Computers and Structures 82 (2004) 271–280 www.elsevier.com/locate/compstruc
Finite element formulation for modeling sliding cable elements B. Zhou, M.L. Accorsi *, J.W. Leonard Department of Civil and Environmental Engineering, University of Connecticut, Storrs, CT 06269-2037, USA Received 20 August 2002; accepted 29 August 2003
Abstract ‘‘Sliding cable elements’’ are developed to solve the general problem of constraining a string of cable elements to continuously pass through a prescribed moving node. These elements can be used for a wide variety of applications and in the current work are used to model various features in parachute systems. The principle of virtual work and total Lagrange formulation are used to derive the element internal force vector, tangent stiffness matrix, and time-dependent mass matrix and body forces. The element equations are implemented in a geometrically nonlinear, transient implicit finite element program. A searching algorithm is developed to track the relative movement of the elements, which is performed each iteration prior to formulation of the element equations. Two problems with analytical solutions are performed to validate the mechanical behavior of the new element and the searching algorithm. Several examples are presented where the sliding cable elements are used in parachute systems. 2003 Elsevier Ltd. All rights reserved. Keywords: Cables; Sliding contact; Parachutes; Finite element
1. Introduction In this paper, a new class of elements called ‘‘sliding cable elements’’ is developed for modeling a variety of problems that arise in parachute applications. Parachute systems consist primarily of a canopy that is modeled with membrane elements and suspension lines, risers, and various control lines that are modeled with cable elements. Parachute systems typically undergo large dynamic motions during their deployment, inflation, and terminal descent. Successful operation of parachute systems often involves feeding various cables through moving points to achieve a specific objective. For example, various reefing systems are commonly used in parachutes to minimize the open shock during inflation [1,2]. Slider reefing uses a special device called a ‘‘slider’’ to achieve this goal [3]. A typical slider is a square piece of
*
Corresponding author. Tel.: +1-860-486-5642; fax: +1-860486-2298. E-mail address:
[email protected] (M.L. Accorsi).
fabric with grommets at each of the four corners. The parachute suspension lines are divided into four groups that are each threaded through one grommet. Before inflation, the slider is positioned at the top end of the suspension lines near the skirt of the folded parachute canopy. During inflation, the canopy opens which forces the slider to move down the suspension lines towards the confluence point at a rate that is controlled by slider drag and friction. Due to the complexity of this process, the design of parachute reefing systems has traditionally been performed using semi-empirical methods [4,5]. Computer simulation provides an effective way to evaluate the behavior of parachute systems and thereby reduce the cost for their design and development. In this paper, a new sliding cable element is developed and implemented in a geometrically nonlinear, transient implicit finite element program for modeling airdrop systems with slider reefing [6,7]. Sliding cable elements are a string of cables that dynamically pass through a prescribed node called the slider point, which can be either still or moving. Within the group, the cable that is passing through the slider
0045-7949/$ - see front matter 2003 Elsevier Ltd. All rights reserved. doi:10.1016/j.compstruc.2003.08.006
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point is called the current active sliding cable, while all other cables are called the current inactive sliding cables. A new three-node cable element is developed to model the active sliding cable, and normal two-node cable elements are used to model the inactive sliding cables. The kinematic assumption for the sliding cable element is that the strain in the active three-node sliding cable is uniform along the element. The principle of virtual work and total Lagrange formulation are used to derive the element internal force vector and tangent stiffness matrix for the active sliding cable. The mass matrix and body force vector are time-dependent for both active and inactive sliding cable elements, due to the relative movement of the slider point within the group. A searching algorithm to identify the current active sliding cable(s) is introduced and performed for every nonlinear iteration. Two numerical examples with analytical solutions are given to validate the new element and the searching algorithm. Two simulations are presented to demonstrate applications of the new element for modeling airdrop systems. The first application involves modeling of slider reefing during parachute inflation, and the second application is a payload re-orientation problem.
Fig. 2. An active sliding cable element.
sliding cables. The primary assumption used in this paper to develop the active sliding cable element is that the strain is uniform along the entire element. This assumption implies that there is no resistance, such as friction, at the slider point. This assumption can be removed by modifying the current work to include these additional effects. 2.1. Element internal force and stiffness
2. Methodology Fig. 1 shows a string of sliding cable element (SCE) consisting of one active three-node SCE passing through the ‘‘slider point’’ and multiple inactive two-node SCE. A special geometrically nonlinear three-node cable element is developed to model the active sliding cables, as shown in Fig. 2. Standard geometrically nonlinear two-node cable elements are used to model the inactive
Fig. 2 shows a definition sketch of an active sliding cable element in its initial and current configurations. Node 3 is the slider point, which divides the element into two straight parts. The fundamental kinematic assumption of the sliding cable element states that the strain is uniform along the element, i.e., the strain in both parts is the same at any time. Applying the principle of virtual work (PVW) and a total Lagrange (TL) formulation, the incremental virtual work done by the internal force is Z dWI ¼ r11 de11 A0 dL ð1Þ L1 þL2
in which e11 is the Green–Lagrange strain, r11 is the second Piola–Kirchhoff stress, and A0 is the initial crosssectional area of the element, which is assumed to be constant over the entire element length. In the TL formulation, the integration is performed over the initial configuration. Because the strain and stress are assumed to be constant along the element, the integration in Eq. (1) is performed analytically dWI ¼ r11 de11 A0 ðL1 þ L2 Þ
ð2Þ
For the three-node sliding cable element, the Green– Lagrange strain is given by Fig. 1. A group of sliding cables.
e11 ¼
g11 G11 ðl1 þ l2 Þ2 ðL1 þ L2 Þ2 ¼ 2 2
ð3Þ
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where g11 and G11 are current and initial components of the metric tensor, respectively [8]. The second Piola– Kirchhoff stress is given by r11 ¼
C1111 e11 Ee11 ¼ G211 ðL1 þ L2 Þ4
ð4Þ
in which E is Young’s modulus. Performing the variation of Eq. (3) yields de11 ¼ ðl1 þ l2 Þðdl1 þ dl2 Þ
ð5Þ
The current element length is determined from the current nodal coordinates ðxi ; yi ; zi Þ as l2i ¼ ðx3 xi Þ2 þ ðy3 yi Þ2 þ ðz3 zi Þ2 ;
ði ¼ 1; 2Þ
ð6Þ
The current nodal coordinates are related to the initial coordinates ðXi ; Yi ; Zi Þ and current nodal displacements ðui ; vi ; wi Þ by xi ¼ Xi þ ui ; yi ¼ Yi þ vi ; ði ¼ 1; 2; 3Þ
zi ¼ Zi þ wi ; ð7Þ
Substituting Eq. (7) into Eq. (6) and performing the variation, gives dli ¼
1 ½Dxi ðdu3 dui Þ þ Dyi ðdv3 dvi Þ li þ Dzi ðdw3 dwi Þ;
ði ¼ 1; 2Þ
ð8Þ
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where b¼
EA0 ðL1 þ L2 Þ3
;
U ¼ e11 ðl1 þ l2 Þ
Note that the b term is constant. Taking the partial derivative of the internal force with respect to the nodal displacement yields the following element tangent stiffness matrix Kt ¼
oFI ¼ bðDC þ UKÞ od
ð12Þ
where oU ¼ ½e11 þ ðl1 þ l2 Þ2 DT ; od 2 3 h1 0 h1 oD 6 7 ¼4 K¼ h2 h2 5 od sym h1 þ h2 9 9 C¼
and 2 2 Dx l2i 14 i hi ¼ 3 li sym
Dxi Dyi Dyi2 l2i
3 Dxi Dzi Dyi Dzi 5 ; Dz2i l2i 3 3
ði ¼ 1; 2Þ
It should be noted that the element equations, as written, are singular when the slider node coincides exactly with either of the end nodes (li ¼ 0). Element equations for these limit cases can be derived analytically. 2.2. Element mass and body force
in which Dxi ¼ x3 xi ; Dyi ¼ y3 yi ; Dzi ¼ z3 zi ; ði ¼ 1; 2Þ Substituting Eq. (8) into Eq. (5), gives the virtual strain term as de11 ¼ ðl1 þ l2 ÞDT dd where Dx1 D¼ l1
Dy1 l1
Dz1 l1
1Dx1 Dx2 l1 l2
dd ¼ fdu1 du3
dv1 dv3
ð9Þ
Dx2 l2
Dy2 l2
Dy1 Dy2 l1 l2
dw1 du2 dw3 gT
dv2
Dz2 l2
Dz1 Dz2 l1 l2
T
dw2
The incremental virtual work given by Eq. (2) can be rewritten as dWI ¼ FIT dd ¼ bUDT dd
ð10Þ
and the internal force vector is FI ¼ bUD
ð11Þ
In a TL formulation, the element mass matrix and body force vector are typically computed only once on the initial configuration. For the sliding cable elements, however, an initially inactive sliding cable may become active, and an initially active sliding cable may become inactive, at any time. Therefore, the element mass matrix and body force vector for sliding cable elements must be updated within the nonlinear iteration. Similarly, the mass distribution within an active sliding cable changes and therefore must be continuously updated. An inactive sliding cable is the same as a two-node cable element, and its element mass matrix is given by
Z qA0 l L 2I I MINACT ¼ qN T N dV0 ¼ 6 l I 2I 6 6 V0
qA0 L 2I I ¼ ð13Þ I 2I 6 6 6 where N are the element shape functions, I is a 3 · 3 identity matrix, q is the density (assumed constant), A0 is the initial cross-sectional area (assumed constant), l is the current element length, and L is the initial element length. The slider point divides an active three-node sliding cable into two straight parts, each with two nodes.
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Therefore, the mass matrix of the three-node sliding cable in the current configuration is obtained simply by assembling the individual mass matrices for each part using Eq. (13) and the current lengths. In the TL formulation, the mass matrix is computed over the original configuration, which is obtained by scaling the current mass matrix by a factor of L=l, where L ¼ L1 þ L2 and l ¼ l1 þ l2 are the initial and current total lengths, respectively. Therefore, the mass matrix for the three-node active sliding cable element is given by 2 3 2I 0 I qA0 l1 L 6 7 0 05 MACT ¼ 4 6 l sym 2I 9 9 3 2 0 0 0 qA0 l2 L 6 7 2I I 5 þ ð14Þ 4 6 l sym 2I 9 9
current configuration of the slider point and sliding elements. Initially, the active sliding element(s) within every group is prescribed as user input, and the initial length of every sliding element is computed and stored accordingly. To search for the current active element for a given slider point S at some time t, every element within that group is first assumed to be active and the assumed current length (l ¼ l1 þ l2 ) of each sliding element is computed and compared with its initial length (L). The one that has the least difference between the assumed current length and initial length is determined to be the current active sliding cable with respect to S. For the numerical examples presented in this paper, this straightforward algorithm performed very robustly.
The element body force, due to self-weight, is formulated in a similar fashion. For an inactive element, Z qA0 L g B ¼ N T f B dV0 ¼ ð15Þ FINACT g 6 1 2 V0
In this section, two problems with analytical solutions are used to validate the sliding cable elements.
where f B ¼ qg is the weight per unit volume, and g is the gravity vector given by
This problem is designed primarily to validate the mass and body force of the sliding elements. As shown in Fig. 3(a), a rope passing over a fixed frictionless pulley is subject to a gravity force only and is released from rest at time zero. The rope has a total length of 100 ft (30.48 m), of which L1 ¼ 40 ft (12.19 m) and L2 ¼ 60 ft (18.29 m) initially. The cross-sectional area A and density q are uniform along the length, and the rope is assumed to be infinitely stiff such that its total length remains constant. Fig. 3(b) shows the free body diagrams of the two pieces of rope on the left and right sides of the pulley. The tension force is the same and the acceleration is equal in magnitude but in opposite directions on both
g ¼ f gX
gY
gZ gT
The body force for an active sliding cable element is, therefore 8 9 8 9 g 0 qA0 l1 L < = qA0 l2 L < = B FACT ¼ þ ð16Þ 0 g 2 l: ; 2 l: ; g 9 1 g 9 1 2.3. Searching algorithm Because the element equations of active and inactive sliding cable elements are different, it is essential to find out within a certain group of sliding elements which element is active before actually formulating the element equations. A searching algorithm is therefore needed to automatically determine this information. Searching is performed on every group of sliding elements, and a finite element model may have more than one group. A group consists of at least one slider point and all the sliding elements that are allowed to pass through the slider point. For a given slider point, there is only one active sliding element in that group at any given time. If more than one slider point is present in a group, the searching must be performed repeatedly for each slider point. There are many potential methods to implement the required searching algorithm. A straightforward method is adopted that predicts the active element based on the
3. Verification problems
3.1. Verification problem 1
Fig. 3. Rope sliding over a fixed pulley.
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sides. The mass of each rope is lumped at its centroid. From Newton’s law, the coupled equations of motion at the current time t are qAl1 g T ¼ qAl1€l1 ð17Þ qAl2 g T ¼ qAl2€l2
agree very closely with the analytical solutions. This is because a lumped mass is used in the sliding cable element and, therefore, the element mass and gravity force at nodes A and B are proportional to the length of the left and right sides, which coincides exactly with the analytical formulation.
where l1 and l2 are the current lengths of the left and right sides, respectively. Uncoupling the two equations by eliminating T and the qA term gives
3.2. Verification problem 2
ðl1 l2 Þg ¼ l1€l1 l2€l2
ð18Þ
Since the total length of the rope is constant and the accelerations of the left and right sides are equal magnitude but opposite direction l1 þ l2 ¼ L1 þ L2 ¼ L ð19Þ €l1 þ €l2 ¼ 0 Substituting Eq. (19) into Eq. (18) yields an ordinary differential equation of motion for the left side €l1 2g l1 ¼ g L
ð20Þ
This ODE can be solved analytically to obtain rffiffiffiffiffiffiffiffi! L 2g L cosh t þ ð21Þ l1 ðtÞ ¼ L1 2 L 2 A single sliding cable element is used to simulate this problem numerically. The fixed pulley serves as the slider point, and the element mass is lumped at the three nodal points. The gravitational acceleration is g ¼ 10 ft/s2 ()3.05 m/s2 ) and the Young’s modulus of the sliding cable element is very large. Shown in Fig. 4 is a comparison between the numerical solution and the analytical solution given by Eq. (21) for the vertical position of the two end nodes as a function of time. It can be seen that the numerical results
Fig. 4. Nodal positions of the sliding rope.
This problem is designed primarily to validate the tangent stiffness matrix and internal force of the sliding cable element, as well as the searching algorithm. As shown in Fig. 5(a), a bead of mass m is suspended on a massless rope and is released from rest at time zero. Driven by its own weight, the bead slides dynamically along the rope without friction. The horizontal and vertical position of the bead is denoted by h and v, respectively, as shown in Fig. 5. Initially, h ¼ v ¼ 0:5 ft (0.15 m), and d ¼ 1 ft (0.30 m) is constant. The rope is assumed to be infinitely stiff such that its total length remains constant at all time. Fig. 5(b) shows the free body diagram of the bead at the current time t. The left and right tension forces are the same, and the current left and right rope lengths are l1 and l2 , respectively. Applying Newton’s law in the x and y directions yields the following coupled equations of motion
€ T cos a2 T cos a1 ¼ max ¼ mh mg T sin a1 T sin a2 ¼ may ¼ m€v
ð22Þ
Since the total length of the rope is assumed to be constant
Fig. 5. Sliding bead problem.
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l1 þ l2 ¼ L1 þ L2 ¼ L
ð23Þ
where L1 and L2 are the initial lengths. At any time, the following geometrical relations apply 8 h v > > cos a1 ¼ < sin a1 ¼ ; l1 l1 ð24aÞ 2d h v > > : sin a2 ¼ ; cos a2 ¼ l2 l2
v2 þ h2 ¼ l21 ð2d hÞ2 þ v2 ¼ l22 ¼ ðL l1 Þ2
ð24bÞ
Substituting Eqs. (23) and (24) into Eq. (22) and uncoupling the two equations by eliminating T gives the following nonlinear ordinary differential equation of motion for the bead 8 > LvðtÞ€hðtÞ > > ¼g < €vðtÞ þ 2l1 ðtÞd LhðtÞ > 2 2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi > > : l1 ðtÞ ¼ L þ 4dhðtÞ 4d ; vðtÞ ¼ l21 ðtÞ h2 ðtÞ 2L ð25Þ Since Eq. (25) cannot be solved analytically, the Runge–Kutta method is used to solve it numerically. Two simulations are performed to solve this problem using the new sliding cable element. The first simulation uses a single sliding cable element, and the second simulation uses five sliding cable elements. In both simulations, the bead serves as the slider point. The gravitational acceleration is g ¼ 10 ft/s2 ()3.05 m/s2 ), and the Young’s modulus of the rope is very large. Figs. 6 and 7 show the horizontal and vertical displacements of the bead calculated from a single sliding cable element, respectively. Figs. 8 and 9 show the horizontal and vertical displacements for multiple sliding cable elements, respectively. For comparison, the analytical solution from Eq. (25) is included on each
Fig. 6. Horizontal displacement of the sliding bead (singleelement).
Fig. 7. Vertical displacement of the sliding bead (single-element).
Fig. 8. Horizontal displacement of the sliding bead (multiple elements).
Fig. 9. Vertical displacement of the sliding bead (multiple elements).
B. Zhou et al. / Computers and Structures 82 (2004) 271–280
figure. It can be seen that both the single-element case and multi-element case yield very accurate solutions. For the multiple elements simulation, the bead moves between two elements several times during the simulation. It was found that localized disturbances occur when the slider point transitions between elements due to a singularity in the element equations (li ¼ 0). These disturbances can cause the numerical results to become unstable. In the current problem, it was found that the use of numerical damping was very effective in stabilizing the numerical solution. A slight decay in the vibration amplitude caused by numerical damping is apparent in the multi-element results.
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Table 1 Material and mechanical properties for slider reefing simulation Young’s modulus (all materials) Mass density (all materials) Poisson’s ratio (all materials) Canopy pressure Drag diameter (canopy gores and suspension lines) Drag diameter (four edges of the slider) Drag diameter (other)
4.32 · 106 lb/ft2 (2.07 · 108 N/m2 ) 0.6 slugs/ft3 (309.23 Kg/m3 ) 0.3 0.1 psf (4.79 Pa) 0.03 ft (0.009 m) 1 ft (0.30 m) 0
4. Simulations of airdrop systems In this section, two examples are given to demonstrate the use of the new sliding cable elements for modeling airdrop systems. 4.1. Slider reefing In this section, sliding cable elements are used to model the slider reefing problem using a half-scale C-9 parachute. The C-9 parachute consists of a 14 ft (4.27 m) diameter canopy and 28 suspension lines of 11.5 ft (3.51 m) long each [2]. The suspension lines meet at a single confluence point that is connected to the payload, which is modeled by a concentrated mass. A single four-node membrane element is used to model the slider with each node representing one of the four slider grommets. The suspension lines are evenly divided into four groups, each running through one grommet. Each suspension line, consisting of eight elements, is treated as a sliding cable group, with the grommet acting as the slider point. Therefore, seven groups of sliding cable elements share the same slider point. The entire model is subject to a gravitational acceleration of g ¼ 32:2 ft/s2 ()9.81 m/ s2 ). The material and mechanical properties of the system is summarized in Table 1. Simulation of the inflation starts from a highly folded initial configuration as shown in Fig. 10. A detailed view of the slider and the suspension lines is also shown. The canopy is pressurized internally with a constant uniform pressure of p ¼ 0:1 lb/ft2 (4.79 Pa) and the system is free to fall under gravity. Velocity-dependent drag forces are applied to the parachute and each of the four edges of the slider. In order to observe the effect of the slider, another simulation is performed that is identical to the first one except it does not have a slider. Fig. 11(a–d) show the side view of the parachute at equal time intervals during the inflation. In each figure, the left plot is from the simulation without the slider, and the right plot is from the simulation with the slider. For the system without slider, the canopy is well inflated
Fig. 10. Folded C-9 parachute and the slider.
at t ¼ 0:50 second, whereas the system with the slider inflates much slower. Nevertheless, at the end of each simulation, both parachutes are very similar to each other in shape, showing that the slider slows down the inflation but does not affect the final inflated shape of the system. Another way to demonstrate the slowing effect is to study the change of canopy radius during inflation. Fig. 12 shows the radial distance from the skirt (node 140) to the central axis of the parachute during inflation for both simulations, and clearly demonstrates that the inflation rate of the canopy with the slider is significantly reduced. 4.2. Payload re-orientation In this example, the payload attached to a parachute system is re-oriented during terminal descent.
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Fig. 11. Side view of C-9 parachute during inflation, (a) t ¼ 0:25 s, (b) t ¼ 0:50 s, (c) t ¼ 0:75 s and (d) t ¼ 1:00 s.
Fig. 12. C-9 parachute canopy radius.
B. Zhou et al. / Computers and Structures 82 (2004) 271–280
Re-orientation of the payload is sometimes required and can be achieved by suspending the payload with temporary support cables and support cables that pass through metal rings attached to the system. When the temporary cables are removed (by pyrotechnic cutters), the payload will be re-oriented based on the configuration of the remaining support cables. The sliding cable element developed here is an ideal tool for modeling these problems. A fully inflated C-9 parachute descending at a terminal speed of approximately 20 ft/s (6.10 m/s) is used in the simulation. The payload is modeled as an 8 · 2 · 2 ft3 (0.91 m3 ) solid box that weights approximately 227 lb (1009.7 N). The entire model is subject to a gravitational acceleration of g ¼ 32:2 ft/s2 ()9.81 m/s2 ). The material and mechanical properties of the system is summarized in Table 2. The inflated configuration and initial payload orientation is shown in Fig. 13. In the initial orientation, the payload is suspended with its long edge positioned vertically by the combined support of two temporary cables and two sliding cables, which both have the confluence point as the slider point. A detailed view of the payload
279
and the sliding cables is also shown. At the beginning of the simulation, the two temporary cables are broken, leaving the two sliding cables to support the payload. The payload becomes unbalanced and starts to rotate, while simultaneously, the sliding cables move through the confluence point. The payload is expected to reach an equilibrium state at which the long edges are positioned horizontally. In this model, however, there is no mass proportional damping and therefore the payload will continue to oscillate about the equilibrium position. It is found that the rotation is quite slow if the payload has uniform mass. To expedite the rotation, an additional point mass is added to each of the four corners of the payload surface that is expected to face the ground at the equilibrium position. The simulation lasts for two seconds. Several side views of the system at selected times during the
Table 2 Material and mechanical properties for payload re-orientation simulation Young’s modulus (all materials) Mass density (sliding cables) Mass density (all other materials) Poison’s ratio (all materials) Canopy pressure Drag diameter (canopy gores) Drag diameter (other)
4.32 · 106 lb/ft2 (2.07 · 108 N/m2 ) 0 0.6 slugs/ft3 (309.23 Kg/m3 ) 0.3 0.72 psf (34.6 Pa) 0.4 ft (0.12 m) 0
Fig. 13. Initial configuration of the C-9 payload re-orientation system.
Fig. 14. C-9 payload re-orientation simulation, (a) t ¼ 0:70 s, (b) t ¼ 0:80 s, (c) t ¼ 0:90 s and (d) t ¼ 1:50 s.
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payload during re-orientation is strongly affected by the parachute, whereas the vertical motion is not significantly different.
5. Conclusions
Fig. 15. Relative horizontal displacements.
In this paper, a sliding cable element was developed and implemented in a geometrically nonlinear, transient implicit finite element program to facilitate the modeling of airdrop systems. A searching algorithm to track the relative motion of the sliding elements is performed at every iteration prior to the formulation of the element equations. Numerical examples were presented to validate the behavior of the new elements. Two application problems were presented to demonstrate the capabilities of the new element for modeling airdrop system. In the current work, it was assumed that the sliding cables move through the slider point with no resistance. This assumption can be removed by modifying the formulation to include some resistance effects, such as friction.
Acknowledgements The material in this paper was based upon work supported by, or in part by, the US Army Research Office under grant number DAAD19-99-1-0235.
References
Fig. 16. Relative vertical displacements.
operation are shown in Fig. 14(a–d). It can be seen that the dynamic rotation of the payload induces a significant horizontal oscillation in the rest of the parachute system. It is also of interest to study the effect of the parachute on the rate of rotation of the payload. In order to do this, a simplified model is used that only has the payload and support cables (i.e., no canopy or suspension lines) and a fixed slider point. Results from this simulation are compared to the previous simulation using the relative movement of one node (node 7) on the payload. Figs. 15 and 16 show the relative horizontal and vertical displacements, respectively, for both simulations. In comparison, the horizontal motion of the
[1] Brown G, Norton B, Lingard JS. Inertially controlled recovery system. In: Proceedings of the 15th Aerodynamic Decelerator Systems Technology Conference, Toulouse, France, 1999. [2] Ewing EG, Bixby HW, Knacke TW. Recovery Systems Design Guide. Technical Report AFFDL-TR-78-151, 1978. [3] Knacke TW. Parachute recovery systems design manual. Santa Barbara, CA: Para Publishing; 1992. [4] Potvin J, Peek G, Brocato B. Modelling the inflation of Ram-air parachutes reefed with sliders. J Aircraft 2001; 38:818–27. [5] Potvin J. Testing a new model of Ram-air parachute inflation. Aeronaut J 1997;101:299–313. [6] Accorsi ML, Leonard JW, Benney RJ, Stein KR. Structural modeling of parachute dynamics. AIAA J 2000;38:139–46. [7] Bathe KJ. Finite element procedures. NJ: Prentice-Hall Inc; 1996. [8] Green AE, Zerna W. Theoretical elasticity. Dover Publication Inc; 1992.