《汽车理论》课程教学课件（英文讲稿）第3章 汽车动力装置参数选择 Decisions for Engine Power and Gear Ratios 3.2 Decision for Minimum Gears Ratio（最小传动比的 选择）

3.2 Decision for minimum gears ratio-imingears ratio=i.i,i.when gears ratio meets its minimumvalue i=l, i=1, → imin=ioMethod:0.377rnrn,I. Calculation:P=0.377u10amaxWamaxig(high-gear) iowhereuamax is the design data;（设计指标)n,istheratingrotatespeed oftheengine（额定转速）( uamax, G → Pemax → engine selection → engine performancecharacteristics→ the corresponding n,note:n,is the corresponding x-coordinate value of thepoint where P,meets its maximum value

Method: 1. Calculation: where uamax is the design data;（设计指标） np is the rating rotate speed of the engine（额定转速） ( uamax , G → Pemax → engine selection → engine performance characteristics → the corresponding np ) max 0 ( ) 0 max 0.377 0.377 a p g high gear p a u rn i i i rn u      gears ratio=ig·i0·ic → when gears ratio meets its minimum value i c=1, ig=1, → imin=i0 note: np is the corresponding x-coordinate value of the point where Pe meets its maximum value. 3.2 Decision for minimum gears ratio-imin

二(P, + P)一Nr1PDee103amax2i02P→andecurveofu101amax3uamax luuuuap3p2plGraph of Power Equilibrium with different i

amax 2 u amax1 u amax 3 u p1 up3 up2 u 03 02 01 i i i ＞ ＞ Graph of Power Equilibrium with different i0 Pe Pe a u 03 i 02 i 01 i 1 ( ) f w T P P   0 e i curveof P    and

3.Evaluating the effect ofio on the fuel economy of vehicleTryto evaluatethree different ioselectioni03ior<Itemsi01<i02<103i02<minimummediummaximumUamaxbestaccelerationworstneutralabilityout-putmaximummediumminimumpowerratioWamax2uamax3bminimummediummaximum-(R+Pa)uamax1nTup3up20upiQminimummediummaximumua

Items i01＜ i02＜ i03 uamax minimum maximum medium acceleration ability worst neutral best out-put power ratio maximum medium minimum b minimum medium maximum Qs minimum medium maximum Try to evaluate three different i0 selection 3. Evaluating the effect of i0 on the fuel economy of vehicle

Conclusions. io is not an idea selection for fuel economyalthough the fuel consumption of io, is the lowest.because:I) The conclusion is drawn with the conditions of constant speedand level road.2)If theevaluationbackground (drive condition is changed intoclimbing the hill with constant speed,the high speed gearratio with io, could not meet the dynamic demand for climbingthe hill, so the transmission gear must be changed into loweronethen b of engine wouldincreased tremendously,whichleads intothe highfuel consumption

• i01 is not an idea selection for fuel economy although the fuel consumption of i01 is the lowest. because : 1) The conclusion is drawn with the conditions of constant speed and level road. 2) If the evaluation background ( drive condition is changed into climbing the hill with constant speed , the high speed gear ratio with i01 could not meet the dynamic demand for climbing the hill, so the transmission gear must be changed into lower one then b of engine would increased tremendously, which leads into the high fuel consumption. Conclusions

Conclusions> For the problemmentioned above somemanufacturesproduced the vehicles with two i selections, the bigger one isselected in bad driving conditions, the less one is selected ingoodconditions.> Usually when vehicle speed gets to its maximum value, thecorresponding n is equal to (0.95~1.1)npratioUamax / Up3%0.5~0.717.5%0.7~0.974%0.9~1.105.5%1.10~1.39Ratio ofcarwithdifferentuUpma

Conclusions  For the problem mentioned above some manufactures produced the vehicles with two i0 selections, the bigger one is selected in bad driving conditions, the less one is selected in good conditions.  Usually when vehicle speed gets to its maximum value, the corresponding n is equal to (0.95~1.1)np . ratio 0.5～0.7 3％ 0.7～0.9 17.5％ 0.9～1.10 74％ 1.10～1.39 5.5％ Ratio of car with different uamax/uP amax P u u/

2.AdjustmentDesign condition: io should be designed to meet thedemand ofvehicle's acceleration performance-Domax (the maximum dynamic factor of vehiclewith highspeedgear).Domax should be more than or equal to 0.05.Tqmax ig i nrC,Au?/ GROmax21.15Where ig- ig-highest; G is design data; CpA could be selectedroughly from its range; u, is the calculation result when thernttorque of engine meets its maximum value-Ttqmax (u. = 0.377Tg

Design condition：i0 should be designed to meet the demand of vehicle’s acceleration performance— —D0max (the maximum dynamic factor of vehicle with high speed gear). D0max should be more than or equal to 0.05. 2. Adjustment 2 max 0 0max / 21.15 tq g T D a T i i C Au D G r            Where ig= ig-highest；G is design data；CDA could be selected roughly from its range；ua is the calculation result when the torque of engine meets its maximum value-Ttqmax ( ). 0 0.377 tq a g rn u i i  

2.Adjustmentnote: nta is the corresponding60Pemaxx-coordinate value of the50point where Ttg~n curve40M16030meets its maximum value-140Ttqmax00(NbuTtgmax:2088导1020nmaxnmin20003000ntq5000np0n/(r-min-l)Adjustment principle:if the calculation resultDomax is less than 0.05 , io should be increased

2. Adjustment Adjustment principle : if the calculation result D0max is less than 0.05 ，i0 should be increased. note: ntq is the corresponding x-coordinate value of the point where Ttq~n curve meets its maximum value￾Ttqmax