Car
Table of Contents
1. Acceleration
We can relate the car's speed to its engine's RPM by the following formula:
\begin{equation} |\vec{v}| = \frac{(2\pi r_{\text{wheel}})(RPM)}{g_{t}g_{d}} \end{equation}where \(g_{t}\) is the gear ratio for the transmission, and \(g_{d}\) is the gear ratio for the differential.
I'm told:
A typical engine accelerates from idling to ~6000 rpm in about 4 secs (\(a\sim 1300\) rpm/s), the driver disengages the clutch to shift gears and the rpm drops to 4500.
The gear ratio for the differential varies depending on the car, but it typically is around
\begin{equation} g_{d}\approx 4.00. \end{equation}We could model the engine's RPM as something like a periodic function:
\begin{equation} f(t) = \begin{cases}\omega t & 0\leq t\leq t_{1}\\ 4500 & t_{1}\leq t\leq T \end{cases} \end{equation}where the shift time is \(T-t_{1}\) on the order of 1 second.
1.1. Typical Transmission
A typical transmission would have something like the following gear ratios:
| Gear | Ratio |
|---|---|
| First | 4:1 |
| Second | 3:1 |
| Third | 2:1 |
| Fourth | 1:1 |
| Fifth | 1:2 |
| Sixth | 1:3 |
Usually the transmissions go up to 4 or so, but this is a fine heuristic.
1.2. Typical Differential
The differential gear ratio is usually around 4:1.
1.3. Typical Tire Size
We usually have tire radius be around 0.400 meters.
1.4. Drag and Rolling Resistance
On the road, rolling resistance is negligible. But drag force is not! (Off the rad, rolling resistant becomes non-negligible.)
1.5. Exact Solution For Drag in 1-Dimension
Actually, we would have Newton's second equation become (for one-dimensional motion):
\begin{equation} \frac{\mathrm{d}v(t)}{\mathrm{d}t} = (2\pi r_{wh})\omega_{n}(t) - c_{d} v(t)^{2} \end{equation}where \(c_{d}\) is the drag coefficient. For cars, \(c_{d}\approx 0.3\) nowadays but before World War 2 cars had \(c_{d}\approx 1\). For motorcycles \(0.5\leq c_{d}\leq 1\).
For simplicity we will absorb the factor of \(2\pi r_{wh}\) into \(\omega_{n}(t)\). We also model \(\omega_{n}(t) = \omega_{n}\cdot t\) as a linear homogeneous function of time.
We can solve this using Bessel functions of the first kind as:
\begin{equation} v(t) = i\sqrt{\frac{\omega_{n}t}{c_{d}}} \frac{J_{-2/3}\left(\frac{2}{3} i \sqrt{\omega_{n} c_{d}} t^{3/2}\right) - C_{1} J_{2/3}\left(\frac{2}{3} i \sqrt{\omega_{n} c_{d}} t^{3/2}\right)}{J_{1/3}\left(\frac{2}{3} i \sqrt{\omega_{n} c_{d}} t^{3/2}\right) + C_{1} J_{-1/3}\left(\frac{2}{3} i \sqrt{\omega_{n} c_{d}} t^{3/2}\right)} \end{equation}where \(C_{1}\) is some constant to be determined by initial data.
We see that \(v(0)\) is indeterminate, but its limit is well-defined and has value:
\begin{equation} \lim_{t\to 0}v(t) = \frac{3^{1/3} (i\sqrt{\omega_{n}c_{d}})^{2/3}\Gamma(2/3)}{c_{d}C_{1}\Gamma(1/3)}. \end{equation}Thus, if we are given the initial velocity as \(v(0) = v_{0}\neq 0\), we can set the limit equal to this, giving us:
\begin{equation} C_{1} = \frac{3^{1/3} (i\sqrt{\omega_{n}c_{d}})^{2/3}\Gamma(2/3)}{c_{d}v_{0}\Gamma(1/3)}. \end{equation}We have been quite cavalier in our computations, since it's entirely symbolic in nature. The proliferation of \(i=\sqrt{-1}\) is alarming, but surprisingly when we plug in realistic values for our parameters…we obtain real values (i.e., the values produced are real numbers).
1.5.1. When Shifting Gears
If we model shifting gears as requiring 1 second of time (or some finite amount of time \(\Delta t\)) where there is no forward velocity \(\omega_{n}(t)=0\), then the solution is quite simple. Our equations of motion
\begin{equation} \frac{\mathrm{d}v(t)}{\mathrm{d}t} = -c_{d} v(t)^{2} \end{equation}has a solution
\begin{equation} v(t) = \frac{1}{c_{d}t - C_{1}} \end{equation}where \(C_{1}\) is determined by initial conditions. If the initial velocity is \(v(0)=v_{0}\) for example, then \(C_{1}=-1/v_{0}\). That is to say, if \(v(0) = v_{0}\neq 0\), then
\begin{equation} v(t) = \frac{v_{0}}{v_{0}c_{d}t + 1}. \end{equation}1.5.2. Complete Solution
We can then use these two general solutions to form piecewise the behaviour of the vehicle as it is shifting gears.
1.5.2.1. Asymptotic Behaviour of Bessel Functions
We can use the asymptotic approximations of \(J_{\alpha}(ix)\) for large real \(x\) (and another asymptotic approximation for small real \(x\)).
As \(x\to\infty\), we have
\begin{equation} J_{\alpha}(ix)\sim\frac{1}{\sqrt{2\pi ix}}\exp\left[-i\left(ix - \frac{\alpha\pi}{2} - \frac{\pi}{4}\right)\right]. \end{equation}For \(x\to0\), we have
\begin{equation} J_{\alpha}(ix) \sim \frac{1}{\Gamma(1 + \alpha)}\left(\frac{ix}{2}\right)^{\alpha}. \end{equation}I'm sure we can invent some function \(f(t)\) such that for \(0\lt t\ll 1\) we have \(f(t)\approx (it/2)^{\alpha}/\Gamma(1+\alpha)\) and for larger \(t\) we have \(f(t)\) approximate the first asymptotic expansion.
2. Actual Solution
The eagle-eyed reader will observe a problem with modeling the contribution of the engine to acceleration. That is to say, we are assuming the RPM increases linearly with time. We would then have
\begin{equation} a_{n}(t) = 2\pi r_{wh}\omega_{n}^{2}t^{2} =: a_{n}\cdot t^{2}. \end{equation}We should instead have the equations of motion be
\begin{equation} \frac{\mathrm{d}v(t)}{\mathrm{d}t} = a_{n}\cdot t^{2} - c_{d} v(t)^{2}. \end{equation}Then we have the generic solution look like:
\begin{equation} v(t) = i\sqrt{\frac{t^{2} a_{n}}{c_{d}}} \frac{J_{-3/4}\left(\frac{1}{2} i t^2 \sqrt{c_{d} a_{n}}\right)-C_{1} J_{3/4}\left(\frac{1}{2} i t^2 \sqrt{c_{d} a_{n}}\right)}{J_{1/4}\left(\frac{1}{2} i t^2 \sqrt{c_{d} a_{n}}\right)+C_{1} J_{-1/4}\left(\frac{1}{2} i t^2 \sqrt{c_{d} a_{n}}\right)} \end{equation}We solve the initial condition by taking the \(t\to 0\) limit and \(v(t)\to v_{0}\), giving us
\begin{equation} v_{0} = \frac{2[-c_{d} a_{n}]^{1/4}\Gamma(3/4)}{c_{d}C_{1}\Gamma(1/4)}. \end{equation}Of course, if I am in error about this, and the correct setup would be to have acceleration due to the engine be linear in time, then the preceding section's scratchwork should be used instead.
2.1. Asymptotic Approximation
If we plug in the asymptotic approximation of \(J_{\alpha}(i x)\) for \(x\gg 1\), we get the "long time" solution
\begin{equation} v_{big}(t) \sim \sqrt{\frac{a_{n}}{c_{d}}}t. \end{equation}The small time asymptotic approximation gives us
\begin{equation} v_{small}(t)\sim i\sqrt{\frac{a_{n}}{c_{d}}}t\frac{2}{3i} \frac{1}{\sqrt{a_{n}c_{d}} t^{2}} \frac{a_{n}c_{d} t^{4}C_{1}\Gamma(5/4) + 2\sqrt[4]{-a_{n}c_{d}}t\Gamma(7/4)}{\sqrt[4]{-ac}t\Gamma(3/2) + 2C_{1}\Gamma(5/4)}. \end{equation}Unfolding the definition of \(C_{1}\) gives us a simplified, more appealing form:
\begin{equation} v_{small}(t)\sim\frac{4}{3}\frac{\Gamma(3/4)\Gamma(5/4) a_{n} t^{3} + \Gamma(1/4)\Gamma(7/4)v_{0}}{4\Gamma(3/4)\Gamma(5/4) + \Gamma(1/4)\Gamma(2/3)c_{d}v_{0} t}. \end{equation}(Assuming, of course, I did my arithmetic correctly.) We can check the units, recall that \(a_{n}\cdot t^{2}\) has units of acceleration, so \(a_{n}\cdot t^{3}\) has units of velocity. The numerator has units of velocity. The denominator is dimensionless, so this looks correct.
Schematically, this looks like (recall that \(c_{d}\sim 1\)):
\begin{equation} v_{s}(t)\approx \frac{a_{n}t^{3} + v_{0}}{4 + c_{d}v_{0}t}. \end{equation}