Description: 本人平时收集的惯性导航算法的论文，对搞导航算法的人很有帮助。
1.易于实现的捷联式惯性导航系统仿真.pdf
2.测量运动物体姿态的三自由度定位算法的研究.pdf
3.基于MEMS 技术的微型惯性测量组合.pdf
4.基于四元数的空间全方位算法研究.pdf
5.捷联惯导积分算法设计（连载二）下篇：速度和位置算.pdf
6.数字磁罗经系统的设计.pdf
7.四阶龙格—库塔法在捷联惯导系统姿态解算中的应用.pdf-I normally collected papers inertial navigation algorithm, the navigation algorithm to engage people very helpful. 1. Easy to realize the strapdown inertial navigation system simulation. Pdf2. Measuring movement of three degrees of freedom objects posture positioning algorithm. Pdf3. Based on MEMS technology, micro-inertial measurement unit. Pdf4. Quaternion-based full-space algorithm study. pdf5. SINS Integral Algorithm Design (Part II) next: the speed and position calculation. pdf6. digital magnetic compass system design. pdf7. Fourth-Order Runge- Kutta method in SINS Attitude operator application. pdf Platform: |
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Author:ddk_ok |
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Description: 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) -First, data on:
1: inertial navigation system that links the north bit of the Czech system. Initial longitude 116.344695283 degrees latitude 39.975172, height h is 30 meters.
Initial speed of v0 = [0.000048637 0.000206947 0.007106781], the same altitude.
2: jlfw for 600 seconds of data, gyroscopes and accelerometers were sampling period of 1/80 sec and 1/80 seconds.
3: Initial attitude angle [0.120992605 0.010445947 91.637207] (pitch, roll, heading, in units of degrees),
jlfw than the power saved information f_INSc (unit m/s ^ 2), angular rate gyro information wib_INSc (units of rad/s), in the order of 1 ~ three lines were east, north, days to the information.
4: The heading angle is positive counterclockwise.
5: Earth ellipsoid long radius re = 6378245 Earth s rotation angular velocity wie = 7.292115147e-5 acceleration due to gravity g = g0* (1+ gk1* c33 ^ 2)* (1-2* h/re)/sqrt (1-gk2* c33 ^ 2)
g0 = 9.7803267714 gk1 = 0.00193185138639 gk2 = 0.00669437999013 Platform: |
Size: 401408 |
Author:袁刚平 |
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