# How much does an electric car consume to keep it at a certain speed?



## Pisoila (Feb 28, 2021)

If the car has reached a high speed, what electrical energy do you consume to maintain it at that speed in the 100Km straight line?

I mean a medium-sized car on the road.

If you set an example:
Audi A4 from the year 2000 whose engine has been replaced with an electric one, 
It goes on a highway close to a straight line 100Km
With a speed of 100 km / h and travels the road in 1 hour

highway to (does not go up the hill does not go down).


You choose the power of the electric motor
It doesn't slow down, it doesn't accelerate. it goes at about constant speed.


Why this discussion? Because it is important to know how much an electric vehicle consumes when we choose an electric vehicle.

We are told to consume less. Not a good answer. I needed "numbers". How less do you consume?


----------



## brian_ (Feb 7, 2017)

Government energy consumption tests include operation in various conditions. The highway tests are close to just driving at a constant speed. Canadian test results are published in various forms including kWh per 100 km, so at 100 km/h that's kWh per hour... or just kW.


https://www.nrcan.gc.ca/sites/nrcan/files/oee/pdf/transportation/tools/fuelratings/2021%20Fuel%20Consumption%20Guide.pdf



A VW e-Golf uses 19.9 kWh/100km in the highway test, and other models use 16 to 30 kWh/100km. So close to 20 kW is a reasonable estimate of consumption at 100 km/h... but if you're interested in what EV uses less energy or just how much energy an EV uses, it's really the consumption averaged over the whole test cycle - not just at a fixed speed - that matters.


----------



## Pisoila (Feb 28, 2021)

~ 20KWh for 100Km




__





Heat values of various fuels - World Nuclear Association


Heat Values of various fuels allowing comparison of different forms of energy




www.world-nuclear.org







[TH]Petrol/gasoline[/TH]
[TD]44-46 MJ/kg[/TD]​
[TH]Diesel fuel[/TH]
[TD]42-46 MJ/kg[/TD]​


Gasoline715-780 Kg/m3
750Kg............ 1m3
750Kg............ 1000 liters

1 Liter =0.75Kg gasoline.... 0.85Kg diesel

20KWh with 30% otto efficiency ....need 20KWh*10/3=67KWh

Gasoline 45MJ/kg=12.5 KWh/kg




__





Convert MJ to kwh - Conversion of Measurement Units


Do a quick conversion: 1 megajoules = 0.27777777777778 kilowatt hours using the online calculator for metric conversions.




www.convertunits.com





12.5KWh..............1Kg
12.5KWh...............(1/0.75)Liters
12.5KWh...............1.33Liters
9.4KWh.................1 Liter

10KWh/L

20KWh is the equivalent of 2 liters of gasoline consumed by the Otto engine.
It would be great if that were the case.

Can a car with gasoline consume 2 liters to go 100 km from the highway? No, consume at least twice as much!

We could "play" with this data. I guess that's why they're public.


----------



## MattsAwesomeStuff (Aug 10, 2017)

brian_ said:


> A VW e-Golf uses 19.9 kWh/100km in the highway test


19.9kwh/60miles = 331 watthours/mile. Not... great. I'd expect a bit better.

Actually, a better phrase would be that it worries me that I'm not planning to carry more battery :/

I was hoping my current stash (~32kwh) would be enough for 200km.


----------



## brian_ (Feb 7, 2017)

MattsAwesomeStuff said:


> 19.9kwh/60miles = 331 watthours/mile. Not... great. I'd expect a bit better.


Perhaps because your expectations are based on the claims of people who test only in favourable conditions? Actually, the e-Golf is not great for its size, and it's an adaptation of a gas/diesel car, so it's probably not optimal. The best results in these tests are from Tesla, I believe not because they are more efficient but primarily because they have optimized the car's programming to the test cycle - it's a long standing practice in the automotive industry.



MattsAwesomeStuff said:


> Actually, a better phrase would be that it worries me that I'm not planning to carry more battery :/
> 
> I was hoping my current stash (~32kwh) would be enough for 200km.


On the bright side, you have lower frontal area and probably narrower tires than the e-Golf.


----------



## Pisoila (Feb 28, 2021)

330Wh/mile
That is a 12V car battery at 60Ah (12V*60Ah=720Wh) It's good for 2.2 miles, a little bit.

When I proposed this subject, I thought of Newton's first law.
Ideally we should not consume energy to keep a vehicle moving in a straight line with constant speed.

This can be seen in space. There is friction on the ground, including air friction.

Let's say then what are the factors that determine this consumption in a straight line.
What can we do to reduce them?


----------



## Pisoila (Feb 28, 2021)

MattsAwesomeStuff said:


> Actually, a better phrase would be that it worries me that I'm not planning to carry more battery :/


That is the problem, not the production of energy but its transport! energy storage.

It has to do with this topic, because the less you consume, the less batteries you need.








Does Over-Inflating Your Tires Improve Gas Mileage?


I ran a little experiment to see if increasing tire pressure had any effect on gas mileage. Check out the article to see how it turned out.




axleaddict.com




I'm not saying that what is written in that article is true or not, I'm just quoting it.


----------



## brian_ (Feb 7, 2017)

Pisoila said:


> ...
> 20KWh is the equivalent of 2 liters of gasoline consumed by the Otto engine.
> It would be great if that were the case.
> 
> ...


Wow, you shuffle a lot of numbers to get to the result. 

I didn't try to follow all of your process, but the energy content of gasoline is assumed in this industry and its regulators to be 8.9 kWh/L, which would be 11.4 kWh/kg at 0.78 kg/L.

If gasoline contains 8.9 kWh/L and 20 kWh is required per 100 km, you can conclude that is somehow equivalent to about 2.3 L or 1.8 kg of gasoline, and that value (in L/100km) is published in the fuel consumption guide, so you don't need to work it out.

That comparison is common, but it is ridiculous because gasoline is not electricity: the gasoline-fueled car must convert the chemical energy in the gasoline to usable mechanical energy, while the electric car is working with very usable electricity. That energy conversion from fuel can be done at 40% efficiency at best with current engines, so the corresponding fuel consumption is at least 2.5 (and more realistically 3 to 4) times as high as that. Indeed, gas-engine cars much cheaper than their EV equivalents are more than that efficient.

If you want a fair comparison of gasoline or diesel to an EV, measure the amount of gasoline or diesel which you would need to feed into an engine running a generator to produce the electricity to distribute via the power utility grid to charge the car. Yes, that's about what gasoline and diesel cars use.


Data note: the heating value of a hydrogen-containing fuel (such as gasoline or diesel) depends on whether the water in the combustion products (the engine exhaust in this case) is a liquid (that's the Higher Heating Value or HHV) or still a vapour (that's the Lower Heating Value or LHV). High-efficiency furnaces can condense the water in their exhaust to collect the heat of vapourization, but a practical engine cannot, so you should use the LHV for engine calculations... for gasoline, that's the 44 MK/kg value.


By the way, I linked to the current (2021) Natural Resources Canada fuel consumption guide, which no longer includes the e-Golf; my e-Golf data was from the 2020 version.


----------



## brian_ (Feb 7, 2017)

Pisoila said:


> 330Wh/mile
> That is a 12V car battery at 60Ah (12V*60Ah=720Wh) It's good for 2.2 miles, a little bit.


In practical terms that doesn't work, because the "60 Ah" car starting/accessory battery can only produce that much output if discharged very slowly (the common test condition is at steady rate for 20 hours). When you try to get it all out in less than an hour, much of the energy is consumed by internal resistance of the battery. But sure... the battery is an energy container.



Pisoila said:


> When I proposed this subject, I thought of Newton's first law.
> Ideally we should not consume energy to keep a vehicle moving in a straight line with constant speed.
> 
> This can be seen in space. There is friction on the ground, including air friction.
> ...


By "ideally", in terms of Newtonian laws, you mean in the absence of an external force. Obviously drag is the external force and reducing it reduces the energy required to move the vehicle. The energy required to overcome drag through a distance is simply the total drag force multiplied by the distance, or for a non-steady-state situation the total instantaneous drag force integrated over the distance.

Millions of people have spent more than a century working on the reduction of all forms of drag; of course we should use the resulting knowledge... as every production car does.


----------



## brian_ (Feb 7, 2017)

MattsAwesomeStuff said:


> 19.9kwh/60miles = 331 watthours/mile.


100 km is 62 miles. If you must use primitive units, it's easier to just convert 199 Wh/km to Wh/mile by converting 199 miles to kilometres (because you need the inverse of km to mile conversion): 320 Wh/mile


----------



## brian_ (Feb 7, 2017)

Note: _Ayrton.980_ is a an advertising spambot, not a real person. There is no point clarifying the nonsensical post by that account.


----------



## sadcar (Mar 18, 2013)

Pisoila said:


> If the car has reached a high speed, what electrical energy do you consume to maintain it at that speed in the 100Km straight line?
> 
> I mean a medium-sized car on the road.
> 
> ...


i have a self built mitsubishi express van, a 24kw/144v dc motor is used but i have higher voltage batteries for more km per charge, to answer your question, flat forward 100kmh needs 120amp @ 155v, following behind a truck can go down to 80 amps @ 100kmh, strong wind against my van will be 160 amps, your driving behaviour will determine the distance you can drive per charge


----------



## brian_ (Feb 7, 2017)

sadcar said:


> i have a self built mitsubishi express van, a 24kw/144v dc motor is used but i have higher voltage batteries for more km per charge, to answer your question, flat forward 100kmh needs 120amp @ 155v...


That's 18.6 kW - good for a DIY van.


----------



## TT-Man (May 28, 2015)

Hi All,
Did some calculations as follows
Hi Girls, This imaginary vehicle going along a theoretical horizontal 100 kM road, what will the power burn be? So let us look at the theory! 
The power burn will be mostly to overcome wind resistance (drag) and rolling resistance. If we take an approximately standard car, work out drag, make an allowance for rolling resistance (not a lot) we can estimate a value to maintain the desired speed. 
The standard car that I have chosen is 2M wide by 1.5M high making 3 sq M, and fudge a factor of 0.85. A modern reasonably aerodynamic 4 door saloon has about a 0.3 Cd value. Using these in the following calculations will give a drag force, from which we can determine the power ‘burn', and energy for the 100 kM drive. 
Power = Force X Velocity. You want to know the power, you have selected the velocity at 100 km/hr so we need to know the force required. 
At a constant speed that force is mostly drag. Drag Force F = Cd . p . ½ V . V . 
A Cd factor ranges from 0.25 low, 0.3 mid modern 4 door, 0.4 SUV. 
Density of air p = 1.225 Kg/ M3 
Area of vehicle normal to direction A = 0.85 x Width x Height (an approximation). 
100 Km/hr is 27.78 m/sec. 
So we have might have: Drag = 0.3 x 1.225 x 0.5 x 27.78 x 27.78 x 0.85 x 2 x 1.5 Drag force = 361.6 Newtons 
Power = F x V = 361.6 x 27.78 = 10045 Watts .... ???? 
10.045 kW Or 13.5 Hp or consumption per hour, nominally 10 kWhr 
This does not take into consideration rolling resistance which is small, but exists. This is perhaps a guide for an ideal constant speed running. You need to consider a factor for head winds, and the power to accelerate the mass of your vehicle up to the 100 kM. I was surprised how much power was needed, unless you accept a very long acceleration time. In my previous calculations a decent hill climbing rate also was a surprise. 
If we do an out and back against a 15 kM head wind in one direction and a 15 kM tail wind in the other we can see what happens to consumption. I will be interested! If I put all the common factors into a constant K, we get:- 0.4686 
So burn against the head wind = k x (27.78 +4.17)squared = 0.4686 x 31.95 x 31.95 = 478.35 N The velocity is still the same so power is F x V = 478.35 x 27.78 = 13,288.6 W (13.3 kWhr)
The return journey will be : K x (27.78 – 4.17) squared = 261.2 N. 
As above Power = 261.2 x 27.78 = 7,256.5 W (7.26 kWhr)
The average = 20545.1 /2= 10272.5 W. (10.27 kWhr) Interesting a surprisingly small increase. Note that the weight of the vehicle is not included in the calculation. This would potentially effect rolling resistance. I suppose adding a nominal 10% or less would cover that. 
Adjusting for other variables would be a linear calculation. Cd, and cross-sectional area would be to divide the power by the existing variable and multiply by the new. 
For example if the Cd was say 0.4 then new power burn = 10 x 0.4/0.3 = 13,333 W 
If CS Area also increased from 3 sq mtr to 4 sq mtr then new power = 13,333 x 4/3 = 17,777 W Now I am rusty on my maths and all the above may be cobblers, but hey Guys, trying to help. Cheers, Joe 
P.S. Hill climbing and acceleration are different sums, and weight (mass) do come into the picture! I take it that someone lives on the flat lands!


----------



## MattsAwesomeStuff (Aug 10, 2017)

TT-Man said:


> Hi All,
> Did some calculations as follows
> Hi Girls,
> hey Guys, trying to help. Cheers, Joe


Did you copy and paste your reply from another reply on the web?

...

Rolling resistance accounts for about 40% of total power required on a typical EV. You can almost ignore it for bicycle weight things, but not cars.

Here's a calculator:



EV Calculator



Just remember to zero out Acceleration unless you're specifically solving for that.

I don't know what "fudge factor" is or why you would include that number versus any other number, but this calculator predicts ~11kW for air resistance (at zero weight) and ~22kW for total power (so, another 11kw for rolling resistance) presuming a 2000kg 4-door car.

If you want to climb a modest hill you'll need to roughly double your power requirements. If you need to accelerate up a hill, triple them.


----------



## brian_ (Feb 7, 2017)

MattsAwesomeStuff said:


> You can almost ignore it for bicycle weight things, but not cars.


Perhaps, because those things have terrible aerodynamics and rock-hard skinny tires.


----------



## TT-Man (May 28, 2015)

Hi Matt & Brian,
The calcs were all my own effort. The formulas were refreshers from the net. I copied from my file on MS Word so limited in math functions ... hand held calculator ... steam tech! ...No a slide rule and logs would be steam tech.
The suggested estimate for rolling resistance came from an observation on the net. I do not believe it to be as large as you suggest, unless on a large ut with fat chunky tyres (tires).
The area calc fudge factor allows for rounding of the nominal rectangle. I chose a 2 x 1.5 giving a nominal 3 sq mtrs. The 0.85 allows for the fact that it is't a rectangle, but rounded.
Remember this is all a bit of a gestimation.
Just a comment about me. Health wise I am now limited in physically activity, plus we have gone back to lock up. So this is an interesting diversion rather than TV..
Cheers all,
Joe


----------



## brian_ (Feb 7, 2017)

TT-Man said:


> The suggested estimate for rolling resistance came from an observation on the net. I do not believe it to be as large as you suggest, unless on a large ut with fat chunky tyres (tires).


Either you misunderstood what you read on some web site, or it was not applicable to typical cars. 

Using the most optimistic value of CRR = 0.01 from the Engineering Toolbox, the rolling drag force is 98.1 N per 1,000 kg of vehicle mass, or 2.73 kW per 1000 kg at 100 km/h. Realistic values are up to twice as high. That's not negligible.


----------



## TT-Man (May 28, 2015)

Hi Brian, 
OK, I shall accept your better sources. So what R.R. at 100 k would you suggest for the example?
Cheers, Joe


----------



## brian_ (Feb 7, 2017)

TT-Man said:


> OK, I shall accept your better sources. So what R.R. at 100 k would you suggest for the example?


How about 0.015?


----------



## TT-Man (May 28, 2015)

Fair enough, so at 100 k/hr and say 1,000 kg (1 tonne) , what is the estimate?
I'm being lasy!
Cheers, Joe


----------



## brian_ (Feb 7, 2017)

brian_ said:


> Using the most optimistic value of CRR = 0.01 from the Engineering Toolbox, the rolling drag force is 98.1 N per 1,000 kg of vehicle mass, or *2.73 kW per 1000 kg at 100 km/h*. Realistic values are up to twice as high.





TT-Man said:


> Fair enough, so at 100 k/hr and say 1,000 kg (1 tonne) , what is the estimate?
> I'm being lasy!


No kidding


----------



## TT-Man (May 28, 2015)

Hi Brian, Yes that is significant power loss at that speed. As a rule of thumb at highway speeds between 25 and 30 %.
Just a by the way, 
I note your maple leaf. In the late 60's I worked for BC Hydro out of Vancouver, with a short spell near Hamilton / Toronto, before repatriating to our little Island.
Take care,
Joe


----------



## LandsPB (Nov 17, 2020)

Hey guys, all this complicated math is great. But I need something more practical for my non German engineering mind... so, I have a fiat 500e. It’s rated at 85 miles max with about a 27kwh battery. So, if I drive 65mph down the highway, I definitely cannot go 85 miles. Maybe 50. so, my question is, does anyone have some kind of graph where I would get the max range out of it, at practical speeds? In other words, if I drove it steady at 45mph how far could I go, or if I drove it steady at 5mph, how far (even tho it would take a long time lol but just for fun would like to see that). If it’s sitting still, of course, it would get infinite charge for 0 miles. But if its driven at 5 mph could it go 200 miles on a charge? etc.?

(and btw, I found that a gallon of gas is 33kwh of energy, so the giant battery pack in this car is literally like having 2/3 of a gallon of gas, for a car that can get maybe 125mpg equivalent. I think that’s pretty cool. Can’t wait for the new technology to take the world by storm)


----------



## brian_ (Feb 7, 2017)

LandsPB said:


> ... so, I have a fiat 500e. It’s rated at 85 miles max with about a 27kwh battery. So, if I drive 65mph down the highway, I definitely cannot go 85 miles. Maybe 50. so, my question is, does anyone have some kind of graph where I would get the max range out of it, at practical speeds? In other words, if I drove it steady at 45mph how far could I go, or if I drove it steady at 5mph, how far (even tho it would take a long time lol but just for fun would like to see that). If it’s sitting still, of course, it would get infinite charge for 0 miles. But if its driven at 5 mph could it go 200 miles on a charge? etc.?


For any electric car, the optimal speed will be very low. Rolling resistance doesn't vary much with speed, and aerodynamic drag increases greatly with speed, so slower is better. If you go too slowly the motor is not in an optimal working range, but slow enough for that to be a concern will be too slow for traffic.

More than half of the energy used to move a car at highway speed goes to aerodynamic drag, so ideally the range at very low steady speed might be more than twice as long as at highway speed.

In practice the improvement is not that much, largely because driving slowly normally means driving in urban traffic with stops and starts and changes of speed, which increase consumption. Governments publish energy consumption values, and in the Canadian 2021 Fuel Consumption Guide the "city" consumption is typically no better than 20% lower than the "highway" consumption.



LandsPB said:


> (and btw, I found that a gallon of gas is 33kwh of energy, so the giant battery pack in this car is literally like having 2/3 of a gallon of gas, for a car that can get maybe 125mpg equivalent. I think that’s pretty cool. Can’t wait for the new technology to take the world by storm)


It's not like that at all, because electricity is not gasoline, and it is very important what form of energy you start with. If you started with gasoline and ran a generator to make electricity (which is essentially what happens if your utility power comes from a generating station which burns coal or natural gas or even oil) only about one-third of the energy in the fuel becomes electricity, so that "125 mpg" becomes 42 mpg... which is pretty ordinary. Some EV enthusiasts like to make this "equivalent" comparison because it does seem impressive, but it's meaningless.


----------



## LandsPB (Nov 17, 2020)

The 33kwh of Energy in a gallon of gas is useful comparison tho, and I don’t think totally meaningless. It can give us a standard to use so the common guy, me, can figure out how much energy the car is using and compare it to another ICE vehicle for example. The speedometer shows, after each drive, what the mileage for that specific drive is, and the economy based on MPGEe which is kind of cool. Oregon is 65% hydro and wind, too, so my power is somewhat better than, say, West Virginia which is probably mostly coal... anyway, I’m still trying to find a graph that would show me how far the car could go at a constant speed, for various MPH’s. Again, mostly for curiosities sake, lol.


----------



## OR-Carl (Oct 6, 2018)

LandsPB said:


> I’m still trying to find a graph that would show me how far the car could go at a constant speed, for various MPH’s.


How about you make the graph youself? 








(I tried to make a nice chart out of my spreadsheet, but it was harder than I thought to make it look right)
Anyway, this is what I have been playing around with for range calculations. The velocities are based on 4th gear at motor RPM intervals of 500. It takes into account motor efficiency, which is 80% at 500rpm, but upwards of 90% in the middle of its speed curve. If you downshifted so you were running higher RPM at 10mph, you could probably add a bit more range. But the difference is pretty stark. Granted, these numbers are for a pickup with a large area and a bad coefficient of drag - they would be a little less extreme for a smaller, more aerodynamic car. 

Here is the equation I am using for Watts, if you want to recreate this sheet: =M4*9.81*N4*O4+(0.6465*P4*Q4*(N4^3))


----------



## brian_ (Feb 7, 2017)

LandsPB said:


> The 33kwh of Energy in a gallon of gas is useful comparison tho, and I don’t think totally meaningless. It can give us a standard to use so the common guy, me, can figure out how much energy the car is using and compare it to another ICE vehicle for example.


But it makes no sense to compare to an ICE vehicle, because the EV doesn't use gasoline. There's no way to say "I have a gallon of gasoline; which car gets me further?", because the EV won't get you anywhere with gasoline. It's not even a cost of operation measure, because gasoline and diesel and electricity all have different prices per unit energy.



LandsPB said:


> The speedometer shows, after each drive, what the mileage for that specific drive is, and the economy based on MPGEe which is kind of cool.


But economy in meaningful energy terms (kWh per distance, or distance per kWh) is much more useful, because it tells you how much electrical energy you need to buy and put in the battery to go that distance.


----------



## LandsPB (Nov 17, 2020)

I still submit its handy or useful to see MPGe. But my more useful calculation is $$$ per mile. I calculate my 500e costs me 2 cents a mile to drive, vs my Ford F-350 diesel which costs me 18 cents a mile. This is the most handy way I can think of to figure out mileage.

Also, I did try to make a graph but wasn’t sure if it’s linear or exponential or whatever, curve because of rolling and wind resistance etc. I’m assuming is a curve, not a straight line...

I noticed on the chart, the car has a transmission. The 500e is just direct drive up to top speed. I suppose around 12,000 rpms (I’ve read). 

the chart looks good! Thanks!

I do know, Oregon has pretty reasonable electricity and is 65% wind/solar/hydro. Gas is almost 3$ a gallon. I also know they gave me a 2500$ Cash rebate when I got the car. I’m into it for actually negative money (I traded 2 motorcycles for it worth 4200$ but in reality I had maybe 1100$ into them plus some labor). So, no matter which way I slice it I’m saving money like a crazy man. And, I LOVE not having to go to the gas station. I actually get slightly annoyed (very slightly heheh) when I have to go fill the truck and our Kia Soul...


----------



## brian_ (Feb 7, 2017)

LandsPB said:


> But my more useful calculation is $$$ per mile. I calculate my 500e costs me 2 cents a mile to drive, vs my Ford F-350 diesel which costs me 18 cents a mile. This is the most handy way I can think of to figure out mileage.


I agree. That's why consumption data should be in the terms that you buy energy: you buy gasoline by volume (gallon or litre) so consumption is volume per distance (or distance per volume); you buy electricity by kilowatt-hour so consumption is kWh per distance (or distance per kWh).

Electrical operation is cheaper than burning fuel largely because electricity is minimally taxed, but motor vehicle fuels are heavily taxed.



LandsPB said:


> Also, I did try to make a graph but wasn’t sure if it’s linear or exponential or whatever, curve because of rolling and wind resistance etc. I’m assuming is a curve, not a straight line...
> 
> I noticed on the chart, the car has a transmission. The 500e is just direct drive up to top speed. I suppose around 12,000 rpms (I’ve read).
> 
> ...


Energy consumption per distance varies as the square of velocity, so the relationship is quadratic, with the second-order term being primarily aerodynamic and the constant term being primarily rolling drag.



LandsPB said:


> I noticed on the chart, the car has a transmission. The 500e is just direct drive up to top speed. I suppose around 12,000 rpms (I’ve read).


All cars have transmissions; an EV usually has a transmission with only one (fixed) ratio. The Fiat 500e is an example of that; it isn't "direct drive" in any sense - it always drives through two stages of reduction gearing, with an overall ratio (motor:wheels) of 9.59:1. That's more reduction than a Nissan Leaf or Chevrolet Bolt, and similar to a Tesla; to correspond to a reasonable top speed for the car with the relatively small tires the motor has to be able to spin reasonably fast. With the 500e's tire size, its top speed of 93 mph corresponds to 1400 RPM at the wheels, and with 9.59:1 gearing that means about 13,420 RPM at the motor... substantially higher than (for instance) a Leaf.

Single-speed transmissions are used in production EVs because the motor works well enough over a broad range of speeds that the extra cost and complication of a multi-speed transmission isn't justified in most cases.


----------



## OR-Carl (Oct 6, 2018)

brian_ said:


> Electrical operation is cheaper than burning fuel largely because electricity is minimally taxed, but motor vehicle fuels are heavily taxed.


Isnt this overlooking the inherent efficiency gains that electric motors have over their ICE counterparts? I looked up fuel taxes here in Oregon, and we pay 36 cents in state tax, plus the 18 cents of federal tax, so about 54 cents per gallon. I paid 2.97 the other day, so that is only 18% tax. With my prius I get at least 40mpg, so, about 7.4 cents per mile, of which 1.3 cents is tax.

If I were to use grid power to charge my electric truck, it would be about 9.8 cents per kwh, which should get me about 3 miles of range. So that is only 3.3cents per mile. Even without tax, the electric car would be almost half the cost to operate as a hybrid, and likely 3 times less than an equivalent gas pickup getting 20 something mpg.


----------



## LandsPB (Nov 17, 2020)

I was going to say, the electric car is supposedly 90% efficient with its energy vs the gas which is about 40%. So yeah, per energy unit, the electric car definitely wins.

again, the gas vs electric. If my battery had 75 kWh (has 26kwh) which would be about 3 gallons of gas worth of energy, I’d be able to go over 200 miles.

btw, oregon stabbed me with a really high registration tax too (300-400$ I think for two years?), since it’s electric. But then, having no idea what the other hand is doing evidently, Gave me a cash rebate of 2500$. Which, any used electric car owner can get, if bought from a dealership.

Also, my battery still has a warranty. But, in about 2 years when it expires I really want to upgrade the battery myself and try and get it to a 200 miles range car.


----------



## brian_ (Feb 7, 2017)

LandsPB said:


> I was going to say, the electric car is supposedly 90% efficient with its energy vs the gas which is about 40%. So yeah, per energy unit, the electric car definitely wins.


Again, they're starting with different energy sources, so there is no reasonable comparison. So, wins what? If I can walk up a few flights of stairs faster than you can run a mile, which one of us "wins"... since we're doing different things?


----------



## brian_ (Feb 7, 2017)

OR-Carl said:


> Isnt this overlooking the inherent efficiency gains that electric motors have over their ICE counterparts?


What efficiency gain? They're starting with different energy sources, so they're not comparable in any way. An electric motor is less efficient than a mechanical transmission, and both are transferring energy from the vehicle's source (battery or engine) to the driveline; pick your terms and you can make anything look good or bad as desired. 



OR-Carl said:


> I looked up fuel taxes here in Oregon, and we pay 36 cents in state tax, plus the 18 cents of federal tax, so about 54 cents per gallon.


Good point - fuel taxes in the U.S. are relatively low, compared to Canada and most other first-world countries. But even then, I doubt that you're paying close to 18% tax on electricity.

EVs do have energy cost advantages beyond tax. Electricity production from fuels is relatively efficient due to the use of large power plants, compared to small engines. Electricity production from other sources uses free energy sources (flowing water, wind, sunshine) rather than fuels which must be purchased or made from resources which must be paid for. And like hybrids, EVs avoid idling consumption and recover some braking energy.


----------



## LandsPB (Nov 17, 2020)

I disagree, respectfully. I love ICE engines and cars, don’t get me wrong. But the energy being used by the ICE cars is 60% being turned into heat and other waste byproducts. With about 40% being turned into useful energy to move the car forward. If we had electric cars with only 40% efficiency we wouldn’t even be having this discussion. The positive nature of them is, they can turn a source of energy into kinetic motion more efficiently than an ICE car. Every time we switch systems (oil out of ground, oil into gas, gas into explosion with heat, driving a by design piston engine with tons of friction) we lose efficiency. There’s a reason ICE engines have radiators and cooling fans. By contrast, wind/hydro/nuclear/coal/naural gas/solar into electricity, stores in a battery and using an electric motor (which has very little drag by design). Skips several steps along the way and makes a more efficient vehicle.

flipping your story about the flight of stairs... if I can run a mile FASTER than you can walk up 2 flights of stairs, then of course I win! I‘ve got evidently a superior lung/cardio system and am therefore more efficiently using them to get a further distance than you up a flight of stairs. So, I can compare the two and it works...


----------



## OR-Carl (Oct 6, 2018)

brian_ said:


> What efficiency gain? They're starting with different energy sources, so they're not comparable in any way.


I feel like this is a bit of a stretch. All energy eventually winds up as heat, right? So why not look at it from the point of view of BTUs or whatever measure of heat you want to use. You can calculate the BTUs that a kWh will produce, same with a gallon of gas once combusted. The bottom line is that the electric vehicle goes father with the same input.


----------



## brian_ (Feb 7, 2017)

OR-Carl said:


> I feel like this is a bit of a stretch. All energy eventually winds up as heat, right? So why not look at it from the point of view of BTUs or whatever measure of heat you want to use. You can calculate the BTUs that a kWh will produce, same with a gallon of gas once combusted. The bottom line is that the electric vehicle goes father with the same input.


But does it have the same input? How do you feed gasoline into an EV? If you feed gasoline to a engine which drives a generator which charges an EV, does the EV go further than a gasoline car?


----------



## OR-Carl (Oct 6, 2018)

brian_ said:


> But does it have the same input?


I agree that what is being discussed is certainly an abstraction - and maybe not a helpful one. But your rebuttal seems to be that gasoline is the only possible power source. Sure, if you have to burn the gasoline in a generator, an electric car will not outperform a gas car. But the whole point is that an electric car can be fueled by any number of sources, any of which are better than burning gasoline to move a car.

Because, lets also not forget that the 33kWh in a gallon of gasoline does not take into consideration the energy used drilling, pumping, transporting, refining, pumping again, transporting again, and then pumping one last time for good measure to get it into your tank. I can put solar panels on my roof, and the electrons will jump right down the wires and into my truck 

So I do agree that MPGe is just a way to make an arbitrary comparison between a new technology with a future, and an old technology that, while still dominant, is on its way out. Eventually nobody will use gas at all, and we can all just get down to talking about kilometers per kWh like civilized human beings. (if we Americans can ever get our act together an adopt a more rational system, that is).


----------



## brian_ (Feb 7, 2017)

OR-Carl said:


> I agree that what is being discussed is certainly an abstraction - and maybe not a helpful one. But your rebuttal seems to be that gasoline is the only possible power source.


No, that's completely opposite to what I am saying! Using "MPG equivalent" assumes that gasoline is the only energy source and that everything is equivalent to it, which is nonsense.



OR-Carl said:


> Sure, if you have to burn the gasoline in a generator, an electric car will not outperform a gas car. But the whole point is that an electric car can be fueled by any number of sources, any of which are better than burning gasoline to move a car.


Gasoline, electricity, uranium-235, moving water... they're all energy sources, and they are very different in their usefulness, so even in quantities with the same energy value, they are of different value in moving a car. As soon as you introduce the term "gasoline equivalent" any comparison is meaningless unless you start with gasoline as the fuel. Use a different energy content (per volume) and you can express the numbers as "diesel equivalent"... and it is only useful if starting with diesel. How about "uranium equivalent"?

The flexibility of energy source is both an advantage of EVs, and a problem. The advantage is that flexibility - any energy source can be used without changing the vehicle. The problem is that electricity doesn't come out the ground - it has to be made from something else. If starting with moving water or wind or sunshine that's easy (electricity is the easiest energy carrier to make), but if starting with fuel or geothermal heat it means going through an engine stage. And however it is made, electricity is heavy and expensive to carry, because batteries are heavy and expensive.



OR-Carl said:


> Because, lets also not forget that the 33kWh in a gallon of gasoline does not take into consideration the energy used drilling, pumping, transporting, refining, pumping again, transporting again, and then pumping one last time for good measure to get it into your tank. I can put solar panels on my roof, and the electrons will jump right down the wires and into my truck


But almost no energy used by EVs comes from rooftop solar panels, and the energy consumption ratings of EVs do not (and cannot) take into consideration the energy used building or operating electrical generation facilities, transmitting and distributing the produced electricity, or even in the inefficiency of charging a battery. Even with your rooftop solar system it doesn't take into consideration the inefficiency of the panels or the losses in wiring and electronics.

Shouldn't your car be rated in kilometres per kWh of incident sunshine? Divide your km/kWh value by five or so. 

Vehicle energy consumption is expressed in terms of the distance travelled and the energy actually provided to the vehicle. That's kWh of electricity, or amount of the specific fuel used by the vehicle, depending on the vehicle.



OR-Carl said:


> So I do agree that MPGe is just a way to make an arbitrary comparison between a new technology with a future, and an old technology that, while still dominant, is on its way out. Eventually nobody will use gas at all, and we can all just get down to talking about kilometers per kWh like civilized human beings. (if we Americans can ever get our act together an adopt a more rational system, that is).


But it doesn't make a useful comparison. If someone made a mistake and used the wrong gasoline energy content so all the "equivalent" values for EVs were twice as good or half as good, would it make any difference to choosing a vehicle? No. We can all use kilometres per kWh for EVs now, and continue to use kilometres per volume of fuel for vehicles that burn fuel.


----------



## LandsPB (Nov 17, 2020)

Electricity in Oregon is 9.8 cents a kilowatt hour x27kwh capacity so it costs me 2.64$ to “fill” my car. And I can drive 60-80 miles.

gas in Oregon seems to be 9 cents a kilowatt hour and I can’t fill my car equivalent (would be about 2/3 a gallon of gas) but with that same amount of energy I can drive about 20 miles (Kia Soul at 25mpg).


----------



## OR-Carl (Oct 6, 2018)

brian_ said:


> Shouldn't your car be rated in kilometres per kWh of incident sunshine? Divide your km/kWh value by five or so.


Yes, _my car_ will have to be rated that way, as that will be the main way I will be able to charge it . I suspect a factor of five will not nearly be enough, as the power will go through a lot of steps. Lets see, Solar panels are about 18%, Charge controller 80%? first battery 95%, inverter 85%, Charger 90%?, trucks battery: 95%. I get just about 10%.

So I will figure I need about 3 to 3.5 kwh of insolation per mile driven . So if the average insolation at the 45th parallel is about 300 W/m2 I would need 10 square meters of sun for an hour to drive a mile.Hmm, thinking about this, I think I will need more solar panels


----------



## Electric Land Cruiser (Dec 30, 2020)

From now on everyone has to calculate the energy of their battery or their gasoline tank in Calories so that we can keep @brian_ happy 

Going back to the OP; simply find the car with the best highway MPG or L/Km and that's the most efficient at speed. Really 1.6L or 5.7L doesn't matter so much a Corvette gets 25mpg on the highway same as a 4 cylinder Audi, everything else is splitting hairs.

Another way to calculate a good EV conversion vehicle is find the coefficient of drag and the frontal area of the vehicle to figure out which vehicles have low air resistance. Coefficient of drag and frontal area are both equally important, most people just think about coefficient of drag.

Ultimately you'll probably find that shorter and narrower = better and more length = better.


----------

