Haris

What is the thing that really gives you lots of pickup and pushed you to the seat hard. Is it all about torque, HP, or just about how fast the car accelerates? I mean like lets say would manual rx8 have more pickup because its faster or would mazda6 have more pickup because it has more torque.

fishsauce

Torque is what you feel. Horsepower is a number derived from torque and rpm.

Someone else may have a better explaination, but thats as simple as I can put it.

Hercules

And to be honest... it's not even right :p

Look for Paul Yaw's explanation, or go to VetteNet and read theirs... it's a combination of horsepower and torque that throw you into a seat, not to mention GEARING (biggest factor) and weight.

Off the line, under 3000RPMs though, the RX-8 is pretty slow but after 3000, or 3500.... power like you won't believe

It's fast enough for me anyway.

Rich

The short answer is that you feel the effects of acceleration, based upon old Issac Newton's second law of motion, Force = Mass Times Acceleration. The "Force" you are pushed back with is directly proportional to the rate at which the car is accelerating.

You exert torque on a typical door when you push or pull it. Torque is what turns the wheels of your car, and that's what pushes you back in your seat. When you leave a car in a particular gear, you accelerate at a rate proportional to the amount of torque the engine is producing. So, in a very real sense, torque is what you feel.

*BUT*, that's nowhere near the whole story. Gearing has a huge effect on the feeling you experience, as does the mass (weight) of the car. The gear ratios multiply the amount of torque at the wheels by some amount. The higher the torque is produced, the more multiplication you can get (again, oversimplified). This is simple to demonstrate. Put your car in second at 2000 RPM, and floor it. Then shift to 5th and floor it at 2000 RPM. The main reason you feel pushed back more in second is that there is a huge gearing advantage in second due to the higher torque multiplication in that gear. Those gears multiply the amount of torque along its path from the engine to the wheels.
This matters a lot! Since the RX-8 and other high revving engines (like the S2000, RSX, etc.) produce their torque high in the RPM range and have high redlines, they can be geared far more aggressive gearing than a big "torquey" engine. Wouldn't it be cool if there was a way to get both the amount of torque and how high in the RPM range it was into one number? Hey, that's what power is! Horsepower = (Torque x RPM)/5252, if you're using the idiotic system we here in the U.S. have been saddled with. If not the 5252 probably goes away if you're using a decent system.

Weight plays a huge role also. Motorcycles feel (and are) so fast because they are so light. A 16 wheeler producing 1,000 lb. feet of torque (just a wild guess) won't push you back, even though it has huge amounts of torque.


So, which is it that you really feel? You feel the force of the seat pushing your back that's accelerating you forward. For a given acceleration, it doesn't matter if you're feeling it from a big torquey V8 or a sportbike with 50 lb. feet. All of those factors - torque, gearing, mass - play into it.

Incidentally, since mass is constant in all of this (unless you count the gas being burned or left the trunk open), force (what you feel pushing you in the back) is proportional to acceleration. You will feel pushed back in direct proportion to how fast you are being accelerated, regardless of how it's produced.

For further edification -
http://www.vettenet.org/torquehp.html

Racer X-8

Good info there, Rich.
I'm gonna quote my college "Internal Combustion Engines" text book here for ya. It's kinda vague in definitions, but with this basic knowledge, other explanations of torque/hp equate, if they're correct in what they say. Rich's reference website falls right in line, cuz what the guy's saying is correct...

"Torque is the measure of the ability of an engine to do work, while power is a measure of the rate at which the work is done. Putting this in another way, the torque determines whether an engine can drive a vehicle through sand or other obstacles, whereas the power determines how quickly the car progresses over the obstacles."

Keep this in the back of your mind as you read Rich's reference.

Schneegz

The force you feel "pushing you back into the seat" is THRUST. The force that pushes the car forward is THRUST.

thrust (in lb) = (torque @ wheels in ft-lb)/(wheel radius in ft)... as long as the wheels are not splipping.

Or, if you want to convert from power to thrust...

thrust (in N) = (power @ wheels in watts [N-m/sec])(60sec/min)/(weel rpm)(wheel radius in meters)

That calculation can be done in HP too, but I don't recall the conversion factors off the top of my head. I like the metric sistem.

So it's all about thrust, or rather, your thrust/weight ratio. More weight means more inertia for the thrust to overcome. Weight bad. Thrust good.

The thrust of the car depends greatly on the drive train, tires, gear ratios, final drive ratios, friction losses in the drive train and so on.

Boozehound

Thrust isn't what you feel. It's acceleration like Rich and Racer X-8 alluded to. Thrust = force. With a constant thrust (force) you have a constant velocity. Like cruising down the interstate at 70 mph. The change in thrust (force) is what you feel = acceleration.

See Rich's comments about where acceleration comes from.


BTW - since thrust is a force, its units are [pounds force] or [newtons]. [in-lb] is a torque measurement (force times a distance).

TJRX8

Getting rear ended by a Hummer!

RomanoM

Like others said it's acceleration you feel, the body registers acceleration through force application.

However, acceleration in not constant in a car.

To picture it, thing of a graph with a curve. On the X-axis there is time and on the Y-axis is instant acceleration. The steeper the curve the greater the kick in *** you get.

Instant acceleration is just the exact acceleration at any point in time.

Or this works too

Schneegz

Wrong.

Thrust = force = (mass)(acceleration)

If force is constant in one direction, acceleration is constant, so you accelerate.

When velocity is constant that means that the NET force = 0. In a car that means that...

Thrust = road friction + aerodynamic drag

or, stated another way...

Thrust - road friction - aerodynamic drag = 0 = no acceleration = constant velocity.


If thrust is constant, but greater than the other forces acting on the car, then the equation becomes...

Thrust - road friction - aerodynamic drag > 0 so acceleration > 0, so you accelerate.

Yes, what you feel is acceleration, but the force that provides that forward acceleration is thrust. When force is constant, acceleration is constant, so you accelerate.


This is physics 101 stuff. I'm a mechanical engineer. Trust me on this. Or don't. Pick up a physics book.

RomanoM

Haris, should try and answer the question more directly.

Is it torque? Power? Gearing? Traction? Yes!

In the reality a vehicles acceleration is a matter of how much torque can you put through the wheels at any given wheel RPM and how much of that is lost in smoke or wheelspin.

But also, it's how fast you can spin the wheels for any given torque. This is where gearing comes in.


Since power is a derivative of force (torque), time and distance (distance and time being wheel RPM), in the end it's power, but power is simply the end result, not the cause.

Torque curves and gearing are the "root cause." The torque curve will dictate where you feel the kick. Turbo cars for instance have a spike in torque where the turbo kick in. While big displacement engines have lots of torque early in the rev range.

Also, I'm assuming mass is equal.

DisneyDestroyer

That's the best DESCRIPTION of torque vs horsepower I've seen.

Boozehound

I'll be damned if there aren't just a few of those around here. You do read profiles don't you?

I'll go along with F=ma, and most of what you said, but to an end, since drag does not increase linearly with velocity (it increases with the square of velocity), it actually takes a non-constant force to accelerate a car through air.

and you're units were still off :p

Engineers and arguments... geez

Schneegz

Aha! Hadn't taken the non-linear nature of drag into effect. You got me.

But THRUST/weight is still the key!:D

And I rechecked my units. Where's the mistake?

Boozehound

If thrust is a force, it won't be [in-lb], that's a torque measurement. It should be something like [lbf] or [N]. But you're right, a thrust to weight ratio is key - that's how they describe performance characteristics of aircraft, and I'm sure they certainly have more than a little to sit you back in your seat.

EDIT: never mind, I read your units as inch-pounds not "in pounds," it would've been the first time I'd heard of an "inch-newton" too...:D

Racer X-8

Interesting you brought that up. The name for that is "jerk".
I'm serious. Stop laughing...
Like a=dvdt, jerk=dadt (in calculus), or it's the slope in the a/t chart, like you explained. Yup, it sure is. It feels like whiplash I think.
Acceleration is what keeps you sunk into the cushion of your seat. It doesn't throw your head back, jerk does that. Ya learn something new every day...
What keeps you sunk into the cushion of your seat is the force (F) on your bod. Since F=ma, m=the mass of your bod & a=the acceleration of your bod, being accelerated by the car you're sitting in.
That acceleration, a=F/m, where m=the mass of everything, the car and everything (including you) in the car.
F=the net force which = driving force minus all drag (resistance) forces (aerodynamic/wind, rolling, gravitational (+/- due to up/down hill), whatever)
The driving force, the tangential force being applied to the tarmac from the driven tire or tires, comes from the simple formula F=T/r, where T=torque being applied to the wheel from the drive train and r=the radius from the center of the wheel to the tarmac.
Torque, T comes from the output of the engine, less drivetrain losses, and that is a function of the engine performance, which is dependant mostly by engine rpm. To get a grasp of that relationship, people refer to the engine power, or, horsepower, and that is hp=Tn/5252, where T=output torque that we're talking about & n = rpm's of the engine. Like the waterwheel, an engine ain't worth a toot in a car if it can't deliver torque above 12 rpm. That's a good analogy as to why horsepower is an important value.

fishsauce

The "force" that you feel when you're pushed into your seat is caused by a change in momentum right? So wouldn't it be correct to call it "impulse?"

Is this what you mean by "thrust" Schneegz?

Racer X-8

No, I wouldn't call it "impulse". Impulse is the product of the force and a time interval . I=FT (lb-sec) or the integral of Fdt when the force F varies in magnitude during the time interval.

You can call it "G's" if you want. 1 G is the force applied on a static body due to the gravitational force of the earth. Acceleration due to gravity = 32.2 feet per second per second, or feet per second squared. If your RX-8 accelerated at 1.0 G's (32.2 ft/sec2), and you weighed 150 lbs, your seat bottom would "feel" 150 lbs (due to gravity) AND your seat back would ALSO feel 150 lbs (due to RX-8). If your RX-8 could turn at 1.0 G's ("side G's"), those side thingy's of your seat would ALSO feel 150 lbs.

Thrust is also a force, although I've only used thrust to describe a force exerted from like a jet engine.

Schneegz

Yeah, I looked at my post again and realised that saying "in pounds" might be confusing, but I wasn't sure how else to say it.

If you ever read one of the British motorcycle magazines you'll notice that as part of their dynomometer runs they show a thrust calculation for each gear along with HP and Torque. It's very interesting to see how some motorcycles that seem a bit down on HP, when compared with their competitors, can put out very respectable thrust numbers due to clever gearing.

The Honda CBR600RR is a good example. It's heavier and not as powerful as its competition, but its acceleration figures are right there with the pack, which makes you go "hmmm..." But then you look at the thrust numbers and it all makes sense. Honda just geared the bike to let the rider make the best out of what the engine dishes out.

It would be good if car magazines would publish similar results. I think it would be an eye opener.

Buger

Hi Haris,

The answers have been mentioned in this thread earlier but let me try my hand at it too. First, there are a few things to remember:

1. Drive wheel torque is what accelerates you
2. HP is a vastly more important variable than engine torque because a higher HP engine allows higher ratio gearing for more drive wheel torque
3. Weight is the factor that affects all areas of performance (acceleration, braking, handling).

The RX-8 will accelerate faster than the Mazda6 because it has more drive wheel torque. The higher HP of the RX-8 allows it to be geared with higher ratio gears. If this weren't enough, the RX-8 is also lighter!

I'lll once again mention the inimitable Paul Yaw :D who wrote a recent article which you can read at: http://personal.riverusers.com/~yawpower/tqvshp.html

Using some of the formulas from his article, below are some simple calculations below:

----------------------------------------------------------------------------------
RX-8 w/6MT

Assuming 20% loss through transmission, estimated rear wheel torque at torque peak (5500 rpms) is:
1st - 3.760 * 4.444 = 16.7094 * 159 = 2656.80 * .8 = 2125.44 ft-lbs

force = rear wheel torque / distance from rear axle to ground (assuming 225/45R18)
force = 2125.44 ft-lbs / 1.08'
force = 1968.0

Acceleration @ torque peak = force / mass

Mass = 3029 + 160 (assuming manual RX-8 w/ options that Mazda estimates 30% of us to buy) + (driver weight of 160)

Acceleration @ torque peak = 1968 / 3189
Acceleration @ torque peak = .62 g @ torque peak in 1st gear
----------------------------------------------------------------------------------

----------------------------------------------------------------------------------
Mazda6 w/5MT

Assuming 20% loss through transmission, estimated front wheel torque at torque peak (5000 rpms) is:
1st - 3.153 * 4.133 = 13.0313 * 192 = 2502.02 * .8 = 2001.62 ft-lbs

force = rear wheel torque / distance from rear axle to ground (assuming 205/60R16)
force = 2001.62 ft-lbs / 1.07'
force = 1870.7

Acceleration @ torque peak = force / mass

Mass = 3243 + 160 (assuming Mazda6 w/MT) + (driver weight of 160)

Acceleration @ torque peak = 1871 / 3403
Acceleration @ torque peak = .55 g @ torque peak in 1st gear
----------------------------------------------------------------------------------

The formulas above do not take into account some variables (like drag, rolling resistance, etc.) but they do a very good job of giving us some numbers to compare different cars with.

As you can see, the 159 ft-lb RX-8 @ it's torque peak will accelerate faster 192 ft-lb Mazda6@ it's torque peak. The truth is that the RX-8 will accelerate at a higher rate at any rpm and will also allow you to stay in gear longer.

The comparatively low engine torque of the RX-8 is something that other car owners may point at to "prove" that their car is better. Many of these owners are just ingnorant though.

Brian

Rich

As always Buger, great job.

Racer X-8

Yessir! Nice number crunchin there Buger.

I like the results too.

Firstly because it is a very good example of the 2nd point that you are making concerning hp & its relationship to gearing, and the 3rd point that you are making concerning weight.

Secondly because, well,,, YAY! WE WON!!!:D