Openng Shock and Shape o he Drag-vs-Te Curve Jean Povn Physcs Deparen, San Lous Unversy, S. Lous MO Conac: povnj@slu.edu 314-977-8424 Talk presened a he 19 h AIAA Aerodynac Deceleraor Syses Conerence Wllasburg, A, May 22-24, 2007
Work unded by Nack Solder Cener Nack, MA U.S. Ary conrac W9124-06-P-1068
The aeral presened here was and wll be publshed n he Journal o Arcra by hs speaker: Unversaly Consderaons or Graphng Parachue Openng Shock acor ersus Mass ao "; JOA, 44, No. 2, pp. 528-538, 2007. "Moenu-Ipulse Balance and Parachue Inlaon: Clusers "; JOA, 44, No. 2, pp. 687-691, 2007. "Moenu-Ipulse Balance and Parachue Inlaon: Ds-eeng"; JOA, 44, No. 2, pp. 691-694, 2007. Moenu-Ipulse Balance and Parachue Inlaon: ocke-propelled Payloads"; o appear n JOA, 2007. "Moenu-Ipulse Balance and Parachue Inlaon: xed-pon Drops "; o appear n JOA, 2007.
Sple esaors or parachue nlaon perorance ax and ll ax nl
Bu here s anoher poran denson o consder: The shape o he drag-vs vs-e curve Jup-o-jup shape varaons on sae canopy slder-reeed spor paraol Anoher ype o shape gure 1.23. Drag evoluon experenced by a Perorance Desgns Sabre 150 paraol slder-reeed [42]. ro op o boo, he drag negral equaon 3.9 has he ollowng values: I = 0.48, 0.50, 0.47 and 0.44.
Is he value o ax conneced o he shape o he curve? You be! How? Use he Moenu-Ipulse Theore MI-Theore
The Moenu-Ipulse Theore Inegraon o Newon s 2 nd law o oon = d + W cos θ d D Moenu change Parachue drag Gravaonal o parachue-payload pulse pulse = descen rae @ end o nlaon = descen rae @ he begnnng o nlaon.e. @ lne srech θ = lgh angle
The Moenu-Ipulse Theore eorulae n ers o ax = d + W cos θ d D = + ax I W cos θ d Inlaon duraon Drag negral; gauges he shape o he drag vs. e curve I = ax D d
The drag negral easures he area under he drag-vs-e curve n uns o ax ll Drag negral ~ ½ Drag negral ~ 1 Mos coon case a hgh ass rao.e. personnel and slow-descen cargo
How abou a low-ass rao? drogues; chues opened n wnd unnels, ec. ME dsk-gap-band parachue nlang n he NASA Aes ull-scale wnd unnel. The drag negral correspondng o hs es s I = 0.23 very ypcal deploy Inlaon begns Inlaon ends
Ds-reeng case long aer reeed nlaon Drag negral ~ ¼. Noe: drag negral 0 as ds ree W/ ds << ½
Solvng he MI-Theore gves ax n ers o he drag negral I Applyng o specc rajecory ypes: Horzonal 1 ax = 1 + g cos θ d I Inlaon duraon ax = 1 I Speed @ end o nlaon roude er ercal ax = I 1 + g Speed @ lne-srech
Solvng he MI-Theore gves ax n ers o he drag negral I 1 ax = 1 + g cos θ d I Iporan noe and are prey uch he sae or any syses Snce he speed a lne srech s osly deerned by launch ac speed In cases where ~ seady descen speeds, whch are prey uch all n he 20-30ps -ballpark I ollows ha he shape o he D -vs- curve s as poran as nlaon duraon n deernng ax :
Wha hs says also: 1 ax = 1 + g cos θ d I The greaer he value o he drag negral,.e. he ore boxy he shape o drag-vs-e curve, he saller he value o ax a slar nlaon es, as I 1 copared o he rangular shape where I ~ ½ On he oher hand, he ore spky he curve, he greaer he ax, as I ¼ or 1/5 Is hs any useul o desgners? Perhaps ore on hs laer
Bu here s ore! Usually ax depends no only on he ax -vs- curve shape bu also on nlaon e, nal and nal speed ec. Bu here wll be suaons where curve shape s he os poran acor
Consder cases where nlaon e s large.e. roude er s donang I g I g g I g I = = + = ax ax 1 Apples o any canopy/reeng ; Inlaon along osly vercal rajecores Dependence on llng e and nal speed has dsappeared!
roude-er donaon s parcularly relevan o clusers o large parachues. Consder he case o a j parachue cluser Express n ers o he nlaon properes o one o he cluser ebers The MI-heore yelds: 3 / 2 1 j j ψ j πc 1 1 2 1 D0 gd0 1 ρ + 2 SC D sd n j 1 2 ll n ll I where j cluser j ax 1 j 1 C D 0 sd = ψ j C D 0 sd S j 0 = js 1 0 j j D n ll j cluser = = ll 1 n j = j 1 ll n ll D 1 0 n 1 0 ll
Coparng wo clusers,.e. a j-cluser and a k-cluser, boh carryng he sae ass: Where l s he ass rao o he enre cluser: l = l ψl 3/2 ρ C D S sd 1-clue 3/2 / = cluser k cluser j cluser k cluser j I I I I j j k k k j j k 2 3 / ax ax ~ ψ ψ
Anoher exaple: ocke-propelled payloads durng nlaon Sar wh he MI-heore usng he denons shown < >= = ± D d + d + W cos θ d d New er rocke pulse Pusher rocke ero-rocke e < + g eecve ass >
Solve or ax horzonal rajecory > < + > < + > < + 1 1 ax e e g I g ± >= < d g e > < + roude-lke er bu assoc d wh rocke propulson
Consder he cases when < >>> g hen < >/ e g ~ 1; or when s large, n whch case he roude er donaes agan. We ge: > < I ~ ax > < + > < + > < + 1 1 ax e e g I g The ore spky he D -vs- curve, he hgher he axu drag g e > < +
Wha s hs good or? a look no he near- uure
AII or AI 2 Arcal Inellgence or Inlaon I we can ndeed undersand and descrbe he lud physcs o parachue nlaon on a copuer, hen uure operaonal parachues wll lkely eaure elecronc canopes wh sensors and conrols o easure lgh condons and on-board copuers ha use he nlaon odels o alor he nlaon process o ee perorance requreens whn syse consrans Carl W. Peerson; The lud Physcs o Parachue Inlaon ; Physcs Today, pp. 32 39, Augus 1993. Operaonal SI descrpons o nlaon ay be a long way o Bu nlaon-unng va he changng o he shape o he drag-vs-e curve ay provde he rs praccal nsance o AI 2 n he shor er
Back o he general resul vercal rajecory ax = I 1 + g Goal: conrol he nlaon process so o reduce ax Prncple: Have he copuer conrol he nlaon process so o yeld he larges possble value o he drag negral; n oher words, produce a ore boxy shape o he drag-vs-e curve so o ge I ~ 1 n coparson o he ypcal I ~ ½ a hgh ass raos personnel chues or I ~ 1/4 a low ass rao drogues The prograng o he copuer usng he MI-Theore s easy enough. Bu how could hs be done echancally? Nex slde
Use load cells on he rsers Use all rae sensor Use acve copuer-conrolled ds-reeng Algorh hgh ass rao syses: - Sar soon aer lne srech - Measure orce a e - Increase reeng lne radus so o keep he drag orce nearly consan we wan he boxy D vs curve; Load cell Copuer-conrolled Snce D = ρ SC D 2 /2 we wan he decrease n 2 be copensaed by an ncrease n drag area CPU gure by Sadeck and Lee paper 2007-2513
ealsc? Only he uure wll ell.
Quesons?