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DERIVATIVES AND ANGULAR VELOCITY VECTOR

Consider a fixed reference frame, A, and a reference frame B. This can be related to a helicopter whose rotor is rotating with respect to the fuselage. If a parameter is given in frame B: Frame B

$\vec{Q} = Q_{x}\vec{i} + Q_{y}\vec{j} + Q_{z}\vec{k}$

The parameter could be for example the position or a velocity of a point on a rotor, given in cartesian coordinates with the origin at the hub. Let us say that we need to know the velocity of that rotor blade location, not with respect to the hub, but with respect to an observer located some distance away, and clearly obtained using natural coordinates system.

The velocity of the point can be denoted as $\dot{\vec{Q}}^{A}$ and it can be expressed in coordinate system A as:

$\dot{\vec{Q}}^{A} = (\dot{Q}_{x}\vec{i} + \dot{Q}_{y}\vec{j} + \dot{Q}_{z}\vec{k}) +( Q_{x}\dot{\vec{i}}^{A} + Q_{y}\dot{\vec{j}}^{A} + Q_{z}\dot{\vec{k}}^{A})$

$\dot{\vec{Q}}^{A} = \dot{\vec{Q}}^{B} +( Q_{x}\dot{\vec{i}}^{A} + Q_{y}\dot{\vec{j}}^{A} + Q_{z}\dot{\vec{k}}^{A})$

So the velocity can be expressed as the velocity in B plus terms associated with the transformation from the coordinate system in B to A. Now, for a fixed orthogonal system like a Cartesian system:

$\vec{i}.\vec{i} = \vec{j}.\vec{j} = \vec{k}.\vec{k} = 1$

$\vec{i}.\vec{j} = \vec{j}.\vec{k} = \vec{k}.\vec{i} = 0$

What about $\dot{\vec{i}}^{A}, \dot{\vec{j}}^{A}, \dot{\vec{k}}^{A}$ ? Recall that the derivative of any vector results in the normal of that vector. Thus, the dot products : $\vec{i}.\dot{\vec{i}}^{A} = \vec{j}.\dot{\vec{j}}^{A} = \vec{k}.\dot{\vec{k}}^{A} = 0$ What about cross products involving different axes? Consider $\vec{i}.\vec{j} = 0$ Taking the derivative with respect to A yields, $\vec{j}.\dot{\vec{i}}^{A} + \vec{i}.\dot{\vec{j}}^{A} = 0$ Similar expressions for the other unit vectors can be obtained. Since we do not know the orientation of A and B, we cannot speculate further on the dot product relations. We can express the derivative of the unit vectors as cross products.

$\dot{\vec{i}}^{A} = \vec{\alpha}\times\vec{i}$

$\dot{\vec{j}}^{A} = \vec{\beta}\times\vec{j}$

$\dot{\vec{k}}^{A} = \vec{\gamma}\times\vec{k}$

By using cross products as follows, we can make sure the two vectors are perpendicular and ensure that the unit vectors remain unit vectors.

$\alpha = \alpha_{x}\vec{i} + \alpha_{y}\vec{j} + \alpha_{z}\vec{k}$

$\beta = \beta_{x}\vec{i} + \beta_{y}\vec{j} + \beta_{z}\vec{k}$

$\gamma = \gamma_{x}\vec{i} + \gamma_{y}\vec{j} + \gamma_{z}\vec{k}$

Then α×i, β×i and γ×i are given by:

$\vec{\alpha}\times\vec{i} = \dot{\vec{i}}^{A} = \alpha_{z}\vec{j} - \alpha_{y}\vec{k}$

$\vec{\beta}\times\vec{j} = \dot{\vec{j}}^{A} = -\beta_{z}\vec{i} + \beta_{x}\vec{k}$

$\vec{\gamma}\times\vec{k} = \dot{\vec{k}}^{A} = \gamma_{y}\vec{i} - \gamma_{x}\vec{j}$

Let us observe what happens with the mixed dot products:

$\vec{\gamma}\times\vec{k} = \dot{\vec{k}}^{A} = \gamma_{y}\vec{i} - \gamma_{x}\vec{j}$