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  1. #1
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    "All About Compression" Boost + NA

    Firstly, i think this tidbit of info should help alot of you....
    Secondly i take no credit for it. It was sourced directly from:
    http://www.hondahookup.com/forums/ar...p/t-21090.html

    ************************************************** **********
    I'm going to write the intro but Ryan Autry (http://www.hondahookup.com/phpBB2/pr...ile&u=9389) wrote a great tid bit about boost and compression ratio

    I'm going to quote

    http://www.chevron.com/prodserv/fuel..._3_rf.htm#SUB3

    for the defnition

    COMPRESSION RATIO
    An important parameter of an engine is its compression ratio. This is defined as the ratio of the volume of the cylinder at the beginning of the compression stroke (when the piston is at BDC) to the volume of the cylinder at the end of the compression stroke (when the piston is at TDC). The higher the compression ratio, the higher the air temperature in the cylinder at the end of the compression stroke. Higher compression ratios, to a point, lead to higher thermal efficiencies and better fuel economies. Diesel engines need high compression ratios to generate the high temperatures required for fuel autoignition. In contrast, gasoline engines use lower compression ratios in order to avoid fuel autoignition, which manifests itself as engine knock, often heard as a pinging sound.



    In layman's terms, Compression Ratio is a comparison. It compares the amount of volume in the cylinder at 2 points.

    BDC, or bottom dead center, is the lowest point the piston reaches in its stroke (up and down motion).

    TDC, or top dead center, is the pistons highest point it reaches in the cycle.

    So, for instance, a compression ratio that states 10:1 means that at BDC, there are 10 units of volume in the cylinder, and at TDC, that volume is compressed to 1 unit.

    As the defnition states, high compression ratios help motors make power. But, with gasoline, there is a point in which air and fuel will ignite spontaneously. This is called knock, pinging, or detonation. Detonation can destroy pistons, bend or break rods, destroy valves, and give you a really big headache.

    To actually calculate compression ratio is very difficult, unless you know specifically all the measurements of your motor.

    I don't know the formula off the top of my head. Sport Compact Car had an article written by Dave Coleman in last month's issue that had the formula in it.

    I'll write down the formula later.

    That's it for now, i gotta go. More to come, and im sure other people will have much to add onto this.


    So, here is Ryan Autry's blurb with FI and CR.
    FYI: Effective CR is also referred to as Dynamic CR.

    One main concern in power production with forced induction is effective compression. Effective compression is the sum of the motors static compression, plus the additional compression added by the forced induction tool. A B18C1 (also B16A) motor will have a higher effective compression than a B18B motor will, on the same boost...therefore, pound for pound, it will make more power.

    The next argument that people usually bring up is that a higher compression is bad for turbocharging. Well, if you understand the concept of effective compression, then you should understand that this statement is entirely incorrect. A higher compression engine makes more power in NA form. So, why do you turbo guys think that a lower compression turbo motor makes more power? Does that make any sense when you really think about it? A turbocharger is a power adder? So why deplete power that was there to begin with? The answer I usually get to that is "So I can run more boost!" Well, sorry to rain on your parade, but more boost does not always equal more power. Check out this mathematical example of effective compression:

    A motor with a 10.0:1 static CR boosting 10psi
    10psi/14.7psi = .68
    .68 + 1 = 1.68
    1.68 x 10 = 16.8 effective CR

    A motor with an 8.5:1 static CR boosting 10psi
    10psi/14.7psi = .68
    .68 + 1 = 1.68
    1.68 x 8.5 = 14.28 effective CR

    Now tell me who is going to make more power? The higher CR motor, or the lower CR motor?

    So, maybe add more boost to the lower CR motor, right? Wrong...

    A motor with an 8.5:1 static CR boosting 13psi
    13psi/14.7psi = .88
    .88 + 1 = 1.88
    1.88 x 8.5 = 15.98 effective CR

    Now you see, even adding 3psi of boost, still does not equal the effective CR of the higher compression, lower boost motor.

    Effective compression is not the only advantage of the B16A/B18C1 either. The B16A/B18C1 has a stronger, better flowing cylinder head. It can rev much higher, making it that much more effective, and it flows great to handle all of the extra volume. The block has oil squirters to help support the bottom end assembly at high RPM. It takes more than a valvetrain upgrade to make a B18B safe at 8k. The higher compression also aids in spooling the turbo faster too.

    Both motors have similar tolerances though. Both motors pretty much top out at around 350-400hp on stock motors, very well tuned. The B18C1 will make it far more efficiently for you though. It takes less boost to do so, it has more safeguards...and the bottom line on any Honda motor is tuning. If it is well tuned, you will be set. That goes for both motors. YOU ARE A FOOL if you think for one second that just because your B18B has a lower compression, you can substitute that for proper tuning.

    A lot of people like to lower their motors compression when they build their motor. I used to think it was a good idea before I understood about tuning, and the positive aspects of compression. In the mathematical representation below, I will show you how a low compression motor must boost more to equal the output of a higher compression, lower boost motor:

    Motor: stock B16A2 boosting 7psi.
    Static Compression Ratio: 10.4:1

    ((boost psi / 14.7) + 1) x motor compression = effective compression

    Stock motor (10.4:1 CR) on 7psi:
    7psi/14.7psi = .47
    .47 + 1 = 1.47
    1.47 x 10.4 = 15.288 effective CR

    Built motor (9.0:1 CR) on 7psi:
    7psi/14.7psi = .47
    .47 + 1 = 1.47
    1.47 x 9 = 13.23 effective CR

    You will lose 2.058 points from your effective compression ratio, this translates to a significant power loss.

    In order to gain back that power, you have to do this:

    Built motor (9.0:1 CR) on 10.5psi:
    10.5psi/14.7psi = .71
    .71 + 1 = 1.71
    1.71 x 9 = 15.39 effective CR

    Add 3.5psi to what you were boosting before, and you should be able to make around the same power as before, granted you haven't done any other kinds of modifications port/polish, cams, etc...

    As you can see, considering all things stay equal (bore/stroke/cylinder head/etc...), you must add 3.5psi to make the motors perform similarly. You just spent about $2,500 to build your bottom end, and make your car slow.

    By now we all should understand the positive aspects of compression, and how when teamed with the faster spoolng turbo, more efficient output, better flowing B-series VTEC cylinder heads, better low end spool time, stock oil squirters, higher redline, etc...you should see that turbocharging B-series VTEC motors is clearly not dangerous, and highly adviseable. I love a good turbo B16A!

    Search Keywords: Static Compression, Dynamic Compression, Compression Ratio, CR, Naturally Aspirated, Turbo, Boost, Spooling, High RPM, VTEC
    Last edited by dsp26; 14-07-2007 at 01:12 AM.

  2. #2
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    Always interesting to read about CR's.

    Can someone explain to me the basics of sea level and atmospheric pressure in regards to it's effect on boost? Obviously the 14.7 is the atmospheric pressure at sea level, but what about when your trying to boost a car at 8,000 feet above sea level with 5psi atmospheric pressure? Anyone know the maths to determine what psi is required to create the same power as a car at sea level? --I tried to work it out myself, but somehow got this which doesn't seem right:

    7psi/14.7psi = .47
    therefore
    .47*5psi=2.35psi

    But my common sense tells me that the number should be significantly higher, not lower to achieve same power?

    Matt

  3. #3
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    just fyi shebangs... atmospheric pressure at 8000 feet is about 10.5 psi, not 5 psi... might help things make a bit more sense

  4. #4
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    The SAE correction for altitude/atmospheric conditions is

    Corrected BHP=BHP X (1-((elevation(feet)/1000) X 0.3))

    But, if a turbo motor is producing 2bar of manifold pressure at sea level and 2BAR of manifold pressure at 5000 feet, then it'll still be making the same numbers. However, your turbo still needs to be able to achieve that boost as the pressure ratio to get to that manifold pressure is going to be higher with altitude. At sea level, you'll have 1 BAR absolute air pressure with 1 BAR of boost (2BAR absolute) so it's a 2:1 ratio but at 6000 feet the air pressure will only be 0.817BAR so it becomes 2.45:1 ratio.

    NA engine are the ones that suffer with altitude. At sea level on a fine day, you might see 1.02 BAR of manifold pressure. But on the same fine day at 6000feet, you're only going to get 0.8BAR or thereabouts.
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  5. #5
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    I read that first post yesterday so I've probably forgotten the part I should be remembering in regards to this post. However, when you get to real high horsepower levels when your pumpin in buckets of boost, isn't it better to have a lower compression ratio so you can fit that extra bit of boost in (including more fuel) to make a bigger bang without messing up anything in the bottom end? Basically, isn't a bigger bang from more air and fuel better than the bang you get from a high cr and less boost in the end?

  6. #6
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    Quote Originally Posted by gnx1987
    I read that first post yesterday so I've probably forgotten the part I should be remembering in regards to this post. However, when you get to real high horsepower levels when your pumpin in buckets of boost, isn't it better to have a lower compression ratio so you can fit that extra bit of boost in (including more fuel) to make a bigger bang without messing up anything in the bottom end? Basically, isn't a bigger bang from more air and fuel better than the bang you get from a high cr and less boost in the end?
    Yes, but don't forget, when you're running low compression the engine is also performing at a much less efficiency. Which is exactly what the article in the first post talks about.

  7. #7
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    Higher CR/boost is fine togethor up until the point the engine becomes knock limited. Then it's pointless.
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  8. #8
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    could you not use then an EBC to tune down Boost at those rpms that knock?

  9. #9
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    It's not a matter of rpm, it's a matter of boost. If it's going to knock due to excessive cylinder pressure then it'll do it pretty much everywhere. Maybe getting a little less with high rpm.

    In which case, why have so much boost/CR to start with?
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  10. #10
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    Quote Originally Posted by EfiOz
    It's not a matter of rpm, it's a matter of boost. If it's going to knock due to excessive cylinder pressure then it'll do it pretty much everywhere. Maybe getting a little less with high rpm.

    In which case, why have so much boost/CR to start with?
    lol.. hhmmm...

    i was under the impression there were Electronic Boost controllers that had variable boost adjustments at rpm points like A/F computers as opposed to the simple dual stage boost controllers....

  11. #11
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    there are, but the point is that you will knock at most (if not all) rpm points where you are making that level of boost.This is because you are reaching the knock threshold of the fuel you are using (the cylinder pressure is enough to ignite the fuel before you add spark).

    So you'll be turning the boost down across the board regardless of how fancy the boost controller is.

  12. #12
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    ^^^i must research this further then....

    i also would have thought colder lower gapped plugs with an improved coil could improve the situation....

    anyways original post with actual physics was there to help others understand... i am now going to research this further and understand it in more depth....

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