Skip to main content

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

See other formats


\ 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