Euler Transformation

The rotational time derivative is the pass key to the invariant formulation of time – phased dynamic systems. As an operator that preserves the tensor characteristic, it depends on a reference frame. Sometimes it is desirable to change this reference frame, for instance from inertial to body frame. Euler’s generalized transformation governs that change of frame.

We approach the derivation of this vital transformation fastidiously in increasing complexity. An ad hoc introduction points to the possible formulation, followed by a heuristic derivation based on the isotropy of space. The eggheads among you are referred to Appendix D, which provides an analytical proof.

The classical Euler transformation is embodied in Eq. (4.34). It transforms the time derivative from frame A to frame B. Comparison with Eq. (4.35) leads to the conjecture that

so that the last term of Eq. (4.35) reflects the vector product [f2BA]B[s]B

Introducing the rotational time derivatives for the two special cases [ds/dt]A = [Dasa and [d. s/dtjft = [DBs]B, we can formulate

[DA5]A = [T]ab([Dbs]b + [£2BA]B[s]B)

The key question is whether this transformation is a tensor concept. As written, it holds only for the associated coordinate systems ]A and Jft. Fortuitously, it can be generalized for any allowable coordinate system. We call it Euler’s general transformation, or just plain Euler transformation.

Theorem: Let A and В be two arbitrary frames related by the angular velocity tensor ПВА. Then, for any vector x the following transformation of the rotational time derivatives holds:

Dax = DBx + ElBAx (4.51)

Proof: The theorem is a direct consequence of the isotropic property of space. lt’ReA is the rotation tensor of frame В wrt frame A, then, because of the isotropic property of space, the rotational derivative of x wrt frame A, DAx can also be evaluated as follows:

1) First rotate x through RBA to obtain Rbax.

2) Take the rotational time derivative wrt the rotated frame, now called B, Db{RBAx).

3) Rotate the result back through RBA into the original orientation. The three steps produce

DAx = RbaDb(Rbax)

The chain rule applied to the right side yields

Da x = DB x + RbaDbRbax (4.52)

If we can show that RBADBRBA = CtBA, the theorem is proved. Let us interchange A and В and execute the same three steps again:

DB x = Dax + RabDaRabx

Adding the last two equations provides

RBADBRBA = – RABDARAB (4.53)

The right-hand side is the desired fl, M because the transpose changes the order of rotation

—RABDARAB = —RbaDaRba = —DaRbaRba and recall the definition of the angular velocity tensor, Eq. (4.47), where

-DARBA~RBA = – ftBA = ftBA

Therefore, Eq. (4.53) is ttBA and indeed Eq. (4.52) proves the theorem:

Da x = DBx + ПВАх

With the Euler transformation at our disposal, we can model many interest­ing kinematic phenomena. First, let us have another look at linear velocities and accelerations and then state and prove several properties of angular velocity and acceleration.

Example 4.8 Relative Velocities

Problem. Refer back to Example 4.6 and Fig. 4.10. Let us assume that the radar lost track of the target, but still receives the missile’s measured target velocity Vj via data link while tracking the missile’s velocity vBr. What is the velocity of the target wrt the radar vB, and how can it be calculated?

Solution. All three quantities are relative velocities. It is tempting to add them vectorially vB = Vj +v", but this is only possible if both the missile and the target were modeled by points only. However, the seeker is measuring the target velocity v" wrt the missile reference frame M. Therefore, the angular velocity of the missile relative to the radar frame enters the calculation. I will derive the proper equation from the basic displacement triangle, employing the rotational time derivative and Euler transformation.

From Fig. 4.10,

Str = Stm +Smr

We impose the rotational time derivative to create the desired relative velocity vR – Drstr

DrStr = DrStm + DrSmr (4.54)

The differential velocity DrStm = vtm is unavailable. Instead, the missile seeker measures Vj = DmStm – The kinship is provided by Euler’s transformation

DrStm — DmStm + nMRsTM

Substituting back into Eq. (4.54) yields an equation, exclusively with relative velocities:

vR = v?+ nMRsTM + V* (4.55)

To implement this equation in the radar processor, we need to know the coordinate system of the measured data. The target velocity is measured in missile coordinates VjM ; the angular velocity tensor, recorded by the onboard INS, is beamed down in radar coordinates QMRR (the R frame is also the reference frame of the INS); and finally, missile position and velocity are measured by the radar in its own coordinates smrr and | ]л. Because ]й is prevalent, we coordinate Eq. (4.55)

first in radar coordinates

followed by transforming the target velocity into missile coordinates

[vr]r = [TfM [<f + [П"*]*[*ш]* + [u*f (4.56)

We have succeeded in deriving the component equation for implementation, pro­vided the coordinate transformation [T]RM from the missile INS is also beamed down to the radar station.

Example 4.9 Relative Accelerations

Problem. Continuing with the Example 4.8, we ask how to calculate the target’s acceleration wrt the radar aR when it is measured by the missile a’j 1

Solution. We start by taking the rotational time derivative wrt frame R of Eq. (4.55) to obtain the desired acceleration aR = DRvR:

Drv$ = Drv¥ + DR(nMRsTM) + DRv% (4.57)

Let us discuss the right-hand terms one at a time.

1) The first term DrVj relates to two different frames. To change it into the desired target/missile acceleration a’j = DMvf, we need to apply the Euler transformation

= DMvf + nMRv%! = a% + ftMRv"

2) The second term Dr($ImrStm) is converted by the chain rule and the Euler transformation to obtain v™:

DR(nMRsTM) = DRnMRs tm + nMRDRs TM = DRnMRsm + nMR(DMsTM + nMRsTM) = DRnMRsrM + nMRv¥ + nMRnMRsTM

The right side consists of the angular acceleration term, one-half of the Coriolis acceleration, and the centrifugal acceleration.

3)

The third term DRvfj is the easy one. It is the missile acceleration wrt the

Collecting terms, Eq. (4.57) expresses the target acceleration in known quanti­ties:

What a formidable equation to implement! If we only could model the missile by a simple point, then ttMR would not exist, and the target acceleration would simply be

This shortcut is made frequently. But do not forget to assess the neglected terms.

These two examples should give you the working knowledge for modeling linear velocity and acceleration problems. We now turn to angular velocities and accelerations. Refer back to Sec. 4.2.3 and recall the definition of the angular velocity tensor Eq. (4.47). Do not forget that it is a skew-symmetric tensor and therefore contracts to an angular velocity vector. With our freshly acquired Euler transformation we are able to prove several properties of angular velocity.