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工廠工藝加工工藝過程卡產(chǎn)品型號CA10B零件圖號831013KG-000產(chǎn)品名稱解放牌汽車零件名稱中間軸軸承支架共 1 頁第 1 頁材料牌號HT200毛坯種類鑄件毛坯外形尺寸26726183每毛坯可制件數(shù)1每臺件數(shù)1備注工序號工序名稱工序內(nèi)容車間工段設備工藝裝備工時/min準終單件010車削 粗車155兩端面金工C630A專用夾具JJ-C016.40020鏜削 粗鏜并半精鏜140內(nèi)孔，倒角C2金工T611A專用夾具JJ-T011.08030銑削 粗銑中間凸平面金工X5030A專用夾具JJ-X010.35040鉆削 鉆中間平面12小孔金工Z5125專用夾具JJ-Z010.10050銑削 銑支架兩側面金工X5030A專用夾具JJ-X021.30060鉆削 鉆側支架12兩小孔金工X61專用夾具JJ-X020.20070鉆削 鉆7.2兩孔金工X61專用夾具JJ-Z020.22100去毛刺 鉗工去毛刺金工110檢驗 終檢質(zhì)檢室描 圖描 校底圖號裝訂號設計（日期）審核（日期）標準化（日期）會簽 （日期）標記處數(shù)更改文件號簽字日期標記處數(shù)更改文件號簽字日期工廠工藝加工工序卡產(chǎn)品型號CA10B零件圖號831013KX-010產(chǎn)品名稱解放牌汽車零件名稱中間軸軸承支架共 7 頁第 1 頁車間工序號工序名稱材料牌號金工010車削HT200毛坯種類毛坯外形尺寸每毛坯可制件數(shù)每臺件數(shù)鑄件2672618311設備名稱設備型號設備編號同時加工件數(shù)臥式車床C630ACM-011夾具編號夾具名稱切削液JJ-C01車床夾具工位器編號工位器具名稱工序工時準件單件6.4工步號工步名稱工藝裝備主軸轉速(rmin-1)切削速度(mmin-1)進給量/mm背吃刀量/mm進給次數(shù)工時機動單件011粗車155mm一端面至52.5mmC630A；硬質(zhì)合金車刀；卡尺188920.32.513.2012粗車155mm另一端面，保證尺寸50mm188920.32.513.2描 圖描 校底圖號裝訂號設計（日期）審核 （日期）標準化（日期）會簽 （日期）標記處數(shù)更改文件號簽字日期標記處數(shù)更改文件號簽字日期工廠工藝加工工序卡產(chǎn)品型號CA10B零件圖號831013KX-020產(chǎn)品名稱解放牌汽車零件名稱中間軸軸承支架共 7 頁第 2 頁車間工序號工序名稱材料牌號金工020鏜削HT200毛坯種類毛坯外形尺寸每毛坯可制件數(shù)每臺件數(shù)鑄件2672618311設備名稱設備型號設備編號同時加工件數(shù)臥式鏜床T611ATK-011夾具編號夾具名稱切削液JJ-T01鏜床夾具工位器編號工位器具名稱工序工時準件單件1.08工步號工步名稱工藝裝備主軸轉速(rmin-1)切削速度(mmin-1)進給量/mm背吃刀量/mm進給次數(shù)工時機動單件021粗鏜140內(nèi)孔至139T611A；單刃鏜刀；卡規(guī)15065.471210.4022半精鏜140內(nèi)孔20087.920.50.510.6023倒角245T611A；單刃鏜刀200902210.04024倒另一面倒角245200902210.04描 圖描 校底圖號裝訂號設計（日期）審核 （日期）標準化（日期）會簽（日期）標記處數(shù)更改文件號簽字日期標記處數(shù)更改文件號簽字日期工廠工藝加工工序卡產(chǎn)品型號CA10B零件圖號831013KX-030產(chǎn)品名稱解放牌汽車零件名稱中間軸軸承支架共 7 頁第 3 頁車間工序號工序名稱材料牌號金工030銑削HT200毛坯種類毛坯外形尺寸每毛坯可制件數(shù)每臺件數(shù)鑄件2672618311設備名稱設備型號設備編號同時加工件數(shù)立式升降臺銑床X5030AXM-011夾具編號夾具名稱切削液JJ-X01銑床夾具工位器編號工位器具名稱工序工時準件單件0.35工步號工步名稱工藝裝備主軸轉速(rmin-1)切削速度(mmin-1)進給量/mm背吃刀量/mm進給次數(shù)工時機動單件031銑中間平面，保證尺寸12mmX5030A；鑲齒套式面銑刀；卡尺10025.121.92.510.35描 圖描 校底圖號裝訂號設計（日期）審核（日期）標準化 （日期）會簽（日期）標記處數(shù)更改文件號簽字日期標記處數(shù)更改文件號簽字日期工廠工藝加工工序卡產(chǎn)品型號CA10B零件圖號831013KX-040產(chǎn)品名稱 解放牌汽車零件名稱中間軸軸承支架共 7 頁第 4 頁車間工序號工序名稱材料牌號金工040鉆削HT200毛坯種類毛坯外形尺寸每毛坯可制件數(shù)每臺件數(shù)鑄件2672618311設備名稱設備型號設備編號同時加工件數(shù)立式升降臺銑床Z5125ZK-011夾具編號夾具名稱切削液JJ-Z01鉆床夾具工位器編號工位器具名稱工序工時準件單件0.1工步號工步名稱工藝裝備主軸轉速(rmin-1)切削速度(mmin-1)進給量/mm背吃刀量/mm進給次數(shù)工時機動單件041鉆12孔Z5125；直柄小麻花鉆；卡尺39214.80.5510.1描 圖描 校底圖號裝訂號設計（日期）審核（日期）標準化（日期）會簽 （日期）標記處數(shù)更改文件號簽字日期標記處數(shù)更改文件號簽字日期工廠工藝加工工序卡產(chǎn)品型號CA10B零件圖號831013KX-050產(chǎn)品名稱 解放牌汽車零件名稱中間軸軸承支架共 7 頁第 5 頁車間工序號工序名稱材料牌號金工050銑削HT200毛坯種類毛坯外形尺寸每毛坯可制件數(shù)每臺件數(shù)鑄件2672618311設備名稱設備型號設備編號同時加工件數(shù)立式升降臺銑床X5030AXM-021夾具編號夾具名稱切削液JJ-X02銑床夾具工位器編號工位器具名稱工序工時準件單件1.3工步號工步名稱工藝裝備主軸轉速(rmin-1)切削速度(mmin-1)進給量/mm背吃刀量/mm進給次數(shù)工時機動單件051銑一側支架，保證尺寸12mmX6030；鑲齒套式面銑刀；卡尺10025.121.92.510.65052銑另一側支架，保證尺寸12mm10025.121.92.510.65描 圖描 校底圖號裝訂號設計（日期）審核 （日期）標準化（日期）會簽 （日期）標記處數(shù)更改文件號簽字日期標記處數(shù)更改文件號簽字日期工廠工藝加工工序卡產(chǎn)品型號CA10B零件圖號831013KX-060產(chǎn)品名稱解放牌汽車零件名稱中間軸軸承支架共 7 頁第 6 頁車間工序號工序名稱材料牌號金工060鉆削HT200毛坯種類毛坯外形尺寸每毛坯可制件數(shù)每臺件數(shù)鑄件2672618311設備名稱設備型號設備編號同時加工件數(shù)臥式銑床X61ZK-021夾具編號夾具名稱切削液JJ-X02銑床夾具工位器編號工位器具名稱工序工時準件單件0.2工步號工步名稱工藝裝備主軸轉速(rmin-1)切削速度(mmin-1)進給量/mm背吃刀量/mm進給次數(shù)工時機動單件061鉆12兩孔X61；直柄小麻花鉆；卡尺38014.320.5520.2描 圖描 校底圖號裝訂號設計（日期）審核（日期）標準化（日期）會簽（日期）標記處數(shù) 更改文件號簽字日期標記處數(shù)更改文件號簽字日期工廠工藝加工工序卡產(chǎn)品型號CA10B零件圖號831013KX-070產(chǎn)品名稱 解放牌汽車零件名稱中間軸軸承支架共 7 頁第 7 頁車間工序號工序名稱材料牌號金工070鉆削HT200毛坯種類毛坯外形尺寸每毛坯可制件數(shù)每臺件數(shù)鑄件2672618311設備名稱設備型號設備編號同時加工件數(shù)臥式銑床X61ZK-031夾具編號夾具名稱切削液JJ-Z02鉆床夾具工位器編號工位器具名稱工序工時準件單件0.22工步號工步名稱工藝裝備主軸轉速(rmin-1)切削速度(mmin-1)進給量/mm背吃刀量/mm進給次數(shù)工時機動單件071鉆7.2兩孔X61；直柄小麻花鉆；卡尺59022.230.420.22描 圖描 校底圖號裝訂號設計（日期）審核 （日期）標準化（日期）會簽 （日期）標記處數(shù)更改文件號簽字日期標記處數(shù)更改文件號簽字日期FP7 2:30 AN EFFICIENT ROBOT ARM CONTROL UNDER GEOMETRIC PATH CONSTRAINTS1 Kang G. Shin and Neil D. McKay Department of Electrical and Computer Engineering The University of Michigan Ann Arbor, Michigan 48109 ABSTRACT Conventionally, robot control algorithms are divided into two stages, namely, path planning and path tracking (or path con- tml). This division has been adopted mainly as a means of allevi- ating difficulties in dealing with complex, coupled robot arm dynamics. Unfortunately, the simplicity obtained from the division comes at the expense of efficiency in utilizing robots capabili- ties. To remove at least partially this inefficiency, this paper con- siders a so!ution to the problem of moving a robot arm in minimum time along a specified geometric path subject to input torque/force constraints. We first describe the robot arm dynam- ics using parametric functions which represent geometric path constraints to be honored for collision avoidance as well as task requirements. Secondly, constraints on input torques/forces are converted to those on the parameters. Finally, the minimum-time solution is deduced in an algorithm form using phase-plane tech- niques. 1. lntroduction During the past several years a great deal of attention has been focused on industrial automation techniques, especially the use of general-purpose robots. Since the purpose of industrial robots is to increase productivity, an obvious question to ask is how robots should be controlled so as to produce as many units as possible per dollar invested. The usual assumption is that fixed costs dominate the cost per item produced, so that it is desirable to produce as many units as possible in a given time. There are a variety of algorithms available for minimum-time or near-minimum-time robot arm controi. These algorithms usually assume that the control structure of the robot has been divided into two levels. The first level is called path planning, and the second level is called path cmfrol or path tracking The usual definition of path control is the act of attempting to make the robots actual position and velocity match desired values of posi- tion and velocity; the desired ,values are provided to the con- troller by the path planner. The path planner receives as input control 11, resolved acceleration control2, and various adaptive techniques3-5. Unfortunately, the simplicity obtained from the division into path planning and path tracking comes at the expense of effi- ciency. The source of the inefficiency is the path planner. In order to use the robot efficiently, the path planner must be aware of the robots dynamic properties, and the more accurate the dynamic model is, the better the robots capabilities can be used. However, most of the path planning algorithms presented to date assume very little about the robots dynamics. The usual assump- tion is that there are constant or piecewise constant bounds on the robots velocity and acceleration 6,7. In fact, these bounds vary with position, payload mass, and even with payload shape. Thus in order to make the constant-upper-bound scheme work, the upper bounds must be chosen to be global greatest lower bounds of the velocity and acceleration values; in other words, the worst case limits have to be used. Since the moments of inertia seen at the joints of the robot, and hence the acce;eration limits, may vary by a factor of three or more, such bounds can result in considerable inefficiency or under-utilization of the robot. To alleviate the inefficiency, this paper presents a solution to the minimum-time robot arm path control problem subject to constraints on its geometric path and input torques/forces. The solution will be in the form of a path planning algorithm, and will take into account the detai!s of the dynamics of the robot arm. The output of the path planner will be the true minimum-time solu- tion, and so will be useful as a standard against which the perfor- mance of other path planning algorlthms may be measured. Note that the problem and its solution considered in this paper are dif- ferent from the near minimum-time control methods in 8,9. The remainder of this paper is divided into five sections. Section 2 describe; a method for making the robot arm dynamic equations more tractable and a method for handling input torque constraints. Section 3 contains a detailed formulation of the minimum-time control problem. In Section 4, the form of the optimal solution is deduced using phase-plane techniques. Sec- tion 5 presents the highlight of this paper, that is, an a!gorithm for generating optima.l(i.e. minimum-time) trajectories. The final sec- tion is a discussion of the significance of the results. some sort of spatisl path descriptor from which it calculates a time historv of the desired Dositions and velocities. The oath 2. Robot ynamics with Constraints tracker thLn takes care of any deviations of the actual position Before delving headlong into the problem of minimum-time and velocity from the desired va:ues. control, consider the behavior of the system to be controlled, that The for dividing the control scheme in this way is that is, a dynamic model of the robot arm. There are a number of ways the process of robot control, if sonsidered in its entirety, is very of obtaining the dynamic equations of a robot arm, i.e. the equa- complicated, Since the dynamics of all but the simplest robots are tions which relate joint forces and torques to positions, veloci- highly and coupled, i,idi, the controller into the two ties, and accelerations. The two most common methods are the parts makes the whole process simpler. h path tracker is fie- Lagrange method and the Newton-mer method. The Newton-Euler quently a linear controller(e,g, a controller), While the method is cornputationally efficient, but is a recursive formulation linearities of robot arm dynamics frequently are not taken into which is hard to deal with in control problems. The Lagrange for- account at this level, such trackers can generally keep the robot mulation, while not computationally efficient, does yield a set of arm fairly to the desired More sophisticated differential equations which are easy to manipulate for robot con- methods can be used, though, such as resolved motion rate trol problems. Since the dynamic equations will be used here only to obtain analytical results, we have used Lagranges method to The work reported here IS supported 10 ,?art by the US AFOSR contract Nc. derive the following robot arm dynamic equations 23 3i F49E23-82-C-0089 and Robot Systems 31 vision, Center for Robotics and Integrat- ed h?aufactur,ng(CRIM), The Universttj of Michigan, Ann Arbor, Michigan. Any opinions, f .togs, and conclusions OT recommendations in this paper are those ot tb authors and do not necessvily reflect the view of the funding agencies. 1449 0191-2216183/0000-1449 S1.00 E 1983 IEEE Authorized licensed use limited to: Nanjing Southeast University. Downloaded on October 23, 2009 at 08:14 from IEEE Xplore. Restrictions apply. (1 a) =,(n)VJ+,j+(n)dvt+,(n) (1 b) where q! =it generalized coordinate v =ith generalized velocity Q =if generalized force Jij = the inertia matrix G, = gravitational force on the it joint ci, = Coriolis force array Kj = viscous friction matrix The Einstein summation convention has been used, and all indices run from one to n inclusive for an n-degree-of-freedom robot. The elements of the inertia matrix J,j are constants of pro- portionality which relate the torque/force exerted on the ifh joint to the acceleration of the jth joint. The entries ci,k of the Coriolis array describe the force felt at joint i due to the veloci- ties of joints j and k. The viscous friction matrix qj gives the frictional force felt at joint i due to the velocity at joint j. Note that this matrix is diagonal, and all the entries are non-negative. The motion of the robot arm will not, of course, be completely unconstrained. In fact, it will later be assumed that the robot arm must be constrained to a fixed path in joint space,2 and that the path is given as a pamrneterized cume. The curve is assumed to be given by a set of R. functions of a single parameter A, so that we are given q* = f i(h), 0 I x A, (2) where A= a parameter for describing the desired path, and it is assumed that the coordinates q vary continuously with A and that the path never retraces itself as X goes from 0 to A, i.e. It should be noted that in practice the spatial paths are given in Cartesian coordinates. While it is in general difficult to convert a curve in Cartesian coordinates to that in joint coordi- nates, it is relatively easy to perform the conversion for individual points. One can then pick a sufficiently large number of points on the Cartesian path, convert to joint coordinates, and use some sort of interpolation technique (e.g. cubic splines) to obtain a similar path in joint space (seelO for an example). Returning to the problem at hand, we may use the parame- terization of the q* and differentiate with respect to time, giving h(O)=O, h(tj)=hm,. where p=h. The equations of motion along the curve (Le. the geometric path) then become h=p (4a) dh dh dh Note that if h is used to represent arc length along the path, then /.L and p are the velocity and the acceleration along the path, respectively. With this parameterization, there are two state variables, i.e. h and p, but (n + 1) equations. One way to look at the system is to choose the equation h=p and one of the remaining equations as state equations, regarding the other equations as constraints on the inputs and on p. However, a better way of obtaining a sin- gle state equation from the n equations given is to multiply the i equation by and sum over i, giving dh Csrtesian space to joint space is needed here. Since any detailed treatment of the 2A path is normally described in Cartesian space and the conversion from mnversion is cut of the scope of this paper, oniy brief remarks on this aregiven in the next pararaph. This formulation has a distinct advantage. Note that the coeffi- cient of p is quadratic in the vector of derivatives of the con- straint functions. Since a smooth curve3 can always be parameterized in such a way that the first derivatives never all disappear simultaneously, and since the inertia matrix is positive definite, the whole equation can be divided by the non-zero, posi- tive coefficient of p, providing a solution for p in terms of h and p. Now there are only two state equations, and the original equations can be regarded as constraints on the inputs and on p (more on this will be discussed later). With this formulation, the state equations become h=p (6a) (6b) Consider now the constraints on the inputs, namely 1% 1 l*(tf)Pf (Sa) x:o)=G, h(tf )=A, (9b) 3.1. Application of the Maximum Principle In order to apply the state constraint Oshh, it is con- venient to add a third state equation. Call the third state v, and let the third equation have the form I/=X1(-h)+(h,-X)Z1:h-X,) (1 0) where , Note that bG, so that the boundary conditions v(O)=v(t,)=G force 1- to be identically zero. But the only way for v to be Identi- cally zero is to have OhIX, thus forcing h to remain in the desired interval. Before doing any further manipulations on the state equa- tions, define the functions U (A) =u, c dh S(A)=G,;h) q dh Again, for convenience the dependence of the above coefficients on h will be omitted in the sequel. Now rewrite the state equa- tions in the following form h=p (1 la) The .U term is a quadratic form reminiscent of the expression for the robot arms kinetic energy. In fact, if the parametric expressions for the Q are plugged into the formula for kinetic energy, one obtains the expression K=Mpz/2. The Q term represents the components of the Coriolis and centrifugal forces which act along the path plus the artificial forces of constraint generated by the parameterization. The R term represents fric- tional components, and S gives the gravitational force along the path. i/ is the weighted input term. Under the above setting the problem MTPP can be converted to the following. Problem MTPP: Find y = (h,u*,v*) and Li by minimizing (8) subject to (1 1 a), (1 1 b), (1 IC), (7d), (sa), and (Sb). meet task requirements 4This is done at the stage of task planning io avoid cclllsion as ne/ as to 1451 Authorized licensed use limited to: Nanjing Southeast University. Downloaded on October 23, 2009 at 08:14 from IEEE Xplore. Restrictions apply. For the MTPP problem the Hamiltonian now becomes 1 Or using the foregoing substitutions, the Hamiltonian is Differentiating with respect to p, we obtain Differentiating with respect to X, pi= - = - - -p aH PZ ai.: d -p- d d-c ah M ah dh dh dh M Finally, differentiating with respect to v, Thus if we are to apply the maximum principle to this problem, we must minimize H in (12b). In addition, constraints (1 la), (1 lb), (1 1 c), (gal, (9b), and (7d) must be met, and H must satisfy the boundary condition. Here, the state vector y is the vector (h,p,v), and we have a sin- gle input term U(tf)=q (Amzx) - df (Ama). In (1 4) we used the dh fact that H does not explicitly depend upon t. Also one can see y(tf) = (Amzx, pl, O)T from the boundary conditions (9), and the condition v(t,)=O. Note that the Hamiltonian functional (12b) is linear in lJ, and 1: is bounded because of the boundedness of u, and dfc in O, Amax. This implies that the optimal solution for I: should fol- low a bang-bang policy. That is, at any point on the optimal tra- jectory the quantity li in (1 2b) must assume its maximum or minimum. The extremum of I can be obtained by maximizing or minimizing itself with respect to the u,. S,ince we may write the equality constraints on the u, as q=g,(A)pth,(h,p), we have dh Because of the bang-bang nature of the control Ii , and because for a given state (A+) the quantity I; is linearly related to /. through the above equation, will also be bang-bang. Therefore p will be equal to either GLB(?,p) or LLB(h,p). Thinking of the three dimensional case again, p must put the input at one of the points where the line formed by the input equality constraints meets one of the sides of the prism formed by the inequalities. If the i-th joint input is on one of the sides of the constraint box, then it is driven at its maximum capability, while the others in general are not. The slowest joint, then, is pushed as hard as it will go while the others exert the proper amount of force to keep the robot arm on the right path. The above discussion is valid whenever the coefficient of the input term is not zero, i.e. when p2 in (1 3a) is not zero. If p2 is zero at isolated points only, then the optimal control is deter- mined almost everywhere. On the other hand, if pz is zero on some interval, we have the following theorem. Theorem 1: If pz is zero on some interval ;tl,t2 (t,S(O)li,(O) then p2(0)O. Proof: It is known that Olhlh, so that at t=tf we must have pl0. Otherwise, since p(t,)=O, we would have to have had p(t)O for some t hmax. But at tf we have ,u(t,)=M-(C;-S)O, this gives U-StO. Now the value of H at time tf is 0, so that If p2(tf)s0, then H(t,)O, so we must have pztZ)O. To determine the sign of p2(0), consider the quantity b(0). By an argument similar to that given above, p(O)O. This gives the result C:-SO. Since the control is bang-bang, we must have U=L:, since otherwise LJ= C7fin and U,-SO. But if U=Umax, then p2 must be less than zero. Therefore pz(0)O. Q.E.D. One consequence of these theorems is that the number of switching points is odd. If the number of switching points were even, then the sign of pz(tf) would be the same as that of p2(0), since s3n(p2:tf)=(-l)msgn(p2(0), where m is the number of slgn changes. 4. Phase Plane Interpretation At this point, it is instructive to look at the systems behavior in the phase plane. The equations of the phase plane trajectories can be obtained by dividing equation (1 1 b) by (1 1 a). This gives It is interesting to note that the total time T that it takes to go from initial to final states is The idea, then, is to minimize this integral subject to the given 1452 Authorized licensed use limited to: Nanjing Southeast University. Downloaded on October 23, 2009 at 08:14 from IEEE Xplore. Restrictions apply. constraints. We therefore want to make p as large as possible, a result which would be expected intuitively. The constraints on have two effects. One effect is to place limits on the slope of the phase trajectory. The other is to place limits on the value of p. To obtain the limits on $, one simply divicies the limits on ,u by b, since %h,gjO. Then the left-hand side of the inequality (1 7b) is a parabola which is concave upward, whereas for (1 7c) it is concave downward. When the parabola is concave downward, then the inequality holds when is between the two roots of the quadratic. If the parabola is concave upward, then the inequality holds outside of the region between the roots(Figure 1 ). Thus in one case p must lie within a closed inter- val and in the other it must lie outside an open interval, unless of course the open interval is of length zero. In that case, the ine- quality constraint is always satisfied and the roots of the qua- dratic will be complex. Since the admissible values of p are those which satisfy all of the inequalities, the admissible values must lie in the intersec- tion of all the regions determined by the inequalities. There are R (n -:)/ 2 inequalities which give closed intervals, so the inter- section of these regions is also a closed interval. The other n (%-:I/ 2 inequalities, when intersected with this closed inter- val, each may have the effect of punching a hole in the interval(Figure 2). It is thus possible to have, for any particular value of A, a set of admissible values for p which consists of as many as n(n+ I)/ 2 distinct intervals. When the phase portrait of the optimal path is drawn, it may be necessary to have the optimal trajectory dodge the little islands which can occur in the admissible region of the phase plane. (Hereafter, these inad- missible regions will be referred to as islands 3f inadr.LF.si.biiZty or just islands) It shoulc! be noted, though, that if there is no friction, then P,j=O. which means tha.7 in the concave upward case the inequality is satisfied for all values of k. Thus in this case there will be no islands in the admissible region. In addition to the constraints on ,LL described above, we must also have p 2 3. This can be shown as follows: if p 0, then the trajectory has passed below the line p = 0. Below this line, the trajectories always move to the left, since p = dX/ cit 9. Since the optimal trajsctorj must approach the desired final state through positive values of p, the trajectory would then have to pass through p = 0 again, and would pass from p 0 at a point to the left of where it had passed from p C to p h,X2 then there must be a zero. If g(h) is not continuously differentiable, assume that no zero exists. Then there are one or more points where g (x) has a discontinuity in its derivative, and a sign change must occur at one or more of these points. If this were not so, then there would have to be a sign change at a point where ?(A) is continu- ous, and hence there would be a zero. Call the first of these points Ad. This being the first (smallest) value of h where a sign change occurs, the change must be from positive to negative. Then at Ad two of the (continuously differentiable) constraint curves which generate g (A) are equal. Call the two active con- straints g, and Y, g1 being active for hhd. Then, smce llmv(X) 0 lim;s(h), we must have h4$ h+h$ But then, for some E we would have Except for higher order terms, this is just g,(Xdtc! hi, contrary to hypothesis. Therefore there must be at least one zero of g(h). The graphical meaning of this theorem is illustrated in Figure 7. Note from the figure that any cusps in 9 (A) must