If you have
- the power and torque curves (Assume the following power/torque curve in this example):

Figure : engine power and torque curves for a car ICE engine of about 150 hp (source carthrottle.com )
- the diameter of the wheel (assume $d_w= 0.7[m]$
then the procedure is the following:
procedure
- you need to find the RPM at the maximum Power.
E.g in this example its 110 kW at 5500 rpm.
- Find the top speed by equating the maximum power to the aerodynamic losses
The top speed is theoretically reached at the maximum power output, and the highest gear.
The top speed can be approximated by equating the car engine power and the losses (i.e. drag) using the following equation:
$$P = \color{red}{F_D} \cdot u = \color{red}{\frac{C_D}{2} A \rho u^2} \cdot u $$
so the top theoretical (see below why this is stressed) speed that engine can reach is:
$$u = \sqrt[3]{\frac{2 P}{C_D A \rho }}$$
where:
- $C_D$ is the drag coefficient assume 0.35 for this example
- $A$ is the frontal cross-sectional area (Assume 2 $m^2$)
- $\rho$ is the air density 1.225 $kg/m^3$
For these value the theoretical top speed is approximately 228.75 kph (63 m/s)
- Calculate the rpm of the wheel for the top speed
The rpm of the wheel $n_W$ are equal to:
$$u = 2\cdot \pi\cdot n_{w} \frac{d_w}{2}\Rightarrow n_{w} = \frac{u }{d_w\pi}$$
where:
- $u$ is the top speed in m/s
- $n_w$ is the rotational speed of the wheel in rpm
- $d_w$ is diameter of the wheel in meters
The result for the example is : $n_w = 275.92[rpm]$
- Calculate the drivetrain ratio
The drivetrain ratio will be given by
$$i= \frac{n_{eng}}{n_w}$$
where:
- $n_{eng}$ are the rpm of the crankshaft when the engine is producing max power (i.e. when top speed is reached)
- $n_{w}$ are the rpm of the wheel at top speed.
for this example:
$$i= \frac{5500}{275.92}= 19.932 $$
If you know the differential ratio and its e.g. $i_d=4$, then from the crankshaft to the differential the overall gear ratio for the top speed should be $i_{ts}= \frac{i}{i_d}$