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Old Dec 15, 2006 | 12:08 PM
  #61  
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Yea, you're right. I didn't actually mean it's determined by RPM, but reading it I see where it sounds like that. I was meaning that in an engine with a given peak RPM, VE will fall off as the engine speed is increased over peak RPM.
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Old Dec 15, 2006 | 12:14 PM
  #62  
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Derricks first of all is wrong.

The theories would model everything thats going on in a boosted situation and the theory of 14.7PSI is in ignoring tempurature. The correct model will do this.

The correct model will have everything. So in gasoline engines theory is correct. In human emotion you cant quite map or theoretically compute, but on gasoline burning engine, or a braking system or a tire, the theory and model can be made to accurately describe what will happen.

And Prodrive. You can win the argument easy, because if we decide to make max effort and max RPM there is no further RPM! So your one case is correct.

But does not answer that for realistic motors, can double the RPM ( funds permitting ) == greater than double the HP?

IE.

The statement was at twice the RPM that greater than 2x the power is not available and I gave several examples which blow this statement out of the water.

And the allusion I make is easy for many without alot of money to understand! Here goes:

1. Say we are talking 5.0 ford and the poor guy really cant comprehend how much his cobra/stock intaked engine is down compared to a max effort. This type of guy thinks 6000 RPM is high.

Say we made cobra/stock intake with porting the limiting factor. That put you at 5500 power peak max right there. And you can max effort a 5500 power peak if cobra intake was the limiter, all you can improve is everything else around it.

Now remove restriction of cobra intake, and unlimited but pushrod valvetrain, contrained at 306 cubic inches and can use canted valves, titanium, whatever you desire, I'm sure you could make a combo that puts power peak squarely at 11,000 RPM.

The model I describe says that if you designed a motor optimized for 5000 RPM ( Cobra intake ). And then optimize it for double the RPM the power is going to more than double!

And in this case and several others ( dual plane carb vs unlimited sheet metal ) double the RPM will more than double the HP.

On the cobra example. You might be lucky if you make 425HP by 5500 RPM.

Go to unlimited, dual tunnel rams, optimized runners, etc you will pass 850HP by 10,000 RPM and if you could center the HP curve at 11,000 RPM it would be well over 900HP.
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Old Dec 15, 2006 | 12:55 PM
  #63  
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Big DEN, so let me just make sure I'm understanding your "model" correctly.

First I did state that IN THEORY 14.7psi above atmo would yield a 100% increase in power. So I'm not wrong. Re-read my post.

Your "model" doesn't include temperature?? Well does it include air density? What else are you leaving out to "optimize" your "model"? Last time I checked there aren't any compressors, superchargers or turbochargers, that offer a 100% efficiency rating. But I guess your "model" uses one of these theoretical compressors. Unless of course you mean increasing the Earth's atmosphere to yield a 23psi air pressure!

"Optimizing" an engine that was designed to turn 5000rpm to turn 11000 rpm is ludicrous! You can't do it! You'd have to COMPLETELY redesign the engine, you'd have to design a NEW engine! 11000 RPM in a pushrod type engine?? You must like to argue, because that isn't possible in this day and age to do efficiently costwise and resource wise.

I tell you what. Go build YOURSELF an engine, put it a car/vehicle, tune it YOURSELF and then try to double its output by doubling its RPM. And every time you grenade an engine you give me $1. And if your pockets are deep enough I'll take those 1 dollar bills and buy an identical vehicle/engine combo then increase its efficiency and airflow leaving RPM unchanged and whoop the **** out of your broken and busted failed 10000 RPM money pit.

You keep with your theories and such, I'll stick to what I know works.
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Old Dec 15, 2006 | 01:15 PM
  #64  
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Big DEN, let me ask you question. Does the crankshaft flex in the block journals during the operation of an internal combustion engine? Meaning does it flex up and down and about while rotating in the engine with pistons being sent up and down the bore in localized areas of its length? Also are connecting rods "stretched" by this same action? Oh and another question for you, what about the pushrods, do they flex and rotate? I guess all these things and more don't occur in your "model" therefore allowing it to be "optimized" for doubling RPM.

I think I already said this once somewhere at this site, this isn't windows, engines aren't plug and play and can't be "optimized". There isn't any firmware in the internal combustion engine.
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Old Dec 15, 2006 | 01:18 PM
  #65  
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"topend" posted the following:

On a forced induction motor, 14.7 psi of boost is considered x2 the engine NA output.
So if a given 350ci engine made 400 hp NA at 6000 rpms. Would the same engine size make 800 hp at 12000 ( with a better flowing induction and exhaust system of course). Would this be right???

-----

OK so the above is the original question. We got on a track o just engine size remaining constant can you double, less than double, more than double the HP at double the RPM.

first of all, Derrick. Are you going to say "well Dennis, after re-reading, your right and this is interesting view"? Not-likely. So like you say what are we arguing for?

I'll continue to present well established technical facts.

Anycase. 11,000 RPM has been done on pushrod motors. Its been done.

A pushrod motor can be designed to make peak RWHP at 11,000 RPM. Its been done.

If our model is as the original poster stats.

A 350CI putting out 400HP at 6000 RPM, will the same engine size make 800HP at 12,000 RPM?

The answer is yes and no. A 350CI if designed properly, with correct heads ( blue thunder/brodix/pro chief/etc ) can make 1100HP before 12,000RPM, and even before 11,000 RPM.

It will be a different, induction, exhust and cam than the one required to make 400HP@6000 RPM

Does 14.7PSI on a boosted app == double the power at atmospheric pressure?

The answer is no, because you have to account for heating of the intake charge AND parasitic losses in turning a supercharger, or slight restriction on exhaust side for a turbo.

Those are the answers.

Not my answers, the technical answers.
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Old Dec 15, 2006 | 01:29 PM
  #66  
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Derrick. I didn't say I have a model, just that it could be modelled pretty dang accurately these days.

On "top ends" original post. He did say that double the RPM that the induction ( heads/intake/cam ) and exhaust could be changed.

I made it alot easier to understand.

Make it 5.0L fords and make a cobra intake the limiting factor. Thatll keep the power peak down under 6,000 RPM. This is the 6,000 RPM model.

for the 12,000 RPM model you can go unlimited on the induction/valvetrain and exhaust.

Will you make double, more than double or less than double by 12,000RPM. You will make MORE than double.

Bottom end was bulletproof and light from the start. The only thing that change is the piston because the higher turning app going to need alot more compression. But were talking: 4 bolt block, best rods, rings, balance job, best balencers and flywheels, best of everything.
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Old Dec 15, 2006 | 01:33 PM
  #67  
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So you're saying you disagree with the technical answers? It's been done but not by "poor souls" like us. NASCAR/NHRA professional engine shops aren't average joes trying to increase power. Those are restricted classes, and in that case increasing RPM is the only thing left. You couldn't double output by doubling RPM in your garage at home and have it live. But you can double engine output by increasing VE and airflow and drive it every single day.
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Old Dec 15, 2006 | 01:37 PM
  #68  
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Derrick.

I usually dont disagree with "technical" facts unless it was made thru a political decision.

You are right then.

On us guys without money, doubling the RPM is only going to equal a broken motor!

And even going 2000 RPM further than we normally do, probably will equal something broke in the valvetrain and only less power.

His question I read was without limits ( us poor guys making regular salaries ). And let you change induction/exhaust unlimited.

On the 14.7PSI. We've all seen and understand why it is why it is.
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Old Dec 15, 2006 | 01:42 PM
  #69  
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Originally Posted by Big-DEN
Derrick. I didn't say I have a model, just that it could be modelled pretty dang accurately these days.

On "top ends" original post. He did say that double the RPM that the induction ( heads/intake/cam ) and exhaust could be changed.

I made it alot easier to understand.

Make it 5.0L fords and make a cobra intake the limiting factor. Thatll keep the power peak down under 6,000 RPM. This is the 6,000 RPM model.

for the 12,000 RPM model you can go unlimited on the induction/valvetrain and exhaust.

Will you make double, more than double or less than double by 12,000RPM. You will make MORE than double.

Bottom end was bulletproof and light from the start. The only thing that change is the piston because the higher turning app going to need alot more compression. But were talking: 4 bolt block, best rods, rings, balance job, best balencers and flywheels, best of everything.
Best of everything still won't make it live and make the "desired" power. You could buy every single part for a NASCAR V8 but who's going to give you the tolerances to assemble it? By 12,000 RPM you'd be eating connecting rods(literally) sooner or later. And that equates to zero power.
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Old Dec 15, 2006 | 01:48 PM
  #70  
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This thread sucks.
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Old Dec 15, 2006 | 01:52 PM
  #71  
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Originally Posted by mzoomora
This thread sucks.


Why? Because none of us have the resources, capital or desire to double output by doubling RPM! That's what nitrous, turbos and superchargers are for! And if you don't like those then build a larger displacement engine!
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Old Dec 15, 2006 | 02:22 PM
  #72  
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There are so many assumptions made in this theoretical discussion, it's
far beyond repair.

If anyone thinks that by doubling RPM, you will achieve double the horsepower
(having twice VE), they have another thing coming.

Friction, pumping losses, efficiency and stability are not equal at 6000 RPM
and 12000 RPM.

Even if that was the case, you would have to assume the torque output is
absolutely flat across the board...which it wont be. By changing engine components
to allow a higher RPM efficiency, you have likely reduced lower and mid range
torque, and most definitely moved the torque peak.

The engine is no longer the same animal.

Believing otherwise defies many physics laws:

"The increase in the internal energy of a thermodynamic system is equal to the amount of heat energy added to the system minus the work done by the system on the surroundings."
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Old Dec 15, 2006 | 02:42 PM
  #73  
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Darrick, with all due respect.

Once again: username "topend" posted the following:

On a forced induction motor, 14.7 psi of boost is considered x2 the engine NA output.
So if a given 350ci engine made 400 hp NA at 6000 rpms. Would the same engine size make 800 hp at 12000 ( with a better flowing induction and exhaust system of course). Would this be right???


This is two questions:

#1 Is 14.7PSI of boost considered 2x the base power level?

The answer is NO, and we are all comfortable with the answer, because to attain 14.7PSI of boost you have heated the intake charge, and have power loss thru turning a supercharger or slight exhaust restriction a turbo brings. So it takes a little more than 14.7PSI to double base power. Most are comfortable with the understanding behind it.

#2. if a given 350ci engine made 400 hp NA at 6000 rpms. Would the same engine size make 800 hp at 12000 ( with a better flowing induction and exhaust system of course). Would this be right???

Number #2 is easy. "topend" says ( with a better flowing induction and exhaust system of course ). So were going unlimted on the 12,000 RPM example. And to do the 6,000 RPM example make it long runner Efi.

A 6,000 RPM motor will be lucky even if it is unlimited to reach volumetric efficiencies of 110%.

However a 12,000 RPM motor can reach volumetric efficiencies of 130%.

So double the RPM in this reasonable case, with unlimited budget, and each optimized for its power peak, the double rpm will greater than double the HP that the 6,000 RPM motor puts out.

#1 is answered

#2 got off track when mention of:
a. Limited budget of most people
b. Assuming engine was built for max rpm to begin with, and doubling based off an already optimized build
c. Inability to draw from sufficient examples to show the argument

To make #2 easy pick 5,000 RPM and 10,000 RPM with unlimited budet or 5,500 and 11,000 RPM with unlmited budget.

To find examples of engines doing 12,000 RPM. The racing pushrod 3.8L's ( ford and chevy ) do it.

To find examples of engines doing 6,000 RPM the factory 3.8L pushrods do it.
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Old Dec 15, 2006 | 05:55 PM
  #74  
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Originally Posted by mzoomora
This thread sucks.
As does many in this section recently. I'm not quite sure when "common knowledge" replaced the cold, hard, laws of physics.

Time for .
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