1
00:00:00,630 --> 00:00:10,510
‫And now, since we have our equations for our Idon, Jadot and K that we can write down our translational

2
00:00:10,530 --> 00:00:13,500
‫equations of motion in the body frame.

3
00:00:14,940 --> 00:00:17,670
‫So instead of writing the equation like this.

4
00:00:19,130 --> 00:00:26,360
‫And we said that this entire thing here was the translational acceleration in the body frame.

5
00:00:27,630 --> 00:00:36,270
‫We can right this entire thing like this, this part stays the same, but this part can be written like

6
00:00:36,270 --> 00:00:36,750
‫this.

7
00:00:37,890 --> 00:00:47,790
‫Where this is IDOT, this is Jadot, and this is K Dot, the derivatives of unit vectors with respect

8
00:00:47,790 --> 00:00:59,130
‫to time, linear algebra allows you to rewrite this yellow part like this, where you, the N.W., they're

9
00:00:59,130 --> 00:01:00,440
‫just like constants.

10
00:01:01,080 --> 00:01:06,900
‫And so when you have this cross product, you can just take this constant and then you can put it here,

11
00:01:07,230 --> 00:01:09,290
‫here and here.

12
00:01:10,110 --> 00:01:11,910
‫And so now you have this form.

13
00:01:12,750 --> 00:01:15,480
‫Remember, again, I'm going to emphasize it.

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00:01:15,870 --> 00:01:19,760
‫This W here, it's a scalar.

15
00:01:20,340 --> 00:01:27,480
‫It's the translational velocity in the Z direction, in the body frame, in metres per second.

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00:01:28,380 --> 00:01:32,370
‫It is part of this Vehbi velocity vector.

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00:01:33,000 --> 00:01:36,240
‫It had three components you the N.W..

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00:01:37,190 --> 00:01:40,970
‫Like this, you see, you've N.W., it's this one.

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00:01:41,880 --> 00:01:51,880
‫But the W vector in the B frame, it's an angular velocity vector in the body frame in radians per second.

20
00:01:52,590 --> 00:01:54,240
‫It's this one here.

21
00:01:54,960 --> 00:02:01,800
‫It tells you the angular velocity is about the body frame, X, Y, Z axis, respectively.

22
00:02:02,790 --> 00:02:06,000
‫And so we can take now this part here.

23
00:02:07,000 --> 00:02:12,800
‫And we can just ride it like this, it's equivalent, right?

24
00:02:12,850 --> 00:02:17,620
‫So you have your eye unit vector J unit vector and K unit vector.

25
00:02:18,690 --> 00:02:29,220
‫And you can rewrite this portion in this form without the unit vectors and let's call it the vector

26
00:02:29,220 --> 00:02:41,160
‫in the body frame X, Y, Z, in order not to confuse it with this one here, because this vector here,

27
00:02:41,850 --> 00:02:46,650
‫you get this when you're either Jadot and Cadart are zero.

28
00:02:46,680 --> 00:02:53,460
‫In other words, when your body frame is not rotating with respect to your inertial frame.

29
00:02:54,350 --> 00:02:59,480
‫You get it when your frame is not rotating in which you're measuring your work.

30
00:03:00,540 --> 00:03:04,410
‫And the other part which is here, we can write it like this.

31
00:03:05,420 --> 00:03:13,490
‫We can just factor out our WB vector here and put it like this and then cross.

32
00:03:14,490 --> 00:03:30,230
‫UI plus the J plus W look like this, or equivalently, we can also write this portion like this, you

33
00:03:30,300 --> 00:03:41,820
‫the W transposed or even in a more compact way, WB Cross and then this thing, we just write it like

34
00:03:41,820 --> 00:03:45,030
‫this, the vector in the body frame.

35
00:03:45,990 --> 00:03:50,400
‫And so this term you get because your body frame is rotating.

36
00:03:50,400 --> 00:03:56,970
‫With respect to the inertia frame, if your body frame didn't rotate with respect to the initial frame,

37
00:03:56,970 --> 00:04:02,970
‫then this yellow term would be zero and you would only be left with this blue part.

38
00:04:03,930 --> 00:04:10,380
‫And so the final form of our translational equation of motion is the following.

39
00:04:11,280 --> 00:04:21,750
‫You have the net force vector equals mass, the acceleration of the velocity vector, assuming that

40
00:04:21,750 --> 00:04:26,490
‫the body frame is not rotating with respect to the initial frame.

41
00:04:27,500 --> 00:04:36,230
‫Plus, the correction factor that takes into account the fact that the body is rotating with respect

42
00:04:36,230 --> 00:04:40,750
‫to the inertia frame, so this term corrects for that fact.

43
00:04:41,790 --> 00:04:48,030
‫The first term assumes that the body frame is not rotating, but in reality it is.

44
00:04:48,360 --> 00:04:55,860
‫So that's why the second term, it's called a correction factor and it corrects for the fact that the

45
00:04:55,860 --> 00:04:59,390
‫body frame actually is rotating with respect to the inertia frame.

46
00:05:00,270 --> 00:05:06,690
‫And so let me also just ride it out in a larger way so that it would be more intuitive for you.

47
00:05:07,780 --> 00:05:10,160
‫This force vector looks like this.

48
00:05:10,540 --> 00:05:16,860
‫And by the way, this is also in the body frame, so it has three elements here.

49
00:05:17,870 --> 00:05:24,440
‫And you can see that it's the body frame, because I'm using small X, Y and Z letters here, and so

50
00:05:24,440 --> 00:05:26,930
‫all that equals mass times.

51
00:05:28,010 --> 00:05:30,050
‫This vector would look like this.

52
00:05:30,330 --> 00:05:39,050
‫Plus, this is the angular velocity vector cross you, the N.W..

53
00:05:40,140 --> 00:05:41,040
‫And there you go.

54
00:05:41,070 --> 00:05:47,100
‫This is your dynamic's equation for translational motion in the body frame.

55
00:05:48,060 --> 00:05:58,590
‫OK, so this equation is just to describe how the drone translates in space, however, if it only translated,

56
00:05:59,040 --> 00:06:08,490
‫then it would only be a three degree of freedom system where your independent variables are X, Y and

57
00:06:08,490 --> 00:06:08,870
‫Z.

58
00:06:09,780 --> 00:06:15,410
‫But the drone doesn't only translate in space, it also changes its attitude.

59
00:06:16,080 --> 00:06:26,340
‫You also have Phi Theta and BPCI Angles as your independent variables, which are oilor angles in our

60
00:06:26,340 --> 00:06:33,300
‫convention and then makes the drone a six degree of freedom system three four position.

61
00:06:34,270 --> 00:06:42,070
‫And three for orientation, and so we also need to derive fundamental dynamics, equations to describe

62
00:06:42,070 --> 00:06:44,010
‫the drones attitude in space.

63
00:06:44,890 --> 00:06:54,220
‫What we have done here was basically force equals mass times acceleration only in the body frame, which

64
00:06:54,220 --> 00:06:56,880
‫rotated with respect to the inertia frame.

65
00:06:57,760 --> 00:07:09,190
‫But now we also have to do moment equals mass moment of inertia times the angular acceleration, and

66
00:07:09,190 --> 00:07:13,960
‫that can also be done in the inertial frame and in the body frame.

67
00:07:14,870 --> 00:07:20,900
‫And we're going to look at both ways and then we're going to end up doing it in the body frame, which

68
00:07:20,900 --> 00:07:24,240
‫also rotates with respect to the inertia frame.

69
00:07:25,100 --> 00:07:27,470
‫And we will do that in the next video.

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00:07:28,070 --> 00:07:29,310
‫Thank you very much.

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00:07:29,330 --> 00:07:30,170
‫And see you there.