based on aerodynamic lift.  Lift and drag are not collinear or opposite, they are typically perpendicular forces, and their magnitudes are only very loosely coupled together, based on the shape of the airflow, its orientation with respect to the airflow, and the relative speed of the airflow.  In most useful airfoils, lift is much, much greater than drag.

Scoop-type turbines, like the Savonius vertical-axis wind turbine, are based on drag, but they have a very low efficiency.  There are other "drag" type airfoils that rely on the difference in drag between the two airfoil surfaces.  All those are less efficient than "lift" type airfoils, and their tip velocity – the radial velocity at the blades – is slower than the relative wind speed.  For "lift" type airfoils, the radial velocity is much higher than the relative wind speed; they rotate faster than the wind.

Thus, the drag power equation has no relevance at all to limits of harvesting power from the wind.  None.  It only describes aerodynamic losses, and certain very poor turbine models, nothing more.

As an example, consider the Gorlov helical turbine I described some time back.  It has three airfoils, narrow wing-like elements, that "coil" around the common axis.  In wind tunnel tests, it has reached \$\kappa = 0.35\$, with drag coefficient on the order of \$C_D \approx 0.05\$).  More interestingly, even the drag force vectors are not parallel to the wind, so a vehicle using one could possibly use the small drag as kind of a keel.  Moreover, it typically rotates much faster than the wind speed relative to its axis.

(In water, according to Bachant and Wosnik's experiments in 2011, Gorlov helical turbine lift to drag ratio can reach 70, i.e. \$\kappa \approx 70 C_D\$.)

I don't know how more clearly that can be stated.  You claim I don't know this or that, only because you live in your own private little world with your private little custom physics.  I do not know those, that's for sure; nobody can.

--- End quote ---

There is only one way that air interacts with an object and that is trough elastic collisions between object and air molecules.

So the same equation applies to any type of wind turbine as well as to any wind powered vehicle and also to any other vehicle when you want to know the power needed to overcome drag.

This will be that equation

P = 0.5 * air density * (area * drag coefficient) * (wind speed - vehicle speed)3

So if it is a propeller type wind turbine you use the swept area of the propeller thus there is no drag coefficient involved or you can consider that as 1 and the vehicle speed is zero if the wind turbine is mounted on the ground.

So for wind turbine that same equation will look like (still the same equation)

P = 0.5 * air density * swept area * (wind speed)3 * efficiency  (basically the first equation you mentioned in an earlier comment).

If the wind turbine is installed on a vehicle that drives say upwind same equation will look like this

P = 0.5 * air density * swept area * (wind speed-(-vehicle speed))3 * efficiency

If it is a vehicle driving direct downwind same equation will look like

P = 0.5 * air density * area * coefficient of drag * (wind speed-vehicle speed)3 this is the wind power available to vehicle ideal case

If vehicle drives upwind the same equation shows the power needed to overcome drag

P = 0.5 * air density * area * coefficient of drag * (wind speed-(-vehicle speed))3
 
So there is no other equation it just looks a bit different depending on what it applies to but is the exact same one in all cases.


Look at energy balance for example.

You agree with drag force and I hope you agree with the Kinetic energy loss or gain (depending on what powers what) is drag force times distance so basically the amount of work done.

KE = Fd * d

Getting back to that 5km/h bicycle example with a 30km/h headwind
distance traveled in one second will be 5km/h / 3.6 = 1.39m/s  so 1.39m

KEvehicle = (0.5*1.2*0.827*(35/3.6)2) * 1.39 = 65.2Ws

Now you need to calculate the energy needed to deal with the air drag
That is almost the same just you need to add the distance air traveled over that same 1 second period

KEair = (0.5*1.2*0.827*(35/3.6)2) * 8.33 = 391Ws

Then you will add this two 391 + 65.2 = 456Ws

So total energy that should be provided by vehicle propulsion for one second is 456Ws so basically the same as what average power needs to be.

Also say you want to be powered by the wind so you do not use the motor just let the wind push the bike your acceleration power will be 391W and will drop as the bicycle speed increases so you will need to integrate.
So say over 1ms that 391W will not change then bicycle gained  0.391Ws of kinetic energy and from that and mass of the bike + rider you get the speed.

Kleinstein:
The interaction of the wind with the vehicle is no way close to 100% efficient - with a standing obstacle it is more like 0%. Even a good wind turbine is not close to 100% efficient. So it does not make sense to base the calculation on the power that the wind could provide with 100% efficiency.

So it is not about an energy or power balance, but about the forces.




--- Quote from: electrodacus on November 29, 2022, 03:13:20 am ---
You agree with drag force and I hope you agree with the Kinetic energy loss or gain (depending on what powers what) is drag force times distance so basically the amount of work done.

KE = Fd * d

Getting back to that 5km/h bicycle example with a 30km/h headwind
distance traveled in one second will be 5km/h / 3.6 = 1.39m/s  so 1.39m

KEvehicle = (0.5*1.2*0.827*(35/3.6)2) * 1.39 = 65.2Ws

Now you need to calculate the energy needed to deal with the air drag
That is almost the same just you need to add the distance air traveled over that same 1 second period

KEair = (0.5*1.2*0.827*(35/3.6)2) * 8.33 = 391Ws


--- End quote ---

The power you calculate for  the air drag is coming from the wind and not form vehicle to drive. It gets obvious for a stationary obstacle - not power needed by the obstacle to keep standing. There is no sudden change from stationaly objects and very slow moving objects. The air drag power is taken from wind and converted to heat.   A wind turbine could use part of that power for other purposes, like driving the vehicle.

PlainName:

--- Quote ---If it is a vehicle driving direct downwind same equation will look like

P = 0.5 * air density * area * coefficient of drag * (wind speed-vehicle speed)3 this is the wind power available to vehicle ideal case
--- End quote ---

This is where you are going wrong. Let's try and fix it...
Once again I will say that previously I would have given you one step and let you agree on it before moving to the next, because otherwise you just jump to the chase and insist it can't be done, skipping everything about how we get there. But life's too short and it's your loss.

So, what we want to know is how much power is available to make the vehicle move (because, ultimately, we need power to make it move faster than the wind is blowing). Your Someone's copied equation can be made simpler thus:

Terms:
  Pu - Useful power. This is what we want to use to move the vehicle
  Vw - Wind velocity. Importantly, this is relative to the ground and in the same direction as the vehicle moves.
  Vv - Vehicle velocity. Again, relative to the ground and in the same direction as the wind

The details (right now) of how Pu is derived from Vw and/or Vv are unimportant - it is sufficient to say that Pu is proportional to Vw - Vv. That is:

    Pu ∝ (Vw - Vv)

Your premise is that when the vehicle is at wind speed there is no power available. That is, Pu = 0. And that's correct for this simple case. What we're interested in is a case where Vw - Vv = 0 and Pu > 0.

Moving on, there's the small problem of the propeller, which is driven by the wheels. The propeller is sucking power, via the wheels, so let's add another term:

   Pv - Vehicle power. This is the power the wheels soak up when driving the propeller.

That changes things thus:

    (Pu + Pv) ∝ (Vw - Vv)

or

    Pu ∝ (Vw - Vv) - Pv

Two things to note here: first that now when Pu = 0, Vw - Vv > 0. Second, Pw is proportional to the vehicle speed since it depends on how fast the wheels turn:

    Pv ∝ Vv

Next, another term is needed for the effect of the propeller:

    Vt = Thrust velocity. This is the speed of the air being pushed backwards relative to the vehicle.

We know from seeing jets take off that the vehicle speed is wind speed plus thrust, and you even agreed earlier in this thread.

    Vv = Vw + Vt

What makes this complicated is that Vt is also proportional to the vehicle speed since the power to turn the prop is derived from the speed of the wheels:

    Vt ∝ Vv

OK, so to fill in the blanks, the usable power is proportional to the wind speed less the vehicle speed less the thrust velocity, minus the power to produce the thrust:

    Pu + Pv ∝ Vw - (Vv - Vt)

or

    Pu ∝ (Vw - (Vv - Vt)) - Pv

Essentially, this means that Vw = Vv - Vt can't occur because of the power Pv used up by the prop. In words, there isn't enough power from the wind to have the vehicle speed less thrust get up to wind speed. But, by the same token, if the thrust Vt is high enough then the vehicle speed Vv can be higher than wind speed whilst the same constraint exists. Pv, thus Vt, is proportional to wheel speed, so the faster the wind the higher these are and at some value (left for someone clever to work out if they can be arsed) we  will find that Vt > Vw when Pu != 0.

Edit: noted that Vt is relative to the vehicle whereas Vw and Vv are reltive to the ground.

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