Originally Posted by JoBy

This is a good example of what a dyno needs.
1) Something to measure the rpm of the roller.
2) Something to measure breaking torque. In this case a load cell measuring force on a lever arm with a known length.

Originally Posted by bjankuski

Horsepower defines the amount of work that can be done, and how quickly it can be done, and that defines the performance of the car. Torque and rpm define the amount of horsepower that is available to do work. It is that simple.

Originally Posted by bjankuski

Horsepower defines the amount of work that can be done, and that defines the performance of the car. Torque and rpm define the amount of horsepower that is availabe to do work. It is that simple.

Matthew Miller is correct.

Originally Posted by Krusty84

Man, I wish a mod would simply delete 2000 word posts to discourage members from posting them. Uggggg.

Originally Posted by Krusty84

Man, I wish a mod would simply delete 2000 word posts to discourage members from posting them. Uggggg.

Originally Posted by jayjones

It's true that power to overcome drag increases with the cube of velocity: Pd = Fd*v (v=velocity, Fd is drag force) and drag force already contains v^2 so drag power = (1/2)*Cd*p*v^3*A
The units have to work out: Power is kg*m^2/s^3 which is why Fd needs to be multiplied by v: kg*m/s^2 * m/s

I don't agree with your 565hp number though. You have to convert everything to SI units (kW and m/s) then scale it by v^3 so …..186.4kW*(89.4m/s / 67m/s)^3 = 442kW which is 592hp

Originally Posted by 69427

Hey sport, am I ever going to get an answer to post # 69?

Originally Posted by jayjones

Lol yeah missed that one on mobile. I agree with your initial calc, I mistakenly used 250hp instead of 240hp. Sorry for the mixup

Originally Posted by MatthewMiller

Just added one phrase for total accuracy. Yes, it is that simple. I feel like motorsports media and more recently car commercials (and even lawn mowers!) have done a real disservice by trying to make this all more complicated than it really is. We see all kinds of fallacious "truisms" that aren't really true, such as "power gives you top speed but torque gets you there," or "power is how fast you're going when you hit the wall, but torque is how far you move the wall." That's all BS.

The absolute numerical value that determines the car's acceleration and top speed (i.e., the number that we should care about) is power.

Originally Posted by MatthewMiller

That is exactly right, and it's exactly my point. Thank you for making it for me in such a short sentence! The fact is, though, that horsepower is the parameter of engine performance that tells you everything you need to know about how fast the car can go. Crank torque literally doesn't mean everything. You just said it yourself: just knowing torque doesn't provide enough information. Crank torque by itself tells us nothing.


Nope, there's just two unknowns: the distance the crankshaft covers and the time it takes to cover it. Remember, power is just force*distance/time. But with crank torque, you only know the force. Or you could say one thing is unknown: rpm (which is just distance/time). The point is, if you know the engine's power, you know exactly how much tractive force the car can put to the ground at any speed, and therefore you know how fast it can go against its total resistive force, or how fast it can accelerate its total mass.


Ummm, yeah...it would be infinite. But back in the real world...


I didn't say all torque is meaningless for all things. I said crankshaft torque is meaningless in predicting vehicle performance. And it is, as you agreed above.


You're absolutely right. I overlooked that you wrote ft-lbs instead of just lbs. My mistake there.


Are you kidding? In the diagram I posted, the pulleys only serve to change the direction of the linear force that moves the block. The pulleys aren't doing any work (it obviously assumes the pulleys have frictionless bearings). The block isn't moving in rotation. It's moving straight up, in response to the linear force that is pulling it that way. Changing the size of the pulleys here would change nothing about the magnitude or direction of the force moving the block. But when you start talking about connecting other ropes and examples of gearing, I see how you tend to get yourself caught up in extraneous aspects that don't actually matter to the example at hand.


It is telling you nothing about how fast the motor can do work, which again was the question at hand. Knowing the force(torque) the engine can apply when stalled does sort of tell us about its potential to move mass, but it tells us nothing about how fast it can move it (again, please see the original question we were trying to answer in this thread). I said "sort of" in the last sentence, because the stall torque of the engine itself doesn't really tell us the potential to just move mass, because we can multiply torque. I can gear any engine of any torque output down to move any mass - it just will move it awfully slowly if I have to gear it down by very much to multiply its force. Which again is the point: if I only know the torque of an engine, I have no idea the speed at which it can apply that torque, and therefore I have no idea how fast it can move a car.


ORLY? Power=Force*Distance/Time. Horsepower=Torque*RPM/5252.

So do you want to stick with this? Because that's very, very wrong. Power isn't just a force, but it most definitely includes force as one of the three key pieces of info you need to find out how fast an engine can move a car. If you think power is not a force, it's no wonder you would think that "torque is everything," and that you wouldn't understand why power is everything.


Well, it's actually not measuring torque - it's measuring tractive force at the contact patch, which is a linear tangent to the drum's circumference. It doesn't know how tall the tire is, so it has no idea how much torque the axle/tire is exerting; nor does it know the rpm at which the axle is spinning. Again, I never said no force matters. I spent a lot of time talking about tractive force, and even linked a definition for it. Force at the contact patch is the Force in F=MA for a car! But the measure that determines the tractive force of the drive tires at any given speed is power. You don't have to know the gearing, the tire diameter, or the actual torque output at the crank to know how much Force you have available to Accelerate your Mass. You just need to know the power and the speed at which the car is traveling. Remember, the power at the crank is also the power at the wheels, so you if you know the speed of the car then you also know the tractive force available to accelerate it or to overcome resistive force.

What I said was that torque at the crankshaft doesn't tell us anything about a car's performance. And one way you know that's true is that an chassis dyno can measure power without knowing the engine's rpm, and therefore without having any idea what the engine's torque actually is. You can put an engine that makes 100hp and 200lb/ft at 2626rpm, another that makes 100hp and 100lb/ft at 5252rpm, and another that makes 100hp and 50lb/ft at 10504rpm. They will all get exactly the same power readout, and the dyno doesn't need the tach sensor connected to determine it. And they all power a vehicle to the exact same terminal velocity. The crank torque doesn't make a bit of difference to the dyno or to the performance of the car.


Seriously, you're reading lots of stuff into my post that I never wrote. I never said an engine doesn't produce torque or that it doesn't need to. Of course it produces force, starting as cylinder pressure which is then translation to rotational force at the crankshaft. You can't produce power without force! When I say power is all you need to know, that includes that force! Because (unlike what you wrote above), power most definitely does include force. My saying torque tells us nothing about vehicle performance isn't me saying an engine doesn't make torque. Come on!


Wrong. Have you been around chassis dynos? If you don't connect the tach sensor, you get your X axis plotted on mph instead of rpm, and you only get a power curve and no torque curve. The dyno doesn't care about engine rpm at all. The tach sensor is an add-on accessory to give torque fanbois something to talk about (okay, I'm being facetious, but only a little bit). If you have the tach sensor connected, the dyno calculates the engine torque from the change in kinetic energy of the drum and the rpm of the engine. It doesn't measure crank torque at all, and it doesn't calculate power by multiplying crank torque*rpm/5252 as you claimed.


It's not tricky at all. It's because the engine is putting out the same power regardless of whatever gearing you install after it. You're tricking yourself because you're comparing acceleration to engine rpm, which isn't the correct measure of performance. The correct measure is acceleration vs mph. That's really, really important to understand here. The car with 3.73s in 4th gear can pull 37% more Gs of acceleration at a given rpm than it can with 2.73s at the same rpm, but it's also going 37% slower when it does it. It doesn't really change how fast it accelerates at the same speed though. The only thing that determines how fast a car can accelerate at any given speed is how much power the engine is sending to the contact patches at that speed. And gearing doesn't change that. People often get confused when the try to imagine which gear they'd be in at what speed, etc. It's much easier to understand this if you imagine a car with a CVT.


This is all a good example of how you're confusing yourself by making this more complicated than it is.


Again, my point was that any chassis dyno of any design measures power at the tires, not the crank. And any of those can tell us the engine's power without knowing its rpm, and therefore without knowing its torque.


All of which tells us nothing about how much power is required to push a given car with given resistive force to a certain top speed...which again was the original question being asked. You do understand that the only reason we have multi-gear or CVT transmissions in racing cars at all is to maximize the power the engine can apply to the tires at any given speed, right? If you want your car to accelerate the hardest at any particular speed, you gear it so that the engine is at peak power rpm at that speed. You don't gear it for peak torque. Not ever. Same for trying to reach a desired terminal velocity: you will gear the car so the engine is making peak power at the desired speed, not for peak torque.


I'm glad you agree with me on that. If only I could convince you that it's engine power that determines the torque at the wheels (or more correctly, the tractive force at the contact patch) then we'd be in total agreement.


This is true.


Oh yeah? Answer this: A car has an engine that produces 500lb/ft of torque at the crank; so how much torque does it have at the wheels, or how much tractive effort at the contact patch, at 100mph? You can't answer that, because without knowing the engine's power, you can't know the gearing required. Since you don't know if the engine is creating 500lb/ft at 1000rpm or 8000rpm, you can't know the torque at the wheels.

OTOH, if I know the power the engine is supplying to the contact patches, then I know exactly how much tractive force is being generated at any road speed, irrespective of engine speed and gearing. I won't know the torque at the wheels unless I also know the tire diameter, but since it's the tractive effort that moves the car (which again is a linear force and not a torque), then it doesn't matter.


But less road speed. At the same road speed, the wheel torque doesn't change (again, think CVT here). Which again is why power is what matters here: it's the only thing that determines how much tractive effort the tires generate at a given road speed.


The question wasn't how to build a car to achieve a certain amount of power or torque. The question was how much power it takes to move a C4 at 200mph. Power is the only thing that matters here. Earlier, we calculated that it should require 565hp to do that. All I need to know is that any engine that can produce 565hp will push the C4 to 200mph, as long as I make sure it's actually outputting 565hp at 200mph.


Yes, of course, which is why I wrote: "Why in the world would anyone trying to set a top speed goal not gear the car to make peak power at the car's projected top speed?" If you're trying to go 200mph and you gear your car to make peak power at 140mph, then you're an idiot. There's no other way to put that. Again, the question was how much power does it take to go 200mph in a C4. The answer to that doesn't depend on rpm or crank torque. It takes however much power it takes, no matter how you get the power.


Which is why power is the only thing that matters in determining how fast the car goes! Why is that? Because power is what determines the tractive force at the contact patches of the driven tires. The more power you have, the more tractive force you have at a given speed. The crankshaft torque you have...doesn't tell us anything about the tractive force at any speed.


Yeah...that's not true. Peak horsepower always occurs at an rpm higher than peak torque - it always occurs after torque output is already dropping.


No, horsepower defines the performance of the engine in a car because it is what determines the tractive force available to accelerate the car at a given speed. You gear the car to allow it to produce the most power possible, not to multiply torque. If multiplying torque were all-important, then you'd gear the car infinitely low (numerically high) to have the most torque possible. The car could pull an infinitely high load with infinitely high acceleration, but it would only achieve an infinitely low speed. There's an obvious reason we don't do that: because speed is actually also important to vehicle performance! Ergo, power is what's important and not crankshaft torque.


Dude, you were talking about tweaking the drivetrain to get torque to the wheels: "Once you have the torque needed then you look at how to get that torque there. Rear gear, tire diameter, RPM range, and transmission gearing all come into play. You build a combination that is delivering enough torque to the rear wheels." Obviously you work on aero trim, suspension, etc. But your goal in setting up the engine is to deliver enough power to the rear wheels so they can create enough tractive force at the road. You are trying to maximize power from the engine, not its torque.


It's not a different animal at all. The truck that can accelerate the heaviest load (or accelerate a heavy load the fastest) is the truck that can apply the most power to the wheels (obviously assuming the rest of the truck remains constant, like chassis and suspension) over the total acceleration run. It's crank torque has nothing to do with that. There are good reasons that low-rpm diesels are used for heavy hauling, but the ability to accelerate the heaviest load isn't one of them. Trucks have granny gears and lots of ratios so that they keep their engines running around peak power rpm for a lot more of their speed ranges. Even more importantly, it's so they don't destroy clutches or torque converters trying to get heavy loads moving from a stop.


This much is correct. Of course they will both achieve the same top speed as long as they are both geared to output peak power at that speed.


This is incorrect. If they both are geared to achieve the same top speed at peak power, then they will both accelerate to that top speed in the same amount of time and distance. Car A has the same horsepower and 33% shorter gearing, but its engine also has 33% less torque at the crank. So with 33% shorter gearing multiplying 33% less crank torque, you get...wait for it...the exact same torque at the wheels as Car B, at all road speeds.


That's just not true. You simply aren't understanding how this works. Your wrong statement above about Car A accelerating to the same top speed faster is proof that you don't understand it. Like said before, the force at the tires is determined by the power being produced, not the crank torque. An engine produces the power required, not just the torque. In other words, it's not just the torque at the crank that matters, but also the rpm at which it produces that torque that also matters. In your example above Car A has only 66% of the torque that Car B has, but they both produce exactly the same tractive force at a given road speed.

OTOH, say you have two engines: Engine A makes 400lb/ft of torque at 3000rpm, and Engine B makes 400lb/ft of torque at 6000rpm. You can't just "adjust your gearing to get the resulting force at speed you need (Power)." If, for example Engine A produces 1000lbs of tractive force at 100mph, then Engine B will produce 2000lbs of tractive force at 100mph. There is no way to gear yourself out of the fact that Engine A only has half the power of Engine B. You can't create a gearset that will allow Engine A to produce 2000lbs of tractive force at 100mph like Engine B does.


Nope. The power is what will determine how fast a vehicle accelerates at a given speed. Period. How it makes the power or what its crankshaft torque peak is won't matter.


No, torque is one of three equally important components of power: force, distance, and time. If either of the first two are zero, you have zero power. If the third is zero, you have undefined power. They all matter just as much as one another.


Nope, you're buying into all those dumb truck commercials. If you have a 20,000lb trailer, the truck that will accelerate it the fastest will be the one that outputs the most power to the tires. If you have a 400hp diesel engine that makes 800lb/ft in your truck, and I have a 500hp engine that only makes 250lb/ft in my otherwise-identical truck (and obviously at a way higher rpm), then I will accelerate that trailer faster than you. Crankshaft torque peaks have nothing to do with how hard you accelerate at a low (or any other) speed, and the mass of the load doesn't change that either. Again, only power determines that.

I'm going to go back and requote my original statement, which you said "couldn't be more backwards and wrong":

This is neither backwards nor wrong. It doesn't matter how the car is producing the required power: rpm and torque (and required gearing) don't matter. If it takes 565hp to push a C4 to 200mph, then the only thing that matters is that the engine produces 565hp at 200mph. The car doesn't care if I'm making 565hp at 1000rpm with 2967lb/ft of torque, or if I'm making 565hp at 20,000rpm with 148lb/ft of torque. It's going to reach 200mph either way.

Originally Posted by TravisSchoech

Matthew,

I have been playing around with a website called gearcal.com , and it seems to be eerily accurate as compared to actual measured 1/4mi time and trap speed.

1988 coupe, 3100lbs w/driver&1/2 tank
600FWHP, 6400RPM
3.45 rear gear, 25.44" tire
2.88:1/1.91:1/1.34:1/1:1/0.86:1 gear ratios (standard trans 88)
The drag coefficient of a 1988 coupe is ~.341
Frontal area of a 1988 C4 is 19.28 sq/ft

Estimated 1/8 mile is 6.66 at 107.01 mph
Estimated 1/4 mile is 10.37 at 134.2 mph (confirmed, as I was bumping the rev limiter in 4th as I passed the 1/4mi cones)
Estimated 1/2 mile speed is 161.97 mph (car ran 149.5mph in the 1/2 with about 100hp LESS than the current setup)
Estimated mile speed is 185.1 mph

Gear1 49mph@6400rpm
Gear2 74@6400
Gear3 105@6400
Gear4 140@6400
Gear5 206@6400

What are your thoughts on this?

Question 1: What do you think the actual TOP SPEED is for the car, or for the sake of this thread, would this car reach 200MPH assuming all STOCK AERO?

Question 2: How bad do you think a 4" cowl induction hood ALONE hurts the aerodynamics of this car?

Question 3: How bad does it hurt the AERO in the 150mph range if I do not have the front air dam (3pc lip) under the car?

Originally Posted by KyleF

I see it now... you just want to argue. Hell we are saying some of the same things a different way, and you still want to argue because you are to stubborn to listen for some reason.. Seems like your main beef is thinking somehow I think a chassis dyno measure torque at the crank. It does't, but it also has nothing to measure if there is not torque coming from the crank.

Originally Posted by TravisSchoech

What are your thoughts on this?

Question 1: What do you think the actual TOP SPEED is for the car, or for the sake of this thread, would this car reach 200MPH assuming all STOCK AERO?

Question 2: How bad do you think a 4" cowl induction hood ALONE hurts the aerodynamics of this car?

Question 3: How bad does it hurt the AERO in the 150mph range if I do not have the front air dam (3pc lip) under the car?

Originally Posted by Andrew669

I have been away for a few weeks and have not read all detail in the threads. I can indicate from my experience that when overdrive is selected in any transmission one is no longer multiplying torque. I see many high torque numbers being displayed that are actually being cut by any gear past 1 to 1 ratio. I still can comment that 460 hp at the crank will push a C4 to 200 MPH estimating final HP at the wheels at estimated 430 HP and Torque at 360 FT pounds of torque at 7,200 RPM, Using a 1 to 1 straight output / input from the motor to the drive shaft to the diff. Any questions are well received.

Originally Posted by MatthewMiller

The bottom line is that the force that accelerates a car or pushes it to its top speed is the tractive force at the drive tires' contact patches. That force is always determined by the power available at the tire, divided by the road speed. The more power available at any given road speed, the more tractive force available to accelerate the car. Conversely, the higher the road speed for any given power output, the less tractive force available. In any max-performance scenario, the gearing's only purpose is to keep the engine as close to peak-power output as possible. It is never more complicated than that, and the torque value at the crankshaft has no relevance to performance.