To double an engine output.
You're comparing power below the peak to the peak. In that case, friction and vibration have a tiny effect on power compared to cylinder filling. The way I understood it, TopEnd was describing two same displacement engines at their power peak. One at 6000rpm, and one at 12,000 and the question was whether power will double with the doubling of rpm. To that question, the answer is no, because friction and vibration increase as the square of rpm.
Al
OK lets do peak to peak.
A combination that is efficient to 4250RPM vs one that is efficient to 8500 RPM. Lets make TQ peaks 3200 and 6400 to make things closer to "fair".
Combination that is efficient to 4250RPM has a E7 like head, small diameter tubular header, and camshaft designed to optimize its efficicny down low.
One thats efficient to 8500RPM has properly sized head, properly sized header, and camshaft designed to optimize its efficiency up higher.
If thats not a fair example, lets do a 5000 and 10000 RPM examples. INduction/exhaust and camshaft each time decide to optimize engines operation over its designed range.
Not trying to knock your knowledge. I just think theres too many cases which go against "HP will not be double at double the RPM because of frictional and pumping losses"
Al
But I believe my example of 5,000 and 10,000 was good. The most efficient 5,000 RPM wonder compared to the most efficient 10,000 RPM wonder with similar bottom end, the 10,000 RPM wonder will be able to have greater than 2x HP potential than that most efficient 5,000 RPM wonder.
Now the difference between the greatest 10,000 RPM wonder of a given displacement and bottom end vs the greatest 20,000 RPM wonder of a given displacement and bottom end is a different question.
So the new question. Say 2.4L f1. Can/induction/exhaust for greatest potential to 10,000 RPM VS the greatest cam/induction/exhaust for greatest potential to 20,000 RPM. Is the 20,000 RPM going to be more than 2x from higher air speeds and stronger resonant tuning effects ( greater natural supercharging to say ), or is the greater friction and pumping losses going to make it less than a 2x affair.
We all can agree on my 5,000 greatest potential vs 10,000 greatest potential on given displacement that the 10,000 RPM optiimized will be greater than 2x over the 5,000 RPM optimized situation.
In higher RPM 10,000 vs 20,000 RPM does friction and pumping losses overcome the greater VE's availabled?
Air speeds aren't 2x faster. Ports are sized to give maximum velocity at the airflow needed without going sonic. When airflow requirement increases as max revs rise for each redesign, the port size increases to deliver more air while keeping velocity close to the same as before. Great questions Den.
Al
Between 10,000RPM and 20,000RPM I cannot comment.
Just wanted to add the disclaimer, none of us have to physically build anything to prove it, we can take the best examples of either situation that others have done and use it for the comparison.
Alin
The 5,000 RPM and 10,000RPM where just two RPM points, it can be 6,000 and 12,000 as well.
5,000 RPM and 10,000 RPM are an example we can pull real world from.
You can optimize a motor for operation up to 5,000 RPM.
You can optimize a motor for operation up to 10,000 RPM.
Just depends on what your trying to do.
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Big-DEN: I have to very respectfully disagree with your position that an engine spinning twice the RPM, all other factors being equal, will MORE than double horsepower, due to the air velocity of the charge filling the cylinders. If I get higher air velocity at higher RPM, allowing higher horsepower, I want to know...
Why does my engine "nose over" past 6500 RPM?
My ports are not going "sonic" they are probably well over 220 cc on my little 346. Also the 305 in my truck likes to "nose over" at about 5 k RPM... With your theory, our engines (any engine) should make an expodential horsepower curve... But they don't.
Sure port velocity goes up as RPM increases, but "air" has momentum... A lot of it... relative to the speed of events in an engine... And the airflow, at least in our piston, poppet valve engines, must stop and start over 50 times a second over 6000 RPM. My personal belief is that no 10,000 RPM engine will make double the power of a 5,000 RPM engine, if both use the same technology and are optimized for their respective peak power RPMs.
I am fine with you respectfully disagreeing. But its going to be best if the actual technical facts come out.
find your best example of say a 5.0L ford - or make it even a 350 chevy - fully optimized for operation to 5000 RPM with power peak centered right at 5000RPM. Any choice of headers any choice of heads, any camshaft and optimized CR for this.
NOw take that 5.0L ford - or make it even a 350 chevy. fully optimize it for operation to 10,000 RPM with power peak centered right at 10,000 RPM. find the best examples.
The best examples on the 10,000 RPM will more than double the HP produced by the 5,000 RPM example.
To make it even more easy to find information, turn it down to 9,000 RPM and 4,500 RPM.
Again, the cases you find that are fully optimized for operation to 9,000 RPM are going to be more than double the HP for a situation that peaks at 4,500 RPM.
You think you going to have more than 350HP out of a 5.0L ford at 4,500 RPM?
What about more than 700HP at 9,000 RPM? Theres many cases with 600HP being found out of 5.0L ford before 7,500 RPM... 700HP is going to be reached before 9000 RPM on optimized combo.
Now on 10,000 RPM vs 20,000 RPM I dont know, but these examples I'm pulling out I am confident I can fully prove it. I don't really want to waste the time.
You're comparing power below the peak to the peak. In that case, friction and vibration have a tiny effect on power compared to cylinder filling. The way I understood it, TopEnd was describing two same displacement engines at their power peak. One at 6000rpm, and one at 12,000 and the question was whether power will double with the doubling of rpm. To that question, the answer is no, because friction and vibration increase as the square of rpm.
Al
Would the engine consume the air/fuel mixture equivalent of 2x the power???
The different RPM ranges have increasing and higher and higher VE ( Volumetric Efficiency ) or cyllinder "overstuffing" per say.
The 20,000 that these F1's are at are nearing 140 VE or 140%!
At around 10,000 RPM your doing in the 120% range
And some good 6000-8000 RPM can exceed 100% to 110% or so.
Theres charts for this.
And the guys are right, frictional and pumping losses increase like that of aerodymics, by the square!
The crank is now going to have to spin 2x as fast. That creates an aerodynamic drag on the rotating assy also. So now the engine also has to fight against that drag also. I'm guessing it would be safe to say X^2 factor also since it would be aero-based drag.
It's also worse in some design's over others. Example being the LS1 vs Ford 351. Narrow skirt vs Wide skirt
You know top-end, I'd have to assume That the air-fuel amount burned increases linearly, but measured power would be lower since the items above would be using that power. If power was increased linearly to RPM, more A/F would have to be burned to keep the power in line.
So let me guess what you think.
Boost or manifold overpressure is free so exactly 14.7 PSI on a boost pressure guage == double a engine output
And that an engine built to be most efficient at 5,000 RPM if built to be most efficient at 10,000 RPM will not more than double the HP due to fricitional losses increasing faster than increased VE.
So let me guess what you think.
Boost or manifold overpressure is free so exactly 14.7 PSI on a boost pressure guage == double a engine output
And that an engine built to be most efficient at 5,000 RPM if built to be most efficient at 10,000 RPM will not more than double the HP due to fricitional losses increasing faster than increased VE.
Will an engine make 2x the power if you spin it 2x the peak RPM? NO. If that were the case then peak wouldn't be where it is. The engine would disintegrate. There are 4 variables you can change to increase power in a gas engine. Airflow, displacement, RPM and volumetric efficiency. Of those four variables, the increasing of RPM yields the least increase in power alone along with displacement. Airflow and VE are the best ways to increase specific power output. It's silly, if you have any understanding of the internal combustion engine, to argue that. Increase airflow, double the airflow you double power while maintaining a given volumetric efficiency. Increasing RPM effectively decreases volumetric efficiency, which is why the torque curve is not linear to redline. Increasing RPM gives less time to fill the cylinder properly, light it off and excavate it. It also increases windage in the crankcase, increases piston side loads and beats the **** out of bearings. Not to mention the harmonics. It would be counter-productive and stupid to even attempt "doubling" power output in such a manner when your application/project isn't limited by displacement, head design/type, etc. To me this isn't a topic for this forum because it makes no sense to even discuss it.
Al

