شبیه سازی غیر خطی زیردریایی

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۱۱ ‏ا ا ات‎ submarine in the dive plane (ar Z,)o- (Z,+ mx)q=Z,U, +(Z,+ mU, + meq" + )۲۷- 3009+ 2۵ 

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forward speed , 1 pitch piteh a: dive plane angle , depth , weight , mass moment of inertia coordinates of center of gravity , coordinates of center of buoyanc, heave force hydrodynamic coefficient , pitch moment hydro 

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W say we want to linearize these equations for a level flight path when the ۱ ae ae MTT eRe ehh ee ‏ا‎ 1 ZuUwo + (W — B)cos% 0, MU wo — (aeW - @p B) cos 0 — (zcW — 22 B) sin > 0 GW 0, U sin @9 + wo Cos A 0 If we assume that the boat is neutrally buoyant 2 = ay and W = B, we have Zyliwo = 0, MyUwo — (2a — 2n)Bsin@o = 0 U sin 69 + wocos@ = 0, from which we can get the nominal position w=d=0, and sin =0, which means %=0, or ۰ 

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پا Em ¢ = Qpa=0, wq = (wo)gt(qo)w =0, sind = (cost) =0, woos) = (—wy sindy)8 + (cosby)w = w ‘The linear equations of motion are then written as (m= Za )ii mag) = Zellw + (Zy + mq + ZU + mag) = Mylw+ (Mg — mag)Uq— (20 — 28) WO-+ ‏تاو‎ 

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10 CCR aE eR RS eee ee cee U is the forward speed (not control). The control is designated by 6; this is .1011 لال لهاك تاج ,= 6- ‎State equations from block diagram.‏ ‎In state space form these are written as‏ 0 9 0 1 0 0 0 ‎ay aa 0 w bu? 9‏ ف ‎so‏ این 7 0 ناويه ناروه 4 0 | 3 0 0 1 2 

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where the coefficients a;;, ; are given by Dy = (m— Zj)(Ly— Mg) — (mag + Z4)(mxe + Mw) , aD, = (ly— Mj)Zo + (mag + ‏ما(‎ ‎aD, = (ly-Mj) ,) + (mag + Z;)(My — mae) , a3D, = —(mag+Z,)W, byDy = (ly— Mg)Z5 + (mag + 25) Ms , Zs) Mw + (mag + M; anD, = (m—Za) dD, = (m—Zq)(M, — mag) + (mag + Mg)(m+Z,) , 43D, = —-(m-Z,)W, beDy = (m= Zu)Ms + (mae + (2۵ + 

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re %- ~ ‏ال‎ 

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0 60 0 3 = و 7 

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SI Ree a SL We ce TSE SOS et) ee nei ابتدا ماتریس 5 را بنحوي محاسبه مي‌کنيم که ماتریس ۸۸ را بتوان بفرم جردن تبدیل نمود: اوه ۳۷ 2۳5 

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۱۶( ۸ Bolandi 

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مق درل ‎Bolandi‏ مستان 1382 = 

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مستان 1382 لد 1:7 

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تال :ماتریس را قطري ‎١‏ ۳ ‎ai > 7‏ ‎Ps ae |‏ 

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{| ۱ LL 

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3 ِ 2 | 0 1 1 ۹4 

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‎N eae‏ ال ‎Se eels CSO Noyce‏ ها معادله مشخصه خود را برآورده مي‌سازد لذا چند جمله‌اي متناظربا معادلات ‎eS EOSy orn ( ITC)‏ 0 ‎ 

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‎٠-2‏ ,: ماتريس انتقال حللت ماتريس را محاسبه ‎۳ ‎Ta ‎0-۰۰۰ ‏ل‎ ME) = 0 * JA) =97.)=0" =a, +00 ‎=u,‏ 2 دن مس — 

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Bolandi 

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۱2/۸ Bolandi FEEDS 

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Example: Consider the submarine linear equations of motion 6=4 3 ‏وتلوره + سثارره + 0ووقوره‎ + 6 ‏وناييه + سدتاريه + ه0ومتويه - و‎ + bol? where we assume a dive plane deflection 6 = —0.2 radians (—11.5 degrees). A simulation algorithin using Euler's integration is as follows © Step 1: Choose integration time step Ai and initial conditions 6, wo, ga. Set 7 = 0. © Step 2: Using the values of 6, w,, q., compute ‏بن‎ , , from the equations of motion. # Step 3: Compute ‎O4+0;-At,‏ = ببرة ‎Wis wy + thy At‏ اك .ون + :و ‎et‏ ‎© Step 4: Set #= + Land go back to Step 2 

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‘Typical results ofthe simulation in terms ofthe pitch angle @ are shown in Figure 8. As with any numerical results, however, the real question is: are they correct? The answer to this borders between art and science, and eee er te PEEPS 1. In this particular simulation we used a time step At = 0.01 seconds. Is this small enough? The easiest way to check this is to reduce (oF increase) Al, say by a factor of 10, and re-run the program. Ifthe results do not change, the above choice for At was good. A more rational way to do the same thing would be to look at the natural time constant of the dynamics of the system. The system poles were found in page 18. It seems that the fastest pole ofthe system has real part 0.5159, and the time constant that corresponds to this is about 110.5 or 2 seconds. This means that it takes a couple of seconds for the boat to “listen” to its dive planes, so At = 0.01 should give very accurate results In fact in this case we could go as far as At = 0.5 and we would still be ees 2. Look again at the system eigenvalues: one of them is certainly dominant, -0.0297, so the response should approximate that of a first order system with a time constant 1/0.0297, or about 33.5 seconds. Now look at the response of the figure: does it take approximately 33.5 seconds to go up to 60% of its final value? 3. ee cee ‏ا‎ ee ‏ا‎ intuition, How about the final or steady state value of the response? This is something we can compute exactly. Reece tens 

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RUC RCO ee Trad etna q=0, ‎aq + dV?‏ + سدتآرره + #وموقوره ‎aU w + agg + b2U76 .‏ + اوی‌دوده ‎Using q = 0, the second and third equations give ‏نارم ‎20 ‏مثآيره + 0ووقؤره ‎a232gp8 + anUw ‎ ‎Substituting 6 0.2 and using the values from page 17 we find ‎ ‎6 = 0.476 radians or 27.3 degree: ‎a result which agrees with the figure. 

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۱ ee ane eee a Ree ۱ ee cc ee aecT ‘moments; in other words the nonlinear equations of motion are 04 ‎U6 ,‏ + وتاوره + سدتآرره + #صذة ومقوره ‎a23zGp sin 8 + anUw + ar2Vq + bU‏ ‎ ‎‘The numerical integration proceeds in exactly the same way as before; the only difference is that here the ‎and Fi ‎‘are computed from the new equations. Typical results are shown in the previous figure where the difference ‏0 مناد ‎ ‎between linear and nonlinear simulations is also shown. Naturally, whenever possible, simulations must be performed for the nonlinear systems since these model the underying physics more accurately. The steady stale value for 8 can be computed from the nonlinear equations inthe same way as before, the algebra is easy in the example case but keep in mind that for general nonlinear equations it may be very dificult, Here we ean find ‎ 

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example time (seconds) (ssox8op) mem o1daw youd 

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pes Pes)