1
00:00:00,090 --> 00:00:06,630
Before we have seen Kubik and fifth order polynomial trajectories based on the calculations of these

2
00:00:06,630 --> 00:00:11,700
two trajectories, we can get higher order polynomial trajectories very easily.

3
00:00:12,300 --> 00:00:18,450
Additionally, we have noted that other degree of polynomial increases smoothness of the trajectory

4
00:00:18,450 --> 00:00:22,920
also increases, but the number of coefficients to be calculated also increases.

5
00:00:23,400 --> 00:00:27,780
So in this lesson, we will try to get benefits of both of these techniques.

6
00:00:28,140 --> 00:00:32,280
How we will mix different degree of polynomial trajectories.

7
00:00:32,580 --> 00:00:39,660
The simplest and popular one is trapezoidal trajectory or with other name linear segment with parabolic

8
00:00:39,660 --> 00:00:40,170
blends.

9
00:00:40,410 --> 00:00:43,800
You will see the reason of this naming as we go further.

10
00:00:45,330 --> 00:00:49,500
Here is the position, velocity and acceleration profiles for this trajectory.

11
00:00:49,890 --> 00:00:52,650
As you can see, it consists of three phases.

12
00:00:52,980 --> 00:00:58,860
The first and third phases are second order polynomials, namely parabola, while the second phase is

13
00:00:58,860 --> 00:00:59,670
just the line.

14
00:01:00,210 --> 00:01:03,990
That's why it is called linear segment with parabolic blends.

15
00:01:04,290 --> 00:01:08,460
If you look to the velocity profile, it is in the shape of trapezoid.

16
00:01:08,730 --> 00:01:12,450
So another name for this trajectory is trapezoidal trajectory.

17
00:01:13,200 --> 00:01:20,520
Another thing you can immediately notice is that while this trajectory provides smoothness in velocity,

18
00:01:20,730 --> 00:01:23,220
the acceleration profile is not continuous.

19
00:01:23,640 --> 00:01:26,790
OK, now let's analyze each segment one by one.

20
00:01:28,470 --> 00:01:31,530
The first phase is second order polynomial function.

21
00:01:31,800 --> 00:01:34,170
It represents constant acceleration.

22
00:01:34,560 --> 00:01:41,240
The time required to finish the first phase is named as acceleration time and denoted by 2A.

23
00:01:41,580 --> 00:01:47,730
You can notice it from acceleration profile, while the second phase is a deceleration phase.

24
00:01:48,570 --> 00:01:52,140
Excuse me, while the second phase is a linear function.

25
00:01:52,380 --> 00:01:55,260
So zero acceleration, but constant velocity.

26
00:01:55,770 --> 00:01:59,100
The third phase is caused them deceleration phase.

27
00:01:59,310 --> 00:02:05,400
It's also a second order polynomial function, and the time required to finish this phase is called

28
00:02:05,400 --> 00:02:09,960
deceleration time, which is generally chosen the same as acceleration time.

29
00:02:10,200 --> 00:02:12,240
So trajectory becomes symmetric.

30
00:02:12,840 --> 00:02:13,230
OK.

31
00:02:13,260 --> 00:02:21,630
As we have see, the 8A is nothing but acceleration time and chosen the same as this celebration time

32
00:02:22,050 --> 00:02:26,400
and T is the overall time required for tracking or executing the trajectory.

33
00:02:26,910 --> 00:02:30,990
T is equal to the difference between final time and the initial time.

34
00:02:31,470 --> 00:02:33,660
We will assume initial time as zero.

35
00:02:33,990 --> 00:02:36,660
Otherwise, you can include them very easily.

36
00:02:37,080 --> 00:02:41,310
Let's deep dive into each phase and analyze equations.

37
00:02:41,430 --> 00:02:43,320
Let's start with acceleration phase.

38
00:02:43,350 --> 00:02:50,190
Namely, time range between zero and T, as this is second order polynomial, orientation, velocity

39
00:02:50,190 --> 00:02:51,060
and acceleration.

40
00:02:51,060 --> 00:02:52,320
We'll have this structure.

41
00:02:53,190 --> 00:02:59,190
The boundary conditions are given as initial position of Q and initial velocity of zero.

42
00:02:59,940 --> 00:03:02,670
But these two conditions are not enough for us.

43
00:03:02,700 --> 00:03:07,800
We need stored condition to be able to solve for a long time coefficients.

44
00:03:08,430 --> 00:03:15,540
Let's analyze this velocity graph as you can see a final velocity for the first phase, namely velocity.

45
00:03:15,550 --> 00:03:21,210
That time T A is equal to the velocity of constant velocity phase or second phase.

46
00:03:22,080 --> 00:03:25,980
So we can write this third and final boundary condition.

47
00:03:26,790 --> 00:03:33,930
Now, as we have enough boundary conditions to solve for unknown coefficients, we can write below equations

48
00:03:33,930 --> 00:03:35,040
in matrix form.

49
00:03:35,040 --> 00:03:36,810
And so for Vector A..

50
00:03:38,070 --> 00:03:43,920
Let's continue with constant velocity phase in constant velocity, phase orientation, velocity and

51
00:03:43,920 --> 00:03:44,670
acceleration.

52
00:03:44,670 --> 00:03:45,720
We'll have this form.

53
00:03:46,590 --> 00:03:50,370
As you can see, acceleration is zero in this phase.

54
00:03:50,610 --> 00:03:52,980
Now let's determine boundary conditions.

55
00:03:53,280 --> 00:03:56,910
We can determine the initial position from Equation 1.0.

56
00:03:57,390 --> 00:03:59,880
This formula is from basic physics.

57
00:04:00,210 --> 00:04:05,730
Well, you can get this equation even from the first phase orientation equation.

58
00:04:07,270 --> 00:04:16,450
Just plug it instead of tea, because the initial time for a second phase is to say no.

59
00:04:16,810 --> 00:04:18,460
Again, analyze this graph.

60
00:04:18,820 --> 00:04:24,250
We need a second boundary condition also to solve for unknown coefficients.

61
00:04:24,850 --> 00:04:31,120
And this is the initial velocity for this phase, namely Cuvee Dot, which is imposed by us.

62
00:04:31,810 --> 00:04:38,380
OK, after we have obtained the boundary conditions, let's go for our long coefficients again.

63
00:04:39,250 --> 00:04:43,900
Not the last phase is deceleration phase, which is similar to the acceleration phase.

64
00:04:44,260 --> 00:04:46,900
Again, let's determine the boundary conditions.

65
00:04:46,990 --> 00:04:50,440
The first one is final position or orientation.

66
00:04:50,980 --> 00:04:53,620
The second one is final velocity, which is zero.

67
00:04:53,980 --> 00:05:00,850
And the third one is initial velocity, which is nothing but the velocity at time.

68
00:05:01,180 --> 00:05:03,250
T a t minus T.

69
00:05:03,490 --> 00:05:04,180
Excuse me.

70
00:05:04,360 --> 00:05:10,060
This velocity is the same as the velocity at the end of the second phase, and this is constant the

71
00:05:10,060 --> 00:05:11,950
velocity that we have imposed.

72
00:05:12,640 --> 00:05:19,720
We cannot determine the unknown coefficients easily after we have analyzed each phase individually.

73
00:05:19,990 --> 00:05:24,730
In the second part, we will take into account some other issues also.
