# How long does it take for an electric car to go from 0 to 60 mph?

I found the freefall motion equation which describes terminal velocity of a falling body, but I can't find a similar equation for a vehicle subject to constant traction force, so I tried determining it by myself, but resulting equation is not plausible, as it shows dozens of seconds needed for a 1600 kg vehicle to go from 0 to 60 mph, so there must be something wrong. I'm using this equation: $$v(x) = v_f \cdot \tanh\left(\frac F {mv_f} \cdot x \right) = v_f \cdot\tanh\left(\frac {T/w } {m v_f} \cdot x \right)$$

• $v_f$ = terminal velocity = $\sqrt {\frac F c} = \sqrt {\frac {T} {wc}}$
• $c = \frac 1 2 \rho C_d A$
• $\rho$ = air density = 1.225 $\frac {kg} {m^3}$
• $C_d$ = air drag coefficient = 0.32
• A = frontal area = 2.19 m$^2$
• T = given torque = 220 Nm
• w = wheel radius = 0.25 m
• m = vehicle mass = 1762.5 kg

Freefall motion equation is:

$$v(x) = v_f \tanh\left( {x\sqrt{\frac{gc}{m}}}\right)$$

with $v_f=\sqrt{\frac{mg}c}$

With above data for the car, I should get around 10s time for 0-60 mph, but I get 63 seconds!

What am I doing wrong?

With above data(1) for the car, I know (2) I should get around 10s time for 0-60 mph, but I get 63 seconds!

What am I doing wrong?

Other literature data:

• Fiat Stilo - 255 Nm, 1488 kg, 11.2 s
• BMW M3 - 400 Nm, 1885 kg, 5.3 s
• Citroen C3 - 133 Nm, 1126 kg, 14.5 s

Literature 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

(1) "Evaluation of 20000 km driven with a battery electric vehicle" - I.J.M. Besselink, J.A.J. Hereijgers, P.F. van Oorschot, H. Nijmeijer

• The units for $Fx/mv$ are m/s and the argument of sine/cosine/tangent functions should be unitless. Perhaps you mean $Fx/mv^2$? – Kyle Kanos Apr 2 '14 at 13:34
• @Kyle Kanos I think it's correct: $\frac {Fx} {mv} = \frac {([kg] \frac{[m]}{[s^2]}) s} {[kg]\frac{[m]}{[s]}}$ – jumpjack Apr 2 '14 at 14:26
• Unless $x$ is measured in seconds, it's wrong. – Kyle Kanos Apr 2 '14 at 14:29
• The freefall equation is complicated because the force changes with distance from the planet. For a car accelerating at constant acceleration just use $v = u + at$. – John Rennie Apr 2 '14 at 14:38
• Are you trying to calculate how long it takes a car to hit 60 when dropped or when accelerated by the engine on a road? You'll need to know the car's acceleration due to the engine like John Rennie suggests. – user6972 Apr 2 '14 at 17:27

You have to split the time domains into the gears needed to reach 60mph. For each gear, there have to be assumptions on the power delivery of the car.

Typically 1st gear is traction limited, so you can assume constant acceleration up to the speed where peak power occurs. The relationship between power speed and acceleration is $P(v) = m \,v\, a(v)$. So run constant acceleration equal to traction of $a_1 = \epsilon \, g$ to speed $v_1 = \frac{P_{max}}{\epsilon g m}$ where $\epsilon$ is traction coefficient (0.4 for FWD, 0.6 for RWD and 0.9 for AWD) representing the peak 'g loading in 1st gear. So in the end of 1st gear the car has parameters

\begin{align} t_1 & = \int_0^{v_1} \frac{1}{a(v)}\,{\rm d} v= \frac{ \frac{P_{max}}{m} }{\epsilon^2 g^2} \\ x_1 & = \int_0^{v_1} \frac{v}{a(v)}\,{\rm d} v=\frac{ \left(\frac{P_{max}}{m}\right)^2 } {2 \epsilon^3 g^3} \\ v_1 & = \frac{ \frac{P_{max}}{m} } {\epsilon g} \\ a_1 & = \epsilon g \end{align}

This is the easy part. Now for the sprint to 60mph. Also use the parameter $w = \frac{P_{max}}{m}$ for the power to weight ratio. From this point on, the acceleration is a function of speed and it can be either: \begin{aligned} \mbox{no air resistance} & & a(v) & = \frac{w}{v} \\ \mbox{with air resistance} & & a(v) & = \frac{w}{v} - \beta v^2 \\ \end{aligned}

The math is simper without air resistance to calculate the parameters for $v_{60} = 60 {\rm mph}$

\begin{align} t_2 & = t_1 + \int_{v_1}^{v_2} \frac{1}{a(v)}\,{\rm d} v= \frac{w}{2 \epsilon^2 g^2} + \frac{v_{60}^2}{2 w} \\ x_2 & = x_1 +\int_{v_1}^{v_2} \frac{v}{a(v)}\,{\rm d} v= \frac{w^2}{6 \epsilon^3 g^3} + \frac{v_{60}^3}{3 w}\\ v_2 & = v_{60} \\ a_2 & = \frac{w}{v_2} \\ \end{align}

So the time to 60 can be estimated as

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

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 $g=9.81\,{\rm m/s^2}$

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

The actual numbers are going to be slower, since less than peak power is delivered most of the time, and there is air resistance and rolling resistance, plus time to shift gears and road grade effects and ...

# Edit 1

For an electric car, given a linear torque curve $T(rpm) = T_{max} \left(1 - \frac{rpm}{rpm_{max}} \right)$ you can create a function of power to weight as function of speed $$w(v) = \frac{P(v)}{m} = \frac{v T(v)}{m} = v( C_0 - C_1 v)$$ given the gearing $rpm(v) = \gamma v$

Now use the acceleration $$a(v) = \frac{w}{v} - C_2 v^2 = \frac{v (C_0 - C_1 v)}{v} - C_2 v^2 = C_0 - C_1 v - C_2 v^2$$

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

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

NOTE: There is constraint that says at terminal speed the motor must make positive torque, or the drag limited top speed must be less than the rpm limited top speed.

# Edit 2

For constant torque $a = a_0 \left( 1 - \left( \frac{v}{v_f} \right)^2 \right)$ and so

\begin{align} t & = \int\limits _{0}^{v}\frac{1}{a_{0}\left(1-\left(\frac{v}{v_{f}}\right)^{2}\right)}\,{\rm d}v\\ & = \frac{v_f}{a_0} {\rm tanh}^{-1} \left( \frac{v}{v_f} \right) \\ & = \frac{v_f}{2 a_0} \ln \left( \frac{v_f+v}{v_f-v} \right) \end{align}

where $a_0$ is the initial acceleration (at zero speed) and $v_f$ the drag limited top speed.

• Why is the constant acceleration of the car related to earth's $g$ and traction? – user6972 Apr 2 '14 at 19:18
• Tire friction. The limit is the tires in the beginning of the acceleration run. If this wasn't true, then you can create a really short 1st gear and put down max. power early on for rocket like acceleration. Unfortunately 1st gear is spaced out to account for available traction. – John Alexiou Apr 2 '14 at 20:51
• Thanks for the help, but as I just added in above comments, I don't want to ignore air drag, actually air drag effect on car acceleration is what I am trying to understand (and it makes no sense neglecting airdrag at 60mph!). And I'm studying electric vehicles, so we can consider constant torque. Actually I have yet to figure out if power or torque or both actually affect acceleration and max speed of a car. – jumpjack Apr 3 '14 at 8:59
• Maybe you should add the phrase electric car in the title to make it clear. – John Alexiou Apr 3 '14 at 12:35
• An electric motor has usually a linear relationship between torque and rpm, but you have to consider also if the cars has gears or not. – John Alexiou Apr 3 '14 at 12:36

I am adding another answer in order to show the steps needed to solve such problems in the general sense.

An acceleration run is split into parts of differing acceleration domains. Given a starting condition for time, distance, speed, acceleration of $t_0, x_0, v_0, a_0$ here is what you can do

1. Constant Torque with Air Drag $t=t_0 \rightarrow t_1$, $v=v_0 \rightarrow v_1$, $$a(v)= a_0 - \beta v^2 = a_0 \left(1-\frac{v^2}{v_f^2} \right) \\ v_f =\sqrt{\frac{a_0}{\beta}}$$ \begin{align} t_1 & = t_0 + \int_{v_0}^{v_1} \frac{1}{a_0 \left(1-\frac{v^2}{v_f^2} \right)}\,{\rm d} v = \\ & = t_0 + \frac{v_f}{a_0} {\rm tanh}^{-1} \left( \frac{v_f (v_0-v_1)}{v_0 v_1-v_f^2} \right) \end{align}
2. Constant Power with Air Drag $t=t_0 \rightarrow t_1$, $v=v_0 \rightarrow v_1$, $$a(v) = \frac{w_0}{v}-\beta v^2 = \frac{w_0}{v_f} \left( \frac{v_f}{v} - \frac{v^2}{v_f^2} \right) \\ v_f = \sqrt[3]{\frac{w_0}{\beta}}$$ \begin{align} t_1 & = t_0 + \int_{v_0}^{v_1} \frac{1}{\frac{w_0}{v_f} \left( \frac{v_f}{v} - \frac{v^2}{v_f^2} \right)}\,{\rm d} v = \\ &= t_0 + \dfrac{v_f^2}{w_0} \Bigg( \frac{1}{3} \ln \left(\frac{v_f-v_0}{v_f-v_1}\right) + \frac{1}{6} \ln \left( \frac{v_1^2+v_1 v_f + v_f^2}{v_0^2+v_0 v_f + v_f^2} \right) + \ldots \\ & + \frac{1}{\sqrt{3}} \arctan \left( \frac{1}{\sqrt{3}} \left(1+\frac{2 v_0}{v_f} \right) \right) - \frac{1}{\sqrt{3}} \arctan \left( \frac{1}{\sqrt{3}} \left(1+\frac{2 v_1}{v_f} \right) \right) \Bigg) \end{align}
3. Linear Torque with Air Drag $t=t_0 \rightarrow t_1$, $v=v_0 \rightarrow v_1$, $$a(v) = a_0 - c_1 v - c_2 v^2 = (a_0 - c_1 v) - \left(\frac{a_0}{v_f^2} - \frac{c_1}{v_f} \right) v^2 \\ v_f = \sqrt{\frac{a_0}{c_2} + \frac{c_1^2}{4 c_2^2}} - \frac{c_1}{2 c_2}$$ \begin{align} t_1 & = t_0 + \int_{v_0}^{v_1} \frac{1}{(a_0 - c_1 v) - \left(\frac{a_0}{v_f^2} - \frac{c_1}{v_f} \right) v^2}\,{\rm d} v =\\ & = t_0 + \frac{v_f}{c_1 v_f -2 a_0} \ln \left( \frac{(v_f-v_1) (a_0 (v_0+v_f)-c_1 v_0 v_f)}{(v_f-v_0) (a_0 (v_1+v_f)-c_1 v_1 v_f)} \right) \end{align}

Good luck!

NOTE: I used a CAS system to do the math for me.

## Appendix

Example calculation with Renault Zoe:

I used $m=1392\,{\rm kg}$ and constant torque up to $v_1 = 10\,{\rm m/s}$, and constant power of $P=63000\,{\rm W}$ afterwards. Final speed is $v_2 = 26.88\,{\rm m/s}$. I had to pick arbitrary top speed of $v_f = 47.0\, {\rm m/s}$

From $w_0 = \frac{P}{m} = 45.26 {\rm W/kg}$, The power at $v_1$ is used to find the inital constant acceleration $a_1 = \frac{P}{m v_1} = 4.52\,{\rm m/s^2}$ ( γ=0.46 traction )

The time to $v_1$ is

$$t_1 = \int_0^{v_1} \frac{1}{a_1 \left(1-\left(\frac{v}{v_f}\right)^2\right)}\,{\rm d}v = 2.244\,{\rm sec}$$

From this point on the acceleration is

$$a(v) = \frac{w_0}{v_f} \left( \frac{v_f}{v} - \frac{v^2}{v_f^2} \right) = \frac{45.26}{v} - \frac{v^2}{2294.0}$$

and the time to 60mph is

$$t_2 = t_1 + \int_{v_1}^{v_2} \frac{1}{a(v)}\,{\rm d}v = 2.244 + 7.553 \approx 9.79 {\rm sec}$$

NOTE: $\int \frac{x}{1-x^3}\,{\rm d}x = -\frac{1}{3} \ln(1-x)+ \frac{1}{6} \ln(x^2+x+1) - \frac{1}{\sqrt{3}} \arctan \left( \frac{2 x+1}{\sqrt{3}} \right)$

• $a_0=C_0=\frac {F_{max}}m$ – jumpjack Apr 7 '14 at 12:00
• $\beta = C_2 = \frac 1 2 \rho C_d A \frac 1 m$ – jumpjack Apr 7 '14 at 12:03
• From second answer, $t_{0-60}=\frac{v_f}{C_0} atanh\left(\frac {27.8}{v_f}\right)$ , but it still gives impossible results of 60 and more seconds to me. – jumpjack Apr 7 '14 at 12:16
• So your assumptions for the the torque curve are not correct then. – John Alexiou Apr 7 '14 at 14:39
• Found the solution to "bad numbers" question in comment above: formulas are right, but I was plotting v and t using WinPlot... which has no atanh() definition, so it actually plotted a*tanh() !! – jumpjack Apr 17 '14 at 7:32

What I was doing wrong was putting manufacturer-provided motor torque directly into the formulas; actually the overall gear ratio must be taken into account, and it results from data taken around on internet that for electric vehicles it is =~8 (dimensionless).

Hence the proper expression to use for $v_f tanh(\frac{F}{mv_f} t)$ is not $v_f tanh(\frac{\frac {T_M} r}{mv_f} t)$ but:

$$v(t) = v_f tanh(\frac{\frac {T_W} r}{mv_f} t)$$

with

$T_W = T_M G$

$T_M$ = motor torque [Nm] (manufacturer given)

G = Overall gear ratio [dimensionless] = $\frac{P_m r}{T_m v_c}$

m = vehicle mass [kg]

$v_f$= terminal velocity = $\sqrt \frac F c$ [m/s]

$F = \frac {T_M G}r$

$c= {\frac 1 2 \rho C_d A}$

$\rho = 1.225$[kg/m^3]

$C_d$ = air drag coefficient [dimensionless] (=~0.3 for cars)

A= cross section area , frontal area [m^2] (~= 2.2 for cars)

Anyway the resulting expression for v(t) is only valid for constant torque, but torque is not constant even for electric vehicles: it's constant only up to 30-40 km/h, then it decreases as $\frac 1 v$ , and expressions for v also taking into account this torque behaviour is yet to be determined.

http://digitalcommons.mtu.edu/cgi/viewcontent.cgi?article=1697&context=etds

"DEVELOPMENT OF THE ECOCAR 3 PROPOSAL AND GUIDELINES FOR MODELING AND DESIGN IN YEAR ONE OF ECOCAR 3 - Tyler B. Daavettila - Michigan Technological University"

$P = \frac{m_e}{2\epsilon_t \underline {t_a}} \left( v_f^2+v_b^2 \right)+\frac 2 3 m g C_{rr}v_f+ \frac 1 5 \rho C_D A_f {v_f}^2$

From this we get:

$t_a = \frac {m_e(v_f^2+v_b^2)}{2 \epsilon_t \left(P-\frac 2 3 mgC_{rr}v_f - \frac 1 5 \rho C_dA_fv_f^3 \right)}$

$m_e$ = equivalent mass (including rotational energy)

$v_b$ = max speed at constant torque

$v_f$ = final speed

$\epsilon_t$ = total powertrain efficiency

$t_a$ = time from 0 to $v_f$ speed

g = 9,81 m/s^2

$C_{rr}$ = rolling friction

$C_D$ = air friction coefficient

$A_f$ = frontal area

$P$ = Motor power

• Equation [11](3) is basically what I proposed. To split the acceleration in two domains and do the integrals. Figure 6.5B clearly shows the constant torque and constant power domains. – John Alexiou Sep 11 '14 at 14:33