Fantastic Read About Induction and Extraction

Hey!
If a 717cc cylinder has 11:1 compression, would not tat add 65.2cc and not 71.7cc, = 782.2cc?

Last time I checked a swept volume of 717 + 71.7 of combustion chamber =788.7 With compression ratio being total cylinder volume (swept plus chamber) divided by the chamber volume, the compression ratio is 11.0:1 (788.7 / 71.7 = 11.0)

I suggest at least getting the math correct. And 100 feet cubed is not 100 cubic feet.

The article is accurate for the most part. The whole exhaust system needs to be considered for tuning.

The issue of collectors is one of definition: What is it and where do you measure its diameter to plug the number into a calculation? A merge collector can be smaller at the throat than at the discharge. Which diameter fits into the "2.75:1" rule, the throat or the exit? How long is its effective length when it is tapered? Old style collectors with the bundled primaries discharging into a tube that fit around them is much more like a resonance chamber on zoomies than most people believe. Therefore, those systems are tuned by primary length alone and have close to zero effective collector length.

Firing order is important as the pulse spacing impacts tuning diameters, lengths and pulse frequency versus the tuned pipe's natural frequency. In some V-8 racing applications, headers have some primaries from both sides collect in common to keep the pulse timing uniform. Some use the "double by Y" or the two into two into one.

What wasn't explained is the difference between steady state, slow and fast rpm acceleration rates on an exhaust system's performance. Any system will respond to varying engine acceleration rates differently. What works really well in first gear may not work quite the same in high gear. If you doubt this, I suggest you run a dyno test on the engine and exhaust system of your choice, only start at red-line and pull the engine rpm back to below 2,000 rpm by varying the load to de-accelerate the engine at the same rate (600 rpm second is common). The torque curve will be different than the one you ran with the engine rpm increasing. Why? This test is not possible on an inertia dyno so don't ask your local tuner to do it for you.

It takes time for a convergence of heat, flow velocity, pipe resonant frequency and wave systems to occur. The time can be very short or in some cases much longer, as in 0.4 seconds or more. Engines and exhaust with greater mass, displacement, and tuned lengths generally respond slower and will show greater disparity between steady state and fast acceleration tests.

So, will you run the salt flats, drag race 660 feet, or zip through the local curves on the way to lunch at your favorite brew pub?
 

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If it is on the Inter-Web IT MUST BE TRUE - How dare you use basic mathematics in public :p

And it has a mass of .0807 - it's weight would depend on the gravitational forces applied.

m=w/g; however, I believe the .0807 is pounds per cubic foot and not mass.
 
At sea level air puts a pressure of 14.7 psi. But that's actually the hydrostatic pressure because of the column of air above being pulled down by gravity.
14.7 - Depends - STP, SATP, IUPAC, ISA or USSA?

OK I love Chemistry, Physics and Maths - I'm a geek - Bite me!.
 
@Speedy
Thanks for your input:
"Last time I checked a swept volume of 717 + 71.7 of combustion chamber =788.7 With compression ratio being total cylinder volume (swept plus chamber) divided by the chamber volume, the compression ratio is 11.0:1 (788.7 / 71.7 = 11.0)"

No intention to be pedantic here, but for my own knowledge . . . I understood the compression area was determined from the swept area (717 cc) and the compression ratio (11:1) and then added.
If what you state is correct, from where is your 71.7 then derived?

"Just to set the record straight, a cube of air 100 feet square will weigh 38 tons!"
This one has me scratching my head.
Here I indeed erred in placing my decimal. 100 cubic feet would be 10' X 10' X10' and not 100' X 100" X 100'.
100 feet cubed is 1,000,000 cubic feet
So then 1,000,000 X .0807 = 80,700 pounds = 40.35 tons.
WOW! I had no idea air was this heavy!
Appreciate the correction.
 
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I simply used the numbers from the author's example. However, if you know the swept volume and the CR you can calculate the chamber volume, or calculate any one of the three knowing the other two factors for that matter.

Because air is everywhere including inside of us, its weight is unnoticed. A fish also doesn't notice the weight of water for the same reason.

The point made about air having mass (weight for shorthand) is that it has inertia. At up to 400 feet/sec velocity during parts of the intake cycle, it can have a significant impact on cylinder filling. Exhaust gas velocity is much higher and approaches Mach 1 across the valve seat at opening. Pressure waves always travel at the local Mach number in both the intake and exhaust streams.
 
Man what I would give to have the knowledge, drive, and capital to start a motorcycle company.

I have so many impractical to one off ideas I want to do/try lol.

Which brings to mind, why don’t bikes have brand associated build houses like the car companies do e.g. AMG, SRT, Shelby, M, JCW etc?
 
I simply used the numbers from the author's example. However, if you know the swept volume and the CR you can calculate the chamber volume, or calculate any one of the three knowing the other two factors for that matter.

Because air is everywhere including inside of us, its weight is unnoticed. A fish also doesn't notice the weight of water for the same reason.

The point made about air having mass (weight for shorthand) is that it has inertia. At up to 400 feet/sec velocity during parts of the intake cycle, it can have a significant impact on cylinder filling. Exhaust gas velocity is much higher and approaches Mach 1 across the valve seat at opening. Pressure waves always travel at the local Mach number in both the intake and exhaust streams.

You da man, Speedy!
Impressive knowledge base indeed!
I acknowledge my error re: air weight calcs, but so far, stand by my critique of the combustion chamber volume and compression ratio calcs of the head cavity cc.
 
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