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Variant ? 15: Carrying capacity of automobile G=75000 N. The maximal speed of automobile Vmax=25 /.
2) Chosen parameters: Accepted the automobile 6? 4. In this connection on axes will be on the back carriage 0,75Ga.? 1?. In the projected automobile is accepted: the carburettor engine with the terminator of revolutions; mechanical transmission (simplicity of a design, service and small capital investments) We accept? =0,9. For the prototype we accept Ural - 377. Radius wheels rk=0,507 m.? 1? The moment of inertia of wheel Jk=31,9 H2.? 1? The moment of inertia of rotating parts of engine Jm=0,62 Hms2.? 1? We accept factor the automobile k=0,9 2/4 [1] and the frontal area of automobile F=4,68 m2. [1] 1 Traction calculation of the automobile
1.1 Definition of full weight of the automobile and selection of trunks Ga=Go+G + (g1+g2) nn (1.1)
Where nn-number of places in the automobile, including a place of the driver; g1 The weight of one person accepted equal 750 [1]; g2 The weight of luggage accepted equal 100 on the person. Ga=149500+75000 + (750+100) 3=227050 Weight on axes (2).
The weight, falling a back axis G2=0,75Ga G2=0,75227050=170288 G1=GaG2=227050-170288=56763 r=0,507 M [1]. The engine petrol.
Full weight the automobile with the trailer Ga=Ga+G (1.2) G=227050+75000=302050 . 1.2 Definition of the external high-speed characteristic of the engine The capacity necessary for overcoming of resistance to movement at the maximal speed /. Nv = (Ga fVmax/1000+kFV3max/1000)/?? ???????????? (1.3) Factor of resistance .
f=f0 (1+13V2/20000), (1.4) Where f0 - the factor of resistance at movement about speed is less 15 25?/with. For highway, with a covering, we accept f0 =0,02 [ 1 ]. f=0,02 (1+13625/20000 0,025 Nv = (2270500,0225/1000+0,94,32253/1000/0,9 = 230,8 kw The maximal capacity
Nmax = Nv / (a? +b? 2-c? 3) (1.5) Where a, b, c - empirical factors [1]. ? - for petrol engines with the terminator of revolutions.? =0,9. [ 1 ] Nmax = 230,8 / (0,9+0,92-0,93) = 235,27 The current value of effective power of the engine Ne = Nmax (a (ne/nN) +b (ne/nN) 2-c (ne/nN) 3) (1.6) The current values of twisting moment of the engine Me=1000 (Ne/ne) (1.7) Results of calculations (1.6) and (1.7) are brought in table 1.1
Table 1.1
Parameters of the external high-speed characteristic of the engine. Parameters rotations of a cranked shaft, ne, 1/80120160200240280320360Ne, 61,3295,85130,38162,98191,70214,6122 9,78235,27Me, ??766,48798,75814,89814,89798,75766,48 718,07653,52 According to table 1.1 the diagram (see fig. 1) is constructed. 1.3 Definition of transfer number of the main transfer U0=rnmax/(UkVmax) (1.8) U0=0,495360/(125) = 7,301 1.4 Definition of transfer numbers of a box of transfers
Transfer number of the first transfer U1=Ga ? maxrk/(MmaxU0?), (1.9) where?? max - factor of the maximal resistance of road; Mmax - the maximal twisting moment of the engine. U1=925500,340,495/(494,616,930,9) = 8,47 Check of coupling of driving wheels from road(dear,expensive) MmaxU1U0 ?? rk? G? ? (1.10) Where G?? - coupling weight of the automobile. G? = G2m2, (1.11) Where m2 - factor of dynamic redistribution of weight of the automobile On a back axis at a traction mode. We accept m2=1,3? 1? ? =0,5 dry asphaltes or .? 1? G? =1702881,3=221374 814,898,477,3010,9? 2213740,60,5 45353,1? 66412,2? Slippings will not be Choice of transfer numbers of a box of transfers Um=n-1? U1n (1.12) Where m-a serial number of considered transfer; n-Number of steps of a box of transfers, not including an overdrive gear. U2=4? 4,913 =4,96 U3=4? 4,912 =2,23 U4=4? 4,911 =1,22 U5=4? 4,910 =1 2 Definition of n??-speed qualities of the automobile 2.1 The traction diagram of movement of the automobile The equation of movement of the automobile a method of power(force) balance: Pm = P? + Pw+Pj. (2.1) Pm=MeUmp ?/rk, (2.2) Where Ump=U0Uk-transfer number of transmission of the automobile. Speed of movement of the automobile is defined(determined) under the formula V=nerk/Ump (2.3) Force of resistance of road
P? =Ga ?? (2.4) Where?? f +i; i - size roads. Force of resistance of air Pw=kFV2. (2.5) We use k=0,9 2/4; F=4,68 m2. [1] Results of calculations (2.2), (2.3), (2.4) and (2.5) are brought in table 2.1 Table 2.1 Traction balance of the automobile. rotations of a cranked shaft, ne, 1/80120160200240280320360Me, 766,48798,75814,89814,89798,75766,487 18,07653,52V1, ?/?0,660,981,311,641,972,302,632,95Pm18 4089,8187630,4389400,7489400,7487630,43 84089,8178778,8771697,62P?? 6042,696044,816047,766051,576056,226061,726068,066075,25Pw1 1,814,087,2611,3416,3322,2229,0336,74V2 1,121,682,242,803,363,924,485,04Pm2 49300,3051376,1052414,0052414,0051376,1 049300,3046186,6042034,99P?? 6045,926052,076060,686071,756085,286101,276119,726140,64Pw2 5,2811,8821,1132,9947,5064,6584,45106,8 8V3 2,493,744,996,237,488,739,972,49Pm3 22131,9923063,8623529,8023529,8023063,8 622131,9920734,1818870,43P?? 6065,416095,936138,666193,596260,736340,086431,636535,39Pw3 26,1958,92104,75163,68235,70320,81419,0 2530,32?? rotations of a cranked shaft, ne, 1/?80120160200240280320360V4 4,556,829,0911,3713,6415,9118,1920,46Pm 4 12138,8212649,9312905,4812905,4812649,9 312138,8211372,1610349,94P?? 6122,166223,616365,636548,246771,437035,197339,547684,46Pw4 87,06195,88348,23544,10783,511066,44139 2,901762,89V5 5,568,3311,1113,8916,6719,4422,2225,00P m5 9933,5710351,8210560,9510560,9510351,82 9933,579306,188469,67P?? 6162,196313,686525,776798,467131,747525,617980,098495,16Pw5 130,00292,50520,00812,501170,001592,502 080,002632,50 According to table 2.1 the diagram (see fig. 2) is constructed. 2.2 A dynamic characteristic of the automobile The dynamic factor of the automobile D = (Pm-Pw)/Ga, (2.6) Results of calculations (2.6) are brought in table 2.2
Table 2.2
The dynamic factor of the automobile. rotations of a cranked shaft, ne80120160200240280320360V10,660,981,31 1,641,972,302,632,95D10,2780,2900,2960, 2960,2900,2780,2610,237V21,121,682,242, 803,363,924,485,04D20,1630,1700,1730,17 30,1700,1630,1530,139V32,493,744,996,23 7,488,739,9711,22D30,0730,0760,0780,077 0,0760,0720,0670,061V44,556,829,0911,37 13,6415,9118,1920,46D40,0400,0410,0420, 0410,0390,0370,0330,028V55,568,3311,111 3,8916,6719,4422,2225,00D50,0320,0330,0 330,0320,0300,0280,0240,019 According to table 2.2 the diagram (see fig. 3) is constructed. The dynamic factor on coupling D? = (Gm2?-Pw)/Ga, (2.7) Where D?? The dynamic factor on coupling. We consider D?? At? =0,2 and? =0,4 Results of calculations (2.7) are brought in table 2.3. ?? D? D? - a condition of unceasing movement of the automobile. Table 2.3 The dynamic factor on coupling. rotations of a cranked shaft, ne, 1/80120160200240280320360D? 1 ( 0,2 ) 0,1470,1470,1470,1470,1470,1470,1460,14 6D? 1 ( 0,4 ) 0,2930,2930,2930,2930,2930,2930,2930,29 3D? 2 ( 0,2 ) 0,1470,1470,1470,1460,1460,1460,1460,14 6D? 2 ( 0,4 ) 0,2930,2930,2930,2930,2930,2930,2930,29 3D? 3 ( 0,2 ) 0,1460,1460,1460,1460,1460,1460,1450,14 5D? 3 ( 0,4 ) 0,2930,2930,2930,2930,2920,2920,2920,29 1D? 4 ( 0,2 ) 0,1460,1460,1450,1450,1440,1430,1420,14 1D? 4 ( 0,4 ) 0,2930,2930,2920,2910,2910,2900,2890,28 7D? 5 ( 0,2 ) 0,1460,1460,1450,1440,1430,1410,1400,13 8D? 5 ( 0,4 ) 0,2930,2920,2910,2900,2890,2880,2860,28 4 According to table 2.3 the diagram (see fig. 3) is constructed. On the first transfer at? =0,4; on the second transfer at? =0,4; on the third transfers at? =0,2, at? =0,4; and on the fifth transfer at? =0,2, at? =0,4 condition of unceasing movement of the automobile is carried out. 2.3 The dynamic passport of the automobile D0=DGa/G0, (2.8) Where D0 - the dynamic factor of the non-loaded automobile. a0=aD0/D, (2.9) Where a and a0 - the scales postponed on axes D and D0. Results of calculations (2.8) are brought in table 2.4 Table 2.4 The dynamic factor of the non-loaded automobile. rotations of a cranked shaft, ne80120160200240280320360D010,660,680,7 00,700,680,650,610,56D020,440,460,470,4 70,460,440,410,37D030,300,310,310,310,3 00,290,270,24D040,200,200,210,200,200,1 80,170,15D050,130,130,130,130,120,100,0 80,06 D? 0 = G? 0 *? / G0a, (2.10) Where D? 0 - the dynamic factor for the non-loaded automobile. D? = G? *? / Ga, (2.11) Where D? - the dynamic factor for the loaded automobile. By results of calculations (2.8), (2.10) and (2.11) the diagram (see rice 4) is constructed. 2.4 the diagram of movement of the automobile
Nm=Ne?? = N? + Nw + Nj, (2.12) Where N? =P?? V/1000, (2.13) Nw=Pw? V/1000, (2.14) Nj=Pj? V/1000, (2.15) Nj=Ne?? - (N? + Nw), (2.16) Results of calculations (2.12), (2.13), (2.14) and (2.16) are brought in table 2.5. Table 2.5 balance of the automobile. rotations of a cranked shaft, ne80120160200240260300360Ne, 61,3295,85130,38162,98191,70214,6122 9,78235,27Nm55,1986,27117,34146,68172,5 3193,15206,8055,19V10,660,981,311,641,9 72,302,632,95N? 13,995,927,929,9211,9313,9415,9617,92Nw 10,000,000,010,020,030,050,080,11Nj151, 2080,34109,41136,74160,57179,16190,7719 3,71V21,121,682,242,803,363,924,485,04N? 26,7710,1613,5716,9920,4423,9027,4030,9 3Nw20,010,020,050,090,160,250,380,54Nj2 48,4176,08103,73129,60151,94169,00179,0 2180,27V32,493,744,996,237,488,739,9711,22N? 315,1222,8030,6138,6146,8355,3364,1573, 33Nw30,070,220,521,021,762,804,185,95Nj 340,0063,2486,21107,05123,93135,02138,4 8132,46V44,556,588,9611,4614,0816,5818, 4620,67N? 427,8340,9557,0475,0495,34116,64135,491 58,84Nw40,401,293,126,2411,0317,6825,71 36,44Nj426,9644,0257,1965,4066,1658,834 5,6016,47V55,568,3311,1113,8916,9019,71 22,0025,00N? 534,2352,6172,5194,42120,53148,33175,56 212,38Nw50,722,445,7811,2819,7731,3945, 7665,81Nj520,2331,2139,0640,9732,2313,4 3-14,52-66,45 According to table 2.5 diagrams (see fig. 5 and 6) are constructed.
2.5 Acceleration at dispersal of the automobile
j = (D-?) g/? p (2.17) Factor of the account of rotating weights of the automobile: ? p=1 +?? Uk2 +? 2, (2.18) Where ??? gJmUo2 ? / (Ga rk2). (2.19) ??? 9,8? 0,687,3012? 0,9 / (302050? 0,5072) =? 0,004; ? 2=g? Jk / (Gark2.) (2.20) ? 2=9,8? 204,16 / (302050? 0,5072) =0,026; ? Jk = Jk1 + Jk2, (2.21) Where Jk1=z1Jk, (2.22) Where z1 - quantity(amount) of conducted wheels. Jk2=1,1z2Jk, (2.23) Where z2 - quantity(amount) of driving wheels. Jk1=231,9=63,8; Jk2=1,1431,9=140,36; ? Jk=63,8 + 140,36 =204,16; Results of calculations (2.17) are brought in table 2.6. Table 2.6 Acceleration at dispersal of the automobile. rotations of a cranked shaft, ne80120160200240260300360V10,660,981,31 1,641,972,302,632,95J11,932,012,062,062,011,931,791,621/J10,520,500,490,490,50 0,520,560,62V21,121,682,242,803,363,924,485,04J21,251,311,341,341,311,251,161, 041/J20,800,760,740,740,760,800,860,96V 32,493,744,996,237,488,739,9711,22J30,5 00,530,540,530,510,480,430,371/J32,011, 901,861,871,942,082,322,73V44,556,289,0 911,3713,6415,5217,0819,00J40,190,200,1 90,180,160,130,090,041/J45,365,085,135, 476,247,7811,3828,17V55,568,3311,0714,1 017,0719,6122,2825,39J50,1150,1180,1110,0920,0630,0250,020,001/J58,718,479,021 0,8418,3440,16126,38? According to table 2.5 diagrams (see fig. 7 and 8) are constructed.
2.6 Definition of time of dispersal of the automobile
