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Electric Volt chassis

Fuel-cell Volt chassis
The Chevrolet Volt moves closer to production

Although there were some previous concerns over the feasibility of General Motors' Chevrolet Volt electric car, it appears the company is ready to charge ahead with production. GM is making the necessary steps to ensure that the Volt makes use of best available battery technology to achieve its project goals -- something that Toyota is having a few problems with right now.

The company announced yesterday that 13 companies sent in proposals regarding the advanced lithium-ion batteries used to power the Volt. When the dust settled, two companies were selected. The first contract was awarded to Compact Power (a subsidiary of LG Chem) while the second contract went to Continental Automotive Systems.

"The signing of these battery development contracts is an important next step on the path to bring the Volt closer to reality," said GM Chairman and CEO, Rick Wagoner. "Given the huge potential that the Volt and its E-Flex system offers to lower oil consumption, lower oil imports, and reduce carbon emissions, this is a top priority program for GM."

The Chevrolet Volt uses a tiny, turbocharged 1.0 liter internal combustion engine (ICE) to charge its onboard lithium-ion battery pack. The ICE is not, however, used to provide propulsion for the vehicle -- that is left to the electric motors. GM says that the Volt can travel 40 miles on battery power alone before the ICE needs to kick in to charge the batteries back up again.

A second version of the Volt is also in the works. GM unveiled a fuel-cell variant of the Volt built on GM's E-Flex architecture in April. This Volt comes equipped with an 8kWh lithium-ion battery pack, three electric motors and a 4kg hydrogen fuel tank. The vehicle is capable of traveling over 320 miles with a fully topped off battery, can zip from 0 to 60 in around 8.2 seconds and has a top speed of 120 MPH.

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Price
By DigitalFreak on 6/6/2007 8:20:11 AM , Rating: 3
Hopefully it will be reasonably priced when it's finally available, and not 25k+




RE: Price
By FITCamaro on 6/6/07, Rating: 0
RE: Price
By Pessimism on 6/6/2007 8:54:31 AM , Rating: 3
Considering they can't even standardize on a battery for a regular car the likelihood of cooperating on battery packs is slim to nil.


RE: Price
By TomZ on 6/6/2007 9:13:47 AM , Rating: 2
quote:
The only thing they could do to cut costs of the battery packs down is come up with one or a few standard battery packs that all the car manufacturers use.

In the article they talked about GM signing "development contracts" with the battery manufacturers. So odds are right around zero of having any kind of interchangeable battery packs.


RE: Price
By h0kiez on 6/6/2007 10:03:24 AM , Rating: 1
Also hopefully it won't come with those giant ass wheels. Would only drain the batteries faster due to the added power that would be needed to turn those monstrosities.


Really? Larger wheels = fewer revolutions of the axle for a given amount of travel...more efficient. Anyway, larger rims are paired with low-profile tires, so the entire wheel/tire package is usually about the same size whether the "rims" are 17s or 20s.

All they do is shift pollution from air to land.

If used by massive amounts of people, they also reduce other countries' reliance on middle-eastern oil. That's a big plus for me.

Not to mention god knows what it'll cost to replace one of these battery packs. I'm thinking somewhere around $4-7000

It's actually more like 2K - 3K based on what I can find online and that's likely to continue to get cheaper based on the obvious economies of scale. Also, as noted in the article here:

http://dailytech.com/Lithium+Technology+Corporatio...

Li-Ion batteries (such as in this Volt) are supposedly going to go strong for at least 150,000 miles. I think most people would buy a new car and send this one to the scrap heap/recycling before they got around to replacing a battery.


RE: Price
By sprockkets on 6/6/2007 11:11:33 AM , Rating: 2
larger tires means more friction and grip to the road but more friction means more load on the engine and thus less fuel efficiency

Here is an easy way of knowing that is true: Is it easier to ride a bike with skinny touring tires or off road tires?


RE: Price
By Erudite on 6/6/2007 12:09:52 PM , Rating: 2
Actually, a better comparison would be is it easier to ride a bike with large tires or small tires (height).

Better yet, consider a push mower. What is easier to push along? A mower with the short plastic wheels, or one with the larger tires?

The only place I could actually find relevant information on the size of the tires said this:

quote:
Other key proportional highlights include a dash-to-axle length that positions the driver far rearward of the front wheels; large 21-inch by 7.5-inch wheels; short front and rear overhangs and departure angles that deliver a sense of taut, compact energy.


I'm not sure if that would really use enough more power to be noteworthy or not.

Quote from: http://www.carbodydesign.com/archive/2007/01/10-ch...


RE: Price
By bldckstark on 6/6/2007 12:32:02 PM , Rating: 3
The major problem with larger diameter wheels is that the farther you get away from the point of power application the more internal friction you create (called hystersis). This includes inertia multiplied by the distance from the axle. This is called a moment arm and is often (incorrectly) equated to torque (but close enough for a tech site). The longer the moment arm, the more energy you must expend to create the same amount of rotational force. Smaller wheels take less energy to turn, unless you can offset the mass by making up for it in other ways. Like using magnesium instead of aluminum or steel. Using an advanced design to eliminate unnecessary material in the wheel (big holes, few spokes) and the like, but usually the realities of manufacturing cancel out all of the fancy design work.

Another way to describe it is through kinetic energy. An earlier poster said that a larger wheel does not have to turn as fast to go as fast. This is true in a circumferential sense. The reality is that it must still turn with the same amount of energy as a small wheel. We still have to put 60 mph of energy into the wheel regardless of the diameter. It takes more energy to make a larger diamter wheel go 60 mph, but less to keep it at 60 mph. The problem is the large difference in attaining speed, and the small difference in retaining speed.

The moral of the story is - bigger wheels = less efficient.


RE: Price
By Ringold on 6/6/2007 1:35:16 PM , Rating: 1
Everybody misses the point, somewhat. The market has shown people don't want geeky girly-man looking cars, they want hip, mean, or just plain cool looking cars that also happen to be really, really fuel efficient. If this thing gets something wild like 60 or 80mpg in the real world do geeks want to have a car that sells like a Honda Insight or a REAL mass-market car that sells like an Accord and gets a couple mpg less but still might be astronomical compared to normal cars?


RE: Price
By emboss on 6/6/2007 3:04:56 PM , Rating: 3
Just wondering (and completely OT), did you learn physics in a non-English speaking place? "Moment arm" is common from non-physics courses, but I don't think I've ever seen angular momentum called hysteresis. Just curious as to where it's called that :)

Interestingly, I actually did a bit of experimentation with this about 10 years ago - measured the inertia of a couple of wheels, and the friction of the drive system on my car at least. I can't remember the exact numbers, but the overall result was that a larger wheel is more efficient is nearly all cases. Things have probably changed a bit in the last 10 years, but I doubt there's been any drastic changes.

A cleaner physics-orientated way of looking at it is that angular motion has many parallels with linear motion - you have (ignoring vector properties for simplicity)
{} torque: angular equivalent of mass, equals force times the perpindicular distance from the axis of rotation.
{} inertia: angular equivalent of mass, equals force times the distance from the axis of rotation (with integration for non-point-masses)
{} angular velocity: angular equivalent of velocity (equal to the speed of rotation around the axis).

Similar to the linear kinetic energy E = 0.5 m v^2, there is an angular kinetic energy E = 0.5 I w^2.

What does this have to do with wheels on cars? Well, if you simply scale up a wheel by a factor of 2, you will increase the inertia by a factor of 8. However, for a fixed road speed, the angular velocity will halve, so the required energy to get the double-sized wheel spinning with the same road velocity is only doubled. It won't even quite be this bad, as mass can usually be removed when the wheel is enlarged.

However - this has nothing to do with *maintaining* speed. The frictional losses (ie: the rate of energy "lost") in a drive system are a function of rotational speed, except for air-spoke resistance which is a function of road speed and also linear with spoke length. By doubling the radius of the wheel, you would drop the drve system friction (by up to a factor of two), while increasing the air-spoke friction (by up to a factor of two). This may not sound like you gain anything, but in cars air-spoke friction is much less than drive system friction (due partially to low spoke count compared to say bicycles). So you actually would significantly drop the overall friction, and hence gain efficiency.

It's only in start-stop driving that the energy required to spin up the wheel has any (significant) effect. Taking a wheel+tyre to be a 20 kg point mass 40 cm from the axis of rotation, and with 45 cm radius to the road surface, gives an equivalent "linear" mass of about 40 kg. Probably significantly less, as I used numbers on the heavy side for my guesses. So, increasing the wheel size by, say 20%, would increase the "mass" (with respect to acceleration) of the vehicle by about 30 kg - pretty trivial in the whole scheme of things.

While there's some situations where the increase in energy required to attain a speed would outweigh the advantages from lower drive system friction, in the majority of cases it's the other way around.


RE: Price
By therealnickdanger on 6/6/2007 4:39:01 PM , Rating: 2
You guys are all talking about bigger wheels and such, but overall wheel/tire height doesn't change all that much between a 17" wheel with thick tires and a 20" wheel with low-pros. It all varies according to tire selection. Wheels of all sizes are getting stronger with less material (therefore lighter).

Tire width is about the only factor here that will significantly affect performance/efficiency. Going from a 205 to a 265 introduces more rolling resistance/friction, not specifically more weight.

Here's a thought: don't hybrids convert "braking" into energy? In this case, perhaps more weight in the wheels would be desirable?


RE: Price
By GTVic on 6/6/2007 5:37:16 PM , Rating: 2
I think it is the weight of the vehicle that matters.

The weight of the wheels/brakes/etc is called unsprung weight. You can improve the quality/performance of the ride by lowering the unsprung weight. Hopefully the braking systems of these cars is further up the drive train which will reduce the unsprung weight compared with disc/drum brakes.

http://en.wikipedia.org/wiki/Unsprung_weight


RE: Price
By masher2 (blog) on 6/6/2007 6:47:53 PM , Rating: 1
A lot of misconceptions in this thread. Larger tires are less efficient...not from 'hysteresis' or more tread contact, but because the tire contains more angular momentum...momentum that is lost when decelerating or turning (which is itself a special case of acceleration). However, as another poster points out, a low-profile 20" tile may not be appreciably larger than a standard 17" tire...and if it uses a low-mass rim, could theoretically have a lower angular momentum.

Another common misconception. The amount of tread of the ground does not affect rolling friction. A larger contact patch means more surface area...but less weight per unit area. The two factors cancel out. The only factors which affect it are vehicle weight and the coefficient of friction, which is determined by tread composition and (to a lesser extend) tread geometry.

This begs the question of why race cars use wider tires then, if it doesn't give them more friction. The simple reason is the larger contact patch contains more rubber to heat up, which reduces the chance of melting the tire....liquid rubber has a very low coefficient of friction.


RE: Price
By kkwst2 on 6/6/2007 11:10:36 PM , Rating: 2
I think you're as misconceived as the rest. Tire friction is more complex than you make it out to be. It has as much to do with how much the tire deforms both locally (the rubber compound) and globally (sidewall deformation, etc.) as it does with the coefficient of friction. Wider tires can be more efficient because they cause less tire and sidewall deformation as the tire turns.


RE: Price
By masher2 (blog) on 6/7/2007 1:10:31 AM , Rating: 1
> "It has as much to do with how much the tire deforms..."

That's encompassed in tire geometry, which I mentioned above. The primary point is that contact patch size does not affect friction. Double the amout of rubber touching the ground, and you halve the contact force per unit area-- frictional resistance remains unchanged.


RE: Price
By emboss on 6/7/2007 3:31:39 AM , Rating: 2
quote:
momentum that is lost when decelerating or turning


Half right - in a car, angular momentum loss during turning is negligible (assuming no loss of traction, where you would have far greater efficiency concerns :) ) as the acceleration is almost exactly along the axis of rotation of the wheel.

Additionally, as another poster pointed out, regenerative braking means that the energy put in to spin the wheel up will be recovered to some degree.


RE: Price
By masher2 (blog) on 6/7/2007 10:46:19 AM , Rating: 2
> "Half right - in a car, angular momentum loss during turning is negligible "

Incorrect. If this were true, a car, once moving, could turn in a circle forever without the engine running.

To turn a vehicle one must angle the wheels obviously. That requires a force, and that force generates frictional losses. The more angular momentum in the wheel, the larger the force required, and the greater the frictional losses.

> " regenerative braking means that the energy put in to spin the wheel up will be recovered to some degree"

Very true. Of course, there are still losses, which means smaller wheels are still more efficient than larger ones.