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Lecture 25: Modeling Diving I

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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

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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)

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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

C7

ankle

knee

hip

L1

T6 elbow

wrist shoulder

C1

The joints

very much not to scale

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head

foot

calf

thigh

pelvis abdomen

thorax

hand

lower arm upper arm

Parts of the diver

neck

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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

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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

Formal assumptions

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Formulation

I have eleven links, hence 66 variables to start

I can confine the system to the x = 0 plane

€

xi = 0, φi =π2

= θi

φ = π/2 moves the body x axis to the inertial y axis θ = 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.

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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

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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 ψs and y5 and z5

(I am using link 5, the abdomen, as my reference link)

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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

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So, my first three qs will be y, z and ψ of the reference link

The others will be angle differences

€

q =

y5

z5

ψ5

ψ5 −ψ4

ψ4 −ψ3

ψ3 −ψ2

ψ2 −ψ1

ψ6 −ψ5

ψ7 −ψ6

ψ8 −ψ7

ψ9 −ψ6

ψ10 −ψ9

ψ11 −ψ10

⇔

abdomen - pelvispelvis− upper leg

upper leg - lower leglower leg - foot

abdomen - thoraxthorax - neckneck − head

thorax - upper armupper arm - lower arm

lower arm - hand

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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.

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€

˙ W = Ti ⋅ ω ii=1

NL

∑ ⇒ Qi =∂ ˙ W ∂ ˙ q i

€

˙ q i = S ji u j = ui

€

˙ p i =∂L∂qi + Qi =

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Mmn

∂qi umun −∂V∂qi + Qi

and, as usual

€

˙ p i =12

Mij ˙ u j +12∂Mij

∂qk u juk

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Put all this together

€

Mij ˙ u j = −∂Mij

∂qk u juk +12

M jk

∂qi u juk −∂V∂qi +

∂ ˙ W ∂ ˙ q i

€

˙ q i = ui

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

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Start from the reference link and proceed distally in each direction

I will have the following torques

€

τ 54,τ 43,τ 32,τ 21τ 56,τ 67,τ 78τ 69,τ 910,τ1011

The first subscript is the “base” and the second the “recipient”

The general work term is

€

τ ij ˙ ψ j − ˙ ψ i( )

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This makes it clear why I chose the generalized coordinates I did

The generalized forces become

€

Q = 0 0 0 −τ 54 −τ 43 −τ 32 −τ 21 τ 56 τ 67 τ 78 τ 69 τ 910 τ1011{ }

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

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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

€

˙ q i = ui

I won’t worry about them for now

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€

Mij ˙ u j = −∂Mij

∂qk u juk +12

M jk

∂qi u juk −∂V∂qi +

∂ ˙ W ∂ ˙ q i

I want the equivalent equations using the method of Zs

We write

€

pi =12

Mij ˙ q j =12

MijSnjun =

12

Mijuj = Ziju

j

remember that S is the identity matrix

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M is pretty complicated in this formulation because we have applied all the constraints

€

M =

• 0 • • • • • • • • • • •

0 • • • • • • • • • • • •

• • • • • • • • • • • • •

• • • • • • • • 0 0 0 0 0• • • • • • • • 0 0 0 0 0• • • • • • • • 0 0 0 0 0• • • • • • • • 0 0 0 0 0• • • • 0 0 0 0 • • • • •

• • • • 0 0 0 0 • • 0 0 0• • • • 0 0 0 0 • • 0 0 0• • • • 0 0 0 0 • 0 0 • •

• • • • 0 0 0 0 • 0 0 • •

• • • • 0 0 0 0 • 0 0 • •

There are 107 nonzero components

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This means 107 nonzero components of Z, each with its own algebraic equation

The gradient

€

∂Zij

∂qkhas 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

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The momentum

The Zs

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Gradients of the Zs

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Those were the actual ones; we need symbolic equivalents.

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

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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