RBE 595: Space and Planetary Robotics Lecture 4

1. RBE 595: Space and Planetary RoboticsLecture 4 Professor Marko B Popovic A term 2019

2. Dynamics: 2D modeling “lift direction” “thrust direction” craftvelocityrelative to inertial frame center of mass (CM) “drag direction” ejected mass velocity relative to craft inertial frame orientation external force Terms “lift direction”, “drag direction”, and “thrust direction” are frequently used in literature however often without being properly introduced.

3. “lift direction” “thrust direction” “Lift direction” is well defined quantity only when notion of “up” and “down” is agreed upon which might not be always the case when considering space conditions. For motion in gravitational field the “lift direction” may be defined perpendicular to craft velocity and within plane formed by gravity vector and craft velocity and with one component of lift direction antiparallel to gravity vector. In the literature sometimes “lift direction” is simply defined antiparallel to gravity vector. However, if there is no gravity these definitions of “lift” are obviously inappropriate. However, one could also define notion of “lift direction” relative to cabin or other part of craft. craftvelocityrelative to inertial frame center of mass (CM) “drag direction” ejected mass velocity relative to craft inertial frame orientation external force

4. “lift direction” “thrust direction” “Drag direction” may be defined antiparallel to craft velocity. However, drag forces are well defined physical quantity and real net drag force direction is typically not oriented in the same direction as “drag direction”. Drag forces in physics are defined as rate of momentum exchange between fluid and craft in direction locally perpendicular to craft’s surface due to relative velocity of surface relative to fluid (and not due to pressure difference). After integration over craft’s surface area net drag force is typically not pointing antiparallel to . Still further net drag force may be zero if there is no fluid surrounding craft. craftvelocityrelative to inertial frame center of mass (CM) “drag direction” ejected mass velocity relative to craft inertial frame orientation external force

5. “Thrust direction” here is antiparallel to ejected mass velocity relative to craft. For rocket type propulsion trust is, according to third Newton law, equal in magnitude and antiparallel to the force applied to ejected mass which is equal to rate of change of linear momentum, i.e. impulse of ejected mass. However, thrust can be also defined relative to cabin or other part of craft. Sometimes all three directions are defined relative to craft such that thrust is antiparallel to drag and perpendicular to lift (with one component antiparallel to gravity vector).In nature insects and birds use flapping wings mechanism to fly. It is interesting to note that “thrust” in nature is mainly obtained by application of drag forces. Same is true for helicopters.In the figure “thrust direction” is antiparallel to and “drag direction” is antiparallel to while lift is perpendicular to . “lift direction” “thrust direction” craftvelocityrelative to inertial frame center of mass (CM) “drag direction” ejected mass velocity relative to craft inertial frame orientation external force

6. Linear momentum of system is sum of linear momenta of variable mass craft and ejected mass and angular momentum of system is similarly One could deduce dynamics of variable mass craft based on external forces and torques and dynamics of ejected mass by using second Newton law for two translational degrees of freedom and single rotational degree of freedom Rate of change of linear momentum of ejected mass is and rate of change of spin angular momentum (angular momentum about CM) of ejected mass is

7. Hence craft dynamics can be expressed by rate of change of craft’s linear momentum and rate of change of craft’s angular momentum In order to solve for craft trajectory one needs to know the ejected mass flow which can be integrated to give the craft’s mass with being initial mass. Also, one needs to know the craft’s moment of inertia ; moment of inertia is affected by missing ejected mass and possibly orientation of craft’s propulsion system (if moving in respect to rest of craft). Finally one needs to know location of craft’s center of mass or more specifically and as craft’s center of mass is also affected by missing ejected mass and possibly orientation of craft’s propulsion system (if moving in respect to rest of craft).

8. In terms of these quantities craft dynamics can be expressed as and This is a system of three second order differential equations corresponding to three degrees of freedom. Being second order one needs two boundary conditions per degree of freedom. These differential equations can be solved for example numerically by rewriting and expressing These equations can be solved for specified corresponding initial conditions and and then integrated to give for corresponding initial conditions . This procedure is a bit more complex in the case of 3D dynamics modelling as rotations about different axis do not commute and order of rotations is important.

9. From 2D to 1D dynamics modelling One dimensional case may be considered if external force, ejected mass velocity and craft velocity are all aligned. Clearly, this is not very realistic case but it may help us elucidate a few characteristic, basic features of craft dynamics. In that case For special case this further simplifies to which can be rewritten as and integrated to give

10. This can be solved with or For this becomes Clearly for craft to accelerate one needs as by definition . For this can be approximated as with both linear and quadratic time terms.

11. In the context of another simplified (and unrealistic) case one may consider , such that Clearly for craft to accelerate one needs

12. In yet another simplified (and unrealistic) case one may consider , then One could solve this equation by rewriting such that Hence and finally This is again the famous Tsiolkovsky rocket equation.

13. Dynamics: 3D modelling Similar to 2D examplelinear momentum of system is sum of linear momenta of variable mass craft and ejected mass however here angular momentum of system has to be expressed in terms of vector quantities Second Newton law applied to system consisting of both craft and ejected mass for three translational degrees of freedom is expressed as and similarly for three rotational degree of freedom Rate of change of linear momentum of ejected mass is and rate of change of spin angular momentum (angular momentum about CM) of ejected mass is

14. Hence craft dynamics can be expressed by rate of change of craft’s linear momentum and rate of change of craft’s angular momentum In order to solve for craft’s center of mass trajectory and craft orientation in respect to inertial frame of reference one needs to know the ejected mass flow , moment of inertia tensor , as well as and defined respectively as point where ejected mass separate from the craft and point where external force is applied. Dynamics can be also stated in terms of multiple s and s such that and Furthermore, similar to 2D model here and in difference to 2D model here This is a system of six second order differential equations corresponding to three translational and three rotational degrees of freedom. Being second order one needs two boundary conditions per degree of freedom.

15. They can be solved for example numerically by rewriting with , i.e. identity 3 by 3 matrix and by expressing and In difference to 2D model here one needs to solve simultaneously for because moment of inertia tensor depends on craft’s orientation such that and with being time dependent moment of inertia tensor in respect to for example initial orientation and with being rotational matrix. And one should note that rotations do not commute in sense that Or in other words application of rotation , with rotation by angle about unit vector , followed by rotation , is generally not equal to application of rotation followed by rotation unless . Angular velocity can be always expressed in terms of instantaneous axis of rotation and instantaneous angular velocity . And because order of rotations matters as it affects moment of inertia tensor one needs to solve coupled equations for both angle and angular velocity.

16. A few hints for Homework 1 We need here 2 component force and 1 component torque (that is moment of force)

17. In the rest frame of rocket the coordinates of say right wing are In the Earth frame the coordinates of right wing are If rocket speed in Earth frame is and right wing angle relative to rocket is then the unit vector perpendicular to right wing is and only perpendicular component of velocity contributes to drag. Thus And torque about the center of mass due to this force is

18. Similarly one could obtain and apply second Newton’s law that is start with boundary initial condition and by numerical integration obtain trajectory for input with q symbolizing percent of throttling. This is now your simplified physical simulator or descending rocket dynamics.

19. In this problem you have 4 control variables for only 3 controlled degrees of freedom. If you try to minimize the amount of fuel you burn then is essence you are introducing weighting factor on you input control variables. One way to think about this problem is to realize that drag forces are quite large when speed is large. So maybe you want to exploit that feature as much as possible at beginning of descent and then switch fully to thrusters when speed substantially drops. (some numbers: at beginning max drag forces can be ~48kN compared to maximal ~56kN from thrusters) You may use PID type of controller. You probably don’t need I part but just P and D. It is unlikely that you could get away with only P part as you need to minimize the impact speed too. But you may try.

20. PID control To introduce and interpret the popular PID control approach in a simplified manner let’s first consider a one dimensional case and ignore time delays. Pstands for the term proportional to error, Figure 5.2, defined as the difference between current and desired position (also called set point or target position) where desired position may be time dependent and may not be known in advance (i.e. one should not assume knowledge of future desired trajectory). Here we used position, but in principle one could use any other physical or abstract quantity.

21. I stands for integral term, i.e. term proportional to integral of error. And D stands for derivative term, i.e. term proportional to derivative of error. The weighted sum of these three terms constitute the PID control law created by control designer used to define for example commanded force sent to actuators Note that commanded force is specified here by only three parameters and error state. The commanded force then affects real motive force in some non-trivial way

22. Quite generally the role of the three terms P, I, and D, depends on the details of the actual physical system. Control designer fine tunes these parameters by trial and error method. Experienced control designer typically quickly recognizes the type of physical dynamics and relative importance of terms. In the PID “kitchen” the P term typically plays the role of a spring that drives the current position towards an equilibrium position related to the desired position and likely cause oscillations about that equilibrium point. The D term typically plays the role of damping that may decrease the amplitude or completely remove these oscillations. The I term plays a role like P, but has a ‘memory’ of previous errors though integration, and it is typically used to regulate the steady state error; one just needs to make sure that there is not much overshoot and that the solution is stable. In most practical cases only two terms (PD, PI, or ID) are sufficient to provide working though non-optimal solution.