Gear Ratio's in a diff
#1
Posted 05 January 2007 - 09:15 AM
i don't quite understand how the ratios work concerning the diff in a mini.
I know that you can alter between higher top speed or a faster pull off......
can any one please explain.....?
Thanks
#2
Posted 05 January 2007 - 09:51 AM
A 3.44 diff will result in a lower top speed but higher acceleration.
Or.....Terminology -
FD - Final Drive (diff ratio)
Calculation for establishing vehicle speed for different final drives.
Formula : 60,000
FD (Final drive/diff ratio) x wheel rev per mile.
Wheel/tyre combinations - revolutions per mile:
145/70/10 = 1176.78
165/70/10 = 1078.47
145/70/12 = 1059.06
155/70/12 = 1030.71
165/60/12 = 1047.27
175/50/13 = 1058.82
These are for common tyre types, and accurate enough for assessment - a combination of using industry standard for theoretical calculation, actually measured assortment of wheels/tyres, and calculated averages! Applying the formula to a modern Mini with 12” wheels gives the following -
60,000 = 60,000 = 18.25 mph per 1,000 rpm.
3.105 x 1059.06 = 3,288.38
So 70 mph is achieved at a whisker over 3,800rpm.
Calculation for gear ratios and transmitted engine rpm
Basic rule to remember is ratio is established by dividing tooth count on driven gear by the tooth count on its driver. To work out overall gearbox ratios you also have to establish the constant ratio.
Constant ratio = tooth count on laygear input gear
This is NOT the fourth gear ratio assumed by many. Fourth gear doesn’t actually exist, as once in top gear the first motion shaft directly engages the mainshaft. Consequently it’s always a 1 to 1 ratio.
Gear ratio = tooth count on driven mainshaft gear X constant ratio.
tooth count on driver on laygear
Example: 1st gear of A+ standard gearbox
Constant ratio = 30 = 1.765
Gear ratio = 31 x 1.765 = 2.066 x 1.765 = 3.647
Calculation for transmitted engine rpm at output shaft
Input gear speed in rpm = engine rpm
transfer (drop) gear ratio
Pinion speed in rpm = input gear speed (rpm)
actual gear ratio
Out put shaft speed in rpm = pinion speed (rpm)
final drive ratio
Right here cometh the maths...
3.11 : 1 means that for every 3.11 turns of the engine ( when in 4th ) then the wheels go round once...
Therefore using the the standard tyre 145/70x12 ( or for that matter 165/60x12 as the rolling distance is only a mm or so different ) approx 1595mm
so to workout the speed for a give engine rpm, use the following
( engine rpm / 3.11 ) * 1595 * ( 60 / 1609344 ) = mph
1609344 is the number of mm's in a mile and 60... minutes in an hour !
so.... at 3000rpm
( 3000 / 3.11 ) * 1595 * ( 60 / 1609344 ) = 57.3 mph
#4
Posted 05 January 2007 - 12:00 PM
So I just put that instead
He also asked how it worked so that give's the explanation I guess....
#5
Posted 05 January 2007 - 01:52 PM
Might try and get my head around it later
Edited by TOMMO0302, 05 January 2007 - 01:52 PM.
#6
Posted 06 January 2007 - 11:58 AM
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