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?
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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) |
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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. |
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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 $ |
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