坦克动力学大作业.pdf
一、坦克学0312070220071201王超坦克动力学作业20071201王超031207021坦克学0312070220071201王超已知条件：1.发动机外特性表 1 发动机的外特性数据nf(r/min)10001200140016001800200022002300Pf(kW)474589697792867918935935Tf(Nm)45314690475447254601438440593882最大功率点时发动机风扇损失功率Ps=116kW。空气滤清器、排气装置的功率损失在合理的范围内自己选取。2.传动简图：齿轮啮合次数 4-6 次。（自己选取）。3.各挡传动比：前传动比：iq=0.68；变 速 箱 传 动 比：ib1=7.353ib2=3.7 83ib3=2.713ib4=1.9 45ib5=1.3 95ib6=1；侧传动传动比：ic=5.3 87。4.车重：战斗全重时质量 M=50 吨。5.履带中心距：B=2.79m主动轮半径：rz=0.318 m。6.主离合器的储备系数为=2.0。7.坦克高（地面至炮塔顶）：2.19m 空气阻力系数：CD=0.5。8.各挡离合器结合时质量增加系数ib1ib2ib3ib4ib5ib63.9031.8231.5391.3941.3041.2579.二挡起步，起步挡加速第一阶段末的发动机转速为其最大扭矩点的转速，并假设起步挡离合器分离时的质量增加系数为 1.2。不考虑其他挡位的加速第一阶段。二、作业1、根据已知条件绘制发动机的外特性曲线。2、根据已知条件做出该坦克的动力特性曲线3、利用 1/a-v曲线计算该坦克的加速特性曲线，并根据在良好路面上 0 32k m 的加速时间对其加速性做出评价。三、要求1.每人独立完成作业；2.作业书写工整，规范，用纸标准；3.使用计算机，交作业时要附程序；4.作业记入平时成绩。2坦克学0312070220071201王超1、根据已知条件绘制发动机的外特性曲线。nf(r/min)Pf(kW)Tf(Nm)10004744531发动机的外特性数据12001400160018002000589697792867918469047544725460143842200935405923009353882发动机的外特性曲线10004531800697469047544725792589474460186791843849354059935500038824000300020001000010001200140016001800200022002300nf(r/min)功率Pf(kW)扭矩Tf(Nm)Pf(kW)6004002000Tf(Nm)2、根据已知条件做出该坦克的动力特性曲线。选取转速计算点：1000、1200、1400、1600、1800、2000、2200、2300，单位为r/min；计算各排挡的车速，利用公式v 0.377 nf rzi，其中 i 为总传动比且i iq ib ic，n单位为 r/min，r 单位为 m。车速单位 km/h：计 算 各 排 挡 效 率 ，=d ch x，其中 冷 却 风 扇 的 损 耗的 功 率P,fs3 n PfsfnfP，所以各转速排气风扇损失的功率（单位为 kW）为：转速10001200140016001800200022002300功率10.98718.98730.15145.00664.08187.903117133.69空气滤清器功率损失取 3%即 28.05kW，排气管功率损失取 1.5%即 14.025kW，所以动力系统的效率为：转速10001200140016001800200022002300效率0.88810.89630.89630.89000.87750.85840.82980.8120而传动系统去五次齿轮啮合，所以效率为 0.985=0.9039，行动系统的效率用经验公式 x=0.95-0.0017v 计算，如下表：一档0.93660.93390.93130.92860.92590.92320.92060.9192二档0.92400.91880.91360.90840.90320.89800.89290.8903三档0.91380.90650.89930.89200.88480.87760.87030.8667四档0.89950.88940.87930.86920.85910.84900.83890.8338五档0.87960.86550.85140.83730.82330.80920.79510.78813坦克学0312070220071201王超六档0.85180.83210.81250.79290.77320.75360.73390.7241计算各档的牵引力 Fj，且 Fj=Tfi /rz,式中 T 单位为 N/m，r单位为 m，F 单位为 N，牵引力如下表：各排挡的空气阻力RkCDAv221.15G，式中 A 单位为 m2，v单位是 km/h。计算各排挡的动力因数 D，D各排挡的数据如下表：FjRk，式中 F、R、G 的单位均为 N。一档 i=3.903nf(r/min)Pf(kW)Tf(Nm)1000474453112005894690140069747541600792472518008674601200091843842200935405923009353882v(km/h)4.45095.3416.23127.12148.01168.90189.791910.237Fj(N)Rk(N)Dnf(r/min)Pf(kW)Tf(Nm)0.75180.75670.75450.74710.73450.71640.69050.67472885573006073038542990092862502660292374282218682.86154.12065.60867.32569.271411.44613.84915.1370.57710.60120.60760.5980.57240.5320.47480.4437二档 i=1.8231000474453112005894690140069747541600792472518008674601200091843842200935405923009353882v(km/h)8.651110.38112.11113.84115.57217.30219.03219.897Fj(N)Rk(N)Dnf(r/min)Pf(kW)Tf(Nm)0.74170.74440.74030.73080.71650.69680.66970.653414646115215315336715049514366413313211847511054910.8115.56721.18927.67535.02743.24352.32457.1890.29290.30420.30660.30090.28720.26610.23680.2209三档 i=1.5391000474453112005894690140069747541600792472518008674601200091843842200935405923009353882v(km/h)12.06314.47516.88819.30121.71324.12626.53927.745Fj(N)0.73350.73450.72860.71770.70190.68090.65290.63621038711076591082621059831009279330082822771844坦克学0312070220071201王超Rk(N)Dnf(r/min)Pf(kW)Tf(Nm)21.0230.26841.19953.81168.10484.08101.73111.190.20770.21520.21640.21180.20170.18640.16540.1541四档 i=1.3941000474453112005894690140069747541600792472518008674601200091843842200935405923009353882v(km/h)16.82620.19123.55726.92230.28733.65237.01838.7Fj(N)Rk(N)Dnf(r/min)Pf(kW)Tf(Nm)v(km/h)Fj(N)Rk(N)Dnf(r/min)Pf(kW)Tf(Nm)0.7220.72060.71240.69930.68150.65880.62930.612733037572375889740337025364710572325323640.89758.89280.158104.69132.5163.58197.94216.340.14650.15130.15160.14780.14020.1290.1140.106五档 i=1.304100047445311200589469014006974754160079247251800867460120009184384220093540592300935388223.4628.15232.84437.5342.22946.92151.61353.9590.7060.70120.68990.67370.6530.62790.59640.5784514115285252704511534828644235389053608679.503114.48155.82203.52257.59318.01384.79420.570.10260.10540.1050.10190.0960.08780.0770.0713六档 i=1.25710004744531120058946901400697475416007924725180086746012000918438422009354059722300935388275.273v(km/h)32.72739.27245.81852.36358.90965.454Fj(N)Rk(N)D0.68370.67420.65830.63790.61330.58470.55050.53153568936426360543472132510295312574423769154.71222.79303.24396.07501.27618.86748.82818.440.0710.07240.07150.06860.0640.05780.04990.0459所以，得到坦克的动力特性曲线：5坦克学0312070220071201王超坦克的动力特性曲线0.70.60.50.40.30.20.10010203040v(km/h)50607080一档二档三档四档五档六档3、利用 1/a-v曲线计算该坦克的加速特性曲线，并根据在良好路面上032k m 的加速时间对其加速性做出评价。首先，根据第二阶段的加速度计算公式，a=g(D-f0)/得到各排挡的数据如下表：1/a二档0.7350.7030.6970.7120.7520.8220.9451.027三档0.9360.8960.8900.9130.9711.0721.2511.375四档1.3351.2771.2741.3181.4191.5961.922.153五档2.1232.0322.0442.1492.3732.7813.5924.246六档4.1283.9574.0714.4765.3407.19512.8321.72v二档8.65110.3812.1113.8415.5717.3019.0319.89三档12.0614.4716.8819.3021.7124.1226.5327.74四档16.8220.1923.5526.9230.2833.6537.0138.70五档23.4628.1532.8437.5342.2246.9251.6153.95六档32.7239.2745.8152.3658.9065.4572.0075.27所以用 matlab（程序见附 1）拟合出二次曲线如图：D6坦克学0312070220071201王超并且得到各排挡加速第二阶段的方程式：2二档：1/a=0.0052v-0.1230v+1.4208三档：1/a=0.0038v2-0.1255v+1.90882四档：1/a=0.0034v-0.1566v+3.0240五档：1/a=0.0043v2-0.2726v+6.2333六档：1/a=0.0181v2-1.6523v+40.1660所以对其取不定积分有：二档：t=0.001733v3-0.0615v2+1.4208v+C1，C1=-3.4295；三档：t=0.001267v3-0.0628v2+1.9088v+C2，C2=-4.5659；四档：t=0.001133v3-0.0783v2+3.0240v+C3，C3=-10.509；g(Dmax f0)2又因为在二档加速第一阶段a，此时加速度 a=4.682636m/s。此f时加速时间 t=12.11/a/3.6=0.72s。所以 v=4.682636t，0t0.72s。加速终了速度为 3.364m/s。在加速第二阶段 t=0.001733v3-0.0615v2+1.4208v-3.4295,3.364v5.53。gf在加速第三阶段a 0，换档时间取 1s，a=-0.326666667m/s。所以f2v=5.53-0.326666667(t-2.8399)，2.8399t3.8399。结束时速度为 5.20m/s.三档加速第二阶段 t=0.001267v3-0.0628v2+1.9088v-4.5659，5.20v7.71。gf在加速第三阶段a 0，换档时间取 1s，a=-0.326666667m/s2。所以fv=7.71-0.326666667(t-6.998542),6.998542t7.998542,结 束 时 速 度 为7.38m/s。在四档加速第二阶段，t=0.001133v3-0.0783v2+3.0240v-10.509，7.38v8.89。当加速结束时，032km/h 加速所用时间为 10.98s有上述方程，用 MATLAB 可得到 032km/h 加速特性曲线（程序见附 2）：通过加速度分析，得出 0 到 32 公里加速时间为 10.98s，可以认为该坦克的加速性接近同时代国际的先进水平，基本能够达到规定的技战术指标。7坦克学0312070220071201王超附 1x=8.651198 10.38144 12.11168 13.84192 15.5722 17.3024 19.03264 19.89775;y=0.735548 0.703885 0.697511 0.712898 0.75233 0.822453 0.945008 1.02783;p=polyfit(x,y,2)p=0.0052-0.12301.4208x1=8.651198:0.3:19.89775;y1=polyval(p,x1);plot(x1,y1)hold ona=12.06321 14.47585 16.88849 19.30113 21.7138 24.12641 26.53906 27.74538;b=0.936429 0.896054 0.890035 0.91378 0.97107 1.07245 1.251911 1.375784;p=polyfit(a,b,2)p=0.0038-0.12551.9088x1=12.06321:0.3:27.74538;y1=polyval(p,x1);plot(x1,y1)hold onc=16.82647 20.19176 23.55706 26.92235 30.2876 33.65294 37.01823 38.70088;d=1.335325 1.277705 1.274391 1.318812 1.41901 1.596582 1.920455 2.153897;p=polyfit(c,d,2)p=0.0034-0.15663.0240 x1=16.82647:0.3:38.70088;y1=polyval(p,x1);plot(x1,y1)hold one=23.46056 28.15267 32.84478 37.5369 42.229 46.92112 51.61323 53.95929;f=2.123406 2.03224 2.044018 2.149611 2.37363 2.781627 3.592154 4.246792;p=polyfit(e,f,2)p=0.0043-0.27266.2333x1=23.46056:0.3:53.95929;y1=polyval(p,x1);plot(x1,y1)hold ong=32.72748 39.27298 45.81847 52.3639758.9095 65.45496 72.00046 75.27321;h=4.128365 3.957901 4.071669 4.4768655.34042 7.195664 12.83805 21.72982;p=polyfit(g,h,2)p=0.0181-1.652340.1660 x1=32.72748:0.3:75.27321;8坦克学0312070220071201王超y1=polyval(p,x1);plot(x1,y1)附 2t=0:0.05:0.72;v=4.683*t;plot(t,v)hold ony=3.365:0.05:5.53;x=0.001733*y.3-0.0615*y.2+1.4208*y-3.4295;plot(x,y)hold onx1=2.8399:0.05:3.8399;y1=5.53-0.326666667*(x1-2.8399);plot(x1,y1)hold ony=5.20:0.05:7.71;x=0.001267*y.3-0.0628*y.2+1.9088*y-4.5659;plot(x,y)hold onx=6.998542:0.05:7.998542;y=7.71-0.326666667*(x-6.998542);plot(x,y)hold ony=7.38:0.05:8.89;x=0.001133*y.3-0.0783*y.2+3.0240*y-10.509;plot(x,y)9
