Calculating linear acceleration from torque in a car [closed]

Question

For fun, I'm trying to play with really simple simulation of a Tesla Model 3. Modeling its acceleration and such. But I've hit a roadblock early on despite validating my math several times.

What I want to do is find the car's velocity at a time $t$ given the motor's torque, wheel radius, and mass.

When I look up values for this car, here's what I get:

• Peak torque: $\tau = 639\ \text{N}\cdot\text{m}$
• Mass: $m = 1800\ \text{kg}$
• Wheel radius: $R = 0.2413\ \text{m}$

Here's how I'm calculating acceleration:

$$ F = \frac{\tau}{R} \\ a = F/m \\ a = \frac{\tau}{Rm} $$

Considering that I'm modeling torque as constant, acceleration will also be constant, which means:

$$ v(t) = at = \frac{\tau}{Rm}t $$

However, when I plug in $t = 3.2$ (the Model 3 accelerates from 0-60mph in 3.2 seconds), the velocity is only $\text{4.7 m/s}$. It should be closer to $\text{30 m/s}$. Where is this discrepancy coming from? Are the official numbers I found simply wrong?

Answer 1

What you need is the whole torque curve (torque vs. speed of the motor).

These are quite complex, but there are distinct stages. First, there is a constant torque, then at about 45 mph the motor transitions to constant power, and over 75 mph to some other form of less power.

To calculate acceleration you have to know the power $P$ produced at speed $v$ and then calculate (in compatible units)

$$ a = \frac{P}{m v} $$

(The above ignores air resistance for now)

Next you can calculate the time to reach a certain speed with $$t = \int \frac{1}{a}\,{\rm d}v$$

or you can calculate the distance needed to reach a certain speed with $$ x = \int \frac{v}{a}\,{\rm d}v$$

1. Constant Torque - The power at every speed $v$ is a straight line with a general equation of $P = c_1 v$. Specifically at speed $v_1$ the power is $P_1$ so $c_1 = P_1/v_1$. The solution is trivial with $$\begin{aligned} t & = \frac{m v}{c_1} & x & = \frac{m v^2}{2 c_1} \end{aligned}$$ this carries up to speed $v_1$ at which point $t_1 = \frac{m v_1^2}{P_1}$ and $x_1 = \frac{m v_1^3}{2 P_1}$.

2. Constant Power - The power at every speed $v$ is constant $P = P_1$. The solution now is $$\begin{aligned} t & = t_1 + \frac{m (v^2-v_1^2)}{2 P_1} & x & = x_1 + \frac{m (v^3-v_1^3)}{3 P_1} \end{aligned}$$ this carries up to speed $v_2$ at which point $t_2 = \frac{m (v_2^2+v_1^2)}{2 P_1}$ and $x_2 = \frac{m ( v_1^3+2 v_2^3)}{6 P_1}$.

3. Diminishing Power - Since we don't know the form of the curve here, and at these speeds aerodynamics become dominant, the analysis stops here.

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Appendix

To help in unit conversions consider the following calculation for acceleration from power and speed

$$\frac{270\,\mathrm{hp}}{4400\,\mathrm{lbm}\cdot 50\,\mathrm{mph}}=4.513\,\mathrm{m/s^2}$$