Invitation only event, but I guess you could ask for an invitation if you are interested. It would be good to have a Tesla or two there. I don't know what EV they will use. I'd guess a Leaf or "i" so Real Power can use their ChadeMo quick charge truck. GSP As a fellow EV supporter and enthusiast, Contour Hardening, Inc.(CHI), manufacturer of the Real Power Mobile Level 3 Quick Charge EV Rescue vehicle, would like to invite you to attend the inaugural EV500 event to be held at the Indianapolis Motor Speedway on Thursday, October 13th, 2011. Note that this event requires registration and can be attended by invitation only. Invitation is attached in pdf form but find event outline below: WHO: Contour Hardening, Inc (CHI), manufacturer of the Real Power Mobile Level 3 Quick Charge EV Rescue Vehicle WHAT: The inaugural EV500 where an all-electric production vehicle will travel a distance of 500 miles…in one day! WHERE: Where else? The famed oval at the Indianapolis Motor Speedway (4790 W. 16th Street, Indianapolis, IN. 46222) WHEN: Thursday, October 13th, 2011. The green flag drops at 9am EST. WHY: To demonstrate that EV technology available TODAY can greatly reduce “range anxiety” and position electric vehicles as a viable means of transportation. To attend this electrifying event, please register by replying to this email or contacting Real Power at 877.670.7325 or sending us a note at [email protected] Thanks for your support and we look forward to seeing you in Indy!

I was fortunate enough to attend this event on Oct 13th sponsored by Real Power Inc.. There were 2 iMiev EV's that were being charged with Real Power EV rescue vehicles. The DC quick chargers were giving the iMiev's approx. 50 miles range in 10 minutes. The cars were circling the track at 58-62 MPH in a stiff breeze and rain. I didn't stay for the entire event but I understand it was completed in 12 hours. I was able to charge my Roadster with the rescue trucks on board Eaton charge station and a J1772/Tesla adapter @ 30 amps. I also was able to charge from the truck using a RFMC and a NEMA 14-50 adapter @ 40 amps straight from the truck's generator. Real Power - News

Did they let you take the Roadster out on the track? :biggrin: (guessing I know the answer but figured I'd ask anyways...)

Thought exercise: Could the Tesla, with its larger battery capactity but slower charging rate, have beaten the iMievs in reaching the 500 mile marker? - Assuming 3 sessions of ~170mi (say 3 hours per driving session) and two 40A recharges @ 5 hours per recharge (best case, avoid top-off), it would take ~19 hours. - Assuming 2 sessions of sloow driving, say 6 hours per session to get 250mi, and a single 40A recharge @ 6 hours for range mode, it would take ~18 hours. Shows how DC fast charging will be the best game in town for long EV road trips, once the infrastructure is available...

No Unfortunately! I'm sure the Indy legal dept put the kabash on that. I was at an Emerging Technology event at Indy earlier this year, and all 100% electric vehicles were allowed on the track for a parade lap, (no passing allowed). Signed liability waivers were required, of course. make that all Plug-in vehicles were allowed on track.

No they did that every time the car left the pits. I think they were emulating an Indy car pit crew. But I suspect the real reason was to save a few watts so the iMiev wouldn't have to start from a dead stop.

I just received this video in an email from RealACpower, and would like to share. Happy Holidays to all, Sam Domain Names, Web Hosting and Online Marketing Services | Network Solutions

The 85kWh Model S will blow away this event next year. Starting from a full pack and with optimal driving speeds, I'm confident that the Model S can drive 500 miles in under 9 hours, possibly under 8 (depending on the speed:Wh/mile and charging characteristics).

With aerodynamic wheels and driving at maybe 35mph, I wonder if they could do it on one charge. Of course if they had their super charger there as well, they could just drive 55 mph and only have to recharge once.

Even if they could do 500 miles on a charge @35 mph, that would take 14.3 hours. Better to drive faster and supercharge once or twice.

500 miles minimizes at 6.9 hrs with 3 stops @90kW to 80% full for 2.2 hours driving 107 mph or 500 miles = 7 hrs with 5 stops @90kW to 50% full for 2.3 hours of charging (full battery at start) Welcome to Google Docs

That made me want to fiddle with math. I got the spreadsheet from the "Roadster Efficiency and Range" blog, and for the Model S, I simply multiplied the Wh/mile numbers with 300/244. Using some weird formula which I wouldn't be sure to be correct, some numbers resulted. Let me know if there were similar calculations and if the results were similar. [EDIT 1/1/2012: Yesterday, my "weird" formula was missing a term, these are now the corrected numbers. Also, the factor 300/244 is by coincidence resulting in exactly 300 miles @ 55 mph, so I am keeping it, even though it should theoretically be calculated as ((85/300)/(55/244)). Since this calculation uses a constant factor to multiply Wh/mile numbers from the Roadster, for each speed, these numbers do not reflect a potentially different percentage of aerodynamic losses, see discussion below.] For very long distances where the initial charge doesn't matter, and the possibility to charge at any lap, I think the following traveling speeds are optimal for the given charging speed (meaning, driving faster would cost more time in charging, than it saves in driving time) : Roadster: 7 kW charging -> 45 mph driving 17 kW charging -> 61 mph driving Model S: 7 kW charging -> 42 mph driving 10 kW charging -> 48 mph driving 20 kW charging -> 60 mph driving 30 kW charging -> 68 mph driving 50 kW charging -> 81 mph driving 90 kW charging -> 100 mph driving So with Supercharging, one can drive as fast as 100 mph without needing more time to recharge than one gains. [EDIT: Yesterday I had computed 83 mph] Driving 500 miles at 100 mph takes 5 hours, and 305 kWh. If one starts with a full battery of 85 kWh, and recharges exactly 220 kWh at 90 kW, that's an additional 2.44 hours. So the total would be about 7.44 hours = 7 hours and 26 min, +/- any mistakes. [EDIT: Yesterday I had computed 7 hours and 10 min, yet again, this is still without trying to reflect differences in aerodynamics other than with a constant factor.] I'm not sure if starting with a full battery modifies the calculation of the optimal speed. It currently seems to me that as soon as one needs to recharge at all, driving faster would cost more time than it saves. But maybe I overlooked something. [EDIT: I am now certain that the optimal speed for a specific charging power is independent of the initial charge, that is, as soon as one needs to recharge at all. If one doesn't need to recharge, there can be enough charge to cover the distance with a faster speed.]

You calculations say that @ 83mph it takes 380Wh/mile = 190kWh/500miles How are you figuring Wh/mile? I've calculated that at 83 mph it would take 423Wh/mile Highway Range Ignorance I've calculated that optimal speed would be 107 mph charging at 90kW 90mph = 554Wh/mile so 500miles * 554 Wh/mile - 85kWh = 192kWh needed to be charged 4.67 hrs = 107mph * 500 miles 2.13 hrs = 192kWh/90kW 6.8 hrs total time Spreadsheet My calcs for Model S are: 7kW charging -> <45mph 10kW charging -> 55mph 20kW charging -> 65mph 30kW charging -> 75 mph 50kW charging -> 90 mph 90kW charging -> 107 mph

Obviously the difference comes in part from your more complex computation of the aerodynamic factor. It'll take me bit of time to look into what you did there. In my calculations, I simply used a constant factor to convert the Roadster numbers into Model S numbers. While that is a more simplistic approximation than yours (which doesn't have the actual numbers either), it surprises me that the resulting calculations for optimal speed are so different. It doesn't seem you have explained how you calculate optimal speed. As a side note, regrading the constant factor I am using: it results in the spreadsheet showing me a 300 mile range at 55 mph, which is what I checked, however it is a good factor only by coincidence. The conversion factor should be calculated as ((85/300) / (55/244)), which falls within 3% of the factor I used, but results only in a range of about 290 miles. This is because the Roadster values from Tesla result only in 239 miles range for the Roadster at 55 mph. Since indications are that the Model S might actually have a range above 300, the factor (300/244) is by coincidence a better factor to use. So I think my results are correct for using a constant factor, which of course is a simplification. I'll take a look at your calculation for the aerodynamics, and whether using them would result in the same optimal speeds that you calculated. (Neither of us posted the method for optimal speed yet).

I utilized my spreadsheet to determine best speed by brute force in 2 mph increments. If I actually approximate a formula for total time and graph it, as well as the first derivative, the minimum is at 107.9 mph ( where the first derivative intercepts zero ). Online Graphing Calculator: 500/x+(500*(.035*x^2+.1778*x+167.29)-85000)/90000 ( x = speed , y = time ) To see charge time curve use: range X = 60 to 120 range Y = 0 to 12 to see intercept of derivative: range x = 107 to 108 range y = 0 to .001 my aerodynamic drag still has some estimations that could alter the curve slightly. But like you I'm assuming that the 85kWh pack will do a lot better than the 300mph even without the aero wheels. My adjustment factor for 60kWh battery pack to 85kWh pack is too large for the little bit of weight difference. (or worse the 40 & 60 kWh packs will not hit their targets :scared: ) 40kWh --> x Wh/mile = f(v) + 24kWh 60kWh --> x Wh/mile = f(v) + 35kWh 85kWh --> x Wh/mile = f(v) + 57kWh these factors 24,35,57 were used to make the curve fit the stated 55mph speeds for the different battery pack sizes. the 85kWh pack should be < 20 lbs heaver than 60kWh pack ( my calcs: 40kWh pack ~ 5051 cells , 65kWh pack ~ 7576 cells , 85kWh pack ~ 7617 cells) 40&60 are 45grams/cells and 85 is 45.5grams/cell) (with ~250lbs difference between 40kWh and 60kWh packs)

after some more formula calculations I've determined the approximate optimal speed formula based on charging rate: q = charge rate in Wh (not kWh) v = velocity to travel for minimizing time (mph) where q> 1000Wh, distance requires charging and it takes no time to stop and hookup charging unit. Model S -> v = (14.286 * q ) ^(1/3) - 0.97 Roadster -> v = (14.286 * q ) ^(1/3) - 2.46 time formulas for each are: t = d/v + (d * (a*v^2 + b*v + c) - s) / q t = time (hrs) d = distance traveled (miles) a,b,c are consumption rate factors s = battery pack starting size (Wh) q = charging rate (W) Roadster a,b,c = .035, .4496, 100.43 Model S a,b,c = .035, .1778, 167.29 In the resulting formulas 14.286 is derived only from "a" (.035), b was eliminated due to it not significantly effecting the formula and simplifying the equation. And in the resulting formula the difference being subtracted is b/(6*a). the non-simplified equation is: (sqrt(q*(27*a^2*q-b^3))/(4*3^(3/2)*a^2)+(54*a^2*q-b^3)/(216*a^3))^(1/3)+b^2/(36*a^2*(sqrt(q*(27*a^2*q-b^3))/(4*3^(3/2)*a^2)+(54*a^2*q-b^3)/(216*a^3))^(1/3))-b/(6*a)

...continued... [EDIT:written before reading your last two messages. I see you meanwhile arrived at 107.9 mph as well] There are two major differences: a) I was missing a term in the method I used yesterday to compute the optimal speed, and I now corrected my numbers in the post above, to avoid confusing new readers with any old numbers. b) You are calculating Wh/mile with your own functions, instead of applying a constant factor to the Roadster numbers, as I do. This may be better or worse, as discussed further down. When I use your functions to compute Wh/mile, I find an optimal speed of 107.9 mph, and a travel time of 6.99 hours (instead of your 6.8 hours, since at 107 mph it seems it would be about 590 Wh/mile, not 554. I can't tell how you arrived at 554Wh/mile, since your spreadsheet is somewhat unreadable to me). I noticed you are using a Cd value of 0.225 to a calculate your aero coefficient. Personally, I'd assume that this would require the optional aero wheels, and that otherwise the Cd will be 0.26 or 0.27, though I guess we don't know yet really. With a Cd of 0.26, the CdA would apparently be higher than the Roadster's, and the resulting optimal speed for 90 kW charging even slightly lower than I calculated above (less than 100 mph). Also you are using a constant value for "RoadsterToSAdjustment", which would be quite accurate if the difference goes into Tire losses, but not if part of it goes into drive train losses, which also increase with the speed and the square of the speed. That again would lower the optimal speed. So I now think that, depending on the CdA value without aero wheels, and on how much of the difference goes into the drive train, the actual values are more or less somewhere between your numbers and my corrected numbers.