# Calculate maximum acceleration of a car from rest (optimum launch)

First disclaimer: I have a background in math, but I'm no expert in engineering or physics. This is the first question I've ever posted and I think it belongs here, but please let me know if belongs somewhere else and I will be happy to move it.

Second disclaimer: Realistically I am aware that the maximum acceleration from rest (or from any given vehicle speed, for that matter) is limited by the force of static friction between the tires and the road, but for this question I'd like to assume "infinite traction" just as a thought experiment. This is mostly a theoretical question anyway. So, here goes:

BACKGROUND

I love cars and I love math so I'm working on a very simple model of vehicle dynamics for fun (at this point it is just modeling basic kinematics). Now please excuse the wordiness, but I'm going to explicitly go through the key points in my thought process leading up to where I'm stuck so that you can correct any inaccurate assumptions of mine...

To calculate an acceleration, we need a force and we need a mass. We can easily calculate the mass of the car by looking up its weight, but what about the force? Well, let's say the engine produces 267 lb*ft of torque at 6000 RPM and 328 HP at 7000 RPM. We can use the torque to calculate the force at the driven wheels, $F_{tract}$, but since this is the force at the wheels, we need to know the torque at the wheels, not the torque at the engine crankshaft. Therefore, one could argue that the quoted torque figure is meaningless, since it doesn't tell us anything useful unless we also know how the car is geared and, moreover, engine torque is going to change as a function of engine RPM. Theoretically, then, let's suppose the engine's torque curve is roughly parabolic and we (maybe) know how the car is geared. Ignoring drag for now, we can calculate the acceleration based on:

$$a_{car}=\frac{F_{tract}}{m_{car}}=\frac{\tau_{wheel}}{r_{wheel}m_{car}}=\frac{\tau_{eng}\gamma_{trans}\gamma_{FD}}{r_{wheel}m_{car}}=\frac{\tau_{eng}\omega_{eng}\omega_{trans}}{\omega_{trans}\omega_{wheel}r_{wheel}m_{car}}=\frac{\tau_{eng}\omega_{eng}}{v_{car}m_{car}}$$

where $\gamma_{trans}$ and $\gamma_{FD}$ are the transmission and final drive gearing ratios, respectively. We want to maximize wheel torque to maximize acceleration, so intuitively we would like a lot of engine torque and a large transmission gear ratio. There is a limit, though, to how high we can feasibly make the gear ratio because, while it amplifies the torque to the wheels, it also means that the engine is spinning faster for any given vehicle speed. The engine can't rev above its redline and, moreover, for a given vehicle speed, we see from the above equation that maximizing wheel torque equates to maximizing the product of engine torque and engine rotational speed, which might not necessarily be maximized at the engine's redline anyway if the engine torque drops off enough from the peak after a certain engine RPM. So how do we maximize this quantity? Well, luckily this quantity is exactly equal to the definition of power, so:

$$a_{car}=\frac{\tau_{eng}\omega_{eng}}{v_{car}m_{car}}=\frac{P}{v_{car}m_{car}}$$

Therefore, to maximize acceleration at any given speed, we need to maximize engine power (which is the same as wheel power, ignoring losses due to friction). This result has important theoretical implications. First, it proves that, although the peak engine torque number is essentially meaningless, the peak engine power number is actually enough to, theoretically, calculate the maximum possible acceleration the car can achieve at any given speed. So it sort of ends the age-old "power vs. torque" debate. Second, in a car with a stepped transmission, it tells you exactly when to shift to maximize acceleration. You want to shift when the power (or, equivalently, wheel torque) in the next gear up equals that in the current gear.

Finally, the equation provides some insight into an interesting question, which really starts to get at the point where I become confused: when does my car accelerate the fastest? Well, let's suppose it has a stepped transmission. You can simulate the wheel torque curves for each gear as functions of vehicle speed, and you'll find that the highest value for wheel torque and hence, acceleration, occurs at the point where the engine is making maximum torque in first gear. But, wait! I thought the peak engine torque number was meaningless and only the peak power number mattered in terms of acceleration! Well, sort of. Before we were talking about theoretical maximum acceleration, and now we are talking about actual (more or less) acceleration, which is limited by the available gear ratios. Within each gear, maximum acceleration occurs when the engine produces peak torque, because this is when maximum wheel torque is realized in that gear. The key is that before we assumed the gear ratio and, hence, the engine rotational speed, could vary for any given vehicle speed, so we could play with both quantities to maximize their product (i.e. power). Now everything is fixed except for engine torque, so that's all we can maximize. When the engine reaches its maximum power output within this same gear, it is now accelerating the car as fast as theoretically possible at that new given vehicle speed. So, in other words, technically the car accelerates fastest in each gear at maximum engine torque, but if it isn't simultaneously making maximum power, this implies that it could be accelerating faster if it were geared differently such that the wheel torque (i.e. power) were maximized. A CVT can be designed to constantly adjust the gear ratio such that, at any given vehicle speed, wheel torque (i.e. power) and, thus, acceleration, is always maximized.

MY CONFUSION/QUESTIONS

AT LAST we get to my point of confusion. This is all well and good, and my simulations work just fine as long as I start my car at idle engine RPM in first gear (i.e. idle vehicle speed). The problem is that my grasp on the theory breaks down when I start the car from rest. Technically if the car is at rest ($v=0$) and is physically connected to the engine, the engine must also not be rotating $(RPM_{eng}=0)$. But that's why manual transmission cars have clutches and automatic transmission cars have transmission fluid (I think that is what an automatic car uses to ensure that the engine can spin while the wheels do not, right?) So if I have the brake on in my automatic car or the clutch in in my manual car, how can I calculate the maximum acceleration from rest? If I appeal to the maximum acceleration in terms of power equation (i.e. what would a CVT do?) it's not defined for $v=0$. The maximum acceleration grows arbitrarily large for smaller and smaller vehicle speed, presumably because you'd need an infinitely high gear ratio to amplify the zero torque the engine produces at 0 RPM in order to make the product equal maximum engine power. I guess that makes sense, but this equation only works when the engine and wheels are physically coupled, right? If I have the clutch in / brake on, there has to be an optimal RPM to rev the engine up to before transferring the power to the wheels to achieve maximum acceleration, but I don't know how to find it. Intuitively it's either going to be where the engine makes maximum torque or where it makes maximum power, but I don't know how to figure out which one it should be. It seems that maximum power is always the answer, but if the transmission is stepped then in first gear it makes the most wheel torque when the engine makes the most torque, so wouldn't I launch from here? But then instantaneously I would want the engine to produce maximum power again, not maximum torque?

That's the meat of my theoretical question, but I also have a more practical question about how to implement this into my model. Let's say I've decided what RPM to launch at. When I dump the clutch, if I assume I have traction, do I assume the wheels instantaneously match that speed and continue with the simulation? This doesn't seem right. How could I model the transfer of speeds (preferably in the simplest way)? I know that, in reality, when you dump the clutch the engine RPMs drop as, I assume, the inertia of the system increases massively and then the wheel and engine rotational speeds "synchronize." This was a long question and I apologize. If anyone actually takes the time to read through it and provide some insight I would really appreciate it. I love thinking about this stuff but I'm stuck on this issue and could really use the help!

SUMMARY/tl;dr

In general, maximum acceleration of a car for a given vehicle speed is achieved by maximizing engine power. This is the principle a CVT uses to maximize acceleration. However, what about the case when $v=0$? The equation for acceleration is undefined, so how can I calculate the maximum theoretical acceleration from rest, assuming infinite traction and given the engine's torque curve and the car's mass? What if I have a stepped transmission and I know the first gear ratio? Do I want to launch the car at the RPM where engine torque is highest, or where engine power is highest?

• Can you edit your question and put in one or two lines a summary and what the actual question is? – Transistor Jun 2 '18 at 16:07
• I'd recommend to summarise your story, or there'll likely be little people taking the time to read and comprehend it. – Bart Jun 2 '18 at 16:09
• Understood! I tried to improve the formatting/organization and added a summary section. – Liam Lang Jun 2 '18 at 16:22
• Doesn't it all come down to S = Ut + 0.5 At^2 ... – Solar Mike Jun 2 '18 at 16:25
• Zeno's paradoxes have been vexing us for nigh on 2500 years. – Phil Sweet Jun 4 '18 at 1:27

The first thing you need to note is that when your engine is rotating freely, i.e. your clutch is disengaged, then it "produces" little to no torque, regardless of the current RPM. The torque rating of the engine at a certain RPM is the torque you need to apply to the output shaft to bring it to an instantaneous standstill. Thus, to mobilise the torque, you need something to resist it.

Thus, with an instantaneous clutch and infinite traction, if you drop the clutch the engine will come to a standstill immediately and stall, and the car will barely move. You'd have the full torque of the engine but only for a split second.

Imagine hitting a punching bag with a baseball bat as hard as you can. It will barely move, but with the same force gradually applied you can move it further.

The second thing to note is that power is the rate at which the engine is generating energy, measured in Watt. 1 Watt is 1 joule per second, and a Joule is the energy needed to move 1 newton a distance of 1 meter (Nm, but not to be confused with torque, which is just a rotational force).

Although your peak power is at a point where the torque is below peak, you are adding energy into the system more quickly, thus "distance increases more quickly" i.e. you accelerate faster.

Imagine hitting the punching bag repeatedly with full force; it will move further and faster with each hit. Now imagine decreasing the force a bit, but hit it more quickly, you'll burn more calories and move the bag quicker.

As you launch your car, you'd like to keep the engine at max power. So start at the RPM for peak power and as you release the clutch, you'll want to increase the gas to compensate for the drop in RPM. You want to do this as quickly as possible without having your wheels spin, so practice by doing this quicker with every launch unit your wheels spin, then turn it down a notch. Obviously the surfacing of the road will also have a big impact on how quickly you can do this.

• I appreciate the insightful answer! The first paragraph was especially interesting to me. When launching an automatic car, does the engine produce torque from a standstill (at the moment before you release the brake and launch)? I guess I just get confused going back and forth between $P=mav$ and $F=ma$ when thinking about the initial acceleration. The first equation is unhelpful because $v=0$ and the second equation implies I want to maximize the force (i.e. wheel torque), although I agree with you that maximizing power makes sense. – Liam Lang Jun 9 '18 at 15:58
• @Liam Lang, F=ma, or more appropriately a=F/m, is true but only if F is applied constantly. If your engine can rotate free from the wheels there is no torque, regardless of transmission type. That's why you need to gradually engage the clutch, to get the car moving without killing the engine. Only when the wheels are in motion and the clutch is fully engaged you can take full advantage of the engine's capabilities. – ChP Jun 9 '18 at 22:35
• Thanks! I've been reading more about modeling friction clutches. Very interesting stuff! The simplest way seems to be to assume the clutch slips during takeoff and transmits all the engine torque. Then you can calculate how long it takes the wheels to accelerate to match the engine speed - at that point the clutch is engaged. I think if the clutch were engaged instantaneously the engine and transmission would spin at a common speed I can calculate by conserving angular momentum. However, that case implies an instantaneous speed change, i.e. infinite acceleration, so I won't use that one! – Liam Lang Jun 15 '18 at 14:37

Not really an answer but too much for a comment. I think you are making this harder than it needs to be.

Assume you start at like 200 rpm and shift at redline.

I think max acceleration will actually be at max horse power (not max torque).

Torque and horsepower are both functions of rpm so this becomes an integral equation.

How to get force is the question as neither torque or horsepower are the correct units.

You also need to account for air friction.

• I appreciate the comment! I wouldn't be surprised if I am overcomplicating things. – Liam Lang Jun 2 '18 at 18:19
• Currently my model does start around 200 RPM and shifts at redline. Max acceleration in each gear occurs at max engine torque, but at max engine power the curve touches the hyperbolic "limiting curve" provided by $a=\frac{P}{mv}$ when I plot them together. I've been back-calculating for the force at each timestep: $F_{tract} [lbs]=\frac{\tau_{eng} [lb*ft]\gamma_{trans}\gamma_{FD}}{r_{wheel} [ft]}$ I will account for friction soon; I just wanted to understand the very basics first. – Liam Lang Jun 2 '18 at 18:29
• @SolarMike So not relevant. – paparazzo Jun 2 '18 at 19:05
• @SolarMike Not interested. – paparazzo Jun 2 '18 at 19:22
• Still not interested. – paparazzo Jun 2 '18 at 19:25