一、数据说明:
1:惯导系统为指北方位的捷连系统。初始经度为116.344695283度、纬度为39.975172度,高度h为30米。
初速度为v0=[0.000048637;0.000206947;0.007106781],飞行高度不变。
2:jlfw中为600秒的数据,陀螺仪和加速度计采样周期分别为为1/80秒和1/80秒。
3:初始姿态角为[0.120992605 0.010445947 91.637207](俯仰,横滚,航向,单位为度),
jlfw中保存的为比力信息f_INSc(单位m/s^2)、陀螺仪角速率信息wib_INSc(单位rad/s),排列顺序为 一~三行分别为东、北、天向信息.
4: 航向角以逆时针为正。
5:地球椭球长半径re=6378245;地球自转角速度wie=7.292115147e-5;重力加速度g=g0*(1+gk1*c33^2)*(1-2*h/re)/sqrt(1-gk2*c33^2);
g0=9.7803267714;gk1=0.00193185138639;gk2=0.00669437999013;c33=sin(lat纬度);
二、作业要求:
1:可使用 MATLAB语言编程,用MATLAB编程时可使用如下形式的语句读取数据:
load D:\...文件路径...\jlfw,便可得到比力信息和陀螺仪角速率信息。用角增量法。
2:分别作出其位置经度纬度坐标曲线图,时间富裕的同学可以加上姿态角和速度的曲线图。
3:以上文件以软盘或电子文档形式交,要附源程序和算法流程图。
load H:\homework1_05\fw
exercise 2
一、the explaination of data
1、 The system is strapdown inertial navigations system which point to the north.
The initial latitude and longitude are 116.344695283 degree and 39.975172degree,
The fight height is 30 m.The initial velocity v0=[0.000048637;0.000206947;0.007106781] and the height are constant.
2、 jlfw record the fight data of 600 seconds .The sampling period of gyro and accelerometer is 0.125 second.
3、 The initial attitude (pitching angle,rolling angle and heading angle) are [0.120992605 0.010445947 91.637207]
the information of force f_INSc(m/s^2) of accelerometer, angle velocity information of gyro wib_INSc(rad/s) is contained in jlfw
the data from one to three row are the record of E,N,U .
4、 Anticlockwise heading angle is positive.
5、 re=6378245;wie=7.292115147e-5;g=g0*(1+gk1*c33^2)*(1-2*h/re)/sqrt(1-gk2*c33^2);
g0=9.7803267714; gk1=0.00193185138639; gk2=0.00669437999013;c33=sin(lat);
二、demands of homework
1、 If you use MATLAB language program,you can get the dates in the following way to get the information of force and angle
velocity information of gyro. load D:\...document path...\fw.
2、 Working out position 、latitude and longitude graph ,and attitudet and velocities gragh if you have enough time.
3、 You should hand in the source program and the flow chart of your arithmetic.
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