Anything producing a maximum acceleration less than 0.3g will feel sluggish. Typical passenger cars can do up to 0.7g easily (full throttle).
$$F_w = \lambda_m ma$$
$$\frac{T_w}{r} = \lambda_m mg\frac{a}{g}$$
$$\frac{\eta\ G_R\ T_m}{r} = \lambda_m W\frac{a}{g}$$
$$T_m = \frac{\lambda_m \frac{a}{g}r}{\eta\ G_R}W$$
In your case (assuming a drivetrain efficiency of ##\eta = 0.85## and a
mass factor ##\lambda_m = 1.38##):
$$T_m = \frac{(1.38) (0.3) (\frac{13}{12}\text{ ft})}{0.85 \times (2.89 \times 4.88)}W$$
$$T_{m\text{ [lb.ft]}}= (0.0374\text{ ft}) W_\text{[lb]}$$
So if you have 258 lb.ft of motor torque, you will be able to accelerate a vehicle weighing 6900 lb at a 0.3g acceleration (from rest, that is). Repeating the same process for a 0.7g acceleration, you'll find your vehicle must weigh no more than 2960 lb.
Grez said:
and "high gear" would be 1.11 and 4.88 and would be for highway and everyday use and should be able 115kmh @ 5krpm and 230kmh @ 10krpm
The typical amount of power you need for highway (mostly due to aerodynamic drag) is about 25 hp. At 5000 rpm, this means 26 lb.ft (or 35 N.m) of motor torque. According to the efficiency map of your pdf file, this is where you are:
(red dot: 115 km/h; blue dot: 200 km/h; green dot: 230 km/h)
As you can see, with this motor, your vehicle would be most efficient at cruise speeds of around 200 km/h, which is also the maximum speed that you can use continuously.
The equation used to find those points on the map is found by equating the motor power to the drag power:
$$P_m = P_d$$
$$\eta T_m \omega = \frac{1}{2}\rho C_d A v^2 \times v$$
$$\frac{\eta T_{m\ [N.m]} RPM}{7122} = \frac{\frac{1}{2}\rho C_d A\left(\frac{1}{3.6}\frac{(115\text{ km/h})}{(5000\text{ rpm})} RPM\right)^3}{746}$$
$$T_{m\ [N.m]} = \frac{(7122)\frac{1}{2}\rho C_d A\left(\frac{1}{3.6} \frac{(115\text{ km/h})}{(5000\text{ rpm})} RPM\right)^3}{746\eta RPM}$$
$$T_{m\ [N.m]} = \frac{\rho C_d A\left( \frac{(115\text{ km/h})}{(5000\text{ rpm})}\right)^3 RPM^2}{9.774\eta}$$
$$T_{m\ [N.m]} = \frac{RPM^2}{793057}$$
Where ##\rho = 1.23 \text{ kg/m³}## and I assume the
drag factor ##C_d A = 0.7 \text{ m²}##
Redoing the previous exercises with a new gear ratio, note that if you could use a higher gear ratio instead of 1.11:1 you would maybe end up in a more favorable zone of your motor where you would gain efficiency at lower speeds while losing nothing at high speeds. You even gain a higher top speed (torque-limited instead of rpm-limited), with only a minor loss in the maximum cruise speed (I imagine you don't need to drive over 185 km/h continuously anyway?). The results are In the following table (some are eyeballed):
| 1.11:1 /[0.833:1]/ (0.555:1) | Speed | Torque | RPM | Efficiency |
| Low Cruise Speed | 58 km/h
[58 km/h]
(58 km/h) | 8 N.m
[11 N.m]
(16 N.m) | 2500 rpm
[1875 rpm]
(1250 rpm) | 63%
[65%]
(67%) |
| High Cruise Speed | 115 km/h
[115 km/h]
(115 km/h) | 31 N.m
[41 N.m]
(62 N.m) | 5000 rpm
[3750 rpm]
(2500 rpm) | 77%
[80%]
(87%) |
| Max Cruise Speed | 200 km/h
[200 km/h]
(185 km/h) | 95 N.m
[127 N.m]
(185 N.m) | 8700 rpm
[6525 rpm]
(4000 rpm) | 93+%
[94%]
(93-%) |
| Max Speed | 230 km/h
[227 km/h]
(245 km/h) | 126 N.m
[210 N.m]
(275 N.m) | 10000 rpm
[7400 rpm]
(5400 rpm) | 92%
[92%]
(92%) |
