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Electroboom: How Right IS Veritasium?! Don't Electrons Push Each Other??
electrodacus:
--- Quote from: gnuarm on July 22, 2022, 04:54:58 am ---Pick up your model and turn the wheels. Make a quick video and show us the wheels won't turn.
--- End quote ---
Picking up the model means the chassis of the vehicle is no longer floating so I can apply a force between wheel and chassis. That is not at all the same thing with applying force at the wheels (input and output) with chassis floating.
You pick up the vehicle with one hand on one wheel and the other on the other wheel (do not touch the vehicle body) just the low part of the wheels.
If one wheel is one palm and the other on the other hand just try and bring the hands together. If you do not allow any slip then you will feel that it is a locked system.
gnuarm:
--- Quote from: electrodacus on July 22, 2022, 05:29:40 am ---
--- Quote from: gnuarm on July 22, 2022, 04:54:58 am ---Pick up your model and turn the wheels. Make a quick video and show us the wheels won't turn.
--- End quote ---
Picking up the model means the chassis of the vehicle is no longer floating so I can apply a force between wheel and chassis. That is not at all the same thing with applying force at the wheels (input and output) with chassis floating.
--- End quote ---
Now you are just babbling. What do you mean when you say the wheels are "gear locked"???
--- Quote ---You pick up the vehicle with one hand on one wheel and the other on the other wheel (do not touch the vehicle body) just the low part of the wheels.
If one wheel is one palm and the other on the other hand just try and bring the hands together. If you do not allow any slip then you will feel that it is a locked system.
--- End quote ---
That is an absurd definition with no purpose. However, with a gear ratio of other than 1:1, you can move the wheels by applying a force between them. When it is 1:1, by definition, they must move together. Not true for any other ratio.
This is BS. You are never going to understand any of this. Or, much more likely, you are just an Andrea Rossi troll.
I'm done with you. I have other things to do in life. All the rest of us understand how the models work. You don't. It's that simple. Bye.
electrodacus:
--- Quote from: hamster_nz on July 22, 2022, 05:29:30 am ---What is the voltage at the midpoint? Directly across from where it is fed? Unless it is 0V...
--- End quote ---
How will you measure the voltage without touching the wire ?
Also you measure the voltage between two points and you did not mentioned voltage relative to what.
If the multimeter is connected to the battery negative and the 1Ohm side is connected there you will measure 2V across that but if the 9Ohm is connected to battery negative you will measure 18V across it.
Your voltmeter is just another wire in parallel with one of them (one of the two halves). Yet since multimeter is maybe around 1MOhm way less power will be dissipated on the voltmeter.
It shows that even to make the measurement you need a wire with high resistance not to influence the conditions to much as current will be slightly larger when you connect the voltmeter.
electrodacus:
--- Quote from: gnuarm on July 22, 2022, 05:39:45 am ---I'm done with you. I have other things to do in life. All the rest of us understand how the models work. You don't. It's that simple. Bye.
--- End quote ---
Have a good life.
hamster_nz:
--- Quote from: electrodacus on July 22, 2022, 05:54:04 am ---
--- Quote from: hamster_nz on July 22, 2022, 05:29:30 am ---
--- 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.
--- End quote ---
What is the voltage at the midpoint? Directly across from where it is fed? Unless it is 0V it will not be symmetrical.
--- End quote ---
How will you measure the voltage without touching the wire ?
--- End quote ---
You have a bigger problem - you stated that "Electric field will also be symmetrical" and then go on to measure the potential at 2V and 18V, showing that it isn't symmetrical at all, and in doing so explaining why different sides are dissipating different powers.
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