Lecture 25: Modeling Diving I

1. Lecture 25: Modeling Diving I

2. What do we need to do? Figure out what and how to simplify Build a physical model that we can work with Once that is done, we know how to proceed although it may be difficult

3. There are lots of models of the human body designed for multiple purposes We are limited to rigid links but we can certainly replace the muscles and tendons with joint torques Let’s restrict ourselves to planar dives (Our experience so far with the bicycle suggests that three dimensions may mean trouble)

4. Planar motion implies bilateral symmetry — treat both arms and both legs as single mechanisms There are several diving models out there with varying numbers of links I’m going to use an eleven link model of my own devising

5. Parts of the diver The joints neck C1 head shoulder wrist hand C7 upper arm T6 thorax elbow lower arm abdomen L1 pelvis hip thigh knee calf ankle very much not to scale foot

6. I number the links in the following order foot lower leg upper leg pelvis abdomen thorax neck head upper arm lower arm hand will be the reference link I need a reference link because the whole mechanism is flying through the air

7. Formal assumptions diver is confined to the plane all the parts of the diver (called links) are rigid the centers of mass of the links are at their geometric centers the diver can apply torques at all the joints consider only the time between when the diver leaves the platform and when he/she hits the water

8. Formulation I have eleven links, hence 66 variables to start I can confine the system to the x = 0 plane f = π/2 moves the body x axis to the inertial y axis q = π/2 rotates the body z axis out of the plane and erects the body y axis There are 33 of these, so the number of variables is now 33.

13. There are no nonholonomic constraints There are ten connectivity constraints relating the centers of mass I consider the links to be connected along their J axes

14. We have a choice to make at this point: apply the connectivity constraints or transform them to pseudononholonomic constraints Two dimensional connectivity constraints aren’t all that bad and there are no nonholonomic constraints, so I’m applying them There are twenty of these: the 2D centers of mass with respect to one reference link This leaves me with thirteen variables: eleven ys and y5 and z5 (I am using link 5, the abdomen, as my reference link)

15. Now it’s time to assign generalized coordinates We could simply plug the variables into a q vector but I want to do something a little different All the joint torques will act on two links at a time with an equal and opposite torque If I define qs as angle differences, this will be easier to apply

16. So, my first three qs will be y, z and y of the reference link The others will be angle differences

17. There are no constraints, so the constraint matrix is null There will be torques at all the joints, so nothing will be conserved There’s no advantage in using a pure Hamilton approach in terms of p and q We can use our normal approach once we realize that a null constraint matrix means that the null space matrix must span the entire space We can use the 13 x 13 identity matrix as S and proceed from there.

18. and, as usual

19. Put all this together We need to be a bit more precise about the rate of work term but it is pretty easy because everything is two dimensional We need to pick signs, assign a convention Index each torque to correspond to its connection

20. Start from the reference link and proceed distally in each direction I will have the following torques The first subscript is the “base” and the second the “recipient” The general work term is

21. This makes it clear why I chose the generalized coordinates I did The generalized forces become one force per variable, which will be extremely helpful when we design our control It means that each Hamilton equation has at most one force in it

22. We’re going to look at control another day Today let’s just focus on the differential (and algebraic) equations The q equations are pretty near trivial I won’t worry about them for now

23. I want the equivalent equations using the method of Zs We write remember that S is the identity matrix

24. M is pretty complicated in this formulation because we have applied all the constraints There are 107 nonzero components

25. This means 107 nonzero components of Z, each with its own algebraic equation The gradient has 537 nonzero components (out of a possible 2197) Operationally we simply charge ahead and calculate everything directly I suppose we know how to find the Lagrangian how to apply holonomic constraints how to assign generalized coordinates

26. The momentum The Zs

27. Gradients of the Zs

28. Those were the actual ones; we need symbolic equivalents.

29. So we have 26 differential equations (thirteen for q and thirteen for u) and 642 accompanying algebraic equations

30. We can see all of this in context in Mathematica but there is nothing to run until we understand the torques and that’s going to wait until next time