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

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hamster_nz:

--- Quote from: electrodacus on July 22, 2022, 03:54:45 am ---Do you think that measuring anything external to wire like electric or magnetic field will help you know how much energy was dissipated to each side ? Since if energy was delivered from outside to the wire surface then you should be able to make measurements related to that.

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If somebody does show you how to do this, without being in electrical contact with the wire, you will just say it proves nothing, and change the goalposts again, or say "but by doing so you change the geometry of the system!". Fruitless.

Unless you say "If anybody can show me a method to determine how much energy was dissipated in each side without making electrical contact to the wires, then I accept that the energy flow is not in the wire" then why bother?

And even if you do say that (which you won't), then you will argue that "electrical contact" is loosely defined, such that the method given is invalid. You will start saying things like "electrical contact includes a capacitor!" or "it's a transmission line", or "but stored energy!"

gnuarm:

--- Quote from: electrodacus on July 21, 2022, 03:35:01 pm ---
--- Quote from: gnuarm on July 21, 2022, 06:57:29 am ---It's not a matter of not "getting it".  I am saying you are dead wrong about that.  There is zero reason to think that is correct.

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Let's look at the extreme case.

a) Wheels are not connected and free to rotate zero friction.   The vehicle will just stay in place while the wheel on treadmill will rotate.
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Not exactly, but close.  If we eliminate friction the inertia of the wheels will cause a force on the bearing of the axle, moving the car.  How much?  That depends on the masses and relative inertia of each.  Eliminate inertia and all extraneous forces, then maybe, yes, the wheel on the treadmill will only turn.

--- Quote ---b) Wheels are welded to vehicle frame.  The vehicle will either stay in place if the wheel on treadmill slips or it will move from right to left if wheel on red box slips.
c) Wheels are connected with a twisted belt (think at infinity symbol shape ∞)  and pulley have the same diameter 1:1 gear ratio. The vehicle will move easily from right to left so in the direction the treadmill moves.
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The twisted belt simply means the wheels turn in opposite directions.  Having a twisted belt connect the two axles, means they turn in opposite directions.  If the treadmill turns, that wheel will turn in that direction, and the other wheel will turn  in the opposite direction.  With a 1:1 ratio, this can not be accommodated without a wheel slipping.

--- Quote ---d) Wheels are connected as in my diagram but pulley are equal gear ratio 1:1.  The vehicle will wheels will be locked no different than point b) so one of vehicle wheels will need to slip.
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You keep screwing up this idea of locked wheels.  YOU HAVE A MODEL.   PICK IT UP AND TURN THE WHEELS.  THEY ARE NOT LOCKED.  It would only be a problem if you put it on a single surface and tried to roll it.  Since the ratio between the wheels is not 1:1, they can't roll on a single surface.  With a 1:1 ratio, they will roll.  This is how 4 WD vehicles work.  They don't use a differential between front and rear wheels, yet, they roll along the ground.

--- Quote ---Hopefully you can agree with all 4 points.
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LOL, hardly.  YOU HAVE A MODEL.  Until you try this with your model, you are ignorant of the facts.

--- Quote ---If you do then the only thing you need to understand is that when gear ratio is say 2:1 the force F2 can not be larger than F1 not until wheel on treadmill slips.
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The gear ratio does not determine the forces between the wheels and the surfaces, just as Newton's third law does not determine them either.

--- Quote ---Newton's 3'rd law applies in this particular case due to the place and direction that  F1 is applied to vehicle.
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Rubbish

--- Quote ---Maybe imagine treadmill moving in the other direction all things being equal as in my diagram. The gearbox will still be locked and one of the wheels will still need to slip.

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Until you learn there is nothing locked about the belted wheels, you are doomed to failure.  Your rocket ship exploded on the launch pad.

What a buffoon!

Pick up your model and turn the wheels.  Make a quick video and show us the wheels won't turn.

electrodacus:

--- Quote from: hamster_nz on July 22, 2022, 04:42:11 am ---If somebody does show you how to do this, without being in electrical contact with the wire, you will just say it proves nothing, and change the goalposts again, or say "but by doing so you change the geometry of the system!". Fruitless.

Unless you say "If anybody can show me a method to determine how much energy was dissipated in each side without making electrical contact to the wires, then I accept that the energy flow is not in the wire" then why bother?

And even if you do say that (which you won't), then you will argue that "electrical contact" is loosely defined, such that the method given is invalid. You will start saying things like "electrical contact includes a capacitor!" or "it's a transmission line", or "but stored energy!"

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You can do a thing as simple as measure the temperature of the wires like with a thermal camera and you will not need to touch the wires but what that camera detects are infrared photons emitted from the wires so that is radiated emission.
You put in 200Ws from the battery in the wires you eventually get 200Ws worth of infrared photons and there is nothing else. 200Ws in and 200Ws out.
Both the magnetic and electric field around the wire's is conservative and it is due to moving charged particles (electrons).

Magnetic field around the wire will be the same around the 9Ohm as it will be around the 1Ohm pipe/wire.
Electric field will also be symmetrical as is based on imbalance of charge and outside surface is the same for both halfs.

Nominal Animal:

--- Quote from: electrodacus on July 22, 2022, 03:54:45 am ---I made it more clear
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No, you defined another system.  I don't care about the answer to your question about your system, I'm saying that you are each talking about different systems, which means there is no common argument at all.

hamster_nz:

--- Quote from: electrodacus on July 22, 2022, 05:09:18 am ---Electric field will also be symmetrical as is based on imbalance of charge and outside surface is the same for both halfs.

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What is the voltage at the midpoint? Directly across from where it is fed? Unless it is 0V it will not be symmetrical.