Advance Mechanics of Machines for M.tech

7/25/2019 Advance Mechanics of Machines for M.tech

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CHAPTER 10

ADVANCED KINEMATICS OF THE PLANE MOTION

This chapter deals with a number of concepts of kinematics which,though quite fundamental and of considerable usefulness in mechanismdesign, are usually omitted from undergraduate curricula.

10 1. The Inflection Circle; Euler Sa vary Equation

As shown in Chaps. 4 and 6 radii of cu rv ature of the paths of movingpoints are of great importance in kinema tic analysis of mechanisms.They are of equal sig nifi cance in synthesis. The only method discussedthus far of determining the path curvature is based on the knowledge ofthe velocity and acceleration of the point concerne d : p = v2/an orp = v3/ V.flu - vua,., . Clearly, as far as mechanisms are concerned, t hisis a roundabout procedure, because here, as in any const ra in ed motion,the points describe paths which are a geometrical property of the system,and thus independent of the actual velocities and accelerations. Theinflection circle, dealt with in this sectio n, offers a direct, purely geometrical solution of the probl em .

Figure 10 -1 shows the fixed and moving polodes, and ?rm whichcharacterize the displacement of the plane m; 0 1 and m are th e centers of

curvature of the polodes at the ir point of contact-the instant velocitypole P 1m-deno ted here by P; Pn is the pole normal, wh ich originates atP and points away from 1 see also Fig. 3-22); p is the pole tangent, thepositive sense of which is obtained by turning Pn through goo in the positive, i.e., counterclockwise, sense .

S is an arbitrary point on m, defined by the ray angle 0,, measuredcounterclockwise from Pn, and the distance PS In the following, dista nces on a ray will be treated as directed line elements, i.e ., taken aspositive or negative, the positive sense being from P to the moving point.

Hence fs = P - S is alway s positive. The overbar will be used toindi cate that the particular quantity is directed.) The positive sensealong a line perpendicular to a given ray is obtained by turning the posit ive ray through goo counterclockwise. Os is the instantaneous centerof curvature of the path of S

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218 KINEMATI CS A;\D DYNAMICS OF PLAN E MECHANISMS

Because of an infinitesimal rotation d


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DV NCED KINEM TICS OF T H E P L N E MOTION 219

With the substitutiondl

- = dd Pm

d)

where d; has the dimension of a length, Eq. c) becomes

Os---+ S 1Os---+ P) P---+ S) cos e. = d,

or0 8 --- P) + P---+ S) 1

Os---+ P) P---+ S) cos e. = d,

and 1 1 ) 1Os---+ P + P---+ S cos e. = [, (10-1)

Usually, the only directly available, or eas ily obtainable, details of themotion of a plan e are the position of the velocity pole and the angularvelocity. The direction of the pole normal, specified by the angle - e

8R a y s

o, OmPn

FIG. 10-2

from a given ray, and the length d, are not known. The objective of thefollowing investigation is the determination of these two quantities, topermit the general evaluation of Eq. (10- 1).

The Inflection Circle. Equation (10-1) shows that points such as St,s2 etc ., on a given ray describe paths which differ in shape in the vicinityof the position under observation. t is therefore reasonable to assumethat there exists on the ray a point I , which passes, a t the given instantthrough an inflection of its path. For such a point, the radius of curva-ture of the path and the distance 01 , --- P are both infinite. Equation

(10-1), applied to I,, reduces, therefore, to

and

1 1P

_ _ _ _ I cos e. = d.P . I . = d, cos e. 10-2)

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220 KINEMATICS AND DYNAMICS OF PLANE MECHANISMS

whi ch shows that on any given ray there really exists one, and only one,inflection point and that the locus of all inflection points in a givenposition of the moving plane m is a circle of diameter d;, centered on p .

and passing through P. This circle is known as the inflection circle.Figure 10-2 shows the inflection circle for the kinematic system ofFig. 10-1.

The following alternative derivation of Eq . 10-2) is interesting.By Eq. 3-17),

However,as = ap. + asp.),. + asp.),

asP.)n = - PS)w.., 2i, = - P-+ S)w,.2

and by Eq. 3-15),

d ap. = w.. ; lp .From Fig . 10-3, as) = ap. cos 8,)i, asp.),.

= w 2d; cos B)i, - P-+ S)w,. 2

Rays

F10 . 10-3

For the inflection point / , , Pr = oo and ar,),. = 0, so that

and0 = w., 2d;{cos 8,)i, - P-+ l,)w,. 2

P-+ /, = d, cos 8,)i, 10-2)

Th e unit vector i, may be omitted because d; is in effect a directed length.The result obtained shows that d,, as defined by Eq. 3-15), is identicalwith d, = dl/d f',.. A more direct proof of this fact is given below.

Euler-Savary Equation. The first form of the Euler-Savary equationis obtained by combining Eqs. 10-1) and 10-2), thus:

1 1Os-+P P - S 1 10-3)P - + I ,

To make this equation more amenable to graphic treatment, i t is trans-formed as follows.

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ADVANCED KINEMATICS OF THE PLANE MOTION 221

From Fig. 10-2,

and

s ~ P = O s ~ S - P ~ S )

P I, = P 8 - 1, 8

Hence 1 1Os 8 - P 8 P ~ 8 P 8) - I - S)and O s - S) I , 8 = P - 8 2 10-4)

which is the second form of the Euler-Savary equation . Since P 8 2is always positive, O s - 8 and I , ~ 8 must both be either positive ornegative, which means that 0 8 and I, are always on the same side of S.

/ To Oc -

Ray c Ray a

Ray hFIG. 10-4

For points 8 inside the inflection circle, I, 8 is negative and, consequently, the center of curvature of the path is beyond S relative toP.Hence such points des cribe paths which are convex when viewed from P,while the paths of points outside the inflection circle are concave. Forpoints 8 on the pole tangent, I , and 0 8 coincide with P. These kinematicpeculiarities are clearly illustrated in Fig. 10-4.

I f the position of the inflection point on a given ray is known, then thelocation of the center of curvature of the path of any point on the ray maybe determined either by computation, by the use of the Euler-Savaryequation, or graphically, by means of the following construction, shownin Fig. 10-iia. Through 8 draw an arbitrary auxiliary line , and select

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222 KINEMATICS AND DYNAMI CS OF PLA N E MECHANISM S

on it at random a point Y. Connect I , with Y line 1), and Y with Pline Through P draw a line parallel to line I intersecting the auxil

iary line at z. Through Z dr aw a line parallel to line II and obtain 0 8 atthe interse ctio n of this line with the ray s.

Ray

Os Rays

p

a) b)

FIG. 10-5

Th e correctness of the co nstr uction is prov ed by cons id ering two sets

of similar triangl es .From the similarity of the triangles S I ,Y and SPZ,

P - + S Z - + S, - y-+ s

a nd from the similarity of the triangl es S0 8 Z and SPY,

Os-+S _ Z - + SP - + S Y - + S

H ence Os-+ S) l ,-+ S) = P-+ 82

Figure 10-5b shows t he same construction for a different relativ e di s-position of the points P, S and / .

10-2. Analyti ca l and Graphical Determination of d;

A moving point S and the center of curvature 0 8 of its path form aconjugat e pair of points.

Equation 10-1) shows that d; and the direction of Pn can be determinedif a) one pair of conjugate point-s and the corresponding ray angle aregiven, or b) two pairs of conjugate points, on different rays, are specified.

Case a. The most usua l example of this case is illustrated in Fig . 10-6.Given are the conjugat e points 0,.. and 0 1 for whi ch the ray angle is zero.

0, is the center of curvature of the path of . See Sec. 3-13 .)

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Analytical Solution. By Eq (10-1),

1 + 1 _. .01-+ p p - + 0 . d;

and d = r - P) P-+ 0,.)0 1 -+ P + P-+ 0,.

With 0 1 - P = p1 and P -+ 0 . = - [> . this rela tion becomes

d = Ptf> .. - -p . - Ptor d = P Pm1

PI p .

(10-5)

(10-5a)

where the plus applies if the two polodes are convex, and the minus if onepolode is convex an d the other concave. The resul t is thus identical withd;, defined by Eq (3-15).

Graphical Solution. The graphical solution, shown in Fig. 10-6,consists in performing t he construction of Fig. 10-5 in reverse order.

FIG. 10-6

Case b. Specified are two pairs of conjugate points, Q and Oo on ray qand S and 0 8 on ray s. Th e given includ ed ray angle 89 = 89 - 8

(Fig. 10-17) .Analytical Solution. By Eq (10-1),

~ P + ~ Q cos 89 = os_:P + ~ s cos 8or oQ: p + ~ Q cos 8q = os: p + ~ s cos 8 q 8q,)Th e above equation is solved for 89 and d is then found by Eq (10-1),applied to Q.

Graphical Solution. The problem may be solved graphically bydetermining separately the inflection points I 9 and I,, using the construction of Fig. 10-5 in reverse order, and then erecting normal s throughthese points to the respective rays. The two normals in t ersect a t I., the

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224 K I K E M A T I ~ :1.:'\D DY:'\.'\MIC':'\ OF PL:\NE ME CHA '>ISMS

t.erminal point of d;. Alternatively, the problem may be solved byBobillier's construction, described next .

10-3. Bobilller's Construction; Collineation AxisBobillier's construction differs from the reverse construction of Fig. 10-5

only in that the auxiliary line and point Z are no longer arbitrary. Herethe auxiliary line coincides with the line through the moving points Q andS and point Z is located a t th e intersection of S with 0Q0s. Because ofthis particular combination of li n e and point, the two inflection points I 9and I . are obtained simultaneously.

Figure 10-7 shows the application of Bobillier's method to the construction of the inflection ci rcle for the coupler 3 of a four-bar mechanism

y

Pole tangent, p1

p 2 4 ~ ~ ~ ~ ~ ~ ~ ~ ~ M / ~ ~ ~ ~ ~ ~z

Pole normal, Pn

Ray b

Constr .uction: Os , 0 , . - Z - P 13 - Y - Ib , I . )

Fw. 10-7

P = P 13 . t can be seen that Z coincides with the instantaneou srela t ive-velocity pole P of the driving and driven links.

Collineation Axis.The

linePZ

in Bobillier' s construction isknown

asthe collineation axis of the rays q and s. Thus in a four-bar mechanism,the line connecting the rela t ive instant centers ? 1 3 and P2 of the physically separate links is the collineation axis of driving and driv en links; i t isalso referred to as the collineation axis of the me chanism.

Bobilller's Theorem. This theorem states that the angle between thepole tangent and one ray is equal t the angle between the other ray andthe collineation axis of the two ray s, both angles being m easured in thesame sense. The usefulness of the t heorem in analysis and synthesis will

be demonstrated later in thi s chapter.The correctness of the t heorem may be verified with the aid of Fig. 10-7.

and


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ADVANCED KINEMATICS OF THE PLANE MOTIO:-< 5

because lines l and P ~ 21 are parallel. Similarly,

1: p P1ala = 1: IbPi3P24

10 4. Hartmann s Construction

Hartmann s method of determin in g t he inflection circle is somewhatindirect, in that it is based on the displacement velocity of the instantcenter . t is applicable without diffi culty in the most complex situations.

From Sec. 10-1 Eq. d),

or

d _ dl _ dl dTi dtp - dT dtp

p

d; w. (10-6)

In Eq. (10-6), w . is the angular speed of the plane m and p the speedwith wh ich the instant cen ter of rotation changes its position along thefixed polode or along the instantaneous pole tangent. p is thereforeca lled the displacement velocity of the instant center. As show n pr eviously, d; is dire cted along the positive pole normal. Consequently, thedirection of Vp along the pole tangent is. obtained by turning d; about Pthrough 90 in the sense of w... Conversely, with p known, the directionof d and thus of the pole normal, is obtained by rotating v P through 90agai nst the sense of . .

In applying Hartmann s method, an arbitrarily assumed value isassigned to the angular velocity w and the corres ponding displa cement velocity of the instant center is determined graphically, with theaid of an auxiliary linkage. I n general, the absolute instant center Pof t he plane whose inflection circle is to be determined is d efi ned by theintersection of two moving rays. The auxiliary linkage consists, therefore, of links which are pivoted to the frame at appropriate points , androtate with the rays, and of blocks which are free to slide along these links,in appropriate directions, but are pin-connected together at P . Thusthe velocity of the pin connection P, in the auxil iary linkage, is identicalwith t he required pole-displacement velocity .

In the following examples, the auxiliary linkages have been drawnseparately, to facilitate the explanations. In practice, however, theconstructions are carried out on the original plan, a procedure whi chsaves time and enhances ac c ur ac y.

Illustrative Example 1 Construct the inflection ci rcle for a movingplane if the velocity pole, one pair of conjugate points (Sand Os), andthe d ir ection of th e pole normal are specifie d.

Solution. The original system is shown in Fig. 10-8 a, and the auxiliarylinkage in Fig. 10-8b. The latter eons ists of the bar s pivoted to the

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226 KINEMATICS AND DYNAMICS OF PLANE ME CHANISMS

frame at 0,, and two blocks q and r pin-connected together at P withq guided along the pole tangent p and r free to slide along s. Th e pointon s coincident with P, is denoted by X,.

The velocity of the pin P, corresponding to the arbitrarily assumedangular velocity w... = } rad / sec, ccw, is the required pole-displacement

o

Os

//

//

/

//

/

//

/

/ b)

-- 0

c)

a)

/

//

/

p ,/

//

/

Pon q and r PX, ons

s

s

--- x == VJf

o

\'m 0.5 rad/sec

\ . \ assumed)

r

Upd 2up 'm

P o n q a n d rq X. on o

FIG. 10-8

0, .

velocity. The input velocity of the linkage is v 8 = w... X P--. S .The velocity of X. is found by proportion, and the velocity of P followsfrom

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Vp = V x + Vpx,

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Figure 10-8c shows the construction for the parti cula r case of specifiedcenters of curvature of both polodes.

Illustrative Example 2 Construct the inflection circle for the coupler 3

of the four-bar m echan ism of Fig. 10-9a. Two sets of conjugate pointsare prescribed: A and 02 B a n d o )

a)

b)

0.5 rad/sec assumed)--

0.5 rad/sec .

Pon 5 and 6}X 2 on 2X 4 on 4

J r ...

y B

FIG. 10-9

Solution Figure 10-9b shows the auxiliary linkage. t cons ist s of thebars 2 and 4, pin-connected to the frame at 02 and o. respectively, andthe blocks 5 and 6, pin-connected together at P with block 5 free to slideon bar 2 and block 6 free to slide on bar 4. The points on bars 2 and 4,coincident with P are denoted , respectively, by X2 and X .

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The input velocities are VA = .o)a X P w - + A) and v s = I) a X P 13 --+ B);vx, and vx, are found graphically , by proportion, and V p follows frQm

and

I2

a)

1

(b)

Vp = V x + V p x ,v , = v , + Vp x

pd 2vp' S

3 0.5 rad/sec(assumed}

c >s 0.5 rad/sec

FIG. 10-10

Pon 5 and6X

4on 4

X 2 on 2

Illustrative Example 3. Construct the infl ection circle for link 3 of theconchoidogr ph (Fig. 10-lOa). (The name of the mechanism derivesfrom the fact that any point on link 3 describes a conchoidal path .)

Solution The auxiliary linkage consists of the extended slider 2, therotating link 4, pivoted to the frame at 04 and the two blocks 5 and 6,pin-connected together at P of which block 5 is free to slide along link 4,




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ADVAN CED KINEMATICS OF THE PLANE MOTION 229

and block 6 along sli der 2. The coinci dent points are again denoted byx2 on 2) and x. on 4). Since link s 3 and 4 form a sliding pair, c l4 = WaFor t his re a so n Vx = WaX (0.-+ x. . Since link 2 is in rec til inea rtranslation, vx, = VA = waX P13-+ A . Vp follows from

andVp = Vx, + Vp x ,Vp = Vx + Vpx,

Illustrative Example 4. Construct t he inflection circle for the bar 3 ofthe mechanism shown in Fig. 10-lla. The mechanism consists of thecrank 2 the bar 3 hinged to the crank a t A, and the stationary surface 1which serves as a guide for bar 3.

Solution. The auxiliary linkage is shown in Fig . 10-llb. I ts elements

are the bar 2, pivoted to the frame a t 02 the bar 4, pivoted to the framea t the point 4 which coincides with the center of curvature M of theguiding surface 1 and the blo cks 5 and 6, pi n -connected together a t P, ofwhich block 5 is free to slide along bar 2, and block 6 along bar 4. Thepoints coin cident wi th P are X2 (on 2) and x. (on 4). Since the auxiliary link 4 remains perpendicular to the link 3 during an elementaldisplacement, c l4 = wa and consequently, Vx = Wa X (O.-+ x. ) . Also,VA= wa X


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230 K l:-;E MATI C8 Al D DYNAMI CS O T PLA: ;E ME C HANI SMS

a)

b)

p

d - - 311p w

Inflection circl e

FIG . lQ 11

Pon 6 nd 6X,on4.X 2 on2

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with block 11 free to slide on bar 10 and block 12 on bar 6. The pointon bar 10, coincident with R is denoted by Xto, and the points onbars 10 and 6, coincident with P are shown as Yto and Ye, respectively.

The input velocities are VB = W6 X Pu --. E) and V x 10 = component of;vRnormal to OtoR . Vp follows from

andVp = Vy, Vpy,

V p = Vy 10 Vpy 10

10-5. The Inflection Circle for the Relative Motion of Two Moving Planes

Thus far the discussion centered around the construction of the inflection circle associated with the absolute motion of a plane. Occasionally

i t is of interest to determine the curvatures of paths traced by the pointsof one moving plane on another plane, also in motion . The inflectioncircle associated with the relative motion of two planes is constructedsimply by considering the kinematic inversion in which the particular

Or d, { rei. m) d, m rei. f o,.P n ~ ~ ~ ~ \ ~ ~ ~ ~ ~ ~ J ~ ~ ~ ~ P ~ ~ ~ = = ~ ~ ~ C < ; Pffrel.m) mre.

d points in the sense of the respective pole normal

FIG. 10-13

reference plane is held stationary, and otherwise following the proceduredescribed in the preceding se ction. I t should be noted that, as shown inFig. 10-13, the two inflection circles involved motion of m relative to f,and motion off relative tom) are polar images of each other, a fact whichemerges from Eq. 10-5) .

Illustrative Example. Construct the inflection circle associated withthe motion of the follower 3 relative to the cam 2 in Fig. 10-14.

Solution. Figure 10-14b shows the auxiliary linkage for the inversion inwhich the cam is held stationary . The linkage is basically the same as theone discussed in Illustrative Example 4 of the previous section and shownin Fig. 10-11. The input velocities for the inversion are

VB = waz X P23 __ B)

and Vx, = Waz X 04 __.X.).

and

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Vp follows from

Vp = Vx, : V p x ,V p = Vx, V p x ,

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DV NCED KINEM TICS OF THE PL NE MOTION 233

10 6. Application of the Inflection Circle to Kinematic Analysis

Since the inflection circle provides the means for a purely geometrical

determina tion of path curvatures, i t becomes possible to analyze complexmechanisms without recourse to the indirect analytical and graphicalmethods of Chap. 6.

3

a)

W3 = 1 rad/secassumed)

w .12 = 1 radfsec

b)

on 5 and 6}X1 onx. on 4

FIG. 10-14

2

d of a rei. 2

4

o

Dlustrative Example. Figure 10 15a shows the skeleton of a certainsteam-valve actuating mechanism whose input crank 2 revolves at constant speed. Required are the angular acceleration of link 4 and theacceleration of C.




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Solution With link 4 straight, as shown, the solution by means ofconventional methods is rather difficult, and the problem is best solvedwith the aid of the inflection circle . After the path curvature of has

1

a)

5

b)

' rad/secassumed)

M

3

02

w 6 =we radfaec

FIG . 10-15

v x ~ w P C )

vx 6 = w, 0 6 P

been determined, either graphically or by computation

[ Os B) lb-+ B = P B 2 )

the problem becomes elementary.With link 4 cu rved, as shown in Fig. 10-16a, the problem may be so l ved

either with the aid o the inflection circle or by the subst itution of t h e


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ADVANCED KINEM TICS OF T H E PL NE MOTION 235

3

a) p

3

1o

b) Kinematically equivalent mechanism

FIG. 10-16

instantaneously equivalent low-complexity mechanism of Fig. 10-16b.The former approach is much less cumbersome.

10-7. Polode Curvature General Case)The shapes of the polodes , which, by rolling on one another, reproduce

a given motion, depend on the characteristics of the constraints. Thenature of this relationship is investigated in the following sections.

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I n Fig. 10-17, the plane m defined by QS is constrained to move in acertain manner by virtue of the fact that the centers of curvature 0 0and Os of the paths of Q and S respectively, are made to traverse the

prescribed lociEQ

andEs. (The locus of the center of curvature of a givencurve is known as its evolu tc .)

I

o, ,I .

, 'PsI

I

/ S

P,

II

II

II

II

a)

c.Q

FIG. 10-17

( a p ) . , . ~Ps

(b)

(a p),.

(ap) ,

>

8.8.

P.(c)

Although the curvature of the fixe polode is a geometrical propertyof the system, it will be determined here indirectly, i.e., by consideringthe displacement characteristics of the velocity pole:

vp2 w . 2d;2Pt = ap),. ap),. (a)

where V p = displacement speed of velocity poleap),. = normal component of its displacement acceleration (not to be .

confused with accelerat ion of point P ,., ap = w . 2d;)

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ADVANCED K KBMATH'S OF T I B PLANE MOTION 237

In order to evaluate Eq. a), a r will be computed with the aid of thesame auxiliary linkage that was used in determining Vp by Hartmann smethod, by solving the equations

andap = ax, apx,ap = ax, apx,

with an arbitrarily assumed w,., which, to simplify the calculations, willbe assumed constant. Hence, in the following, dw/dr = 0.

The conventions relating to the positive and negative sense along agiven ray, and at right angles to it, were laid down in Sec. 10-1. Since,in Fig. 10-17, which forms the basis of the subsequent derivations, allrelevant directed lines are positive, the diacritical overbar will be omitted

in the calculations. t will be reintroduced in the final expressions.The component of ap, perpendicular to 0 8 -+ S = p8 , consists of two

parts, viz., the tangential component of ax, and the Coriolis componentof P relative to X,:

(ap) 0 . p p a = (ax.) ,+ aPx,)..,.

ax.) 1 = dvx./dr is computed as follows:

vx, = Wp8 0s---+ X,)

because wp8Ps = w,.r 8 = vs; and

= wm Os---+ X,) rsPs

( )_ dvx, _ dvx, dl _ dvx, _ d dvx,

ax, 1 - dr - dl dr - V p dl - w,. i dl

b)

(c)

= w 2d [rs d Os---+ X,) _ Os---+ X,)rs dps Os---+ X, drs] d)' Ps dl Ps 2 dl Ps dl

The three derivatives in Eq. d) may be expressed in terms of the givenquantities in the following manner:

Since X, and S are fixed points on the ray s,d Os---+ X,)

dlBut from Fig . 10-la,

d Os---+ S)dl

dps = Pa du

[dps is positive because 0 ~ - - - +S ) > (0 8 -+ S) .) From Fi g . 10-lb,Os-+ P) du = dl cos 8,

Hence d Os---+ X,) dps p,, cos edl = dl = 0 s --- p

Also, from Fig. 10-lb,

- sin 8,

e)

(f)

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[drs is negative because (P --+ S ) < (P--+ S).] With

andOs-+X, = Os-+P = s rs

(Os--+ S) - (Os--+ X, = P--+ S = rssubstitution of Eqs. e) and (f) into d) yields

( ) 2-1 [r s 2 Ps Ps - r s . ]ax, , = ' ' 2 cos 8, - sm 8,Ps PS - Ts PS

The Coriolis component aPX.) ... is calculated as follows:

(apx.) .. = 2wp,PPX,

From Fig. 10-17,

Vpx, = Vp sin 8,

Hence ( ) 2rs . 2d 2rs .apx, ..,. = - WmV SID 8, = w,. ; - SID 8,Ps Ps

Equations g) and h) may now be substituted into b), giving

(2 3- )rs p,. rs - Ps .ap).,.,PP = w,. d, 2 _ _ cos 8, + _ sm 8,

Ps Ps - rs Psor

and similarly,From Fig. 10-17c,

Hence

From Fig. 10-17c,

ap)PPPB = Wm 2d;Ksap).,.,PPQ = W,.2d;KQ

Wm 2d;K S W,. 2d;K QaP = cos (8, - o - -co_s ;;( 8 ~

ta li = Ko cos 8, - Ks cos 8qn Ks sin 8q - Ko sin 8,

( ) . li '' 2d,Ks . liap n = ap SID = 8 /i) SIDcos

_ 2d Ks tan li- Wm , cos 8, + sm 8, tan liwhich, with Eq. (l), reduces to

( )_ 2d Ko cos 8, - Ks cos f q

ap n - w,. . 8 - 8 )Sin q ,

Equation a) combined with (m) yields the final result

d, sin 89

- 8,)Pt = Ko cos 8, - Ks cos 8q

wher e the coefficients K 0 and Ks are defined by Eqs. (i) and j).

(g)

h)

(i)

(j)

(k)

(l)

m)

(10-7)

The radius of cu rvature p1 of the fixed polode, as determined b yEq. (10-7), may be positive or n egative . A positive result shows tha t




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p1 = 0 1 - t P points in the direction of the positive pole normal, whichmeans that 0 1 and the inflection circle are on opposi te sides of the poletangent.

10-8. Polode Curvature (Special Case); Hall s Equation

Equation (10-7) is an expression for the polode curvature in the mostgeneral case . Consider now the specia l case where the path curvatures ofboth moving points, denoted in the following by Y and Z, ha ve instan-taneous ly stationary values; that is , dpyjdl = dpz/dl = 0. I t is obviousfrom Eq. e) that to satisfy this condition, the evolutes E y and Ez musthave cusps a t the corresponding instant; that is, p,, = Pz = 0.

Equation (10-7) reduces, therefore, to

_ d; sin IJ. - Ov (n)P = 3 f - - 3ry - -

z _ pz sin IJ, cos IJ

_ PY cos IJ, sinpz PY

With the introduction ofk = s ~ nIJ. 0 )

sm IJ. - 0 )

the expression for Pt is further simplified, and

2dp - - ~ :- 1 1 k) ( f - 1 1 - k

2d;- ~ ~ ~[

3 ? - t Z 1 ] 1 k) [3 P - t Y 1] ( k)Oz - t Z Qy - t Y

(o)

(10-8)

This important equation was first repor ted in the literature by A. S. Hall. 1

In the particular case of stationary path curvatures, the polode curvature may also be expressed in terms of a ratio


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240 KINEMATIC S AND DYNAM I CS OF PLANE ME C HAN I SMS

H enced R [ Or -+ P Or-+ P ) ] 2drj = - p - y - P - + 7) 2 tan ~ =


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If, in Eq r}, cos 9v is expressed in termiS of tan 9v, and the lat ter interms of diR/dTJ and IR by Eq q}, then the following expression results:

[ffi2(


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Hall s equation may be used directly for the computation of the curvatureof 1r1:

2dP =

[3 ( ? - + B) - 1 ] 1 k) [3 P-+ A) - 1 ] 1 - k)

where

O. - +B 02-+A

k = sin e. e2)sin e. - e2

10.10)

10-11)

Illustrative Example. D etermine the polode curvatures for the coupler3 of the four-bar mechanism of Fig. 10-20.

/

//,

//

/

13

//

//

/

/ 1

/

//

/

//

//

Fro . 10-20

Solution. Here P-+ B = 1.3 in., 0 4 -+ B = - 2 . 3 in., P-+ A = 1.7 in. ,

and 02-+ A =1 . 2

in.Construction of the inflection circle yields the following parti culars:d, = 4.22 in. , e = 61, e = 346 = - 14. Hence, by Eq. 10-11 ),

k = 0.755and by Eq. 10-10),

P = -2 .45 in .

The radius of curvature of 1rs is calculated by Eq. 10-5), with m = 3and f = 1:

4.22 = P3(-2.45)iis 2.45

and jis = - 1.55 in.

The polodes in the vicinity of ?13, with the centers of curvature denotedby 0.,1 and O.,s, respectively, are shown in Fig . 10-20.




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10-10 . Polo de Curvature in the Four-bar Mechanism; Relative Motion ofthe Output and Input Links; Determination of the Output Angular

Accelerationand Its Rate

of ChangeAn in spection of Fig . I0-19 b r eveals tha t Hall s equat ion , with appro

priately modified symbols, may be used to de termine the cur va t ur e of t hepolode 1rz, associated with t he motion of t he output link 4 relative to t heinpu t link 2:

2d.z =[

3 P-+ B _ 1] 1 + k) + [ 3 P - + O.) _ 1] _ k)A- B Oz- o

1 0- 12)

wherek = sin a + fJ1)

sin fla - fJ1)10- 13)

I t should be noted t ha t, s in ce t he discussion is conce rn ed with the m otionof link 4 relative to link 2, the positive sense alo ng bars 1 an d 3 is these nse from P = Pz to t he correspon ding points on link 4, i.e ., fr o m P to0 and from P to B respect iv ely.

Equation 10 -9), with a suitably alte r ed notation, is also applicable inthis case. Comparison of Figs . 10-1 8 and 10-19b sho ws tha t drp2 = -d . ,

and tha t 02 corresponds to 0y an d o to Y Hence


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so that ( ) 8 0 4


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ADV.-\: >CED KINEMATICS OF THE PLANE MOTI 0:-1 24 ;)

The diam eter of the inflec t ion circle may be determined graphi cally orby calculation, using the Euler-Savary equation in the form

P 8) 2 = 0 8 ~ S) I, S = Os S [(P S - d cos ll,]Hence (?24 ~ 04) 2 = (02 O.)[(P2 ~ 04) - d, cos ll1)

where P ~ o = 6.57 in.0 2 0 4 = 4.00 in.

A ~ B = 1.87 in.P24 ~ B = 5.20 in.

With these values:By Eq. (10-14),


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link, the pole tangent is perpendicular to the fixed link, and by Bobillier stheorem, the collineation axis is perpendicular to the coupler.

Carter Hall Circle. Another kinematic property which may be

ascribed to the collineation axis is the following .f

a family of four-barmechanisms is constructed on the same fixed base line 0 2 - 0 4, whosemembers have the sa me instantaneous velocity ratio


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ADVANCED KINEMATICS OF THE PLANE MOT ION 247

2d;~ 2 = ~ = ~ ~

3 P z . ~B _ 1] [ 1 _ s i n~ 3+ ~ > ]

A ~ B sin ~ a - ~ ~ )

+ [3 P2t O.) _ 1] 1 + s n ~ a +~ ~ ) ] b)Oz ~ 0 sm ~ a - ~ t )

The Euler-Savary equation, applied to link 1, yields

from which

(Pz, ~ 0, 2 = 02 ~ O.)[ Pz---+ O.) - d; cos 61]= Oz ~ O,)[(Pz---+ O.) + d; sin ~ d

Pz. ~ o (Pz. ~ o.) + d; sin ~ IO z ~ o . P z . ~ o .

and, in a similar manner, for link 3,Pz---+ B (Pz, ~ B + d; sin ~ aA---+ Pz---+

(c)

(d)

Substitution of Eqs. (c) and (d) into (b) and rearrangement of terms yieldsthe following expression:

3(Pz --- O,)d; s i n ~ ~cos ~ a sin 3 sin ~ a - ~ ~ )P z . ~ B= ~ ~ ~ ~ 7 ~ ~ ~ ~

(Pz ~ 04) sin ~ - ~~

-~ ) + 3 d ;s i n ~ ~C O S ~ ~sin ~

(e)f now the right-hand side of Eq. (e) is substituted for Pz ~ i n Eq. a),

the expression is simplified to

By Eq. (10-14),

Hen ceOz ~ Pz. =


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248 KINEMATICS AND DYNAMI CS OF PLANE MECHANISMS

Equation s 10-14 ) to 10 -16) show that mecha ni sms whi ch h ave the samein stantaneous values of CR, CR', an d CR have, ipso facto, t he same inflectio ncircle and the same polode cu rvature. H ence, for such a family of

mechanisms,P24 P u--=-:....._-=--.:.: = co nst = d.

cos ~10-20)

whic h i s the equation, in polar coordinates, of a cir cle having the character isti cs s ta t ed above.

f p2 is eliminated between Eqs. 10 -15) and 10-19), then, after so m ele ng thy and tedious algebraic operat ions, t he following expression isobtained whi ch relates CR to q uanti t ies dir ectly determinable byconstruction :

Illustrative Example. Fo r the mechanism of Figs. 10-21 and 10-22,0 2 - 0. = 4 in ., R = 0.39, 81 = 104.5, and d. = 5 .6 in.

Wi th these values, Eq. 10-21) y ields CR = 0.7 2, which agrees with theresult obta in ed in the illustrative example of Sec. 10-10.

10 12 . The Circling-point Curve General Case)

The last item to be di s cussed in this chapter is the circling-point curve,or cubic of stationary curvature, the locu s of all those points on a movingbody wh ose paths, a t th e instant considered, have stationa ry curvature .Th is curve, too, is of importan ce in synthesis, and particula rl y in thede sign of multilink mechanisms with prolonged dwell requirement s forthe outpu t lin k .

Let Y denote a point on the circl in g-point curve, and apply the EulerSa vary equation to it:

orOy----+ Y) Iy Y) = P Y 2

Oy Y)[ P - Y) - d, cos Oy = P----+ Y) 2

which, with th e substitutions y Y = py and P----+ Y = fy, becomes

a)

As in Sec . 10-7, the overbar indi cating dire cte d line elements will beomitted in the calculatio n s, but reintroduced in the final ex pr ession .)

f the path curvature of Y is to be instantaneously s ta t ionary, then

dpy = 0dl

where l = elemen tal displacement of the pole P.




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Differentiation of Eq. a) yields, therefore ,

[dry d d;) d . d011 ] 2 dryPr dl - dl cos ~ , sm 0 dl = ry dl b)

From Fig. 10-18,dry

- sin 011 c)i f=

dO. d., d ~d)l dl dl

d ~e)l PI

andd1J Ty d {J,. ry 1 f)- - -dl py dl py d;

The relation d, = dljdqJ,. = upjw ,., was d e rived in Sec. 10-1.)

Henced0 11 Tydl = py d - p; g)

Introduction of the expressions c) and g) into b) leads to

py sin 011 - d ~ ; )cos 011 d; sin 011 : ~~ - P:J - 2rr sin 011 h)Finally, elimination of py between Eqs. h) and a) yields the equation ofthe c ircling-point curve in polar coordinates, with the variables f y and 011 :

sin 20 f y = 2 2 d; . 2 d;)

3d; - PI sm 011 - 3d; dl cos Oy

_ s in 20 At sin ~ N cos 0

10-22)

where M and N are co ns tants in t he phase considered.As a preliminary step to t h e evalua t ion of Eq . 10-22), i t is necessary

to determine the instantan eous rate of growth of the diameter of theinflection circle with the phase of the moving body; that is, d d;) Idl. Theco mputation becomes f easible if, as shown in Fig. 10-17, the path curvatu r es of two moving points are given.

The following ex pressions , releva n t to t he present investigation, werederived in Sec. 10-7, with the assumption w,.. = const:

Equation m), normal co mponen t of the pole-displacement acceleration:

)_ 2d KQ cos o - Ks cos Oq

ap n - w,. . 0 - 0 )l n q

i)

Equation l), angle between pole-displacement a cce leration and pole-

Digitized by oglOriginal from



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displacement velocity : = KQ cos 9 - K 8 cos 99won u K . K .s sm 99 - Q sm 8

with KQ and s defined by Eqs . i) and j) of Sec. 10-7 .From Fig. 10-17b,

a ) = ap),. = w 2d s sin 99 - q sin 9P

1 tan 6 ' sin 9 - 9,)

How ever,dvp dvp dl dvp

ap), = dr = dl dr = dl Vp

Since Vp = w. .d and w . is constant,

dvp d d,)i f = w,. dl

Finally, the combination of the last three equations yields

d d,)dl

s sin 9 9 - KQ sin 9sin 9 - 9,)

FIG. 10-23

j)

k)

Z)

(m)

10-23)

Normally, as shown in Fig . 10-23, the circling-point curve is a loopedthird-order curve whose branches cross a t right angles at the instantcenter and then continue to infinity in an asymptotical approach to a linewhich makes an angle 9.. with the pole normal. 9., is obtained byequat ing to zero the denominator in Eq. 10-22):

Digitized by oogle

9.. = arctan Z) 10.:24)

Original fromUNIVERSITYOF MICHIG N


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ADVANCED KINEMATICS OF THE P LANE MOTION 251

The asymptote i tself does not pass through the instant center. Its offsetn is determined as follows:

n = lim ry sin 8., - 8 ~ )9 . - s .

1. sin8 ~ sin 8., - 8 ~ )

- m- 9 . - 9 . M sin ~ N cos 8

which is indeterminate for 811 = 8.,. However i the numerator anddenominator are differentiated separately wit h respect to 811 , the limitingvalue is found to be

sin 28.,n = ~ ~ ~ ~ ~ ~N sin 8., - M cos 8., (10-25)

The circling-point curve degenerates into a circle and a straight linewhen ever d, ha s an extreme value. By Eq. (10-22 ), i d d,)/dl = 0,N = 0 also, and Eq. (10-22) reduces to

(2 COS 811 , Ory - M s1n ~ =

where y (2 cos 8 1 /M = 0 = equation of a c ircle of diameter 2/Mcentered on pole normal and passing throughinstant center P

sin 8 ~ = 0 = equation of a straight line coincidentwith pole normal

10-13. The Circling-point Curve for the Coupler of a Four-bar Mechanism

In a four-bar mechanism the two coupler hinges are obviously pointson the circling-point curve. Hence if their polar coordinates are substituted into Eq. (10-22), two independent equations are obtained fromwhich the constants M and N can be calculated. With M and N known

the circling-point curve is easily constructed.Illustrative Example. Draw the circling-point curve for the coupler of

the four-bar mechanism of Fig. 10-24.Solution. The positions of t he pole tangent and pole normal are found

by Bobillier s theorem and the polar coordinates of A and B are determined by m easurement:

r = 4.27 in.ra = 5.75 in.

e = 145.58a = 108.1

These two pairs of values, substituted into Eq. (10-22), yield

M = -0.0273 and N = 0.2 H\0

\Vith these constants points on the curve are located as shown in detail

Digitized by Co ogleOriginal from



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for p.oint C whose polar angle o = 125. By Eq. 10-22),

sin 250 .r = -0.0273 sin 125 + 0.2460 cos 125 = 5 .44 In .

Figure 10-24 shows clearly t he d egree to w hi ch the actual path of apoint on t he circling-po in t curve may be approximated in the vicinity ofthe phas e considered by the osculation ci rcle. For the poin t C the two

Osculation

circleat

C-..\s

c

.

4

_J

P ~ l ' ? ' l ' : / ' 7 ? , . , . , . , . , . , . ; . , . , . , . . , . , ~ 01

F 1 . 10-24

arcs are pra ctically indis t ingui shable from one another for about 70 ofinput-crank rotation. Hence , i two links 5 and 6 were added to themech anism as shown the output link 6 would remain sensibly a t rest during this period.

The location of the hinge between links 5 and 6 whi ch coincides withthe center of cu rvature of the path of C may be found either graphicallyafter constructing the inflection circle, or by caleulation as follows :

Digitized by oogle Original fromUNIVERSITYOF MICHIG N


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ADVANCED KINE M ATICS OF THE PLANE MOTION 253

T he Euler-Savary equation, applied to A, yields the diamete r of t heinflection circle:

P1a - A 2 = 0 2 A l a - A) = 0 2 A)[ P13- A) d, cos a]

where P 13 - A = 4.27 in.2 - A = 1 00 in.

B. = 145.5giVIng d, = 16.96 in.The Euler-Savary equat ion, applie d next to point C, yields the requiredradius of curvature:

P 1 a - C 2 = O c - C [ P u - C d , cos 8,]

where P 1 a C = 5.44 in.d, = 16 .96 in.8 = 125

giving O c - C = 1 95 in .

Advance Mechanics of Machines for M.tech

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