This old document: http://google.brand.edgar-online.com/EFX_dll/EDGARpro.dll?FetchFilingHtmlSection1?SectionID=7150512-1202928-1209281&SessionID=Cx2VHv6UTlKsQP7 says: "...TESLA shall order not less than 2,400 units of Product..." So, lets say 1464, based on highest NA VIN reported, is about the number of North American units produced...? That would mean 2400-1464=~936 total for Europe+Asia to satisfy the contract? I think the highest European VIN we saw advertised for sale recently was around #799. So, does that mean there are ~137 gliders (not 330) waiting around? I guess time will tell... Just idle speculation reading tea leaves here. Note, there may have been some small VIN sequence gaps, but they are probably covered by the extra EP/VP/Founders... still 1464 is approximate.
Wikipedia says 2,500 gliders total from Lotus... Is it confirmed that Tesla ordered an extra 100 beyond the original contract?
Depends to what level of proof you mean by confirmed (e.g., I haven't seen a link to a contract). But yes, like a zillion times in the media last June.
Well, then, I guess we could place bets to see if the European VINs ever get over 1000 before they run out of gliders.
It does seem "officially" to be 2,500: http://sec.gov/Archives/edgar/data/1318605/000119312511167522/d8k.htm So, yeah, maybe there are at least a couple hundred Roadster/gliders still yet to be customer delivered and Europe/Asia VIN #1000 to show up someday. I wonder if there will be another "final five" kind of thing with some special edition versions...?
Congrats. And that is an interesting solution to making a ramp. Tip the whole truck. I do not ever recall seeing it done that way. I have always seen an additional fold down ramp or the flatbed tilt.
Just for fun and killing time waiting for our cars, lets calculate the weight difference of a full roadster battery.: For your home, 120V, a kilowatt hour is an 8.33 amp device running for an hour. 8.33 amps/sec = 8.33 coulombs/sec 8.33 x 60 x 60 = 29,998 coulombs/hour 29,998 x 6 x 10^18 = 1.8 x 10^23 electrons. I believe the maths right. An electron's mass is 9.11×10−31 kg 9.11 x 10^-31 x 1.8 x 10^23 = 1.64 x 10^-7 kg or .16 micrograms x 53 Kwh for a roadster = 8.69 micrograms
But it shouldn't even be that difference, you don't add any electrons to the battery, you separate them from the cathode and store them in ions in the anode, it's the act of traveling across the battery that causes the current to flow. Remember, in order for the current to flow, it must be connected at both ends, so electrons flow from one part of the battery, over the wire, to the other.
Of course on average there is no net change of electrons. They're just being moved from one side of the cell to the other.
I think the formula you are looking for is E=MC^2. So 53kWh -> 1.908×10^8 J (joules) C -> 299,792,458 m/s 1.908E8 / (2.998E8)^2 = M (in kilograms) I get about 2.13 ug (micrograms). Or according to Wikipedia the weight of a human ovum. Or about 6 grains of sand. Orders of magnitude (mass) - Wikipedia, the free encyclopedia That has powers in reference to kg so it is hard to see correctly.
Seems to be a bit of a "rite of passage" to go through that thought exercise. Weighty Matters Involving Electrons | Blog | Tesla Motors Anyways, you can calculate the weight of the electrons in the pack, even considering that they are "internally reused".
Congrats! Is there something special about the car that is only available in the "final five"? Care to share the last 4 digits of your VIN#?
The Los Angeles store this afternoon that there are now 3 remaining new Roadsters in the U.S. 2 are in the Los Angeles store -- medium green and a red. The red one is one of the final 5.
The red one (VIN 1463, the second to last US VIN), is now in Menlo Park, all nice and shiny. No-one knew why the previous deal fell through. Yes, that's Model S Alpha-1 in the background. I'll start another thread with some pics of its interior.