SpeedGeek

Force at the wheels has nothing to do with rpm. It depends only on torque (and gearing).

A Honda S2000 makes the same maximum power as, say, a 3-liter turbodiesel Audi. But the Audi will 'force' you back in your seat much harder than the Honda. The Audi feels more powerful even though it's not. What the human body feels is force (torque), not power. It doesn't matter what speed or rpm you're doing, what you feel is purely torque.

Now consider these 3 equations for power:

Power = force X velocity (linear).
Power = torque X rpm (angular).
Power = rate of doing work.

The 1st is good for understanding top speed.
The 2nd will tell you how much power your engine is making at any given point in the revband.
The 3rd is useful for explaining acceleration.

For acceleration, power is what counts, not torque. To understand this, I like to think of power purely in terms of energy, ignoring torque or revs. Power is the "rate of doing work". Consider that the engine's job is simply to convert fuel energy into kinetic energy of the vehicle. The more 'powerful' the engine, the faster it is able to use fuel energy to accelerate the vehicle's mass. Thus acceleration is about power, not torque or revs. That is why, for maximum acceleration, you always want to keep your engine close to it's maximum power, not its maximum torque.

As for top speed, consider that power = forceXvelocity. Or velocity = power/force (where force is aero drag pushing back against the vehicle). From that equation, it's simple to see that top speed is entirely dependent on the engine's power vs. drag. Neither weight nor torque nor gearing affect top speed (in theory).

allegretto

Here is the example I was taught and I felt it made a lot of sense. It is ripped directly from Richard Feynman, one of the ten top five Physicists of the 20th Cent.

Two exactly identical twin brothers, Moe and Joe are sitting on the beach admiring a beautiful lady in a swimsuit (those of you who are familiar with RF know why he makes examples like that). They are motionless and transfixed by her beauty. The air is perfectly still. They are both exerting a Force (F) upon the sand which is MG. But since they are motionless, we now know that F has nothing to do with motion. F is simply a "push" and says nothing about what is happening or in what direction it is happening if there is motion. This is where the trouble with misunderstanding begins. Most folks erroneously say if a given F is acting on a M then there must be an A, and they are correct. However A is only the ratio of a stated F and a stated M. The real M will act in response to the "Net F" which is the sum of all F's and often folks don't consider the other F's

Now let's introduce another term, Work (W). It can be defined by F displacing an M a distance. Suppose Moe and Joe decide to go up a flight of stairs. Moe walks steadily up the staircase and it takes him 14 sec. Joe, ever the impatient one, runs up the stairs in only 7 sec. Both men have done exactly the same amount of work.

Finally lets introduce the term Power (P). It is defined as an F doing W over an interval of time. In the above example Joe has expended twice the P of Moe.

Force (TQ) feels good when applied in the proper vector, Power makes it go fast and wins races.

The great thing about Feynman was the way he taught with examples even I could easily understand...

allegretto

I like that.

In other words the downforce vector is only down and has no effect on the lateral force. Is that what you mean?

insite

exactly! we separate the forces into their components. with aero, we get the extra lateral grip from the extra weight on the tires, but our lateral force at any given speed hasn't changed since our car still weighs the same amount.

insite

wow; just picked up on the power vs. torque argument. isn't that a fun one that never dies!

allegretto

The only thing I'd add is that P is always a scalar

Yargk

I skipped a step. Force at the wheels is a function of torque, gearing, and your tire's rolling diameter. At a given engine speed and ground speed the gearing is fixed. So the gearing is a function of rpm and ground speed. So at a fixed ground speed, force at the wheels is a function of torque and rpm. I think it's useful to see it this way because given two cars with 100 foot pounds of torque (constant vs. rpm) but one with a redline of 5000 rpm and the other 10000 rpm, then at a given speed one can be geared to be running at ~4500 and the other ~9500 and the latter will have 9.5/4.5 times the force at the wheels. But this is equivalent to saying it has 9.5/4.5 times the power.

C.J. Ichiban

you guys are so off topic

SpeedGeek

True, thanks. I've edited that line.

SpeedGeek

True, CJ, but I can't help myself correcting a technical inaccuracy when I see one (as allegretto did to me).

C.J. Ichiban

haha it's all good.

SpeedGeek

You are definitely confused, but I'm not sure how to unconfuse you.

Force at the wheel is never a function of rpm, only a function of torque and gearing. Ok, in the real world it's a function of rpm in as much as engine torque varies with rpm, but mathematically, or if tq remains constant vs. rpm as per your example, force at the wheel has nothing to do with rpm, or even with power.

In your example, the only reason one car will push against the road twice as hard as the other is because it's geared twice as low. The fact that it's revving twice as high is true but irrelevant. All that matters is torque and gearing.

Consider, instead, just the car that revs to 10k, with torque constant through the entire revband. Now imagine selecting say 3rd gear at 1k rpm and flooring it. It will immediately apply a force to the road and accelerate. But here's the thing - as it accelerates, the revs climb and the power climbs, yet as long as you stay in that gear, the force to the ground will not vary. At 10k rpm, the engine is revving 10 times as high as at 1k rpm (duh) and the engine is making 10 times as much power. Yet the force to the road remains constant. So force to the road depends purely on torque and gearing, not rpm or power.

allegretto

oh shyte, are you gonna call the Mods???

the horror......the...hor-or

Yargk

EDIT: I thought of a way to explain this in a more simple manner. If you gave me a car, the force at the wheels it produces is ONLY a function of gearing and torque. BUT, if you gave me a car with a fixed engine and variable gearing the force at the wheels would be different if I chose different gearing and I would choose that gearing depending on the torque vs. rpm of the engine. This is worth considering for those who think torque is what matters in a motor because this idea directly shows that the force at the wheels in a given car is related to where the torque is made in the rev range ONLY because this influences the gearing decisions by the engineers. If you still think I'm confused, read on. Otherwise, spare yourself the pain.

If you set up conditions such that your gearing is a function of rpm, then of course force at the wheels is a function of rpm since it's a function of gearing. I am free to DEFINE gearing in terms of rpm.

So let me derive a function for force at the wheels in terms of peak power rpm ASSUMING optimal gearing.

Example:

2 cars miata and an s2000

Speed 50 m/s

wheel diameter 1 meter for simplicity
wheel circumference 3.14 meters

redlines 6000 rpm and 8000 rpm for miata and s2000

assume each car makes 100 foot pounds of torque at all rpms

wheel rpm = 15.9 rps *60 seconds/minute=955

DEFINE gearing for each car such that they will be sitting at their power peak while traveling at 50 m/s, this power peak is at 6000 rpm in the miata and 8000 rpm in the s2000.
(I call this instantaneous optimal gearing, it should be clear that I'm only analyzing a single moment in time at 50 m/s)

gearing miata = 6000/955 = 6.28

gearing s2000 = 8000/955 = 8.373

optimal gearing at 50 km/h general if constant torque at all rpm = redline/955 (this is where I define gearing as a function of peak power rpm which happens to be redline in my example.

torque at the wheels = torque at crank * gearing
force at wheels = torque at wheels in foot pounds / radius of wheel in feet

force at wheels = torque at crank * gearing / radius of wheel in feet

= 100 foot pounds * (redline/955) / 1.64 feet

There you go, if you assume optimal gearing, force at wheels is a function of peak power rpm.

I think this is a reasonable thing to consider because the gearing for cars with powerbands at different rpms will be different. Assuming this different tailored gearings just allows us to see the connection between the power output of a car and how much force it puts to the ground. This is only an exact function when I DEFINE optimal gearing, but as a rough idea it's useful as well. The reason it's rough is that car manufactures try to make optimal gearing but you don't have infinitely many gears so aren't at the perfect rpm for every ground speed. But for two well geared cars at the same speed this approximation gives a rough idea of force at the ground.

Finally really this example is just to show people who insist that force determines acceleration, that the force at the wheels is in fact closely related to power if you're in the right gear. Now I give up.

SpeedGeek

Correct. The simplest way to understand this is to eliminate gearing altogether and just imagine driving a CVT car. Slam your foot to the floor in a CVT and the revs immediately jump to max power, not max torque. The reason for this is exactly what you describe above; for max acceleration you want max possible force to the ground. Intuitively, that may seem like you want to be at max torque - where the engine is providing its maximum cranking force. But gearing between the engine and the road conspires to reduce that crank force before it reaches the asphalt. But at the higher rpm of max power, the CVT will have lowered the gearing accordingly (same ground speed but higher revs = lower gearing). So while the crank is indeed now producing less twist, the lowered gearing actually results in a higher force at the wheels. Therefore faster acceleration.

I think all this confusion is because people intuitively think of power in terms of force. The problem with such thinking is that speed, rpm and gearing totally distort the force vs. power picture for our brains.

That is why I always urge car folks to think of power in terms of "the rate of using energy" rather than anything to do with force or torque. Simply imagine the engine to be a black box whose job it is to convert fuel energy to kinetic energy by accelerating the car's mass. This makes it simple to understand that for maximum acceleration, you want the engine running at it's peak power, because that is where it will be burning the most fuel.

There is, however, an obvious caveat to this. If two cars with regular gearboxes race, identical in every way except the shape of their torque curve, the one with the fatter torque curve will win, even if they have the same peak horsepower. That's because during in-gear acceleration, before reaching maximum power, the higher-torque car will be burning more fuel, thus making more power, as it climbs up the revband. So even if it's peak power is the same, it would be accelerating the car faster than its less-torquey opponent during in-gear acceleration.

In a CVT, there is no in-gear acceleration, and the torquier car would thus lose its advantage. All that matters would be peak horsepower.

Whew! This is almost as tiring as watching the markets.