subatomicparticle

finally someone is making sense. I may have been a little off on the 8/srt-4 comparison. Let's try a different scenario what rg is telling me is that identical cars in every way except one has 300 hp and 300 ft. lbs will have no advantage over a car with 300 hp and 50 ft. lbs. And in regards to rg's comment about not backing uti because they teach that torque is an important number. Since they are backed by porche, bmw, mercedes, audi, international, ford, cummins, etc. I don't think that they really are going to care.

zoom44

You know.. You are all a bit daft!

RX8SaxMan

Most likely the 300 ft. lbs torque car will be faster (in a drag race scenario) and that does not contradict anything RG said. See, the car with 300 torque will reach 300hp @ 5252rpm, whereas the 50 torque car will not reach 300hp until 31,512rpm. Because the powerband is so much narrower on the high torque car, it's average horsepower is higher, and therefore faster. The low torque car is going to spend quite a bit of time in low-hp territory while its revving up to 31.5k rpm. HOWEVER, if you can keep both cars in their optimal RPM ranges, there will not be a difference between them. Hence why Formula 1 cars manage to be so damn fast, despite their low torque, and also why nobody drag races them

Bottom line is, it all has to do with horsepower, not torque....just like RG has been saying.

RX8SaxMan

Just to clarify, I was assuming these theoretical engines had constant torque across the powerband. This obviously isn't realistic, but it makes the point easier to illustrate, since it seems as though some people are having a hard time understanding the dynamics of an engine.

rotarygod

If both cars make the same average horsepower within the same usable rpm range and of course are geared the same, weigh the same, etc, they will accelerate at the exact same rate of speed. You guys need to get over torque. You DO NOT need to know the word torque even exists when it comes to determining rate of acceleration. It is completely irrelevant. However...

You DO need to know torque if you are designing engine blocks, transmissions, drivetrain components, etc as this will help you design around the twisting forces that are trying to tear it all apart. Torque is a completely worthless and irrelevant value when it comes to determining how fast your car can go and how fast it can accelerate to that point but you need to know it when determining how strong stuff needs to be built. Ever wonder why a diesel engine is built so robust compared to gasoline engines of the same horsepower levels? Someone is going to say it's based on compression ratio but that's not it at all. That's a very small part of it but not nearly enough to justify how overbuilt those engines appear to be. Diesels also don't run on detonation so don't go there either. It's all about torque on engine components. The same horsepower at a lower rpm is far more twisting on the engine block and drivetrain and you need to design for this. This is why F1 engines are built out of aluminum and can be so light. They make great power but they don't produce much torque. That's where torque is a need to know number. It's not in determining acceleration. Modern diesel engines such as VW's are slowly being built lighter and lighter and out of "weaker" (for lack of a better word) materials. Why? They make less torque. How? They rev higher than they used to and their powerbands are shifted upwards as a result. There is less torque being exerted on the engine components. Many of these engines are still cast iron but the trend is changing.

Remember an 850 hp F1 engine revving up near 18K rpm makes somewhere around 250 ft lbs of torque. If they had engines that revved to only 8K rpm and made 850 hp, at that rpm, it would make somewhere around 550 ft lbs of torque. If the only difference were the engines, and the transmission gear ratios were different to account for the powerband differences, and everything else stayed the same from a weight standpoint, which car would be faster? Neither! They'd be the same speed. Why? Torque didn't tell you jack! The car with the higher torque is not faster. It can't be. Only horsepower matters from a performance standpoint as only horsepower moves you.


Keep in mind I didn't specifically say UTI was a bad school. I said that one of 2 things was happening with the instructor. One was that he was being told something I wasn't saying which is why he disagreed with me. The other option was that he was being told what I've been saying but is just plain wrong about what he understands. I did make it perfectly clear that I think he is probably just not correctly being told what I'm saying which is why he disagrees. However if in fact he does get the info correct but still doesn't agree with me, only then would I say UTI is crap. I doubt it is and I've just explained what I think has been happening. Clear yet or do I need to draw a picture or perhaps a pie chart of probable outcomes?

subatomicparticle

the f1 scenario your math is correct but the car with more torque would absolutely have a better 1/4 mile time but yes they would have the same top speed and mid and high end acceleration characteristics but you absolutely need torque to get you going. you said yourself that torque is a twisting force kind of like your wheels turn it goes back to the leverage thing you have a horse with a plow (1 horsepower) trying to pull a large rock and is stuck use a large lever (to increase the torque) and it will pull the rock out because when torque is applied to a wheel it is in essence leverage.

subatomicparticle

Thanks I imagine you went from the 4.44 if you have any other info on how this all worked out for you I would like to hear it.

RX8SaxMan

Not if the car with less torque had a transmission optimized for its higher powerband. A transmission is essentially a torque multiplier.

rotarygod

The car with more torque wouldn't accelerate any faster as torque does no work. I'm not sure how often I have to keep saying it. Torque does no work. Torque does no work. Torque does no work. Torque does no work. Torque does no work. If it does no work, it can't get you moving or accelerate you. Why? torque does no work! Each car in my example of course would have to be geared around their respective powerband (horsepower). Torque can't just get you moving and then miraculously hand off somehow to horsepower. It doesn't work that way. The same thing that gets you moving off the line is the same thing that accelerates you and that is pure and simple horsepower. Forget you've ever heard the word torque as it's as relevant to vehicle acceleration as a ham sandwhich to keeping an airplane up. ONLY horsepower is relevant.

You bet it's a twisting force. However we aren't using it because that twisting force is work. It isn't. It's leverage. We are gaining leverage by changing output vs input speed. You may need leverage to get you moving but it's not the leverage that's moving you. It's the force acting upon that leverage that is. You can have all the leverage (torque) in the world but have no input horsepower acting on it. What would happen? Nothing. Why? Torque does no work! It's leverage. That's it. Go sit on a see saw alone and let me know when you suddenly go up. That fulcrum isn't going to get you moving as it's not doing any work.

You admit torque is leverage and that's good. However torque isn't horsepower and without horsepower nothing is going to move. Torque does NO work! It takes work to get your car moving. Torque allows you to have leverage that will permit work to be done through the input of horsepower but without that horsepower nothing is going to move anywhere. The whole point of a transmission is to get the engine rpms up to the max horsepower range.

RX8-Frontier

No it's not. It's a FORCE applied at a DISTANCE... ("Leverage" is just a distance)



Technically speaking... horsepower does NO work...

Horsepower is a RATE at which WORK gets done. It's not the actual WORK itself...

(But if you wanted to find out how much WORK got done, you'd use the HORSEPOWER curve for your calcs)



If 'ya get right down to it... WORK actually is physically done by TORQUE (fuel/air/sparky go boom, applying a force at a distance to make things go spinny...). However, the RATE at which that WORK gets done is HORSEPOWER (a force applied at a distance during a period of time).

ACCELERATION is the result of WORK being done to the car... So it's pretty simple, really... Do you want to talk about FORCES, or RATES? You have to talk apples and apples, here... So talking RATES means you talk about HORSEPOWER and the RATE OF ACCELERATION.

So if you want to know how FAST WORK will get done, you want to talk about HORSEPOWER. TORQUE won't tell you how quickly the WORK can be done; only how much FORCE is gonna' be added at any second.

RG:

Unfortunately, you're not quite there when talking about what is doing the actual WORK. HORSEPOWER can't do work; it's a RATE at which WORK gets done. TORQUE is the actual force, applied at an actual distance, and if the car is turned on, then it's being done over an actual amount of time. This is the WORK being done by the engine.

BUT, that doesn't tell you how quickly it's being done; for that, you need to talk HORSEPOWER (which you already know, and some others haven't quite gotten there, yet.), which adds the last physical property, revolutions of the engine, into the conversation.

Everybody else, pay attention!

RG is 100% right that HORSEPOWER is what's important in discussing acceleration rates. Torque is NOT a RATE, and therefore is NOT going to give you any information regarding how fast a car accelerates!

rotary.enthusiast

I like your new title RG

Some basic physics equations:

F = m*a
W = F*d
P = W/t
v = d/t

Let's get power and acceleration into the same equation shall we?

P = W/t = (F*d) / t = F*v = m*a*v

Solving for a, we get:

a = P / (m*v)

So if we know power, mass, and velocity, we can compute instantaneous acceleration. Assuming we don't have a car that magically changes mass over time, acceleration will be greatest at any given velocity at maximum horsepower.

This just so happens to be when torque at the wheel will be greatest as well. Let's simplify our equation a bit more:

a = P / (m*v) = W / (m*v*t) = W / (m*d) = (F*d) / (m*d) = F / m

Hey look, we're right back at Newton's second law. Looks like all we really need to compute instantaneous acceleration is force and mass. If that's true, then based on what we figured out earlier, the force generated at the wheels will be greatest if we have a transmission that puts us at our horsepower peak at any given speed. Can we all agree that the force applied to the ground is proportional (not equal) to torque generated at the wheel? I'd rather not get into angular accelerations and moments of inertia because I'll probably have an aneurysm

If I'm still being retarded, somebody please correct me

Mazmart

iTrader: (12)

This is so funny. I was literally JUST speaking about the topic of the relationship of torque to hp (5 mins ago) and only this weekend had to explain it to a good car ghuy friend of mine. Torque could be described as 'potential' work whereas hp is work performed.

Paul.

rotarygod

Are you saying that torque at the wheels is greatest where maximum horsepower occurs? That's not correct. Here's why. Let's say our engine does 175 hp (141 ft lbs) at 6500 rpm. That's it's horsepower peak. Let's say it's peak torque is 155 ft lbs (103 hp) at 3500 rpm. The car is definitely going to accelerate it's fastest around the horsepower peak. We know this. If all we want to know is torque at the wheels, time becomes irrelevant. All we need to know is gear ratios, or in other words, torque multiplication. It's always going to be a fixed number x the torque of the engine. That means that in a 1:1 ratio, it's number in, same number out. Let's say we've got a 4.00 rear end. This means that there is 564 ft lbs (141 x 4) at the wheels at our hp peak but 620 ft lbs (155 x 4) at our torque peak. Plug in any gear ratio. It still works the same. That means greatest acceleration is not where the torque peak occurs. Not even at the wheels. Why doesn't the vehicle accelerate fastest at peak torque?

rotarygod

That I'll gladly concede to!

rotary.enthusiast

No, I'm saying that at any given speed, if you could be at the horsepower peak then the torque at the wheels would be the greatest due to the mechanical advantage the transmission give you. If you're stuck in a single gear ratio, then the torque to the ground will be greatest at the torque peak, and you will indeed accelerate faster there. Horsepower tells you where acceleration will be greatest at a specific velocity. It's all in the equations above. Correct me if I'm wrong, but I don't think I am this time around.

EDIT: the math based on your example:

a = P / (m*v)
n = velocity at 6500 RPM in whatever gear you want

at 3500 RPM: a = 103 / (m*(3500/6500)*n) = 103*6500 / (m*3500*n) = 191 / (m*n)
at 6500 RPM: a = 175 / (m*(6500/6500)*n) = 175 / (m*n)

Acceleration at 3500 RPM in the same gear is greater.

RX8-Frontier

Yes. You're being retarded.

Peak torque and Peak Horsepower don't ever coincide at the same point in time, at the same RPM. Maybe you could make a theoretical one that would, but I'm not going to sit and do the math to figure it out.

Basically, torque typically peaks much, much, MUCH sooner than horsepower does in an engine, because if the engine RPMs keep increasing after the peak torque value, then the horsepower still increases.

Just snagging a generic image off the internet shows my point:



The torque started decreasing, but because the RPMS kept increasing, the Horsepower kept increasing, to a point when the RPMS no longer increased fast enough to make up for how much torque was dropping; that that point, horsepower begins to drop, too. But the point is, the peak torque/horsepower come at different times.

rotary.enthusiast

Torque at the wheels, not at the fly-wheel. If you're at peak horsepower, then the torque generated at the wheels is greatest for whatever velocity you are traveling.

RX8-Frontier

Torque at the wheels is 100% dependent on transmission & differential gear ratios, and wheel size. Those 3 things are picked to attempt to keep the engine in its powerband (powerband is horsepower, not torque... apples to apples, ok?). BUT, that doesn't mean they do a bang-up job of it; it'll never be perfect. Change anyone of those 3 things, and all the sudden the theoretical goes out the window. Even something so small as overinflating/underinflating the tires, or even tire wear over the life of the tire will change it.

RGs math/example a few pages back show that Horsepower to the wheels is the same as Horsepower at the flywheel (assuming no drivetrain losses), regardless of the gear variations.

AND... Since gears are nothing but different size levers, all they will do to the torque curve is scale it by whatever the ratio is. So the torque at the wheels will be numerically higher, but will be the EXACT same curve. So if the maximum torque at the flywheel occurs at 3,000 engine rpm, the maximum torque at the wheels is still at 3,000 engine rpm; it'll just be a numerically higher number. The horsepower, however, will keep on increasing as the RPM increases, so no, even at the wheels, the maximum torque does NOT occur at the same time as maximum horsepower.

Nice try, though.

rotary.enthusiast

You're missing the velocity component. Go back up and read my edited post that shows some math using RG's 2nd gen RX8's numbers.

RX8-Frontier

velocity of the car doesn't matter. You're mixing velocity of the car with angular velocity of the engine in your formulas.

Make your life easier.

Look at the horsepower/torque curves. Yes, engineers are trying to select the right gear ratios to get a high amount of force to the tires when the engine is in its peak horsepower. BUT, it can't be exact, because no matter what you do, the engine's peak torque doesn't happen at the same time the peak horsepower does. And scaling the torque up or down isn't gonna' change that.

rotary.enthusiast

I'm going to use RG's number as an example again.

Let's say you're moving along at 30MPH at 3500 RPM (the torque peak of the engine). What if you were going 30MPH at 6500 RPM? The engine makes less torque at 6500 RPM, but you've changed the mechanical advantage by 6500/3500.

Pick any speed... if you have a gear ratio (be it because of transmission gear, diff gear, or wheel size, it doesn't matter) that puts you at the engines HP peak, then the torque delivered to the ground at that point in time is the greatest that it can possibly be for that speed.

RIWWP

iTrader: (2)

Curious... (i really am, not challenging, just inquiring as to why. You guys are smarter than I am)

Why can't the HP and TQ peak be the same? I've never seen a dyno that show they are, but if we are talking math here, then isn't it mathematically possible for the 'curve' for both of them to be flat enough and steep enough that they are both continually increasing? If so, then each of the lines would have their peak at the same point: 'redline'.

For example, take the graph above. Imagine a redline at 3500rpm. Both torque and horsepower peak at the same time then.

The only thing I can think of that pushes against it relates to the expansion rate of the combustion, it pushes harder the slower the piston/rotor is traveling. (/side point: Also doesnt that mean that there is a max RPM where the rotor/piston is already traveling away from the point of combustion at the speed of expansion of that combustion?)

PS: I am also curious when the realization comes across that god > enthusiast

rotary.enthusiast

HP = TQ * RPM / 5252

The only time HP and RPM are equal is at 5252 RPM. If your HP peak is at 5252 RPM, then they would be the same

Ouch

RIWWP

iTrader: (2)

(Meant that as a friendly joke )
I really wish I had a graphing tool on hand. I can see a circumstance where HP and TQ are truely (As you say) equal at 5,252. Prior to that TQ is higher, after HP is higher. But both lines are both on an increasing angle (on the graph) before AND after, the entire way to redline. Nothing says that TQ must be decreasing after 5252 that I have seen. Just that TQ must be lower than HP after 5252.

Kinda like this: (pick your own scale ) Obviously no dyno chart looks like that. but the question is, is it possible? Seems to me that it is. Put the redline anywhere you want, it will be peak for both HP and TQ, while still conforming to the /5252 math.

I'm, talking math here, not engines. Real world isn't this smooth, i get that, but then real world engines run on math, so it makes me wonder.

rotarygod

^Yes it's possible. It almost never happens though but that doesn't make it impossible.