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**Stingray** · Oct26-06, 03:29 PM

Now let's calculate an acceleration. I'm going into some messy detail, but assume that you always have sufficient traction. More things can be taken into account, but this will then get complicated very quickly.

Let's say that the static mass of the car is

, and that including rotational inertia gives you $LaTeX Code: m_{\\mathrm{eff}}(g)$ . g is the gear ratio (a positive integer). Now say that the power available at the wheels is $LaTeX Code: P(\\omega)$ . Here, $LaTeX Code: \\omega$ is the rotational speed of the engine (i.e. rpm, but possibly in nicer units). Suppose that the actual gear ratios are $LaTeX Code: \\alpha(g)$ (first gear would be largest), the differential ratio is $LaTeX Code: \\beta$ , and the tire radius is

.

In general, various driveline losses will change depending on gear ratio, but I'll ignore that. They also depend weakly on vehicle speed and the temperatures of the air and various oils, etc etc. On top of that, standard dyno measurements have several systematic errors in them, but I'm just going to say that the dyno measurements are precisely $LaTeX Code: P(\\omega)$ .

Assuming that the clutch and tires aren't slipping significantly, the (rotational) speed of the engine and the (linear) speed of the car are directly related to each other:

$LaTeX Code: v(\\omega, g) =\\frac{\\omega r}{\\alpha(g) \\beta} $

This can also be inverted, so $LaTeX Code: \\omega = \\alpha \\beta v/r$ . Now the mechanical power supplied by the engine goes into accelerating the car and overcoming aerodynamic and other losses. Group the latter conditions into $LaTeX Code: P_{\\mathrm{drag}}(v)$ , so one ends up with an acceleration

$LaTeX Code: m_{\\mathrm{eff}}(g) a = \\left[ P(\\alpha(g) \\beta v/r) - P_{\\mathrm{drag}}(v) \\right]/v $

This equation can be solved directly to give the time required to accelerate between two given speeds in a single gear:

$LaTeX Code: t = \\int_{v_{i}}^{v_{f}} \\left(\\frac{ m_{\\mathrm{eff}} v }{ P - P_{\\mathrm{drag}} } \\right) dv $

Generalizing this to include shifting is straightforward, but an important point is that torque never enters this equation. You could write it in terms of the torque the engine is producing, but it would be more complicated. The interesting thing to note that is that the acceleration at a given speed is always proportional to the power delivered.

Also, if the drag is negligible, the above equation can be written a little differently:

$LaTeX Code: t = m_{\\mathrm{eff}} \\left( r/\\alpha \\beta \\right)^{2} \\int_{\\omega_{i}}^{\\omega_{f}} d\\omega / \\tau(\\omega) $

The two limits in the integral are the different rpm's corresponding to different velocities, and $LaTeX Code: \\tau(\\omega)$ is the torque "at the engine." This equation is nice because the integral depends only on properties of the engine. The driveline is irrelevant. This is as close as you can get to the handwaving idea that improving "average torque" over the rev range makes a faster car.

Edit: I've edited this a bunch of times to fix and add things.

Oct26-06, 03:29 PM	Last edited by Stingray; Oct26-06 at 03:56 PM.. #11
Stingray Stingray is Offline: Posts: 629	Now let's calculate an acceleration. I'm going into some messy detail, but assume that you always have sufficient traction. More things can be taken into account, but this will then get complicated very quickly. Let's say that the static mass of the car is , and that including rotational inertia gives you $LaTeX Code: m_{\\mathrm{eff}}(g)$ . g is the gear ratio (a positive integer). Now say that the power available at the wheels is $LaTeX Code: P(\\omega)$ . Here, $LaTeX Code: \\omega$ is the rotational speed of the engine (i.e. rpm, but possibly in nicer units). Suppose that the actual gear ratios are $LaTeX Code: \\alpha(g)$ (first gear would be largest), the differential ratio is $LaTeX Code: \\beta$ , and the tire radius is . In general, various driveline losses will change depending on gear ratio, but I'll ignore that. They also depend weakly on vehicle speed and the temperatures of the air and various oils, etc etc. On top of that, standard dyno measurements have several systematic errors in them, but I'm just going to say that the dyno measurements are precisely $LaTeX Code: P(\\omega)$ . Assuming that the clutch and tires aren't slipping significantly, the (rotational) speed of the engine and the (linear) speed of the car are directly related to each other: $LaTeX Code: <BR>v(\\omega, g) =\\frac{\\omega r}{\\alpha(g) \\beta}<BR>$ This can also be inverted, so $LaTeX Code: \\omega = \\alpha \\beta v/r$ . Now the mechanical power supplied by the engine goes into accelerating the car and overcoming aerodynamic and other losses. Group the latter conditions into $LaTeX Code: P_{\\mathrm{drag}}(v)$ , so one ends up with an acceleration $LaTeX Code: <BR>m_{\\mathrm{eff}}(g) a = \\left[ P(\\alpha(g) \\beta v/r) - P_{\\mathrm{drag}}(v) \\right]/v<BR>$ This equation can be solved directly to give the time required to accelerate between two given speeds in a single gear: $LaTeX Code: <BR>t = \\int_{v_{i}}^{v_{f}} \\left(\\frac{ m_{\\mathrm{eff}} v }{ P - P_{\\mathrm{drag}} } \\right) dv<BR>$ Generalizing this to include shifting is straightforward, but an important point is that torque never enters this equation. You could write it in terms of the torque the engine is producing, but it would be more complicated. The interesting thing to note that is that the acceleration at a given speed is always proportional to the power delivered. Also, if the drag is negligible, the above equation can be written a little differently: $LaTeX Code: <BR>t = m_{\\mathrm{eff}} \\left( r/\\alpha \\beta \\right)^{2} \\int_{\\omega_{i}}^{\\omega_{f}} d\\omega / \\tau(\\omega) <BR>$ The two limits in the integral are the different rpm's corresponding to different velocities, and $LaTeX Code: \\tau(\\omega)$ is the torque "at the engine." This equation is nice because the integral depends only on properties of the engine. The driveline is irrelevant. This is as close as you can get to the handwaving idea that improving "average torque" over the rev range makes a faster car. Edit: I've edited this a bunch of times to fix and add things.