‫Welcome back.
‫In this video, I will give you an example that should really make clear what we need those rotation
‫mattresses for.
‫Let's take our example in which we rotated pie over to radiance about each axis and let's use the R,
‫X, Y, Z convention.
‫So you have your inertial frame here and you also have your body frame here that is perfectly aligned
‫with the inertia frame.
‫Let's imagine that we have our drone here that has this body frame attached to it.
‫Remember, when dealing with orientation, then it does not matter how the origins of the frames are
‫positioned with respect to each other.
‫What matters is how the axis of the frames are oriented with respect to each other.
‫So we align their origins.
‫And so here you can see that the inertial and body frames are perfectly aligned with each other.
‫And let's say that the drones angular velocity vector that has three components P, Q and R Radians
‫per second.
‫These are the three components of the angular velocity vector.
‫Let's say that each component has a value of one radians per second.
‫So we can say that Omega with respect to the body frame, which is our angular velocity vector, which
‫has these components, has these values for each component and of course it's in radians per second.
‫Remember P q r r the angular velocities in the body frame, you measured them in the body frame.
‫And so I'm going to switch to green now.
‫And if P equals one radians per second, well then that's your pride and your one would be here then
‫that would be.
‫Q again your one would be here and that would be R again, your one would be here.
‫And so we can try to draw our resultant vector, which would be this one, the purple one.
‫That's the result.
‫And vector in three D has three components, one in the body frame X dimension one in the body from
‫Y dimension and one in the body frame Z dimension.
‫And this purple vector will that's your Omega.
‫With respect to the body frame, it's this one here in the coordinates of this resulting vector in the
‫body frame is one one and one.
‫Now a question.
‫What would the coordinates of this vector be in the inertial axis?
‫Well, in this case, this is your inertial frame and this is your body frame.
‫You can see that they're equal to each other in terms of orientation.
‫Therefore, if you are, which are measured in the body frame, are one one one, which are also the
‫coordinates of this result vector, then the coordinates of this resultant vector in the inertia frame
‫is also one one one.
‫So that's pretty easy.
‫But now let's rotate about the inertial x axis by plus PI over two radians.
‫So now our body frame is like this.
‫However, the P Q are angular velocity components.
‫They are still one radians per second since p q are variables are measured in the body frame, then
‫this would be your P in the body frame X direction, this would be your Q in the body frame Y direction
‫and this would be your R in the body from Z direction.
‫And so that means that this would be your resultant vector, this would be your omega be the coordinates
‫of Omega B are still one, one and one because this Omega B is your angular velocity vector in the body
‫frame.
‫And so you measure it with respect to the body frame.
‫By the way, the angular velocity vector is usually called small.
‫Omega, but because it looks like W. at some point in the course, I accidentally call it W. but I'm
‫still talking about the same thing.
‫So either I call it Omega or W. is still the same angular velocity vector in the body frame.
‫I'm just telling you that so that you would not get confused.
‫But OK, so now what would the coordinates of this omega vector be in the inertial frame?
‫Of course, it's easy to see what it is.
‫You can clearly see that in the inertia frame it's X equals one because your piece here, then Y equals
‫minus one because your R is here.
‫But now in the inertial frame, R is in the negative Y direction.
‫This is your positive Y here and your R is in the negative both direction.
‫And your Z of course is one.
‫But let's use rotation matrices to find that answer.
‫Since we rotated plus pi over to radians about the inertial x axis, we take our R X matrix which looks
‫like this.
‫And by the way, C means cosine and X means sine and then we plug in plus power to radians into these
‫angles here, into our FY angles.
‫If we do that then cosine pi over two is zero sine power, two is one, the minus sign power two is
‫minus one and then cosine power two is zero.
‫And so our accordionists in the inertia frame equals r x times our coordinates in the body frame then
‫our r x matrix for power to radians times the actual values of P are and of course P Q are.
‫They are always measured in the body frame and if you multiply this matrix by this vector, then you
‫will get one minus one and one.
‫You see through this rotation matrix we found the coordinates of this omega B vector in the inertia
‫frame and just like you expected, you see X equals one, just like here, Y equals minus one just like
‫here, and Z equals one just like here.
‫But let's go further.
‫Let's rotate pi over to Radians about the inertial y axis.
‫So this rotation would also be pi over to radians like this.
‫Now our body frame is like this and again P Q are equal one radians per second.
‫Now this one would be P, this one would be Q and this one would be R because P, Q are measured in
‫the body frame.
‫And so that means that the resulting vector would be something like this.
‫That would be your Omega B again, it's easy to see what kind of coordinates this vector has in the
‫inertia frame.
‫In the body frame it's one one one.
‫But in the inertia frame it's one minus one and minus one.
‫You can see that this vector has a positive component in the X direction, a negative component in the
‫Y direction, and then a negative component in the Z direction because Q points in the positive direction
‫then are points in the negative Y direction and then P points in the negative Z direction when we talk
‫about the inertia frame.
‫But again, let's use the rotation matrices.
‫In this case, the inertia frame coordinates equal, the matrix are Y times the Matrix R X and then
‫coordinates in the body frame.
‫So it's our Y times, our X times this vector here.
‫So R Y equals this one.
‫But since Seeta equals Pi over.
‫To radiance, then this matrix will be like this cosine pi over two will be zero, then zero.
‫Here, here, this one will be minus one.
‫Then you will have zero one zero.
‫And then here you will have one zero and then zero.
‫And now you have to compute this thing here.
‫So are Y times are X equals.
‫This is our Y, this is our X.
‫So again, this is our Y and our X.
‫And so if you multiply them together and I'm just going to write it here, OK, this will be their product,
‫that's your R Y times, our X.
‫And so now you write it here again, that's your r y times are X times this vector.
‫And if you perform the matrix vector multiplication then you will get here one minus one and minus one.
‫And as you can see, you got what you expected, your omega vector that has the coordinates of one one
‫one in the body frame in the inertia frame.
‫However, they are one minus one and minus one.
‫And finally, let's rotate about the inertial Z axis by plus pi over to radians here.
‫I would ask you to do it yourself after you have rotated pi over to Radians about the inertial Z axis
‫peak, you are will still be one one and one radians per second.
‫Remember p q r r angular velocities in the body frame X, Y and Z respectively.
‫So your task is to use rotation matrices to find the coordinates of this omega B resultant vector in
‫the initial frame after you had rotated pi over to radians about the initial Z axis, the zero rotation
‫matrix r z.
‫It looks like this.
‫So try to use the same procedure and we will look at the solutions in the next video.
‫Thank you very much.