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Suppose that we start up the ship s engines and get moving We accelerate faster and faster, with the goal of eventually reaching speeds that are nearly the speed of light Suppose that we manage to accelerate to a sizable fraction of the speed of light, and then we shut off the engines so that we are coasting through space You ask, Relative to what are we moving This, as we shall see, is an important question! For now, suppose that we measure speed with respect to Earth We measure the time it takes for the laser to go across the ship and back again We are riding along with the laser, the mirror, and all the luxuries of a spacecraft that is only 3 m (about 10 ft) wide We find that the time lag is still exactly the same as it was when the ship was not moving relative to Earth; the oscilloscope still shows a delay of 20 ns This follows directly from Einstein s axiom The speed of light has not changed because it cannot The distance between the laser and the mirror has not changed either Therefore, the round trip takes the same length of time as it did before we got the ship moving If we accelerate so that the ship is going 60 percent, then 70 percent, and ultimately 99 percent of the speed of light, the time lag will always be 20 ns as measured from a reference frame, or point of view, inside the ship Let s add another axiom to Einstein s: In free space, light beams always follow the shortest possible distance between two points Normally, this is a straight line You ask, How can the shortest path between two points in space be anything other than a straight line This is another good question We ll deal with it later in this chapter For now, note that light beams appear to follow straight lines through free space as long as the observer is not accelerating relative to the light source Relative motion does not affect the straightness of light rays (As we will see, acceleration does).

ib de nition of b in eqn (1025), and the de nition of the gyro measurement in eqn (108), the gyro measurement error is modeled as b = xg g ib with xg = Fg xg + g Substituting eqn (1056) into eqn (1055) yields = Rn xg Rn g Rn n b b b in (1057) (1056)

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an outer join, you should be certain that it is necessary to obtain the correct desired result in your reports If you need an outer join, you also need to know which side of the relationship you always want returned If in some instances you want one side and in other instances you want the other side, you will need to alias the query subjects and model two separate relationships on two separate sets of query subjects You can also model a full outer join by specifying 0n cardinality on both sides of the link, which would return all entities from both query subjects whether matches existed or not If you are trying to model a full outer join involving two fact tables related through a common dimension table, note that Cognos 8 will handle such situations using a stitch query, which does not require that you model any of the relationships as an outer join See the discussion on stitch queries in the next section

The vector n , derived on p 52 and reprinted in eqn (1051), is the in rotation rate of the navigation frame relative to the inertial frame represented in navigation frame Eqn (1057) shows that the lack of position and velocity information a ects the AHRS accuracy by the accumulation of attitude error at rates related to and In the description that follows, n in is treated as a white noise term driving the error state model In fact, this term is not white or even stationary Improved handling of this term could lead to improved performance The accumulated error is intended to be estimated and removed by the accelerometer and magnetometer aiding sensors

Imagine now that we are outside the ship and are back on Earth We are equipped with a special telescope that allows us to see inside the ship as it whizzes by at a significant fraction of the speed of light We can see the

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