# Max speed as a function of engine power

My question started out as finding the maximum speed of a go kart, taking into account only the drag forces as force opposing the motor.

I've done some investigation to find:

$$F_{drag}=\; \frac{\rho v^{2}\ C_{x} A_{f}}{2}$$

Where $C$ = drag coefficient, and $\rho$ is the density of air.

Further, with $F_{motor}=\dfrac{P_{motor}}{v}$ and the condition that at maximum speed, the acceleration will be 0, net force will be 0 as well, so $F_{motor}=F_{drag}$.

Combining equations, $F_{drag}=\dfrac{P_{motor}}{v}$, so $\dfrac{P_{motor}}{v}=\; \dfrac{\rho v^{2} C A_{f}}{2}$

Simplifying gives $P_{motor}=\; \dfrac{\rho\cdot v^{3}C\cdot A_{f}}{2}$

Solving for $v$ yields:

$$v=\; \sqrt{\dfrac{2P_{motor}}{\rho C A_{f}}}$$

Converting to units of miles per hour for $v$, and horsepower for $P_{motor}$:

$$v=\; 2.2 \sqrt{\dfrac{HP_{motor} \cdot 745\cdot 2}{\rho\mbox{C}A_{f}}}$$

From this pdf, I've found values of $\rho = 1.2 \frac{kg}{m^{3}}$, $C=.8$, and $A_{f}=.57$.

So! Questions:

1. Are my equations correct? I'm particularly concerned about the third root portion, given that drag is a quadratic. Doesn't connect in my mind, so perhaps my substitutions are faulty?

2. Are the values I have for those constants for ideal values, (i.e. perfectly enginneered kart, very unlike what I'd be able to construct myself). If so, by what general percent will my values be different?

3. Finally, is it unrealistic to call all forces other than drag insignificant? If so, by what general percent will my values for speed be too high?

For reference, this calculation has a 5 hp engine at a max speed of 52 mph, 15 hp at 75 mph, and 35 hp at 100 mph.

• Well, $P=W/t=Fd/t=Fv$, so the third power makes sense, right? Which is actually what you wrote already. The constant make it more of an engineering question, which is considered off-topic here. – Bernhard Aug 8 '14 at 20:53
• – alemi Aug 8 '14 at 21:36
• Oops, sorry about that. Well, forgetting about the constants, the equation seems to be in order then? – Jackson Aug 8 '14 at 21:55