- #1
Johan M
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- TL;DR Summary
- I would like to confirm if the sign convention for the forces present in a model is correct. It entails the linear and angular momentum for a simplified rocket model. The rocket has airfoil shaped flaps whose lift force can be used to adjust the rocket orientation
Hi everyone . I had came across a simplified simplified rocket lateral dynamics model :https://github.com/build-week/hover...n-scripts/design-scripts/jet_vane_speed.ipynb . It has vanes at the exit which generate lift force and can control the rocket orientation. In it, linear and angular momentum equations are present for the current orientation:
Fj: engine thrust
Lv: lift force
Dv: drag force:
alpha: vane angle of attack
theta: pitch over angle of the rocket
The angular momentum:
[IMG alt="$$
L_v r_v = I \ddot{\theta}
$$"]https://render.githubusercontent.com/render/math?math=L_v r_v = I \ddot{\theta}&mode=display[/IMG]Linear momentum:
[IMG alt="$$
-(F_j - D_v) \sin\theta + L_v \cos\theta = m \ddot{x}
$$"]https://render.githubusercontent.co..._v \cos\theta = m \ddot{x}&mode=display[/IMG]
I don't seem to understand the sign convention for the lift force generated by the vanes in these equations: at the current angle of attack, the lift force Lv would be in a south-west direction. In the equations, it seems the author took it in the north-east direction. In other words, should the equations instead read as:
-[IMG alt="$$
L_v r_v = I \ddot{\theta}
$$"]https://render.githubusercontent.com/render/math?math=L_v r_v = I \ddot{\theta}&mode=display[/IMG]
-(F_j-D_v)sin(theta)-L_vcos(theta)=m*xdottDoes anyone happen to know why he did otherwise? Any advice is appreciated
Fj: engine thrust
Lv: lift force
Dv: drag force:
alpha: vane angle of attack
theta: pitch over angle of the rocket
The angular momentum:
[IMG alt="$$
L_v r_v = I \ddot{\theta}
$$"]https://render.githubusercontent.com/render/math?math=L_v r_v = I \ddot{\theta}&mode=display[/IMG]Linear momentum:
[IMG alt="$$
-(F_j - D_v) \sin\theta + L_v \cos\theta = m \ddot{x}
$$"]https://render.githubusercontent.co..._v \cos\theta = m \ddot{x}&mode=display[/IMG]
I don't seem to understand the sign convention for the lift force generated by the vanes in these equations: at the current angle of attack, the lift force Lv would be in a south-west direction. In the equations, it seems the author took it in the north-east direction. In other words, should the equations instead read as:
-[IMG alt="$$
L_v r_v = I \ddot{\theta}
$$"]https://render.githubusercontent.com/render/math?math=L_v r_v = I \ddot{\theta}&mode=display[/IMG]
-(F_j-D_v)sin(theta)-L_vcos(theta)=m*xdottDoes anyone happen to know why he did otherwise? Any advice is appreciated