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In order to not overcrowd the picture, for the complementary two shadow components only their projections on the lower base circle, (1 , 5 ) and (6 , 10 ), are shown Unlike the RR arm, each point in W -space of our RP arm has only one arm solution That is, there is one-to-one mapping between W -space and the corresponding C-space, as compared to the one-to-two mapping in the case of the RR arm Because of this, and also because virtual lines in W -space are always simple curves, the virtual boundaries in C-space are also simple curves Recall that this constitutes a necessary condition for the basic path planning procedures (Section 33) In general, each virtual boundary in C-space of the RP arm consists of a combination of three distinct segments: 1.

A curve formed when the front or the rear end of link l2 follows the boundary of the actual obstacle; for obstacle B in Figure 524b this segment passes through points 7-8-9 2 A vertical straight-line segment formed when points of the arm body other than the arm endpoint touch the obstacle while passing around it; in Figure 524b this consists of lines 7-6-12 and 9-10-11)..

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of the desired output; and the derivative feedback component enhances the system stability and reduces oscillation due to control. Equations characterizing the control law here will come out as (see Ref. 7), n0,1 = K01 ( 1 1d ) + L01 e1 + H01 1 e 1 = 1 1d and n1,2 = K12 ( 2 2d ) + L12 e2 + H12 2 e 2 = 2 2d (2.17) (2.16)

ME: What CRAIG: Yep I got bold when I should have sat on my profits Think about how I made the money in the first place: I planned a trade using all sorts of research I planned the entry with knowledge of important support and resistance areas I didn t put all my trading capital into the first position Instead, I built it up over time It took me over two months of research to put the trade together I d been piecing together my plan for weeks before I ever sat at the trading desk at Ernest Wellington That s why I was so confident about it The planning that I put into it ME: And the other trades CRAIG: I kept buying and selling the EUR/JPY I thought it had some magical quality that it had been built just for me and was my new best friend.

Steady-State Analysis. Assume that the commanded angles ( 1d , 2d ) are constant, the disturbances are constant, and the control system is stable. System stability implies that eventually all time derivatives approach zero, which, using Eq. (2.14), gives

wire [32* Nb-1:0]

1 = 1d 2 = 2d 1 1 m2 l2 g2 cos( 1d + 2d ) + n2,3 e2 = L12 2 1 l2 cos( 1d + 2d ) f2,3y l2 sin( 1d + 2d )f2,3x L12 1 1 L12 e2 + m1 + m2 l1 g2 cos 1d e1 = L01 2 + + 1 l1 cos 1d f2,3y l1 sin 1d f2,3x L01 (2.18)

Analysis of these equations shows that steady-state tracking will occur regardless of disturbances. Note that the value of error e1 must compensate for the value

wire [32* Nk-1:0]

So I took oversized positions, trying to buy it up to 13100, but I couldn t stay in the positions for long because I was trading huge amounts of capital from the start, instead of building the position over time I had no idea what I was doing ME: Sounds like you were trying to get back the position you had before you closed it CRAIG: Right But I couldn t buy the pair at 11900 anymore! When I closed the big trade, I lost the right to hold the EUR/JPY at a superduper discounted price So when I started trading again, I wanted to get all that lost profit, and took monster-size positions What a mistake ME: Why didn t you just go back to planning big, long-term trades all over again CRAIG: Right I should have done the same preparation all over again.

of e2 but not vice versa, which is understandable given the links sequential connection. It can be shown that if the gains L01 and L12 are zeroes, then joint values 1 and 2 differ from their desired values by amounts proportional to the disturbance values and inversely proportional to the feedback gains K01 and K12 . If, in turn, both gains K are zero, then the system s steady state is independent of the commands and re ects only the balance of gravitational and disturbance forces. This of course indicates the importance of the feedback.

wire [3:0]

Dynamic Stability Analysis. This is done to verify the stability hypothesis. Suppose that n2,3 = 0, f2,3x = 0, and f2,3y = 0. Assume a small initial error 1 (t),