N O T I C E 

THIS DOCUMENT HAS BEEN REPRODUCED FROM 

MICROFICHE. ALTHOUGH IT IS RECOGNIZED THAT 

CERTAIN PORTIONS ARE ILLEGIBLE, IT IS BEING RELEASED 

IN THE INTEREST OF MAKING AVAILABLE AS MUCH 

INFORMATION AS POSSIBLE 
SC,,^^S

KINEMATICS OF FOLDABLE DISCRETE SPACE CRANES

by

Adnan H. Nayfeh

Department of Aerospace Engineering

and Engineering Mechanics

University of Cincinnati

Cincinnati, OH 45221-0070

ABSTRACT

Exact kinematic description of a NASA proposed prototype foldabledeployable discrete space crane are presented. A computer program is

developed which maps the geometry of the crane once controlling parameters

are specified. The program uses a building block type approach. In which J

it calculates the local coordinates of each repeating cell and then combines t

them with respect to a global coordinates system.

(NASA-CE-176360) KINEMATICS CF FOLDABLE

N86- 13735

UISCBEIE SPACE CRANES F

inal Report

(Cincinnati Univ.)

A A

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Introduction

The coming years will see an increasing utilization c,• ace through

the use of very large, lightweight structures constructed in .;. • oit for such

purposes as antennas, energy collection platforms and work stations. With

the recent successful flights of the Space Shuttle, such projects are one

step closer towards becoming reality. It will be most desirable that these

structures be made of repeating sub-assemblies In order to aid in the ease

of construction, minimize fabrication costs, and reduce complications in

analysis.

Under our NASA Grant NSG 1185, we have been investigating various

methods of designing and analyzing the geometry of discrete large space

structures. Our previous extensive work has dealt with both the areas of

design and analysis. Most recently, we completed and published results on

establishing the dynamic behavior of large space structures in the forms of

periodic trusses and frames.

For the purposes of in-space construction and subsequent reparis, large

space tools are required. It is most desirable for these tools to be kept

packed and to be deployed only whenever they are needed. An important

example of such tools are the deployable crane 'large arm) and telescopic 

supports as illustrated in Figure 1. This tool has the mobility of full or 

partial deployment and to move in any direction in order to reach desired

points in the host structure.

The mechanism and control of the deployment of this tool is very

essential. This specific NASA crane-prototype can be deployed and

maneuvered in various directions by changing the lengths of members in its

bays. To know how much change in these lengths is required to achieve the

2
Figure 1 Crane Geometry

3
required configuration is essential. This is also important to know for the

control of the vibrations of the crane.

In this report we analyze the kinematics of the crane configuration

which result from the changes of the lengths of the bay members.

Specifically, we developed a computer program capable of calculating the

coordinates of each node of the crane resulting, from member length changes.

This information is very essential in the utility of the crane. The new

configuration kinematics can also be fed to a vibration program to calculate

the dynamic characteristics of the new configuration, if found necessary.

Structures Description

The specific crane structure under consideration consists of repeating

unit cells as shown in Figure 2. Each unit cell constitutes a truss

substructure with nine nodal points as illustrated in Figure 3. In this

unit cell the lengths of the rods connecting the nods 4, 5 and 6 are

variables and hence control the overall shape of the cell and ultimately the

total structure. On the other hand all remaining rod members have constant

lengths. Specifically, members (1,2), (1,3), (2,3), (7,8), (7,9) and (8,9)

have constant equal lengths denoted by G, and aside from the variable member

lengths (4,5), (4,6) and (5,6), all remaining members have the constant

equal lengths B.

Summary of the Analysis Procedure

In order to give complete description of the kinematics of the crane

with arbitrary number of unit cells, we use a building block type approach

consisting of the following steps:

(1) By choosing a local coordinate system (on the single unit cell

level) we calculate the coordinates of its' nine nodes.

4
w

1

0

0

- Figure 2 Repeating Unit Cell

iL

^Cess

fit

Figure 3 Projected Geometry

5
(2) These coordinates are then referred to a preassigned global

coordinate system (for the total structure).

(3) If the nodal global coordinates of the (n-1)th cell are known,

then the plane of its' top three nodes (7, 8 and 9) will define

the base for the nth cell. This suggests the use of coordinate

transformations to define such a base.

(4) Orthogonal coordinate transformations are constructed for the top

plane of each cell as a base for the next cell.

Note: A major fact in constructing the local coordinates of each

unit cell is that the plane of its 7, 8 and 9 nodes is a mirror

Image of the plane of the nodes 1, 2 and 3; the mirror being the

plane of the nodes 4, 5 and 6.

The floor of this unit cell is the equilateral triangle with side

lengths G and having the nodal coordinates:

Node 1: (0,0,0) (1)

Node 2: (G,0,0) (2)

Node 3: (G/2, 2 

G,0) (3)

Now, let us arbitrarily assume that points 4, 5 and 6 have the vertical

heights (coordinates) Z 4 - H4 ; Z 5 - H5 and Z 6 - H6 . If we now designate the

projected lengths of the members 2-5, 1-6 and 1-4 by B25' 

B16 and B14,

respectively, then one concludes that

B( 

B

2 _ H2)112

25 5

B

(B2 _ H2)112

14 4

(4)

(5)

6
F

Bt6 (B2 

j H22)1/2

(6)

where B is a constant.

Furthermore, by inspection one can immediately deduce the coordinates

of the point 4 as

Node 4: (2' -[B14 -(•5G)2]112 H4) (7)

Let us now concentrate on finding the coordinates of point 5. From

Figure 3 we see these to be

Node 5: ( G + B25c06::e

25' B

25 since25 , H5 ) (8)

using the trigonometric manipulations

sinC^6 5 = sin(3- - e25)

= sin

21r 21r

- cos3'^sine25

= 2 ose25 + 112 sine

yields

1

G 1 

LB2 - (G/2)21 5 2

since r3

= 2 2B + 2 B

(9)

25 25

Hence,

Y5 = B25 since

25

1

3

G + 2(B25- (G/2)2)2 (10)

and by 31milar arguments we see that

cosce^5 = cos(3'r ' e25)

= cos 

3

ose25 + sin 

3! 

sine

25

7
- 2cose25 + 2 `ine25

1

_ 1 G + 3 CB25 - (G/2)2]2

2 28

25

2 B25

(11)

Hence,

x5 - G + B25 cosCe25

1

3G* 2 [B25 - (G/2) 2 ] 2 (12)

By symmetry and inspection of the coordinates of point 5 We can deduce

the coordinates of point 6 as

Node 6:

1

2

x6 - = 2 [B 16 = ( G/2) 2] (13)

1

Y6 - 7 G + 2 CB16 = ( G/2) 2 ] 2 (14)

and

Z 6 - H6

Remarks

Notice that from (10) and (12) that

x5 - r Y5(15)

and from (13) and (14)

x6 - 1 - r3 Y6

(16)

8
Calculations of the Lengths D

45 , D46 and D56:

In what follows we shall normalize our length G to be unity.

Accordingly, by inspection we have the further relations

X4 + Y4+ Z4 - B2 (17a)

X6 + Y6+ Z6 - B2 (17b)

( X5 - 1 ) 2 + Y5+ Z5 - B2 (170

We also have

D45 - ( X5 - X4 ) 2 + ( Y5 - Y4 ) 2 + (Z5 _ Z 4 ) 2

(18a)

2 2

5 5

S

- X

2 +Y2 +Z+X4+Y2

4+Z4

- 2(X4 X5 + Y 4 Y 5 + Z 4 Z 

5 )

- (X5 ; 1) 2 + 2X5 = 1 + Y6 + Z5 - 2(X4 X5

+ Y

4 Y5

+ Z4Z5 ) + B 0 8b)

- 2B2 + 2X5 -1 - 2(X4 X5 + Y4 Y5 + Z4Z5)

where we used (17c), hence

D45 1 2B2 + 1 - 2[X5 _ x

4 x

5

Y

4 Y

5

i Z 4 Z 5 ] (19a)

- 2[X5 (1-X4 ) - Y 4 Y 5 - Z 4 Z 5 ) ( 19b)

which from (7), where now X4 - 2'

x

- 2[ 52 - Y4 Y5 - Z 4 Z5 1(19c)

Hence,

X

2 [D25 - 2B

2 + 1 3 -

 2

5 - Y

4 Y

5

- Z4Z5

which, by using the relation (15) can eliminate X5 to get

9
1ED45 - 2B2 + 1] - Y5 ( 2 . Y4 ) - Z4Z5

which can be rewritten as

1+D2

B2 - ( 2 45 ) - Y5 ( Y4 = 2 ) * Z4Z5

Let now

1 + D2

A 45 - B2 ' ( 2 45 )

Then we have

A45 - Y5 (Y4 - 2 ) + Z4Z5

Let further

* r

Y4 - Y4 2

Then

A45 - Y5 Y4 + Z 4Z5(24)

By similar arguments one can replace subscript 5 by 6 and immediately obtain

the relation

A46 - Y 6 

Y 4 + Z426

where

A B2 - (1 + D46)

46 - 2

Finally from the relation

D56 - (X5 - X6 ) 2 + (Y5 - Y6 ) 2 + (Z 5 - Z6)2

X 5 + Y5 + Z5 + B2 - 2(X5 X6 + Y5 Y6 + Z5Z6)

(20)

(21)

(22)

(23)

(25)

(26)

10
one gets,

D56 - 2B2 + 2X5 - 1 - 2(X5 X6 + Y5 Y6 + Z5Z6)

thus, we define

A B2

_ (1 + D5 6)

56 - 2

and get

(27a)

A 56 - X5 (X6 - 1) + Y 

5 Y 6 + Z5 Z 6 (27b)

which, if the relation (16) is used, finally yields

A56 - - r X5 Y6 + Y6 Y5 + Z5 Z 6 (270)

Now, if the relation (15) is used to eliminate X5 we finally get

A56 - Z 5Z 6 - 2Y5 Y 6

(28)

Method of Solution (The Inverse Problem)

We indicated earlier that if H H and H are known then all of the 

V 5 6

coordinates of the points 1, 2, 3, 4, 5 and 6 will be known. Unfortunately,

what is known a priori are not the H 4 . H5 and H6 , but D 45 , D46 and D56 . In

what follows we shall develop solution methods whereby if the lengths D 450 

E

y

D46 and D 56 are given then we can solve the inverse problem and find H 4 , H5

and H 6 and hence all the coordinates of the nodes. This will be done by

simple iteration on the value of H 4 :.a will be demonstrated below.

Let us assume that D 45 , D 46 , D Cj and H 4 are given. Then in the

relations (2 1 1), (26) and (28), the parameters A 45 , A 469 A 56' Y 4 , Y4 and (Z4

H 4 ) will be known. Furthermore, since from (10), Y 5 can be written in

11
terms of Z 5 - H5 and from (14) Y 6 can be written in terms of Z6 - H6 , we can

easily et that we have only the two unknowns H5 and H6.

Specifically, we now use the relations (24) and (10) and get

1

A45 - 

Y4[ 4 * 2(82 5 - .25) 2 J + L4Z5

or

-

* 1

A 45 - Y4 (B2)25 - .25 + Z4Z5 Y4

Now, from (4)

B2

B2 - HS

5 - B2 - Z5 

we get

2

4(A45 - 4 - Y 4 Z 4 Z 5 ) 2 - Y4 [B2 - Z? = .251

J

which can be further rearranged in the form of a quadratic equation for Z5

is

(4Z4 + Y 4 2 )Z5 - bF 45Z 4 Z 5 + 4F45 - Y4 2 (B` - .25) - 0 (29)

where

F45 - A 45 - 4 Y4 (30)

Notice in (29) that if Z 4 - H4 is known and since B is known then Z 5

is the

only unknown. Hence, given H 4 there is a unique value for Z5.

By interchanging 5 and 6 similar relation to (29) can be constructed

for point 6 as

4 + Y

* 2

(4Z ) Z 6 - 8F46Z4Z6 + 4F4 6 - Y4 2 (.B 2 - .25) - 0 (31)

12

J
where

F 46 - A 46 -

124 Y4 (32)

Thus, given Z 4 - H4 then there is a unique solution for Z6.

Since, if Z 4 is known, we can calculate Z 5 and Z 6 uniquely, the

condition that fixes the exact value on Z 4 is the relation (28).

Specifically one has to iterate of Z 4 until the calculated values Z 5 and Z6

via (29) and (31) satisfy (28).

Now equations (29) and (31) admit the two solutions

1

2

5(1,2) - [T2 i (T2 - T 1 T 3 ) 2 ]/T 1 (33)

1

26(1,2) - [W2 t (W2 W1 W3 ) 2 ]/W1

(34)

where

T 1 - W1 - 4Z2 + Y

n42

T2 - 4F4524

T 3 - 4F4 5 - Y4 2 ( B 2 - . 25)

W2 - 4F4674

W3

 - 4F2 6 - Y4 2 ( B2 - . 25) (35)

Corresponding to the values (33) and (34) we can calculate new values

for Y5 and Y6 which we shall refer to as K 5 and K6 , respectively. Thus, we

have (see (10) and (14) with G - 1)

_ 1

KS - 3 + 2(V2 5

- .25) 2

(36a)

13
1

K6 - 4 + 2-(V2 .25)2 (36b)

where (see also (4) and (6))

1

V25 - (B2 -

z50))2 (37a)

1

V16 - ( B2 - 

Z6(1))2 (37b)

From (33), (34), (36a) and (36b) we construct the function (see 28)

S 56 ' z

50) z 60)- 2K5 K6 (38)

where the idea here is to iterate on Z 4 - H 4 until the value of S56

converges to the value of A 56 as calculated via (27). Once this convergence

takes place we will have the right values of Z 5 and Z 6 (i.e. namely H5 and

H6 ) and we can then proceed to calculate the correct coordinates of points

4, 5 and 6.

Calculations of the Coordinates of Nodes 7, 8 and 9

By inspection one can easily find that the nodes 7, 8 and 9 are the

mirror images of nodes 1, 2 and 3 respectively with respect to the plane of

nodes 4, 5 and 6. To this end, using some algebraic manipulation we thus

get:

X i+6 - X i - 2niPi

Y i+6 - Y - 2niPi

z i+6 - z - 2n i P it i - 1, 2, 3 (39)

where

n, are the components of the unit normal to the plane of the nodes 4,

5, 6 and P i is the dot product between the normal n and the vector

14
connecting the nodes i and 4. This completes the local coordinates of the

repeating unit cell.

Building Block Results

With the coordinates of nodes 7, 8 and 9 now known, their plane defines

the base for the next unit cell. This is done by introducing a linear

transformation of the original coordinates according to

Xi = BijXj

(40)

where Xi are the new coordinates and 

aii 

are the direction cosines between

Xi and X^. These values of the transformation tensor can be constructed

from the normal to the plane of the vortices 7, 8 and 9. Once this is done

then a building block approach yields the results for an arbitrary number of

cells as is documented by the accompanying computer program. In the program

a sample calculation of five unit cells is included and the coordinates of

the vortices are listed in local and global coordinate systems.

15
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