Slam the lid on it, unless you want to befancy.
I should imagine it has something to do with Making Money. 
That’s a separate matter. The standard hardware could still be in the car (possibly in a module that can be removed for service or transplant to another vehicle), with only an ordinary heavy-duty extension cord running from outlet to vehicle.
NO.
the risk may be low with Level 1 (120V 12A,) but with Level 2 (240V 30A) and the potential for 240V 80A chargers, you do not want to disconnect the cord while current is flowing. part of the reason for the logic in the charging station is so if the car is actively charging and the user pulls the cord, the sense pin(s) disconnect first so the station can turn off the charging pins before they break contact.
anyway, to address the specific question in the OP, the cords are robust enough where simply having a tire pressing down on them with part of the weight of the car isn’t likely to cause any harm. However, the act of actually rolling the tire onto the cord will eventually abrade the insulation and cause problems.
I would think that a cable with a built in disconnect would cost less than $500.00.
These batteries charge at 19000 Watts?
yeah, because they’re going to sell so damn many EV charging connectors that they’ll be rolling in profit.
what is it with people that the instant they see a price higher than they expect, they jump right to the conclusion that it’s a “profit making scheme?”
see my previous replies as to why it’s not a “low cost universal item.”
and LOL at the “back when things were built to last.”
Shirley, you can’t be serious. If one wheel supports 900 lbs, how does the internal air pressure of the tire factor into this? Would a flat tire (zero air pressure) be weightless?
Even so, it’s a business, everything is a ‘profit making scheme’. Without profit, they’d go out of business. Then everyone gets mad at them because of all the employees they put out of work.
If you get a high-power charger installed in your garage, or if you’re at a public charging station that supports it, yes. And there are other standards that allow for even higher charging current - Tesla is building some Supercharger stations that support 100 kW charging.
Look here:
Contact area = load carried by tire/tire psi
Therefore psi = load carried/contact area.
Less air pressure (same weight) leads to greater contact area. More weight (same psi) leads to greater contact area.
The pressure against each square inch of the wheel with an inflated tire supporting the weight of this car is 38 pounds per square inch. Each square inch of tire contact area supports 38 pounds. If I inflated to a higher psi then there would be less tire contact area with the road supporting the same weight.
If the tire was flat then the pressure would be against the rim, and lots of pounds per square inch on those rims as they have little surface area compared to the tires.
And for kicks, my tire contact area is likely an ellipse with axes of 4.35 and 6.96 inches (if it follows the same 1.6 ellipse rule that airplane tires do.)
I have no complaint with your using the tire pressure to compute the area that makes contact with the ground, but I fail to see how it relates to the weight of the object the wheel supports. If you had a rigid wheel with 38psi internally, but with one square inch contacting the pavement, it’s the weight of the car that presses on the ground, not the psi inside the wheel that is important. Double the contact area, for whatever reason, and you will halve the vertical pressure per square inch, but you do not change the weight of the car.
The internal pressure affects the contact area, but not the total weight on the ground.
Unless I’m misunderstanding you, you are adding apples to oranges.
Not all the weight is on the cord. That is the point. You don’t have 900 lbs on the cord. 90% of your 900 lbs is on the ground. The rest presses against the cord.
Or in other terms, 25% of the weight of the car is on that tire, and a fraction of that presses against the cord.
Exactly. But DSeid is mixing the internal tire pressure with the weight:
(bolding mine)
No, it doesn’t.
Oh I see what you mean, yes. Sorry, I am right before bed time.
The pressure on the ground is 38 pounds per square inch.
Here ,you answer this questio:. The contact area with the ground of an object is about 24 square inches. The object weighs 900 pounds. What is the pressure that the object is exerting on the ground?
I’m thinking that while this would work well on flat ground, the pressure would be uneven on an uneven surface. You’re going to have a small gap of tire between the extension cord and the ground that isn’t actually contacting anything, yet the air in the tire is still pressing out. That force is going to be transferred to nearby parts of the tire that are touching something, namely the extension cord.
Without verifying your numbers, and assuming my calculator is operating normally, 900 lbs / 24 inches is approximately 38 lbs/inch.
Which has nothing to do with the internal tire pressure.
The numbers are coincidental, but the coincidence is meaningless. If I increase the internal tire pressure to 48 PSI, does that mean that the pressure per square inch on the ground has doubled?
Good points. Running across a pebble would increase local pressure a lot.