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

--- Quote from: electrodacus on July 10, 2022, 01:23:26 am ---
--- Quote from: Naej on July 10, 2022, 12:42:49 am ---I'm not.
Mass flow is close to the density*area*relative wind speed.
(Not equal because when you have a relative wind speed equal to 0, you can still get some mass flow on your propeller)

The wheel. Perhaps the guy is some gears connected to the wheel.
And the ball stops moving relative to the ground, it loses its kinetic energy.

--- End quote ---

So you changed now to relative wind speed but you need to do the same on the other part of the equation.

--- End quote ---
I didn't change anything and I don't need to change anything.

--- Quote from: electrodacus on July 10, 2022, 01:23:26 am ---There is the vehicle kinetic energy that the guy could have used but that means reduction in vehicle kinetic energy which also means reduction in speed.

--- End quote ---
Yes. But the recoil means you accelerate the car and it is larger.

--- Quote from: electrodacus on July 10, 2022, 01:23:26 am ---I guess the confusion comes from the fact that you think that using power taken at the wheels (breaks the vehicle) can result in more power output at the propeller (propulsion power) witch is obviously not allowed by energy conservation law.

--- End quote ---
No. What I think is that you get more force, but you don't see the difference between force and energy so  ::)

--- Quote from: electrodacus on July 10, 2022, 01:23:26 am ---Then just write it in the simplest form and we will put that to the test (see what prediction it makes in a few different cases and see if that matches reality).
To my understanding and your latest modifications this is what you say all speeds relative to ground.
Pw = 0.5 * air density * area * (wind speed)2 * (wind speed - vehicle speed)

--- End quote ---
First, it wasn't changed. Second, the formula about kinetic energy was proven a long time ago.
So where do you disagree then? Either it's the amount of kinetic energy per kg, or it's the mass flow.
electrodacus:

--- Quote from: Naej on July 10, 2022, 08:15:51 am ---I didn't change anything and I don't need to change anything.

--- End quote ---
If you say so. The relative wind speed was just wind speed before. But not relevant anyway, I just need a final equation you think describes what happens in reality.


--- Quote from: Naej on July 10, 2022, 08:15:51 am ---Yes. But the recoil means you accelerate the car and it is larger.

--- End quote ---
So will it be able to "recoil" without energy storage. There is no chance to put more power in to propeller than you take at the wheel.


--- Quote from: Naej on July 10, 2022, 08:15:51 am ---No. What I think is that you get more force, but you don't see the difference between force and energy so  ::)

--- End quote ---
What has force to do with anything ? Looking at force alone is a bad idea and that is what likely got Derek in trouble.
The one that needs to see the difference between force and power in not me.
Take the case of vehicle at 2x the wind speed (say wind speed is 10m/s and vehicle speed 20m/s)
Apply a 10N force at the wheel for 1ms (with the generator) and the power you take out ideal case will be 10N * 20m/s = 200W and say you do so for 1ms just so that we can also talk about energy and you have 200W * 0.001s = 0.2 Joules = 0.2Ws
Now take this energy 0.2Ws as that is all you have and apply it for 1ms to the propeller again ideal propeller 100% efficient at propulsion
What you get in therms of propulsion is exactly what you got out by slowing down the vehicle.
So 200W applied at propeller for 1ms will be 20N * (10m/s-20m/s) = -200W so -0.2Ws back into increasing vehicle kinetic energy the negative sign is your choice you can have that for generated power or for propulsion power is non-relevant is just to show direction is different putting in or taking out from the vehicle stored kinetic energy.
So looking at force it is 2x as large but due to speed being half the amount of thrust power and so the amount of kinetic energy you put back in vehicle is the same as the amount you took out in ideal case. In real case it will of course be way less.
 


--- Quote from: Naej on July 10, 2022, 08:15:51 am ---First, it wasn't changed. Second, the formula about kinetic energy was proven a long time ago.
So where do you disagree then? Either it's the amount of kinetic energy per kg, or it's the mass flow.

--- End quote ---

Please just confirm that the one I wrote is what you mean. If is not write the one you think is correct. Just one equation do not split it in two parts.
Pw = 0.5 * air density * area * (wind speed)2 * (wind speed - vehicle speed)
PlainName:

--- Quote ---Please just confirm that the one I wrote is what you mean. If is not write the one you think is correct. Just one equation do not split it in two parts.
Pw = 0.5 * air density * area * (wind speed)2 * (wind speed - vehicle speed)
--- End quote ---

I already did that way back, and you've yet to say precisely what's wrong with it (except you 'think it's wrong'), nor have you produced an equation that takes into account all aspects of the device. The one you keep quoting is the wrong one for this situation. As long as you keep fooling yourself about this, you will never get out of your rut.
electrodacus:

--- Quote from: dunkemhigh on July 10, 2022, 05:02:52 pm ---
--- Quote ---Please just confirm that the one I wrote is what you mean. If is not write the one you think is correct. Just one equation do not split it in two parts.
Pw = 0.5 * air density * area * (wind speed)2 * (wind speed - vehicle speed)
--- End quote ---

I already did that way back, and you've yet to say precisely what's wrong with it (except you 'think it's wrong'), nor have you produced an equation that takes into account all aspects of the device. The one you keep quoting is the wrong one for this situation. As long as you keep fooling yourself about this, you will never get out of your rut.

--- End quote ---

I provide the one that is correct many times. You can have it in the same format:
Pw= 0.5 * air density * area * (wind speed - vehicle speed)2 * (wind speed - vehicle speed)

Or the cleaner one that I have posted maybe hundreds of times
Pw = 0.5 * air density * area * (wind speed - vehicle speed)3

Once Naej confirm it is what he means we can test the equation at different vehicle speeds and see what it predicts and if the prediction is what we observe in reality. Spoiler alert it will not.
But maybe my last replay where I explained that higher force does not mean higher power and so no increase in speed even for ideal vehicle will convince him.
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