How to evaluate the acceleration performance of a car is believed to be of great concern to every owner and prospective owner. In fact, the acceleration performance of automobile technical performance index is only a reference value. Many people know the relationship between force, mass and acceleration, but the acceleration performance of a car is related to many factors. Some netizens hope to find the definite relationship among the torque of the automobile engine, the volume of the automobile and the acceleration, which is actually very difficult, because these three parties cannot represent the whole problem, and simple calculation contains many errors.

Generally speaking, under the same vehicle weight, the greater the maximum torque of the engine, the better the acceleration performance of the vehicle. Under the same engine torque, the smaller the vehicle weight, the better the acceleration performance. However, many comparable factors are ignored here.

1, the torque of the engine changes with the change of speed. Therefore, the maximum torque of a car is often marked at the same time as the speed. For example, the maximum torque of a car is 150 Newton meters (4000 rpm), and the maximum torque of a car is 150 Newton meters (4500 rpm), which is also the maximum torque of 150 Newton meters. At the same engine speed, the two cars accelerate.

2. The maximum torque index corresponds to the engine speed rather than the vehicle speed. The power output by the engine must be decelerated and twisted through the transmission system, and then act on the driving wheels to generate the force needed for automobile acceleration. The transmission systems of different vehicles are different, so the acceleration characteristics are not necessarily the same when the maximum torque of the engine is the same.

Not all the power of the engine is used to accelerate the car. The force F in the formula F=ma is the resultant force, including road resistance, wind resistance ... maybe the gravity that needs to be overcome to increase the potential energy of the car.

Because there are so many factors at work, and the calculation should be carried out in a way that netizens can understand, I can only answer this question on an imaginary basis: imagine that the car starts to accelerate at a uniform speed on a flat road, the combined effect of all resistances at any time is equivalent to 0. 1 of the weight of the car, the resistance at any time is 180 degrees with the driving direction of the car, and the engine speed at any time is the same.

If the required speed is accelerated from 0 to 65,438+000 km/h within 65,438+00 seconds, the required acceleration can be calculated as 2.778 (m/s/s) according to V =at, and if the vehicle mass is 65,438+0 tons, the required average driving force can be calculated as 2,778 Newton according to F=ma, and it is considered that

Because the length of the acceleration section is S=at2/2= 138.9 meters, the work consumed in the whole acceleration process is FS=524764.2 joules and the power is 52,476.4 watts. If the engine speed is always 4000 rpm (which is practically impossible), the required torque can be calculated as 52476.4/(4000 * 2 * 3.14159/60) =125.3 (Newton meters).

If the previous assumptions remain unchanged, it can be seen from the above calculation process that the required torque is directly proportional to the vehicle weight. That is to say, if the vehicle weight increases or decreases, the required torque will increase or decrease in proportion. Other car weights can be calculated by netizens themselves.