Two IC Engines: Comparing Torque Curves

[B565]

Hello there , hopefully someone can shed some light on this for me . let's say you have two IC engines . all variables are the same . Engine A has a completely flat torque curve at 500’ lbs to 550’ lbs . Engine B has a torque curve that starts out at 450’lbs but gradually makes its way to 650’ lbs. What engine is going to accelerate faster from a complete stop ? And why ?

 

[Baluncore]

Welcome to PF.
Initially, the 500'lb engine will accelerate faster from zero, because the 450'lb engine has less torque.
You have not specified the RPM at which the torque crossover occurs, or the time for the comparisson.

 

[sophiecentaur]

What engine is going to accelerate faster from a complete stop

I always have a problem with this sort of question because a true comparison is only possible if they both have 'ideal' (continuous ratio) gearboxes. Under those conditions, (ignoring the first few hundred RPM) they can both be made to develop their maximum Power. After all, it's Power times time that gives the amount of Kinetic Energy the car acquires. I imagine that car propulsion techniques are very 'practical' with pragmatic design so the OP assumes certain things about the situation.

Torque is very important in practice but, without knowing more about those two engines and transmissions, you can't be sure of an answer to this one.

PS Torque: I am not sure what a 'flat torque curve' means - except to imply that there are no actual departures from a torque curve with constant slope (not horizontal - aka flat) over those speeds. I realize that torque has what you feel as pressure from the back of the seat but I don't see how constant acceleration gives the fastest mean speed increases. At the lower speed, there should be more torque available if the gearing is doing all it can.

 

[B565]

Thank you for the great information.

 

[B565]

Ok , let me try to rewrite this question, 2 IC engines same weight car , transmission and gearing. Everything the same . Car A starts off at 400lbs of torque at 4000 rpm and goes to 410lbs of torque in .300 of a second. Car B starts off at 400lbs of torque at 4000 rpm and goes to 525lbs of torque in .300 of a second. Which car accelerates faster in .300 of a second ? Both have the same fluid torque converter & leave upon release of a transbrake valvebody.
Thanks

 

[jack action]

So Car A & B both start at the same RPM & torque - therefore also the same power - and one increases torque faster than the other. Here comes your first glitch in your "everything the same" statement: Either both engines increase their RPM at the same rate - meaning the one with the highest torque has more power - or both engines have the same power - meaning that the one with the highest torque has a lower RPM. It is impossible that everything stays the same when changing the torque.

It is meaningless to try to link torque & acceleration in a moving vehicle, within a given velocity range. For more info: When Vehicle Power Dictates Acceleration

 

[sophiecentaur]

I understand that you have been subjected, via 'the trade' to many engineering inaccuracies. It has confused you. To get to grips with this, you need to unlearn quite a lot so that you can distinguish between fact and BS.

and one increases torque faster than the other

How can that be happening?

Which car accelerates faster in .300 of a second ? Both have the same fluid torque converter

I don't understand your scenario. If both the torque and speed are increasing, then the engine output power must be increasing. With optimal gearing, the engine would be operating at maximum power all the time so your model is not optimal or even properly described (afaics) because an increase in
The acceleration will mean an increase in kinetic energy per second and that should be the power developed. You should be using the right units for torque (lbs,ft)
If the engine is working optimally then it will be developing its maximum power at all times (torque times revs). Your figures of (400 X 4000) and (525 X 4000) tells you that the second car would have higher acceleration because there's more power available.

It's true that it's the results of torque that you 'feel' (pressure on your back from the seat) but it's always down to the actual power output of the engine. The gearbox does its best to make sure that the engine is running at the optimum speed so it can deliver its power at the optimum revs. The torque is actually an intermediate quantity.

 

[Baluncore]

Since everything else is defined as being identical, the only explanation is that the two IC engines must have different power curves.
Maybe car B had greater acceleration, or maybe the car B handbrake was still on.

 

[jack action]

and one increases torque faster than the other

How can that be happening?

Very easily. Take this extreme case:

• Engine A's RPM goes up and torque goes down, and car accelerate;
• Engine B's RPM goes down and torque goes up, but car decelerate.

You assume both RPM increases at the same rate, but it is not really specified. The end point of my post is that it is impossible that everything else stays the same when you change the torque. Your assumptions are good as anyone else's.

If both the torque and speed are increasing,

Nobody stated the speed is increasing. This is the problem with the question.

Going back to the question in post #5, let's imagine that the RPM stays the same in both cases (isn't that the hypothesis of the question, everything the same?), then if the torque increases to any value, the acceleration will be zero in any case, because the RPM did not increase. But, of course, this can only happen if something changes, like a gust of wind increasing the aerodynamic drag. The moral of the story is still that you cannot change the torque without changing something else.

We can play this game as long as we want, because linking acceleration directly to torque in this case is pointless. Acceleration can only be linked to power output and mass (inertia), for a given velocity.

 

[sophiecentaur]

Nobody stated the speed is increasing.

I understood that from the following:

Car A starts off at 400lbs of torque at 4000 rpm and goes to 410lbs of torque in .300 of a second. Car B starts off at 400lbs of torque at 4000 rpm and goes to 525lbs of torque in .300 of a second. Which car accelerates faster in .300 of a second ?

Something I have a problem with, in this sort of question is that it's always numbers and not formulae. The above statement implies that the throttle pedal is pressed harder (??) so does that happen without any increase in speed? Can we do any actual calculations without knowing the increase in KE? I see the problem here as insufficient data and imprecise description.
I think the handbrake comment further up the thread must be relevant.

 

[Lnewqban]

Ok , let me try to rewrite this question, 2 IC engines same weight car , transmission and gearing. Everything the same . Car A starts off at 400lbs of torque at 4000 rpm and goes to 410lbs of torque in .300 of a second. Car B starts off at 400lbs of torque at 4000 rpm and goes to 525lbs of torque in .300 of a second. Which car accelerates faster in .300 of a second ? Both have the same fluid torque converter & leave upon release of a transbrake valvebody.
Thanks

All this happens in around 20 turns of the crankshafts.
What makes engine B increase its output torque 31% during the same short period of time as engine A increases its torque only 2.5% in 20 turns?

In my humble opinion, that is only possible if there is big difference in how open each throttle is.
If that is true, engine B is sucking much more air/fuel mix, which increases its mean combustion pressure, which increases its delivered torque, which increases the forward force of car B’s wheels.

As both cars have identical masses, transmission reduction rates, etc., car B would have burned more fuel and pushing forward a little harder and increased its velocity over car A a little.
There is not much delta-acceleration to measure in 0.300 of a second; perhaps a good portion of the initial extra energy is used in deforming the tires and heating converter or clutch.