planetside

The TL-S is the first car I've ever owned where I'm actually interested in the numbers. The problem is, I don't know what the numbers really mean, and what they mean in a practical sense. Maybe you folks can enlighten me somewhat. I'm at Edmunds.com just doing some reading on specifications, and the numbers that interest me are the following:

260hp @ 6100 rpm
232 ft-lbs. @ 3500 rpm

What is it that I can glean from the above info? I am a total car enthusiast newbie, and about the only thing I know about my car is that it has 260hp. Other than that, I have no idea how to read the numbers, or know what they mean on the road (or in theory, even).

Compare the above data for the TL-S to the Lexus GS300:

220hp @ 5800 rpm
220 ft-lbs @ 3800 rpm

I'd appreciate any kind of lesson on such statistics. Thanks!

Danny

juniorbean

iTrader: (5)

Well, to learn about HP or Torque and how they apply... check out these links:

http://www.howstuffworks.com/horsepower1.htm

http://www.howstuffworks.com/fpte3.htm

That is a great site.. so those links should help.

Another thing you will learn is that paper stats don't tell the whole truth. There are a lot of variables when driving that can make your car perform better or worse then what you read. Sure, your car always has 260hp at the engine... but it may perform differently depending on the temperature, humidity, wind, weight in the car, etc.

AcuraTLFan

For the most part this is crank horsepower, the power the engine will produce. It's not the wheel horsepower, the power which is measured at the wheels on the road. WHP is usually 15-20% less of the CHP.

260 HP @ 6100 RPM, means that 260 HP will be achieved once you are at 6100 RPM. You don't get 260 HP from 0 RPM- 6100RPM, the power will increase linearly with RPM...it will reach it's peak then it will decline.

It works the same way with torque. You will acihieve 232 ft-lbs of torque starting at 3500 RPM to 5500 RPM. Anything below 3500 RPM or higher then 5500 RPM you will get less then 232 ft-lbs of torque. Torque is your accleration curve. What i mean by that is looking at a graph of the torque curve, taking the slope of the points is how you will accelerate.

If you have the 2002 Acura TL-S Brouchure you get from the dealer...i still keep mine. It's on page 8, it's a graph of the HP and Torque curve. The definition of slope is change in Y over change in X, meaning the difference in the vertical direction divded by the differnce in the horizontal direction. The higher the slope the steeper the line. So looking at the graph, when you are at 500RPM to 1250RPM you will have the best accerlation(highest slope), from 1250RPM to 3500 RPM you will have better accl. (higher slope),from 3500RPM to 5500 RPM you will have good constant accel. (zero slope) and anything after that your accel willl start to fall.

*Physics review: Accleration is a vector, meaning you have both magnitude and direction. So constant velocity with changing direction will result in acceleration. You can have increasing velocity and have constant or ZERO acceleration...this is what happens when you are at 3500 to 5500 RPM*

So you look at the numbers...and ask yourself...how can 232 ft-lbs move a 3500 lb car? Good question. That is where Gearing and Gear Ratio comes in. Ever wonder what those number are?

For an example take a look at a 18 speed mountain bike. You have 3 gears in front and 6 gears in the rear. There are 18 combintions of 3 x 6 gears = 18 speed. If u are at the smaller gears it will give you alot of torque it is harder for you to pedel....but there's a limit at how fast you can go...once you pedaled as fast as you can you realize that you can go anyfaster. So you shift you to the next gear, which will be easier to pedal and you can go faster. What happens when start off using both of the biggest gears...you find out that you can be pedaling as fast as you can and your not moving very fast. Smaller gear,less teeth; bigger gear, more teeth. This is essentually how gearing on cars work. But we only have 5 gears.

TL-S
I : 3.563
II :1.551
III :1.021
IV :0.653
V :0.470
Final :4.428

The gear ratios are essentually "Multipliers" they multiply the torque that we have to the desired amount that we need. Our first gear ratio is 2.563 and our final drive is 4.428. So you multiply 2.563 and 4.428 = 11.348. So your multiplier is 11.348 times your torque 232 ft-lbs = 2,632 ft-lbs. of torque avaible to push the 3500 lb car. As your gearing moves up, the multiplier decrease, since the car is already moving it doesn't need alot of power pushing it. In fifth gear you're only using 482 ft-lbs.

This is where redline also comes in. Ever wonder what redline is? Redline tells you, you gotta be careful...your engine is working at a critical point. Ok...lets say your crank shaft has 18 teeth on the gear, and your first gear has 6 teeth on it...how many times does the drive shaft turn for every revolution of the crank shaft? 18/ 6 =3 revolutions. So everytime your crankshaft turns once, the drive shaft turns 3 times . So the gear ratio is 3:1. So for the TL-S our first gear is 2.563 means that 2.563 revolution of the drive shaft for a revolution of the crankshaft. In fifth gear, it's .0470 turn of the drive shaft for a single turn in the crank shaft. So when you redline the first gear and shift to the 2nd gear the RPM drops because now there are less teeth to turn and requires the crank shaft to turn less. Over drive gears which are gears <1, our 4th and 5th gear, .0653 and .470 respectively. The Driveshaft turns less times then the crankshaft...give you less power. (It depends more then just the number of teeth, also depends on the diameter of the gear to fit the number off teeth you want also)

Final Drive Ratio: This is the ratio the drive shaft make to turn the WHEEL one time. So 4.428 means, the driveshaft turns 4.428 times for every turn in the Wheel. If you have different size tires, 18" or 19" you just changed ur Final drive ratio. So in first gear, the crankshaft turns 11.34 times to turn the driveshaft 4.428 times to turn the wheel 1 time.

On to the MAX top speed of each gear. the equation is: Wheel speed = (redline rpm * tire diameter)/(336 * drive ratio).

In one revolution of the wheel, a linear distance equal to the circumference of the wheel will be covered. Every minute, the distance covered will be engine speed (rpm)/drive ratio times the circumference (pi* diameter). So in one hour (the factor, 60), the distance covered, (60*pi*diameter*engine speed)/drive ratio = speed.

Our diameter measurement, in inches , must be converted to mile (1 mile = 5280 feet, 1 feet = 12 inches). So we have,
speed = (60*pi*diameter*engine speed)/(12*5280*drive ratio)
or,
speed = (redline rpm * tire diameter)/(336 * drive ratio).

That's why we can't stay in first gear forever, even though that's the gear with the most power. If we changed the final drive ratio lower, we could go faster in that gear, but we would accel. slower because we have less torque now. So it's basically a design perameter that has compromise/tradeoff.

Ever heard of the pharse "Low end torque at high speeds"? Well, that's what every car maker would like, but it's not possible.

kennelm

If you have the 2002 Acura TL-S Brouchure you get from the dealer...i still keep mine. It's on page 8, it's a graph of the HP and Torque curve. The definition of slope is change in Y over change in X, meaning the difference in the vertical direction divded by the differnce in the horizontal direction. The higher the slope the steeper the line. So looking at the graph, when you are at 500RPM to 1250RPM you will have the best accerlation(highest slope), from 1250RPM to 3500 RPM you will have better accl. (higher slope),from 3500RPM to 5500 RPM you will have good constant accel. (zero slope) and anything after that your accel willl start to fall.

Are you defining acceleration as the change in velocity of the entire car with time?

If so, I would think that the best acceleration can be achieved at the area of the torque graph where it is the highest. That is, along the flat part between 3500 and 5500 rpm. That's why engines with a broad, flat torque curve are better than ones with a peaky one. I suspect the greatest acceleration cannot be achieved when the RPMs are below 3500.

Have I misunderstood your point?

AcuraTLFan

Yes, accerlation is the change in velocity over the change in time.

The intervals with the greatest accel. is the intervals of the graph where the line has the most slope. Looking at the graph, the slope from 500 to 3500 is greater than the slope from 3500-5500.

So as your Velocity increase so does your RPM...that is covered over an interval of time. So the change in Velocity over the change in RPM which translate to time.

What this whole thing means is that you will feel a greater pull when you floor it at 500 to 3500, then you would feel at 3500 to 5500. You will feel a strong pull when you start, and the pull will start to die down and stay constant

A flat torque curve is more desirble...but a flat curve will result in constant accerlation but at a higher value. A car with a peak value only reaches that point and starts to decrease, you don't want that.

You can take a Integra, it has a torque curve that will contiune to climb, it will peak and start to decrase. So you will feel a strong pull and you will continue to feel the same strong pull until you peak, but at a lower value

Here's an example



The first graph is position vs. time...so as time increase you move farther away from where you started....it starts off slow and it increase and it stays constant then stops.

The second graph is the velocity vs. time, which is the derivative of the position vs. time graph. It's bassically the slope of the line. So starts off slow and then steps on the gas real fast and then keeps the gas constant at a certain value, let's say 60 MPH, then he slows down, then stops

The third graph is the accel. vs. time, which is hte derivative of the velocity vx time graph. It's basically the slop of that line. Starts off with constant accel, then increase in accel as the slope of velocity increase...but the slope of the velocity is constant , so there is constant accel and the accel curve flatens out. NOw you're at 60 MPH, although your velocity is higher, your accelration is decreasing. As long as you stay at 60 MPH , there no accel, and accel is at ZERO. NOw you start pressing the brakes...your velocity decrease and now you're deccelrating. Once your velocity is constant again, your accel is ZERO again.

nmbcatch21

ah i see another caluclus man here. I just finished Calc 3 and am taking advanced differential equations right now.

AcuraTLFan

Differential Equations...this must be one of the easiest, but hardest class.

Beiruty

wait till you go to partial diffrentials, surface and mulitdimensional integratls, Closed and open systems, system equations, open closed systems, feedback control systems... fluid dynamics, and state systems, Finite Element Analysis... and the list goes on... Markovian Queues and the Theory of Queuing systems...

Can we add field theory, Stohaistic Analysis, Random Processes and the worst of all...the theory of Chaotic Systems...

AcuraTLFan

Beiruty, you just showed me that i know nothing..heheh

Just finised feedback control systems....complex for such a simple concept.

Learn Differential equations well, you'll use this alot later on

Beiruty

how is your root locus, and what about your stability? I accept at least a "marginal" one with 5 degrees of safety margin...

Beiruty

Oh... I forget to add Digital Signal Processing and some laplace and Fourier transformations to boot.. and do not forget the s and z space... before junpimg to hyper space...