# Full text of "NASA Technical Reports Server (NTRS) 19930081548: Analytical determination of control system pulley-axis angles"

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\ CLASSFO DOCUMENT dieNotfoool Mnn of Mm Uwttd State* wfckin ta«wni of dW Eaptcnxc* Act, USC 30*31 tf»_ 1 II ■■ llf I ■■ fcl. ■ _| 1 - ^ I ■ , m t || 1X3 TlfT""""" O* wll IWMMIOO 4b to «a OBwdMHneci _ •bob to eloHiSod any pottoot io aW aa^itaxy ood Mm FodonJ GovenuMot who bar* iluidu. buJ !■ TJaftnl °k»ki$ c« lii t in -jscKSsn who fm Bcccomy Bax n» TECHNICAL HO TBS NATIONAL ADVlSbRY COi&llSDEB ?0H AERONAUTICS No. 827 ANALYTICAL DETERMINATION OP CONTROL SYSTEM PULLEY -AXIS ANGLES By I. 3. -Driggs Bureau of Aeronautics, Navy Department Washington September 1941 3 1176 01425 7415 NATIONAL ADVISORY COMMITTEE JQR AERONAUTICS TECHNICAL NOTE NO. 827 ANALYTICAL DETERMINATION OF CONTROL SYSTEM PULLEY- AXIS ANGLES ' By I. H. Driggs * The ideas developed in this paper are presented as a means of saving the designer's time and as a method of re- duction of control-system friction "by an accurate calcula- tion of pulley-axis angles without the errors and. di f f icult checking incident to any graphical method. The growing realization of the necessity for greater care and refinement in the design, of control systems is justification for the more accurate analytical methods given. Saving in layout and checking time is also of im- portance, although probably secondary to the greator re- finement obtainable by the' use of a calculating machine even in the hands of inexperienced personnel. It is sug-. gested.that the formulas given here are worthy of careful trial by any engineering " department that is not already using an equivalent method of design. Two mathematical laws are omployod in the derivation of the formulas for pulley-axis angles to give -correct alinoment with control cables and therof ore avoid "binding, and to reduce friction. Those laws are: 1, Any two intersecting straight lines determine a single piano, 2, The first partial differentials of tho equation of a surface are proportional to the direc- tion cosines of the normal to that surface at any point chosen for investigation. Since any pulley, to operate properly, must have its center plane coincident with tho plane containing the cables loading around it, the location and angles of the pulley axis, with reference to the coordinato planes, are tho values required to properly construct and locate a pulley bracket. The equation of tho plane of the two cables (and of tho pulley) is first determined and then ttACA Technical Hot© No, 827 the direction cosines of the pulley axis found from the first partial derivatives of this plane., To express math- ematically: f (x,y,z) = Ax + By + Cz = 0 This is the form for the equation of a plane surface passing through three points, ono of thorn tho origin of coordinates , 3f =A , 2i aBl S£=0 Sx 3y 3z Tho valuos A, 3, and C are proportional to tho direc- tion cosines; L, M, and K of tho normal to tho piano, f (x,y,z) . Then A M = 3 V = / , 2 Ta _a / , a Ta Ta" / . a Ta Ta v A + B. + 0 vA+B+0 vA+B+O If the coordinates of each end of two intersecting lines (or cables) are: P x (x lf y x , z x ) and P 3 (o, o, o) and P 2 (x g , y 2 , z 3 ), respectively, then the equation con- taining these three points and the two straight lines or cables is (fig, l): ac(y x z 2 - y 2 z l ) + y(z x y a - x x z a ) + z (xj. y 2 - y x x a ) - o Then A = y x z a ,y a zj. B = Zj, x 3 - x x z a G ~ x % y a - y x x a It is to he noted that this equation is true only if one of the three points used to dot.ermlne the plane is taken as tho origin and tho coordinatos of .the other two points measured: therefrom. (Sco fig. 1.) Positivo direc- tions and angles are also .defined in this sketch. 2TACA Technical IToto Ho, 827 3 It should "bo noted that the points P x , P 3 ., P 3 must lie on the cables, that is, -in figure i.- P 3 is the in- tersection of the lines P x - P 5 and P a - P 3 . This point is not the center of the pulley. The sketch (fig. 2) illus trates this point. Some confusion regarding the Bigns of the angles 9, \|/, and 0 may ho avoided if the direction cosines L, M, and N are considorod to he the coordinates x n , y n , and z n of a point on the normal" to tho pulloy plane unit dis- " tanco from tho origin. The praper angle of this normal may be laid out on the drawing board by choosing any~coh- venient length - .say, 10 inches - as the unit, distance and laying out the coordinates of the point x n , y^, and s. n . Since • • . . ..... 3c n = 10L y n = 10H z n = ION If it is desired to compute the angles for purposes of greater accuracy, the ahovo graphical method suggests the following: tan 8 = 2. L tan V = "k I V 10. tan $ = It may be desired to find the angle between the two cables in plane containing them for purposes of layout of cable guides, etc. If this angle is a, then cosoC= + 7] + 7x + Zj. 3 ) (ac £ + 7a The coordinates of the center of tho pulloy may also b 4 2TACA Technical Koto So. 827 determined by similar analytic means but no simple expres- sion can be readily found, Bince the values of x a , y a , and z ft (the coordinates of the pulley center) depend upon the solution of three simultaneous equations. It is believed that sufficient accuracy can be obtained if a combination graphical and analytical solution is employed. A layout is made to scale similar to figure 3, in which the two cables intersocting at tho center of coordinates are shown in all throe views. Bisoctors of tho angles bo- twcon the cables are then drawn in each view by the usual graphical construction. We know that these three bisectors must contain the center of the pulley axis, which must like- wise be equidistant from tho two cablos, Tho above solu- tion for tho truo included angle, a, botwoon the cablos allow a solution to bo roachod oithor graphically or an- alytically for tho truo, distances from the arAs of coordi- nates (intersection of cables) . Analytically R & = S = radius of- pulley to cable sin % center line If the direction .cosines of the bisector as laid out above are determined, then tho coordinates x a , y a , and z a can be calculated. The direction cosines of this bisoctor are found by choosing any point, such as P-^ (fig. 3), on tho lino and projecting through tho throe viows. If tho co- ordinates of this point ar-o x- D , y-j, , and z^, , then r fb H -r- ' - / / 3 2 2 \ </ <*- D + ^b + z b > Nv = V ( x b + ^b + z b > Then 1TACA Technical 'JTote No. 827 5 y a = cL M-jj Example ffigure 3 is a line sketch that might "be made up giv- ing the locations of three points through which cables must pass for a given airplane installation for use in finding the pulley-axis angles. C h o o»e the center point, P 3 , as the origin. (Hoto: Any ono of the throe points might have "beon taken but it is somowhat cloarer to placo tho origin at the point of intersection of the two cables where the pulley is to "be located.) *1 = -40 in. x a = 20 in. 7i = 25 in „ -8 in. z i = 10 in . Z 3 = 40 in. A = y 1 z 2 - y 2 z x = 25 X 40 + 8 X 10 = 1080 B = z x x a - x x z a = 10 X 30 + 40 X 40 = 1800 0 = x x y 2 _ y x x 2 = 40 X 8 - 25 X 20 = -180 Then J 'a 2 ■+ B S + O 3 = 2104 The equation of the plane containing the two cables of figure 3 is thon I080x + 1800y - 180z = 0 or 6x + lOy - z = 0. L = VttS - 0.5138 x n '= 5.138 in. 2104 M = 2104 = ,856 ° *B = 8,56 in * N = = -.0856 Kn ' = -.856 in. 6 NAQA Technical -Soto No. 8 27 tan 8 = -0.1867, -tan \|/ = 0.600, tan 0 = -10.00 8 = -9° 30' ■ \|/ = 31°- 0 = 95° 40' *i x a + y x y a + zi z a cos a = ■ ■ „ , J 2 2 3 \ / I 2 2 C \ f / * x + y x ) + y a + z 2 ) -40 X 20 - 25 X 8 + 10 X 40 (J(^40)* + (25 ) S + 10 S ) (J20*~+ (-8) S + (40) o >00 48.2 X 45.5 -600 n 008 a = 2190 = - 0 * 274 a = 105° 50' + Lot E = 1,50 in, - radius of pulloy to cablo contor Then " ' "' ~ - "■ ' d = = — _ 1#886 in sin, 52° 40' 0.7951 Tho bisoctors of the angles botweon tho cablos aro drawn and tho point, P^, is choson at random, Tho co- ordinates of this point aro scalod as: x D - =-11 ,8 y b = 13.75 = 25,00 * L b = ~}}' & - = -0.375 D 30.95 » M-h = 3- g * 75 = 0.4245 0 30.95 iTAGA Technical Note No. 827 7 - Htfe - °- 808 X], = -1.886 X 0.375 = -0.7075 y-jj = 1.886 X 0.4245 = 0.801 Z D = 1.888 X 0.808 = 1.525 Prom the above example, the steps necessary for the olution of this problem may he listed as follows: 1. Obtain a lino skotch, similar to figuro 3, giving the locations of tho intersection point of the two cablos for which a pulloy bracket is to be dosignod and ono other point on each cable. This skotch should bo fully dimensioned to ap- proximate scalo from any convonient pianos, as shown in figuro 3. 2« Choose tho origin at tho intersection of tho two cables, P 3 , in figure 3, and then determine and tabulate x x , y lt z x and x 2 , y 2 , z 2 with due regard to signs. 3. Calculate A, B, and C ast A = 7i z 2 - y 2 z x B = z x x 2 - x x z 2 C = x x y 2 - y x x 2 A +3 + C and then, L, H, K as: L = — A _ V A +3 + C M - g~ - V A + B + C H = / ,2 „3 _3 V A + B 4- C 8 NAOA .Technical Ifote No . 827 5, Determine 3c ni -.y 0 ,- and sJ n from formulas 1 1.0L y n - = 10M z n , = ■1QB . * These dimensions are then laid out to determine axis angles in three views as the angles may be computed from: tan 6 = N/L tan \j/ = L / M tan 0 •' = • M/N • .6, If tho angle <x betwoen tho two 'cables is dosirod, compute from: cos a (A 8 + yi a + Zl a ) (A s 2 + y s a + z? s ) find a from a table of trigomotric functions. '.* ■ • 7. Bisect tho angles botwoen ' cables in three views on scale layout, choosing any point, such as P-jj (fig. 3) on this bisector. 8. Detorraino tho coordinates of this point x^, y^ , and z-jj. 9. Find distance from cable intersection to .pulley axis along this bisoctor in true vlow from for- mula: a- s sin § . . 10, Find direction cosinos of bisoctor as x b L b = V x^ a + y h s + z b E NACA Technical Soto No. 827 H- : ^= / 2 a 2 y^b + + z b z h Nb = 3=b + 7b + z b 11. Pind coordinates of pulley center as; *a = . d Lb ^a = d Mb z a = d *b Bureau of Aeronautics, Navy Department, Washington, D. .0., Septembor 17, 1941. NACA Technical Note No. 827 Positive directions and angles are defined in this figure. Y + Figure 1