3.4 Decisions for the i。 (gear ratio)of other gears(withgivenigiandio)(numberofgears)nAccelerationPerformance个FuelEconomy1+PPUM↑→nl(overcomeresistancedependonreservepower)+PPUM ↓→n↑ (overcome resistance depend on shiftgear)+ heavy truck and off road vehicle →itmax/itmin ↑, n ↑
PPUM ↑→ n↓（overcome resistance depend on reserve power） PPUM ↓→ n↑（overcome resistance depend on shift gear） heavy truck and off road vehicle →i tmax/i tmin ↑，n ↑ (number of gears)n↑， Acceleration Performance ↑， Fuel Economy ↑ 3.4 Decisions for the ig（gear ratio) of other gears （with given ig1 and i0）
3.4 Decisions for the i。 (gear ratio) of other gears(withgivenigiandio)Two assumption:I)Enginealways works between n,and n2.2)There is no speed change of vehicle and at the minute when thegearis beingchanged.And the speed of vehicleughas a lineal relationshipwith revolutionspeedofengine nwithgiven vehicleandgearratios.u60Pemax5040MXd16030140Tumak(w-N/bu100208060104020?3000"tqnmax0"min 20005000"pn/(r-min-b)
Two assumption: 1) Engine always works between n1 and n2 . 2) There is no speed change of vehicle and at the minute when the gear is being changed. And the speed of vehicle ua has a lineal relationship with revolution speed of engine n with given vehicle and gear ratios. 3.4 Decisions for the ig（gear ratio) of other gears （with given ig1 and i0）
IVPeuIIIuua34a43uu=a23a32uua12a21nnn
n1 n2 Pe I II III IV a a 34 43 u u a a 23 32 u u a a 12 21 u u n ua
rnrn.: ai2 = 0.377, Ua21 = 0.377(number of gears)could be3nn29calculated from i.. . Ua12 = Ua21 = A.1ng282and q;n23g2u.=Ua32Usuallyq shoulda23ni92g393be lessthan1.7~1.8 .nq<1.7~1.83计ion=1=q=n/Sgn
• (number of gears) could be calculated from ig1 and q； • Usually q should be less than 1.7~1.8 . 2 1 12 21 g1 0 g2 0 0.377 , 0.377 a a rn rn u u i i i i q 1.7 ~1.8 g1 g 1 1 g g1 , 1 n n n n i i q if i q i 2 2 3 1 2 1 23 32 g2 g g1 g2 g3 2 g2 g3 g4 1 3 . g g a a i n i n i i i n n n u u A A i i q i i i n 2 1 12 21 g1 g 1 2 2 2 1 g g a a n n u u A A i i i n i n
>i, according to geometric proportion is convenient forcombination of basic transmission and auxiliary transmission+basic transmission: n=5, qbasic=q2,ig1=q8、igμ=q、igm=q4、igiv=q?、igv=l +auxiliary transmission: n=2, qauxiliary =qigの=q、ig@=l 。+overall ratio: n=10, qoverall =q→igi=q、ig2=q8、ig3=q、ig4=q、igs=q5、ig6=q4ig=q、ig8=q2、igg=q、ig1o= 1
basic transmission：n=5，qbasic =q 2 ， igⅠ=q 8 、igⅡ=q 6 、igⅢ=q 4 、igⅣ=q 2 、igⅤ=1。 auxiliary transmission：n=2，qauxiliary =q ig①=q、ig②=1 。 overall ratio： n=10，qoverall =q ig1 =q 9 、ig2 =q 8 、ig3 =q 7 、ig4 =q 6 、ig5 =q 5 、ig6 =q 4 、 ig7 =q 3 、ig8 =q 2 、ig9 =q、ig10 = 1。 ig according to geometric proportion is convenient for combination of basic transmission and auxiliary transmission
AnalyzingItis hard forthegears to be designed intothefixedgeometric proportion with each other.Eachgearhas its own usage efficiency,generallyhigher gears have higherusage efficiency which leadsto higher output-power ratio and lower fuelconsumption aswell.This means we'd better to make our vehicle to workon the highgear as much as possible.Actually there is speed decrease when changing thegear, and the speed of vehicle decrease greatly with thehighgear
• It is hard for the gears to be designed into the fixed geometric proportion with each other. • Each gear has its own usage efficiency, generally higher gears have higher usage efficiency which leads to higher output-power ratio and lower fuel consumption as well. This means we’d better to make our vehicle to work on the high gear as much as possible. • Actually there is speed decrease when changing the gear, and the speed of vehicle decrease greatly with the high gear. Analyzing
ConclusionAccording to the analysis above,it is not necessary toarrange the gear ratios with geometric proportion.If we takethefactor of speed decreaseinto account inthe idea graph, the speed decrease would bring engineto work away from its optimal work range(n,~n),which is not what we want.For both meeting the demand of optimal range andtaking thefactor of speed decrease into account, theu.~n line of eachgear should be changedintothisway:O
• According to the analysis above ,it is not necessary to arrange the gear ratios with geometric proportion. • If we take the factor of speed decrease into account in the idea graph, the speed decrease would bring engine to work away from its optimal work range(n1~n2 ), which is not what we want. • For both meeting the demand of optimal range and taking the factor of speed decrease into account, the ua~n line of each gear should be changed into this way: 1 2 3 234 . g g g g g g i i i i i i Conclusion
ExamplesJetta:38CA1091:POLO:3.45.一g111i=7.640,=3.455.1gl35gl= 1.94.= 4.8358282=2.095,18g2= 2.857,ix = 1.89537=1.387,g3=1.37.g383元27=1=1.337.ig5of= 1.026,32g4i=1.03.gl=1.580,g2=1.69284= 0.813.i31g511g234~：0.85ii'g2gl=1.649,=1.510g5'g440g3 =1.508,=1.417.~1i1ig3g2'g4'g5'glg2=1.416=1.778..i11g3g45=1.337.=1.352= 1.262g2g31-ig4g6g5g3'g4= 1.330.=1.2111igsg4
Examples 1 2 3 4 5 1 2 2 3 3 4 4 5 : 38 3.45, 11 35 1.94, 18 37 1.37, 27 32 1.03, 31 34 0.85, 40 1.778, 1.416, 1.330, 1.211. g g g g g g g g g g g g g Jetta i i i i i i i i i i i i i 1 2 3 4 5 6 1 2 2 3 3 4 4 5 5 6 1091: 7.640, 4.835, 2.857, 1.895, 1.337, 1, 1.580, 1.692, 1.508, 1.417, 1.337. g g g g g g g g g g g g g g g g CA i i i i i i i i i i i i i i i i 1 2 3 4 5 1 2 2 3 3 4 4 5 : 3.455, 2.095, 1.387, 1.026, 0.813, 1.649, 1.510, 1.352, 1.262. g g g g g g g g g g g g g POLO i i i i i i i i i i i i i
Inquiries after-classWhatis the C curve?
Inquiries after-class • What is the C curve? 9