Category Modeling and Simulation of Aerospace Vehicle Dynamics

3.2.2.1 Go back and re­view Figs. 3.4 and 3.5. We choose the preferred coordinate system of the inertial frame triad i, i2, and І3 (see Fig. 3.12). The Iі inertial axis is aligned with the

з1

vernal equinox, and the 31 axis with the north pole. The I’ and 21 axes lie in the equator. The heliocentric coordinate axes are tilted by the obliquity of the ecliptic є. Let us build the TM [T]HI of the heliocentric wrt the inertial coordinate systems with the help of Eq. (3.5)

[hxV

1Ы1

VhY

where the [hi]1, і = 1, 2, Зате the base vectors ofthe heliocentric triad coordinated in inertial axes. By inspection of Fig. 3.12, you should be able to verify

(3.11)

Notice the pattern of the TM. The 1 appears in the first column and row indicating that the transformation is taking place about the 1 direction without change in coordinates. The diagonals are the cosine of the transformation angle, a fact verified from the direction cosine matrix Eq. (3.6). The remaining off-diagonal elements are the sine of the angle, again verifiable by Eq. (3.6). You only have to decide where to put the negative sign. A simple rule says that the negative sign appears before that sine function, which is above the row containing the 1. If that row is on top, as in our example, imagine continuing rows in the sequence 3-2-1.

An alternate rule focuses on the positive sign. Inspect Fig. 3.12. The new axis that lies between the original axes indicates the row with the positive sine function. In our case 2H lies between 21 and 31; therefore, the second row of the matrix carries the positive sine.

As promised, I peel off the connections between the piercing points and have drawn them in Fig. 3.13. To help you in the transition, I still show the coordinate axes. With practice you can soon do without them.

By earlier agreement we use only right-handed coordinate transformations. Transforming the ]’ system about the 1′ axis, we bring the 21 axis through the angle £ to establish the 2H axis by a right-handed motion with the index finger

3′ 3E
l

pointing in the I’ direction. This transformation [T]HI is symbolically expressed as ]H verbahzing transformation of coordinate system H wrt I through

the angle є. Sometimes I may also use the expression, “transforming from I to H through the angle є” although the first form is preferred. Taking the transpose of the TM [f]H/ changes the sequence of transformation to [T]m, through the negative angle —є, or symbolically ‘ <— ]я.

Coordinate systems are pervasive in computer simulations of aerospace vehicles. In contrast to frames with their base point and base vectors, coordinate systems have no physical substance. They are just mathematical schemes of relabeling the coordinates of tensors. However, we have seen, if base vectors are expressed in pre­ferred coordinate systems, they take on a particularly simple form. This relationship invites us to display geometrically the direction and positive sense of the coordi­nate axes. Remember, however, that the coordinate axes do not have to emanate from the base point, although it will be convenient for us to do so most of the time.

We are already acquainted with the triads of the heliocentric, inertial, Earth, and body frames. Over these triads we superimpose the preferred coordinate systems with the same names. The axes are labeled 1,2,3, rather than x, y, z, withacapital superscript indicating the associated frame. We let the coordinate axes pierce the unit sphere and connect the piercing points to create surface triangles like orange peels. After some practice it will not be necessary to draw the axes any longer. The sketches of the orange peels will suffice to help you visualize the coordinate systems. At least that’s what happened to me. Professor Stuemke, University of Stuttgart and formerly Peenemuende, was an orange lover and demanded from his students to think of coordinate systems as orange peels. Having dealt over the past 40 years with many coordinate systems, I am thankful that he did not waver in his devotion.

Rather than introducing individual coordinate systems, I pair them up and show mutual relationships that lead to coordinate transformations. We begin with the two most important reference systems.

Coordinate systems are related by coordinate transformation matrices that re­label the coordinates of a tensor. We will dissect them and examine their ele­ments more closely. Two representations will help us to better understand their composition.

3.2.1.1 Base vector representation. Introduce x, a first-order tensor, and any two allowable coordinate systems ]A, R with their transformation matrix [Твл. From the definition of a first-order tensor, Eq. (2.3), we have

[x]B = [T ]BA[x]A

Furthermore, introduce the triad ai, аг, and аз (see Fig. 3.8), associated with the frame A. Decompose x into the vectors of the triad

x = xfai + хАаг + хАаз

x/’. / = 1,2, 3 are the coordinates, and х/’а,-, і = 1, 2, 3 are the components of the vector. Now express x in ]s coordinates:

[х]в = х?[аі]в +х£шв +х*[а3]в

This equation can be written as a scalar product

(3.3)

If we select ]A as the preferred from all of the associated coordinates systems, then the vector x is coordinated directly as

MA =

and therefore Eq. (3.3) becomes

WB = [[fli]B ШВ [a3]B][x]A

Comparing this equation with Eq. (3.1), we find a representation for the coordinate transformation matrix

[T]BA = [[a i]B [n2]B [я3]в] (3.4)

Its columns consist of the frame A base vectors, coordinated in ]B, and predicated on using the preferred coordinate system for frame A.

We can reverse the derivation and decompose vector x into the triad b,b2, and b2 of frame В

x = xBb + xBb2 + xBb2 and coordinate it in the ]A system

[x]A = xflbtf +x2>2]A +xB[b3]A = [Ybrt [b2]A [Ьъ]А}[х]в

Solving for [x]B and comparing with Eq. (3.1) yields another representation for the transformation matrix: [bi]A

[b2]A

[Ьз1А

The rows are the base vectors of frame В transposed as row vectors and coordinated in the ]A system. Both presentations can be summarized in the following schematic with tij the elements of [T]BA

[bi]A

mA

[b3]A

In summary, the coordinate transformation matrix consists of base vectors coor­dinated in their preferred coordinate systems. Frequently, I will abbreviate the expression “coordinate transformation matrix” simply by TM.

3.2.1.2 Direction cosine transformation matrix. There is another inter­pretation of the elements of TM. If we write out the coordinates of Eq. (3.4),

we can interpret each element as the cosine of the angle between two base vectors. The first subscript indicates the b triad and the second subscript the a triad. Let us check it out using the a® term (see Fig. 3.9). To calculate the cosine of the angle between the base vectors b and a, take their scalar product and express them in the ]® coordinate system

cosZ(ft],ai) = [51]s[a1]B = [1 0 0] a® = a®

Indeed, a® is the cosine of the angle between the two base vectors. You can also start with Eq. (3.5) and reach the same conclusion. In general, for any b, and we can calculate the cosine of the angle according to the formula

cos l(bi, ak) = аЦ = tik’, г = 1,2,3; к = 1, 2, 3 (3.6)

Because the elements are the cosines between two directions, the TM is also called the direction cosine matrix, a term quite frequently assigned to the TM of the body wrt the geographic coordinates.

Example 3.1 Tracking Coordinate System

Problem. Figure 3.10 shows a missile В being tracked by a search radar A. The onboard computer must determine the tracking radar’s coordinate system and the TM [T]BA of the missile’s coordinates wrt those of the tracking radar. Available are the gravity vector [g]s and the displacement vector of the radar wrt the missile both measured in body coordinates. Some additional information is known about the coordinate system of the tracking radar. Its 1A axis is parallel and opposite of the displacement vector and the 2A axis is horizontal (into the page).

What are the equations that the missile processor has to execute?

Solution. We start by formulating the radar’s base vectors in missile axes. Then it is just a matter of using Eq. (3.4) to calculate the transformation matrix. The first base vector is the unit vector of the negative displacement vector

I I

The second base vector is obtained from the vector product

в = [SAB]B[g]B

|[ЗДВЫВ1

where the upper case [5дв]в indicates the skew-symmetric matrix of the lower case displacement vector sAbr. The third base vector completes the triad

We program the following matrix with three columns for the missile computer:

[Jab]b [£ав]ВЫВ

клві |[ЗДВЫВ1 1^В||[ЗДВЫВ1

As the missile continues on its trajectory, the and [g]s coordinates change

(why does [g]s also change?), and the calculations have to be continually up­dated.

3.2.1.3 Properties of transformation matrices. As we model and simu­late aerospace vehicles, we will be manipulating many coordinate transformations. It is, therefore, important for you to have a good understanding of their properties.

Property 1

Transformation matrices are orthogonal.

Proof: Use the fact that the scalar product is the same in two coordinate sys­tems. Introduce any vector* and any two allowable coordinate systems ]A andH. Formulate the scalar product in both coordinate systems and then use Eq. (3.1):

mA[x]A = mB[xf = mAmBA[T]BA[x]A

The outer equations form

[x]A([T]BA[T]BA – [£])[x]A = 0 Because [x]A is arbitrary,

[T]ba[T]ba = [£] (3.7)

which expresses the orthogonality condition

[f ]BA = ([£]M)_1 QED

Property 2

The determinant of the TM is ±1.

Proof: Take the determinant of Eq. (3.7):

ітЯАтм! = i[£]i
|[ffA||[£]M| = |[£]| = 1
|[£]M|2 = 1

|[Г]ВА| = ±1 QED

To maintain the right handedness, we choose |[£]ЯА| = +1 only.

Property 3

Taking the transpose of the TM corresponds to changing the sequence of trans­formation, i. e.,

Proof: Premultiply Eq. (3.1) by (T]BA,

[TfA[ x]B = [T]BA[T]BA[x]A = [x]A
[x]A = [ffA[xf

Because in Eq. (3.1) ]A and ]B are arbitrary, we can exchange their positions:

[xf = [T^xf

Comparing the last two equations yields [Т]м = [T]AB QED.

Property 4

Let]A, ]B, and]c be any allowable coordinate systems, then [Г]СА = [TfB[TfA, i. e., consecutive transformations are contracted by canceling adjacent superscripts.

Proof: Let us apply Eq. (3.1) three times:

Property 5

Transformation matrices are not tensors.

Proof: If [T]BA were a tensor, it would have to be invariant under any coordinate transformation. Introduce a TM between any two allowable coordinate systems, say [TfD. According to Sec. 2.2.2, an entity like [T]BA is a second-order tensor if it maintains its characteristic under the transformation

[j’jM__ jC1^

Clearly, the right and the left sides are completely different types of matrices. We cannot contract the superscripts of the right-hand side to conform to the left side. Therefore, [Y JfiA is not a tensor.

You will avoid errors in modeling if you remember three rules (Rules 1^1 are in Chapter 2).

Rule 5: [T]BA is always read as the TM of coordinate system f with respect to (wrt) coordinate system ]A.

Rule 6: The transpose of the TM reverses the order of transformation ТВЛ = 17′ AB.

Rule 7: In transformation sequences, adjacent superscripts must be the same.

Example 3.2 Multiple Coordinate Transformations

Problem. The antenna A of an unmanned aerial vehicle В images the target T and sends the information to a satellite tracker S (see Fig. 3.11). The uplinked information consists of the following TMs: [T]rA, TBA. and [T]BS. How does the satellite processor calculate the TM of the target coordinates ]7 wrt its own coordinate system ]s?

Solution. String the TMs together according to Rules 6 and 7:

і’у’jSA j-y jSiS’

Notice that the sequence of adjacent superscripts is maintained if the transposed TM is replaced by Rule 6:

j-yjT’iS

These strings of TMs are quite common in full-up simulations. Naturally, you do not multiply them term-by-term, but leave the manipulations to the computer.

You may have noticed in Fig. 3.11 that the coordinate axes of the satellite do not meet. I offset them intentionally, to emphasize that coordinate systems have no origin and do not have to emanate from a common point. They are completely defined by direction and sense.

With deeper insight into the structure of TMs, we are prepared for the multitude of special coordinate systems and their associations. I will introduce the most important ones next, but defer some to later sections as they naturally occur in the context of specific applications.

The significance of coordinate systems is their ability to enable numerical cal­culations of symbolic equations. After all, modeling of aerospace vehicles finds its fulfillment in simulations, and computers can only chew on numbers and not on symbolic letters.

As you build your simulations, you will require many types of coordinate sys­tems. Let me list those that are the most important ones: heliocentric, inertial, Earth, perifocal, geographic, local level, velocity, body, stability, aeroballistic, and relative wind coordinate systems. Others arise as applications require it: gimbal, sensors, nozzle, target coordinate systems, etc.

With such a confounding multitude it is understandable that order had to be established by standardization. In Germany, the LN Standard 93002 has been in use for many years. The U. S. has lagged behind. In the past I had to rely on a sole U. S. Navy document3 for aeroballistic modeling. In 1992 AIAA published, in collaboration with the American National Standards Institute, the Recommended Practice for Atmospheric and Space Vehicle Coordinate Systems.4 Of course, over the years, many textbooks have served as references as well: Etkin,5 Bate et al.,6 Britting7; and more recently, Pamadi,8 Vallado,1 and Chatfield.9

As we make the transition from frames to coordinate systems, the preferred coordinate system, defined earlier, will play an important role. If a triad has been defined for a frame, it is most convenient to pick from the infinite number of associated systems one that lines up with the base vectors.

Just like frames refer to each other—establishing their relative position— coordinate systems are related by coordinate transformations. Let us briefly re­view (see Chapter 2): coordinates are ordered algebraic numbers that are related to the Euclidean space by coordinate systems and relabeled by coordinate transfor­mations. We employ only right-handed Cartesian coordinate systems. Before we detail the most important coordinate systems and their transformations, we will discuss the properties of coordinate transformation matrices.

Without reference points and frames we could not model positions or motions of aerospace vehicles. Reference frames in particular have captured the interest of astronomers over millenniums. Just recall the argument whether the Earth, the sun or the stars should constitute the primary reference. In recent times, the launching of satellites and interplanetary travel of spacecraft have made it a matter of prac­tical importance to clearly understand reference frames and coordinate systems. Vallado1 gives an up-to-date account in his book Fundamentals of Astrodynamics and Applications.

We will concentrate on those frames that are of primary importance for the modeling of aerospace vehicles. The Sun-centered (heliocentric) frame serves all planetary space travel, whereas the Earth-centered frame suffices for Earth satellite trajectory work. Both are considered inertial frames in the Newtonian sense; the choice depends on the application. For Earth-bound flights the rotation of the Earth can often be neglected. Under these circumstances the Earth frame itself becomes the inertial reference. Even the vehicle’s body can become a reference frame for rotating turbine blades, propellers, or gimbaled seeker heads. We shall define these frames now in detail using the concept of base point and base vectors.

3.1.2.1 Heliocentric frame. For all of humanity the sun is the major frame of reference. It separates day and night, the seasons, and the years. It is the main source of energy, and its gravitational pull keeps the Earth and the planets on their

elliptical paths. As the Earth revolves around the sun in one year, it encircles the ecliptic plane. The astronomers speak of the mean ecliptic, which averages some minor periodic fluctuations. Because the Earth is tilted about 23.5-deg from the ecliptic, we experience the four seasons. Figure 3.4 visualizes the yearly cycle for you.

The sun is not solid but gaseous and possibly liquid. In the midst of all of the explosions, there are no reference points mutually at rest; therefore, we have to create them artificially. We pick the gravitational center as the base point H. The third base vector A3 is normal to the ecliptic, pointing upward, based on the Earth’s orbital direction and the sense of the vector abiding by the right-hand rule. We need at least one more base vector to fully define the triad (the third one follows from the orthonormal condition). The choice is the first base vector h, which points to the First Point of Aries; or at least where the star constellation was during Christ’s lifetime. Today it points in the direction of the constellation Pisces.

How is that direction defined? At the first day of spring (vernal equinox), position yourself at the center of the Earth. As you look out, you see the intersection of the equator and the ecliptic lining up with the sun. If you had been bom 2000 years ago, you would have seen the constellation Aries beyond the sun—an impossible feat with human eyes. The sign indicating Aries is a modified Greek T, resembling a ram.

Summarizing, the heliocentric frame is modeled by the base point H and the three base vectors h,h2, and й3, with h 1 pointing to T, й3 being the normal of the ecliptic plane, and h2 completing the triad. If you ever get to travel to Mars, you

would become very familiar with this heliocentric frame. However, back at planet Earth, even as astronaut of the International Space Station, you would more likely be dealing with the geocentric-inertial reference frame.

3.1.2.2 Geocentric-inertial (J2000) frame Now let us concentrate on the Earth as we view it in Fig. 3.5. The most useful inertial frame is collocated with the center of the Earth, but its orientation remains fixed in the ecliptic. This is a good example of why we have to distinguish between location and orientation of a frame. The location of the frame is given by the displacement vector Sm of the center of the Earth I wrt the center of the sun H. Its orientation is described by the base vectors i’i, t’2, and 13.

To define the vector i’i, we have to carry out a Gedanken experiment (“Gedanken” is German for “thought”). Imagine a nonrotating shell around the Earth with the etched trace of the equator. The ecliptic, projected on the shell, is the path of the sun during one year. It will intersect the equator at two points. Of particular interest is the point when the sun crosses the equator in spring. This point is called the vernal equinox, already introduced in Fig. 3.4. We align with this direction the first base point *1. It points at the constellation Aries and therefore is fixed with the stars. The remaining two base vectors are easily defined. The Earth’s axis of rotation serves as direction and sense for І3, and t’2 completes the right-handed triad.

Unfortunately, the equatorial plane, just as the ecliptic and the axis of rotation, moves very slightly over time. We can only speak of a truly inertial frame if we refer to its position at a particular epoch. Therefore, the astronomers defined the J2000 System that is based on the Fundamental Katalog, FK5.1 This system is the best realization of an ideal inertial frame. It is the foundation for all near-Earth modeling and simulation. We will therefore refer to it just as the inertial frame.

3.1.2.3 Earth frame. Look around you and take notice of the Earth (see Fig. 3.6). Its particles form the Earth’s frame. The base point E is at the Earth’s center, and the triad consists of the base vectors e , e-i, and £3. One meridian of the Earth assumes particular significance. It is the prime meridian that traces through the Royal Observatory at Greenwich, a suburb of London. Its intersection with the

equator establishes the penetration point of the first base vector e. The Earth’s axis of rotation serves as the direction and sense for the third base vector Є3, and Є2 completes the triad.

The prime meridian serves as origin for measuring longitude (positive in an easterly direction), whereas latitude is surveyed from the equatorial plane (positive in northerly direction). These reference lines are fixed on the Earth and ideal for locating sites on the Earth’s surface.

3.1.2.4 Body frame. Aircraft, missiles, and spacecraft are of primary con­cern for us. If they can be considered rigid bodies, they are represented by a frame, the so-called body frame. Although this frame is usually not used as a reference, it is nevertheless important for modeling location and orientation of vehicles under study. Its base point В coincides with the c. m., and the base vectors b,b2, and Ьз are aligned with the principal axes of the moment of inertia tensor.

Sometimes other directions are more relevant. For instance, the b vector for an aircraft is more likely aligned with some salient geometrical feature, like the tip of the nose or the zero-lift line at a particular Mach number (see Fig. 3.7). The aircraft designer has even adopted the terminology of his colleague, the ship architect, and calls it the waterline. However, for all vehicles bj is parallel to the second principal moment of inertia axis.

b,

For aircraft, the positive sense of the b base vectors is out the nose, bj out of the right wing, and b3 down, completing the triad. For missiles with rotational symmetry, any direction for b2 and Й3 is a principal axis. Therefore, control fins or geometrical marks may be used to fix them.

If elastic phenomena need be modeled, a body frame is still required as a ref­erence for the bending and vibrating modes. This frame could be defined as co­inciding with the vehicle under no-load conditions; or, in the case of wing flutter, the fuselage alone could be chosen. Sometimes the whole vehicle is divided into many rigid body subframes, and one of them is chosen as the primary reference.

3.1.2.5 Summary. In this section I have introduced four important frames. Starting first with the heliocentric frame of reference for interplanetary travel, then zeroing in on the geocentric inertial frame for orbital trajectories, we eventually come down to Earth to define the Earth frame, which serves as reference for most atmospheric flight. In addition, the body frame is of particular significance. It models the position and orientation of the vehicle, which we want to simulate.

Table 3.1 summarizes the four frames. Notice the significance of the ecliptic and the equatorial planes. Their normal unit vectors define the direction of three base vectors (the Earth’s spin axis is normal to the equator). You may be puzzled by the fact that the center of the Earth is the base point for both the inertial and the Earth frames. Indeed they coincide. Of all the points of the Earth frame, the center is the only point shared with the inertial frame.

You probably cannot wait any longer to meet the more familiar coordinate systems. Their time has come. I only hope that you keep in mind their fundamental difference with frames.

Let A and В be two frames containing three noncollinear points А і, A2, A3 and B, B2, B3, respectively (see Fig. 3.1). The position of frame В wrt frame A is determined by three displacement vectors Sb, a,, і = 1, 2, 3. Six of the nine vector coordinates are independent, i. e., a frame has six DoF. Are you surprised to encounter six rather than three degrees of freedom in the Euclidean three-space? Frames have the property of location and orientation, whereas space has only the attribute of location. That explains the difference. I will define these frame properties in more detail.

3.1.1.1 Location of a frame. Let A and В be two points of frames A and B, respectively (see Fig. 3.2). The displacement vector sba determines the location of frame В relative to frame A. We call the two points base points. A displacement vector can only relate to one point in each frame. It fixes the location of the base point, but leaves the frame the freedom of orientation. Yet remember that the frame points are mutually unchanging. Two more noncollinear and nonplanar points would define the complete location and orientation of the frame В wrt frame A. However, we chose three additional points such that they form the endpoints of an orthonormal vector triad and thus define the orientation of a frame.

3.1.1.2 Orientation of a frame. Build a triad from a set of three orthonor­mal base vectors (see Fig. 3.3):

a, яг, Я3 where я, a, = | j for) = j ; * = 1 > 2, 3; j = 1, 2, 3

with magnitudes one, which connect the base point A with three other points of a frame A. The orientation of a frame is given by this triad.

Just like location requires a reference point, so orientation needs a reference frame. A displacement vector models location, whereas orientation, as we shall see in Sec. 4.1, is portrayed by the rotation tensor. Both, location and orientation determine the position of a frame. A triad, with its base point and three vectors, is sufficient to define a frame.

You may interject that in Fig. 3.3 you recognize the familiar coordinate axes system. But not so! A triad is a physical entity, consisting of points and vectors, and is not a coordinate system. However, now we can finally bridge the gap. Among the coordinate systems associated with each frame, there is a particular one that coordinates the base vectors in the simple form [a i]A = [1 0 0], [«г]’4 = |0 1 0], and І аз ]д = [0 0 1]. It is called the preferred coordinate system and is the most important one among all the possible coordinate systems associated with a frame.

Recall the definition from Chapter 2: A frame is an unbounded continuous set of points over the Euclidean three-space with invariant distances and which possesses, as a subset, at least three noncollinear points. The inertial frame is such an unbounded set of points. We will give it a precise definition shortly. The Earth, although bounded, has also a frame associated with it. Theoretically, the Earth frame extends beyond the confines of the geoid, but when we refer to the points of the Earth’s frame, we remain on the Earth. A similar approach is taken with the body frame. Strictly speaking, the body must be rigid to be modeled by a frame. For elastic modeling it is common practice to divide the body into finite, rigid elements, each of which is represented by a frame.

We must be able to identify at least three noncollinear points of a frame. Oth­erwise, the frame may not occupy the three-dimensional manifold of space. In particular, we may pick the three points such that, if connected with a fourth base point, they establish a triad that defines the position of the frame completely.

Chapter 2 introduced frames and coordinate systems. Frames are models of physical references, whereas coordinate systems establish the association with Euclidean space. Both entities are important elements of aerospace vehicle dy­namics. Frequently, they are presumed to be the same, but I will maintain a carefiil distinction throughout this book, heeding TruesdelFs warning, quoted earlier, that frames should not be regarded as a synonym for coordinate systems.

This chapter will expand our understanding of these concepts. Important frames of reference, such as inertial, Earth, and body frames will be introduced. The triad of base vectors will emerge as a keystone to define their location and orientation. It will bridge the chasm between frames and coordinate systems with the so-called preferred coordinate systems.

Coordinate systems are the spider web of simulations, providing structure, di­rection, and focus. They structure the Euclidean space into application-specific associations, establish sense of direction, and focus on numerical solutions— the sustenance of any simulation. However, they can also lead to bafflement and mystification and may ensnare the careless user. (I once developed an air combat simulation that dealt with 24 different coordinate systems.) Clarity of definition is essential. We shall deal with a host of systems: inertial, Earth, geo­graphic, local-level, perifocal, velocity, body, gimbal, relative wind, stability, aero – ballistic, and some more—hopefully all of the coordinate systems you will ever need.

A plane of symmetry has the characteristics of a mirror. The left side repeats the right side. Any aircraft exhibits this reflectional symmetry—what ever happened to the oblique wing? Even right-hand maneuvers can be reflected into left-hand maneuvers just by geometrical manipulation. The tensor that makes this happen is called the reflection tensor M.

What are the characteristics of this tensor M that reflects the vector t into t’ by the mirror plane with unit vector и (see Fig. 2.20)? First, project t onto и with the projection tensor P = ий to get r = Pt and then derive t’ from the vector triangle

t’ = t – 2r = t – 2Pt = (E – 2P)t

The reflection tensor of the mirror plane with unit normal к and unit tensor E is therefore

M = E – 2km (2.26)

It is not only symmetric but also orthogonal, as we can demonstrate by

MM = (E — 2 ий)(Е — 2км) = E — 4кк + 4мм km = E

The reflection tensor plays an important role in Chapter 7, where we will sort out the existence of higher-order aerodynamic derivatives of airframes exhibiting reflectional symmetry. If the mirror plane is oriented in body coordinates }R such that its unit normal has the coordinates [й]в = [0 1 0] (right wing of aircraft), then the reflection tensor has the coordinates

We conclude that the reflection tensor changes the sign of the second coordinate, but keeps the other two coordinates unchanged.

Example 2.9 Application of Reflection Tensor

Problem. An aircraft (see Fig. 2.21), with a canted twin tail, executes a push­down maneuver. What is the resultant force of both control surfaces / if the force on one surface is/,? Derive the equations in invariant form and then introduce body coordinates ] .

Solution. We use the reflection tensor to determine the force on the other control surface and add both together

/ = /і + /2 = /1 + Mfx = (E + A#)/i = 2(E – P)fl

With the unit normal of the mirror plane being к

This is the desired result in symbolic tensor notation. Introducing body coordi­nates for the unit normal [й]в = [0 1 0], and for the force on the right control surface [/J® = [0 /12 /13] yields the result

We find that the horizontal force components cancel, and the vertical component is doubled by the second surface.

2.4 Summary

If you are reading this, you have persevered until this chapter’s end. We are indebted to the physicists of the 18th and 19th centuries for the foundations of clas­sical mechanics. Its axiomatic treatment puts our modeling tasks on a sure footing. Simple Cartesian tensors in Euclidean three-space are the symbolic language, and their realization as matrices by coordinate systems are the fodder for computers. I hypothesized that points and frames suffice to model flight mechanics—a statement that still needs verification. Some of the cherished traditions of vector mechanics had to be abandoned. Coordinate systems have no origins, and the radius vector has no place in our tool chest.

Then, with the help of tensor algebra we assembled some basic operations. The scalar, vector, and dyadic products are essential for general modeling tasks, and they are applied to specific geometric problems. Some of them we readied for the toolbox: straight line, plane, normal form of plane, plane projection tensor, and reflection tensor.

Needless to say, but worth emphasizing, we just got started! There is so much more for you in store. Although we already introduced frames and coordinate systems, we need to dig deeper. I shall attempt to clearly delineate their distinctly separate purposes in the next chapter.

References

‘Hamel, G., “Die Axiome der Physik,” Handbuch der Physik, Band 5, Springer-Verlag, Berlin, 1929, Chap. 1.

2Noll, W., “On the Continuity of the Solid and Fluid States,” Journal of Rational Mechan­ical Analysis, Vol. 4, No. 1, 1955, p. 17.

3Truesdell, C., and Noll, W., “The Nonlinear Field Theories in Mechanics,” Handbuch der Physik, Vol. III/3, edited by S. Fluegge, Springer-Verlag, Berlin, 1965, pp. 36, 41, and 42.

4Ricci, G., and Levi-Civita, T., “Methodes de Calcul Differentiel Absolu et Leurs Appli­cations,” Mathematische Annalen, Vol. 54, 1901.

5Einstein, A., “Die Grundlagen der Allgemeinen Relativitaetstheorie,” Annalen der Physik, Vol. ‘4, No. 49, 1916, pp. 769-822.

6Duschek, A., and Hochrainer, A., Tensorrechnung in Analytischer Darstellung, Vols. I, II, and III, Springer-Verlag, Berlin, 1968, 1970, and 1965.

7Wrede, R. C., Introduction to Vector and Tensor Analysis, Wiley, New York, 1963.

8Betten, J., Elementare Tensorrechnung fuer Ingenieure, Vieweg, Brunswick, Germany, 1977.

Problems

2.1 Scalar triple product. Show that the scalar triple product V = (Xy)z = yXz is the volume contained within the three vectors x, y, and z by using scalar and vector products. For the body coordinate system ]B the vector coordinates are [x]B = [8 -2 -3], [y]B = [3 14 1], and [z]B = [3 -2 10] m. What is the value of V?

2.2 Helicopter landing aid (Example 2.6). Refer to Fig. 2.15, the helicopter landing patch, to visualize the following numerical measurements: [sb0h]g = [-22 -10 121], [Svh]g = [—12 16 124], and [їсзд]° = [9 -5 116]. Calculate the displacement vector of projection В wrt the helicopter H, {sHH]G with the equations derived in the example. (Numbers are in meters.)

2.3 Parallel lines. Determine the equation of a line [spR]R = usup„]k -f – [.?/>„«]B that intersects the point Po, given by its displacement vector [sp0r]r = [1 2 3]. This line is parallel to another line [sqr]r = u[l 1 1] + [1 0 0]. The coordinate system ]R is arbitrary (Solution: [Spr]r = [и + 1 и + 2 и + 3]).

2.4 Intersecting lines. Determine the point of intersection I, [5/л]в, of the two straight lines with the same reference point [5pr]b = и[1 1 l] + [0 1 0] and [sqr]b = r[0 1 1] + [1 1 0]. (Hint: Two lines intersect if there exist и and v such that [sbr]r = [«ек]в.) Complement the sketch by a three-dimensional scaled figure. The coordinate system ]R is arbitrary (Solution: [5/r]r = [1 2 1]).

2.5 Closest distance to a line. Determine the point P*, Sp*r, on the straight line SpR = uS(jp0 +spoR that has the shortest distance from R. (Hint: Find u* such that SprSpr is minimized.) Derive the general equation first, then apply it to the example [sm?]r = u[0 2 l] + [0 0 1]. Make a three-dimensional sketch. The coordinate system ]R is arbitrary (Solution: [л>.д ]л = [0 —2/5 4/5]).

2.6 Closest distance to a plane. Determine the point P* of the plane sPR = usup0 + vsVp0 +sPop that has the shortest distance from R. Procedure: Let u* and v* be the parameters of the point P* in the plane that is closest to R and show that

_ S VPos UP0S PqrS VP0 — s VPos VP0S paRS UP0 W — 2

s UPos UP0S VPos VP0 — (Svp0SUP0)

_ S UPos VPpS PqRS UPq — s UPQS UPqS p0rS yp0

s UPos UP0S VPos VPo — (s VP0S UPa)

by minimizing sprspr with respect to и and v.

2.7 Projected angle on focal plane array. A building is imaged on a focal plane array. Two unit vectors t and t2 model the sides of the building, emanating from one of the rectangular comers. The normal unit vector of the focal plane array is u.

(a) Derive the tensor equation that allows you to calculate the angle ф between the two sides as seen on the focal plane array.

(b) Calculate ф from the numerical values:

ft]G = [l 0 0]

[h]G = l 0 1 0]

[u]G = [0.648 0.3 0.7]

(Solution: ф = arccos(j^pl^), where N = E — ми).

2.8 Miss vector in target plane. Given the target plane P and its associated coordinate system ]p with its I p, 2P – axes parallel to the surface. The point target T is contained in this plane. The missile’s last two calculated positions are at В and

B. Derive the miss vector [л д. у ]р with B* the penetration point. By interpolating the straight line [s5b], you should obtain

[sa*r]P =

SBB) 3

Furthermore, derive the intersection time tB. with the target plane, if the time tB and the integration interval At are given (assume constant velocity):

You will need these two equations for any simulation that models the intercept of a plane. Air-to-ground and air-to-ship missiles are typical examples. For most of these applications, the linear interpolation of the intercept point is adequate. You will find them implemented in the CADAC simulation CRUISE5, Subroutine G4.

в

t— 2P

/ /’ B>"’

3P/ 2P

2.9 Closest approach of two lines. Determine the displacement vector sB-r> of the closest approach of a missile c. m. В and a target T by linear interpolation of the last two integration steps. Assume that both velocities are constant during the integration interval Af. First, derive the time of closest approach t* = tB + r by minimizing subject to the elapsed time r,

f* _ t_ _ д?(жвд —Stt)(sBe —ste) (sbb — sTf)(sbb ~ sTf)

then derive the miss vector of closest approach:

x

Sb*T* = — (*BB — s7t) + SBE — sTE T

These equations are indispensable for an air-to-air intercept simulation. They cal­culate the terminal miss distance, which determines the probability of kill, for a particular fuse and warhead. You will find them implemented in the CADAC simulations AIM5, SRAAM5, and SRAAM6 in their respective Subroutines G4.

2.10 Tilt angle of maneuver plane—project. An aircraft executes a coordi­nated banking maneuver (zero sideslip angle) with bank angle фвв and generates the normal load factor acceleration aN by the angle of attack a. Its orientation is given by the base vectors t, G, and t$ and its velocity of the c. m. T wrt to Earth E by vf. The gravitational acceleration g, acting on the aircraft, points in the direction of the unit local vertical /3 and lies in the vertical plane v§, /3. The load factor plane, subtended by aN and the base vectors t and /3, differs from the maneuver plane by the gravitational acceleration g.

(a) Derive the equation for the tilt angle фм in tensor form between the maneuver plane and the vertical plane. (Hint: vf is contained in both planes). Write down all of the steps.

(b) Develop the matrix equations to calculate the tilt angle. Consider as given the bank angle фвв, the normal load factor n, mass m, reference area S (wing loading w = mg/S), normal force derivative C, vn, aircraft velocity V, heading angle фуь, and flight-path angle 6VL. Write down all matrix equations that must be programmed in (d).

(c) Calculate, before you do (d), the tilt angle фм for the following numerical values:

Фвь = 40 deg; V = 250 m/s n = 3; iJ/vl = 0

w = 3249 N/m2; 6VL = 10 deg
CNa = 0.0523 deg-1

Sketch the attitude of the aircraft from the rear looking forward and indicate the approximate magnitude of the angles фм, фвъ (Solution: фм = 84.1 deg).

(d) Write a computer program that automates the calculation of the tilt angle. Consider as input: V, xj/VL, &vl, n, фвь, and the other variables as parameters. Use the matrix utilities UTL. FOR of CADAC as much as possible. As output, record the tilt angle, angle of attack, and the direction cosine matrix |T]fiL. First run the test case of (c) for verification, then change the following values from the baseline for Case 2: jrVL = 90 deg; and Case 3: n = 7; and Case 4: V = 350 m/s. Provide the source code and the numerical results of all five cases.

з

With the definitions of planes in plain view, we can address the task of projecting vectors into planes. Your simulation may require the velocity vector of an aircraft be projected on the ground or the silhouette of a missile be imaged on a charge – coupled device (CCD) planar array. For all of these situations, the plane projection tensor will be a useful tool.

We met the line projection tensor P earlier. It is formed by the dyadic product of the unit vector и, P = и m. According to Eq. (2.21), P produces the vector r from t by projecting it onto u: r = Pt. Now, и does not only establish the direction of a line, but also the unit normal of a plane (see Fig. 2.19).

The challenge is to find the projection tensor N that projects vector t onto the plane given by и. The projected vector is labeled s. From the vector triangle we derive and substituting Eq. (2.21) forr

s = t — Pt = (E — P)t = (E — uU)t

We define the plane projection tensor with E the unit tensor and и normal of the plane

N ~ E — ий (2.25)

Just like the line projection tensor, the plane projection tensor is symmetric.

Example 2.8 Focal Plane Imaging

Problem. An aircraft is imaged on a focal plane array. To simulate that process, we need to develop the equations that project the aircraft’s silhouette on the focal plane. We keep it simple by modeling the perspective of the aircraft with the displacement vectors of the tip, stem, right wing tip, and left wing tip wrt the

geometrical center C, tBlc, 1в2сЛвъс, and te„c – The displacement of the aircraft center C wrt the focal plane center F is given by tCF and the orientation of the planar array by the unit normal vector u. Separation distance and optics reduce the scale of the projections on the focal plane by a factor /. Determine the aircraft attitude vectors sBlc, $в2с, sb3c, and Sb4c and the displacement vector sCf in the focal plane. (To practice, make a sketch.)

Solution. Subjecting the displacement vectors to the plane projection tensor N = E — ий and reducing the magnitude by / produces the image

S BiC = fNtBlc, s B2C — fNtg2c, Sb3C = fNtB3C, s IhC = fNtBic

and the displacement of the aircraft from the focal plane center

scf = fNtCF

For building the simulation, the vectors have to be converted to matrices. Most likely, the aircraft data are in geographic coordinates ]G, and the image should be portrayed in focal plane coordinates ]F. Therefore, a transformation between the two coordinate systems [T]GF will enter the formulation.