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*To*: type3@vwtype3.org*Subject*: Re: Fw: Fwd: Re: [T3] too much mind time*From*: sml214@casbah.it.northwestern.edu*Date*: Wed, 15 Aug 2001 21:13:21 -0500 (CDT)*Cc*:*Reply-to*: s-laws@northwestern.edu*Resent-date*: Wed, 15 Aug 2001 22:13:25 -0400*Resent-from*: type3@vwtype3.org*Resent-message-id*: <"XpHJnD.A.A7E.Dxye7"@client4.hv-ywh.com>*Resent-sender*: type3-request@vwtype3.org

Hey Jim- > I have always assumed that what was being tuned was the > frequency of consecutive pulses down the pipe. If I've got this wrong > please straighten me out. If I'm right, the cam duration has nothing > to do with this, it would just be a matter of changing the Fourier > distribution of frequency components above the fundamental, but > you seem to be after something completely different. Quick note: everything I describe in this response is at best a fairly accurate approximation. For dead-on accurate results, one must play with expensive engine simulation software... anyway... We can idealize the runners' scenario by saying this: there are two very different events that occur in the runners. One is the actual intake of air. The other is the time in between actual intakes. For the runner length calculation, we'll assume that the former lasts a fixed number of crank degrees and, more or less, sharply ends. This isn't a bad approximation if the effective (i.e. at 0.050" lift) cam duration is used. The latter lasts the remaining of the 720 degrees. So, in the calculation I performed, this is 210 and 510 degrees, respectively. The other factor in the idealization is that immediately after the intake valve sharply closes (thanks to our idealized "digital" cam :-), a pressure pulse is generated at the back side of the valve. This "positive" (i.e. higher than mean plenum pressure) pulse travels to the plenum, is reversed, travels back to the valve as a "negative" (i.e. lower than mean plenum pressure) pulse, reflects, travels back to the plenum and is reversed again, then travels back to the intake valve as a "positive" pressure pulse. This scenario is what I termed the 1st harmonic. The 2nd harmonic is when this sequence happens twice and the 3rd is thrice, etc. A graphical way to look it is this: If you measure pressure right behind the intake valve during that 510 degree period, it will look like a cosine wave offset by the plenum pressure (i.e. the "y" of the cosine wave has the plenum pressure added to it and "x=0" is when the intake valve closes). The total timescale goes from x=0 to x=A. A=[(720-ECD)/360]*[60/W] where ECD is the effective cam duration in degrees, W is the engine speed in RPM, and x is in seconds. A quarter-wave is one travel through the runner. So, T=4*L/V where T is the period in seconds, L is the effective runner length (i.e. runner length plus port length plus 1/2 the diameter of the runner) in inches, and V is the speed of sound in hot air in inches per second. The goal, obviously, is to make that timescale an integral multiple of the period of the wave. So, A=N*T where N is our reflection value. Result: V*(720-ECD)=24*W*N*L Let's plug this in: L=20.625 in V=15000 in/sec ECD=210 degrees For N=1, W=15455rpm. For N=2, W=7727rpm. For N=3, W=5151rpm. Of course, there are deviations from this idealization. Quite frankly, the intake valve doesn't sharply open and close and the pulse isn't suddenly generated... but it's not too far off... > I've read this several time and I still haven't managed to get it > through my head what you're doing. You're going to have to break > this down into first year concepts, I guess. Are you just concerning > yourself with events that happen withing the duration of a single > pulse and ignoring the relation of each pulse to it's neighbor? Yes. My idealization assumes that the actual time the intake valve is open pretty much resets the whole game. > > BTW, interestingly enough, at a runner air velocity of 180 ft/sec. (a good > > number for a rough maximum speed...), a 1.25" runner cooresponds to about > > 5100rpm if the engine is at 75% VE. > > > > But then again, when was the last time you still got any decent amount of VE at > > 5100rpm on a stock cam with stock heads? > > Define VE (Volumetric Efficiency?) please. You're correct. Q=A*V=VOL*VE*T, where flow is Q, area is A, velocity is V, cylinder volume is VOL, volumetric efficiency is VE, and time (i.e. the amount of time the intake valve is open) is T=[ECD/360]*[60/RPM]. Again, this is an idealization at best, but I've read that with this idealization, estimating V=180 ft/s actually corresponds somewhat well to dynoes. Take care, Shad > > - > ******************************* > Jim Adney, jadney@vwtype3.org > Madison, Wisconsin, USA > ******************************* > > ------------------------------------------------------------------- > List info at http://www.vwtype3.org/list or mailto:help@vwtype3.org > >

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