I'll try to break this down as best I can, because a long time ago I also was of the opinion that horsepower was what mattered, and not torque. However physics just didn't agree, so eventually I learned how it really worked.

Let's start with the simplest thing... what makes anything move? Aplpied FORCE. For a car, it's torque. NOT Horsepower, and let me demonstrate that with an example from airplanes. Suppose we want to make a plane that can take off vertically, how do we go about doing that? Well first we figure out how much the plane will weigh, let's say 30,000 lbs. How much force do we need to apply to make it take off vertically? Anything over 30,000 lbs. Why? Because the resistance to movement of the plane imparted by gravity will keep it right there unless we overcome that force with a greater one of our own. Now if we applied 29,000 lbs of force to the plane in lift, it still wouldn't move. We've applied a force, but have not accomplished any work. So in order to accomplish work, we must provide a superior force which can then accomplish that work. And what if we want to lift the plane faster (more work over the same time period)? We apply a greater force than before, thus increasing the ratio of thrust to weight. Now we're getting somewhere, and if all goes well getting there in a hurry. If we continue to apply the same amount of force we will continue to accelerate the plane as it lifts upward, until the force imparted reaches equilibrium with the forces resisting our motion (which now is both gravitational force and aerodynamic drag).

In essence, its exactly the same with a car. Though the resistance to motion is quite different in nature, its the same overall thing we are doing. We move a car by applying a force (aka torque), and we can later sum up how much work we've done by stating it as horsepower. But what is moving the car is the same thing that moves anything, applied force. And the more force applied, the more we can move it in a given period of time. It's only after we have moved it that we even have horsepower, because this sums up what has already been accomplished. Horsepower cannot move a car, it can only tell you how much you've already moved. Nor can it accelerate a car, becuase it is the force applied that defines how quickly it can move.

Yet another example: Imagine a car accelerating down the 1320. It drives through the traps at 100 mph, but in essence what you've done is move down the track while hitting an infinite number of speeds between 0-100 mph. Let's say you are at 8.357 mph, how do you get to 8.358 mph in the least amount of time? By applying the maximum amount of force possible to the vehicle, thus moving it down the track and accelerating at the same time. If you broke the track down into these little slivers of speed, you'd see that what makes the car move at all (much less accelerate) is the force imparted by the drivetrain. Its a summation of force over time, but as you see what causes the vehicle to move and accelerate is not time based, it's based on force imparted.

Last example: We still want to go from 8.357 mph to 8.358, but this time we'll introduce gearing. We'll take two identical cars with different tuned engines, and gear each to make maximum torque at 8.357 mph. Both cars are identical except the engines, where engine A makes 200 lb/ft of torque @ 5000 RPM, and engine B makes 200 lb/ft of torque at 8000 RPM. If both have a 26" tall tire, the tire needs to spin exactly 108.04 times a minute to run that speed. So in order to gear down the engines to this speed, the overall gear ratio for engine A would be 46.125:1. Engine B would want an overall ratio of 74.047:1. Having already covered torque multiplication earlier, we can see this extra RPM ability of engine B is having a HUGE effect on applied force. Engine A's effective torque to the wheels is 9,225 lb/ft, while engine B is pushing 14,809 lb/ft! That's over 60% more effective applied force, even though both engines are making the same amount of torque. And that, ladies and gentlemen, is what all this talk is about.

Effective torque to the wheels is what moves a car. Force imparted equals movement and accelerative ability, and the more force you can apply the faster you can move and accelerate (given unlimited traction). I've also shown how important gearing becomes assuming target RPM points are identical based upon desired torque output, in that the engine which makes torque at the higher RPM has a huge advantage in torque multiplication. And those engines, well the one that makes the same torque at a higher RPM makes more horsepower. Why? The torque output is the same, but the rate at which it produces that torque is enhanced and, taken into proper account with gearing, makes all the difference in the world. Now, does that clear things up?

 

You know what, you got me on something . I really don't know my calculus. The derivative of velocity is acceleration. Becasue I forgot most of calculus, I don't know how to take that any further to show my point.

Let me try another method to show how acceleration is related to energy:

We agree that the speed of an object can be expressed in terms of energy, correct?

v = sqrt(2KE * m) ---solved KE formula in terms of v

Can we agree that velocity can also be expressed in terms of acceleration, such as:

v = v0 + at (initial velocity + acceleration * time)

If not, then:
http://hypertextbook.com/physics/mec...ns/index.shtml
This link shows some other ways that velocity can be expressed in terms of acceleration.

Simply substitute v in the original formula for v0 + at:

v0 + at = sqrt(2KE * m)

Rearrange, and:

a = (sqrt(2KE * m) / t) -v0 / t) or,
KE = 1/2m(v0 + at)^2

You have acceleration expressed in terms of energy, or energy in terms of acceleration.

This link will help drive my point further:
http://theory.uwinnipeg.ca/mod_tech/node30.html






[Modified by Lsos, 8:51 AM 10/17/2002]

 

V = Vo + at
is only valid when acceleration is constant. Anytime you have have a varying acceleration, that kinematic equation will not apply.

In order for it to be valid for your substitution into KE, acceleration must be constant for the given system. And you are trying to prove that acceleration will not be constant right?

I wish I could crank out how to determine the amount of KE or ME the engine produces w/ respect to engine speed, torque, time and/or HP, but I can't quite think of that now. Lemme get back on that, thou I'm guessing it increases exponentially as well with whatever variable that controls it.

 

I see what both of you are saying, and it makes perfect sense. More torque to the wheels is equal to more acceleration. That's stimple high school physics right there.

My problem is that while acceleration can easily be shown in terms of torque, I just demonstrated how it can be shown in terms of power. That's simple thermodynamics right there.

So which one is it? From my experience in a b16 crx, I'd say power. A 5%, 10%, even 30% difference in torque cannot explain the outright kick to the back at 7000rpms compared to virtually nothing at 2000rpms. The b16, although lacking in torque, has been praised many times for a linear torque curve.

Texan, I still don't know what your answer to this would be: When would a car accelerate faster in the same gear, considering its engine has a perfectly linear torque curve of 100lb-ft from 0 to 10000rpms.....at 1 rpm or at 9000rpms?

If you say at 1 rpm, how would you defend against the thermodynamics argument?

 

Whatever it is I'm trying to prove, I just showed you how acceleration is related to kinetic energy.

Constant acceleration or not constant, the two are very much intertwined. If acceleration is in fact not constant, simply divide the acceleration curve into a million pieces and use my formula for each one. It can be done.

The relationship between KE put out by an engine and horsepower is that it is the same exact thing.


[Modified by Lsos, 9:24 AM 10/17/2002]

 

KE and Velocity are related, but the fact of the matter is that the basic rules of acceleration and force will still hold true. The reason why KE grows exponentially because when you calculate it, you are calculating the total sum of all the values of velocity over a given interval times the mass of the object, usually starting w/ zero so the 1/2 mVo² term is zero. Thus while it may appear that you have _____ joules of KE for a given speed, which seemingly is 4x the amount of KE for something half the speed, you have to remember that amount KE does not direct the acceleration of the system. It is a measure of how much total velocity the system has experienced over a given interval (with reference to the mass as well). So regardless if 8000rpm is a higher engine speed or higher vehical speed w/ much more KE than say 2000rpm, it does not mean that it is accelerating much faster.

The engine does not put an amount of KE equal to 1/2 mv² into the system at any given moment. Which I think is what you might have thought earlier. The measure of the engine's input at any given moment would be the input of torque.

Also, are you sure you are converting HP to joules properly? HP by definition increases linearly w/ engine speed (and vehical speed), while KE increases exponentially w/ vehical speed. I dont see how you can convert them linearly by multiplying w/ a factor of time.

For your proposed situation, if there is no inertial or frictional losses or air resistance, the car w/ a flat torque curve will accelerate constantly at 1rpm or 9000rpm. In real life, there is a given amount of inertia and friction that an engine must overcome to move the car, that is why the car usually feels sluggish taking off, regardless if the torque curve is flat.


[Modified by TypeSH, 10:57 AM 10/17/2002]

 

I'm not saying that the amount of KE a system has dictates its acceleration. I'm saying that the amount of KE a system is GAINING dictates its acceleration. Like my water example, it's the amount of heat that the water is gaining that dictates its change in temperature, just like the amount of KE a car is gaining dictates its change in velocity.

As a car gains speed, it gains KE. That KE has to come from somewhere. It comes from the engine, which by definition converts some sort of energy into kinetic energy.

If a 1000kg car is traveling at 100m/s, it has 5000000 joules of kinetic energy. That energy came from somewhere, and unless the car went downhill or something terrible happened, it came from the engine. If the car got to 100m/s in 10 seconds, then that means that the engine has to have put the 5000000 joules of energy into that car, in 10 seconds. The engine has to have been producing an average power of 500000j/s, put in another way, 670hp.

If the engine was producing any less than an average of 670hp during that 10 seconds, then what we have just done is created energy.

Once again, let me reference this link to show that if I am in fact crazy for stating all this, I'm not the only one
http://theory.uwinnipeg.ca/mod_tech/node30.html
http://www.aapt.org/tpt/pdf/april02/...ez_april02.pdf
http://www.geocities.com/roadratccc/vel2hp/vel2hp.html

By the way, what are you talking about with inertial losses? Inertia is the only reason it takes torque or horsepower or anything to accelerate an object with mass. In fact, inertia, by definition, is a measure of an object's resistance to acceleration. No inertia is no resistance, and a 0-60 time of...0 seconds.


[Modified by Lsos, 5:41 PM 10/17/2002]

 

"If the engine was producing any less than an average of 670hp during that 10 seconds, then what we have just done is created energy."

Absolutely, and nobody disagrees with this. If you re-read my original post, this is exactly what I was getting at. No single number is more important to gauging the ability of a car to accelerate than horsepower, BUT at the same time it has nothing to do directly with the force behind acceleration. It is a derived number from that force, expressed as that force over time. Aka the rate of force output. Now, to put this into final perspective, let's take one final example.

Engine A produces peaks of 200 lb/ft of torque @ 4000 RPM, and 250 hp @ 7000 RPM (187.57 lb/ft of torque @ 7000 RPM).
Engine B produces peaks of 150 lb/ft of torque @ 6000 RPM, and 250 hp @ 10,000 RPM (131.3 lb/ft @ 10,000 RPM)

Both cars have CVT transmissions this time, instead of conventional forward gearing. Now in previous examples I've definitely shown how torque is the motive force for acceleration, and given that you are in a single gear ratio throughout the RPM range, this by definition means that (because of the single multiplying factor) the force available for acceleration will exactly mirror that of the torque curve, albeit multiplied. Wherever peak torque is reached one will also have the most force available for acceleration, otherwise stated as the maximum point of accelerative ability for that gear. BUT, if we now use a CVT transmission capable of picking exactly the gear ratio we want for any given speed, well let's see how that math works out taken at both the torque and hp peaks of these two motors. Both are again using a 26” tall tire.

Engine A @ 20 mph:
Gear ratio @ 4000 RPM torque peak: 15.47:1
Effective wheel torque: 3094 lb/ft
Or…
Gear ratio @ 7000 RPM hp peak: 27.07:1
Effective wheel torque: 5077 lb/ft

Engine B @ 20 mph:
Gear ratio @ 6000 RPM torque peak: 23.21:1
Effective wheel torque: 3481 lb/ft
Or…
Gear ratio @ 10,000 RPM hp peak: 38.67:1
Effective wheel torque: 5077 lb/ft

Now, if this doesn’t explain it Lsos, I can’t think of what will. Things to take away from these examples:
-more torque at a lower RPM point, meaning a lower mechanical advantage for our gearing, can still equal a lower effective output at the wheels
-equal horsepower ratings, regardless of RPM point or torque required to produce that horsepower, will always equal the same effective torque to the wheels at a given speed and given properly optimized gearing.
-if we had a CVT in our cars, we would not want to have it hold peak torque RPM for maximum accelerative ability, we would want it to hold at peak horsepower output RPM.

So what this should finally bring into focus is how absolutely important horsepower is to defining the accelerative ability of a car, and also how it is useless in finding the force behind acceleration in a given gear ratio (you must derive the torque value to calculate). In our CVT cars we wouldn’t want to hold the engine at peak torque and adjust gearing accordingly, because getting the maximum amount of work done over time is all about horsepower output (which you see in the effective outputs). And that horsepower output does just what I said it does in my first post, it nicely packages torque and potential gearing into a single number, thus accurately guesstimating a vehicle’s ability to accelerate while also not having anything directly to do with accelerating it.

Now if you disagree with this, please stop throwing up math equations for my sake and explain in plain english how your car accelerates, using a single gear ratio, at its horsepower peak faster than at its torque peak. Because as far as I’m concerned, this is absolutely and positively how it all works.

 

I believe TypeSH was disagreeing with this, although it might have been a misunderstanding.

I intend to now show how your examples, which I understand and admit I intuitively want to leave as fact, actually lead to a disagreement with the above quote.

Imagine an engine with a flat torque curve of 100lb-ft from 0 to its redline. No gearing, just straight to the 2ft wheels. According to you and the torque-at-the-wheels-argument, it should accelerate a car at the same rate throughout its rev range. That's 100lb-ft of torque, which makes for 100lbs of accelerative force. Is this correct? Please stop me at any moment and tell me where I'm straying off.

So here we supposedly have 100lbs of force acting on the 100kg car.

After 1 second, the car should have reached 4.4m/s

The thing is that if it's traveling that fast, the car is carrying 968 joules of energy. You stated that nobody disagrees that this requires an average of 1.3 horsepower from the engine. However, the engine in that car is only turning around 2rpm! It has only started to make .04 horsepower! So here's two things that disagree with each other, and from what I understand you agree with both of them.

So that's the piece of the puzzle that doesn't fit for me. As long as that piece does not fit, I can't accept that a car's acceleration follows it's torque curve, and not its horsepower curve.
I'm sorry for all the numbers and lack of plain English, but that's what I need to do to show my case. My point is that when mirroring the acceleration of a car after it's torque curve, the numbers just don't add up.

Don't ask me why. I honestly don't know. All I know is that your simple F=ma equation has been accepted by physics, but so have my thermodynamic equations. For some reason they don't arrive at the same answers. Please don't throw any more examples at me now, because while they help you prove your argument, they don't disprove mine. Instead of building your case, I ask you now to destroy mine.



[Modified by Lsos, 7:17 AM 10/18/2002]

 

I will spend some time this weekend to go over the relationships between all of them... I am still not sure if you are converting the HP to joules correctly, or using them in the right context. I will spend more time to go over what you are saying this weekend.

just a note, for that 4.4m/sec after 1 sec example u used...
2ft = ~0.61m, diameter I'm assuming you mean?
so 0.61*pi = 1.9163m circumference of the thing.

for 4.4m/sec, the wheel/engine will have to revolve 2.29609 times per SECOND to maintain that speed.

You need to multiply that by 60 to get Revs Per Minute, so rpms should be 137.7654, not 2 rpm.

w/ that rpm, i have 2.623 HP produced, more than the required 1.3HP you said is required ( i didnt check the math), so i would say the car is still accelerating.

 

"Imagine an engine with a flat torque curve of 100lb-ft from 0 to its redline. No gearing, just straight to the 2ft wheels. According to you and the torque-at-the-wheels-argument, it should accelerate a car at the same rate throughout its rev range. That's 100lb-ft of torque, which makes for 100lbs of accelerative force. Is this correct? Please stop me at any moment and tell me where I'm straying off.

So here we supposedly have 100lbs of force acting on the 100kg car.

After 1 second, the car should have reached 4.4m/s

The thing is that if it's traveling that fast, the car is carrying 968 joules of energy. You stated that nobody disagrees that this requires an average of 1.3 horsepower from the engine. However, the engine in that car is only turning around 2rpm! It has only started to make .04 horsepower! So here's two things that disagree with each other, and from what I understand you agree with both of them."


You stated the vehicle has a 2' tall tire. That measn it has a circumference of 75.3982", and as you stated your engine is turning 2 RPM after that one second of applied force. Assuming it started out at 1 RPM, that means your vehicle has covered exactly 1.885" of ground in said second ((1.5 RPM/60) x tire circumference)). Which means it's velocity is .0479 m/s at the end of that second. In order for the vehicle to travel a 4.4 m/s, the engine would need to be spinning at 137.851 RPM at the end of that second. I took a look at your numbers, and I think I found where you goofed (or perhaps I'm goofing things up further).

You are converting pounds-feet to newtons with a conversion factor of 4.4482217. That's all fine and good, if you initial measure is in pounds. It's not; it's in pounds-feet, and that conversion is 1 lb/ft to 1.3558 Nm (aka joules). You can't convert pounds/ft to Netwons because a Newton is a measure of force over time, and torque is not time based. The actual amount of energy put into this object is 135.58 joules per second, which doesn't really help us either, we need Newtons! However, we can convert pounds-feet to kilograms-meter (1 = .138255), which just happens to be 1 newton when applied per second (thank you for picking that time frame). So here's how the math should work:

A = F/M
Where F = force in Newtons (1kg/m per second per second)
M = mass of object in kg
A= m/s

And so using these corrected figures, we get A = 13.8/100, which is .138 m/s. Convert meters to inches with a multiplier of 39.370079 and we get 5.433 inches per second, or 8.575 e-5 mph. Our 220 lb car is now screaming along at about the pace grass grows, but its engine has moved on up to the busy pace of 4.323 RPM. In the next day or so, it should reach redline. See why gearing is so important?

Now, how am I wrong here? And even if my math is flawed in applying F=ma, yours is definitely off too. So is there a third party who can confirm or deny any of this?

 

You got me on the rpms
I appreciate you guys taking the time to go through my equations, and I apologize for the mistakes.

Oh... another mess up, in my example the car started at 0rpms. I did not mention that.

So 1 second doesn't show my point. However, if you only increase the time past 1.8 sec, the crossover will occur. At 4 seconds, you'll see that the car will be carrying ~9000 joules, requiring an average of 12 hp. At that point the engine is only producing 7.3 hp. Keep increasing the time and the energy deficit wil grow exponentially.

A Newton is not force/ time. It is simply force. 4.448 Newtons = 1 Pound.
http://www.themeter.net/conv6_e.htm

You also need to realize that torque is not energy. The units get confusing because both are expressed in force and distance. However, in energy it is referring to how far a force has pushed something, expressed in force * distance. In torque it is referring to how much force is pushing on how long a lever, expressed in something like force - distance. That's a dash in there, not a minus, not a *, not a /. Those are two very different things. You cannot express torque in joules. You simply say Newton meters or foot pounds, but never joules.



[Modified by Lsos, 10:42 AM 10/18/2002]

 

lemme see if some graphs and general formula/graphs can help explain this a bit better...

you guys mixing and matching SI units w/ american units is very very confusing, esp w/ the chance for error in converting and also in converting properly by maintaining the proper units.


------------
if we say that acceleration is constant for a given car, the velocity graph or function will be a line, lets call it y = x to make it simple.

Get a piece of paper and draw y = x cuz i dont want to do this in photoshop.

Now draw vertical lines at x = 1 and x=2.

now, lets analyze this situation.

At x = 1, we have a given velocity thats already there, the BEFORE. And then point 2 is the new velocity after a given period of time, the AFTER.

Now if you were to plug in the kenetic energy equation of

KE = 1/2 mV² (AFTER) - 1/2 mVo² (BEFORE), the figure you get you would say is the amount of energy difference between the initial (BEFORE) velocity and the final (AFTER) one right? So you would say that that KE figure is the amount of energy required to accelerate that car over a given interval of time? So the car would need to average a __ corresponding HP figure over the interval in order to properly accelerate as I say it would. At least that is what I believe you are trying to prove here for us.

However, I dont think that is the case. Draw a horizontal line at the point where the vertial x = 1 crosses w/ your y = x velocity function. At the point of x = 1, the car has already has some velocity and some motion in it. The energy required to increase the speed from the intial to the final, I believe is the area of the triangle formed by that horizontal line you drew, x = 1 and x = 2.

The rectangle formed by the horizontal line you drew, the x-axis, x=1 and x=2 can be ignored because it is the energy already in the system from the interval of 0 to 1. It was already there when you began measuring at x=1. The car does NOT have to add that amount of energy.

That is why even thou KE grows exponentially w/ respect to a linear increasing velocity, it does not mean it required an expoential amount of energy added to increase the velocity linearly.

I hope this can clear it up a bit...


-edit-
addition:
Now if you were to make the interval of comparision infinately smaller and with: 1) each interval cut having a rectangle area representing the amount of KE that is already present in the system at the begining point that you are analyzing 2) and the triangle area representing the amount needed to increase the velocity from intial to final, then it might also be clearer to visualize my point.


[Modified by TypeSH, 11:10 AM 10/18/2002]

 

You're right, my conversions were wrong Lsos. So I went about it a different way (simpler IMO because there's very few conversions).

gForce = wheel thrust/vehicle weight (all in pounds)
or g = 100/220.4 = .4537g
1g = 32.174 fps/s
So this vehicle is able to accelerate 14.6fps/s, and should have moved 14.6 feet thus far.

So it's going 4.4m/s, which would equal 990.16 joules, which we then convert again with you supplied math (from those links) to horsepower by dividing by 746. Which equals 1.327 hp. Now over that one second the engine has also gone from 0 RPM to 139.42 RPM, which we should average to find average horsepower over the second. Which equals 69.71 RPM, converted to horsepower equaling 1.327 hp.

And I'm spent!


[Modified by texan, 12:36 PM 10/18/2002]

 

Torque is basically the power it takes to turn your driveshaft, it is measured in ft/pds. Horsepower is basically how fast a motor can make torque. So a high torque/low horsepower motor, such as a diesel, is great for towing, it can pull alot, just not very fast. The reason hondas are able to go fast is because they are so light, and they don't have to pull alot, which is good because they are relatively low torque motors.

 

Type SH, I don't think that's the way it works at all....

"That is why even thou KE grows exponentially w/ respect to a linear increasing velocity, it does not mean it required an expoential amount of energy added to increase the velocity linearly."

How can energy in a system increase faster than it is being inputted? That's a direct violation of the 1st Law of Thermodynamics, and the reason why this whole discussion is going on.

In the graph you describe, first off the area under the curve is how far the object has traveled. Second, at point 2 you cannot ignore the rectangle formed by the horizontal line. It is not already in the system. At point x=1, the area under the curve (a small triangle) is in the system. That's it. At point two, another small triangle plus the rectangle you ignored is added.

To verify what I said:
http://www.angelfire.com/home/montav.../SampleLab.htm

And like I said, your concluding statement "That is why even thou KE grows exponentially w/ respect to a linear increasing velocity, it does not mean it required an expoential amount of energy added to increase the velocity linearly.' is absolutely false.

texan. like I mentioned previously, I goofed with the 1 second example and apologize for the inconvenience. However, since your calculations seem to match mine, we are on the same page. I ask you to keep everything the same except increase the time to anywhere past 1.8 seconds. You can probably do it quickly in your head while looking at your last post.
You'll see that the horsepower deficit grows exponentially as you increase the time.



[Modified by Lsos, 10:02 AM 10/19/2002]

 

You know what guys, I need to reiterate something. I fully agree with Newton and that F=ma. I fully agree that a higher torque at the wheels must produce a higher acceleration.

However, you have not made me budge from the fact that energy must be conserved as well. If my equations are correct, then it means that I agree with two things that don't agree with each other. That is obviously crazy.

I do have a theory that could tie it all together, though. It's just a theory that makes everything make sense to me and keeps me from going crazy, though, and could very well be wrong.

Perhaps when the torque of an engine is given at an rpm, it's only there when there is no acceleration present. That is, there's a resistive torque that's just as great. Could it be possible that once that resistive torque is taken off and the car / engine starts to accelerate, the torque drops? What I'm saying is once the torque causes the car to accelerate, the torque can't keep up with the car? This would be sort of like not being able to push on a wall as hard when the wall is falling away from you.

Think of a bike and putting all your weight on the pedal with the brakes depressed. The pedal is heavily stressed and puts a lot of torque on the sprocket. The instant the brake is let go, the pedal falls away and is not as stressed anymore, even though you're trying to put all your weight on it. Your weight can't fully keep up with the pedal, and the torque you're applying to the sprocket drops as a result.

Does this only make sense to me?

 

gForce = wheel thrust/vehicle weight (all in pounds)
or g = 100/220.4 = .4537g
1g = 32.174 fps/s
So this vehicle is able to accelerate 14.6fps/s


So after 2 seconds it's going 8.9m/s, which would equal 3960.5 joules, or 1980.25 watts, or 2.65hp. Now the car's engine is also spinning at 278.84 RPM, which averaged out over the 2 seconds would be 139.2 RPM, which equals 2.65hp.

After 10 seconds the car would be traveling at 44.5 m/s, which would equal 99,016.06 joules, or 9901.6 watts, or 13.27 hp. The engine is now spinning at 1394.198 RPM, which averaged over the 10 seconds would be 697.1 RPM, which equals 13.27hp.

Where is the inconsistency again Lsos?

 

There was a couple more glitches I found in my calculations. One was that I wasn't using a precise enough conversion for meters to feet, making for a small % error, and the other was that I divided KE carried by the car by a wrong time....and not to waste anymore of your time and answer your final question....there is no inconsistency.

It looks like the piece of the puzzle that was missing to me was there all along, I was just trying to put it in upside down. Both thermodynamic and F=ma equations seem to yield the same answer. I think most everything makes sense to me now. Something still doesn't feel right to me though, such as why I remember the b16 having close to nothing at lower revs if it supposedly has a flat torque curve. I might have to test that with an accelerometer or something before I get the idea out of my mind.

Nevertheless, because I can't think of any more logical arguments to back my original claim, I have to say you guys were right. As far as I now see, a car accelerates at a rate that exactly matches its torque curve.

Thanks for bearing with me texan and Type SH.

Type SH, I apologize for dismissing too fast and never responding to this argument:

It explained my thermodynamics arguments, and if I had thought about it deeper this discussion would have ended a long time ago.

Come to think of it, this statement by Daemione

could very well explain my possibly exaggerated b16 memories.


[Modified by Lsos, 7:08 PM 10/21/2002]

 

just one thing to add... about my y = x line example w/ the rectangle,

I was wrong to say that that rectangle can be ignored. what i was thinking was when looking at that interval of 1 to 2, when looking at the HP curve (which would also be a line since torque is constant) for the same time interval, it would also have a rectangle which accounts for the KE/ME needed for that rectangle. I was focusing on the change in HP and change in KE from 1 to 2 (both would be triangles).
but good debate, it got me reviewing and applying physics concepts I learned in the classroom.


[Modified by TypeSH, 5:43 PM 10/21/2002]