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Engine Index Ratings (EIR) By Cyrus


cyrus

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Engine Index Ratings

 

I just came up with EIR to evaluate the best possible engine availble. Tired of specific out put balonney, etc. Well to rate an engine overall: EIR = (HP x combined mpg)/weight

 

 

ie...

EIR of Stock LS1 = (325hp x 21 mpg) = 17.06

----------------

400 lbs

 

Lets run some more motors to see what is best.

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Good points. We will have to assume the economy as installed in a 240Z. And since we are evaluating engines we should use the engine weight. Now the rotoary with turbos is 255 stock and 400 pounds with twin turbos........ We need a stock and modified class.

 

 

 

EIR of Stock 94 Rotary = (255hp x 20 mpg)/400 = 12.75

 

 

EIR of Stock 94 LT1 (corvette) = (300hp x 22 mpg)/510 = 12.94

 

EIR of Stock 94 Supra Turbo = (320hp x 22 mpg)/590 = 11.9

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Guest bastaad525

I think I've got a better idea....

 

 

I've always thought it makes no sense to talk about an engines capabilities in hp and torque PEAKS

 

Rather, just like everyone always talks about power under the curve and area of the curve truly determining speed, I think motors should actually be rated by their TOTAL power. Maybe take a sampling of their HP in 500 or 1000rpm increments, from idle to redline, and either just give a total or divide by the number of increments you used to get the average power or something like that, something that really tells you how strong the motor is and would point out the weaknesses of really peaky motors with high peaks but sucky curves...

 

I'm not genius so dont know what all would have to go into this equation but I think someone could come up with a way of doing this that would very accurately be able judge exactly how fast any car would be.

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Guest KEINoze

The area under a curve? Sounds like a calculus problem! Anybody taking calculus?

My last class was 4 years ago . . . don't remember much about it now.

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area under a curve is given by the integral from x1 to x2,

 

so if x= rpm

 

and y = horsepower

 

then

 

y(x1 to x2) = the integral from x1 to x2, d(x).

 

I can write it out without symbols........

 

ahh nevermind

 

back to the earlier post the purpose it evaluate engines. Aux has a point in that his engine differs from stock a lot. So he could cal the EIR for stock and for modified.

 

I thought of this when comparing a "little" 4.6 mustang engine "big" 5.6 LS1, the ford look about double the size and make way less power. So I want to be able to quantify that. I was reading an article automotive engineering and GM looks at MPG x HP = index

 

from page 100

car hp x mpg = index

c6 400 x 22.6 =9040

GT2 porsche 477x 18.2=8981

viper 500x 15.5 = 7750

911 turbo 415 x 18.2 = 7553

911 3200 x 20.6 = 6592

murcielago 580 x 10.8 = 6264

s2000 240 x 22.7 = 5448

z4 225 x 23.6 = 5310

360 400 x 12.7 = 5080

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In the ideal world let's consider two cars assuming:

 

Equally weighted cars (including engine and drivetrain)

Perfect transmissions that will keep the RPM exactly at whatever your HP peak RPM is consistently with no shifting.

Perfect traction (no slipping).

Equal Aerodynamics.

 

Car 1 has a 4L V8 with 300 peak HP at 5000RPM it's transmission will keep running at exactly 5000rpm

Car 2 has a 2L I4 with 300 peak HP at 6500RPM it's transmission will keep running at exactly 6500rpm

 

Two fast cars right? Which one is faster? Or are they equal?

 

Remember Torque = HP X 5252/RPM (Ever notice on every dyno curve you have ever seen that HP and torque always cross at 5252? Not a coincidence.)

Also HP = Torque X RPM/5252

 

Results:

 

Car 1 has 315 ft-lbs of torque at 5000RPM

Car 2 has 242 ft-lbs of torque at 6500RPM

 

They have the same power rating, but car 1 has more torque and will always win.

 

However, it is probable that Car 2 has a higher torque than car 1 at 6500RPM, so run them both at 6500PRM and Car 2 wins.

 

 

In the real world: (IN GENERAL)

 

Car 1 has a heavier drivetrain and engine and needs to be stiffer(heavier) to handle the extra torque.

 

Car 2 has a higher peak and narrower powerband and will require a higher 1st ratio to get the torque to the road and more frequent shifting (more gears) as a result which wastes time.

 

Car 1 will have a higher torque across the entire range of RPMs and a wider powerband would require fewer gear shifts.

 

Car 2 will have better gas mileage because it can be made lighter and at lower RPMs the engine can be made to run more efficiently.

 

So to truly compare one car to another we need weight, gear ratios, final drive ratio, the complete powerband curve, what tires they are using (size, brand and model), their aerodynamic signature including downforce and drag (assuming drivers are equal), and one heck of a complex empirically based formula taking all these into account. :icon8:

 

This could be done and we could make a simple number to tout your ride, but with unequal drivers, you could still lose. :icon52:

 

So to avoid this complexity we could just compare HP and be done with it. :)

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In that "perfect world" example, the car which peaks at 6500 rpm would be equal to the car which peaks at 5000 rpm, since your "rules" state that they would be kept at peak rpm at all times. This means that the 6500 rpm car would be geared 1.3 times higher (numerically) than the 5000 rpm car at all times to compensate for the rpm differences ( 6500 / 5000 = 1.3 ), which means torque at the wheels remains the same for both cars at all times:

 

300 HP / 5000 RPM * 5252 = 315.12 LB/FT with 1:1 gearing = 315.12

300 HP / 6500 RPM * 5252 = 242.4 LB/FT with 1.3:1 gearing = 315.12

 

Of course, like you said, its only hypothetical and the real world is more complex.

 

Another great example of this, as well as an excellent explanation of hp/torque is given on this site: http://www.v8914.com

You have to read ALL the way through the page to get the answer.

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I like engine rating index, idea I did a similar rating a while ago on excel. I included a number of other variable so it was somewhat subective, but it does give you some trends.

Bastaad525, I agree that the area under the curve is much more important than the peek horsepower. In fact that is how we characterize engine modifications on the SAE car. There is a relatively simple way to approximate area under curves where you don't have a linear equation. It is called a Riemann Sum. I am shure some of you are aware of this technique, but for those who are not here goes.

 

Prof's cap:

With an undetermined area under a curve such as a horsepower curve we can estimate the area by constructing an number of rectangles. The curve should be broken up into a number n, of diffrent intervals. These rectangles should be constructed with a hight of the horsepower curve halfway between the start and end of each interval. Now all you have to do is to sum all your rectangles and now you have an approximation of the area under the curve.

 

Whew, I think I broke a sweat, too much math for a friday...

So for your pleasure I made a spreadsheet to take the work out of it and uploaded it into the download section.

 

TL1000R-(1157*30)/150=231

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OHHHHHHHHHH my goodness!

 

BrandsonZ question was perfect setup of what was CRAZY280Z's grand slam! I have been trying to get that point across to some many people. If all conditions were as brandsonZ says AND weights were equal AND the engines were geared as Crazy280Z says, the cars will run exactly the same as proveable in physics. Brillant!!!

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I thought if their powers were equal they shoudl be equal, but my reasoning was flawed so I was confused. NOW I understand and will most probably forget by tomorrow. But yes, finally someone gets it.

 

HP is a POWER!!! POWER is the energy you need to GO. Torque and HP are related so saying "let's concentrate on torque" will only confuse. HP is the key becasue it takes RPM into account, torque does not. 100ft-lbs at 5000rpm is very different than 100ft-lbs at 2000, but 100 hp at 5000 is exactly the same as 100 hp at 2000. Right? :icon8:

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Why does something so simple have to be so complex?

 

 

I just thought about it all day and came up with this scenario.

 

you have an engine 4000RPM torque max and 6000RPM hp max.

 

If you keep the speed constant...

You have more "power" at the HP max then the Torque max because...

 

Say your transmission is setup such that when you are in say 3rd at 6000RPM (hp max) and you shift to 4th down to 4000RPM (TORQUE MAX). Because you gearded down to get back to the torque max you actually lost torque at the wheels because even though you had less torque at the engine in 3rd at 6000RPM you have less mechanical advantage in 4th at 4000RPM.

 

Can we find 7 other ways to rehash this?

 

The final answer on this is use HP always to describe your engine, even if it is a 1.0L with 1000hp or a 10L Diesel with 1000 HP . :-D

 

But that 1.0L engine better have a CVT attached to it.

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