Gearing Question

[Fred444]

If you have a flat power curve of say 100 BHP all the way through the rev range...

Will that car accelerate from a set speed faster in 4th or 5th gear?

I'm saying 4th, but don't actually know the reason why.

How do you work out how a gear effects the power at the wheels? Is that right, its power at the wheels being geared down here?


PS. I'm looking into this because I'm trying to work out the optimum shift points on a diesel that has a very flat power curve.

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It would accelerate exactly the same, if you disregard any difference in drivetrain efficiency between the two gear pairs.

One way of expressing power is through torque and angular velocity: P=M*w (W = Nm * rad/s).
Another way is through force and velocity: P=F*v (W = N * m/s).
Now the latter will easily show you that for a given P (100 BHP) and given v (the speed the car is currently driving) you'll always get the same tractive force F, and thus the same acceleration.

If you want to know your optimum shift points, set up a table (i.e. in excel) of your engine torque vs. rpm. Now you can translate this torque curve to a torque curve at your driven wheels for any given gear ratio.
Divide your engine rpm by the total gear ratio to get wheel rpm.
Multiply your engine torque by total gear ratio to get wheel torque.

If you do this for say, 5 gear ratios, you end up with a graph which has 5 torque curves, plotted against your rear wheel angular speed. The point where two subsequent torque curves cross eachother, is the point at which you need to shift up.
Covert wheel angular speed to actual car speed by taking into account your chosen tyre circumference.

[Fred444]

Wow, thats an awesome reply. Thanks!!!



EDITED - I was confused about BHP and torque, but now understand that bit.

[Fred444]

These are my torque figures...

Code: Select all

RPM   Lbft
2000   260
2250   280
2500   290
2750   280
3000   265
3250   250
3500   245
3750   240
4000   230
4250   210
4500   190
4750   170
5000   120


My gear ratios are:

3.77
2.1
1.32
0.98
0.84
0.75

Gear ratios multiplied by torque:

Code: Select all

RPM   Gear 1   Gear 2   Gear 3   Gear 4   Gear 5   Gear 6
2000   980   546   343   255   218.4   195
2250   1056   588   370   274   235.2   210
2500   1093   609   383   284   243.6   217.5
2750   1056   588   370   274   235.2   210
3000   999   557   350   260   222.6   198.75
3250   943   525   330   245   210   187.5
3500   924   515   323   240   205.8   183.75
3750   905   504   317   235   201.6   180
4000   867   483   304   225   193.2   172.5
4250   792   441   277   206   176.4   157.5
4500   716   399   251   186   159.6   142.5
4750   641   357   224   167   142.8   127.5
5000   452   252   158   118   100.8   90


Does that mean the shift points are where the highest number in the next gear crosses over?

i.e. pretty much this....

1-2 4750
2-3 4500
3-4 4250
4-5 3750
5-6 3250

Is that right? changing to 6th at 3250 seems way too low??

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What's your final drive ratio? And what's your tyre circumference?

You should get a table for each gear, that looks like this:

wheelrpm (first column):
(rpm = 2000/(1st gear ratio * final drive ratio)

wheelspeed (you can replace the wheelrpm column with this one, or just keep them both)
v = (rpm/60)*circumference

wheeltorque:
(260*1st gear ratio * final drive ratio)

Now complete this with a row for each of your 13 rows, and make a table for each gear.

What you got now is a table showing wheel torque vs wheelspeed for each gear. The thing you'll notice is that (obviously) 1st gear will run from i.e. 1 m/s till 15 m/s, whilst 5th gear will run from maybe 20 up till 80. Now use excel to create a plot that has speed on the x-axis, and wheel torque on the y-axis.

You'll get something like this:


Now each crossing of subsequent curves is a optimum shift point (disregarding differences in gear efficiencies and shift-time).