Engine technology thread.

[Ryephile]

I might as well add this from my short time with the BRZ earlier this week. The O2 sensors are 4-wire heated narrow-band. Are there any other GDI cars that use narrow-band? Perhaps I'm too used to GDI + turbo having WBO2 these days. It's not the end of the world, I'm just a bit disappointed....it definitely shows the low-cost nature of the car.

[serialk11r]

^ darn, that sucks From what I've read, wideband and narrowband operate quite differently, so is it difficult to switch in a wideband and have the ECU cooperate?

[Ryephile]

No dice. Narrowband are 0-1V signals that oscillate around 0.5V, and wideband O2 sensors produce a 0-5V linear-relationship signal. Not compatible whatsoever.





What this means is the car will run open-loop at WOT, which isn't a bad thing [cars have been doing it forever], but not super-precise like closed-loop WBO2 100% of the time.

[serialk11r]

Okay this is kinda random and stupid but I went on Wolfram Alpha and found that somewhere between a rod:stroke ratio of \sqrt{14.3}/2 and \sqrt{14.4}/2 (this is about 1.89ish) the acceleration at BDC does not have that little "lump" (acceleration has only 1 local min/max around there). If I had Mathematica still installed I could've had it solve in general form but oh well. Fun fact!

[OrbitalEllipses]

Originally Posted by serialk11r

Okay this is kinda random and stupid but I went on Wolfram Alpha and found that somewhere between a rod:stroke ratio of \sqrt{14.3}/2 and \sqrt{14.4}/2 (this is about 1.89ish) the acceleration at BDC does not have that little "lump" (acceleration has only 1 local min/max around there). If I had Mathematica still installed I could've had it solve in general form but oh well. Fun fact!

If you give me the code for Mathematica I can run it for you, we have licenses for the program. I've never used it, though.

[SkullWorks]

There is lots of good content out there regarding the rod stroke ratio. I find serialk11ler's response to clearly define what I know about Rod:stroke, longer rods allow for slower acceleration away from TDC at the beginning of the intake stroke, reducing load on Small end of rod, and allowing more time for peak pressure to build as the piston starts to travel down the bore on the power stroke, there is a slight increase of maximum piston velocity (increasing friction) at mid stroke, but it is momentary and only requires consideration if you are already pushing the 5000 in/min speed that is usually referenced as the max desirable.

[Grip Ronin]

The FA20 uses dual AVCS (on both intake and exhaust cams). But that's essentially the same as VVTI. One profile with an oil pressure actuated cam gear to advance or retard the cam's timing.

this sounds very similar to i vtec found in the honda k series, they use this cam advancement feature, really screams when it hits and allows tuners to bring out alot more power from it

[serialk11r]

Originally Posted by Grip Ronin

this sounds very similar to i vtec found in the honda k series, they use this cam advancement feature, really screams when it hits and allows tuners to bring out alot more power from it

Not really, they're very different. Honda has both a system that will vary the cam timing, and 2 lift profiles to improve volumetric efficiency across a large range of rpm, which this lacks. The "VTEC kicked in yo" is not because of a cam advance, but because the engine switches to a long duration cam lobe that is meant for maximum power. The low duration cam lobe could make more power if they wanted it to, but they kept it mild to presumably reduce emissions and fuel consumption.

[Want.FR-S]

Originally Posted by serialk11r

Crank angle and time are interchangeable at constant speed, please don't get nitpicky now. And sorry I couldn't resist being smart on that last comment, but please...

Well.. it depends on how you look at this. If you look at the *position* of the piston in a constant rev engine and plot it over time, you get a quasi sine wave (like it but not it). However, the *motion* of the piston is definitely linear. So your statement is simply incorrect with all the fancy terms.

If you said your observation or measurement of such the piston motion over time is quasi-sine wave, I have no problem with it. A mistake is a mistake.

[serialk11r]

Whatever, I said quasi- simple harmonic. Sure, sinusoidal is the wrong term. You get the point. It's obvious that the piston is limited to travel in 1 direction, and normal people associate "sinusoidal" with "simple harmonic". Fancy terms? Not really... jesus.

[Want.FR-S]

Oh Boy, Why do I have to do this again....

Originally Posted by serialk11r

The piston's path is "linear", but its motion is quasi-simple-harmonic (aka sinusoidal) with respect to time. [Explained before, its motion is linear but its measurement over crank angle at constant rev, or time, is quasi-sine wave]

The rod adds some complexity to this, most obviously decreasing it around TDC. The Wikipedia article varies stroke and keeps rod length constant, if you change the rod length instead, you'll see that it's increasing BDC peak acceleration. [Up to this point, it is correct.]

This can be proved simply based on the formula itself. At BDC (crank angle = 180 degree), the function sinA= 0 while cosA = -1. The acceleration equation becomes:

X" = -r(-1) - r^2/l - r4*0*1/l^3
= r - r^2/l
= r (1 - r/l)

So if you increase l, the term r/l is reduced so that (1 - r/l) increase, and thus the acceleration value increase and vice versa. Simple as that.

The reason is the piston's postion at a given crank angle is the vertical distance between a circle of radius = stroke/2 and a tangent circle of radius =rod length, vertical meaning perpendicular to the common tangent of said circles. An infinitely long rod would make the piston's motion perfectly sinusoidal. [This is where you screwed up with fancy terms and thing. How does the position of piston have any relationship with longer rod that causes faster acceleration at BDC.]

Actually, suppose at TDC, if you draw a circle at r = stroke/2 from crank center, and you draw another circle from the wrist pin of the rod connecting the piston with radius = rod length. These two circles touches, or tangent, at the crank pin. However, once you start cranking the engine, these two circles start intersecting one and another at the crank pin. The piston and wrist pin continue their linear motion along the line, while the crank pin moves around the crank circle. The crank circle remains the same while the rod circle continue to intersect until BDC, where two circles touches, or tangent, again. And then they start intersecting again until to TDC.

Now, notice that the center of the rod circle, at wrist pin, *always* travels along the line. This means that the circle of the rod *always* travel along the line. It never bounce up, or down. If It had bounce up or down to create a sinusoidal motion, the center of the circle must be moving up or down. That was not the case here with finite-length rod, and that is never the case with infinite long rod.

serialk11r, if you want to explain something, make sure you explain it right with enough detail and use the common terms. Sometime using fancy word does not mean it is more correct.

[serialk11r]

There's nothing wrong with what I wrote, the vertical distance between the circles is going to be the distance that the piston moves since you're just translating the circle up by that amount. Anyways, it seems like you get what's going on, so what's the problem? Why do you keep trying to point out things that I did "wrong" that are either just nitpicking at word choice or not wrong at all? Longer rod with the same stroke gives lower peak acceleration...this is pretty obvious.

And I already said many times, sinusoidal means simple harmonic, I don't know why you keep trying to argue that I'm oh so wrong because the piston is clearly limited in travel by the bore. I know that, and so should everyone else, it's OBVIOUS.

The reason why I brought in the tangent circle drawing was it's an easy way to visualize the differences. A lower rod length reduces the speed of the piston around the bottom of the stroke, and increases it around the top. That's just the easiest way to understand the difference for me, so I shared it.

By the way, "circle" , "radius", "perpendicular", "tangent" etc. are not fancy words. They are basic geometry terms.

[serialk11r]

Anyways back to your original question, what are the benefits. Newton's 3rd law says there's a reaction force to whatever force is going through the rod, and the rod ends are free bearings so there is no torque along the rod, thus the sideloading force is determined by the rod angle and the piston's net force (it's the tangent of the angle times the net piston force). Since the tangent is increasing along the relevant interval, the sideloading force at any crank position besides 0 and 180 is greater because the rod angle is greater.

The other thing would be that since the piston is travelling slower at TDC, the pressure increase wrt crank angle is greater, and you have a higher peak cylinder pressure. As the engine speed increases, the charge has less and less time to burn. I suspect this is the main reason for going with a longer rod, as a minor change in rod ratio doesn't actually change peak forces that much.

Anyways, lower friction is good, higher peak pressure is good.

[Dimman]

Factory cars are coming with widebands these days?