Sure, Along with tire diameter and the effects of drivetrain loss.

I never said that. All I said was that TQ accelerates the car. Gears don't apply force to the tires on their own - they just multiply (or divide) the input force (engine torque).

Agreed - this is what HP is all about, the rate at which you apply a force.

I didn't think it worked that way. Acceleration is derived from Force/Mass. Now since the force in question is the TQ produced by the engine, please explain to me how either of these two scenarios will lead to greater acceleration then the other - I believe they will be equal as long as you don't consider any TQ values outside of this 1000 rpm range.

I still believe that acceleration is dependent on force applied, and not power generated. A motor cannot increase the rate at which the force is applied without applying the force (torque). The TQ is what allows the motor to spin faster and faster, thus "generating" more HP in the process.

 

 

Correct me if I am wrong, but there is an old saying "hp wins races".

I was wondering this... IF RPM is revolutions in one minute...

The higher HP average will always win (given that it is in the shifting range)

Why? My thought... 310ft-lbs @ 4000rpm is not as much "total torque" to the ground as 250@6000rpm... Doesn't that mean that the latter is pounding the floor with a force of 250 6000 times per minute? That's kind of what a hammer drill does now does it? Ultra sound breaking etc??? Yet if you put an S2000 with 240HP on a camaro it will run like crap, simply to heavy. Or even worse, put a LT1 into a MACK or something like that...
I always thought and felt that torque was the way to go to move mass but its a combination of both???

More torque with the same HP will yield better or quicker results since you will get to your peak HP quicker than with lower torque... I think... Correct me, this has been a never ending discussion..

 

The distinction I was trying to make was between flywheel and "delivered" torque, or flywheel * gear multiplication. Since gear multiplication *isn't* constant (either shifting gears or changing your rear end gears) Flywheel torque isn't the only issue here. In the context you used I assumed you were talking about flywheel torque (as I had seen nothing to indicate otherwise).





Okay, so we have 2 engines, both have the same area under the torque curve, but the second engine has greater area under the hp curve. Say

Engine (1) 4000-5000 rpm goes from 450ft-lbs to 400ft-lbs of torque, linearlly

Engine (2) 4000-5000 rpm goes from 400ft-lbs to 450ft-lbs, linearly.


Both have the same area under the torque curve, but engine 2 has more area under the hp curve.

Now to avoid calculus lets simplify this to a single point - one engine that makes, say, 425ft-lbs at 4000 rpm and one that makes 425ft-lbs at 5000 frpm. They both have the same torque, only hp is different.

Now let's look at gearing. Let's say at an arbitrary linear velociy engine 1 was at a 1:1 final ration, so it was giving the tire 425 * 1, or 425 ft-lbs of torque.

Now engine two, at the same velocity, will need a 1.25 :1 gear ratio to be at it's rpm point (both putting out the same torque). So to the rear wheel it will be giving out 425 * 1.25, or 531.25 ft-lbs.

Since, as you point out, Acceleration = Force / Mass, one can see the second instant will result in a greater instantaneous acceleration.

Now take a look at our original problem - for each "slice" of the rpm band you could perform the same equation, and then average all the accelerations. Our torque terms will be identical, but since in engine two the torque will occur at a higher rpm point the gear multiplication will be greater, so the delivered torque will be greater.




No, because torque isn't power - it's the rotational analouge of force. The higher hp car would "out-power" the lower hp car.

Now a lower *torque* car can out accelerate a higher torque car - given that we are talking about flywheel torque. See my example above.




You mean HP curve, right? In my heads+cam setup I shifted at 6800 in first and dropped down to about 5700 or so. My torque peaked at 4800 and had long since been dropping. If I shifted to hit the rising portion of my torque peak I would be shifting at 5500 or so - and it should go without saying that that would slow me down .




If you built a peaky vehicle then I would argue you have not properly definted your operating range (if a peaky vehicle isn't what you wanted). If you drop down to a low power value after the shift, then you either need tighter gearing, or to re-think your "operating range".

(and again, it's hp that is the issue when determining acceleration of a vehicle, not torque unless we are limited to only 1 gear and can't change the rear end ratio).




That is to a certain extent the point and the idea. You then use gearing to take advantage of the greater rate at which you produce torque (even if it's less flywheel torque) - and so the value that actually moves your car is the product of flywheel torque * gear multiplication.


Gear multiplication is going to be dependent on the rpm you produce the torque (engine velocity) - so an analogue would be acceleration is proportional to torque * engine velocity.

You might note that it can also be written that

power = force * velocity, or

horsepower = force * velocity

so here horsepower is directly proportional to acceleration for the vehicle system.


Chris


Chris

 

OK this is starting to get a little beyond me.... but please confirm this.

Suppose the HP of a vehicle holds steady from 5250 rpm to 10500 rpm at 200 hp. This would imply that by 10000rpm, the TQ to the ground would have halved from 200 lb/ft at 5250 rpm to 100 lb/ft at 10500 (since RPM has doubled, TQ must be halved to get the same HP).

You are saying that the car will accelerate just as hard at 5k rpm as 10k rpm even though the TQ has halved because the HP is the same? We can ignore all gearing issues since no shifting is occuring, and let's forget about drivetrain loss for this one. Obviously the force has halved since the TQ has halved. Would you say that the RPM has doubled, thus the rate at which the force is applied to the ground has doubled, therefore maintaining acceleration?

And Highlander, the saying is that "Horsepower sells cars, Torque wins races" - but if what Chris is saying is true, then Horsepower should win races too

I'm glad this hasn't turned into a flame fest - this is interesting stuff

 

That is my point... Torque is dependant on cylinder pressure, you will get peak torque when you get peak cylinder pressures...

Cylinder pressure is dependant on fuel, air, spark and explosion. It should accelerate as quick even torque has gone down since at 10krpm the engine will pass through it a lot more cycles of fuel, air, spark and explosion although that mixture is not as powerfull as before...

THat is how I see it and get it... You need torque to move mass... That is for certain.. and even if you have lots of HP if there is no torque behind it, it will be hard to move the car and it will take longer to get to HP peak..

 

Draco: Yep - pretty much! Take a second and look at the "ignore gearing" part though - if both vehicles are at the same roadspeed then the one at 10500rpm will actually be at a steeper rear end ratio - so though he produces half the torque he will have twice the gear multiplication (vs. the 5250 vehicle) - so in effect the same torque will be applied to the rear wheels.

On you same vehcile that made 200hp from 5250 to 10500 lets look at random points. Say 5250

he makes 200hp at 5250 so that is 125ft-lbs of torque. Let's see he is currently at a 1:1 ratio at a given velocity X. That is 125ft-lbs to the rear wheel. Look at the 10500 case. He will have to be at twice the gear ratio to have the same rpm at the same roadspeed, so though he only makes 1/2 the torque (62.5 ft-lbs) it is multiplied by a 2:1 drivetrain ratio, so he ends up with 125 ft-lbs delivered also.

Your acceleration = delivered torque = flywheel torque * drivetrain ratio.

If we are keeping hp constant then flywheel will vary as a function of rpm, so if we say the hp is a constant, y, then torque can be written as HP*5250/RPM = torque.

So acceleration = delivered torque = HP * 5250/RPM * gear mutiplication.


Now let's determine our gear multiplication. If we say it has to be at 1:1 for a given arbitrary speed at 5250 rpm, then the actual gearing would be a function of the RPM at which the hp was made, or (RPM / 5250 * 1) : 1 gear ratio, or just RPM / 5250. So if we substitute that in for the gear multiplication term we get


acceleratioon = delivered torque = HP * 5250/RPM * RPM / 5250

the 5250 and rpm terms cancel out and so you get

acceleration = delivered torque = HP

Now admittedly this is only a limited case, and I convienently ignored units, but the proportionality still holds.

If we had only 1 gear and no ability to change it then acceleration would be directly related to the torque curve - but since this is not the case, acceleration will be determined more closely by the hp curve. The more ratio's you have, etc. the more closely acceleration will mimic the hp curve. With a CVT transmission, ignoring any mechanical losses, best performance would come from maintaining a constant max hp rpm point.


Chris

 

Hold on a sec, I wasn't trying to describe two different cars moving at the same speed where different gearing can be used to equalize acceleration... I was picturing a single car accelerating from 5250 RPM to 10500 rpm. I am arguing that as the car accelerates from 5250 to 10500, if the HP stays the same, then the TQ is halved by 10500rpm. Since the TQ is halved, I believe that even though the HP is the same at 10500 rpm, the acceleration will drop to half what it was at 5250 rpm. Therefore HP cannot be used to determine acceleration without factoring in RPM (thus you end up at Torque again). Again, I am ignoring gearing since this is 1 car staying in 1 gear the whole time.

Maybe we are saying the same thing here and dancing in circles. Or maybe I just don't get it

 

 

wow this has turned into some interesting stuff. im glad i took cal II before leaving school.

lets concentrate on one gearing. 3.42s on a six speed car. sure you can play with the gearing to help out, but im running that gear and will most likely not change unless i can get 3.73s.

so with that in mind the car that has the higher rwtq average will be faster even if they have the same rwhp averages correct? after all torque is what moves you. this is also considering they have similar torque curves. you could have two averages the same but no torque up top on one and torque all the way across for the other. the one with it all the way across will be faster.

thanks grygst for the specific info. and good luck. Trey

 

If you look at the oldsmobile motors of the late 60's and 70's they had like nothing for horse but double the torque!! They were bad a$$ for their time.. So I am one who agrees that torque is king but horse follows it to a degree.. stroker motors also follow the torque rule since they produce a lot of it when you increase the stroke. I could have 900 horses coming out of a corner or going thru a strip but if you had more torque you would wipe me out in less then a few seconds!!

 

But think about it like this - to go down the track you have to accelerate over discrete intervals - so both cars would accelerate over the 0-1mph interval, 1-2mph interval, 2-3mph interval, etc.

From the example above I have proven that at any instantaneous velocity the higher hp setup can produce greater acceleration.

Now take the above reasoning and reduce the trip down the strip, etc. to a series of discrete velocities. At each point the higher hp car will have greater acceleration, so over the entire interval it will have greater acceleration overall.








Mostly correct - I did point out previously that if gearing is fixed and both cars accelerate then acceleration will follow the torque curve directly. As to your example if you don't change the rear end gearing (like I did) then you are forcing the higher hp car to accelerate over a range that begins at twice the velocity - not really a apples to oranges comparison.

Think about it this way though - both cars start from a stop, with totally fixed gearing - the car that makes more torque lower in the rpm range will *also* be putting down more hp since you have artifically limited the rpm range by giving static gearing.

So again, the higher hp car wins - even though the other car can make the same hp, it looses because it isn't in the rpm range where it makes that power.

Now if you allow gearing optimization so they both are in an rpm range where they put down the same horsepower then they will both accelerate the same (over the same velocities).

Saying a car is a high hp engine doesn't mean anything if it isn't working in the rpm range where it makes that hp - the engine is only "worth" as much hp as it is currently producing.



Let's look at a 1/4 mile, where Vi = 0mph, and velocity = your trap speed

Work will equal the final KE since Vi = 0, so KEi = 0.

Work = Force * Distance

Force = Mass * Acceleration

Work = Mass * Acceleration * Distance

Acceleration = Work / (Mass * Distance)

Power = Work/Time

Acceleration = Power / (Mass * Distance * Time)

So just taking average power, mass, distance, and time you can calculate average acceleration. Note no RPM information above. You can derive for different terms, but this was the simplest one that met your requirements.


Note that you can't do the same with torque only


I would still re-iterate that you need to look at it from the perspective of delivered torque, or flywheel torque * gear multiplication. If you are "torquing" the rear wheels with more torque, you will be accelerating faster (assuming traction). Whether the magnitude of that torque is derived from low gear multiplication and high flywheel torque or low flywheel torque and high gear multiplications is irrelevant - it is just the magnitude of that value that matters. The HP function corresponds to the product of gear multiplication and flywheel torque - as it goes up so does the potential for this value.

Chris

 

Hmmm,

This is a interesting thread because it leaves me in a spot where I agree and disagree with the Old man and Chris.

I really don't think this is a thing about just TQ or just HP, it's actually both. In racing they teach you to shift the car in the spot where you have the most average TQ+HP, which at different shift points will follow the law of dimishing returns. That's why there is a point at which you can shift at too high of a RPM because the RPM span that you are using is not the optimum for average HP+TQ.

Now drag racers have always wanted more gear, and more RPM. More RPM allows more gear which usually means a harder launch, which results in a faster ET (if the power can be put down) but there will still be a given MPH for the car, because there is a given amount of HP and a given amount ot weight. Think about that for a sec, how many times to do see a guy gain MPH when swapping gears say from 3.42 to 4.10 (unless he is going thru the traps at way to low of a RPM) but he will pick up ET. Gears also pay into the law of diminishing returns.

How about the actual rear wheel numbers? The efficentcy of the gears has alot to do with the rwhp number right? Most times you see a guy drop rwhp numbers when he installs a numericaly higher gear.

How about another place where gears play an important role. Cirlce track racing. Most saturday night classes have a given carb, which pretty much limits your RPM range for them.(but all motors have something that limits them from making more HP at a higher RPM) So say you have a motor that can't really spin past 7000rpm without the HP peak dropping off pretty bad. So say at peak I have 450hp @ 6800, and @ 7000 I have 430, 7200 I have 390hp. Now a chart would really helps this, but you get the point. Will the car still accelerate at 7200 even though it's past it's HP peak and down about 60hp? Yep it will, that's because the HP required to accellerate the car is less than the amount produced by the engine and gearing. The reason a Geo Metro can even move at all with 50hp is that it doesn't take that much to move the car, we accelerate faster because we have alot more HP/TQ than what is needed to move our cars. All you have to do is have more Hp/TQ than what is needed to maitain that speed to accelerate the car. The important thing to look at is the value of the acceleration. The accelleration rate in the above example will be lower than it was @ 6800 vs the 7200rpm, but it will still be postive. If guys put too much gear in a car then the engine will always be way to high in the TQ/HP curve and have a slower average accelleration rate. There is an example on how this can happen farther down.

Gear really isin't used to multiply the the TQ/HP, even though that's what it does. (I understand that's confusing) Gear is used to pick the RPM range that you want to run the car in. I have an example to help explain this, it's out of my head since the Skip Barber racing book that it came from is at my buddies house. At least that's the way I look at it, and I think Old SStroker is the same way. (BTW thanks for that book OS)

Here is your engine's HP curve. (it's only HP because I don't want to convert the HP numbers to TQ)

4500 360
4750 370
5000 385
5250 400
5500 415
5750 430
6000 435
6250 440
6500 445
6750 430
7000 395
7250 365

Now let's say all of these have the same gear ratio and we need to find the best shift point.

Our tranny when we shift drops us 1500rpm, so we have to find the best 1500rpm to run it in.

if we shift at HP peak of 6500 then we have the range from 5000 to 6500. That gives us a average power of 421.4.

if we shift at 6750, we have a average of 427.8.

if we shift at 7000, we have a average or 427.1.

if we shift at 7250, we have a average of 420.0.

So shifting at HP peak would be wrong here, because it doesn't have the highest average power. Going slightly past it gives us the best average power for the car, if we use the 5250 thru 6750 RPM band. That should be where we shift at.

Now let's throw the rear gear in this equation now. To make it simple we will say it's a cirlce track car, and we only use one gear.

I can choose from three different gears. (I'm going to leave numbers out of this for now, because it's late) But like the tranny in your F-body each one will will have a different speed range for the same RPM range. i.e. 1st will go from 0-45mph in a 0-6000rpm band, while 2nd will go from 0-70mph in that same RPM band.

Now on a cirlce track I exit the corner at 45mph and accelerate to 100mph before the next corner. Now each rear gear I get to choose from will have a RPM band that I have to use. The numerically higher the gear (4.10 vs a 3.42) will require that I work in a wider RPM range than a numerically lower gear. So the 4.10 will need 2500rpm to go from 45-100mph, vs the 3.42 will need 1500rpm, and a 3.73 will need 2000rpm.

So now we look at the HP curve above and pick the best areas to run each gear.

the 4.10's with the 2500rpm span would like 4750-7250
the 3.73's with the 2000rpm span would like 5000-7000
and the 3.42's with the 1500rpm span like 5250-6750

That gives us average power numbers for each gear of:
4.10's = 376.0 avg HP
3.73's = 419.4 avg HP
3.42's = 427.8 avg HP

So you think now that the 3.42's will be the best gear right, because they make the most average power.

Now let's throw the gear in there to get the rwHP numbers.

4.10 x 376.0 = 1541.6 avg rwhp
3.73 x 419.4 = 1564.3 avg rwhp
3.42 x 427.8 = 1463.0 avg rwhp

Looks like the 3.73's give you the most average power at the wheels.

Now this is all simplified because there are other things at work here.
-Traction is one, if you can't get the car to bite then it's not actually getting that full HP value at that RPM hence the average will be lower.
-Higher numeric gears eat up more power thru the driveline so the multiplication factor is lower.
- The rate at which the motor accelerates will effect it's HP/TQ output. The faster it accelerates the less HP/TQ it produces. You can see this on a engine dyno, but looking at chassis dyno runs in different gears will show you that the lower gears make less HP/TQ due to this (and the inefficentcy of those gears) Now taking out rotating inertia helps this, so light mass and low inertia cranks, engine internals, flywheels, dampers, driveshafts, clutches all make more power because the motor doesn't doesn't have to speed up more mass, which eats up HP. In a drag car most of that stuff just hurts launch on a clutch/flywheel setup.

ughhh, back to work I need to get some stuff done before I go home and go to bed, so hopefully I get to go to the racetrack today.

This might not all be 100% but it's the way I see it.

Bret

 

O.k a few more points to add.

A tranny will always drop a certain RPM from shift to shift if the ratios in the tranny all stay the same, so a 1500rpm drop on a 3.42 rear gear is still a 1500rpm drop.

With that being said, different motors like different TQ curves because there is different RPM ranges each one has.

A powerglide with a 1.48 1st and a 1.00 2nd (which is all it has) is different from the 2.48 1st, 1.48 2nd, 1.00 3rd and .70 4th 4L80E. Then throw in the T56 with 6 gears and you really get somewhere. the more gears (excluding overdrives) the narrower the TQ/HP range that the motor has to be built for, unless you have a stupid high stall converter.

O.k. I'm done

 

I agree with most of your saying, and I don't think any of your examples really disagree with my point that you want to maximize the area under the horsepower curve - you just went into some specifics about how that was achived.

Your first statement I would take issue with though, specifically:



How are you defining "average HP+TQ" - are you actually meaning the average of the sum of the two, or is there some sort of weighing factor, and if so what is it?

Also I can't think of any reason why you would combine the two terms, what is the basis for that?

Let's take two cars - one that makes 500hp from 4000-6000 rpm (flatline). The other makes 500hp from 8000-10000rpm.

They both come out of a corner at the same linear velocity, but the second car has twice the gearing of the first car, so it exits at 8000rpm while the first exits at 4000rpm. Their acceleration will be identical within the given rpm band since they will both be applying the same torque to the rear wheels (the second will make half the flywheel torque, but have twice the gear multiplcation)

But in that instance, however you weight it, the first car would have a *much* greater "factor" if torque is directly involved in the equation, in an average HP+torque sense, etc. Yet this doesn't lead to any additional performance, so I would submit that this isn't infact a valid metric.




This is where you get in trouble using HP/TQ interchangeably as you have been - gearing does not multiply HP at all - only torque values. Yes you use gearing to pick your rpm range, but in your example you used it based on the power curve of the car, so essentially you were using it to optimize torque multiplication - at this level it's a matter of semantics though - the gearing *does* multiply torque, allowing you to take advantage of the ability to do work at a greater rate (even if the quanta of work is smaller).


I am in total agreement with your examples and conclusions though - other than the above two statements I don't see where you are disagreeing with anything I said? In fact I noticed in your examples you used average hp to maximize acceleraiton potential, not worrying about torque at all (or average HP+TQ)



Chris