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60 CHAPTER 6. CONSTRAINED RIGID BODY DYNAMICS Figure 6.1: Illustration of index sets, complementarity conditions and pivoting. Thick lines correspond to index sets — constraint i belongs to an index set if point (λi, ai) lies on the thick line associated with the set. Circles indicate “singular points”. 0 ai ✻ i ∈ NH ✎☞i ∈ C ✍✌ ✻ ✎☞ ✍✌ λlo i 0 λhi i ✲ i ∈ NL ✲ λi The algorithm starts with letting λ = 0 (hence a = b) and then takes c iterations to update λ. At the i-th iteration it computes λi so that (6.18) would hold for constraint i and in doing so it adjusts λj, 1 ≤ j < i so that the conditions for constraints j, that were processed at previous iterations, would remain obeyed (constraints j, j > i are ignored at the i-th iteration). When all c constraints are processed, all conditions are obeyed and λ contains the value we were solving for. Index Sets Index sets NH, NL, C are defined as a mutually disjoint subsets of constrained indices {1, . . . , i}, where i is the number of the current iteration, that characterize what conditions of (6.18) are currently obeyed by constraints j ≤ i. We will say that the constraint (index) j is • clamped and belongs to a set C if λ lo j < λj < λ hi j ∧ aj = 0. • non-clamped hi and belongs to a set NH if λj = λ lo j ∧ aj > 0. • non-clamped lo and belongs to a set NL if λj = λ hi j ∧ aj < 0. • non-clamped and belongs to a set NC = NH ∪ NL if either j ∈ NH or j ∈ NL. Whenever updating λi, a changes as well and it must be ensured that each j, j < i remains in one of the index sets so that (6.18) would not be violated for j < i — j either remains in its current index set or transfers from an index set to another “adjacent” set (this is called pivoting). Cases where λj = λ lo j ∧ aj = 0 or λj = λ hi j ∧ aj = 0 are “singular” and indicate that j should be transferred to a neighboring index set (see (6.1)). For example if j ∈ C and λj reaches λlo j , j is transferred to NH5 . Constraint j can not be in NH (NL) if λ lo j = −∞ (λ hi j = +∞). Constraint j can not be in C if λlo j = λhi j = 0, such a constraint can be ignored. Index sets and the pivoting process is illustrated by diagram (6.1) specific to a constraint i. Point (λi, ai) in the diagram is restricted to lie on one of the three disjoint lines that correspond to index sets NH, C, NL and the pivoting corresponds to transfers of (λi, ai) from one line to the other neighboring line when a “singular point” is reached. 5 Due to the definition of NH, ai must also be tweaked so that ai > 0.
6.2. ACCELERATION-LEVEL CONSTRAINTS 61 Iteration The algorithm loops over i, at iteration i it computes λi and ai while maintaining the complementarity conditions on λj and aj for j < i. Indices j > i are ignored at the i-th iteration. Let us look at the i-th iteration more closely. When the i-th iteration is started, λi = 0 and ai = Ai,C∪NC · λC∪NC + bi. If (λi, ai) already lies on a line of the complementarity diagram due to constraint i then (6.18) already holds for i, index i is assigned to the corresponding index set and the algorithm can proceed to the next iteration. Let us assume that (λi, ai) is not on a line. Our goal is to “drive” i to NH, C or NL. We will do so in multiple steps. At each step we compute a vector ∆ λ, the adjustments to λ, that will make (λi, ai) “move towards” a line in the diagram. This will be looped until (λi, ai) ends up on a line. Index i will be then assigned to the corresponding index set and the iteration will finish. To ensure that conditions for constraints j < i are not violated while taking a step during iteration i, we will require that clamped indices remain clamped and non-clamped indices remain non-clamped or a “singular point” is reached. If (λj, aj) reaches a “singular point”, j is transferred to the neighboring index set. We will now derive what ∆ λ should look like. • Since we are not interested in constraints j > i at the i-th iteration, ∆ λj>i = 0. • If ai > 0 we would like to decrease λi so that i would eventually reach NH, hence ∆λi should be a negative non-zero value. If ai < 0 we would like to increase λi so that i would eventually reach NL, hence ∆λi should be a positive non-zero value. However we do not know what the exact value of ∆λi should be. To tackle this problem we will write λdir = ∆λi = ∓1 (a λ “direction”) with a note that we will actually add ∆ λ · scale to λ and scale > 0 (a scaling factor) will be determined later. • If ∆ λ · scale was added to λ, a = A · λ + b would change by ∆a · scale = A · ∆ λ · scale. From the requirement that clamped indices remain clamped or reach “singular points” and are transferred to NH or NL we get a condition ∆aC = 0 and from the requirement that non-clamped indices remain non-clamped or reach “singular points” and are transferred to C we get ∆ λNC = 0. Putting these two requirements together we get that ∆ λC = −A −1 CC · AC,i · ∆λi. This is easy to show. Since we work with matrix slices we can assume (without loss of generality) that C = {1, . . . , k} and NC = {k + 1, . . . , i − 1}. Constraints j > i are ignored at this iteration and to simplify writing we can assume that this is the last iteration. Then, utilizing the requirement ∆λNC = 0, ∆λi = λdir and letting ∆λC = x, ⎛ A = ⎝ ACC AC,NC AC,i ⎞ ⎠ ∆ ⎛ λ = ⎝ x 0 ⎞ ⎠ ANC,C ANC,NC ANC,i Ai,C Ai,NC Aii ⎛ λdir ⎞ ∆a = A · ∆ ACC · x + AC,i · λdir λ = ⎝ ANC,C · x + ANC,i · λdir ⎠ Ai,C · x + Aii · λdir and since it is required that ∆aC = 0 we get that x = ∆λC = −A −1 CC · AC,i · λdir. We thus have ∆ λC = −A −1 CC · AC,i · λdir, ∆ λNC = 0, ∆ λi = λdir, ∆ λj>i = 0 (6.19)
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Univerzita Karlova v Praze Matemati
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Contents 1 Introduction 1 1.1 Backg
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CONTENTS vii 8 Figure Library 103 8
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2 CHAPTER 1. INTRODUCTION data from
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6 CHAPTER 1. INTRODUCTION
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8 CHAPTER 2. DIFFERENTIAL EQUATIONS
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122 BIBLIOGRAPHY [16] Katsuaki Kawa
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