## Gearing Question

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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.
Fred444
0

Joined: 14 Sep 2011

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.
spacer
1

Joined: 1 Nov 2009

Wow, thats an awesome reply. Thanks!!!

EDITED - I was confused about BHP and torque, but now understand that bit.
Last edited by Fred444 on Wed Jan 04, 2012 7:57 pm, edited 1 time in total.
Fred444
0

Joined: 14 Sep 2011

These are my torque figures...

Code: Select all
`RPM   Lbft2000   2602250   2802500   2902750   2803000   2653250   2503500   2453750   2404000   2304250   2104500   1904750   1705000   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 62000   980   546   343   255   218.4   1952250   1056   588   370   274   235.2   2102500   1093   609   383   284   243.6   217.52750   1056   588   370   274   235.2   2103000   999   557   350   260   222.6   198.753250   943   525   330   245   210   187.53500   924   515   323   240   205.8   183.753750   905   504   317   235   201.6   1804000   867   483   304   225   193.2   172.54250   792   441   277   206   176.4   157.54500   716   399   251   186   159.6   142.54750   641   357   224   167   142.8   127.55000   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??
Fred444
0

Joined: 14 Sep 2011

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).
spacer
1

Joined: 1 Nov 2009