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Hello everyone, here is the "Car Story Collection" at the top of the Purple City, welcome all the friends to watch the wonderful content of this issue.
In this issue, I'm going to tell you something new.
Dynamic characteristics of automobile driving (Part 2)
7. The method of the car's climbing ability
Hill climbing ability refers to the ability of a car to overcome the slope of the road while driving. At constant speed, dv/dt=O, so Pk=F1+Fw+F, F=Pk-1(F+Fw)
P,=M·I·i,·m
Because:
F=W·f·cosar=W·f (since the value of a is small, it can be roughly assumed that casa = 1)
Fw=C,AV2;F.=W·sina
Fw=21.15
So:
W.sina=Me·ik·lo·n-w.f-CĐAv.2
a=arcsin(M·i·i·) /r-W·f-CAV2121.15
The maximum climb of the car is in the gearbox first gear, but the maximum climb of the direct gear should also be noted, otherwise the car in the direct gear, encounter a smaller slope to often shift gears, which affects the average speed of driving, the driving resistance curve under various slopes can be seen in Figure 1-5.
The climbing ability of each gear is shown in Figure 1-6.
8. The method of automobile acceleration ability
By the formula: Pk=W.f +W.i+ CoAV2_ (W+ AW) dv
可知:wF=M-ikion-W·f-W·i-CoAv2
According to the above equation 12, the torque value M on the external characteristic curve of the engine is substituted point by point according to the speed, and after conversion, the acceleration curve of each gear can be obtained when driving on the flat road, see Figure 1-10.
Some vehicles due to the difference in gearbox structure, the first gear AW is very large, the result is sometimes the acceleration of the second gear will be greater than the first gear.
Since the value of acceleration is not easy to measure, the acceleration time is generally used to indicate the acceleration ability of the car.
If driving in direct gear, the time required to accelerate to 80% Vmm at the lowest stable speed is expressed.
The simple calculation is described below:
t= V1' Ldy
From kinematics, it can be seen that the acceleration a = dv / dt, so dt = l / a dv, that is, the time in the acceleration process of speed from V to V2 is equal to the integral of 1/a and dv.
1/a is the reciprocal of the acceleration, as long as according to Formula 12, according to the speed Va to find a = dv / dt, you can find its reciprocal 1 /a, and then you can draw the acceleration reciprocal 1/a with the speed V change curve (see Figure 1-11a), you can use the graphical method to find the acceleration time.
[(W+AW) see tables (1-2), (1-3), (1-4)].
When performing the plot integration, the velocity interval can be divided into several intervals, (often taken as 5Km/h) and the areas A1, A2, A3 are determined, respectively. etc., as shown in Figures 1-11b.
rule:
t,=A1/3.6ab (s) 。
t= (A1+A2) /3.6ab (s) 。
t= (A1+A2+A3+....+An) /3.6ab (s)。
Note: On the coordinate diagram, amm is used to represent 1Km/h, bmm is used to represent 1s2/m, (1Km/h) × (1s2/m) = 1s/3.6.
By marking the resulting acceleration time in the t-V coordinate system according to the corresponding speed coordinates, the highest acceleration time map can be obtained, and the acceleration time map from the beginning of the 1st gear can be continuously shifted to the highest level (see Figure 1-12).
According to general statistics, when using a gear, it should be able to produce an acceleration of 1.7 to 2.0m/s2, and an acceleration of 0.25 to 0.5m/s2 when using high-grade, which is also applicable to trucks.
9. Backup driving force
In Figure 1-8, two road resistance curves are shown, namely two curves of flat road (a=0 and slope (a=al)."
There are also three gears of drive force curves.
The maximum speed is at the intersection, Vmax.
When the gearbox is driven on a flat road with three gears at the speed V, the throttle does not need to be fully opened, so it has a residual or backup driving force, which is equal to the figure shown (Pk3-F1).
If driving on a ramp (a=a1), there is no backup drive force, i.e. the throttle is fully opened to achieve equilibrium.
(i.e. Pk3=F2).
However, when driving with the second speed V1, there is a reserve force on the slope (P2-F2).
So the lower the gear, the greater the driving force.
The greater the backup drive.
However, when it should be changed from low-grade to high-grade without changing, the car's backup driving force remains for a long time, and the engine is running at a high speed under low gear, resulting in a substantial increase in fuel consumption and cannot be increased, this phenomenon is called "drag gear" by the driver, which should be avoided as much as possible.
In order to overcome the driving resistance, the car generates a driving force Pk(N) to move forward, and at the same time at a certain speed Va (Kw/h), so that the required driving power Nk(Kw) should be:
Nk=P V /3600 (Kw) .. ①
At this time, due to the mechanical efficiency of the drive train n, the power output of the engine is: N = N / n = Pk · V /3600·n .... ②
Because: P1 = rolling resistance + air resistance + uphill resistance + acceleration resistance CoAV?
e-+W+ (W+AW) /g· (dv/dt)=W+21.15
Rolling resistance power N = WV/ 3600
Air resistance power N=C·A· V.3/21.15×3600
Uphill resistance power N=W· V/3600
Acceleration power N= (W+AW)/g· (dv/dt)· V/3600
So: engine output power N = (N + N + N + Na) / CAV?
=(W.f+-+W·i+ (W+AW)/g· (dv/dt))· V。 (1/3600n)
This is the balance between the engine output power and the driving resistance power of the car.
Figure 1-13 is to express the output power of the engine and the power consumed by the driving resistance of the car in longitudinal coordinates, and the horizontal coordinates indicate the speed of the vehicle.
Due to the different gearbox gears, the relationship between engine speed and vehicle speed varies depending on the speed ratio.
Low speed in low gear, the speed range occupied is narrow, such as n1, high speed is high, and the speed range occupied is wide, such as nm.
However, the power size of the engine in each gear (only a little difference in mechanical efficiency) and the morphology of the power curve are basically unchanged, so the power curves of each gear are arranged horizontally on the coordinate diagram (high-end moves to the right).
At the same time, the resistance power curve of the car is drawn on the coordinate diagram.
When driving at constant speed on a flat road, the resistance power generally includes rolling resistance power N, and wind resistance power Nw, which is converted to the demand for the engine and consumes power (N/N)/n, which is the superposition of two kinds of resistance power, a function of vehicle speed at low speed, mainly overcoming wind resistance at high speed, and the power is a three-time function of vehicle speed, which is a curve with increasing slope as the speed increases.
As can be seen from the figure, the speed corresponding to the intersection of the engine power curve and the resistance power curve at the highest level is the maximum speed of the car on a goodly straight road is Vamax.
At the highest level, the speed corresponding to the maximum power of the engine is V, which is generally equal to or slightly less than the maximum speed.
Therefore, the selection of engine power should first be based on the requirements of the maximum speed of the car.
When the climbing ability and maximum acceleration capacity in the car parameters (such as: transmission ratio, etc.) have not been clear, due to too many factors are often difficult to determine, so it is reasonable to use the maximum speed to select the engine power, because the maximum speed also reflects and includes the acceleration ability and climbing ability.
The high speed of the car must require the engine to have a large power, the greater the power, then at low speed, the acceleration ability and climbing ability of the car are also stronger.
The range of power selection can be seen in Table 2-1.
10. Engine backup power during car driving
When the car is driving at V isometric speed on a good road, the resistance power of the car is the bc segment (see Figure 1-13), the throttle is only partially open, so on the vertical line of V, the ab segment is the backup power of the car, which can be used to accelerate or climb, at this time, the engine power required to maintain the car is not large, the throttle opening of the engine is small, when climbing and accelerating, the driver increases the throttle, so that all or part of the backup power plays a role, the larger the backup power of the car, the better the power performance.
The backup power can be expressed by the formula N, = N- (N + N)) / n, that is, the ac-bc in the figure, (ac is the full power that the engine can emit at the speed of V', that is, N).
When accelerating there is no climb, so i = 0, the result of the acceleration at different speeds is:
a=(W+AW)Va-[N-(N+Nw)/n]
When climbing a hill without accelerating, the gradient that can be climbed at different speeds can be approximately calculated as:
i=36007[N-(N,+N)/7]wWa
The above is to use the concept of backup power to analyze the dynamic performance of the car, for example, the higher the speed, the greater the resistance encountered, the product of resistance and speed (the power consumed by the resistance) is greater, so there is not enough engine power, and it is impossible for the car to drive at high speed.
Similarly, the acceleration performance of the car and the speed of climbing need backup power, so the power performance of the car mainly depends on the rated power and a certain speed power that the car engine can output (see Figures 1-14 and 1-15).