哈工大导航原理大作业
哈工大导航原理大作业-标准化文件发布号：（9456-EUATWK-MWUB-WUNN-INNUL-DDQTY-KII
《导航原理》作业
(惯性导航部分)
一、题目要求
A fighter equipped with SINS is initially at the position of ?35 NL and ?122 EL,stationary on a motionless carrier. Three gyros X G ,Y G ,Z G ,and three accelerometers, X A ,Y A ,Z A are installed along the axes b X ,b Y ,b Z of the body frame respectively.
Case 1:stationary onboard test
The body frame of the fighter initially coincides with the geographical frame, as shown in the figure, with its pitching axis b X pointing to the east,rolling axis b Y to the north, and azimuth axis b Z upward. Then the body of the fighter is made to rotate step by step relative to the geographical frame.
(1) ?10around b X
(2) ?30around b Y
(3) ?50-around b Z
After that, the body of the fighter stops rotating. You are required to compute the final output of the three accelerometers on the fighter, using both DCM and
quaternion respectively,and ignoring the device errors. It is known that the magnitude of gravity acceleration is 2/8.9g s m =.
Case 2:flight navigation
Initially, the fighter is stationary on the motionless carrier with its board 25m above the sea level. Its pitching and rolling axes are both in the local horizon, and its rolling axis is ?45on the north by east, parallel with the runway onboard. Then the fighter accelerate along the runway and take off from the carrier.
The output of the gyros and accelerometers are both pulse numbers,Each gyro pulse is an angular increment of sec arc 1.0-,and each accelerometer pulse is g 6e 1-,with 2/8.9g s m =.The gyro output frequency is 10 Hz,and the accelerometer ’s is 1Hz. The output of gyros and accelerometers within 5400s are stored in MATLAB data files named gout.mat and aout.mat, containing matrices gm of 35400? and am of 35400? respectively. The format of data as shown in the tables, with 10 rows of each matrix selected. Each row represents the out of the type of sensors at each sample time.
The Earth can be
seen as an ideal sphere,
with radius 6368.00km
and spinning rate
rad/s 10292.75-?, The
errors of sensors are
ignored, so is the effect
of height on the
magnitude of gravity.
The output of the gyros
are to integrated every
0.1s. The rotation of
geographical frame is to be updated every 1s, so are the velocities and position of the figure. You are required to:
(1)Compute the final attitude quaternion, longitude, latitude, height, and east, north, vertical velocities of the fighter.
(2)Compute the total horizontal distance traveled by the fighter.
(3)Draw the latitude-versus-longitude trajectory of the fighter, with horizontal longitude axis.
(4)Draw the curve of the height of fighter, with horizontal time axis.
二、Case1解答
2.1 方向余弦阵法
（1） 绕Xb 轴转过ψ?=10?
????
? ????-??=????? ??-=10cos 10sin 010sin 10cos 0001cos sin 0sin cos 0001?????C
（2） 绕Yb 轴转过?=30θ
????
? ?????-?=????? ??-=30cos 030sin 01030sin 030cos cos 0sin 010sin 0cos θθθθθC
（3） 绕Zb 轴转过?-=50ψ
????
? ???-?-?-?-=????? ??=1000）50（cos ）50（sin -0）50（sin ）50（cos 1000cos sin -0sin cos ψψψψψ
C 所以变换后的坐标
X E Y C C C N Z ?θψζ???? ? ?= ? ? ? ???
??
由于初始时刻有 009.8E N ζ???? ? ?= ? ? ? ?????
所以计算得
三个加速度计的输出分别是 ,/4504.42x s m A -=
,/6027.22y s m A -=
s m A /3581.8z =2
计算程序见附录一
2.2 四元数法
（1）绕Xb 轴转过ψ?=10? i 2sin 2cos
q ???+= （2）绕Yb 轴转过?=30θ
j 2sin 2cos q θ
θ
θ+=
（3）绕Zb 轴转过?-=50ψ
k 2sin 2cos q ψ
ψ
ψ+=
则合成四元数
)2
sin 2(cos )2sin 2(cos )2sin 2(cos k j i q q q q ?+?+?+==ψψθθ??ψθ?
合成四元数的逆
1123q p i p j p k λ-=---
由公式 q N E q Z Y X ????
? ??=????? ??-ξ1 计算得
三个加速度计的输出分别是 ,/4504.42x s m A -=
,/6027.22y s m A -=
s m A /3581.8z =2
由两种计算方法的计算结果可以看出，方向余弦阵法和四元数法的计算结果是一致的。
计算程序见附录二
三、Case2解答
3.1程序流程图