Nomogram for calculating the technical and operational performance of road transport for the transport of productsA Nomogram is a graphical representation of the functional relationship between several variables (for example, vehicle performance)
It is used to find the numerical value of one of them from the given values of other indicators.
The principle of nomograph is the following. Let us assume that for a given indicator determining the daily volume of traffic Q , m and W, mkm, the distance of the loaded trip/f, km, (Figure 2.13). Then a ruler is superimposed on the corresponding points of the scale A 1 (the daily volume of cargo transportation, t) and A 2 (lrp is the distance of the loaded carriage), and at the intersection of the scale A 3 with the ruler we get the required value - the number of ton-kilometer output (t · km = W).
Fig. 2.13. Determination of the sought indicator
Consider the use of a motor nomogram using specific tasks.
Task 2.1. Calculate some technical and operational performance of the car on a pendulum route when transporting Q = 500 tons of first class cargo. Carrying capacity qа = 5 т. The distance of transportation l = 20 km, the average technical speed of the car V т = 25 km/h, unloading t pp = 40 min, working day T n = 9 hours
Fig. 2.14. Auto-transport nomogram:
l - distance of transportation, km; V m - technical speed, km/h; t д - time of movement, min; t pr - idle time for loading and unloading, min; t c - time of the ride, min; n about - the number of turns, flights; T n - dress time, h; γ is the load utilization factor; Q the number of tons transported by a car; q n - the carrying capacity of the car (nominal); W - the amount of generated t • km; l gr - distance of the loaded trip; G - the volume of transportation, t; A - number of cars; q a is the practical carrying capacity of a car
1. Determine the time of the car in one turn. To the nomogram (Fig. 2.14), we apply a right-angled triangle so that one of its legs is directed to the distance scale mark ( l gr), and the right angle touches the velocity curve at the point corresponding to 25 km/h. The desired result will be at the intersection of the second leg of the triangle with the time scale and will be 96 mm (Figure 2.15).
2. The time spent by the car for one run (& pound; s) is set as follows. From the 96 min mark, time is set aside on the time scale (Figure 2.16) for 40 minutes of loading and unloading. We get 136 minutes.
Fig. 2.15. Definition of driving time
Fig. 2.16. Determining the turnaround time
3. Determine the number of revolutions of the car on the Pob route for the day of work. We apply the ruler in such a way that it intersects the scales l dв - t e at the point corresponding to 136 min, and the scale T н at the point corresponding to 9 hours (dressed time). On the P06 scale, we find the desired value - 4.0 (Figure 2.17).
4. We find the actual load of the car. We superimpose the ruler on the scale γst at the point 1.0 and on the scale q n at point 5 (load-carrying capacity of the car) (Figure 2.18). At the intersection of the line connecting the scales γst and q n with the scale q 1 we find the result of 5.0 tons. In this case, the nominal load capacity of the car and its loading coincide, use of load-carrying capacity for a cargo of the first class is equal to unit.
5. The daily output of a car in tons is determined as follows (Figure 2.19). We put the ruler on the scale of the actual loading of the car q 1 at the point 5.0 and on the Pob scale at the point 4.0. The result is obtained at the point of intersection of the line with the scale Q = 20 tons.
Fig. 2.17. Determination of the speed of the car
Fig. 2.18. Determining the actual car load
6. Daily output of the car in ton-kilometers. We superimpose the ruler on the scale l gr at the point 20 km and on the scale Q at the point 20. The required value lies at the intersection of the line with the scale W and will be 400 ton-km (Figure 2.20).
Fig. 2.19. Definition of daily output, t
Fig. 2.20. Determination of daily output, t · km
7. Determination of the running number of vehicles for the implementation of the transportation plan. We put the ruler on the scale G at point 500 (the given volume of transportations) and on the scale Qd at the point of 20 tons. The required result will be obtained at the point of intersection of the line with the scale A and equal to 25 units (Figure 2.21).
Fig. 2.21. Determination of the number of cars
Task 2.2. Define some metrics when carrying a shipment by an annular route.
Fig. 2.22. Ring route scheme
In Fig. 2.22 the ring route A-B-V-G-A is shown. In sections A-B and D-V, the vehicle moves with cargo, and B-B and G-A-run sections of idling. It is necessary to organize the work of cars q = 4 tons with the following data: on a section A - B of 15 km length 200 tons of cargo are transported with the coefficient of using the load capacity of 0.9, the technical speed is 30 km/h, the loading and unloading - 26 min. On section B-V of 5 km length the car moves without cargo at a speed of 35 km/h. 180 km of cargo is transported on the section V-G with a length of 20 km with the coefficient of utilization of the vehicle's carrying capacity - unit, the technical speed is 25 km/h, the idle time for loading and unloading for the journey is 47 minutes. On the section G - A, the car moves without cargo at a speed of 20 km/h. The time in the attire is 14 hours.
1. We determine the driving time of the car in some sections (Figure 2.23), using the part of the motor transport nomogram showing the vehicle's running time ( t dv), the distance ( l ) and the technical speed ( V T ). Given that the movement time on the site is given for the run in both directions, the value found by the pattern is divided by two and we get the required time. For example, in section A-B, the driving time (see Figure 2.22) is 60 minutes (60: 2), and the desired time is 30 minutes.
Fig. 2.23. Determining the driving time of the vehicle in certain sections
The time in the outfit is determined by Fig. 2.24.
The number of turns of the car on the route during the stay in the outfit is by overlaying the ruler on the scale t e = 184 and the scale T n at point 14. The desired result is obtained on the scale Zo6 and will be 4.6, rounded 5 revolutions (see Figure 2.24).
Calculate the time required to complete 5 revolutions. We overlay the ruler so that it intersects on the scale the point t e = 184 and Zo6 = 5. The desired result at the intersection of the line with the scale of T. It is 920 minutes, or 15.3 hours.
Pic . 2.24. Defining the time in the outfit
Fig. 2.25. Determination of the daily output in section AB -
Let's calculate the daytime production in section A-B (Fig. 2.2e). Determine the load of the car (qnγ), where q n - the car's carrying capacity is 4 tonnes; γ - coefficient of load capacity utilization - 0,9. Therefore, the car load is 3.6 tons (4 0.9). If the car makes 5 revolutions per day, then the daily output will be 18 tons. Such a system of calculation on the section B-G, only the daily output will be 20 tons (Figure 2.26).
The development of the car in ton-kilometers on section AB (Figure 2.27). We put the ruler on the scale l gr at the point 15 and Q dn at the point 18. The desired value will be obtained on the scale W dn at the point of intersection of it with line - 270 t/km.
The development of the car in ton-kilometers on section B-D. We superimpose the ruler on the scales l gr and Qdn at the point 20. The required value will be obtained on the scale Qdn at the point of intersection with the line-400 t • km (see Figure 2.27).
Fig. 2.26. Determination of the daily output at section B-D
Fig. 2.27. Determination of daily output, t · km
Fig. 2.28. Definition of the operational number of cars
The operational number of cars is determined by the section with the lowest volume of traffic, i.е. along section B-D. We superimpose the ruler on the scale Qdn, at points 20 and 180. The required value will be found at the intersection of the line with the scale A, and will be equal to 9 (Figure 2.28). Therefore, 9 vehicles are needed to complete the transportation plan.
We overlay the ruler on the scale A, at the point 9 and Qdn at the point 18, then on the scale (Qpl, the line will intersect point 162 (see Figure 2.14).
Thus, the work of the car on the A-B-V-G-A route will be designed to carry 162 tons of cargo in the section AB and 180 tons in the section B-D. For removal of 36 tons of cargo in the section AB it is necessary to organize a pendulum route with a reverse unloaded run and perform the calculation in the sequence indicated in the solution of the first example.
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