Calculating 0-60 mph time for a vehicle

[jumpjack]

After figuring out how to calculate Cd and Crr coefficients, I'd like to continue my "vehicle physics exploration": I'd like to determine how much it takes to go from 0 to 60 mph depending on engine power. Or does it depends on torque? By sure it also depends on air drag, which depends on speed, which makes things very complex.

 

[TumblingDice]

After figuring out how to calculate Cd and Crr coefficients, I'd like to continue my "vehicle physics exploration": I'd like to determine how much it takes to go from 0 to 60 mph depending on engine power. Or does it depends on torque? By sure it also depends on air drag, which depends on speed, which makes things very complex.

This sounds like a cool way to add some fun to physics. The 0-60 vehicle calculation could become a never-ending quest without some limitations on what you expect to calculate. Setting a reasonable target keeps it fun, and you can add new pieces in the future from your list.

At the track there are a bunch of factors to complicate further. Transmisson gearing and ratios and clutch performance, for example, if you want to gear for best torque. Pavement, temperature, tire compound, size, pressure, and more decide how much of your power is lost to friction/heat/slippage.

Maybe you'd consider doing this for an electric vehicle like Tesla, which would remove multiple gear/transmission/shifting from the equation?

 

[Borek]

Power alone gives you estimate of the shortest required time. Real time will be always longer.

 

[jumpjack]

@Tumblingdice,
We can take into account just airdrag in a first instance.

@borek , your assumption is already included in my question...

 

[Borek]

It is not an assumption, it is the energy conservation.

 

[jumpjack]

I don't need results which do not take into account airdrag.

 

[Borek]

I think you missed the point of my post. You stated

I'd like to determine how much it takes to go from 0 to 60 mph depending on engine power.

and all I said is that it is impossible - you need to know more than just power. Power alone can give an estimate of the lower boundary, but the real time will be always longer.

 

[jumpjack]

Ok, it "looks like" $v(t)=v_f * tanh(\sqrt {cF}* \frac t m)$ ,where $v_f = 27.777 m/s$ (100 kph) and $c= \frac 1 2 \rho C_d A$.
Now how do I relate vehicle power to F? Maybe I should instead use vehicle torque... but wouldn't F then depend on wheel radius, which I do not know?

 

[jumpjack]

Oh, and let's suppose both power and torque constant, I'm interested on electric vehicles mostly.

 

[jumpjack]

No ideas?

 

[AlephZero]

Oh, and let's suppose both power and torque constant.

You can't make both assumptions at once, unless the speed is also constant.

Given your apparent lack of knowledge about basic mechanics, it's hard to give any constructive advice, except "learn some basic mechanics first".

 

[jumpjack]

You can't make both assumptions at once, unless the speed is also constant.

Given your apparent lack of knowledge about basic mechanics, it's hard to give any constructive advice, except "learn some basic mechanics first".

Just help if you can, don't argument.
Thanks.

I reword my question:
I have a vehicle with given power and torque, say 100 kW / 100 Nm motor. Known it's Cd,frontal area , Crr (rolling friction coefficient) and mass, which is the minimum time it can achieve to go from 0 to 60 mph?

 

[BiGyElLoWhAt]

well this is what I would do:
∑F=ma;
So
$F_{propulsion}-F_{drag}-F_{rolling friction}-F_{whatever else} = m_{car}a_{car}$

$F_{propulsion}$ is due to the tires spinning, which stems from the torque, right? we don't really care about the "direction" of the torque (it's a pseudo-vector), so we don't necessarily need R x F, we can just use RFsin(Θ), but Θ=pi/2, so sinΘ=1,
From using T = RF, we get $F_{propulsion}= \frac{T}{R}$.
so...
$\frac{T}{R} - F_{drag} -...=m_{car}a_{car}.$

$\frac{∑F}{m_{car}}=a_{car}$
integral(from 0 to t) a dt = $a_{SomeConstantFromNonDragTerms}t - inegral(from 0 to t)\frac{1}{2}ρC_{d}v^2A_{crossection}$= velocity change from time 0 to t. The tricky part is that v is a funtction of t, so you can't just treat it like a constant, I'm not entirely sure how to write v correctly to give you a good answer... soooo... let's act like physicists and forget drag? lol.
Actually I am curious to see what people have to say about this, any ideas to help finish this off?

 

[jumpjack]

Did you mean this?
$\int_0^t a dt = a_{SomeConstantFromNonDragTerms}t - \int_0^t\frac{1}{2}ρC_{d}v^2A_{crossection}$= velocity change from time 0 to t.

 

[jumpjack]

I don't know if it can help, anyway in this thread we determined the solution of equation motion when no traction force is acting on the vehicle (coasting down).

$ma = -mgC_r - \frac 1 2 \rho C_d S v^2$
$m \dot v = -R - D$

$R -mgC_r$
$D = \frac 1 2 \rho C_d S v^2$

If we put $D = d v^2$, we have $m \dot v = -R - dv^2$ and so we have $$ \dot v = -{R \over m} - {d \over m}v^2 $$ So let $$ A = - \frac R m $$ and $$ B = \sqrt {\frac d R } $$ then $$ \dot v = A(1 + B^2 v^2) $$

This post explains how to solve it to:

$$ v = \frac 1 B tan(A B T + \arctan (B v_i))$$

Adding traction force:

$ma = -mgC_r - \frac 1 2 \rho C_d S v^2 +F_t$
$m \dot v = -R - D +F_t$
$\dot v = -{R \over m} - {d \over m}v^2 + {F_t \over m}$

New differential equation to solve:

$\dot v = A(1 + B^2 v^2) + C$

 

[jumpjack]

$\dot v = -{R \over m} - {d \over m}v^2 + {F_t \over m}$

I could put the equation into form $A(1+B^2v^2)$ to be able to solve it like previous one:



$\dot v = (-{R \over m} + {F_t \over m}) - {d \over m}v^2$

"Extracting" $-{R \over m} + {F_t \over m}$ :

$$\dot v = (-{R \over m} + {F_t \over m}) (\frac {-{R \over m} + {F_t \over m}} {-{R \over m} + {F_t \over m} } - \frac {d \over m} {-{R \over m} + {F_t \over m}} v^2) $$


$$\dot v = (-{R \over m} + {F_t \over m}) (1 - \frac {d \over m} {-{R \over m} + {F_t \over m}} v^2) $$

$$\dot v = (-{R \over m} + {F_t \over m}) (1 + \frac {d \over m} {{R \over m} - {F_t \over m}} v^2) $$

$$\dot v = (-{R \over m} + {F_t \over m}) (1 + \frac {d } {{R } - {F_t }} v^2) $$

So if we put:

$A = -{R \over m} + {F_t \over m}$
$$B^2 = \frac {d } {{R } - {F_t }}$$
$$B= \sqrt {\frac {d } {{R } - {F_t }}}$$


we get our familiar equation:

$\dot v = A(1 + B^2 v^2)$

which as said solves to:

$$ v = \frac 1 B tan(ABT+arctan(Bv_i))$$

but with different A and B:
$A = -{R \over m} + {F_t \over m}$ ..... vs ...$A = -{R \over m}$


$B= \sqrt {\frac {d } {{R } - {F_t }}}$ ..... vs ... $\sqrt {\frac d R}$

But this must be wrong, because speed should be increasing in case of applied traction force!

So where is the error?!?

 

[craigi]

But this must be wrong, because speed should be increasing in case of applied traction force!

So where is the error?!?

Not necessairily. You can apply traction and still lose speed, when going up a steep slippy hill for instance.

I didn't read the full thread, but you might want to check that you picked the right B when you squarerooted it.

 

[jumpjack]

I also found the solution for free fall motion eqauation:
$F = ma = -cv^2 + mg$

where $c = \frac 1 2 \rho C_d S$

solves to:

(1) $v=\sqrt{\frac{mg}{c}}tanh(\sqrt{\frac{gc}{m}}t)$

In place of mg I have traction force and rolling friction, which sum to $F_t-mgC_r$:
$F = ma = -cv^2 + F_t-mgC_r$

Grouping weirdly :-)

$F = ma = -cv^2 + m(\frac {F_t} m -gC_r)$

If this is correct, than I have just to replace $g$ by $(\frac {F_t} m -gC_r)$ in (1) to get my solution:

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

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

$v=\sqrt{\frac{(F_t -mgC_r)}{c}}tanh(\sqrt{(\frac {F_t} {m^2} - \frac {gC_r} m ) c}t)$

$v=\sqrt{\frac{(F_t -mgC_r)}{c}}tanh(\sqrt{(\frac {F_t} {m^2} - \frac {mgC_r} {m^2} ) c }t)$

$v=\sqrt{\frac{(F_t -mgC_r)}{c}}tanh(\frac 1 m \sqrt{(F_t -mgC_r ) c }t)$

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

$v=\sqrt{\frac{(F_t -mgC_r)}{\frac 1 2 \rho C_d S}}tanh(\frac t m \sqrt{(F_t -mgC_r ) \frac 1 2 \rho C_d S })$

I think for $F_t$ I should use $\frac T r$, with T = engine torque and r = wheel radius.

$$v=\sqrt{\frac{(\frac T r -mgC_r)}{\frac 1 2 \rho C_d S}}tanh(\frac t m \sqrt{(\frac T r -mgC_r ) \frac 1 2 \rho C_d S })$$



Now it's "just" a matter of extracting t from this mess to be able to calculate 0-60 time given Cd, Cr, S and T!

(but of course only if my math is right!)

 

[jumpjack]

I simplify again to try to extract t:

$$v=\sqrt{\frac{Q}{c}}tanh(\frac t m \sqrt{Q c})$$
$Q=\frac T r -mgC_r$
$c=\frac 1 2 \rho C_d S$

In linear form:

v = sqrt(Q/c) * tanh( (t/m) sqrt(Qc))

y = sqrt(Q/c) * tanh( (x/m) sqrt(Qc))

This should solve to:

$$t = \frac {m * atanh \frac {v} {\sqrt{\frac Q c}}} {\sqrt {cQ}}$$

Now back to impossible form :-)

$$t = \frac {m * atanh \frac {v} {\sqrt{\frac {\frac T r -mgC_r} {\frac 1 2 \rho C_d S}}}} {\sqrt {(\frac 1 2 \rho C_d S) (\frac T r -mgC_r)}}$$

$$t = \frac {m * atanh \frac {v * \sqrt {\frac 1 2 \rho C_d S}} {\sqrt{\frac T r -mgC_r }}} {\sqrt {(\frac 1 2 \rho C_d S) (\frac T r -mgC_r)}}$$

$$t = \frac {m * atanh (v * \sqrt \frac {\frac 1 2 \rho C_d S} {\frac T r -mgC_r } )} {\sqrt {(\frac 1 2 \rho C_d S) (\frac T r -mgC_r)}}$$

 

[jumpjack]

In planar form:

t=M*atanh( v*sqrt((0.5*1.22*X*S)/((T/(d/2))-MGR)) ) / sqrt((0.5*1.22*X*S)*((T/(d/2))-MGR))
M=mass (kg)
v = speed (m/s)
X = Cd = Cx = Air Drag coefficient
S = Frontal Area (m2)
T = Torque (Nm)
d = wheel diameter (m)
G = 9.81 m/s2
R = Cr = Rolling Friction coefficient

 

[jumpjack]

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

I found confirmation to my formula in this book;
§2.6.1.1, formula 2.11:
$v(t) = \sqrt \frac {K_1}{ K_2} tanh (\sqrt {K_1 K_2} t)$

It matches with mine considering that book defines (p.30):
$K_1 = \frac F m - gC_r$
and
$K_2 = \frac 1 2 \rho C_d A * \frac 1 m$

So in my equation I have:
$K_1 = \frac Q m, Q= mK_1$
$K_2 = \frac c m$ , $c = m K_2$
$Qc = m^2 K_1 K_2$
$\frac Q c = \frac {K_1}{K_2}$

My formula for 0-60 time:

$$t_{60} = \frac {m * atanh \frac {27.7} {\sqrt{\frac Qc}}} {\sqrt {Qc}} = \frac {m * atanh \frac {27.7} {\sqrt{\frac {K_1}{K_2}}}} {m\sqrt {K_1K_2}} = \frac {m * atanh (\sqrt{\frac {K_2}{K_1}}27.7 ) } {m\sqrt {K_1K_2}} = \frac {atanh (\sqrt{\frac {K_2}{K_1}}27.7 ) } {\sqrt {K_1K_2}} $$

Book's formula:

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

But both using my formula and looking at book's example I get impossible times to get to 60 mph: 60-90 seconds rather than 10-20.

What is it going wrong in this math?

 

[RedRook]

What values are you using with units. It is hard to check where you might be going wrong, if I don't know the details of the object you are trying to make calculations for.

 

[RedRook]

I think your biggest mistake is probably where you started using T / r with T = engine torque, and r = wheel radius. The wheel radius alone is not enough, because the torque on the wheel is dependent both on the gear ratio and the velocity of the car. Unless you are assuming that power is on a never ending climb to the heavens, torque needs to be an equation in relationship to velocity. You can't keep torque as a constant value unless you increase power.

Power = Work / Time

The torque needed to produce work over time at a specific RPM changes as the RPMs change:

Power = Torque * RPM / (Unit conversion constant)

You can't keep torque as a constant unless you are assuming your engine is constantly increasing power no matter what speed you are at. If your engine is constantly changing power then it has a limit. Gasoline engines can get better as their crank speed increases, but even gasoline engines eventually hit a power peak, and then you are going to have to decrease torque at the wheel by shifting gears or watch the torque drop in the same gear as has to happen if power doesn't increase. You might get away with using an average torque, but then it is going to have to be the average torque at the wheel, not the engine. Torque at the engine needs gear ratios to determine how much force is applied in relationship to the wheel radius. Torque at the wheel is directly related to the wheel radius, but it is based on both gear ratio and power loss moving power from the engine to the wheel. Keep in mind to figure out average torque at the wheels you are going to need to know the power being sent to the wheels in relationship to time as well as the vehicle velocity in relationship to time. You still end up needing an equation for torque.

What might be easier is going back and determining how exactly you want to define force at specific times. That F in itself needs to become F(t), and my personal opinion is that power is easier to work with in that equation, because it doesn't rely on the gears as much. A power in first gear is going to apply very similar power compared to second gear at the wheel. A gear ratio 2:1 vs. 1:1 applies different torques at the wheel by a factor of 2. Torque is more important if you are trying to figure out whether or not the car has enough torque to overcome the high friction of the vehicle at rest, but working it back into power is easier when dealing with the transfer of force from the engine to the wheel.

 

[jbriggs444]

Power = Torque / Time

Power = Torque / Time [taken to rotate through one radian]

This assumes you are using a system of units where torque and energy share the same units, e.g. both in foot-pounds or both in kg meter/sec²

I expect that RedRook is perfectly well aware of the above.

 

[jbriggs444]

Well, I'm starting with the assumption that Torque is being used as a measurement of work, which makes since because it is force applied over a distance.

Torque is computed as a vector cross product. Force times perpendicular distance.
Work is computed as a vector dot product. Force times parallel distance.

They share the same units (more or less) but they measure different quantities. They are connected by the fact that if you exert a given torque through a rotation angle of one radian, the torque and the resulting work are numerically equal.

 

[jumpjack]

Just consider electric vehicles: constant torque, no gears, motor efficiency =~100% .

In a second instance we could investigate Power/Torque correlation in electric motors.

 

[jumpjack]

Some literature data (for internal combustion engine cars):
Fiat Stilo: 1488 kg, 255 Nm, 11.2 s
BMW M3: 1885 kg, 400 Nm, 5.3 s
Citroen C3: 1126 kg, 133 Nm, 14.5 s

Data for electric cars:
kg W Nm sec to 60mph
Chevrolet Volt 1715 63 130 9,0
smart fortwo electric drive 900 55 130 12,9
Mitsubishi i-MiEV 1185 47 180 13,5
Citroen zEro 1185 49 180 13,5
Peugeot iOn 1185 47 180 13,5
Toyota Prius Plug-in 1500 60 207 10,7
Renault Zoe 1392 65 220 8,0
Renault Fluence Z.E. 1543 70 226 9,9
Nissan leaf 1595 80 280 11,9
Toyota RAV4 EV (US only) 1560 115 296 8,0

 

[craigi]

Some literature data (for internal combustion engine cars):
Fiat Stilo: 1488 kg, 255 Nm, 11.2 s
BMW M3: 1885 kg, 400 Nm, 5.3 s
Citroen C3: 1126 kg, 133 Nm, 14.5 s

Data for electric cars:
kg W Nm sec to 60mph
Chevrolet Volt 1715 63 130 9,0
smart fortwo electric drive 900 55 130 12,9
Mitsubishi i-MiEV 1185 47 180 13,5
Citroen zEro 1185 49 180 13,5
Peugeot iOn 1185 47 180 13,5
Toyota Prius Plug-in 1500 60 207 10,7
Renault Zoe 1392 65 220 8,0
Renault Fluence Z.E. 1543 70 226 9,9
Nissan leaf 1595 80 280 11,9
Toyota RAV4 EV (US only) 1560 115 296 8,0

If you're talking electric cars, the power curves are a lot flatter. As I understand it, torque is constant upto very high revs and power increases linearly. So perhaps you need the constant acceleration solution after all.

 

[jumpjack]

Yes, traction is constant, but air drag is not, hence the equation which uses K1 and K2. But why doesn't it work?!?

 

[jumpjack]

I got this answer from stackexchange, but I had to fix and complete it due to several errors:

no air resistance:
$$a(v) =\frac F m = \frac {\frac P v } m = \frac P {vm} = \frac P m \frac 1 v = \frac w v$$

with air resistance:
$$a(v) = \frac{w}{v} - \frac 1 2 \rho C_d A v^2 = \frac{w}{v} - C_2 v^2$$
$$C_2= \frac 1 2 \rho C_d A \frac 1 m$$

$$w(v) = \frac {P(v)} m = \frac {vF(v)} m = \frac {v F_{max}(1-\frac{rpm(v)}{rpm_{max}})} m = \frac {v F_{max}(1-\frac{\gamma v}{rpm_{max}})} m = \frac {vF_{max}} m - \frac{\gamma v^2 F_{max}} {m \times rpm_{max}} = C_0v - C_1v^2$$

$$a(v) = \frac{w}{v} - C_2 v^2 $$

$$\boxed {a(v)= C_0 - C_1v - C_2 v^2} $$

$C_0=\frac {F_{max}} m$

$C_1=\frac{\gamma F_{max}} {m \times rpm_{max}}$ Torque (and force) dependence from speed

$C_2= \frac 1 2 \rho C_d A \frac 1 m$ Air drag

$F_{max} = \frac {T_{max}} {r_w}$
$r_w$ = wheel radiusIf Torque is constant vs speed (as in electric cars):

$$a(v) = C_0 - C_2 v^2$$

(rolling friction not yet taken into account)

Without air friction (not what I am looking for, but useful for comparison and for the example):

$${ t_{60} = \frac{P_{max}}{2 m \epsilon^2 g^2} + \frac{m v_{60}^2}{2 P_{max}} } $$

Example

A $m=1200\,{\rm kg}$ car with peak power $P_{max} = 160\,{\rm hp} = 119,000\,{\rm W}$ goes to $v_{60} = 26.9\,{\rm m/s}$. Traction is $\epsilon=0.4$ and $g=9.81\,{\rm m/s^2}$

$$ t_{60} = \frac{ \frac{119,000}{1200} }{2 \times 0.4^2 \times9.81^2} + \frac{26.9^2}{2 \frac{119,000}{1200}} = 3.23 + 3.63 = 6.86 \, {\rm sec} $$

---------------

Going on:

$$ a(v) = C_0 - C_1 v - C_2 v^2 $$

With direct integration you have

$$ t_1 = \int_0^{v_1} \frac{1}{a}\,{\rm d}v = \int_0^{v_1} \frac{1}{C_0-C_1 v-C_2 v^2}\,{\rm d}v = \ldots$$

With the parameter of top speed $a(v_f) = 0 \$ ==> $v_f = \dfrac{\sqrt{C_1^2+4 C_0 C_2}-C_1}{2 C_2}$ and the dimensionless parameter $\zeta = 2-\frac{C_1 v_f}{C_0}$ the time to speed is

$t(v) = \frac{v_f}{C_0 \zeta} \ln \left(1+\zeta \frac{v}{v_f-v}\right)$ Variable torqueFor constant torque, $C_1=0$ and $\zeta = 2$ , hence:

$$t(v) = \frac{v_f}{2 C_0 } \ln \left(1+ \frac{2v}{v_f-v}\right) = \frac{v_f}{2 C_0 } \ln \left( \frac{v_f+v}{v_f-v}\right)$$

$\boxed{t(v) = \frac{v_f}{2 C_0 } \ln \left( \frac{v_f+v}{v_f-v}\right)}$ Constant torque

$v_f = \dfrac{\sqrt{C_1^2+4 C_0 C_2}-C_1}{2 C_2} = \frac{\sqrt{4 C_0 C_2}}{2 C_2}$
$v_f= \sqrt \frac {C_0}{C_2}$ AirDrag-limited top speed for constant torque$$\boxed{t_{100kmh}= \frac 1 {2 \sqrt{C_2C_0}} ln{ \left( \frac{\sqrt{\frac {C_0}{C_2}}+27.8} {\sqrt{\frac {C_0}{C_2}}-27.8 } \right)} }$$
$$\boxed{C_0=\frac {T_{max}} {rm} }$$
$$\boxed{C_2= \frac 1 2 \rho C_d A \frac 1 m}$$
T = torque (Nm)
r = wheel radius (m)
m = vehicle mass (kg)
$\rho$ = air density = 1.225 kg/m3
$C_d$ = air drag coefficient (dimensionless, around 0.30 for cars)
A = frontal area (m2), around 2,2 for cars

Note: later I'll double check steps...

 

[russ_watters]

Yes, traction is constant, but air drag is not, hence the equation which uses K1 and K2. But why doesn't it work?!?

I wouldn't bother with drag here. While drag is a significant fraction of your engine load when cruising at 60 mph, it is an insignificant fraction of engine load when accelerating to 60mph. A car that gets 30 mpg is putting out only about 17 hp in cruise.

The fact that there are multiple drivers (tire friction, torque or power, gear ratio) and multiple levels of complexity would lead me to approach this problem with a numerical analysis in Excel. That way, you can start off simple and add complexity later, without losing what you already have. And Excel can figure out on its own, with if-then statements, when the transition between traction limited and engine limited acceleration happens and when the gear shifting happens.

 

[jumpjack]

I need an equation which takes into account drag and friction.

 

[craigi]

I need an equation which takes into account drag and friction.

The differential equation you have in the last post, with acceleration equal to the sum of terms proportional to velocity and velocity squared and a constant pretty much covers it. It's not going to be easy to compute real world coefficients without a wind-tunnel and telemetry though.

A real car loses energy to the environment in many different ways. Laminar air flow, turbulent air flow, road noise, engine noise, heat, etc. The equation you have been given for drag is a decent approximation for something like a lorry traveling at high speeds. At lower speeds for a car it's not going to be a good approximation. Cars are designed aerodynamically, to take advantage of this.

You can't realistically derive the contribution from these terms from first principles, but I'd expect there to be a spectrum of terms proportional to different non-integer powers of velocity. Even if you did find away to compute the coefficients for each term, you wouldn't solve such a differential equation analytically. A better approximation would be to use an empircially determined power to raise the velocity in the drag term to, which depends on the aerodynamics of the car. You can throw all your other loss terms in here too.

As already suggested, it is known that it is relatively safe to approximate these terms away under full acceleration and low-to medium speed, for a car, but as you approach maximum speed, those terms become equal to the acceleration term provided by the engine.

 

[russ_watters]

I need an equation which takes into account drag and friction.

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

And because of the fact that you have to account for traction and shifting, I'm not sure what you are asking is even possible.

 

[yuiop]

I wouldn't bother with drag here. While drag is a significant fraction of your engine load when cruising at 60 mph, it is an insignificant fraction of engine load when accelerating to 60mph. ...

I agree.

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

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

where m is the kerb weight in Kg, P is the max power in kW and T is torque in Nm.

This is a table for performance of electric cars obtained from an earlier post with some modifications and additions:

Data for electric cars:
kg, kW, Nm, top mph, sec to 60mph*, predicted $t_{60}$

Smart fortwo electric drive 900, 55, 130, 78, 12.9, 13.5
Mitsubishi i-MiEV 1185, 47, 180, 80, 13.5, 13.4
Citroen zEro 1185, 47, 180, 80, 13.5, 13.4
Peugeot iOn 1185, 47, 180, 81, 13.5, 13.4
Renault Zoe 1392, 65, 220, 84, 13.5*, 12.4
Renault Fluence Z.E. 1543, 70, 226, 84, 13.7*, 12.8
Nissan leaf 1595, 80, 280, 93, 9.9, 10.3
Toyota RAV4 EV 1560, 115, 296, 100, 8.0, 8.2
Tesla S 1999, 225, 430, 120, 5.9, 5.9
Tesla S Perf 2108, 270, 440, 125, 5.4, 5.6
Tesla S Perf Plus 2108, 310, 600, 130, 4.2, 4.2

The Chevrolet Volt and Toyota Prius have been removed from the table as they are hybrid cars with internal combustion engines and the Tesla models have been added to increase the range of data in the table. It should be noted that it is not always clear if the manufacturers include battery weight in their quoted kerb figures and legislation does not appear to have caught up with how the kerb weight of electric cars is defined.

* The Renault 0 to 60mph figures have been highlighted as they are actually figures for 0 to 62mph and because there is some variation on the figures quoted depending on the source. These are the sources I used: http://www.motorline.co.uk/renault/pdfs/fluence-ze.pdf and http://www.carfolio.com/specifications/models/car/?car=342397

Just consider electric vehicles: constant torque, no gears, motor efficiency =~100% .

In a second instance we could investigate Power/Torque correlation in electric motors.

While the lack of gears for electric cars simplifies the calculations, there is still an effective gear ratio in the drive train from the motor to the wheels and this varies from model to model. This is not taken into account in my simplified formula, as this information is not readily available (but I haven't looked that hard). I am also assuming that not all electric cars are direct drive. Maybe none are? Anyone know?

Including the top speed in the calculations would help the accuracy as it gives an idea of how efficient the aerodynamics of the car are and the overall gearing of the car, but if you are going to the trouble of finding the top speed on a test track, you might as well test the 0 to 60 mph times while you are there.