qqwqeqwrqwqtq

LOL, great analogy. As for even tempered, not so sure about that.

......and I breath a sigh of relief while pouring my CFL/Halogen mixer down the drain..

.....gazing in disappointment at my wasted CFL/Halogen mixer, I chamber a round into my P226.....



I would skip that for now and try a little reading comprehension.

Mr. Sir

Yes, it is this as well, but this is just the math, not the theories that prove that math. Consider this; torque is based on Volumetric efficiency (how much air you can get into the cylinder). For now, torque can be the measurement of power for 1 rpm (technically it's not even an rpm based value, but pressure value. To be able to compare it to horsepower, which is rpm dependent, it can be defined as the power produced in a single rotation (a single rotation at 4400 rpm is more efficient than a single rotation at 1000. At least in an ls based engine with stock or larger cam)). So given torque is based on how much air is in the cylinders, and that hp = tq*rpm/5252, it would stand to reason that horsepower is a measure of how many times you can replicate a given torque value.

For example:
A 6.2 liter engine with 100% VE (volumetric efficiency) will require 6.2 liters of air per revolution (at atmospheric pressure). For now, we'll assume 100% VE at all rpm. Given the earlier definition of torque, this means your torque curve will be completely flat, and (for this argument) we'll say it makes 620 ft/lbs. So, as rpm increases, since the engine needs to fully fill the cylinders, the amount of air being sucked into the engine increases. We'll pretend the engine redlines at 6000 rpm, where it produces 708.30 hp.
Now take a 4.8 with 100% VE. It requires 4.8 liters of air per revolution, and (theoretically) produces a flat 480 ft/lbs. For it to produce 708.30 hp, it must spin 7750 rpm.

Now to calculate air mass:
6.2 liters of air (per rotation) x 6000 rpm = 37,200 Liters of air per minute at 1 bar (atmospheric pressure), or 43,896 grams of air
4.8 liters of air (per rotation) x 7750 rpm = 37,200 Liters of air per minute at 1 bar (atmospheric pressure), or 43,896 grams of air

It doesn't reach the same mass flow, otherwise it'd be making the same hp. Remember the 4 page argument we had about trying to get the mass flow on the 6.2 to equal the 4.8? Look at my previous post to square away any doubts about how it effects horsepower.

Fbodyjunkie06

iTrader: (10)

Mr. Sir you do realize that the Godfather of LS engines has come into this thread 3 times and told you indirectly that you have no clue what you are talking about right? And I am not talking about Intm8d although he knows probably just as much as Kurt(427) does.

You are an idiot sir.

05HD

Check this out- http://en.wikipedia.org/wiki/Horsepower

slow67

iTrader: (2)

Wow....you just calculated air VOLUME not air MASS. I want a lightbulb also.

qqwqeqwrqwqtq

I see what you are saying but your example was offered in response to this-

Which is correct given the context is was stated in (small turbo maxed out on a larger engine)-

Not likely, but I do get smarter every time Kurt is kind enough to share some knowledge with me

Mr. Sir

yes, correction; 37,200 liters of air per minute at 1 bar (or 43,896 grams of air).
I'll correct the earlier post as well.

...That was the Mr. Sir of yesterday. I'm an enlightened chap now.

...I did. It's kind of informative, but was there any specific part you wanted me to look at?

05HD

I'm still trying to screw the lightbulb in instead of eating it.

Basically, you have horsepower and torque confused. There is no such thing as "700 hp worth of air". All there is is 700 ft/lbs worth of air. If an engine (picture a healthy big block chev) happens to suck that much in at exactly 5252 RPM and magically convert it at zero loss, it can be called "700 hp worth of air" on that particular engine at that particular RPM but, it still really isn't because another engine (picture a Cummins turbo Diesel) that pulls the same exact amount of air at 2000 RPM it is only going to make 700 x 2000 / 5252 = "266 horsepower worth of air".

Know what I mean?

Mr. Sir

...no. If it sucks in 43,896 grams of air at 2000 rpm, it'll make 700hp and 1838.2 ft/lbs.

Think of it this way:
6.2 liter engine, 100% VE at all rpm.

at 500 rpm, it'll require 3658 grams/minute of air, and produce 620 ft/lbs.
at 1000 rpm, it'll require 7316 grams/minute of air, and produce 620 ft/lbs.

It's the hp that goes up with air mass, not torque. (as torque is a measure of force exerted over a distance, say 1 rpm of distance)

What may be 700 ft/lbs worth of air at 3000 rpm will only fill the cylinders half way at 6000, so VE (torque) will be cut in half, even though hp remains the same.

05HD

What? No. You have it backwards. If an engine magically makes the same torque at every RPM, that would be because it is ingesting the same mass of air per revolution at every RPM speed. AKA, forget about "grams per minute" and start thinking about grams per cylinder per cycle.

Here- http://en.wikipedia.org/wiki/Torque

What is a RPM of distance?

Mr. Sir

Alright, you have the correct train of thought (it's the same idea, but from a single rotation vs rpm viewpoint). I like the continuous flow/rpm viewpoint better, since hp is expressed as the amount of air that moves through the engine in a given period of time, which was very relevant earlier in this thread. So you're right; there's no such thing as 700 hp worth of air. What I meant was 700hp worth of air mass flow.

The difference in ideals is (yours) air mass per revolution vs (mine) air mass flow per revolutions per minute, as it's more commonly denoted.

...nothing. My eyes were impaired and thought it looked correct. Or perhaps I was considering one my other ideas and let my mind wander and mix up the thoughts. I need sleep, I wasted all night reading about damn pressure ratios.

MIGHTYMOUSE

iTrader: (59)

During the order I stated clearly that I wanted the turbo to CARE and Jose at FI made it happen and the results speak for themselves; I don't need to know the details.

I can't post the link because they are not a sponsor, but if you go to amazon dot com and search 'care'; the second result has some good deals.

05HD

Could you please pass the light bulbs? I'm hungry now.

MUSTANGBRKR02

iTrader: (91)

http://youtu.be/5hfYJsQAhl0

This is all that I can think of....

427

iTrader: (3)

I wish!
I do have something coming that I believe you will like, it involves a new dyno room design and something I have always wanted to do.

KurtYou know how long it will take to clean the food out of this keyboard!


Kurt

MUSTANGBRKR02

iTrader: (91)

Your welcome.

It just seems to fit.

JS01

iTrader: (10)

Let's throw another wrench into this. Would like to pick the brains of some of the experts in this thread.

In theory which car will get down the drag strip quicker? (Equal weight for comparison purposes)


Car A.) 700rwhp powerband from 3000-6000rpm

-Or-

Car B.) 800rwhp powerband from 7000-8000rpm

Mr. Sir

Here's your thought process:
In 1 revolution, a 6.2 liter engine ingests 7.316 grams of air (6.2 liters at 1 bar). Therefore, it produces 620 ft/lbs of torque. It doesn't matter how many rpm the engine is operating at, so long as it's ingesting 7.316 grams of air (W amount of air) per revolution, it'll produce 620 (X) ft/lbs. (which is true)

Care to expand on that?

If you think differently, let me know. (and I KNOW if an engine consumes Y grams of air at a constant rate and in a 1 minute time frame, it will produce Z amount of horsepower, regardless of rpm. I'm pretty sure you tried to disprove that earlier.)

qqwqeqwrqwqtq

Whichever one has the smallest intake restrictor

05HD

I didn't have any extra lightbulbs (damned CFLs never burn out) so, I started drinking bourbon.

You can't relate airflow to horsepower. Engines don't make horsepower. They make torque. How fast they make that torque is how we get to your obsession, horsepower.

What was your question again?