100 miles?

[garygid]

And, they could use the USB port to connect for data retrieval, diagnostics, error codes, etc.
That could make it more difficult for independent mechanics to diagnose the Leaf.

[garygid]

It is unlikely they would use the bluetooth "connection", due to both lower speed and access security.

[EVDRIVER]

I doubt it's going to be too tough to crack every control system and data point in this car, right down to the EPS module.

[garygid]

EVD...,
What makes you think "cracking" the Leaf system will be easy?

Have you cracked the 2010 Prius?

[EVDRIVER]

EVD...,
What makes you think "cracking" the Leaf system will be easy?

Have you cracked the 2010 Prius?

Have been abe to get it done on Prius and others. Most CAN BUS and other related systems are not that difficult to access or control for practical use.

[Bicster]

I believe the battery will be considered an emissions component. In any case, you can be assured the car will have an OBD2 port. It would be pointless for Nissan to make proprietary diagnostic tools for this vehicle, when a software update for the dealer's existing tool would do the job.

[EVDRIVER]

I believe the battery will be considered an emissions component. In any case, you can be assured the car will have an OBD2 port. It would be pointless for Nissan to make proprietary diagnostic tools for this vehicle, when a software update for the dealer's existing tool would do the job.

Most likely. One nice feature on screen would be the ability to enter ones kWh for various charge times. Since the vehicle should know how many kWh it takes in during charging at these times that then could be translated to real time cost per minute or mile based on the way one is driving. Even if consumption is monitored post charger the numbers can be adjusted for charger efficiency.

[TimeHorse]

Okay, back to the original question: what is the expected range of the Nissan Leaf. We know that under EPA*4* test LA4 that it has a total range of 100 mi, assuming ideal conditions. So, I'm going to repeat some of the math I did on Facebook but not all of it, so don't be frightened.

What we know is:

Range: 100 mi @ LA4
Battery Potential Energy: 24 kWh (86.4 MegaJoules)
Tires: P205/55R16 (205mm width, 55% ratio width to height, 16" rim -> 0.31595 meters wheel+tire radius*)
Average speed of LA4: 19.6 mph**
Maximum speed of LA4: 56.7 mph
Total distance of LA4: 7.45 mi

Since we know 7.45 mi is covered in LA4 and this represents the cycle under which 100 mi is achieved, that one LA4 represents 7.45% of the total driving range of the Nissan Leaf and thus 7.45% of the battery capacity:

Total Energy Used for LA4: 1.788 093 kWh or 6,437,136 Joules

Now we know what LA4 constitutes from this: http://www.epa.gov/nvfel/methods/uddscol.txt

Specifically, we have a table that gives us the speed used in 1 second intervals.***

The question is, how can we, if possible, use that to calculate what the expected range of the Leaf is under the more realistic US06*3*, HWFET*2* or EPA75?

I think our best bet is to try and match the Leaf to the Tesla curve. The trick is, I don't know what polynomial we're trying to fit, never mind the point where the slope becomes 0. Any thoughts?

I tried calculating the Moment of Inertia for the Nissan Leaf but I now think there was a flaw in the way I summed energies. So I think curve fitting is the best bet, since we know, in general, any electric car should follow the same curve Tesla uses, with a basic scaling factor. I'm sure it's more closely related to angular acceleration, perhaps angular acceleration * the moment of inertia to get the net energy transfer. If I knew the torque required for each speed, I could work backward by calculating the power over the 1-second interval from the torque and angular velocity, which in turn would give me the energy drain (1 Watt per Second = 1 Joule of Energy).

Anyway, I hope my Moment of Inertia calculations are very wrong since I came up with 50.5 mi range under US06, never mind 60.7 for HWFET, since I need a solid 67.

Of course, if Nissan would just publish these numbers...

Any thoughts? Thanks.

-------

* This is arrived at as follows: the 16" wheel rim diameter is halved and converted to meters to get the wheel radius: 16*0.0254/2 = 0.2032 m; add this to the tire height radius: 205mm = 0.205 m width * 55% or 0.55 = 0.11275 m. Sum the wheel radius and the tire height to get 0.31595 m total wheel-tire radius.

** Notice in the Tesla curve the optimal energy/speed ratio is around 19.6 mph!

*** The FTP, or EPA75, is LA4 + LA4[0:504], i.e. LA4 plus the first 505 seconds of LA4 repeated, which includes a second burst of 55+ mph driving in the test (the first burst peaks at second number 240 or exactly 4 minutes in). It replaces UDDS, which is the official name for LA4. EPA75 can be measured in 0.1 second intervals and therefore can give a finer grain of test data as well: http://www.epa.gov/nvfel/methods/ftp10hztable.txt

*2* HWFET or HFET is the Highway Fuel Economy [Test] Driving Schedule and also can be measured in 0.1 second intervals: http://www.epa.gov/nvfel/methods/hwy10hztable.txt

*3* US06 is known as the aggressive driving schedule, though it best matches my driving pattern because I do a lot of highway driving on my daily commute at hours where traffic is light: http://www.epa.gov/nvfel/methods/us06col.txt

*4* For all the EPA test driving suites, please visit: http://www.epa.gov/nvfel/testing/dynamometer.htm

[TimeHorse]

Yep, specifically look at the speed/range chart on their blog.



If you assume that the Tesla's ~240mi range is the same as the Leaf's 100mi range, that gives you an idea of what speeds you can travel for how far. Just divide the range number on the chart by 240 to get some real rough approximations.

Ah, but here's the rub: again, recall that that LA4's average speed is 19.6 mph, which on the Tesla chart comes to a maximum range of 410 mi. Scale that down to the Leaf: at 51 mph (my average speed) you're down to 260 mi, a drop of 63.4%, meaning you could expect the leaf's range to sink to 63.4 mi if you started driving 51, and then drove that for a distance of 63.4 you'd run out of fuel. Of course, I said Average speed, there's a lot of faster and a lot of slower actually under real circumstances. So clearly, even the Tesla doesn't go around boasting a range of 410 mi because they'd be laughed out of business, at the very best!

However, the Roadster efficiency and Range post does say that it achieves 244 mi under EPA "dynamometer". Technically, this could be HWFET or US06 or EPA75 or even SC03, the Air Conditioning supplement. But I agree, it's safe to assume they mean LA04 since most electrics look best in that light. So I agree, in principle, your numbers looks sound. I just wish I could get that curve as either a polynomial function or a spreadsheet of values to play with...

[TimeHorse]

The power curve strikes me as the most useful graph. Not that I think I could map that to the Leaf directly, though Drees's first-level analysis is pretty good. Instead, I think the best thing to note here is what we know about the curve. At higher speeds, the drag increases as the square of the velocity. Thus, the power required and the energy used should, as speed increases, be dominated by the square of that speed. This is reinforced by the commentary that follows this curve. In fact, I think it's probably "good enough" to assume the power-velocity equation is of the form:

Code: Select all

P(w) = a*w^2 + b*w + c

Where w is the angular velocity of the wheel (since we know the tire radius, we can easily convert between speed in mph and the wheel rotational velocity in radians per second by a factor of 1.414 907 422 060 45 hr-rad/mi-sec, exactly) and the Power is calculated in Watts. Here, a, b and c are unknowns constants, where a is some type of Moment of Inertia over Time in kg-m^2/s, b is some kind of Torque in N-m and c is some base power in Watts.

Of course, the specific units of a, b and c are irrelevant. The important thing here is that we have 3 unknowns we need to solve.

Again, we know what LA4 entails, we know it uses 6,437,136 of the 86.4 Joules of total battery capacity. We know it's 1370 or 22:50 seconds long (much of which is idle, hence the 19.6 mph average speed). Since we know the Energy and the time, we can compute the average power used but that doesn't really tell us anything AFAICT. However, the reverse should be straight-forward to calculate: if we know how much power is being used at a given 1-second interval we can calculate the energy by multiplying by 1 second.

So, in effect:

Code: Select all

sum(P(w)[instant]*1sec)  = 6,437,136 Joules

This translates to:

Code: Select all

a*sum(w^2) + b*sum(w) + c - 6,437,136 Joules = 0

Now, I can compute sum(w^2) and sum(w):

Code: Select all

sum(w) = 37,949.797 930 052 3 rad/sec

Code: Select all

sum(w^2) = 1,643,140.091 306 37 rad^2/sec^2

Hence:

Code: Select all

37,949.797 930 052 3 rad^2/sec^2 * a + 37,949.797 930 052 3 rad/sec * b + c - 6,437,136 Joules = 0

It would also be safe to assume if no angular velocity is achieved, the power should be 0. However, I don't think from a mathematical point of view that this is a safe assumption in solving the polynomial. If we assume when w is 0 that the Energy must be 0, that would mean c must be equal to the LA4 energy. That seems kind of strange to base it on that so I have a feeling there is something more to the constant factor.

So what we're still left with is 1 equation and 3 unknowns! What I need are two more equation to solve this. Any ideas?