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Technical Document 2643May 1994

Mathematical Analysisof Actuator Forces in aScissor Lift

H. Spackman DTIC

<'-O 94-27990

Approved for public releame; diseVbulon is unlimted.

9 4 # 3 3 0 :06 N, QU LV CI- i ~D 5

Technical Document 2643May 1994

Mathematical Analysisof Actuator Forces

in a Scissor Lift

H. Spackman

4

NAVAL COMMAND, CONTROL ANDOCEAN SURVEILLANCE CENTER

RDT&E DIVISIONSan Diego, California 92152-5001

K. E. EVANS, CAPr, USN R. T. SHEARERCommanding Officer Executive Director

ADMINISTRATIVE INFORMATION

This work was performed for the U.S. Army, Army Missile Command, RedstoneArsenal, AL 35898, under program element 0604370A and accession numberDN308274, by the Systems Engineering Branch, Code 536, of the Naval Command,Control and Ocean Surveillance Center (NCCOSC), RDT&E Division, San Diego, CA92152-5001.

Released by Under authority ofJ. P. Bott, Head D. W Murphy, HeadSystems Engineering Branch Advanced Systems Division

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1.0 ABSTRACT

In 1985. NCCOSC began development of a tele-operated vehicle as part of theU.S. Marine Corps' 'nd-Air Tele-Robotics Systems Program. One of the requiredvehicle components i rigid, light-weight, and compact lift mechanism capable ofdeploying a surveilluice package 10 feet above the vehicle bed. The lift mechanism thatwas eventually built and implemented was a 3-level scissor lift. In order to analyze theforces throughout the lift structure, a set of mathematical equations was derived. Fromthese equations it was discovered that prudent placement of a lift's actuator can signifi-cantly reduce the forces required of the actuator and the stress levels in the adjacent scis-sor members. The purpose of this paper is to present the equations that were derived foranalyzing the actuator forces. Using these equations, a designer can quickly determine theoptimal locations for mounting an actuator and the resulting forces.

2.0 - DUCTION

In reference (1), equations are deri, ,d fo- determining the reaction forces through-out a scissor lift. To facilitate analysis, reference , _') divides the problem into two parts. Inthe first part, equations for a basic scissor structure - a scissor structure with no actuatorsand with all four bottom joints pinned to "ground" - are werived. In the second part. equa-tions for calculating the actuator forces are derived. A later section dis.asses how theseequations can be used to calculate the reaction forces throughout the scisso. structure.

The actuator force equations in reference (1) are derived assuming taj: one or bothof the actuator ends lie along the longitudinal axis of the scissor members. In practice, thisis seldom the case because increased mechanical advantage and improved fitment can beobtained by offsetting the actuator ends. An attempt was made to derive equations that arevalid for any actuator placement, but the solutions were not found until after publication ofreference (1). The purpose of this paper is to present the equations for the more generalcase.

3.0 DERIVATION OF EQUATIONS

As in reference (1), the actuator force equations will be derived assuming conser-vation of energy and quasistatic equilibrium. The first assumption implies that frictionalforces are negligible, and the second that acceleration is negligible. In the material thatfollows, an equation referred to as the fundamental equation is first derived. This equationexpresses actuator force as a function of the derivative of lift height to actuator length, i.e.,

F =f (dh

Following this. equations for dh/dl are derived for the two possible methods of mountingactuators. Substituting the equations for dh/dl into the fundamental equation gives thefinal result.

3.1 DERIVATION OF THE FUNDAMENTAL EQUATION

An n-level lift is shown in figure (1) along with the possible loads that can beapplied to the top of the lift and some dimensional nomenclature. Of the six possibleapplied loads (see figure (1)) only Hx and Hy result in work as the lift elevates, and Hxresults in work only if the platform translates in the x-direction as the lift elevates.

Because the lift itself is a significant part of the load, its weight must be includedin the derivation. Let B equal the total weight of the lift, B, and Bz the components of B inthe x, and z directions respectively, and By the component of B in the negative y direction.As with the applied loads, only B, and By result in work as the lift elevates. If the lift'smass is evenly distributed then the weight of the lift can be modeled by placing By /2 at thetop of the lift and B,,/2 at the translating side of the lift. The validity of this statement isillustrated as follows.

Consider the mass shown in figure (2). The total potential energy of the mass canbe approximated by

E = lim "gPuwAvy,Am -- 0

where g = acceleration due to gravityp = density.

Passing to the limit gives

h

E = Jgpuwvdv.

Now, if the mass of the object is evenly distributed over its height then

Wpguw= h

where W = total weighth = height.

2

y

HMy Hx

Hxx

Figure 1. Nomenc LEVEL a

3

S~ LEVEL i

,•====• •.•LEVEL i+1

Figure 1. Nomenclature.

h

U w

Figure 2. Arbitrary Mass.

Making the appropriate substitution results in

h

E = Wy4dy

0

W h2hy Y 10

wit2

Now, if the block height increases but the total weight stays the same then

AE = E2-Et

- (h2 - hi)2

This is the desired result.Figure (3) shows a four level lift with the loads that result in work as the lift ele-

vates. Remember that Hx contributes only if the platform is attached to the joints thattranslate in the x-direction as the lift elevates. In most designs this is not the case, and Hxwould be disregarded.

Now, the assumption of conservation of energy requires that work in equals workout. In other words, if F = actuator force and 1 = actuator length then

1I hI u1JFdl = , J +-!,)dA- (H, +du

Taking the derivative of both sides with respect to 1 results in

F _ + nx + ( .)x 1d (EQ 1)

But an expression for duldl in terms of dh/dl can be found as follows. From figure (1) it isclear that

h = n(D2 -u2) 1/2

Hy + By/2

0 0

0

0_ 0 _--- Hx + Bx/2

0

0 0

Figure 3. 4-level lift with loads that result in work.

6

dh I n (D2 - u2) 1 /2 (-2u)ddl 2 dl

-nu dii(D 2 _- it2) 1/2 dl

-n dutan 0 dl

Solving for du/dl gives

du tan0dhdl ndl

Substituting this last equation into equation I results in

F- + + n .0 (EQ 2)

This is the fundamental equation.Having derived the fundamental equation, the next step is to derive expressions for

dh/dl as a function of actuator placement and 0 (see figure 1). Two options exist formounting actuators. The first option is to attach both ends of the actuator to scissor mem-bers. In this case both ends of the actuator translate as the lift elevates. The second optionis to attach one end of the actuator to a scissor member and the other end to a fixed point.In the next two sections equations for determining dh/dl are derived for each option.

3.2 DERIVATION OF dh/dl FOR OPTION 1

Figure (4) shows an n-level Aft at several stages of deployment. The fully retractedand fully extended positions are, of course, impractical but are included to show usefulterminology. The x-y coordinates of the offset point on the positively sloping member oflevel i are given by

x = x0 cos0 + x90 sin0

y = y90 sin0+y 0cos0.

(Notice that x90 is negative in figure (4)). However, these formulas are also valid for offset

7

o LEVEL 1

0

X00

o LEVEL i

o 000

oo Y

yo v 0 LEVELn

Sy xx 90

FULLY RETRACTED PARTIALLY EXTENDED FULLY EXTENDED

Figure 4. n-level lift at vi us stages of deployment.

8

points associated with positively and negatively sloping members of any level. Now given

any two points A and B. the distance between the points is given by

12 = (xB x.4) 2 + (.B -..- ) 2

= [ (XBo-X.40 ) cosO + (xB90 - X49o) sinO] 2 + I (.BQO - '.49o) sin0 + (YBo -- YAo) cosO] 2

(acosO - bsinO) 2 + (csinO + dcoso) 2 (EQ 3)

where

Ui = XBO- Y = YB90--YAQ.90

b = xB90 - X4 9 0 d YBo-YAO •

Taking the derivative of / with respect to h gives

21d [2 (acosO + bsinO) (-asinO + bcosO) + 2 (csinO + dcosO) (ccosO - dsinO) I LOh dh'

and solving for Lhresults indI

2 1dh

dl 2 (acosO + bsinO) (-asinO ý- bcos0) + 2 (csinO + dcosO) (ccosO -dsinO)

Solving for 1 in equation (3) gives

I = [ (acosO + bsino) 2 + (csinO +dcoso) 2]11 2 .

From figure (1) it is clear that

h = nDsin0 (EQ 4)

dh = nDcosOdO

Making the appropriate substitutions results in

dh ; nscosO [ (acosO + bsino) 2 + (csinO+dcoSO) 2 ] 1/2 (EQ 5)(acos8 + bsinO) (- asinO + bcosO) + (csin0 + dcose) (ccosO - dsinO)

Equations (2) and (4) together give the final result.

3.3 DERIVATION OF dh/dl FOR OPTION 2

The equation for dh/dl for the 2nd option is derived using a slightly differentapproach. Figure 5 shows the incremental displacement of an offset point Q (presumablywhere one end of the actuator is attached). The coordinates of point Q are given by

XQ = XaQCos 0 + XQ90sin0 (EQ 6)

YOC = yo 0 Sin0flu +.Q0 (EQ 7)

Taking the derivative of both equations results in

dx( ) _Q0sin0 + XQo 0COS0

dO

"dy() =.YoS0-Yo0sin0 (EQ 8)dO QCOOYSfl

dy dYQ' dxQ'

Now, dytQ can be found by dividing -- C by -d-. Doing so results indxo d

dyQ ) YQ9O cos 0 - YQO sin 0

dx, -XQ0 sin 0 + XQgCos0

An equation for ý (see figure (5)), then, is

( YQ9ocosO -YQosinf0""a(xQ0 sin 0 - Xo9ocos0J"

10

PATH OF ACTUATOR END

1\1dll

dxQ

/ /\

LONGITUDINALAXIS OF ACTUATOR

Figure 5. Incremental displacement of actuator end.

11

From figure 5 it is clear that

dv- = siný, (EQ 9)ds

and that

ds = ___

di sin 7i-- (EQ 10)

cos (71 + 90)-1

cos( +)

Taking the derivative of EQ (4) with respect to yQ results in

dh = nDcosO dOdyc, dyQ

But, from equation 6 we have

d_.A =Idv C Y0 90cosO-yQOsinO

which when substituted into the previous equation results in

dh = nD (EQ 11)dýV( YQ9O - YQ0tan0

12

Now, we can combine equations 9, 10, and 11 to obtain

dh A dh d4v) dsC dl dyv ds dl

-nDsin

(YQ90 -.Vo0 tan 0) cos (4? +

If the other end of the actuator is attached to a fixed point, P, then

(Yo-YPIatanxo.-xp

In summary

Xo = Xo0Cos 0 + XQ90 sin0

YQ = YQ90sin0 +YQocOSO

4 = atan (YQ--YP)xo - Xp

(Y090 COS 0 - Y00 sin 0-S=atan(Y, • OYoSfl

XQ0 sin0 - XQ 90 cos0)

dh A -nDsin•dl (yQgo -yQ0tan 0 ) cos ( +4)"

4.0 CONCLUSION

A critical component of scissor lift design is the placement of the lift's actuator(s).Prudent placement can reduce the force required of the actuator(s) and reduce stress levelsin the scissor structure. By assuming conservation of energy and quasi-static equilibrium,it is possible to derive an equation that gives actuator force as a function of the derivativeof lift height to actuator length, i.e., F =.fidh/dl). This equation is referred to as the funda-mental equation in this paper, and the result is equation (2) of section 3. 1. Having foundthis equation, the next step is to derive an equation for dh/dl as a function of angular dis-placement of the scissor members. However, there are two possible options for mountingan actuator, and the derivation of dh/dl is different for each option. The first option is toattach both ends of the actuator to scissor members, and the second is to attach one end of

13

the actuator to a scissor member and the other end to a fixed point. The derivations of dl

dl for the two options are given in section 3.2 and 3.3 respectively. By combining the fun-

damental equation with the appropriate equation for dh/dl, the designer can quickly deter-

mine the optimal locations for mounting the actuator.

14

Foirm APProvedREPORT DOCUMENTATION PAGE OMB No. 0704-0188

Pu~ic reporting burden for th collection of informalin is estin',ed to average 1 hour per response, inckjding the time for revinwing istructions, searhing existing Ma sources. gathenng andmaintaining the data needed, and completing and revi•eng the collactionof inromation. Send cornrrents regarding this burden estie or any other aspeca otthis collection of i nfomlaion. includingsuggeatiorn for reducing the burden, to Washington Headquarters Senrices, Directorate for Infornation Operations and Reports, 1215 Jefferson Davis Highway, Suite 1204, Arlington, VA22202-4302. and to the Office of Management and Budget, Paperwork Reduction Protect (0704-0188), Washington, DC 20503

1. AGENCY USE ONLY (Leam blank) 2 REPORT DATE 3 REPORT TYPE AND DATES COVERED

May 1994 Final

4 TITLE AND SUBTITLE 5, FUNDING NUMBERS

MATHEMATICAL ANALYSIS OF ACTUATOR FORCES IN A SCISSOR LIFT PR: CH58

6. AUTHOR(S) PE: 0604370AH. Spackman WU: DN308274

7. PERFORMING ORGANIZATION NAME(S) AND ADDRESS(ES) a. PERFORMING ORGANIZATION

Naval Command, Control and Ocean Surveillance Center (NCCOSC) REPORT NUMBER

RDT&E DivisionSan Diego, CA 92152-5001

9- SPONSORING/MONITORING AGENCY NAME(S) AND ADORPSS(ES) 10. SPONSORING/MONITORINGAGENCY REPORT NUMBER

U. S. ArmyArmy Missile Command TD2643Redstone Arsenal, AL 35898

11. SUPPLEMENTARY NOTES

12a DISTRIBUTION/AVAJLABiLITY STATEMENT 12b. DISTRIBUTION CODE

Approved for public release; distribution is unlimited.

13. ABSTRACT (MJrnsum 200 words)

In 1985, NCCOSC began development of a teleoperated vehicle as part of the U.S. Marine Corps' Ground-Air Tele-RoboticsSystems Program. One of the required vehicle components was a rigid, light-weight, and compact lift mechanism capable ofdeploying a surveillance package 10 feet above the vehicle bed. The lift mechanism that was eventually built and implementedwas a 3-level scissor lift. In order to analyze the forces throughout the lift structure, a set of mathematical equations was derived.From these equations, it was discovered that prudent placement of a lift's actuator can significantly reduce the forces required ofthe actuator and the stress levels in the adjacent scissor members. The purpose of this paper is to present the equations that werederived for analyzing the actuator forces. Using these equations, a designer can quickly determine the optimal locations formounting an actuator and the resulting forces.

14. SUBJECT TERMS 15. NUMBER OF PAGES

Surrogate Teleoperated Vehicle (STV) 21Teleoperated Vehicle (TOV) 16. PRICE CODE

telerobotics unmanned ground vehicle

17. SECURITY CLASSIFICATION 18. SECURITY CLASSIFICATION 19. SECURITY CLASSIFlCATION 20. LIMITATION OF ABSTRACTOF REPORT OF THIS PAGE OF ABSTRACT

UNCLASSIFIED UNCLASSIFIED UNCLASSIFIED SAME AS REPORT

NSN 750-01-280-5500 Standard form 298 (FRONT)

21 a NAME OF RESPONSIBLE, NOIVIUAL 21 b. TELEPHONE (inck-e/ eA Co0d) 21c. OFFICE SYMBOL

H. Spackman (619) 553-6110 Code 536

NSN 754001 -210-810 Sta04=f Ionl 2O (BAC1)

INITIAL DISTRIBUTION

Code 0012 Patent Counsel (1)Code 0274 Library (2)Code 0275 Archive/Stock (6)Code 50 H.O. Porter (1)Code 53 D.W. Murphy (1)Code 532 C. D. Metz (3)Code 535 M. R. Solorzano (1)Code 536 J.P. Bott (1)Code 536 H. M. Spackman (15)Code 581 B.L. Croft (1)

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