# Finding instantaneous speed in MPH given acceleration, RPM and gear ratio

I am still trying to figure out how to (semi) accurately model instantaneous speed after having found acceleration. I have found that at higher RPMs, the resultant acceleration will be lower. I was told that the boost in speed comes from changing the gear ratio due to a changing gears at higher RPMs. Is this true? If so, what is the algebraic relationship between these independent variables?

• You also need to know the radius or diameter of the tires and the power and torque curves of the engine. This is non-trivial. – dmckee --- ex-moderator kitten Jun 10 '12 at 20:13
• I do have the power and the torque curves of the engine, assuming a constant horsepower, as well as the effective radius of the tire. – dtg Jun 10 '12 at 20:14
• "assuming a constant horsepower," Part of the point is that you can't assume constant horsepower. You have to understand how much the engine is putting our just then. – dmckee --- ex-moderator kitten Jun 10 '12 at 20:28

In conjunction with the answers already given, and @dmckee's comment, this is why it is difficult:

So your calculations need to take into account the power output variation across the range of engine speeds you are interested in.

(image from http://www.fordscorpio.co.uk/flatspot.htm)

• Please find another image, as the website you linked to seems to be infected by a trojan horse: Troj/Iframe-HX. – Alexander Jun 11 '12 at 9:37
• There you go @Alexander - I couldn't find any specific issue with the previous website, but happy to change. – Rory Alsop Jun 11 '12 at 10:29
• Thanks for the change, now my virus scanner is happy again here. – Alexander Jun 11 '12 at 15:32
• I see now that I need to find a way to calculate the horsepower over time as another independent variable. I may just abandon this model now as I am not sure how to recalculate horsepower due to the change in RPM. My current model for calculating torque uses a constant horsepower. Since torque is dependent on horsepower, I am not sure how to calculate horsepower over time since rearranging the equation HP would be dependent on torque... – dtg Jun 11 '12 at 16:34

The relationship between speed $v$ and acceleration $a$ is $$v(t) = v_0 + \int_0^t a(t)\, dt$$ If you know the acceleration you have to integrate it from the point in time where you know the starting speed $v_0$. So effectively you sum up the accelerations over short time intervals $\Delta t_i$: $$v =v_0 + \sum_i a_i \, \Delta t_i$$ This all requires a known acceleration over time. Keep in mind dmckee's comment though. Neither the horsepower, nor RPM, nor gear ratio are constant during the acceleration of a car.

In principle you can calculate the acceleration from the car specs, but this can get complicated as you need to know a lot about the whole powertrain.

Engine rot. speed $\omega_E$ is linked to wheel speed $\omega_W$ via the gear ratio $\gamma$ such that $\omega_E = \gamma\, \omega_W$. Wheel torque $T_W$ is linked to engine torque $T_E$ also via the same gear ratio, such that $T_W = \gamma\,T_E$.

You can transfer the wheel values into vehicle linear values via the wheel radius $R$.

NOTE: The total gear ratio $\gamma$ includes the differential ratio times the gearbox ratio.

Once you have established the vehicle acceleration as a function of speed $a(v)$ then use the following relationships to get displacement and time.

$$x_2-x_1 = \int_{v_1}^{v_1} \frac{v}{a(v)}\,{\rm d}v$$ $$t_2-t_1 = \int_{v_1}^{v_2} \frac{1}{a(v)}\,{\rm d}v$$

why?

$\frac{v}{a}\,{\rm d}v = \frac{v\,{\rm d}v}{{\rm d}v/{\rm d}t} = v\,{\rm d}t = {\rm d}x$

$\frac{1}{a}\,{\rm d}v = \frac{{\rm d}v}{{\rm d}v/{\rm d}t} = {\rm d}t$