Author Topic: Newton's third law problem.  (Read 26663 times)

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Online Kleinstein

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Re: Newton's third law problem.
« Reply #200 on: November 28, 2022, 08:19:55 pm »
The equation / expression alone does not prove anything:  one has to use the formulas in the right context. Just because a formula applies to one context does not mean it is also correct in another context.

An important point constantly wrong by electrodacus is assumung a balance of power to decide if a vehicle can move. The seems to be his starting point leeding to lots of stupid mistakes and claims.
The more relevant point is the balance of forces.
 
 

Offline fourfathom

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Re: Newton's third law problem.
« Reply #201 on: November 28, 2022, 08:40:57 pm »
You do realize you try to describe an overunity device ? If vehicle is direct down wind aboe wind speed.

But I didn't describe an over-unity device.  If I had, the device would accelerate with zero (ground-referenced) wind speed.  Nobody is claiming this, and no experiments have shown this, because this *ISN'T* over-unity.

Your cherished equation for "wind power available to a wind powered vehicle" applies to an isolated wind-generator mounted on a moving vehicle, or mounted on the ground, assuming the equation is correct, which may be the case.  Yes, of course, at zero relative windspeed that wind-generator will deliver no power.  But that's not the situation, and your equation does not apply.

Stop thinking of the propeller as a wind-generator.  That's not how it's operating in the DDWFFTW situation.  The propeller provides the propulsion -- it's a "pusher prop".

Please go back to the wheels, gears, and two surfaces model.  Try to understand how the gears are not locked.  Otherwise you are doomed.
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Offline electrodacusTopic starter

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Re: Newton's third law problem.
« Reply #202 on: November 28, 2022, 08:41:57 pm »
The equation / expression alone does not prove anything:  one has to use the formulas in the right context. Just because a formula applies to one context does not mean it is also correct in another context.

An important point constantly wrong by electrodacus is assumung a balance of power to decide if a vehicle can move. The seems to be his starting point leeding to lots of stupid mistakes and claims.
The more relevant point is the balance of forces.

Yes you are correct in the fact that formula needs to be applied in the correct context also.

So here is the simplest test that shows what I'm claiming.
Have an electric vehicle (simpler that way) even an ebike drive directly upwind within a known wind speed and see what power is needed to start moving.

All of you seems to be saying this below is power needed by the motor (correct me if you think something else) when vehicle wants to move upwind:

 P = Fd * vehicle speed

I (and others) say:

P = Fd * (wind speed - (-vehicle speed)) = Fd * (wind speed + vehicle speed).

The difference is so large that it will be easy to measure.

This calculator can be used for the equation I agree with https://www.electromotive.eu/?page_id=12&lang=de

A day with 30km/h winds should be easy to find (right now at my location is 50km/h with 70km/h gust)

Anyone that biked with a 30km/h wind gust will know what a struggle that is so with my prediction power needed to be provided to drive at 5km/h is
We use 0.827 as effective projected area as default in that calculator 

Fd = 0.5 * 1.2 * 0.827 * (35/3.6)2 = 46.9N  (we all agree with this)

Then power needed for propulsion I say
P = 46.9 * (35/3.6) = 456W  a bit less than what that calculator will show 554W as that takes rolling resistance and drivetrain efficiency in to account.

But you say
P = 46.9 * (5/3.6)  = 65W and that is such a large difference that is very easy to verify.






Offline Circlotron

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Re: Newton's third law problem.
« Reply #203 on: November 28, 2022, 09:17:48 pm »
If you are staying at zero speed then no work is being expended. A force maybe, but if there is no movement then no power is being used.

Have you ever had an electric motor providing a force / torque without using energy ? Or even simpler an electromagnet providing a force without using energy.
Even you using your arm to provide a force will be using energy even tho there is no movement and yes that energy will be dissipated as heat.
 
A superconducting motor or electromagnet would continue to supply a non-moving force without any ongoing power being consumed once initially energised. For that matter, so would a permanent magnet. I have seen that “supplying continuous power without movement” argument used by the free energy from magnets crowd.

As far as your arm needing energy to produce a non moving force, that’s true, but you could also get that force by for example driving a wedge into a narrow space. Once you have stopped pressing it in you are no longer putting energy into it but it continues to exert a force against an object. Are we to assume it continues to dissipate energy into this object indefinitely? A force, certainly, but there is no movement, so where is the energy?
 

Online Kleinstein

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Re: Newton's third law problem.
« Reply #204 on: November 28, 2022, 09:19:01 pm »
Driving 5 km/h against a 30 km/h head wind is relatively easy, if you have suitable low gears (e.g. montain bike). The difficult point is more to keep the balance at slow speed, if gusty.
5 km/h is about walking speed and when walking 30 km/h head wind is not hard either.

Providing 460 W is in contrary only possible for a short time for a top athlet.


 

Offline electrodacusTopic starter

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Re: Newton's third law problem.
« Reply #205 on: November 28, 2022, 09:46:32 pm »
A superconducting motor or electromagnet would continue to supply a non-moving force without any ongoing power being consumed once initially energised. For that matter, so would a permanent magnet. I have seen that “supplying continuous power without movement” argument used by the free energy from magnets crowd.

As far as your arm needing energy to produce a non moving force, that’s true, but you could also get that force by for example driving a wedge into a narrow space. Once you have stopped pressing it in you are no longer putting energy into it but it continues to exert a force against an object. Are we to assume it continues to dissipate energy into this object indefinitely? A force, certainly, but there is no movement, so where is the energy?

Opposing magnets and springs can store energy. So there is a potential stored energy. So that wedge you mention takes advantage of materials elastic properties.
You do miss the point that none of your examples will work without brakes.
You have a vehicle on wheels with no brakes that needs to keeps his relative position on the ground using an electric motor.

The main question is the one currently discussed with Kleinstein. That is how much power is needed to advance at any low speed against the wind.
There is an equation for that and we disagree on which one is the correct one.
If the correct one is the one I claim it is then that alone is enough proof that no vehicle can drive upwind without energy storage.

Offline electrodacusTopic starter

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Re: Newton's third law problem.
« Reply #206 on: November 28, 2022, 09:57:46 pm »
Driving 5 km/h against a 30 km/h head wind is relatively easy, if you have suitable low gears (e.g. montain bike). The difficult point is more to keep the balance at slow speed, if gusty.
5 km/h is about walking speed and when walking 30 km/h head wind is not hard either.

Providing 460 W is in contrary only possible for a short time for a top athlet.

I know all this about walking speed and the 300 to 500W sustained by a human is possible for at least a few minutes that is why I selected the values for speed.
The difference is just huge 460W to 560W when frictional losses are added is very different from 65 to 150W you may be claiming.
As an example that calculator will say 85W for 5km/h with no wind so your claim is that wind makes almost no difference when that is not the case.

Also your theory is not consistent.
For example if you release the brakes and let the wind pushing you what will the power provided by the wind be with wind at 30km/h ?
Changing gears will not help with amount of power needed.
If you change in a lower gear yes the force you need to apply at the pedal is lower but the speed will be higher so exactly the same amount of power will be needed.

You will say that since bicycle is at 0km/h the wind power available to accelerate the bicycle will be zero.
While I say is almost 300W witch again correspond to what you will observe in reality.
« Last Edit: November 28, 2022, 10:00:20 pm by electrodacus »
 

Online Nominal Animal

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Re: Newton's third law problem.
« Reply #207 on: November 28, 2022, 09:59:18 pm »
Here's some of the actual physics related to this phenomena, for anyone interested:

Let \$v_v\$ be the vehicle speed with respect to ground, and \$v_w\$ be the wind speed with respect to ground.

For a vehicle that is in continuous contact with ground (through e.g. wheels), the amount of power available from the wind is
$$P_\text{in} = \frac{\kappa}{2} \rho A_\text{in} v_w^3 \tag{1}\label{1}$$
where \$\kappa\$ is the extraction efficiency, up to about 0.593 in theory (Betz's coefficient), and up to about 0.475 in existing devices; \$\rho\$ is the density of air; and \$A_\text{in}\$ is the cross-sectional area of the extraction device. 

In addition to the ground friction losses, the amount of power lost via drag is
$$P_\text{out} = \frac{C_D}{2} \rho A_\text{out} \lvert v_w - v_v \rvert^3 \tag{2}\label{2}$$
where \$C_D\$ is the drag coefficient depending on the shape of the vehicle (varies between approx. \$0.02\$ for airfoils to \$0.5\$ for golfballs and similar), and can be different for \$v_w \gt v_v\$ than for \$v_w \lt v_v\$ (i.e. front-to-back drag coefficient can be and often is different than back-to-front drag coefficient); and \$A_\text{out}\$ is the cross-sectional area of the vehicle.

The only situation where using \$P_\text{out}\$ to model the amount of energy available from airflow to a vehicle makes any sense at all, is when that vehicle is in flight and only uses airflow propulsion, and \$C_D\$ is used to denote the energy extraction efficiency \$\kappa\$, noting that there is absolutely no physical reason or mathematical relation tying \$\kappa\$ to \$D_C\$.

For example, existing airfoil-type wind turbines, ignoring the tower, have \$A_\text{in} = A_\text{out}\$, \$v_v = 0\$ (stationary), and \$\kappa \gt C_D\$.  In current commercial wind turbines, \$C_D \lt 0.1\$, with \$\kappa \gt 0.4\$, i.e. \$\kappa \approx 4 C_D\$.  If they had \$P_\text{out} = P_\text{in}\$, they could not produce any power at all.  It is the difference between \$\kappa\$ and \$C_D\$ that allows a stationary object to extract power from wind.

The reason there is no \$v_v\$ term (typo fixed) in \$\eqref{1}\$ is that the fact that the contact to ground, even if rolling, allows the mechanism to balance static forces with forces against the ground.  A similar effect with the keel is extensively used in sailing; the keel having very little drag in the direction of travel, but enormous drag in the perpendicular direction, allowing the keel to be used to balance the forces exerted by the wind to the vehicle.  In recent decades, hydrofoils are being used more and more, lifting the vessel upwards, reducing drag (\$C_D\$) due to smaller cross-sectional area \$A_\text{out}\$, without increasing the total drag much at all.

In very rough terms, a sufficiently clever land vehicle can balance the forces from the wind with static forces through the contact with the ground.  Such mechanisms will have some small losses, but at least in theory, they can be reduced without any known lower limit.

While we currently believe the upper maximum limit for \$\kappa\$ is indeed about 0.593 as described by Betz's limit (at essentially \$C_D \approx 0\$), there are no known physical limitations for the ratio of \$\kappa\$ to \$C_D\$: that is just an engineering problem.

Indeed, the Mythbusters experiment by dimpling a Ford Taurus with over a thousand golfball-like dimples, increasing the fuel efficiency from 26 to 29 MPG, shows that many widely used vehicles have an unreasonably large \$C_D\$.  This affects what humans perceive as normal, and results in quite normal but efficient mechanisms obeying Newtonian physics to defy human intuition.  Those interested in such quirks, may find the misconceptions related to Bumblebee flight informative.
« Last Edit: November 28, 2022, 11:16:06 pm by Nominal Animal »
 
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Online Kleinstein

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Re: Newton's third law problem.
« Reply #208 on: November 28, 2022, 10:08:57 pm »
Mechanical power is force times (inner product if using vectors) speed. So at a low speed there is little power needed. How much depends on the motor and gearing. With a suitable low gear the power can be very low and the motor internal friction may be enough. To provide some braking  (intentionally avoid the word power here).  Common languish english seems to use the word "power" also for a few different things than the physical power.

One can use this also to convert physical power back to the force. So the force is power divided by the speed.
I think it should get obvious that the idea of an essentially fixed power even at low speed leads to a contradiction, as this means the force about doubles when you half the speed.
So there must be some wrong point leading to deverging large forces: The formula or power = force times speed is universally accepted, sometimes even used as definition. So the wrong  point is assuing the essentially (vehicle much slower than the wind) constant power from the kinetic energy of the wind as the power needed to move the vehicle.

 

Online Kleinstein

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Re: Newton's third law problem.
« Reply #209 on: November 28, 2022, 10:12:31 pm »

You will say that since bicycle is at 0km/h the wind power available to accelerate the bicycle will be zero.
While I say is almost 300W witch again correspond to what you will observe in reality.

The mechanical power is force times speed. So how on earth can you get 300 W with 0 speed as one factor.
A limited force and zero speed naturally gives zero power - where is the problem ?
 

Offline electrodacusTopic starter

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Re: Newton's third law problem.
« Reply #210 on: November 28, 2022, 10:16:45 pm »
Here's some of the actual physics related to this phenomena, for anyone interested:

Let \$v_v\$ be the vehicle speed with respect to ground, and \$v_w\$ be the wind speed with respect to ground.

For a vehicle that is in continuous contact with ground (through e.g. wheels), the amount of power available from the wind is
$$P_\text{in} = \frac{\kappa}{2} \rho A_\text{in} v_w^3 \tag{1}\label{1}$$
where \$\kappa\$ is the extraction efficiency, up to about 0.593 in theory (Betz's coefficient), and up to about 0.475 in existing devices; \$\rho\$ is the density of air; and \$A_\text{in}\$ is the cross-sectional area of the extraction device. 

In addition to the ground friction losses, the amount of power lost via drag is
$$P_\text{out} = \frac{C_D}{2} \rho A_\text{out} \lvert v_w - v_v \rvert^3 \tag{2}\label{2}$$


OK so you agree with the correct equation for power available to any wind powered vehicle. Keep in mind that first equation is only for a stationary vehicle.
All you did is add the efficiency while I took the ideal case with 100% efficiency to be generous.

Glad you find the equations somewhere and agree with what I'm saying.

Offline electrodacusTopic starter

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Re: Newton's third law problem.
« Reply #211 on: November 28, 2022, 10:17:58 pm »

The mechanical power is force times speed. So how on earth can you get 300 W with 0 speed as one factor.
A limited force and zero speed naturally gives zero power - where is the problem ?

Wind speed is not zero is 30km/h
Vehicle speed is zero.

Offline electrodacusTopic starter

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Re: Newton's third law problem.
« Reply #212 on: November 28, 2022, 10:21:08 pm »
So the wrong  point is assuing the essentially (vehicle much slower than the wind) constant power from the kinetic energy of the wind as the power needed to move the vehicle.

That is exactly the case when vehicle drives upwind. Once you understand that you will understand all that I was saying.

To give you maybe a more intuitive example.

Imagine the vehicle is colliding with balls say 8 balls of 1.2kg each per second. That amount of lost kinetic energy will need to be replaced by the motor in order to maintain speed.
« Last Edit: November 28, 2022, 10:24:58 pm by electrodacus »
 

Offline IanB

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Re: Newton's third law problem.
« Reply #213 on: November 28, 2022, 10:22:58 pm »
Common languish english seems to use the word "power" also for a few different things than the physical power.

I'm not sure how it translates in German, but in English technical language, thermodynamics has the two concepts of work and heat (first law: ΔU = Q - W). Mechanical work is force times distance, while heat relates to the molecular energy of materials. It gets confusing with power, because power applies to the rate of doing work, and also to the rate of transfer of heat (there are no different words for mechanical power and thermal power). Thus, you can have a 10 kW gas furnace that does no work at all.

Because of this, careful thermodynamic arguments tend to use Work and Heat as the terms of reference in order to be precise.
 

Online PlainName

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Re: Newton's third law problem.
« Reply #214 on: November 28, 2022, 10:27:37 pm »
There is a simple question.

What is the wind power available to a wind powered vehicle ?

My answer (and not just mine) is this https://scienceworld.wolfram.com/physics/DragPower.html

That equation alone is what proves "your" explanation of how this vehicle's work wrong.

No it doesn't. Let's just assume that the equation is correct (there is some doubt) - you're not applying it appropriately. Specifically, you're ignoring the thrust from the propeller. There is thrust (it's a turning propeller, after all) so where does that figure in your equation?

Quote
All you need to do to prove me wrong is do an experiment that shows that equation is incorrect.

Pointless. There are now a large number of such experiments showing that the thing works - pick any of those and there is your proof. All you're saying there is "I want you to waste your time and then I'll find some dubious reason why I'll still ignore the evidence." If the equation says to you that those experiments that work can't actually work, then either the equation is wrong or you're using the wrong one.
 

Offline IanB

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Re: Newton's third law problem.
« Reply #215 on: November 28, 2022, 10:28:21 pm »
Keep in mind that first equation is only for a stationary vehicle.

No, don't keep that in mind. The formula is also for moving vehicles. Stop this ridiculous behavior of changing what people say and then pretending to agree with them.
 

Offline electrodacusTopic starter

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Re: Newton's third law problem.
« Reply #216 on: November 28, 2022, 10:47:33 pm »

No, don't keep that in mind. The formula is also for moving vehicles. Stop this ridiculous behavior of changing what people say and then pretending to agree with them.

Where in that equation do you see the vehicle speed ?

Offline IanB

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Re: Newton's third law problem.
« Reply #217 on: November 28, 2022, 10:51:34 pm »
Where in that equation do you see the vehicle speed ?

Also, stop this ridiculous behavior of asking snarky and patronizing questions.
 

Offline electrodacusTopic starter

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Re: Newton's third law problem.
« Reply #218 on: November 28, 2022, 10:52:37 pm »
No it doesn't. Let's just assume that the equation is correct (there is some doubt) - you're not applying it appropriately. Specifically, you're ignoring the thrust from the propeller. There is thrust (it's a turning propeller, after all) so where does that figure in your equation?

The equation is correct.
We are discussing the simpler upwind version.
The propeller trust in the direct downwind version can only be lower than the braking at the wheel witch powers that propeller. 

Pointless. There are now a large number of such experiments showing that the thing works - pick any of those and there is your proof. All you're saying there is "I want you to waste your time and then I'll find some dubious reason why I'll still ignore the evidence." If the equation says to you that those experiments that work can't actually work, then either the equation is wrong or you're using the wrong one.

"The thing works" When did I claim it does not ?
It just does not work the way you think it does.

Offline electrodacusTopic starter

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Re: Newton's third law problem.
« Reply #219 on: November 28, 2022, 10:54:32 pm »

Also, stop this ridiculous behavior of asking snarky and patronizing questions.

I think that is a perfectly normal question to ask. If you think my questions are patronizing then that is your problem not mine.

Offline fourfathom

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Re: Newton's third law problem.
« Reply #220 on: November 28, 2022, 11:02:42 pm »

No, don't keep that in mind. The formula is also for moving vehicles. Stop this ridiculous behavior of changing what people say and then pretending to agree with them.

Where in that equation do you see the vehicle speed ?

If you had bothered to read Nominal's post for content, you would have seen this:
Quote
The reason there is no vw term in (1) is that the fact that the contact to ground, even if rolling, allows the mechanism to balance static forces with forces against the ground.

He should have written "no vv term", (there is a vW term in equation 1, but no vv), but from context the meaning should have been obvious.
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Online Nominal Animal

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Re: Newton's third law problem.
« Reply #221 on: November 28, 2022, 11:15:14 pm »
OK so you agree with the correct equation for power available to any wind powered vehicle.
Finally!  Yes, thank you.  Have you finally realized your error?

Keep in mind that first equation is only for a stationary vehicle.
Nope, you need to learn to read.

All you did is add the efficiency while I took the ideal case with 100% efficiency to be generous.
No, I wrote down the correct equations.

What you do, is take the amount of drag a vehicle experiences, and somehow equate that with the available power.
That is beyond stupid, replacing the entire field of aerodynamics with a singular concept, "drag", that somehow explains all phenomena.

You must be a troll, or someone on the autistic spectrum with severe learning and understanding difficulties.  Nothing else can explain this denseness.

Any wagers on which one it is, anyone?
 

Offline electrodacusTopic starter

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Re: Newton's third law problem.
« Reply #222 on: November 28, 2022, 11:15:28 pm »

If you had bothered to read Nominal's post for content, you would have seen this:


I had read his entire post. I know what I was asking.
He is getting confused by the two equations and how they apply.

If you look at the original post here the main discussion is about the direct downwind verizon as that is simpler to debunk.

There is only one equation that describes wind power available to a vehicle for any type of wind powered vehicle but applies also to a stationary wind turbine and also to power needed to overcome drag.

They are all one and the same thing as is all about elastic collisions between an object and air particles.

So a vehicle driving at 120km/h in a day without wind will need the same amount of power to overcome drag as a vehicle driving at 20km/h with a 100km/h headwind.
It is always about the wind speed relative to vehicle and if vehicle is stationary you will remain only with the wind speed in that equation.

It is irrelevant if air, vehicle or both move is all about the relative speed between the two.

So if a heavy car at 120km/h hits a stationary lightweight motorcycle the energy of the impact will be the same as if motorcycle drove at 100km/h and collided head on with the car driving at 20km/h. 

Offline electrodacusTopic starter

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Re: Newton's third law problem.
« Reply #223 on: November 28, 2022, 11:23:15 pm »
No, I wrote down the correct equations.

What you do, is take the amount of drag a vehicle experiences, and somehow equate that with the available power.
That is beyond stupid, replacing the entire field of aerodynamics with a singular concept, "drag", that somehow explains all phenomena.

You must be a troll, or someone on the autistic spectrum with severe learning and understanding difficulties.  Nothing else can explain this denseness.

Any wagers on which one it is, anyone?

That is correct the equation is universal it can calculate the power required to overcome drag and that is also the same with ideal case available wind energy.

That is not stupid it is the reality.

If you have a stationary vehicle and release the brakes wind will accelerate that with the amount of power provided by that single equation.
Also if you want to know what is the amount of power to overcome drag so same vehicle wants to move upwind the equation will provide minimum power required to start moving assuming drag is the only loss.

I guess you need to do the experiment as you do not understand the concept that from any reference frame the result will be the same so it is irrelevant if air particles collide with vehicle or vehicle collides with air particles.

Online Nominal Animal

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Re: Newton's third law problem.
« Reply #224 on: November 28, 2022, 11:31:37 pm »
He is getting confused by the two equations and how they apply.
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?
 


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