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