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

--- Quote from: electrodacus on July 14, 2022, 09:41:39 pm ---Everything you responded in the previous post was wrong so there is no place for me to start.
You just live in a fantasy world and get basic physics concepts wrong.
And yes claiming that the 100W input mechanical power taken at the wheel becomes 300W at propeller is pure fantasy with no proof in real world.
And since for some reason you believe Derek's description of how this works you need to come up with this sort of phantasy else the math will not work out.

--- End quote ---
Funny that you understand a 100W compressor can move 300W heat, but you don't understand that a 100W engine can move 300W in kinetic energy/s.
 :-//
electrodacus:

--- Quote from: Naej on July 14, 2022, 10:29:04 pm ---Funny that you understand a 100W compressor can move 300W heat, but you don't understand that a 100W engine can move 300W in kinetic energy/s.
 :-//

--- End quote ---

Air particles can directly transfer kinetic energy to vehicle no "pump needed" but that is the case when vehicle speed is below wind speed.
When vehicle speed is higher than wind speed vehicle will collide with the air particles so the air particle kinetic energy will increase while vehicle kinetic energy will decrease (by a factor of 1:1 in ideal case).

So using the 1.2kg ball analogy when that hits the vehicle it will transfer the kinetic energy to vehicle.
So say ball speed 10m/s and vehicle speed 4m/s both relative to ground.
Say vehicle is 100kg

Vehicle kinetic energy before collision  0.5 * 100 * 42 =800Ws
Ball kinetic energy before collision 0.5 * 1.2 * 102 = 60Ws

Kinetic energy transferred from ball to vehicle
(0.5 * 1.2 * (10-4)2 = 21.6Ws

Over a period of 1 second 6 of this balls will collide with the vehicle as relative to vehicle balls move at (10-4)m/s

To check the above against the Wind power equation
Pw = 0.5 * 1.2 * 1 * (10-4)3 = 129.6W
So over one second this will be 129.6Ws and is the same value as (6 * 21.6Ws)
Both way of making the calculation gives the same result.


When vehicle speed equal wind speed there are no collisions so zero wind power thus no acceleration possible (this is a ideal case with no friction).
 

Then when vehicle speed is higher than air speed the vehicle will bump in to air particle or 1.2kg balls spaced 1m apart.
If you go below the ball wait for the ball to be above and hit that with a bat then the energy you put in to the bat to hit the ball can not put in more energy in to accelerating the vehicle than you took at the wheel.

Say why not put that energy you got at the wheel in to the front wheel (a wheel is more efficient than a propeller but maybe you do not agree with that).
You may say but vehicle moves at say 15m/s while air only moves in the same direction at 5m/s
Let say the front wheel is on a different surface a treadmill that moves backwards at 5m/s to simulate the air.
If you take 100W for one second at the back wheels so 100Ws worth of energy you took that from vehicle kinetic energy (you slowed down the vehicle but again not something you agree with).
So 100W at 15m/s means a brake force of 6.66N and if you want to apply 100W to the front wheel that is on a surface moving at 5m/s relative to vehicle you need a force of 20N
You may look at the forces and think that vehicle will accelerate but that will not be correct as the force is applied to a different medium traveling at different speed.
Naej:

--- Quote from: electrodacus on July 14, 2022, 11:36:25 pm ---Say why not put that energy you got at the wheel in to the front wheel (a wheel is more efficient than a propeller but maybe you do not agree with that).
You may say but vehicle moves at say 15m/s while air only moves in the same direction at 5m/s
Let say the front wheel is on a different surface a treadmill that moves backwards at 5m/s to simulate the air.
If you take 100W for one second at the back wheels so 100Ws worth of energy you took that from vehicle kinetic energy (you slowed down the vehicle but again not something you agree with).
So 100W at 15m/s means a brake force of 6.66N and if you want to apply 100W to the front wheel that is on a surface moving at 5m/s relative to vehicle you need a force of 20N
You may look at the forces and think that vehicle will accelerate but that will not be correct as the force is applied to a different medium traveling at different speed.

--- End quote ---
Yes you got it! Brake force of 6.66N, and accelerating force 20N so the vehicle accelerates!
Or do you disagree with Newton's law too?
electrodacus:

--- Quote from: Naej on July 15, 2022, 12:09:30 am ---Yes you got it! Brake force of 6.66N, and accelerating force 20N so the vehicle accelerates!
Or do you disagree with Newton's law too?

--- End quote ---

Not sure you read fully what I wrote.
is 6.66N against the ground with vehicle relative to that drives at 15m/s
and is 20N applied against a surface that moves at just 5m/s relative to vehicle.
So vehicle will not accelerate as there are 100W and same 100W out. Is ideal case so it will maintain the same speed else it will slow down with this conditions.
If it was 6.66N brake against ground and 20N to accelerate against ground then it will accelerate but 20N at 15m/s is 300W and so you need some stored energy to add the missing 200W.
gnuarm:
I see the lunacy continues.  I get it.  After all, it is a full moon! 

Arrruuuuhhh! 
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