Robot Motion Modelling

Chapter 3 Project / System Design

3.6 Robot Motion Modelling

This project is aimed to study the multi-mobile robot control framework.

Therefore, the robots are low fidelity because it is not necessary to consider the physical or mechanical part of robots to study the multi-mobile control framework. But, it's still necessary to consider some physical constraint such as the velocity of robots in a different situation because it might affect the system performance significantly. For example, robots need to slow down if it needs to make a turn, it cannot exceed a certain velocity due to the hardware limitation. For simplicity, we assume all the robot motion is linear motion. The list below shows the information that required to capture the effect of robots' motion:

• These exists a maximum acceleration for each robot, a_max.

• These exists a maximum deceleration for each robot, d_max.

• These exists a maximum velocity for each robot, v_max.

• The velocity of a robot should be zero before it makes a turn.

• If a robot is not allowed to move to next grids, it should stop at the last grid it can reach.

Due to the discrete-event simulation is used, we skip the transition state when a robot is travelling from one grid to another grid but focus only on the event when the robot reaches the next grid. Thus, robots are considered jumping from one grid to another after certain amount of time 𝑡, and the velocity model of the robot is then used to calculate what the exact value of 𝑡. Besides, during the interval time 𝑡, both current grid and next grid is blocked. This situation is because the moving robot occupied part of the current grid and next grid. Since one grid can only occupy one robot, then current grid should be blocked until the robot completely move to next grid.

The velocity model is described as follows. We assume robots always try to move as fast as possible, and it always used maximum acceleration to increase its velocity and maximum deceleration to decrease its velocity. The objective for robots is to minimize the time needed for a robot to move from start vertex g₀ to end vertex g_n. Apart from that, the robot needs to stop at 𝑔_𝑛, that means it should slow down at certain position in its moving path, let say g. Also, if possible, the robot needs to increase its

speed in order to reduce the time needed for it to reach g_n, but note that the maximum velocity is limited by v_max.

Now, we can reformulate this situation in another way. Let t₀ denotes the time when the robot reaches g₀, t₁ is the time when the robot reaches maximum velocity but if the robot never reaches maximum velocity then t₁ is infinity, t₂ is the time when the robot starts to slow down its velocity, 𝑡_𝑓 is the time when the robot reaches g_n. Besides, we let v₀ denotes the initial velocity when the robot reaches g₀, and let s_g denotes the distance from one grid to another. Figure 3.3 shows how to get the value of 𝑡₁ and 𝑡₂ and Figure 3.4 shows the algorithm used to calculate the time needed for a robot from one to another grid.

After that, we can use the method shown in Figure 3.3 to know the value of t₁, t₂. Every time the reserved grids of a robot change, such as a grid is released or success to request a new grid, we can recalculate the t₁ and t₂ to make sure the robot can move in maximum speed subject to it able to stop at the last reserved grid or stop at the reserved grid before it makes a turn. On the other, it is necessary to know what is the transition time for a robot moves from one grid to another and this is the objective to design the algorithm shown in Figure 3.4.

Let:

1. Assume velocity of robots are not limited.

2. 𝑡^′: time spent in acceleration.

3. 𝑡^′′: time spent in deceleration.

4. Total time needed: 𝑡 = 𝑡^′+ 𝑡^′′. 5. Total distance move: s.

Robots required to decrease their velocity to 0 once they reach last grid in their path.

Thus, 𝑡 = 𝑡^′+ ^{𝑎𝑚𝑎𝑥𝑡′+𝑣0}

𝑑_𝑚𝑎𝑥 , and s =¹

2(𝑣₀𝑡′ + 𝑎_𝑚𝑎𝑥𝑡′²) +¹

2(𝑎_𝑚𝑎𝑥(𝑡 − 𝑡^′′)²)

= ¹

2(𝑣₀𝑡′ + 𝑎_𝑚𝑎𝑥𝑡^′2) +¹

2(𝑎_𝑚𝑎𝑥(^{𝑎𝑚𝑎𝑥𝑡′+𝑣0}

𝑑_𝑚𝑎𝑥 )²)

Value of s is known, thus, by solving the equation above, we can get the value of 𝑡′ (always take the positive and real number), and therefore both t and 𝑡^′′ can be solved. However, this calculation is not always correct due to the first assumption. In fact, robots are limited by speed v_max. Therefore, we need to validate whether the robot exceeds the speed v_max.

A robot does not exceed the speed 𝑣_𝑚𝑎𝑥 if 𝑎_𝑚𝑎𝑥𝑡^′+ 𝑣₀≤ 𝑣_𝑚𝑎𝑥. If true, we can use the previous solution but if it is false, we need to introduce a new variable 𝑡^′′′ which indicate the total time the robot is remaining its maximum speed. Then, the 𝑡′, 𝑡′′, 𝑡′′′ are calculated by 𝑡′ = (𝑣_𝑚𝑎𝑥– 𝑣₀)/𝑎_𝑚𝑎𝑥

𝑡′′ = 𝑣_𝑚𝑎𝑥/𝑑_𝑚𝑎𝑥

𝑠’ = 𝑠 −¹

2(𝑣₀+ 𝑣_𝑚𝑎𝑥)𝑡^′−¹

2(v_max)𝑡′′

𝑡′′′ = 𝑠/𝑣_𝑚𝑎𝑥

Figure 3.3

Time needed for robots to acceleration / deceleration / remain maximum speed.

INPUT: 𝐚𝐦𝐚𝐱, 𝐝𝐦𝐚𝐱, vnow, vmax, tnow, 𝑡1, 𝑡2, 𝑠

OUTPUT: Time needed for robot to reach next grid; New velocity.

CALC_TIME_NEEDED () 1. IF t_now< t₁

2. result = Max ( Solve t : s = v_nowt +¹

2𝒂_𝒎𝒂𝒙𝑡² ) # Ignore Complex Number.

3. IF t_now+ 𝑟𝑒𝑠𝑢𝑙𝑡 > t₁

4. t^′= 𝑡_𝑛𝑜𝑤+ 𝑟𝑒𝑠𝑢𝑙𝑡 − t₁

5. 𝑣^′= v_now+ a_max𝑡^′

6. 𝑠^′= 𝑠 −¹

2(𝑣𝑛𝑜𝑤+ 𝑣′)𝑡^′

7. t, v_new= CALC_TIME_NEEDED(a_max, d_max, v’, v_max, 𝑡₁, t₁, t₂, 𝑠’) 8. RETURN t + t^′, vnew

9. ELSE

10. 𝑣_𝑛𝑒𝑤= 𝑣_𝑛𝑜𝑤+ 𝑎_𝑚𝑎𝑥𝑡^′ 11. RETURN 𝑟𝑒𝑠𝑢𝑙𝑡, 𝑣_𝑛𝑒𝑤 12. END IF

13. END IF

14. IF t₁≤ t_now< t₂ 15. result = s/v_max 16. IF t_now+ result > t₂

17. 𝑡^′= 𝑡_𝑛𝑜𝑤+ 𝑟𝑒𝑠𝑢𝑙𝑡 − 𝑡₁ 18. 𝑠^′= 𝑠 − 𝑣_𝑚𝑎𝑥𝑡^′

19. 𝑡, 𝑣_𝑛ew= CALC_TIME_NEEDED(𝑎_𝑚𝑎𝑥, 𝑑_𝑚𝑎𝑥, 𝑣_max, 𝑣_𝑚𝑎𝑥, 𝑡₂, 𝑡₁, 𝑡₂, 𝑠’) 20. RETURN 𝑡 + 𝑡^′, 𝑣_𝑛e𝑤

21. ELSE

22. RETURN 𝑟𝑒𝑠𝑢𝑙𝑡, 𝑣_𝑛𝑜𝑤 23. END IF

24. END IF 25. IF 𝑡_𝑛𝑜𝑤≥ 𝑡₂

26. result = Max ( Solve t : 𝑠 = v_now𝑡 −¹

2𝑑_𝑚𝑎𝑥𝑡² ) 27. 𝑣_𝑛𝑒𝑤= 𝑣_𝑛𝑜𝑤− 𝑑_𝑚𝑎𝑥𝑡^′

28. RETURN 𝑟𝑒𝑠𝑢𝑙𝑡, 𝑣_𝑛e𝑤 29. END IF

Figure 3.4

Algorithm to calculate time needed for a robot to move from one to another grid.

3.7 Multi-Mobile Robot System Design