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The question which bothers me for some time is the following: at higher speeds the dissipated power (heat) inside the battery (Pb with units in e.g. kW), due to the internal resistance of the battery, is, according to my calculations, proportional to the sixth power of the EV speed (v), so (constant K1 mainly depends on particular EV model):

Pb = K1*v^6

This means that if the EV travels at 60 mph (96 km/h) and the dissipated power in the battery is for example 1 kW, the dissipated power at 70 mph (112 km/h) will be 2.5 kW and the dissipated power at 50 mph (80 km/h) will be 335 W.

By the way, the cumulative dissipated energy in the battery on the fixed travelled distance (Eb with units in e.g. kWh) is proportional to the fifth power of v:

Eb = K2*v^5

If the temperature of protective case for the battery is more or less equal to the environment temperature (Te), the final temperature difference (after long long time!) between the battery (Tb) and the environment should be proportional to the battery dissipated power:

Tb – Te = K3*v^6

So, for example, if the battery temperature at 60 mph is 6oC higher than the environment, at 70 mph it will be about 15oC and at 50 mph about 2oC higher.

Note that short-term temperature changes at some fixed travelled distance depend on the fifth and not the sixth power of the speed. Therefore, if the battery is at environment temperature and we travel some shorter distance at 60 mph and the battery temperature rises for 6oC, at 70 mph, on the same distance, it will rise for about 13oC and at 50 mph for about 2.4oC.

Therefore, even slight changes in EV velocity can result in larger changes of dissipated power inside the battery (and hence in larger temperature changes). Don’t drive too fast!!

By the way, the above dissipated power expression is derived from the following relations:

- At higher speeds the EV power is proportional to the third power of velocity (by the way, the average consumption (in kWh / mile or km) is proportional to the square of velocity). The current through the battery can be assumed to be proportional to the power delivered to the EV motor (due to more or less constant voltage of the battery), so the current through the battery is proportional to the third power of the EV velocity.
- Due to internal resistance of the battery, the dissipated power in the battery is proportional to the square of the current. Therefore, the dissipated power in the battery is proportional to the sixth power of the EV velocity.

Note that the above derivations are approximations and I did not want to make the main idea too complicated.
 

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By the way, the above dissipated power expression is derived from the following relations:

- At higher speeds the EV power is proportional to the third power of velocity (by the way, the average consumption (in kWh / mile or km) is proportional to the square of velocity). The current through the battery can be assumed to be proportional to the power delivered to the EV motor (due to more or less constant voltage of the battery), so the current through the battery is proportional to the third power of the EV velocity.

- Due to internal resistance of the battery, the dissipated power in the battery is proportional to the square of the current. Therefore, the dissipated power in the battery is proportional to the sixth power of the EV velocity.
When you multiply powers you add their exponents not multiply them. ;)

So if you have speed^3 as your speed vs power relationship and I^2*R as your current vs dissipation relationship in the battery, (with R and battery voltage presumed constant) that means battery dissipation is proportional to speed^5 not speed^6, so you're a whole power out.

Also, power consumption with speed only comes close to approximating the 3rd power at very high, UK illegal speeds because power consumption is a combination of factors with aerodynamics being the only one that is to the 3rd power, and only starting to dominate well past 50mph.

I direct you to the following article that goes into great detail about the power consumed by an EV at different speeds using actual numbers and graphs for the original Tesla Roadster: (Figure 7)

Can EVs handle the distances we drive? • A Study

If you look at what goes into power consumption the major contributor at speeds over 50mph is of course aerodynamic losses, which increase with the square of the speed in terms of energy consumption per mile, or the cube of the speed in terms of instantaneous power consumption.

All the other losses are lower order powers than the aerodynamic losses. The ancillary losses are pretty constant regardless of speed, tyre rolling resistance is pretty much linearly proportional to speed, and drive train losses are proportional to less than the square of the speed but a bit more than linearly proportional like the tyres.

So once you combine those together you get the dark blue curve in figure 7 whose power (order) varies significantly with speed. For example comparing 50mph to 70mph you'll see that the power consumption doubles - roughly to the 2nd power not 3rd power, (which would be a 2.744 times increase) so over the range of 50-70mph total power dissipation increase of the battery would be to the 4th power not 6th power as you're proposing.

Still a rapid increase but not nearly as bad as your post is suggesting. (Fortunately!)

If the temperature of protective case for the battery is more or less equal to the environment temperature (Te), the final temperature difference (after long long time!) between the battery (Tb) and the environment should be proportional to the battery dissipated power:

Tb – Te = K3*v^6

So, for example, if the battery temperature at 60 mph is 6oC higher than the environment, at 70 mph it will be about 15oC and at 50 mph about 2oC higher.
Ignoring for the moment that some of your numbers are wrong, you're also calculating temperature rise wrong.

Final temperature rise (after a new state of equilibrium between heat input and heat loss is attained) can't be calculated without knowing the total thermal resistance from the cells to ambient air - we don't know what that is, and I don't see you showing it in your calculations anywhere so I don't know where you're plucking specific temperature rise figures from. Without the thermal resistance you can't calculate a specific final temperature rise.

Temperature rise is also slowed by the thermal mass of the batteries themselves, as it takes a certain amount of power dissipation to raise the temperature of a certain mass at a certain rate even ignoring heat loss to ambient that would try to slow that process down.

It turns out that its actually battery thermal mass that limits temperature rise of the cells during "normal" driving with EV's that don't have active cooling, meaning that if you do keep pushing the car hard for a long time the temperature will eventually and inevitably get to unacceptably high levels - there isn't sufficient cooling (thermal resistance to ambient is too high) to sustain high output on a permanent basis, and the car relies on the fact that you have to drive a long time dissipating power in the battery to get it to unacceptable temperatures due to high battery thermal mass.
 

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When you multiply powers you add their exponents not multiply them. ;)



It turns out that it
While everything you say is right... Respect to the OP for highlighting the huge effect of speed on temperature and efficiency. And describing it well enough to allow the maths errors to be quickly identified and amended.

Perhaps explaining why the autobahn racers are posting problems while most of us are doing long journeys without issue...
 

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I like maths, but after 1 pint of Devon Red Cider, my head is a little fuzzy (I’m a lightweight) doesn’t this just mathematically just prove what I know from experience of my Leaf 40, or my old Leaf24, but not to such an extent, that the faster I drive the car, the quicker the battery warms up and the less efficient the car is at higher speeds? Or have I missed the point.
 

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When you multiply powers you add their exponents not multiply them. ;)

So if you have speed^3 as your speed vs power relationship and I^2*R as your current vs dissipation relationship in the battery, (with R and battery voltage presumed constant) that means battery dissipation is proportional to speed^5 not speed^6, so you're a whole power out.
No it doesn't.

You are mixing dimensions there. One is speed cubed and the other is current squared. You don't add apples and bananas.

The OP is working on the principle that power demand is velocity cubed. As
power == v x v x v, and
current == power, then so
current x current == (v x v x v) x (v x v x v)
 

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No it doesn't.

You are mixing dimensions there. One is speed cubed and the other is current squared. You don't add apples and bananas.

The OP is working on the principle that power demand is velocity cubed. As
power == v x v x v, and
current == power, then so
current x current == (v x v x v) x (v x v x v)
You're right, my mistake. :oops:

However the other factors I discussed I believe still apply and still mitigate the severity of the problem.

Specifically:

1) Power use doesn't approach increasing by velocity cubed until you're already well past legal speed limits in the UK. This is because only aerodynamic losses go up with velocity cubed, these are summed with other losses that dominate at speeds lower than about 50mph that increase with much lower powers - some like tyre rolling losses only go up with the first power, eg linearly, and some are constant irrespective of speed.

This means in the critical 50-70mph range where we're discussing potential driving speeds to alleviate rapidgate, the power of the curve is roughly p=v^2. At lower speeds the power curve drops considerably (compare 30-50 on the graph) while it increases asymptotically towards p=v^3 at very high speeds.

While the graph I presented is specifically for the Roadster and will differ somewhat for other EV's, I'd also point out the original Roadster is not particularly aerodynamic - I believe it's Cd is 0.35.

2) Instantaneous power dissipation in the battery is only meaningful in calculating the temperature rise if the thermal mass was low and the thermal resistance to ambient was also low - meaning that it could reach a state of thermal equilibrium of heat dissipation vs heat loss to the environment in a relatively short time - eg within a driving session.

This is something that you would expect of say a transistor on a heatsink in a power supply, where there is some thermal lag from the mass of the heatsink but it reaches equilibrium with the ambient environment in a relatively short time and thus reaches a final temperature rise over ambient of Power dissipation x Thermal resistance.

From what I've seen this is not the case on an EV with no active battery cooling such as a Leaf. (or my Ion, which has no active cooling whilst driving, only during rapid charging)

Here the thermal mass of the batteries is huge - hundreds of Kilograms, and the thermal resistance to ambient air is so high that it might as well be in a thermally insulated blanket.

In this scenario temperature rise is realistically not bounded and relies on both the limited possible driving time before the battery runs out, and the BMS in the car to cut the charging rate or power use during driving to set an upper limit on the temperature.

Temperature rise over ambient now becomes Battery power dissipation * Time, or total energy instead of power. So in this scenario the driving time is just as important as the power dissipation as driving faster means more power dissipation but for less time.

So now it's the Wh/mile vs speed graph we need to look at instead of the Power vs Speed graph, as that will tell us the total energy consumed over a fixed length journey at different speeds.

If we compare 50 and 70mph we see total power consumption at 50 is 200Wh/mile and at 70mph it's 300Wh/mile, or a 1.5x increase. However battery dissipation is roughly the square of the power output of the battery due to I^2*R, so that becomes a factor of 2.25.

So if we were to superimpose the Roadster energy consumption profile shown in the graphs onto a car with no battery thermal management like a Leaf (since we don't have similar consumption graphs for a Leaf handy) then driving at 70mph to complete a fixed length journey would cause the temperature of the cells to rise 2.25 times more than driving at 50mph.

We still don't know what the actual temperature rise will be of course, for that we need to know distance travelled, thermal mass of the batteries, total battery internal resistance and so on.

But we know how much worse it would be at higher speeds and could work that ratio out between any two different speeds simply by dividing the two Wh/mile figures from the two speeds on the graph and then squaring the result to account for I^2*R.

On a car with any sort of thermal management that can actually cool the batteries while driving all bets are off of course and the simplistic approach above would not be applicable.

Do you see any mistakes in my calculations or reasoning ?
 

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@DBMandrake I believe you are quite right in your assertion that we can discard the assumption of cubic power, as other loading will apply. The power^2 term still applies for the battery, and therefore it could actually be potentially worse than that.

The question which bothers me for some time is the following: at higher speeds the dissipated power (heat) inside the battery (Pb with units in e.g. kW), due to the internal resistance of the battery, is, according to my calculations, proportional to the sixth power of the EV speed
I understand where you are coming from and actually based on the analysis I suppose you are doing it is even worse than that.

For example, at 50A drawn I think a 60Ah LMO pack will have an internal resistance of around 0.1 Ohm while at 100A it might be 0.2A (illustrative numbers of around the right magnitude).

So at 50A x 375V = 17.5kW, say 65mph, you should end up with 50^2 x 0.1 = 250W heating, but at 100A x 350V = 33kW, say 75mph, you get 100^2 x 0.2 = 2kW.

The next question to ask is whether this is significant. For a water cooled system, clearly this is trivial. Water cooled ICE systems will happily reject 10s of kW all day long, possibly 100's if autobahn cruising is anything to go by.

The real crunch here is inactively cooled packs; I am going to assume 0.75J/g/K on average for a pack, and assume a 300kg pack. An input of 2000J per second would therefore be a heat rise of 0.009K/s. What I don't know now is the thermal conductivity of the pack and the likely heat rejection of the pack by radiation and passive air convection.

Assuming it is on average the same temperature as the environment then we can discount radiative cooling. Therefore, after 60 minutes of 70mph driving with the pack starting at 20C, this estimate would suggest a pack temperature rise of 32K to 52C.

However, once the temperature is above 40C or so I would expect radiative cooling effect provided that the ambient is low. Like I say, no passive cooling can take place until the pack is higher in temp than the surroundings. But assuming a total pack area of 5m^2 and bidirectional emissivity of 0.5 then the outward power flux will be 5.67x10^-8 x 0.5 x 5 x (310^4-290^4)* = 300W.

*(This is an approximation on bidirectional black body radiation flux, it is more complicated than that but we are interested in OOM.)

So for these higher powers I think radiative cooling helps but is not going to do much for you. This goes to 0.0075K/s battery pack rise, and if the starting point is 20C and the max is 45C this gives you an hour of driving at 33kW continuous.

There will be other cooling routes such as the ducted heat flux down the power cables and into the inverter, which will be water cooled. I suspect this would take more heat out than the radiative cooling. The convective heating to the underside of the car will be more significant again, so overall, just an educated guess now, we're looking at 1kW or so of cooling.

Let's now move it up to top speed of 90mph, I think this will be around 45kW continuous with acceleration pluses to 80kW. The average in quadrature (according to the square of the speed) I would imagine to be around 55kW. So we now have something like (55kW/330V)^2 x 0.3 = 8kW of battery heating with (being favourable for higher temps and faster air flow) 2kW of cooling, net 6kW.

Battery temp rise would therefore be in the range of (6000kW/0.75)/3600 = 0.027K/s, or ~1000s for a 25K heat rise, about 20 minutes.

6kW would be nothing for a water cooled system. A system that draws in cold air from the cabin (like Soul) will be limited by the power of the HVAC of a couple of kW, so at 75mph the Soul has half a chance of keepng its pack cool.

There is only one conclusion; high speed distance driving in an EV with no battery pack water cooling is a no-no. It's beyond the laws of physics. You have to drive slowly in EVs with no battery cooling if you want to drive for long.
 

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There is only one conclusion; high speed distance driving in an EV with no battery pack water cooling is a no-no. It's beyond the laws of physics. You have to drive slowly in EVs with no battery cooling if you want to drive for long.
This has been shown in practice by experimentation. But there might be two drive schedules available to achieve the same result.

For instance, in a Leaf 40 if you drive for 1.75 hours at the motorway limit, after a couple of stops then the next full charge could take as long as 1.25 hours. A total of 120 miles in 3 hours - average 40mph.

But if instead, you drive at 50mph for 2.25 hours then that next full charge might only take 0.75 hours. A total of 120 miles in 3 hours - average 40mph.

The difference being that the second drive regime might be able to be sustained for a longer distance.
 

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No it doesn't.

You are mixing dimensions there. One is speed cubed and the other is current squared. You don't add apples and bananas.

The OP is working on the principle that power demand is velocity cubed. As
power == v x v x v, and
current == power, then so
current x current == (v x v x v) x (v x v x v)
I saw it differently. The power demand for driving is partly proportional to speed squared. Power = current x voltage so assuming constant voltage, current is partly proportional to speed squared.

Assuming constant internal resistance of the battery, this means heating power within the battery (internal resistance x current squared).

As current is partly proportional to speed squared, this means heating power is partly proportional to speed ^ 4.

Depending whether the battery reaches equilibrium (one extreme) or has zero heat loss (other extreme), and we travel finite distances, this may be reduced to somewhere between speed cubed and speed ^ 4.

I travel long distances and seem to reach equilibrium, so at my speeds I am sticking to speed ^ 4 being closest to my case.

Also, assuming I drive three hours at 60mph per charge (about my experience) on a flat road (not my experience) then I am using average 13kW-ish. The heating effect of 50kW charging, if the internal resistance is the same for charging as discharging, would be 50^2 compared to 13^2, so 2500:169, or roughly 15 times worse per minute. Rapid charging is challenging for batteries!

That's my take on it, any advance on this?
 

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Equilibrium:- Assuming 1kW passive heat flux from the pack, from the above OOM estimates, then the battery current would be around I^2 x ~0.2 = 1000 == I = 45A, so 350V x 45A = 15kW. This is around 60 to 65mph at 3.8~4mi/kWh.
 

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You're right, my mistake. :oops:

However the other factors I discussed I believe still apply and still mitigate the severity of the problem.

Specifically:

1) Power use doesn't approach increasing by velocity cubed until you're already well past legal speed limits in the UK. This is because only aerodynamic losses go up with velocity cubed, these are summed with other losses that dominate at speeds lower than about 50mph that increase with much lower powers - some like tyre rolling losses only go up with the first power, eg linearly, and some are constant irrespective of speed.

This means in the critical 50-70mph range where we're discussing potential driving speeds to alleviate rapidgate, the power of the curve is roughly p=v^2. At lower speeds the power curve drops considerably (compare 30-50 on the graph) while it increases asymptotically towards p=v^3 at very high speeds.

While the graph I presented is specifically for the Roadster and will differ somewhat for other EV's, I'd also point out the original Roadster is not particularly aerodynamic - I believe it's Cd is 0.35.

2) Instantaneous power dissipation in the battery is only meaningful in calculating the temperature rise if the thermal mass was low and the thermal resistance to ambient was also low - meaning that it could reach a state of thermal equilibrium of heat dissipation vs heat loss to the environment in a relatively short time - eg within a driving session.

This is something that you would expect of say a transistor on a heatsink in a power supply, where there is some thermal lag from the mass of the heatsink but it reaches equilibrium with the ambient environment in a relatively short time and thus reaches a final temperature rise over ambient of Power dissipation x Thermal resistance.

From what I've seen this is not the case on an EV with no active battery cooling such as a Leaf. (or my Ion, which has no active cooling whilst driving, only during rapid charging)

Here the thermal mass of the batteries is huge - hundreds of Kilograms, and the thermal resistance to ambient air is so high that it might as well be in a thermally insulated blanket.

In this scenario temperature rise is realistically not bounded and relies on both the limited possible driving time before the battery runs out, and the BMS in the car to cut the charging rate or power use during driving to set an upper limit on the temperature.

Temperature rise over ambient now becomes Battery power dissipation * Time, or total energy instead of power. So in this scenario the driving time is just as important as the power dissipation as driving faster means more power dissipation but for less time.

So now it's the Wh/mile vs speed graph we need to look at instead of the Power vs Speed graph, as that will tell us the total energy consumed over a fixed length journey at different speeds.

If we compare 50 and 70mph we see total power consumption at 50 is 200Wh/mile and at 70mph it's 300Wh/mile, or a 1.5x increase. However battery dissipation is roughly the square of the power output of the battery due to I^2*R, so that becomes a factor of 2.25.

So if we were to superimpose the Roadster energy consumption profile shown in the graphs onto a car with no battery thermal management like a Leaf (since we don't have similar consumption graphs for a Leaf handy) then driving at 70mph to complete a fixed length journey would cause the temperature of the cells to rise 2.25 times more than driving at 50mph.

We still don't know what the actual temperature rise will be of course, for that we need to know distance travelled, thermal mass of the batteries, total battery internal resistance and so on.

But we know how much worse it would be at higher speeds and could work that ratio out between any two different speeds simply by dividing the two Wh/mile figures from the two speeds on the graph and then squaring the result to account for I^2*R.

On a car with any sort of thermal management that can actually cool the batteries while driving all bets are off of course and the simplistic approach above would not be applicable.

Do you see any mistakes in my calculations or reasoning ?
Internal resistance of the battery is not fixed, it varies with temperature. So drive at a steady 70 mph and the battery will start to warm up and the internal resistance will reduce. However it is not enough to compensate for the extra loss due to the larger load current but it does help a bit. But it is partly why folk get better than usual ranges on motorways in the hot (for the UK) summer weather. Conversely with a cold battery in a UK winter spell the losses to internal battery heating when driving on lighter loads (i.e. below 50 mph) with a high internal resistance will be in the same sort of ballpark as driving at 70 mph on a hot UK summers day with a warmed battery above ambient. The killer in terms of battery efficiency is to drive fast in winter on a battery that is initially cold and is slow to warm (assuming no thermal management) because the external ambient temperature is low.
 

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The question which bothers me for some time is the following: at higher speeds the dissipated power (heat) inside the battery (Pb with units in e.g. kW), due to the internal resistance of the battery, is, according to my calculations, proportional to the sixth power of the EV speed (v), so (constant K1 mainly depends on particular EV model):...................................

Pb = K1*v^6.....................................

Eb = K2*v^5..................................

Tb – Te = K3*v^6........................................

Note that the above derivations are approximations and I did not want to make the main idea too complicated.
:D:D:D

Holy guacamole ! I'll bet you're doing all this in your head as you drive along aren't you ? I thought I was weird calculating range, miles to go, miles in hand, point of no return etc. Feeling a lot better now.

But seriously - what would it take to have an app (or a display in the dash) that tells you what speed to drive at and when to charge and for how long in order to minimise journey times depending on all the variables mentioned above ?
Does it exist already ?
 

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But seriously - what would it take to have an app (or a display in the dash) that tells you what speed to drive at and when to charge and for how long in order to minimise journey times depending on all the variables mentioned above ?
Does it exist already ?
Yes.

Just get a piece of paper and write "65mph [max]" on it!

The quickest option when driving/charging is to plan a route and stops that minimise the number of stops. In theory if you can charge faster than you can consume electrical power then that would be quicker but in practice the act of stopping seems to soak up so much time that it is simply the number of times you stop that dominates the calculation.
 

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It's just I squared R isn't it? More current = lots more heat!
Only to a rough first approximation because R is variable.

Internal Resistance = A function of (Temperature, Current draw, State of Charge , State of Health)

To which one has to add the fixed resistance of the interconnects and connectors.

The largest variation in internal resistance is due to the temperature which gives rise to about half the range reduction experienced in winter time driving. The rest being road conditions e.g. wind and wet and use of heating.

For a fuller model one would consider the internal impedance rather than just the internal resistance
 

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Discussion Starter #19
As I already stated in my first post, the derivations are approximations, since I did not want to make the main idea (and results) too complicated.
Even if taking into account some of the imperfections, like “not exactly the sixth-order relation”, it is obvious that even a relatively small increase of speed may still result in a relatively large battery temperature change.
Out of curiosity I have played with the data for 30 kWh Leaf on GreenRace (GreenRace ) at different speeds on a highway (averaged in both directions, just in case) for different speeds and have added 1 kW accessories power.
The calculated consumption (on 100 miles) is shown below (it still seems quite large consumption in my opinion, probably more suited to winter season):
vehicle_consumption_mph.png

We can see that the function is not linear, so the motor power over speed is apparently higher than the second order, by the way. The calculated motor power (in kW) vs. speed is the following:
vehicle_power_mph.png


The relative dissipated battery heat (power), according to the dissipated power at 60 mph, is then the following:
heating_power_mph.png


As can be seen, the dissipated power at 70 mph is not 2.5 times higher than at 60 mph, but it is still close to 2-times higher.
Concerning the relative accumulated dissipated heat at some fixed travelled distance (relative to 60 mph), the results are below:
heating_energy_mph.png

Of course, as already stated in some posts above, the problems are somehow alleviated by decreased internal battery resistance at higher temperatures, but also slightly worsened by the fact that battery current is not exactly proportional to the consumed EV power due to voltage drop on battery internal resistance (each % of voltage drop will have to be replaced by % of current rise and therefore by additional 2% of battery dissipated power). I expect that dropping internal resistance prevails.
I would like to repeat again that the original post was made just to make you aware that some relations are much more non-linear than they appear at first sight. Is it the sixth or fifth power in practice it does not change the main message: don’t exaggerate with speed. Even a relatively small drop in speed might be helpful (and vice-versa, of course!).
 

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Discussion Starter #20
Since the forum is international and in most countries (including mine) are using metric system, I am attaching the graphs with km/h (some of them relative to 100 km/h). I have to do that in this separate post, since forum does not allow more than 6 attached pictures.
vehicle_consumption_kmh.png


vehicle_power_kmh.png

heating_power_kmh.png

heating_energy_kmh.png
 
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