torque vs hp for 360 replacement

There must be something magic going on inside the CVT.

 

Maybe Harry Potter designed the original CVT.


I think everyone else here keeps forgetting:

Maximum acceleration in a given gear occurs at peak tq.
Maximum accleration at a given speed occurs at peak hp.

A CVT allows the effective gearing at any given speed to match PEAK HP, NOT PEAK TQ to obtain max acceleration.

 

QUOTE]

This statement is true for CVT drivelines and not true for gear driven transmsiions. Since the cars we are talking about are the later, the sentance is not relevent to the discussion at hand.[/QUOTE]


And as we have transmissions with more and more gears, i.e Mercedes 7 speeds, that more closely approach a CVT. Where then does the magical line cross?

I call major BS on your statements and ask this simple question? Why haven't you taken me up on my previous offer and gotten your free Modena?

To restate the offer more clearly:

A simple challenge to you: If you can get a reputable physicist to truthfully deny this statement I will GIVE you my Modena:

Knowing nothing more than the engines maximum horsepower and the weight of the car, you can calculate the maximum potential acceleration, 1/4 mile e.t. and trap speed of a Ferrari Modena (or any other car).
Knowing nothing more than the engines maximum torque and the weight, you CAN NOT.

P.S. I always believe in someone who puts his money where his mouth is. Are you willing to do the same?

 

The first statement is ture, the second is misleading.

I took a previous thrust diagram and added a single verticle line in orange at 60 MPH. This line is above the rev range of 1st gear, in the meat of the powerband of 2nd gear and at peak TQ in 3rd gear. It is easy to see that of the 6 gear ratio choices, that 2nd gear delivers the most thrust. 1st gear leads to engine destruction, and 3rd through 6th gears have noncompetitive thrust. This a completely Torque based acceleration statement!

Maximum acceleration occurs in that gear which has higher/highest HP at any chosen speed (even if that speed is nowhere close to peak HP)! This is also a torque based statement, it just hides TQ as HP = TQ*RPMs/5252.

It should also be clear that above the RPM of peak HP you are accelerating fster than if you were the next higher gear--as long as your engine can handle the revs.

 

Audi did offer a CVT gearbox a couple of years ago, at least here in Europe. I think it was introduced in 2000 or 2001. Do a search on google for Audi CVT and you will find info about it.

 

Your charts are a little hard to read and really not necessary for the basic discussion. Simple pure physics doesn't need them, although they can be interesting.

I'll expound a little on the statement I made that you think is misleading:

At any given speed, a car will accelerate hardest if it is geared to be at peak engine hp at that speed, NOT peak engine tq. Maybe you don't realize that at that peak hp, the engine will be turning more rpm's than it would at peak tq. Through the wonders of gearboxes and torque multiplication you will therefore be developing greater tq at the rear wheels.

By knowing the engines peak hp and tq and at what rpm's, and the gear ratios, you can do simple calculations to derive the tq and rpm at the drive wheels. Play with the variables all day long and you will always get the same answer. Peak acceleration at any given car speed occurs at peak engine hp, not tq.

Big torque numbers at low rpm just mean relatively greater hp at low rpm because, again, torque is merely a COMPONENT of HP.

You can generate massive tq all day and do no work whatsoever. Just stand there on your two foot breaker bar on the lug nut of your car wheel and you can develp 400 lbs ft of torque for hours on end with absolutey nothing happening ,while you sip your tea. Apply 400 hp to the wheel and I gaurantee you something big will happen and your tea will be spilled.

Torque in and of itself does not imply that ANY work at all is being done, only when it is combined with rpm (another component of hp) does it mean ANYTHING regarding the acceleration of a car.

Again, why haven't you taken me up on my offer?

 

Mr Janne,

I have already corrected my mistake ... read back a few posts, and I did a search and found a site that discussed CVTs. What I find interesting is that Audi's own site makes no mention of it? ... which is where I originally went.

It is important with these threads that you read the whole story otherwise we keep going round in circles

Pete

 

After all this, Im still confused as to why Audi would program their CVT to spin to power peak rather than torque peak if it would make the car slower.

 

I'm done with this thread. I appreciate that everyone has remained civil, that in itself speaks volumes for the people involved in the conversation.

My offer remains open. And if any one of you truly thinks I'm wrong about it you will look awfully silly if you don't take me up on it. If you're correct you will have yourself a Modena, If not I'll be driving it for a long time, and you will all look the fools.

 

It seems to me that thats what I recall about torque too....there is no specified rate associated with it. When we are talking about moving a certain mass from point A to point B, there is going to be a certain amount of time associated with that process. Thus, a rate of doing work, also known as power, seems to be what is significant.

Moving the 3000 lb car 1320 ft takes the same amount of work regardless of how long it takes, whether that be 10 seconds or 10 hours. Doing that work in 10 seconds will certainly require a larger power input however.

 

Wrong a gearbox does not turn power into torque, it multiples the torque that the engine produces.

BTW: Power is a factor of torque not the other way around, remember:

HP = T*RPM*5252

Also what drives the car is a force ... whether you like it or not. It does not matter about time, as that force is there and constantly changes, etc. but it is the force that counter acts the resistences and makes the vehicle move, ie:

If F (with the car) = F (against the car) we have constant velocity.

Thus when the car is stationary we have a balanced force diagram, and the car has no force pushing, and none against and we have the constant velocity of 0 mph.

When the car is cruising at 100 mph the engine is providing enough force to equal the resistance forces to enable the object (vehicle) to maintain that constant velocity. Again time is not a factor, it could be for 1 week, or 1 second, or 0.000000000000000001 of a second.

To accelerate a car, ie. increase velocity over time, F (with the car) must be greater than F (against the car). Acceleration is measured as an instantaneous value ... as again it is (especially in the case of a car) hardly ever constant, due to increasing resistence, etc. and other REAL world values.

Again remember that acceleration is an instantaneous measurement ... while to you keep forcing on time and work, etc.

Quite simply at time X the engine produces Tx, that gets multiplied to the rear wheels to get Txr, and thus produces Fx at the ground for that instantaneous snap shot of time.

Now going back to your challenge. Just because you have a lovely little formula that approximates the 1/4 mile time using power, does not mean that torque does not provide acceleration. Remember Power is Torque multiplied by rpm and thus time ... and as you are interested in the elapsed 1/4 mile time ... then you need to calculate Torque and add in time. Thus if you took your lovely little formula and extraporated it you could get back to the Torque component.

Also regarding the misleading statement:

Thus look at Mitches maths:

[quoteE= 1/2*m*v**2
dE/dt = 1/2*m*(dv**2/dt)
dv**2/dt = 2*v*dv/dt
dv/dt = a
dE/dt = P
Therefore P = 1/2*2*m*a = m*v*a
Therefore P/v = m*a
but since f = m*a and P/v = m*a then
P/v = f = m*a[/quote]

The end formula makes that statement appear correct, because we have:

P/V = m*a

But what we really have is Power divided by Velocity equaling Force. Divide Power by Velocity (thus a component of time) and you return to an instantaneous measurement, which will be a value based on Torque.

Pete

 

Anyone have an answer to this?????

 

Acceleration is an instantaneous value in G's. We are not talking about how long it takes to get to say 60 mph ... we are talking about acceleration RATES.

Different kettle of fish, and I can see where the confusion exists.

Your car taking 5 seconds to get to 60 mph, has involved many different acceleration rates. When it first launched the acceleration would have been higher and then it would have tapered off as it closed in on 60 mph.

Thus thinking about it as a 5 seconds to 60 mph is the AVERAGE acceleration over time, not the REAL acceleration rate.

Pete

 

But, I thought average acceleration rate was what we were trying to maximize!

 

I see this will be make the answer to this debate ... still thinking about it though

Pete

 

No.

The question is: Where is the maximum acceleration point in accelerating a car in a particular (constant) gear?

This is not the average, but the theoretical instantaneous acceleration, and this is at peak torque. It might be only by 0.01 of a G, but it is there, not at peak power.

Thus put a G force indicator in your car and the maximum value will be recorded for a split second as the engine passes peak torque. As you drive a Ferrari it will maintain a high value right up to peak power. If you drove my Toyota people move you would see it tail off very quickly AFTER peak power.

I think we are not thinking apples with apples. Some of us are being theoretical, others are thinking of 0 to 60, or 1/4 mile times ... big difference.

Pete

 

I have never indicated otherwise. Your statement writen in the form of TQ:

Peak acceleration at any given speed does occur at the highest TQ multiplication ratio of the gear box that does not exceed the RPM limits of the engine.

See, you can say the same thing and only use TQ and end up with the same answer. Except this is much broader than the HP version of the same statement and can be used to show why big TQ engines are faster than little TQ engines of the same power, and why CVTransmissions are more efficient than gear box transmsisions. Your statement does not extend to answering these issues, TQ does.

The difference is that TQ is still sensical when the object to be accelerated is at rest (V=0) whereas for HP this is not the case (division by zero troubles). Therefore, I submit that since both TQ and HP can be used most places, but TQ can be used in all places (even when V=0), that TQ is really the correct starting point for measurements of acceleration and work. The equations of yesterday also indicate the completeness of this logic. If there is a flaw it will be found in the equations, otherwise, you simply remain wrong or incomplete at best.

Here is the point of confusion: HP is a component of TQ not the other way around!

Here you are having the P/V = m*a when V=0 problem. HP at a standstill does nothing either!

Can you explain how a gasoline engine produces TQ without spinning at any RPMs? In other words, TQ and RPMs are inseprable in automotive engines, and thereby, TQ causes acceleration to happen, HP is just the measurement of what happened, TQ caused it to happen.

TQ does the work, HP gets the credit..............

 

It certainly will!

And, it will also prove all the manufacturers who use CVTs wrong! They should have "stalled" the CVT to torque peak and not power peak.

A little birdy tells me that they are not that stupid

 

Actually I do not expect to prove that. What I do instead is to prove why the rev past peak torque, and how CVTs work.

There might be other reasons that they rev passed peak torque ... like the balanced engine thingy, ie. if the car hits extra resistence and slows it will slow the engine to peak torque and thus fight against this extra resistence. If they sat it at peak torque, then if say you got a huge blast of wind (from a passing truck), etc. the car would slow dramatically as it would drop below peak torque and have to struggle back up. Thus I am thinking it has to do with efficiencies and economy, as lets face it the fastest acceleration figure is NOT the manufacturers prime concern, is it?

Pete

 

A CVT operating at peak HP accelerates faster than operating anywhere else! It operates as if it were accelerating as the black line on the thrust curves (the 1/V line). That this happens is because the gear ratio is constantly changing. Under constantly changing gear ratios, it actually is better to operate at peak HP than anywhere else (for accelerative purposes).

Since the gear ratio is constantly changing, it interacts with the calculus in the equations as another variable that needs to be differentiated, and thus, alters the resulting differential equations, and therby, it changes the computation of maximum acceleration.

This is unique to the CVTs

 

There are numerous articles on the net which discuss this with a good amount of detail and Im sure their words will be better than mine.

Heres one of my favorites--

http://www.autospeed.com/cms/A_0755/article.html

 

Coudn't resist one more post. This is the silliest thing you've ever written Mitch, you should work for Saturday Night Live.

READ THE LINK ON THE PREVIOUS POST BY 4SFED 4

ALSO....STOP POSTING AND TAKE ME UP ON MY OFFER!

 

Yes and again if you are interested in average accelerations, time comes into it. Notice how car A (on your link) has actually the highest acceleration figure ... but only for a short time. This is because it has the most torque at the rear wheels. Thus if you took an instantaneous value Car A would win ... but in the real world as it has no power over time it looses.

Again instantaneous acceleration versus average acceleration (or lets say maintaining acceleration). To maintain acceleration you need power as you need to be able to rev out in that same gear.

Thus it is still 100% true that the maximum acceleration figure a car will ever see will be in first gear at peak torque, as the link clearly shows.

Again we are talking different things. I am not disagreeing with you at all that to maintain acceleration you need power. BUT F = m*a is all about instantenous values ...

I am not sure how to explain what I am saying better than what I have already said but one more try:

1. The most G forces an accelerating car will ever see will be in first gear (due to maximum torque multiplication) and at peak torque ... for a split second.
2. To obtain quick 0 to 60 times (assuming you stay in the same first gear) you need high torque to get going and high power to be able to rev out in that first gear.
3. To obtain quick overall acceleration (using all the gears available) you need high torque to get going, and high power to be able to rev out in each gear ... so the torque multiplication is the most effective. Obvously if you cannot reach past peak torque, when you change up you will be way below peak torque and the engine will struggle to accelerate the car.

Thus my definition of power is the engines ability to maintain an acceptable torque level as it revs to its mechanical limits. Remember all dynos measure torque and a flat torque curve demonstrates the engines ability to continue to make torque as it revs out.

Pete
ps: Take this comment from that site:

Why would you change the final drive ratio to compensate for a lack of torque? ... to increase the torque multiplication! Why are we even interested in the lack of torque? ... you guys have said torque does not matter, as we have all this power ... because torque is what provides the force that accelerates the car. This is why a high revving powerful engine (remember my definition of power, the engines ability to produce torque at high rpm) is the best as you can gear it to take advantage of the large rev range.

This is also why trucks are so slow, because they have a very small working rev range. If a truck could rev out ... you would be wasted by a 70 tonne truck ... er, maybe not

 

Thanks for providing the link ... explains it nice and clearly.

I guess the reason a CVT revs out to peak power is it is still compensating for a lack of torque and the CVT transmission as a whole is still a gear ratio ... just a constantly changing one

Pete

 

Thus, at any given speed, the power peak provides the most acceleration

It works out that way since (usually) the torque drop off (%-wise) from the rpm where torque peaks and the rpm where power peaks is alot less than the gain one gets from the muliplication of not having to change gears. So, you come out ahead "planting" yourself at the power peak.

Heres my mindset---

Assume a sample engine makes 200 ft lb at 3500 rpm (torque peak) and 175 ft lb at the power peak of 7000 rpm. Thus, there is 200 hp at 7000 rpm. Torque at the crank dropped 25 ft lb, or 12.5% from it peak value to its value at the power peak.

Now...look at transmission ratios. Lets ignore tire size and FD ratio since they are fixed and dont change with each gear (generally). Lets say that 1st gear is 4:1, and 2nd gear is 2:1. and 3rd gear is 1:1 (just to pick easy numbers).

Shifting gears from 1-2 costs a 50% reduction in wheel torque. Revving out to the power peak only costs 12.5% reduction in wheel torque. Thus, you want to hold that gear as long as possible (i.e. until there would be more power available in the next gear.)

In fact, that is a standard test on a dyno. One runs the car through the gears to redline in each. The curves are then plotted as hp versus speed. Wherever the curves cross is the ideal shift point. If they dont cross, shifting at redline is best.