I think if you spent as much time with a basic physics test as you spent typing your posts, it would be beneficial.

Duke

This is where you are confused.

Torque at the rear wheels determines acceleration.

Torque from the engine passes through a gearbox and a rear axle and is increased based upon those gear trains.

If I have an engine that produces 100 lb.-ft. torque at 6000 rpm, I can use a transmission with 2:1 gears, as an example so that my output speed is 3000 rpm and the torque to the pavement is 200 lb.-ft.

If I have an engine that produces 100 lb.-ft. torque at 3000 rpm, this same engine will put only 100 lb.-ft. torque to the pavement with 1:1 gears. The output speed from the transmission, 3000 rpm, is the same for both engines.

But, the engine that produces the torque at 6000 rpm will accelerate the vehicle at twice the speed of the engine that produces the torque at 3000 rpm.

The engine that produces 100 lb.-ft. torque at 6000 rpm has twice the horsepower as the engine that produces 100 lb.-ft. torque at 3000 rpm.

I spent a thousand hours at Argonne Nat'l Lab during grad school, though my PhD is in neurobiology. I spent about 100 unofficial hours at Fermi Lab during this same period. Your response gave me no indication of what understanding or thoughts you have or don't have regarding instantaneous velocitie's, or instantaneous value's of the other metrics and how they relate in relevant ways to your remarks on xxx as a function of time. Time spent typing posts, good point. I could have shortened it considerably. Leaving out any condescending(sp?) remarks kept it a bit shorter.

Brian,
I agree with you as regards gearing and "torque at the rear wheels determining acceleration". I'm sorry I didn't make it clear that gearing, frictional losses, etc were left out so as to speak of torque in a stripped down sense. I should not have said the torque curve falls atop the acceleration curve exactly, but it does, I believe, follow the torque curve in general form and has a much different relationship, in form, to the hp curve. And, yes, I agreed with Watt's equation so agree with your last sentence. Since seeding the Watt's equation with a value assigned to one or the other, but not both, of torque and horsepower, we automatically can generate the same graphs, regardless of which variable we provide with a seed value, this tells us that hp & torque are so intrinsically linked so as to be cautious in making too bold of statement regarding either. I was guilty of that.
....separate thought: I think, without re-reading my post, I said it makes no difference at what rpm the engine produces maximum torque, it is that rpm which produces maximum acceleration; this is not so with horsepower. The gearing variable, with no good reason not to do so, only to simplify the discussion, I left out.

This discussion is now well beyond hopeless. If one does not understand basic 400 year old physics, which includes the concepts of work, energy, and power in addition to Newton's Laws you'll just never "get it".

You can pretend you're Aristotle and engage in a soliloquy of "natural philosophy", but the bottom line is that all this was codified nearly 400 years ago and has been taught in basic physics courses at universities and high schools ever since.


Duke

It makes a significant difference at what rpm produces maximum torque. The higher the rpm at which the maximum torque is produced.........the faster the acceleration of the vehicle.

The higher the rpm at which the maximum torque is produced......the greater the horsepower.

You're still hung up on engine torque........it's the torque to the pavement that pushes the vehicle.

05 e320 cdi made more torque (@lower rpm) and less hp (lower rpm) than the gasser e320.
The cdi was faster to 60 but slower in the 1/4.
Will someone explain the physics as to why the cdi was faster?:D

Acceleration is a function of the time-average thrust force at the tire/pavement interface. And since you know from this thread that thrust force = HP/V you can determine the thrust force at any speed/gear if you have RWHP and gearing data. The higher the average thrust force, the more kinetic energy the vehicle will gain.

Unless you have a CVT that allows the engine to rev to the power peak and stay threre, the thrust force will vary with applied power, which is a function of the shape of the power curve and gearing including the intergear ratios.

Turbo diesels tend to have flat torque curves, so they make more power at low revs than a naturally aspirated SI engine of equal peak power that typically has narrower torque bandwidth, so the TD would probably be quicker off the line (which could be offset by turbo lag).

Once you reach peak revs in first gear and the transmission shifts power falls off, so average power through several gears is a function of both gear spacing and the shape of the power curve, and the vehicle that can deliver the highest AVERAGE power, not the one with highest peak power, will accelerate faster.

Duke

A twin engine airliner is cruising at 35,000 feet at a true airspeed of 500 MPH. From available technical data the pilot knows that the power setting he selected yields exactly 10,000 pounds of thrust from each engine.

How much horsepower is each engine producing?

With the information in this thread and a calculator, it's a fairly simple calculation.

Anyone feel free to respond, and please go through your logic and calculations for the benefit of others.

My intent is education.

Duke

Ramblin Wreck to the Rescue
 

F= 2 x 10000 lbs. = 20000 lbs.
Velocity = 500 miles/hr x 1 hr/3600 sec x 5280 ft/mile =733 feet/second

:. HP = 20000 lbs x 733 ft/sec / 550 (ftlb/sec/hp) = 26655 hp :beer: Go Jackets!

And by the way, real Germans still measure their car's output in PS, not Kw.

26655 ÷ 2 engines = 13327.5 hp / engine.;)

Excellent! And I'd like to offer some further discussion, as jet engines offer an opportunity to show a simpler perspective.

At steady speed and altitude the thrust of the engines is exactly equal to total drag force, so the "thrust horsepower" is simply F x V, with suitable attention to dimensional homogeneity.

Ultimately what accelerates a car is thrust at the tire pavement interface, but since an automotive engine's available power is dependent on shaft speed, and gearing establishes the relationship between shaft speed and vehicle velocity, how does this relate to drive thrust?

Very simply, Horsepower = F x V.

If you do chassis dyno pulls and plot HP versus MPH you can use the above relationship to determine the drive thrust curve, and it could be easily programmed into dyno software. Available drive thrust is much more useful for vehicle performance analysis. Most guys focus on "horsepower", even though they really don't understand what it is, and they obsess over peak horsepower even though it's seldom achieved in driving.

Do the test in all gears and you have a complete characterization of maximum available drive thrust at any speed in any gear.

If you have drag force data available, which is determinable from various road and lab tests, you can determine how much excess thrust force is available, over and above total drag at any speed and in any gear, to accelerate the vehicle, and acceleration is given by Newton's Second Law, a=F/m.

This is how drag racing simulation programs work, and they are fairly accurate for a wide variety of cars.

Jets are simple because their output is measured directly in pounds of thrust, so we don't have to go though all the convoluted calculations with power at a shaft and gearing to determine available drive thrust.

The Toyota Prius has a CVT, and if you floor the throttle from a dead stop the engine revs up to a constant speed, and it accelerates more like a light airplane than a car because thrust force is maximum at initial acceleration. The initial high available force to accelerate the car is reduced steadily as speed increases and a greater proportion is required to overcome drag, so acceleration moderates with greater speed. It's somewhat complicated by the electric assist, but the electric motor delivers near constant power at WOT regardless of engine speed as long as the batteries are at full charge. There is probably also some variation of power with speed due to variation of CVT efficiency.

The "ideal car" would have a high efficiency (at all speeds) CVT. At any speed, if you want maximum acceleration you just floor the throttle. Revs would increase nearly instantly to peak power and peak power would delivered to the rear wheels continuously - no need to downshift and wait for engine revs to get into the peak power range to achieve maximum drive thrust and acceleration.

The Prius CVT does not allow the engine to rev to the peak power revs, initially. It stablizes somewhat above peak torque. This is probably to reduce, noise, wear, and keep BSFC at a low level. At peak power speed most automotive SI engines have higher BSFC than near the peak torque speed because friction horsepower increases with the square of speed, so most automotive engines achieve best full load BSFC in the vicinity of peak torque speed. So the Prius will not allow access to peak power at normal speeds in order to lower fuel consumption.

At high speed the Prius engine revs may close in on peak power revolutions, but I only did 0-60 tests. With the Prius modest power it will probably only achieve around 100 MPH, and it would probably take nearly two miles to achieve, which is typical of low drag, low power cars.

High powered high drag cars like F1 achieve their considerably higher top speeds in much shorter distance - probably no more than about a mile with a high downforce setup.

Duke

I might add that a gas turbine powered airliner falls under this scenario. The engine achieves maximum power within 10 seconds and the acceleration of the airplane is at its maximum within 10 seconds of brake release.

It's not possible to use the thrust formula to compute horsepower in this instance because the airplane is not at constant speed.

As the airplane builds speed, the thrust from it's engines continually decreases while, simultaneously, the force due to friction across the fuselage and wings increases. At the end of the runway, the thrust available for acceleration of the airplane is less than half of the thrust available when the airplane is sitting with locked brakes.

I find airliners fascinating in this regard........and calculation of their available horsepower at any given point in the acceleration of the airliner to be quite a challenge.

Good job guys!
 

I am really impressed! That calculator must be red hot from all this punching!

Now, seriously. Max horsepower and max torque are intrinsic and independent characteristic of an engine. Because of the relationships stated here many times already, you cannot have one without the other. A car with very low torque will take a long time to accelerate to a high speed no matter how much power is available. It will stop accelerating when the rpms (and velocity, P = drag x V) are such that max power has been reached. Conversely a car with high torque and low power will accelerate very quickly to a low speed, the one at it will reach max power (as before). So, in mi humble opinion, Torque will determine the max possible acceleration, power will determine the maximun speed.
Mi $0.0002

JL

Unfortunately, not.

A motorcycle is a classic example of a machine with relatively low torque but very high horsepower. Small engine that revs very high. With proper gearing, it puts massive torque to the pavement and accelerates very quickly.

I'll take a vehicle with high horsepower and low torque and beat the pants off your vehicle with high torque and low horsepower, provided that I get a decent set of gears to work with. Provided that the gears are available, higher horsepower will win the race every time.

Duke explained that the gears are not always perfect, so, it's possible for a vehicle with lower horspower and slightly higher torque to be beaten by the converse, but, it's not the norm.

The converse is also true.