Torque curves for B18B1

ffernandez · Newbie Joined: Apr 15, 2003 Posts: 2 Location: Costa Rica

I recently found on the internet some equations to solve in order to find the optimal rpms for making each shift (1th to 2nd, 2nd to 3rd, and so on).

I have each gear ratio (Y80), the final ratio, the max HP@RMP and the max torque@RPM, BUT I NEED THE TORQUE CURVE. Could somebody help me?

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TKO · Newbie Joined: Mar 26, 2003 Posts: 14 Location: USA

Hi ffernadez,

Thats a pretty tall order. I have looked online before for dyno results, b/c I too have a B18b1 but i couldn't find one. I wish i had a torque curve for my car, but I don't. I think if you really need that info your're going to have to take your car to the dyno. Its not that expensive, well ok it kind of is but it can make all the difference. I am interested in this equation you speak of. I am always trying to go a lil faster

ffernandez · Newbie Joined: Apr 15, 2003 Posts: 2 Location: Costa Rica

This formula disregard transmission losses which is supossed to be 10%, but despite of that, it is very helpful.

Torque x overall gearing
----------------------------- = TWDS (torque on the wheel drive shaft)
Radius of driving wheels

I´ll show you the numbers of the example they use when explaining the best shif points:

Torque: 228 lb/ft @ 4800rpm
Fifth gear ratio: 0.868:1
Fourth gear ratio: 1.086:1
Final ratio: 3.440:1
Overall gearing(in 4th): 3.740:1
Overall gearing(in 5th): 2.910:1
Radius of driving wheel: 1.007 ft (assumed 1ft)

So, in 4th MaxTWDS =

228 x 3.74
------------ = 853 lb, at 4800rpm
1

They are trying to prove this is not the best shift point to 5th, because if you do so, the engine revolutions would be

2.99
--------x 4800 = 3837rpm, where the torque is 219 lb/ft
3.74

and the driving force or TWDS would drop to

219 x 2.99
------------ = 658 lb
1

and they can conclude this because they took this 219 lb/ft from this particular "engine torque curve".

Just for the records, the final result is that the ideal shift point would be around 6620 rpm where driving forces in 4th and 5th would be equal.

Hope you enjoy it and hope the text editor don´t mess the formulas.