Calculating 0-60 mph time for a vehicle

[RedRook]

Ok, you might be right about the gear box. I couldn't find the spot on the Renault site that gave the transmission details, so I looked on other sites. I went back to those sites, and I saw they also mentioned a petrol engine. They are obviously talking about a different car. When I finally got to the brochure on Renault, I found that it has a continuously variable transmission. So it doesn't have one gear. It has always changing gears, so that would certainly make sense for a constant power. What you are saying is entirely possible with a CVT. The constant power characteristic wouldn't be coming from the fact that this is an electric car though. A petrol car with a CVT can have constant power as well. The constant power comes from the transmission.

 

[jumpjack]

This equation predicts a 0 to 60 mph time of 7 seconds for the Tesla S performance and 7.2 seconds for the Performance Plus, yet the Plus has significantly more power and torque and the same weight. Something not right there. Tesla claim 5.4 seconds for the Performance and 4.2 seconds for the Plus.

Yes, this is not the only wrong formula: also the one which takes into account air drag gives impossible results (more than 60 seconds to go from 0 to 60 mph!)

There's no point in getting cross when someone points out that this is not possible - by the definition of Power from Torque and speed. You would be better to stop at this point and get your definitions right before moving on.

Ok, I add some clarification and set some basic hypotheses from which to start from.

This is the typical output of an electric car (brushless motor):

This is the actual, "physical" torque/speed curve for a dc/brushless motor:

Unfortunately, you can't exploit the whole curve from 0 rpm (stall torque) to no-load speed (= 0 Nm torque), because at stall torque you would have something like THOUSANDS of amperes in your motor, which, guess what, would melt it; so a "rated torque" is fixed by electronic (first picture).

When you read "CVT" for electric cars, they just mean "automatic gear", i.e. you have a gear stick with just three positions: Drive, Neutral, Rear ("standard" people does not need technical details about how CVT is accomplished: if by complex mechanics or just by an electric motor directly placed into the wheel (hub motor), or an electric motor directly connected to differential gear).
I know how actually "electrical CVT" work because I drove 6 different electric cars by myself, I own an electric scooter declared by manufacturer as "CVT", and I also wrote a book on this topic.

This said, I think that to solve the original question, we should really simplify the problem.
In a first instance, let's ignore rolling friction: in worst case, it can increase the 0-60 time, but we currently have too LONG time with all equations we found, so it does not help in finding right equation.
Then, let's consider torque as constant from 0 to 60 mph: again, once we'll add real torque/speed data, we'll get a LONGER time, and again we are not looking for a longer time, 60 seconds to go from 0 to 60 mph are already too many!

I think we should work&think about the actual torque applied to wheel and the actual contribution of air drag to the force expression and speed equation, because there must be something wrong here.

We could also fix a lower limit to the 0-60 time be means of energy balance: a 1500 kg car needs amount of energy to go from 0 to 60 mph which is given by:
E = 0.5 * 1500 * 27.9 * 27.8

How do we get 0-60 times for the final-energy vaue?

That is specifically what we were talking about and it does match the torque and power curves for many electric cars very well. The Fluence Z.E, for instance, has flat torque upto about 25 mph, followed by flat power up to about 70 mph.

Where did you get these data?

 

[jumpjack]

Once we'll find a "best case" result for 0-60 time, we could take into account the real torque/speed curve of an EV:

http://img534.imageshack.us/img534/146/005elecnogears.jpg

Compared to an ICEV curve:
http://img841.imageshack.us/img841/772/004icegears.jpg

Of course this will give longer times, but as I said, we are currently trying to REDUCE time resulting from our current formulas, because even the worst car requires st most 30 seconds to get from 0 to 60 mph, but we get 60+ seconds with our (wrong) formulas.

http://www.energeticambiente.it/vei...re-elettrico-per-veicoli-3.html#post119396037

 

[jumpjack]

Additionally, this site provides a tool which allows calculating needed power to keep constant speed taking into account both rolling friction and air drag:
http://ecomodder.com/forum/tool-aero-rolling-resistance.php

 

[jumpjack]

May I ask why? What is the purpose of this inquiry?

To be able to check if manufacturers data about acceleration are plausible; for example I read that Start Lab Open Street (a neighbour electric car) is capable of accelerating from 0 to 50 km/h in 3.3 seconds, which I know it's impossible, but only because I drove several vehicles like that one, and I measured that they require up to 20 secs if they carry lead battery and 6 seconds if they carry lithium battery.
I'd like to be able to figure out which real performances of a vehicle are without need to test it...

 

[jumpjack]

I agree.

Here is my simplified formula (for electric cars) obtained by finding the best fit in Excel:

$t_{60}=\frac{2.79 m}{P^{0.46}T^{0.72}}$

This is an interesting formula... but how did you get to it? "Best fitting" what? And how?

 

[jumpjack]

Found some actual charts:

 

[yuiop]

We could also fix a lower limit to the 0-60 time be means of energy balance: a 1500 kg car needs amount of energy to go from 0 to 60 mph which is given by:
E = 0.5 * 1500 * 27.9 * 27.8

How do we get 0-60 times for the final-energy value?

Basically we equate the input energy over the acceleration period to the final kinetic energy, so;

$Pt = \frac{1}{2}Mv^2$

\Rightarrow t = \frac{1}{2}\frac{Mv^2}{P}

If P is the maximum power and if this could could be delivered right from zero mph then the time given by the above formula represents the absolute theoretic minimum acceleration time, but in practice, as you know, the power has to ramp up with the revs and velocity of the vehicle and so we have to find a reduced value for the power representing an averaged figure and taking losses into account.

This is an interesting formula... but how did you get to it? "Best fitting" what? And how?

Basically by trial and error and using the solve function in the Excel software. I have gathered together some more data for electric cars and found a better formula. The new equation is:

$t = \frac{1}{2}\frac{M v^2}{P}+\frac{1}{2}\frac{M^{0.48} v^2}{P^{0.48} T^{0.48}}$

where v is the final velocity (26.8 m/s), P is the maximum power output in Watts, T is the maximum torque in Nm and M is the Kerb weight plus 75 kgs to take account of the driver's weight. The first term represents the bare minimum possible time and the second term represents additional time due to power ramp up, friction and how the the torque curve relates to the power curve. The table below shows an extended table of electric cars and their 0-60 mph times together with the predicted time (t60 column) and the percentage error of the predicted value from the published time in the yellow columns.

The average absolute deviation of predicted values from the published values is about 6% which is not bad considering the very wide range of vehicle weights, power and torque outputs in the table. Generally acceleration is considered to be mainly a function of power to weight ratio, but it can be seen that some cars have faster 0-60 times than other cars that have a higher power to weight ratio, so torque has to be taken into account.

I have found an exception to the above formula where the error margin is unacceptable. It is the million dollar RIMAC Concept One super car which has 1088 horse power (!) and 1600 Nm torque. The predicted 0-60 time is 1.5 seconds whereas the actual time is 2.6 seconds. I am assuming that the RIMAC does not have sufficient traction to transfer all of that extreme power to the tarmac efficiently despite having 4 wheel drive. Actually, this video would suggest that traction IS an issue for the RIMAC. A lot of the energy is going into smoking the tires and melting the tarmac. Still, 2.6 seconds is not to be sneezed at!

To be able to check if manufacturers data about acceleration are plausible; for example I read that Start Lab Open Street (a neighbour electric car) is capable of accelerating from 0 to 50 km/h in 3.3 seconds, which I know it's impossible, ...

Your neighbours claim may be plausible. he is saying it take 3.3 seconds to go from 0 to 31 mph. Since the kinetic energy is proportional to v^2 the time to get to 31 mph (50 kph) is much less than half the time it takes to go from 0 to 62 mph (100 kph). I would say that 0-31 mph in 3.3 seconds would be consistent with 0 to 62 mph in about 11 seconds which is mid range in the above table.

 

[jumpjack]

Your neighbours claim may be plausible.

No it's not: I forgot mentioning it has a 4 (yes, four) kW electric motor! :-)


Thanks for the new equation and table.

About the RIMAC Concept One... it's not easy to keep wheels in contact with the road when they try to give 1.9 g acceleration to the car! (27.8 m/s / 1.5 = 18.53 m/s2)
Even 2.6 secs mean 1g!
I think it's not very comfortable for passengers... ;-)

 

[jumpjack]

The new equation is:

$t = \frac{1}{2}\frac{M v^2}{P}+\frac{1}{2}\frac{M^{0.48} v^2}{P^{0.48} T^{0.48}}$

[...]
The average absolute deviation of predicted values from the published values is about 6%

Maybe we could try taking into account the non-constant curve to get better results:
http://www.renault.com/fr/innovation/gamme-mecanique/images_without_moderation/courbe-zoe.jpg
We could suppose that given torque is available only for first 30% of speeds w.r.t maximum declared speed, than conisder linear (or quadratic) decrease of torque, or consider instead power values: 0 to Pmax in first 30%, constant P=Pmax for remaining 70%.

Specifically for this picture:
First 30%: T=220 Nm, P=6.3x
Last 70%: T=0.022 x^2- 5.2x + 357, P = 63000W

But torque at maximum speed is not usually an available data, so quadratic approximation of torque is not usually available and we must rely only on power curve.

 

[RedRook]

Another way of finding torque at the wheel at top speed is looking at power of the forces resisting acceleration. I have to get to work, so I'll just say that the equations for A and C that you posted in the other thread are exactly what I think is correct. You can replace C_RR with a more general coefficient that represents the overall friction rather than just rolling friction, and the power loss from this friction should be directly related to speed. You normally have the top speed posted. The drag coefficient is normally posted, and if you can't find the frontal area affected, one approximation equation I see floating around using a modern car's height and width from the front is:

Area = ((H * W)/4)^2 * \pi

Top Speed = \sqrt[3]{P_M / 2A+\sqrt[2]{(P_M / 2A)^2+(C/3A)^3}}+\sqrt[3]{P_M / 2A-\sqrt[2]{(P_M / 2A)^2+(C/3A)^3}}

Those two equations combined should let you come up with your more generalized coefficient of friction. Then the power of the drag + the power of the friction should roughly equal the power at the engine. The power of drag alone should roughly equal the power being put out by the wheels, and the torques at the wheels can be determined by figuring out RPM's that will be determined by tire diameter that could lead to your force output at the wheels. Also keep in mind that a tire measured in milimeters and rim size measured in inches has a diameter equation of:

(((Width * (Aspect Ratio / 100) * 2)/25.4) + Rim size

The 25.4 comes from the millimeters in an inch. You'll likely have to play with this number depending on what units your tire width and rims diameter are measured. Those are simply the units I'm used to. Basically this is all based on the idea that acceleration stops when the power of resisting forces equals the power of accelerating forces. I have to get to work have fun.

 

[yuiop]

Maybe we could try taking into account the non-constant curve to get better results:
http://www.renault.com/fr/innovation/gamme-mecanique/images_without_moderation/courbe-zoe.jpg

Actually, I found the power curve diagrams useful to make an improved predicted 0-60 time. Here is how it works. Looking at the curve for the Zoe it can be seen that the power is a linear function of the speed up to a critical speed ($v_c$) which in this case is about 30 km/h or 9.16 m/s. It is fairly easy to show that when the power is directly proportional to the speed, that the acceleration is constant and all the standard equations of motion apply, within the speed range $0<=v<=v_c$. Using the first equation of motion:

$V=at$
$\rightarrow t=V/a$
$\rightarrow t=M V/Force$

$\rightarrow t_c=\frac{M V_c^2}{P_{max}}$

Note that this is twice the time predicted by the usual formula $t=\frac{1}{2}\frac{M V^2}{P_{max}}$ because the average power over this phase is half the maximum power. For the latter part of the curve from the critical speed $V_c$ to the final speed $V_f$ , where the power is constant ($P_{max}$) and the time is given by the usual relationship:

$t_{cf} = \frac{1}{2}\frac{M V_f^2}{P_{max}} - \frac{1}{2}\frac{M V_c^2}{P_{max}}$

Combining the two times above to gives the total theoretical acceleration time:

$t_{60} = \frac{1}{2}\frac{M (V_f^2+V_c^2)}{P_{max}}$

For the Renault Zoe this works out as 9.8 seconds.

This time makes no allowance for drive trains losses, drag, dynamic losses etc. Using a simple assumption of 15% power loss to encompass all these losses the equation can be written as:

$t_{60} = \frac{1}{2}\frac{M (V_{f}^2+V_{c}^2)}{0.85 P_{max}}$,

where M is the kerb weight plus driver weight in kgs, P is the maximum power in Watts and V is the velocity in m/s. This gives an estimated 0-60 time of 11.5 seconds which is pretty close. Applying the same formula to the 5 electric cars that we have power torque curve data for, gives an average absolute deviation of 5.6% from the published values for this admittedly small data sample.

The data set including $v_c$ is shown in the table below together with the predicted values and percentage errors.

The above results suggest the Tesla cars have less losses due to friction, etc than the Renault electric cars. For completeness the diagram below shows power and velocity as a function of time for the Renault Zoe:

It can be seen with power plotted as a function of time, that the average power output over the acceleration period is somewhere between 85 and 90% of the maximum power for a typical electric car. The velocity curve is linear to Vcrit as per the magenta line and then follows the blue curve after that.

if you have some more power/torque curves for electric cars, it would be interesting to see how they compare.

 

[jumpjack]

From chart I calculated that for Zoe we have, for $v<v_c$, P = k * v (k = 6900 = 63000 W / 9.16 m/s).
But, given that it must also be P = F * v , and given F = 220 Nm... shouldn't I get k=220 Nm ?!?

 

[yuiop]

From chart I calculated that for Zoe we have, for $v<v_c$, P = k * v (k = 6900 = 63000 W / 9.16 m/s).
But, given that it must also be P = F * v , and given F = 220 Nm... shouldn't I get k=220 Nm ?!?

220 Nm is the torque (T) and and not the force (F) in Newtons.

The force in the example you give is F = P/V = 63000 W / 9.16 m/s = 6877 N so you are right that F = k = Constant for $V<=V_c$.

The Force can also be found from F=Energy/distance = E/s.

First we need to know the time to accelerate to Vc which is given by t_c = M V_c²/P_max = 1.955 seconds. (The time to Vc in the diagram in the previous post is slightly longer because I was using a value of 85% power to take drive train losses into account.)

The distance traveled in that time is s = V_c*t_c/2 = 9.16*1.955/2 = 8.954 metres.

So F = E/s = (1/2)M V²/s = 0.5*1468*9.16²/8.954 ≈ 6877 N.

The acceleration in this part of the curve is constant and is given by a = F/M = 6877 N / 1468 kgs = 4.685 m/s² and ##t_{c}# is given by t_c = V_c/a which is 9.16 m/s / 4.685 m/s² = 1.955 seconds which confirms the earlier result.

 

[jumpjack]

220 Nm is the torque (T) and T = F * r where r is a 'leverage' ratio.

The force in the example you give is F = P/v = 63000 W / 9.16 m/s = 6877 N = k = Constant for $V<=V_c$

P.S. I am still editing this post. Got a bit confused somewhere :P

yes, me too. :-)
Actually I forgot wheel radius, so if T=220 Nm we should consider F = 220/0.3 = 733 Nm, but we're still far from 6800! And by a factor ~10, which is ~same factor between actual 0-60 times (>6 secs) and results I obtain (>60 secs).

I've been suggested the maybe torque is given at motor rather than at wheel, but I don't think the gear ratio is 1:10!

 

[gmar]

About the RIMAC Concept One... it's not easy to keep wheels in contact with the road when they try to give 1.9 g acceleration to the car! (27.8 m/s / 1.5 = 18.53 m/s2)
Even 2.6 secs mean 1g!
I think it's not very comfortable for passengers... ;-)

You know that feeling is probably the whole point of buying a stupidly fast car in the first place. Why would you NOT wish to experience it? :-)

 

[yuiop]

yes, me too. :-)
Actually I forgot wheel radius, so if T=220 Nm we should consider F = 220/0.3 = 733 Nm, but we're still far from 6800! And by a factor ~10, which is ~same factor between actual 0-60 times (>6 secs) and results I obtain (>60 secs).

I've been suggested the maybe torque is given at motor rather than at wheel, but I don't think the gear ratio is 1:10!

Actually, the overall gear ratio might be something like 10:1. In this chart for the Tesla Roadster that you posted in #57, the overall gear ration is quoted as being 8.28:1 and the revs at 100 mph are about 12000! (Much higher than a geared petrol engine).

From the specifications in the Tesla website, the single speed Tesla Roadster Sport (red curve in the above chart) reaches maximum power at 4400 rpm which equates to a critical velocity of 39 mph or 17.4 m/s on the above chart. This is slightly earlier than the speed that the torque drops off from maximum in the chart.

With Vc = 17.4 m/s, a kerb weight of 1335 kg and maximum power of 235 kW, the 0-60 time predicted by my equation (assuming 15% drive train losses and a 75 kg driver) is 3.8 seconds while Tesla claim 3.7 seconds. Pretty good eh!

The outside diameter of a 225/45 R17 tyre as used on the Roadster is 634 mm giving a radius of 0.317 m. The rpm of the wheel at Vc is 17.4/(2*Pi*0.317)*60 ≈ 524 rpm. The engine speed at Vc is about ≈ 4400 rpm (from the chart) giving a overall gear ratio of 4400/524 ≈ 8.4:1 which is close to the 8.28:1 ratio quoted by Tesla.

 

[jumpjack]

So I must assume that, if not stated differently, a vechicle torque is given at motor and that it's final gear ratio is ~8?
I wonder if there is a database of final ratios available around...

I found this page for Nissan Leaf:
http://livingleaf.info/2010/11/nissan-leaf-electric-motor-and-transmission/

Summary:
Final Drive Ratio:7.9377
Max motor speed: 10390 RPM
Max car speed: 94 mph (151 kmh)
Tyres diamater: 24,9" (31 cm radius)
60 mph = 837 RPM
Curb Weight= 3355 lbs (1535 kg)

From here:
http://techcrunch.com/2010/11/02/nissan-releases-final-specs-on-the-leaf/
Max Power = 80 kW
Max torque = 280 Nm

Curves:
http://img834.imageshack.us/img834/8247/leaftorquehp.jpg

This variant has 107 kW and 187 Nm and goes from 0 to 60 in 0.7, has Cd=0.28 and mass 3291 pounds:
http://www.topspeed.com/cars/nissan/2014-nissan-leaf-ar161509.html

--------

Renault Zoe is sold with tyres having radius 305-310mm ( http://myrenaultzoe.com/index.php/zoe-description/wheels-tyres-and-range/)
65 kW @ 3000-11300 rpm 8http://webcache.googleusercontent.com/search?q=cache:uSE2wA2QwfwJ:www.carfolio.com/specifications/models/car/?car=342397+&cd=2&hl=en&ct=clnk&gl=it)
220 Nm @ 250-2500 rpm

---------
i-Miev has 47 kW between 3000 and 6000 rpm and 180 Nm from 0 to 2000 RPM ( https://www.sia.org.au/downloads/Divisional/ACT/i-MiEV_presentation.pdf)
Tyres are:
Front: 145/65R15 (28cm radius)
Rear: 175/55R15 (29cm radius) (drive)
Mass is 1100 kg
Final gear ratio 7.065
0-100 in 15.9 secs
Max speed: 130kmph
-------
Ford Focus Electric
0-62mph 11.7secs;
Top speed 85mph;
Kerbweight 1700kg;
Power 143bhp (107 kW);
Torque 184lb ft (250 Nm);
Gearbox Reduction box with ratio of 10:1

 

[jumpjack]

For completeness the diagram below shows power and velocity as a function of time for the Renault Zoe:

How did you get the equation to plot the v(t) graph?

 

[jumpjack]

New data:

Honda Fit EV (http://automobiles.honda.com/fit-ev/specifications.aspx)
Power:
92 kW @ 3695~10320 (sport mode)
75 kW @ 3695~10320 (normal)
47 kW @ 3695~10320 (economy)

Torque:
189 lbft (256 Nm) @ 0 - 3056


Gear Ratios:
1st: 2.185
Reverse: 2.185
Final Drive (axle): 3.688
Overall: 8.05828

Tyres:
P185 / 65 R15 86T (31 cm radius)

Curb weight: 3252 lbs (1473 kg)

0-60 time: 8.7 s

Max speed (limited): 91 mph



$$v_{vehicle} = \omega_{wheel} r = 2 \pi f_{wheel} r = 2 \pi \frac {RPM_{wheel}}{60} r = 2 \pi \frac {\frac {RPM_{motor}} {G_{overall}}}{60} r = 2 \frac {\pi}{60} \frac {RPM_{motor}} {G_{overall}} r = 0.105 \frac {RPM_{motor}} {G_{overall}} r $$

"Typical" value:
$$ 0.105 \frac {RPM_{motor}} 8 0.31 = 0.004 RPM_{motor} = \frac 4 {1000} RPM_{motor}$$

For Honda Fit EV:
v = $\frac 4 {1000} 3056$ ~= 12 m/s = 43 km/h = 26.8 mph

 

[jumpjack]

$v=\sqrt{\frac{m(\frac {F_t} m -gC_r)}{c}}tanh(\sqrt{\frac{(\frac {F_t} m -gC_r)c}{m}}t)$

Can also be written as:

$v(t) = \sqrt \frac {K_1}{ K_2} tanh (\sqrt {K_1 K_2} t)$


$K_1 = \frac {F_t} m - gC_r$
$K_2 = \frac 1 2 \rho C_d S * \frac 1 m$

$K_1 = \frac Q m$

$K_2 = \frac c m$

$Q= mK_1$
$c = m K_2 = \frac 1 2 \rho C_d S$

$Qc = m^2 K_1 K_2$
$\frac Q c = \frac {K_1}{K_2}$

And, above all:
$$F_t = \frac {T_w} r $$
$T_w$ = Wheel torque <> EngineTorque
r = wheel radius - Typical value = 0.31 m

but we must take into account Overall Gear Ratio:

$T_w = T_e G$
G = overall gear ratio - Typical value for EVs = 8

hence:

$$F_t = \frac {T_e G} r$$

For typical values:
$$F_t = 26 T_e $$

$$v=\sqrt{\frac{m(\frac {\frac {T_e G} r} m -gC_r)}{c}}tanh(\sqrt{\frac{(\frac {\frac {T_e G} r} m -gC_r)c}{m}}t)$$

For typical values:

$$v=\sqrt{m\frac{ (\frac {26 T_e} m - 9.81 C_r) }{\frac 1 2 \rho C_d S}}tanh(\sqrt{ \frac{(\frac {26 T_e} m - 9.81 C_r)(\frac 1 2 \rho C_d S)}{m}}t)$$



But this behaviour is valid only as long as $v<v_c$, with:
$v_c= \frac 4 {1000} RPM_{MaxTorque}$

After $v_c$, $T_e$ is no longer constant but it decreases as $\frac K v$... and I have yet to determine K value.

 

[jumpjack]

So, considering these new expression for K1:

$K_1 = \frac {F_t} m -gC_r = \frac {\frac {T_e G} r} m -gC_r$

and same of previous for K2:

$K_2 = \frac 1 2 \rho C_d S \frac 1 m$Terminal Velocity (theoretically maximum achievable speed) is:
$v_t = \sqrt{\frac {K_1}{K_2}} = \sqrt{\frac { \frac {\frac {T_e G} r} m -gC_r}{\frac 1 2 \rho C_d S \frac 1 m}} =\sqrt { 2\frac { T_e G - 9.81 \times rmC_{rr}}{r \rho C_d S}}$
(but I get 122 m/s for Zoe, maybe because this formula assumes constant torque "for ever" rather than just up to $v_c$ ; v=122 m/s means 440 km/h, but to achieve such a speed by applying constant force, a final power of 770 kW would be needed!)

And time-to-60mph is:

$$t_{60} = \frac 1 {\sqrt{K_1 K_2}} atanh(\sqrt{\frac {K_2} {K_1} } v_{60}) $$

Which expanded is:
$$t_{60} = \frac 1 {\sqrt{( \frac {\frac {T_e G} r} m -gC_r) ( \frac 1 2 \rho C_d S \frac 1 m)}} atanh(\sqrt{\frac {( \frac 1 2 \rho C_d S \frac 1 m)} {( \frac {\frac {T_e G} r} m -gC_r)} } v_{60}) $$

 

[jumpjack]

And I eventually found K in T =$\frac K v$ !
$T=T_m$ ($0<v<v_c$)
$T(v) = \frac {P_m r} G \frac 1 v$ ($v>v_c$)

$P(v)=\frac {P_m} {v_c} v$ ($0<v<v_c$)
$P = P_m$ ($v>v_c$)

I also found that I can calculate overall gear ratio in this way:

$$G = \frac {P_m r}{T_m v_c}$$

As seen above, it allows calculating torque at wheels rather than at motor.

Formulas in previous posts have been edited, revised and fixed multiple times, but they should ok now!

 

[jumpjack]

It is worth noting that as long as torque is constant, the expected terminal velocity is $\sqrt(\frac{K_1}{K_2})$, but once torque becomes decreasing, expectations are very reduced! terminal velocity becomes $$\sqrt[3]\frac{\frac {P_m} m}{\frac 1 2 \rho C_d A \frac 1 m} = \sqrt[3] \frac {2 \times P_m}{\rho C_d A}$$
which is way lower than previouse one!
(53 m/s vs 122 m/s for Renault Zoe)

To calculate terminal velocity you've just to equal acceleration to 0 and taking into account two different cases for constant and variable torque:

$a(v) = a_0 - \beta v^2 =0$
==> $v_f = \sqrt{\frac{a_0}{\beta}}$

$a(v) = \frac {w_0}{v} - \beta v^2 =0$
==> $v_f = \sqrt[3]{\frac{w_0}{\beta}}$

$a_0 = \frac {GT_e} {rm}$
$\beta = \frac 1 2 \rho C_d S \frac 1 m$
$w_0=\frac {P_m} m$

 

[jumpjack]

If I consider $$v=\sqrt{\frac F c}tanh(t\frac{\sqrt{cF}}m)$$ I can visually see that $$t_{62}=6.3 secs$$ , but if I use inverse formula $$t=\sqrt{\frac{m^2}{Fc}}atanh(v\sqrt{\frac c F})$$ I get 13.4 secs, almost double the time, why?

 

[jumpjack]

I considered $$c=\frac 1 2 \rho C_d A$$ and $$F =\frac{T_e G} r$$ , being G=overall gear ratio ~=8

 

[jumpjack]

If I consider $$v=\sqrt{\frac F c}tanh(t\frac{\sqrt{cF}}m)$$ I can visually see that $$t_{62}=6.3 secs$$ , but if I use inverse formula $$t=\sqrt{\frac{m^2}{Fc}}atanh(v\sqrt{\frac c F})$$ I get 13.4 secs, almost double the time, why?

Solved:
I was plotting using WinPlot... which unfortunately has no definition for atanh(), hence it plotted a*tanh() !
Above formulas are correct, and the final formulas for v(t) and t(v) are:


$$v(t) = \sqrt{\frac {T_eG}{rc}} \tanh \left( \frac t {cT_eG}{rm} \right)$$


$$t(v) = \sqrt \frac {m^2r} {cT_eG} atanh \left( v \sqrt \frac {cr}{T_eG} \right) $$

$T_e$ = Engine torque [Nm]
G = total gear ratio (~8 for EVs) (=$\frac {P_mr}{T_m v_c}$)
r = wheel radius [m]
c = air drag constant (=$\frac 1 2 \rho C_d A$)
m = vehicle mass [kg]
$P_m$ = Maximum power
$T_m$ = Maximum engine torque
$v_c$ = critical speed (Torque is constant and maximum for $v<v_c$, Power is constant and maximum for $v>v_c$)

P(v) and T(v) behaviours:
$T_e(v)=T_m$ ($0<v<v_c$)
$T_e(v) = \frac {P_m r} G \frac 1 v$ ($v>v_c$)

$P(v)=\frac {P_m} {v_c} v$ ($0<v<v_c$)
$P(v) = P_m$ ($v>v_c$)

Unfortunately, v(t) and t(v) formulas above are valid only for $v<=v_c$.