Calculating 0-60 mph time for a vehicle

[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 $v<=v_c$.