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Wanxiang May Give 2012's Fisker Karma a Relaunch
New submitter sumit sinha notes recent reports that Tesla may soon be joined again by Fisker in the world of high-end, all-electric car makers. According to a Reuters story, the Fisker Karma in something very close to its previously available form may be offered for sale again sometime soon. Says the article:
The "new" Karma that California-based Fisker, acquired by Wanxiang earlier this year, is rushing to finish is based largely on the 2012 model, said the people, who asked not to be identified. Wanxiang's top U.S. executive said in February the Karma would be reintroduced within a year. "It will have to be nearly identical to the 2012 model, or it would need to go through (safety) testing and certification again," a person close to Fisker's suppliers said. "I don’t think they want to put a lot of engineering into it either, as well as probably use up some of the old parts that are in inventory."
Close, but not exact,: Fisker does not plan to simply reintroduce the 2012 Karma, a source close to Fisker said. “Not 100 percent identical," the person said. "The new Karma will be different in many key areas. It will have noticeable upgrades." He declined to provide details. Using the 2012 Karma design could present problems given it has older features and technologies. "You're not buying something that's considered 'state of the art' necessarily," the supplier source said. "It's a big hurdle to overcome."
New submitter sumit sinha notes recent reports that Tesla may soon be joined again by Fisker in the world of high-end, all-electric car makers.
The Fisker Karma is not an "all-electric car." It has an electric drivetrain with a gasoline range extender. The article itself makes this quite clear:
The Karma, a hybrid-electric vehicle equipped with a small gasoline engine that kicks in when its on-board battery is depleted, previously had a starting price of around $100,000.
If you could try to make more accurate article summaries in the future, that'd be great. Thanks.
Anyone who could afford a Karma and wanted electric would have already bought a Tesla S or Roadster.
The only way they'll sell some is to greatly upgrade the interior to Tesla Model S-type screen and software.
Help! I'm a slashdot refugee.
Best present a young couple of recent college grad can give themselves would be an E-car.
Looooooool. Recent college grads do actual interesting things, like go on a road trip or go skiing. EVs are great for dads who drive from home to job day in day out.
Electric cars would work if you have a short commute and a garage where you can charge overnight/every night. But they won't work if the commute is too long or you don't have a convenient place to charge or you commute using public transportation. Plus you will need a second car or use rentals for pretty much anything besides the daily commute. That really limits the market.
Don't listen to Wanxiang, they're probably just jerking us off.
Never understood why anyone would buy this over a Tesla. The Fisker was heavier and way more expensive. If I wanted a range-extended hybrid-electric luxury car, I'd get the Cadillac ELR instead.
At least in California HOAs cannot restrict EV chargers and must approve them. I have a coworker who bought a Tesla model S who lives in a condo. He had no problem getting a charging outlet installed though in his case the parking was under the building and access to power was not difficult. I think in the years to come apartment complexes will start installing support for EVs.
Already some places like San Jose are requiring that all new parking lots run conduit and whatnot to make it easy to install EV chargers in the future.
Many workplaces are also installing EV charging stations in the Bay Area. The problem is that even with this there typically aren't enough so they keep expanding them.
Some apartment complexes are also installing chargers.
This post is encrypted twice with ROT-13. Documenting or attempting to crack this encryption is illegal.
No thank you, but China's reputation has been to make it worse in the name of making it "cheaper".
They've done it themselves, and do it to about every brand they touch.
Lenovo? They have the opposite of the Midas Touch - everything they touch becomes worse (Thinkpads, servers, etc.).
The GM H2/H3? It's not even a Suburban.
Buick? At least you could get a decent one before China was prioritized. Now it's Opels, Daewoos, and cut-down I4 mysterymeat cars everywhere.
Geely? They've devalued the Volvo brand in ways that no other country would dare.
No thank you, but I'll pass on something that had problems *before* China got involved.
Twitter supports and protects racists - by smearing their critics with the "Hate Speech" label.
And if that couple has two cars, and if they want new cars, then it makes sense for one of those two cars to be electric.
My commute is 20 miles round-trip. My wife's is 40 miles round-trip. Nearly every type of store or business that we do regular business with is along our commutes. We could easily commute with an electric without running out of range, even if we had to go ten miles out of our way to do so, and even if we couldn't plug-in at our places of employment, though I expect that the positive press from the charging station would make that pretty likely.
Do not look into laser with remaining eye.
moms will never get EVs because they have to schlep kids around everywhere and be ready in an emergency to go get the kids. they want real cars not toys. boring dads will commute in them, and say it saves them pennies per mile net.
No, Jane tried to use an equation that only calculates radiative "power out" when Jane needs to use an equation for heat transfer that calculates radiative "power out minus power in".
Jane's accounting for "power out" without including a term for "power in". That's not A = A, it's A = 0 because one of the terms has been ignored. It's led Jane to the absurd conclusion that electrical heating power doesn't depend on the cooler chamber wall temperature. If that's the case, then how did we detect the 2.7K cosmic microwave background radiation with warmer detectors? How do uncooled IR detectors see cooler objects? Again, why is Venus hotter than Mercury?
This is complete gibberish, Jane. Power radiated in from the chamber walls needs to be accounted for using one term. Power radiated out from the source needs to be accounted using another. Once again, accounting for power flowing in doesn't violate the second law of thermodynamics or somehow imply net energy transfer from cool to hot, no matter how many times Jane wants to assert that nonsense. However, failing to account for power flowing in does violate conservation of energy, because power in = power out through any boundary where nothing inside is changing.
Completely backwards, as usual. I've already shown that my solution keeps electrical heating power constant. Once again, Jane's solution halved the electrical heating power. Jane didn't notice this because he calculated net transfer incorrectly, which led him to the absurd conclusion that Jane was only off by about 0.1% when Jane was actually off by ~100%.
Good grief, Jane. How did the Sky Dragon Slayers brainwash you into endlessly regurgitating this nonsense? Once again, radiation is absorbed by any surface with absorptivity > 0. Jane's either hopelessly confused about the very term "NET" which he keeps capitalizing, or Jane/Lonny Eachus has betrayed humanity by deliberately spreading civilization-paralyzing misinformation.
Again, how do Slayers think we detected the 2.7K cosmic microwave background radiation with warmer detectors? How do Slayers think uncooled IR detectors see cooler objects? Again, why do Slayers think Venus is hotter than Mercury?
Again, Jane/Lonny Eachus actually means that he intends to show where mainstream physics "went wrong" according to the Sky Dragon Slayers. There are many ignorant, stupid physicists that Jane/Lonny Eachus needs to educate: Prof. Brown, Dr. Joel Shore, the American Institute of Physics, the American Physical Society, the Australian Institute of Physics, and the European Physical Society, etc.
Does Jane have the memory of a goldfish? Of course Jane has argued with these other physicists. Jane personally asked Prof. Brown about Sky Dragon Slayerism, but wasn't able to "educate" him. Lonny Eachus personally asked Dr. Joel Shore about Sky Dragon Slayerism, but wasn't able to "educate" him. And now Jane/Lonny Eachus fantasizes that these physicists agree with his Sky Dragon Slayerism? Maybe Jane/Lonny Eachus should read those exchanges again, and notice that Prof. Brown and Dr. Shore told Jane/Lonny Eachus the same things I am. That's because Prof. Brown, Dr. Shore and I are simply reiterating elementary mainstream physics.
How bizarre. The whole reason Slayers deny that an enclosed source warms is because that implies greenhouse gases can't warm the surface:
That's why Jane, Dr. Latour and the rest of the Slayers disagree with the American Institute of Physics, the American Physical Society, the Australian Institute of Physics, and the European Physical Society.
Again, how did we detect the 2.7K cosmic microwave background radiation with warmer detectors? How do uncooled IR detectors see cooler objects? Again, why is Venus hotter than Mercury?
If Sky Dragon Slayers could answer these questions without resorting to gray Oreos or basketball player gloves, physicists might take the Slayers more seriously.
It's not that interesting that Prof. Brown decided to ignore Jane/Lonny Eachus, given that he later said:
"Wow, Joel, I gotta say (after reading some of the replies on this thread) that this really is pointless. These folks have no conception of the FIRST law of thermodynamics, let alone the second. The argument for warming doesn't even require mentioning the SBE, it only requires the first law, the second law, and a monotonic relation between temperature difference in ANY channel and the rate of energy transfer in that channel, subject to very broad constraints.
But seriously, just a waste of time. When people just make stuff up and reject the contents of ELEMENTARY textbooks on the subject because they just don't like the conclusion those contents lead to, how can you argue with them? If somebody tries to solve the light bulb problem while pretending that it doesn't primarily cool via radiation and completely ignoring radiation, what can you do?
Get them to say "oops"?
Never happen. It's a religious issue, not a scientific one."
In other words, Prof. Brown gave up trying to educate Slayers like Jane/Lonny Eachus because it's a "waste of time."
That's odd. Just yesterday Jane had no argument with Dr. Shore. Now Jane claims that Dr. Shore "FUCKED UP" his physics.
Physicists have "FUCKED UP" their physics, and only the Slayers can save the day! Or maybe the Slayers are crackpots. How could anyone tell, unless maybe Dr. Shore explained that:
"Actually, the idea that radiation goes only from the warmer to colder objects is an invention of the Slayers. It appears nowhere in the physics literature. I don't know about the exact history of our understanding, but my physics textbook from 1983 (Serway, "Physics for Scientists and Engineers", after introducing the law P = sigma*A*e*T^4 says
"A body radiates and also absorbs electromagnetic radiation at rates given by Eq. 17.11. If this were not the case, a body would eventually radiate all of its internal energy and its temperature would reach absolute zero. The energy that the body absorbs comes from the surroundings, which also em
This isn't a quantum effect. The reason IR detectors measure DIFFERENCES, not absolute radiation, is because electrical heating power = (e * s) * (Ta^4 - Tb^4). If that weren't true, there would be no way to detect this difference, so uncooled IR detectors wouldn't be able to see cooler objects. And we couldn't have detected the 2.7K cosmic microwave background radiation with warmer detectors. But we did! How?
Once again, Jane insists electrical heating power = (e * s) * (Ta^4). Once again, Jane's ridiculous equation doesn't just say there is no net "radiative power in" from cooler to hotter. Jane's wrongly saying the source absorbs no radiative power at all.
If Jane would reconsider conservation of energy and include a term for "radiative power in", then Jane could honestly say he was only claiming that no net radiative power is absorbed by the source. Until then, Jane's equation claims that no radiation is absorbed by the source at all. And since Jane seems to think he's only saying no "NET" radiative power is absorbed, Jane will probably never be able to recognize his error, let alone correct it.
Once again, it's not included in the "radiative power out" term of heat transfer equations because it's included in the "radiative power IN" term.
I'm having supreme difficulty getting your concept through my head because it's Sky Dragon Slayer nonsense. The fact that more heat
That reference shows the object (i.e. chamber wall) temperature has an effect on the temperature controlled cavity (i.e. source). Which Jane denies:
No, that's Sky Dragon Slayer nonsense. If radiation from the cooler walls really had no effect on the heat source whatsoever, the IR thermometer wouldn't work because the cooler object temperature would have no effect on the temperature controlled cavity whatsoever.
When the source temperature is held constant, its required electrical heating power is an IR thermometer.
Here's one way to see that: draw a boundary around a heated aluminum source. It's heated by constant electrical power flowing in. Aluminum cold walls at some unknown temperature T4 also radiate power in. The source at 150F (T1 = 338.7K) radiates power out. At steady-state, power in = power out. Using the equation which neglects reflections:
electricity = (e*s)*(T1^4 - T4^4)
If the required electrical heating power is 82.1 W/m^2, then the chamber wall is at absolute zero (-459.7F).
If the required electrical heating power is 55.6 W/m^2, then the chamber wall is at 0F.
If the required electrical heating power is 27.8 W/m^2, then the chamber wall is at 90F.
If the required electrical heating power is 0.0 W/m^2, then the chamber wall is also at 150F.
If the source needs to be refrigerated to stay at 150F, the required electrical power is negative. The same equation can be used to determine the chamber wall temperature, regardless of whether it's warmer or cooler than the source.
That's why when the source temperature is held constant, its required electrical heating power is an IR thermometer. At least, it's a thermometer when using mainstream physics. But Jane's equation is:
electricity = (e*s)*T1^4 (Jane's equation)
Since Jane's equation doesn't depend on the chamber wall temperature, uncooled IR detectors can't see cooler objects in Janeland. And we couldn't possibly have detected the 2.7K cosmic microwave background radiation with warmer detectors. But we did! How? This must be inexplicable to Slayers who are brainwashed into believing that:
No. Again, mainstream physics shows that electrical heating power gradually decreases to zero as the chamber wall temperature increases. That's how uncooled IR detectors can see cooler objects.
It's beginning to seem like we disagree about the meaning of the term "NET".
1. Can we agree that net heat transfer through a boundary around the source = "radiative power out" minus "radiative power in"?
2. Can we agree that net heat transfer always contains terms in both directions?
3. Can we agree that just because "radiative power out" > "radiative power in", that doesn't mean "radiative power in" = 0?
If we can agree on all those points, that's great. Maybe this will help Jane write down a simple equation describing the electrical heating power required t
Charming. As I just explained, IR detectors don't have to depend on quantum effects. Classical mainstream physics allows a temperature-controlled source to detect IR from the cooler chamber walls as follows:
electricity = (e*s)*(T1^4 - T4^4)
If the required electrical heating power is 82.1 W/m^2, then the chamber wall is at absolute zero (-459.7F).
If the required electrical heating power is 55.6 W/m^2, then the chamber wall is at 0F.
If the required electrical heating power is 27.8 W/m^2, then the chamber wall is at 90F.
If the required electrical heating power is 0.0 W/m^2, then the chamber wall is also at 150F.
If the source needs to be refrigerated to stay at 150F, the required electrical power is negative. The same equation can be used to determine the chamber wall temperature, regardless of whether it's warmer or cooler than the source.
If you don't deny that some radiation is absorbed, then it should be very easy to write down a simple equation describing the required electrical heating power (not the radiative power out) of a blackbody source.
Jane, if we can't agree on the meaning of the term "NET", why are you still capitalizing the word "NET"? Screaming the word louder and louder is unlikely to be productive.
1. Can we agree that net heat transfer through a boundary around the source = "radiative power out" minus "radiative power in"?
2. Can we agree that net heat transfer always contains terms in both directions?
3. Can we agree that just because "radiative power out" > "radiative power in", that doesn't mean "radiative power in" = 0?
If Jane answers "no" to any of those three yes/no questions... why?
If you're so confident that you're right, why not prove it by taking a few seconds to write dow
Jane probably won't write down an equation describing electrical heating power for a blackbody source, so I'll try to guess at Jane's reasoning.
Draw a boundary around the blackbody heat source:
Jane's power in = electrical heating power + radiative power in from chamber walls
Jane's power out = radiative power out from source + radiative power from chamber walls, re-emitted back out
At steady state, Jane's power in = Jane's power out:
electrical heating power + radiative power in from chamber walls = radiative power out from source + radiative power from chamber walls, re-emitted back out (Jane's equation?)
Jane, is that your equation for required electrical heating power? By "A = A", are you saying "radiative power in from the chamber walls" = "radiative power from chamber walls, re-emitted back out"?
Again, Jane's gray body equation has to reduce to the black body equation when emissivity = 1, so this is a way to check Jane's work. But since Jane seems convinced that checking his work is "lying" let's write down both equations simultaneously.
Draw a boundary around the (gray or black body) heat source:
Jane's power in = electrical heating power + radiative power in from chamber walls
Jane's power out = radiative power out from source + radiative power from chamber walls, re-emitted back out
At steady state, Jane's power in = Jane's power out:
electrical heating power + radiative power in from chamber walls = radiative power out from source + radiative power from chamber walls, re-emitted back out (Jane's equation?)
Jane, is that your equation for required electrical heating power? By "A = A", are you saying "radiative power in from the chamber walls" = "radiative power from chamber walls, re-emitted back out"?
Now use the Stefan-Boltzmann law to describe the radiative terms, one at a time. First for Jane's gray body:
Because "radiative power in from chamber walls" is emitted by graybody walls at temperature T4, the Stefan-Boltzmann law says:
gray electrical heating power + (e*s)*T4^4 = radiative power out from source + radiative power from chamber walls, re-emitted back out (Jane's equation?)
Is that what you're saying, Jane?
Now for Jane's black body check:
Because "radiative power in from chamber walls" is emitted by blackbody walls at temperature T4, the Stefan-Boltzmann law says:
black electrical heating power + (s)*T4^4 = radiative power out from source + radiative power from chamber walls, re-emitted back out (Jane's equation?)
Is that what you're saying, Jane?
Since Jane probably won't even say yes or no, I'll keep trying to guess at Jane's reasoning. Now the next term for Jane's gray body:
Because "radiative power out from source" is emitted by the graybody source at temperature T1, the Stefan-Boltzmann law says:
gray electrical heating power + (e*s)*T4^4 = (e*s)*T1^4 + radiative power from chamber walls, re-emitted back out (Jane's equation?)
Is that what you're saying, Jane?
Now the next term for Jane's black body check:
Because "radiative power out from source" is emitted by the blackbody source at temperature T1, the Stefan-Boltzmann law says:
black electrical heating power + (s)*T4^4 = (s)*T1^4 + radiative power from chamber walls, re-emitted back out (Jane's equation?)
Is that what you're saying, Jane?
I have to guess at your reasoning because what you've said doesn't make any sense.
I have to guess at what Jane meant by this, because it's not in equation form. In physics, statements in equation form are easier to analyze.
Draw a boundary around the (gray or black body) heat source:
Jane's power in = electrical heating power + radiative power in from chamber walls
Jane's power out = radiative power out from source + radiative power from chamber walls, re-emitted back out
At steady state, Jane's power in = Jane's power out:
electrical heating power + radiative power in from chamber walls = radiative power out from source + radiative power from chamber walls, re-emitted back out (Jane's equation?)
Jane, is that your equation for required electrical heating power? By "A = A", are you saying "radiative power in from the chamber walls" = "radiative power from chamber walls, re-emitted back out"?
Once again, to calculate "electrical heating power" you need to use a heat transfer equation which accounts for power in and power out. That's because power in = power out through any boundary where nothing inside is changing. Once again, the equation Jane's using is only valid for "radiative power out" which is completely different than "electrical heating power". That's why I'm starting with the principle of "conservation of energy" and trying to understand what Jane's saying, in equation form.
Jane, if you don't agree with the "power in" and "power out" that I've tried to glean from your rants, just fill in the following blanks like I did. It'll be much faster than accusing me of dishonesty, fraud, and libel.
Jane's power in = ?
Jane's power out = ?
Jane responds.