Electroboom: How Right IS Veritasium?! Don't Electrons Push Each Other??

[electrodacus]

So you agree that you can get much more propulsion out of the propeller than what you took from the wheels, thanks to the wind?
Nice, now you can understand why you can go faster than the wind 

Have you read my comment ?
I took 100W from the wheels to deliver 70W worth of propulsion.
The other 900W worth of propulsion is done by the wind not the propeller
But 970W total propulsion is still less than 1000W total propulsion if I will not have involved the wheels and 70% efficient propeller.

[Naej]

So you agree that you can get much more propulsion out of the propeller than what you took from the wheels, thanks to the wind?
Nice, now you can understand why you can go faster than the wind 

Have you read my comment ?
I took 100W from the wheels to deliver 70W worth of propulsion.
The other 900W worth of propulsion is done by the wind not the propeller
But 970W total propulsion is still less than 1000W total propulsion if I will not have involved the wheels and 70% efficient propeller.

So you agree that you can get much more propulsion (970W) than what you took from the wheels (100W), thanks to the wind (900W) ?
And the car can go much faster than the wind.

[electrodacus]

So you agree that you can get much more propulsion (970W) than what you took from the wheels (100W), thanks to the wind (900W) ?
And the car can go much faster than the wind.

Wind was 1000W worth of propulsion before you involved the wheels and propeller so you think 970W is more than 1000W ?

[Naej]

So you agree that you can get much more propulsion (970W) than what you took from the wheels (100W), thanks to the wind (900W) ?
And the car can go much faster than the wind.

Wind was 1000W worth of propulsion before you involved the wheels and propeller so you think 970W is more than 1000W ?

If you remove the propeller you have 1000W of wind power above your car, and 0W in propulsion… because it's above your car.
If you add the propeller, geared to the wheels you get (at least according to you) 970W in propulsion, and removing the 100W in the wheels you get a nice acceleration.

[electrodacus]

If you remove the propeller you have 1000W of wind power above your car, and 0W in propulsion… because it's above your car.
If you add the propeller, geared to the wheels you get (at least according to you) 970W in propulsion, and removing the 100W in the wheels you get a nice acceleration.

You can remove the propeller and replace with a sail of equivalent area to get 1000W of wind power that accelerates the vehicle.
Yes propeller looks like a variable area sail based on rotational speed and design starting from the area of the blades when it starts and ending with the max area equivalent to the swept area of the propeller.
You get to max equivalent area way before vehicle gests to wind speed and at that point there is no wind power available no matter how large the area is.
So if you had not stored energy while below wind speed then vehicle could not exceed wind speed.

The vehicle (blackbird) can exceed the wind speed only because of energy storage (pressure differential in this case but can be done in many other ways) and that also means it will have a limited period that the vehicle can spend above wind speed directly proportional with amount of stored energy and inversely proportional with the amount of friction losses.

[Naej]

If you remove the propeller you have 1000W of wind power above your car, and 0W in propulsion… because it's above your car.
If you add the propeller, geared to the wheels you get (at least according to you) 970W in propulsion, and removing the 100W in the wheels you get a nice acceleration.

You can remove the propeller and replace with a sail of equivalent area to get 1000W of wind power that accelerates the vehicle.

Wrong.
If the car speed is larger than the wind speed, then with a 1000W wind power, a sail can only brake, for example with 500W (for example).
While with a propeller you get propulsion (970W according to you).

[electrodacus]

Wrong.
If the car speed is larger than the wind speed, then with a 1000W wind power, a sail can only brake, for example with 500W (for example).
While with a propeller you get propulsion (970W according to you).

If the car speed is higher than wind speed direct down wind there is no wind power available to accelerate that vehicle.
P_w = 0.5 * air density * area * (wind speed - vehicle speed)³

[Naej]

Wrong.
If the car speed is larger than the wind speed, then with a 1000W wind power, a sail can only brake, for example with 500W (for example).
While with a propeller you get propulsion (970W according to you).

If the car speed is higher than wind speed direct down wind there is no wind power available to accelerate that vehicle.
P_w = 0.5 * air density * area * (wind speed - vehicle speed)³

Wrong.
P_w approximately equal to 0.5 * air density * area * wind speed² * |wing speed - car speed|
Do you disagree with the famous kinetic energy formula 1/2*m*v^2 or do you disagree with the mass flow?

[electrodacus]

Wrong.
P_w approximately equal to 0.5 * air density * area * wind speed² * |wing speed - car speed|
Do you disagree with the famous kinetic energy formula 1/2*m*v^2 or do you disagree with the mass flow?

I do not disagree with kinetic energy formula. I disagree with the formula you invented for wind power.
Will you say that your formula is universal for any type of wind powered vehicle ?

[Naej]

Wrong.
P_w approximately equal to 0.5 * air density * area * wind speed² * |wing speed - car speed|
Do you disagree with the famous kinetic energy formula 1/2*m*v^2 or do you disagree with the mass flow?

I do not disagree with kinetic energy formula. I disagree with the formula you invented for wind power.
Will you say that your formula is universal for any type of wind powered vehicle ?

So you disagree with air mass flow approximately equal to area * density * relative speed?
Interesting. Why? What's your opinion on the subject and can you substantiate it?

It applies to all vehicles where you have air flowing through an area. So don't apply this to sails.

[Alex Eisenhut]

[electrodacus]

So you disagree with air mass flow approximately equal to area * density * relative speed?
Interesting. Why? What's your opinion on the subject and can you substantiate it?

It applies to all vehicles where you have air flowing through an area. So don't apply this to sails.

I can write the equation less simplified like this
P_w = (area * air density * (wind speed - vehicle sped)) * (wind speed - vehicle speed)² * 0.5  =  0.5 * air density * area * (wind speed - vehicle speed)³
Can you recognise the mass flow ?

This equation (witch is the correct one) applies to all vehicles including those using sails or wind turbines or anything else that you may think off. It is universal.

[Naej]

So you disagree with air mass flow approximately equal to area * density * relative speed?
Interesting. Why? What's your opinion on the subject and can you substantiate it?

It applies to all vehicles where you have air flowing through an area. So don't apply this to sails.

I can write the equation less simplified like this
P_w = (area * air density * (wind speed - vehicle sped)) * (wind speed - vehicle speed)² * 0.5  =  0.5 * air density * area * (wind speed - vehicle speed)³
Can you recognise the mass flow ?

This equation (witch is the correct one) applies to all vehicles including those using sails or wind turbines or anything else that you may think off. It is universal.

So you agree with the mass flow equation? (Except that you invented a negative mass flow  )
But then you changed the kinetic energy part? For some reason.
Your formula is correct in the car reference frame. And in the car reference frame the car has no kinetic energy. Zero, nada, nothing. However the ground is FULL of kinetic energy.
It's much more intuitive to look in the ground reference frame, where wind power is what I said and the car has kinetic energy. And the car can get energy from the wind.

[electrodacus]

So you agree with the mass flow equation? (Except that you invented a negative mass flow  )
But then you changed the kinetic energy part? For some reason.
Your formula is correct in the car reference frame. And in the car reference frame the car has no kinetic energy. Zero, nada, nothing. However the ground is FULL of kinetic energy.
It's much more intuitive to look in the ground reference frame, where wind power is what I said and the car has kinetic energy. And the car can get energy from the wind.

You do have a big problem understanding energy conservation so not quite sure how I can explain anything to you as all of physics depends on understanding this first law of thermodynamics.
Saying things like more than 100% efficiency is possible and "However the ground is FULL of kinetic energy."
Changing reference frames if done correctly will not change anything so same result no matter what reference frame you use.
So you can not say that my equation is true in some reference frame but not in another. It is either true in all reference frames or in none.

As for negative sign for power or flow it just shows the direction.
When you charge a capacitor you have one direction for current and opposite direction when discharging. You need to differentiate between the two with a sign.
Here you are either charging or discharging the stored kinetic energy of the vehicle.

A vehicle stationary relative to ground will have zero potential energy relative to ground but if there is wind it will have some potential energy relative to air.
And at the other extreme a vehicle traveling at wind speed will have zero potential energy relative to air but some potential energy relative to ground based on speed relative to ground and the mass of the vehicle.

Without using energy storage the vehicle (no matter the design) can only have a speed relative to ground between zero and wind speed and nothing outside this range so no negative speed (direct upwind) and no higher than wind speed.

[Naej]

So you agree with the mass flow equation? (Except that you invented a negative mass flow  )
But then you changed the kinetic energy part? For some reason.
Your formula is correct in the car reference frame. And in the car reference frame the car has no kinetic energy. Zero, nada, nothing. However the ground is FULL of kinetic energy.
It's much more intuitive to look in the ground reference frame, where wind power is what I said and the car has kinetic energy. And the car can get energy from the wind.

You do have a big problem understanding energy conservation so not quite sure how I can explain anything to you as all of physics depends on understanding this first law of thermodynamics.
Saying things like more than 100% efficiency is possible and "However the ground is FULL of kinetic energy."

Alright what's the kinetic energy of 1kg of ground if the car is moving at 10 m/s ?

Changing reference frames if done correctly will not change anything so same result no matter what reference frame you use.

True. But you're so confused by the car reference frame that you don't understand that ground has kinetic energy in this frame.
So it's better to stay in the ground reference frame.

So you can not say that my equation is true in some reference frame but not in another. It is either true in all reference frames or in none.

Wrong. The kinetic energy in air depends on the speed, which depends on the reference frame.

As for negative sign for power or flow it just shows the direction.
When you charge a capacitor you have one direction for current and opposite direction when discharging. You need to differentiate between the two with a sign.
Here you are either charging or discharging the stored kinetic energy of the vehicle.

A vehicle stationary relative to ground will have zero potential energy relative to ground but if there is wind it will have some potential energy relative to air.
And at the other extreme a vehicle traveling at wind speed will have zero potential energy relative to air but some potential energy relative to ground based on speed relative to ground and the mass of the vehicle.

Sure, if you have a flow of -1 capacitor/s containing 1J then you have -1W of capacitor power.

But then it means that with -1000W of wind power, I can take 300W with a wind turbine. Because the sign does not matter. 

Without using energy storage the vehicle (no matter the design) can only have a speed relative to ground between zero and wind speed and nothing outside this range so no negative speed (direct upwind) and no higher than wind speed.

You must have asserted this 500 times without any proof, but it's still wrong.

[electrodacus]

But then it means that with -1000W of wind power, I can take 300W with a wind turbine. Because the sign does not matter. 

You must have asserted this 500 times without any proof, but it's still wrong.

Sorry I need to give up on you. Not sure how you can get rid of your wrong understanding of physics and be able to start from scratch.
It will be a waste of my time to continue to argue with you when you get fundamental things wrong. Unless you understand those there is nothing I can do for you.

Sure you can rotate the wind turbine to face the other direction when wind direction relative to vehicle changes and yes you will be able to generate 300 or 400W for 30 or 40% efficient turbine but your vehicle will be slowed down by at least the equivalent of 300 or 400W so your vehicle will slow down.

The direct downwind faster than wind blackbird is a very different vehicle. You need to stop and modify the vehicle you can not use the same vehicle for downwind or upwind.
But say you take the upwind version the one where propeller is used as a generator  and push that to above wind speed directly downwind it will just not work the way you think and it will slow down as soon as you stop pushing.

I was thinking electricity is more of your field so how come you did not went with the 200W 20V 10A CC-CV lab power supply analogy.
Show me how you can get 30V or 15A without using energy storage. Or how you can get any negative voltage without energy storage.

[Naej]

But then it means that with -1000W of wind power, I can take 300W with a wind turbine. Because the sign does not matter. 

You must have asserted this 500 times without any proof, but it's still wrong.

Sorry I need to give up on you. Not sure how you can get rid of your wrong understanding of physics and be able to start from scratch.
It will be a waste of my time to continue to argue with you when you get fundamental things wrong. Unless you understand those there is nothing I can do for you.

Sure you can rotate the wind turbine to face the other direction when wind direction relative to vehicle changes and yes you will be able to generate 300 or 400W for 30 or 40% efficient turbine but your vehicle will be slowed down by at least the equivalent of 300 or 400W so your vehicle will slow down.

Again, this is only correct when there is no wind.

But say you take the upwind version the one where propeller is used as a generator  and push that to above wind speed directly downwind it will just not work the way you think and it will slow down as soon as you stop pushing.

You gave no argument for this whatsoever. But sure.
The only thing you have is the wind power, which you don't know how to interpret, and wrongly believe that ground kinetic energy is 0 in the car reference frame.
You also don't know how to apply any principle of thermodynamics.

I was thinking electricity is more of your field so how come you did not went with the 200W 20V 10A CC-CV lab power supply analogy.
Show me how you can get 30V or 15A without using energy storage. Or how you can get any negative voltage without energy storage.

Here is my power supply analogy.
One long wire is at 20V (it's the wind), another is at 0V (ground).
Make a device which goes fast, say 10 times faster than electron drift speed.

[hamster_nz]

Show me how you can get 30V or 15A without using energy storage. Or how you can get any negative voltage without energy storage.

Just sayin'....

Or just swap the meter probes around...

[electrodacus]

Here is my power supply analogy.
One long wire is at 20V (it's the wind), another is at 0V (ground).
Make a device which goes fast, say 10 times faster than electron drift speed.

Mech
P = F * v

Electric
P = V * I

Both cases 200W = 20 * 10

I asked to provide the circuit you will use at the output of the power supply so that any of those values can be higher than the max available.
If you use a resistor divider you can provide me with any voltage output between 0V and 20V but nothing below or above that same as you can get a direct downwind, wind powered vehicle at any speed between 0 and wind speed but not above or below unless you use energy storage.

And I will really love to see how you can get more than 100% power output after that circuit.   

[electrodacus]

Show me how you can get 30V or 15A without using energy storage. Or how you can get any negative voltage without energy storage.

Just sayin'....

Or just swap the meter probes around...

Nice try.  You did not changed anything other than the way you interpret the data.
I will have a voltmeter at the output of the power supply and one after your circuit with both having the negative probe connected to GND
So both voltmeter will measure the same -20V no inversion done by your circuit.

[hamster_nz]

Show me how you can get 30V or 15A without using energy storage. Or how you can get any negative voltage without energy storage.

Just sayin'....

Or just swap the meter probes around...

Nice try.  You did not changed anything other than the way you interpret the data.
I will have a voltmeter at the output of the power supply and one after your circuit with both having the negative probe connected to GND
So both voltmeter will measure the same -20V no inversion done by your circuit.

  

[Alex Eisenhut]

[rfeecs]

To convert 20V to 30V, power an LED with 20V.  Put enough solar cells in series to produce 30V with the LED powering them.

Energy conversion.  No energy storage.   

Are we back to electrons pushing each other now?

[electrodacus]

To convert 20V to 30V, power an LED with 20V.  Put enough solar cells in series to produce 30V with the LED powering them.

Energy conversion.  No energy storage.   

Are we back to electrons pushing each other now?

Will think more about what energy conversion will mean as an analogy.

But as you are interested in electrons will turn this argument around and say the same thing use an LED to illuminate a solar panel and then you may say that energy transfer was done outside the wires through photons.
But you did not actually transported electrical energy but you converted to electromagnetic radiation (photons) those traveled through air or vacuum and then were converted back to electricity.
Will you accept this as electrical energy travels outside the wires?
O will you say energy traveled outside the wire but not as electrical energy?

What about the closed loops like the LED and battery? or solar cells and some resistive load?
What happens if I put my hand in front of the LED ?  Will energy transfer completely stop? How come nothing similar is possible in Derek's experiment?

Maybe I should have started by asking what is your position regarding electrical energy transfer. Is it done through wires? or outside the wire? 

[rfeecs]

Maybe I should have started by asking what is your position regarding electrical energy transfer. Is it done through wires? or outside the wire?

I’m a microwave semiconductor engineer.  I use whatever model works for the given problem.