Strojniški vestnik - Journal of Mechanical Engineering 53(2007)2, 114-126 UDK - UDC 629.735.4:533.6 Strokovni članek - Speciality paper (1.04) Matematični modeli dinamike helikopterskega letenja Mathematical Models of Helicopter Flight Dynamics Dragan Cvetkovič1 - Duško Radakovič2 ('University "Union", Serbia; Federal Bureau for Measures and Precious Metals, Serbia) Helikopter se razlikuje od drugih prometno transportnih sredstev, ne le po svoji sestavi temveč tudi po možnostih svojega gibanja. Helikopter se lahko premika navpično, lahko lebdi v zraku, lahko se vrti na istem mestu, lahko se premika naprej in v stran, a lahko tudi kombinira vse te premike. Zaradi tega sta modeliranje in testiranje dinamike helikopterja zelo zapleten problem. Trenutno se problemi dinamike helikopterskega letenja v glavnem rešujejo z uporabo sodobnih računalnikov. Čeprav so za mnoge zahtevne probleme računalniki neizogibni, ne omogočajo razumevanja fizikalne plati problema. Na srečo je veliko problemov v zvezi s helikopterji mogoče analizirati brez preveč zapletenih izračunov in navadno je mogoče priti do preproste formule. Čeprav niso ustrezne za preračune, te formule, pri konstrukciji helikopterja, omogočajo zadovoljivo interpretacijo potrebnih aerodinamičnih in dinamičnih pojavov. Helikopter spada v skupino zračnih sistemov in njegovo tradicionalno modeliranje se deli na: a) prostorsko geometrijo in kinematiko in na b) dinamiko togega telesa in dinamiko zraka, skozi katerega se giblje. V zadnjem času, so se razvili naslednji modeli: c) elastični model, odvisen od aerodinamičnih sil, d) model pogonskega sistema, e) model hidravličnih in drugih izvršilnikov za aerodinamično krmiljenje, ß model obnašanja pilota, g) model sistema navigacije in h) model problema vodenja. Matematični model, opisan v tem prispevku, se nanaša na a) in b). © 2007 Strojniški vestnik. Vse pravice pridržane. (Ključne besede: helikopterji, dinamika letenja, matematični modeli) The helicopter is a specific form of traffic-transportation, not just in terms of its structure but also in terms of its possibilities for motion. The helicopter can move vertically, hover in the air, turn around, move forward and move laterally, and it can also perform combinations of these movements. As a result, modeling and testing helicopter dynamics is a very complex problem. The problems in helicopter flight dynamics are mostly solved with the aid of modern computers. Though inevitably, with many complex problems, computers do not make it possible to understand the physical nature of the problem. Fortunately, many problems related to helicopters can be analyzed without overly complex calculus, and usually it is possible to obtain simple formulae. Though not suitable for calculus, these formulae, when designing the helicopter, enable a satisfactory interpretation of the required aerodynamic and dynamic phenomena. The helicopter belongs to the group of aerospace systems and its traditional modeling may be divided into: a) three-dimensional (space) geometry and kinematics, and b) rigid-body dynamics and the fluid dynamics through which it moves. Recently, the following models were developed: c) the elasticity model in intersubordinance with a fluid, d) the propulsion system model, e) the hydraulic model and other actuators that achieve aerodynamic control, ß the pilot-behavior model, g) the navigation-system model, and h) the beacon-problem model. The mathematical model described in this paper is related to a) and b). © 2007 Journal of Mechanical Engineering. All rights reserved. (Keywords: helicopters, flight dynamics, mathematical models) 1 GIBANJE ROTORSKIH KRAKOV 1 MOTION OF ROTOR BLADES Da bi razumeli dinamiko letenja helikopterja in določili dinamične momente in sile, ki delujejo na helikopter, je nujno potrebno najprej raziskati To understand the flight dynamics of helicopters and determine the dynamic moments and forces that act upon the helicopter, it is a necessity to pre-investigate 114 Strojniški vestnik - Journal of Mechanical Engineering 53(2007)2, 114-126 gibanje nosilnih krakov rotorja. Iz velikega števila različnih helikopterjev smo izbrali helikopter z enim rotorjem, katerega kraki so povezani na glavo rotorja s členkom, okoli katerega se prosto gibljejo. Treba je povedati, da so kraki trdno povezani z glavo rotorja. 1.1 Enačbe mahanja kraka Krake rotorja vzamemo kot togo telo. Vodoravni členek je na razdalji eR od osi vrtenja. Kotna hitrost gredi je Q=konst, a krak maha s kotno hitrostjo dßdt. Vzdolžna os kraka je vzporedna vztrajnostni osi kraka in gre skozi členek(sl. 1). Na sliki 1 je R dolžina kraka, kot #je položaj kraka. Po zapletenih preračunih, dobimo enačbe gibanja krakov: 1.2 Enačbe zaostajanja kraka Predpostavimo, da je />0 in da se krak giblje naprej glede na navpični členek za kot zaostajanja L Navpični členek je na razdalji eR od osi vretena. Koordinatni sistem postavimo enako kot v prejšnjem primeru. Slika 2 prikazuje poenostavljeno skico za določanje zaostajanja kraka. Iz tega izhaja enačba zaostajanja kraka: the motion of the supporting rotor blades. From a vast number of different types of helicopters, we chose the single-rotor helicopter that has its blades coupled with the main rotor by a hinge about which they can move freely. It should be noted that there are also rotors that have the blade connected in a fixed manner to the hub. 1.1 Equations of blade flapping Rotor blades are regarded as a rigid body. The horizontal hinge is placed at a length eR from the rotation axis. The shaft rotates at an angular speed W=const, and the blade flappers at an angular speed of db/dt. The axis that passes through the blade is parallel to the axis of inertia of the blade and passes through the hinge (Fig. 1). In Figure 1, R represents the length of the blade, b represents the flapping angle of the blade. Following some complex calculus, the equations for blade flapping are obtained: (1) (2) (3). 1.2 Equations of blade throwback It is assumed that b=0 and that the blade is moving forward in relation to the vertical hinge by the throwback angle amount x. The vertical hinge is placed at a distance eR from the shaft axis. The coordinate system is positioned as in the previous case. Figure 2 presents a simplified scheme for determining the blade throwback. From this the equation for blade throwback follows: Jxy0Q cosß + Jx(-ß)Q. sinß = 0 -2 JyQ.ß sinß = Mz Sl. 1. Prikaz mahanja kraka Fig. 1. Explanatory drawing for blade flapping Matematični modeli dinamike helikopterskega letenja - Mathematical Models of Helicopter Flight Dynamics 115 Strojniški vestnik - Journal of Mechanical Engineering 53(2007)2, 114-126 Sl. 2. Prikaz zaostajanja kraka Fig. 2. Explanatory scheme for blade throwback x +W2e x -2 W b b =Mz /Jz && & (4) Če je kot med položajem kraka in smerjo letenja y=Q.t, sledi: d2Ç If the azimuth angle is described as y=Wt, then it follows that: dy2 + s C-2ß db Mz 1.3 Enačba vzpenjanja kraka Vzemimo, da sta kota mahanja in zaostajanja enaka nič. Korak kraka je kot med tetivo profila kraka in ravnino glave rotorja, označen kot 9k. Na sliki 3 vidimo koordinatni sistem, povezan s krakom. Enačbe gibanja kraka okoli vzdolžne osi so: dy JzW2 1.3 Equation of blade climb (5) It is assumed that the flapping and the throwback angles are equal to zero. The blade step is the angle between the blade cross-section chord and the plane of the hub, designated as qk. Figure 3 shows the coordinate system attached to the blade. The equations of blade motion about the longitudinal axis are: qk +W2 qk = Mx /Jx && -2 Jz W q&k sinqk = Mz J W q& cosq - J q& cosq =0 (6) (7) (8) Sl. 3. Koordinatni sistem na profilu kraka Fig. 3. Coordinate system at the blade cross-section 2 ROTORSKE SILE 2 ROTOR FORCES Za projekcijo sil lahko uporabimo naslednje osi: os v smeri vlečne sile rotorja, os rotorskega diska, ki je pravokotna na ravnino rotorja, to je na ravnino, kjer ležijo konci krakov in os gredi. Ko izberemo eno izmed teh osi, bosta preostali dve osi v koordinatnem sistemu pravokotni nanjo, usmerjeni bočno, oziroma proti repu helikopterja. Običajno se komponenta sile v smeri izbrane osi To project the forces the following axes may be used: the control axis, the rotor disc axis (which is normal to the rotor plane, i.e., to the plane on which the blade tips reside), and the shaft axis. Once the axis is chosen, the remaining axes of the coordinate system will be normal to it and pointed laterally, i.e., to the tail of the helicopter. The force component along the chosen axis is normally referred 116 Cvetkovic D. - Radakovic D. Strojniški vestnik - Journal of Mechanical Engineering 53(2007)2, 114-126 imenuje vlečna sila, komponenta sile proti repu se imenuje süa H, a komponenta sile, usmerjena bočno, se imenuje sila Y. Če komponente sile označimo brez indeksiranja, menimo da se nanašajo na os v smeri včene sile. Indekse “D” in “5" uporabljamo, če se komponente nanašajo na os rotorja oziroma na os pogonske gredi. Ker sta kota mahanja in zaostajanja ponavadi majhna (do 10°), lahko napišemo: to as the tow force, the force component pointed towards the tail is called the H force, and the force component pointed laterally is said to be the Y force. If the force components are designated without subscripts, it is assumed that they are determined relative to the control axis, whereas the subscripts "L>" and “S” are used when they relate to the rotor axis, i.e., the shaft axis. Since the flutter and mount angles are usually small (to 10°), a relation between these components can be obtained: 2.1 Vzdolžno ravnotežje sil 2.1 Longitudinal equilibrium of forces Kot 5, je vzdolžna amplituda ciklične Angle Bx is the longitudinal amplitude of a cyclic spremembe koraka kraka; kot au je kot med gredjo in change in the blade step; angle au is the angle be- osjo rotorskega diska. Po obsežnih izračunih dobimo tween the shaft and the axis of the rotor disc. After izraz za vzdolžno amplirudo ciklične spremembe extensive calculus the expression for the longitudinal koraka kraka: amplitude of cyclic change in the blade step is obtained: M -G-fR + H-hR + Ms-a1 B1 T-hR + M (9) Za e=0, lahko vzamemo da je Ms=0 in Mf=0. Ker je T=G, sledi: For e=0, we can say that Ms=0 and Mf=0, and since T=G, it follows that: B=_l + K 1 hG D f M — cost- —H------ G h GhR (10) (11) Za enačbo (10) imamo preprosto fizikalno Equation (10) has a simple physical pojasnilo: amplituda vzdolžnega cikličnega krmiljenja interpretation: the amplitude of the longitudinal krakov mora imeti tako vrednost, da postavi smer cyclic control must have such a value in order to rezultirajoče sile rotorja skozi masno središče. position the direction of the resultant rotor force through the center of mass. Vodoravna os Horizontal axis SI. 4. Skica vzdolžnega ravnotežja sil Fig. 4. Drawing for determining the longitudinal equilibrium of forces Matematični modeli dinamike helikopterskega letenja - Mathematical Models of Helicopter Flight Dynamics 117 Strojniški vestnik - Journal of Mechanical Engineering 53(2007)2, 114-126 2.2 Prečno ravnotežje sil 2.2 Lateral equilibrium of forces Kot A, pomeni amplitudo prečne ciklične spremembe koraka kraka rotorja: Angle A1 represents the amplitude of lateral cyclic change in the blade step of the supporting helicopter rotor: G-W + Ms-b1+Trhfl G-hR + M (12) Če vrednost A, uvrstimo v ustrezne enačbe, dobimo vrednost köta f ki določa lego trupa. By replacing the value A, in the corresponding equations, we obtain the value Of angle f which determines the position of the fuselage. Tt | G-fR + Ms-b1+Tt-htR G G-hR + M (13) SI. 5. Skica prečnega ravnotežja sil Fig. 5. Drawing for determining the lateral equilibrium of forces Če je M=0 in h=h, kar pogosto lahko vzamemo, sledi: If M=0 and h=h, which can often be assumed, it follows that: f1 , h kar pomeni, da je glava rotorja navpično nad masnim which means the rotor hub is positioned vertically središčem. Vse vrednosti izračunanih kotov so tako above the center of mass. All the values of these imenovane uravnovešene vrednosti. determined angles are the so-called trimmed values. 3 NELINEARNI MATEMATIČNI MODEL DINAMIKE LETENJA 3 NON-LINEAR MATHEMATICAL MODEL OF THE FLIGHT DYNAMICS Matematično modeliranje helikopterskega gibanja je izredno zahtevna naloga in zato je nujno privzeti številne predpostavke in približke. Za analizo dinamičnih značilnosti helikopterja ni nujno treba, razen v izjemnih primerih, poznati gibanja Mathematical modeling of a helicopter’s motion is a very complex task and, therefore, it is necessary to introduce a series of assumptions and approximations. Knowledge of the motion of the individual helicopter blades is not necessary for investigating the dynamic 118 Cvetkovič D. - Radakovič D. Strojniški vestnik - Journal of Mechanical Engineering 53(2007)2, 114-126 posameznih krakov. Za definiranje sil in momentov pri motenem letu je dovolj opazovati rotor kot celoto. Zaradi velikega števila različnih helikopterjev, v tem prispevku analiziramo helikopter z enim samim rotorjem, katerega kraki so s členki pritrjeni na glavo rotorja. Kakor je že bilo rečeno, je helikopter zmožen različnih gibanj in bi zato bilo zelo težko narediti matematični model za kombinacijo vseh gibanj. Vzeli bomo, da je helikopter vzletel in da leti premočrtno. Komponente hitrosti helikopterja pri nominalni vrednosti in premočrtnem letenju so: W, W in W. Koti spremembe smeri % nagiba ^in vzpenjanja éz veljajo dokler so velikosti motenj v dovoljenih mejah. Na sliki 6 je predstavljena shema helikopterja s premičnim koordinatnim sistemom, vezanim na njegovo masno središče, a na sliki 7 je blokovni diagram helikopterja. Po uvedbi nekoliko predpostavk, na primer da: • je masa helikopterja konstantna, • je helikopter togo telo, • ravnina xz simetrijska ravnina, • so kotni prirastki A ^ A6>, A0 majhni in tako dalje; pridemo do nelinearnega matematičnega modela z odkloni v obliki: characteristics of the helicopter, except in a special case, but rather for defining the forces and moments in a disturbed flight it is sufficient to view the rotor as a whole. Because of the large number of different helicopters, in this paper a single-rotor helicopter that has its blades connected to the hub by hinges was studied. As mentioned before, the helicopter can perform different movements and it would be very difficult to make a mathematical model that would combine all those movements. It is assumed that the helicopter is airborne and in straightforward flight. It is necessary that the helicopter, during its straightforward flight, has the following velocity components, Wx, Wy, and Wz, at nominal values, and the angle of turn y, the angle of roll f, and the angle of climb q, as long as the intensity of disturbance is within permitted limits. Figure 6 presents a schematic of the helicopter with a floating coordinate system tied to its center of mass, and Figure 7 presents a helicopter block diagram. After introducing a series of assumptions, such as: • the helicopter mass is a constant value, • the helicopter is a rigid body, • the 0xz is a plane of symmetry, • the angle increments DY, Dq, Df are too small, and so on, we come to a non-linear mathematical model with deviations in the form: Sl. 6. Shema helikopterja Fig. 6. Schematic of helicopter Sl. 7. Blokovni diagram helikopterja Fig. 7. Helicopter block diagram Matematični modeli dinamike helikopterskega letenja - Mathematical Models of Helicopter Flight Dynamics 119 Strojniški vestnik - Journal of Mechanical Engineering 53(2007)2, 114-126 d(AWr) ir, . x . ^—i = -[f1(AWx,AWz,Aß,u1,u2)- mgcosr )A0 dt mL v 7 d(AW) lr / x dt mL v 7 i----Z = A# dt d(A0) i ... jv x z z l l' dt J d(AWy) \v f , ^ = —\fAAWA0,Aiy,ui,u4) + WzNmAiy + mgcosTA0 + mgsmTAii dt mL y d(A^) d(D f&) dt D f& dt J -[/5 ( Wy,A,Aifs,u3,u4 ) +JxzAif/ d(Ay d(D y&)= 1 dt J dt D y& -[f6 (kWy,AeAY,u„u4 ) +JxzA(j> (14) (15) (16) (17) (18) (19) (20) (21) (22) Kjer so: • u=AB1 - amplituda ponovitvene spremembe koraka v vzdolžni smeri (za vzdolžno gibanje), • u=A0Q- sprememba skupnega koraka kraka rotorja helikopterja (za vzdolžno gibanje), Where: u1=DB1 – the amplitude of the cyclic change in step in the longitudinal direction (in terms of longitudinal motion), u 2=Dq 0 – the change of the collective step of the helicopter rotor blade (in terms of longitudinal motion), SI. 8. Shema prečnega gibanja Fig. 8. Schematic for lateral motion 120 Cvetkovió D. - Radakovió D. Strojniški vestnik - Journal of Mechanical Engineering 53(2007)2, 114-126 • u=AAl - amplituda ponovitvene spremembe koraka v prečni smeri (za prečno gibanje) in • u=A0t - sprememba skupnega koraka repnega rotorja (za prečno gibanje). Shematska diagrama predstavljamo na slikah 8 in 9. SI. 9. Shema vzdolžnega gibanja Fig. 9. Schematic for longitudinal motion • u=AAx - the amplitude of the cyclic change in step in the lateral direction (in terms of lateral motion), • u=A0t - the change of the collective step of the tail rotor (in terms of lateral motion). Schematic diagrams are presented in Figures 8 and 9. 4 LINEARIZIRAMMATEMATIČNI MODEL DINAMIKE LETENJA 4 LINEARIZED MATHEMATICAL MODEL OF THE FLIGHT DYNAMICS Dokazano je, da v tehniki lahko uporabimo s sprejemljivo natančnostjo linearizirane matematične modele pod pogojem, da imajo fizikalne veličine majhna odstopanja od nominalnih vrednosti. Nelinearni matematični model dinamike letenja helikopterja ni primeren za določanje splošnih rešitev v analitični obliki, čeprav se problem rešuje s sodobno računalniško tehnologijo. Zaradi sprejetih predpostavk bodo izstopne vrednosti, vstopne vrednosti in vektor stanja tako za vzdolžno kakor za prečno gibanje: In technical applications it has been shown that, with an acceptable accuracy, linearized mathematical models may be used under the condition that the deviations of the physical quantities from their nominal values are small. A nonlinear mathematical model of the helicopter’s flight dynamics is inadequate for finding general solutions in an analytical form, even though the problem is solved with the aid of modern computer technology. The outcome of the adopted presumptions is that the output values, input values, and the vector of state for both the longitudinal and lateral motion will be: X = (X1...X9 x=(x...x u = [u ...u (23) (24) (25) Vektorska enačba stanja za linearizirani matematični model z brezrazsežnimi veličinami, odstopanji, to je veličinami stanja, je enačba (26). Enačba izstopnih veličin je (27). The vector equation of state for the linearized mathematical model with non-dimensional quantities, deviations, i.e., the quantities of state, is shown in Equation (26). Equation (27) presents the output values. Matematični modeli dinamike helikopterskega letenja - Mathematical Models of Helicopter Flight Dynamics 121 Strojniški vestnik - Journal of Mechanical Engineering 53(2007)2, 114-126 X a 11 a 12 a 13 a 14 0 0 0 0 a 21 a 22 a 23 a 24 0 0 0 0 0 0 0 1 0 0 0 0 a 41 a 42 a 43 a 44 0 0 0 0 0 0 0 0 a 55 a 56 0 a 58 0 0 0 0 0 0 1 0 0 0 0 0 a 75 0 a 77 0 0 0 0 0 0 0 0 0 0 0 0 0 a 0 a 0 x + 0 0 X 100000000 010000000 001000000 000010000 00000 1 000 [000000100 X 0 0 00 (26) (27) Čeprav so tukaj prikazana v skupni matrici, je treba poudariti, da so vzdolžna in prečna gibanja ločena, ker je to bil eden izmed pogojev za izpeljavo tega matematičnega modela. V enačbah (26) in (27), so enačbe za vzdolžno gibanje predstavljene v prvih štirih vrsticah matrike, medtem ko preostalih pet vrstic pomeni enačbo stanja in enačbo prečnega gibanja. V enačbi (26) uporabljamo naslednje označbe: In addition to the way this is presented, in the form of a common matrix, it should also be noted that the longitudinal and lateral motions are separated, because this was the condition for deriving this mathematical model. In Equations 26 and 27, the equations for the longitudinal motion are presented within the first four rows of the matrice, while the remaining five rows present the equation of state and the equation of lateral motion. The designations used in Equation 26 are: C* 1-i/ij, a a 12 a 13 a 59 =W* a23 = -mc sin t a24 = WxN+zq a5 a41 =mu + mivzu a42 =mw+ m^zw a4 01 Z ö9 X ö9 Z -mc cos Z" O14 = x a21 = zu = y v a56 =mccosT a58 = mc sin = -milmc sin z a44 = mq + m. I Wm + zq b =mz +m b =mz +m 11 B 1 1 0 0 11 0 0 K=yA K=ys, K={lA+nAh/h) C* K={le,+ne,h/h)c* K={ne,+le,h/h)c* a75={nX/h H) C* a77={lp+npi*z/h) C* a95={nv+lX/h) C* a97={np+lpL/h) C* a99={nr+lrixz/iz) C* b93 ={nA + lAixz/iz) C* a79={lr+nrixz/ix) C* Na slikah 10 in 11 so predstavljene sheme lineariziranega matematičnega modela v vzdolžnem in prečnem gibanju. 5 REZULTATI PROGRAMA Figures 10 and 11 present schematics of the linearized mathematical model in a longitudinal and lateral motion. 5 PROGRAM RESULTS Program je testiran na primeru helikopterja z enim samim rotorjem, ki ima krake členkasto vpete na glavo rotorja. Helikopter je opisan z naslednjimi vstopnimi podatki: teža helikopterja G=45042N, količnik pokritja rotorja 5=0,058, polmer rotorja fl=8, lm, koeficient višine glave rotorja A=0,25, količnik upora (5=0,013, število krakov glavnega rotorja b=4, masa kraka m=79,6kg, količnik napredovanja rotorja The program was tested on the example of a single-rotor helicopter for which the main rotor blades are tied to the hub over hinges. The helicopter is described by the following input data: helicopter weight, G=45042N; rotor abundance degree, 5=0.058; rotor radius, tf=8. lm; hub height coefficient, Ä=0.25; drag coefficient, (5=0.013; number of blades of the main rotor, b=4; blade mass, m=79.6kg; rotor 122 Cvetkovic D. - Radakovié D. Strojniški vestnik - Journal of Mechanical Engineering 53(2007)2, 114-126 SI. 10. Shema lineariziranega matematičnega modela pri vzdolžnem gibanju Fig. 10. Schematic of a linearized mathematical model when in longitudinal motion SI. 11. Shema lineariziranega matematičnega modela pri prečnem gibanju Fig. 11. Schematic of linearized mathematical model when in lateral motion /y=0,3, vzgonski gradient profila a=5,65 l/rad, hitrost konca kraka Qtf=208m/s, masno središče kraka x je na 45% radija kraka R, oddaljenost členka kraka'od gredi pa je 0,04R, gostota zraka na višini letenja (100m) /7=1,215 kg/m3. Za vzdolžno gibanje je matematični model v vektorski obliki: X X = A -0,0509 0,1323 -0,0734 0,1216 -1,2525 0 0 0 0 6,512 12,1 0 operating mode coefficient, m=0.3; gradient of lift, a=5.65; velocity of blade top, WR=208m/s; distance of blade mass center coefficient, xg=0.45; distance of hinge from shaft, eR=0.04R; and air density at flight altitude (100m), r=1.215 kg/m3. For longitudinal motion the mathematical model in vector form is: X +b u 0,00263 0,3 1 -0,844 X X2 X, x x+ 0,1344 0,066 0,3578 -0,9477 0 0 -28,329 17,88 Matematični modeli dinamike helikopterskega letenja - Mathematical Models of Helicopter Flight Dynamics 123 Strojniški vestnik - Journal of Mechanical Engineering 53(2007)2, 114-126 Enačba izstopnih veličin je: X "1 0 0 0" 0 1 0 0 0 0 1 0 0 0 0 1 The equation at the exit is: X kjer je/where is Xi Matrika A linearizkanega modela helikopterja za prečno gibanje je: X i2 X i3 Matrix A of the linearized model of the helicopter for lateral motion is: -0,15125 0,0734 0 0 0,3 0 0 10 0 -64,42 0 -2,948 0 1,378 0 0 0 0 1 55,297 0 0,413 0 -1,64 X = A X + B u X = yX5 X6 X7 X8 X9 J u = \u u 1 6 SKLEPI 6 CONCLUSIONS Da bi bil matematični model dinamike helikopterskega letenja, strogo določen, bi moral biti sestavljen iz sistema nelinearnih, neustaljenih, parcialnih diferencialnih enačb. Da bi te enačbe poenostavili, smo vzeli nekaj predpostavk. Zanemarili smo elastične lastnosti helikopterja, da bi helikopter analizirali kot togo telo in tako eliminirali razpršitev parameterov. Poleg tega smo zanemarili porabo goriva in s tem neustaljenost, ki bi se pojavila zaradi časovne spremembe mase helikopterja. Glede na to, da ima helikopter šest prostostnih stopenj, smo zaradi poenostavitve vzeli, da se gibanje da ločiti na vzdolžno in prečno gibanje in da se ta gibanja analizirajo posamično. Povdarimo, da se matematični model helikopterja nanaša na helikoptersko premočrtno gibanje s hitrostjo W. Matematični model, ki bi obsegal vsa gibanja helikopterja, vključno z vzletom in pristankom, bi bil veliko bolj zapleten. Vpliv resonance in vibracije se prav tako zanemari. Zaostajanje kraka se tukaj tudi ne upošteva, ker drugače kotna hitrost kraka v ravnini rotacije ne bi bila več nespremenljiva. Ločene analize posamičnih gibanj krakov so velika poenostavitev, zato ker obstaja velika medsebojna odvisnost med gibanji kraka. Če bi ta gibanja ne bili ločili, bi bilo nujno analizirati stabilnost vseh gibanj kraka. Postavljanje koordinatnega začetka v vztrajnostno središče omogoča odstranitev nekaterih vztrajnostnih momentov, tako da se The flight dynamics mathematical model of a helicopter that would be strictly determined would comprise a system of non-linear, non-stationary, partial differential equations. To simplify these equations we introduce a number of assumptions. Ignored are the elastic characteristics of the helicopter so the helicopter can be thought as a rigid body and, in this way, the dispersal of parameters is eliminated. Also, fuel consumption is disregarded and so is the non-stationarity due to the temporal change in helicopter mass being eliminated. Because the helicopter has six degrees of freedom, for simplification it is assumed that the motion can be separated into longitudinal and lateral motions and that they can be investigated independently. It should be noted that the mathematical model of the helicopter relates to the helicopter’s forward motion at velocity W. A mathematical model that would incorporate all the motions of a helicopter, all together with takeoff and landing, would be far too complicated. The influence of resonance and vibration is also ignored. The blade throughback is also ignored in this paper, because if this was not the case the blade-angle velocity in the plane of rotation would no longer be constant. A separate study of the individual motions of blades is a great simplification, because there is an interdependency of all the blade motions. If the motions are not separated, then it is necessary to analyze the stability of all the motions of the blade. The choice of the coordinate origin in the center of inertia makes it possible to eliminate certain moments of inertia so the Euler equations can be 124 Cvetkovič D. - Radakovič D. Strojniški vestnik - Journal of Mechanical Engineering 53(2007)2, 114-126 Eulerjeve enačbe lahko poenostavijo. Opazovanje simplified. Viewing the rotor as a whole eliminates rotorja kot celote odstrani potrebo po preučevanju the need for investigating the motion of an individual posameznih gibanj krakov. V tem je v veliko pomoč blade. This is made much simpler by the introduction uvajanje osi rotorskega diska in osi v smeri vlečne of the rotor disc axis and the control axis. sile rotorja. The determination of the aerodynamic Določanje aerodinamičnih odvodov je derivatives is related to a series of approximations. povezano z vrsto približkov. Treba je poudariti, da It should be noted that, besides assumptions in the poleg predpostavk pri modeliranju, uporabljamo modeling, mathematical simplifications were also tudi matematične poenostavitve (na primer, made (for example, omitting small values in the izpuščanje zanemarljivo majhnih vrednosti iz equations) which could not have been derived in enačb), ki jih ni mogoče prikazati v obliki the form of an assumption due to their meaning, predpostavke, ker je njihov pomen tesno povezan which is tightly related to a specific equation. z določeno enačbo. It is possible to determine projections of the Kot izstopne značilnosti je mogoče določiti position vector with respect to the non-moveable projekcije vektorja lege v nepremičnem koordinatnem coordinate system tied to Earth instead of using sistemu, povezanem s tlemi namesto projekcij hitrosti projections of the helicopter’s velocity with respect helikopterja v premičnem koordinatnem sistemu. Ta to a moveable coordinate system such as the exit problem bi se rešil s projiciranjem hitrosti helikopterja characteristics. Projecting the helicopter velocity na nepremični koordinatni sistem, nakar bi se onto a non-moveable coordinate system and then integrirale projekcije hitrosti po času z začetnimi integrating the velocity projections over time with pogoji. the initial conditions may solve this problem. Nadaljnja analiza matematičnega modela se A further analysis of the mathematical model lahko usmeri na raziskovanje dinamičnih in can be made in order to investigate the dynamic and statičnih lastnosti in na določanje primernega static properties, and to determine the control that krmarjenja, ki bi helikopterju zagotovilo zahtevano would guarantee the object to execute the required dinamično obnašanje. dynamic behavior. 8 LITERATURA 8 REFERENCES [I] W. Z. Stepniewski (1984) Rotary-wing aerodynamics, New York. [2] M. Nenadovič (1987) OAK - Elise i propeleri, Beograd. [3] M. Nenadovič (1978) OAK - Aeroprofili -1, deo, Beograd. [4] M. Nenadovič (1977) OAK - Opšti deo, Beograd. [5] M. Nenadovič (1982) Osnovi projektovanja i konstruisanja helikoptera, Beograd. [6] A. R. S. Bromwell (1976) Helicopter dynamics, London. [7] W. Johnson (1980) Helicopter theory, London. [8] G. Saundres (1972) Teorija leta helikoptera [prevod sa engleskog], Beograd. [9] Gesov & Myers (1952) Aerodynamics of the helicopter, New York. [10] A. K. Martinova (1973) Teorija nesuščego vinta, Moskva. [II] M. L. Mil’ (1973) Vertoljoti, Moskva. Matematični modeli dinamike helikopterskega letenja - Mathematical Models of Helicopter Flight Dynamics 125 Strojniški vestnik - Journal of Mechanical Engineering 53(2007)2, 114-126 Naslova avtorjev: Dragan Cvetkovič Univerza "Union" Fakulteta računalniških znanosti Knez Mihailova 6/VI 11000 Beograd, Srbija Authors’ addresses: Dragan Cvetkovič University "Union" Faculty of Computer Science Knez Mihailova St. 6/VI 11000 Belgrade, Serbia Duško Radakovič Zvezni zavod za mere in dragocene metale Mike Alasa 14 11000 Beograd, Srbija Duško Radakovič Federal Bureau for Measures and Precious Metals Mike Alasa St. 14 11000 Belgrade, Serbia Prejeto: Received: 27.2.2006 Sprejeto: Accepted: 22.6.2006 Odprto za diskusijo: 1 leto Open for discussion: 1 year 126 Cvetkovič D. - Radakovič D.