LeonV

It helps to relieve the mind... I really miss working on problems involving topics like the flow of energy. I'm currently in injection molding, and while satisfying at times, it's not what I want to do (but who the hell am I to complain, especially at this point in time!). This was a fun distraction.

Last week, I replaced the custom 100 air jets from the O3 etube with 140s. This flatlined my AFR trace at WOT! No more getting richer as rpms climb. This did make low-rpm transition a bit leaner but I'll be tuning that out with the hypojets, which control low-speed, low-throttle opening operation. Next plan: - Smaller hypojets, either the H20 or H18 should do the trick. - Drilling out my pump jets to .62mm, I still need a slightly fatter shot when I get on it. By optimizing (leaning out, in my case) all the other circuits in the carburetor, the pump shot needs to do more work to keep the mixture proper now that everything isn't masked by running overly rich.. - Re-sync and re-tune idle

Well there's your problem, the car's too quiet! I bet the exhaust on your Z tends to drown out a lot of that noise. I was in heavy traffic on the commute home last week and shut off my Z when I was on a downhill (yes, don't try this at home...). I heard a strange clicking noise that sped up with the car that I'd never heard before because all the other noises mask it. I'm sure a different, less worn, transmission or a bearing replacement will quiet it down but by how much I don't know. I don't think new fluid will help much if it's topped up and has been replaced fairly recently. Is "years ago" a couple of years or 15 years?

Call Isky with your specs and requirements and have them regrind a cam for you. It's cheaper and more effective. Win-win.

Cost/benefit: is it worth it? Like others have said, it's hard to tell in the video, but the noise sounds similar to the typical noise a Z-transmission makes. We can't hear it in person, so we can't really tell how "bad" it is. If you're really concerned and don't want to/can't do the work yourself, get a quote from a shop and go from there. I can't tell how loud the noise really is, so I'd try to sell it as is, and if someone was concerned then be up-front with them. It's probably typical noise, but it's going to cost $X from Y shop to replace the bearings if deemed necessary. The transmission will make some noise no matter what.

Oh, you'd be surprised, Julio! Don't get me started, just hope no one mutters, "backpressure"...

Finally a decent explanation of what you're saying with the math behind it (besides just yelling F=MA, F=MA!). I think this discussion is very good and beneficial, as it encompasses quite a few things. No need to hurtle insults, I grasp math and physics just fine. I'll agree that this is a way to look at it, but my viewpoint works just as well. Either weight can be unchanged (and inertia changes) thus acceleration changes because of a different amount of power available at the wheels, or power at the wheels doesn't change but effective mass does, thus does acceleration. In the end, both cases accomplish similar tasks. I prefer the "effective mass" method since it nicely shows where and when inertia is most prominent and, in some situations, can be a better descriptor for what's going on. Other times, the "more available power" method can describe a situation more clearly. It depends on what one is looking for. Sometimes, it can be tough to see someone else's viewpoint when you think only you're "right". Sometimes, no one is really "right" but both are seeing things from a different prespective that "works" for them. This is not a jab at you, Steve, but rather a commentary about this discussion.

Countershaft bearing(s). You'll have to figure out if the cost/benefit is worth it. Before you delve deeper, have you replaced the transmission fluid?

Thanks for the shout, but Phil (Blue) takes the credit for the actual Tech Tips! Removing the spindle pins is much nicer with a puller tool, especially a sturdy one, and a lot of heat into the strut housing around the pin.

Very clever! I will say that the current logo looks good and is perfectly fine.

P = T * RPM. RPM represents the rate at which torque is being applied, i.e. rotational mechanical power. Notice no time involved, but a UNIT of time is. Yes, but this is true whether the force is used to rotate or translate mass, e.g. removing the interior increases "available" force to motivate the car. See above. Clearly, but that's not what I was getting after. By your argument, you are making that equivalent to saying you've "freed up" power at the wheels. Can I say that I've got more power to the wheels if I remove my spare tire? Am I correct in saying that? Easy. F=ma. Since inertia is an equivalent mass (you've agree with this), mass increases and thus acceleration decreases for a constant force, or torque if you like that better. Nope. See above. Sort of... Well, that is the logic behind my post. If you've followed what I said and agree, then there is no point for further discussion.

Nice photos, Mike!

If that's the goal then you need to develop a metric to test performance. Since your stated goal is to decrease 1/4 mile times, the easiest and most logical thing to do would be to go to a test-and-tune night at a drag strip and actually try out different combinations, etc. If the strip is too far for you, then things are tougher. It's not very easy, effective, or recommended to tune your car on public streets. I'd give my car a tune-up, put together a plan and head out to the strip for a day.

Power need not be anything. It's all about how we define it. We're not pulling a trailer or a railcar, so drawbar power is irrelevant. There are all sorts of power measurements and calculations and it's up to the user to choose which one fits their model. Yes, force * distance is the same UNIT for energy and work, as well as torque. Energy and torque are intertwined but torque is NOT. As you say, power = force * distance / time, so when we're talking about a rotating wheel being powered by a rotating engine, the pertinent equation becomes Power = Torque * Angular Displacement / Time. You're still not getting it. You're describing total power consumed. Stop thinking about time. Time is out of the picture. Picture a car accelerating and imagine you can freeze time. Again, total, instantaneous power is Power = Torque * RPM. It makes no difference at any discrete torque and RPM value whether you're accelerating or not. How do you explain an increase in vehicle acceleration due to removing non-rotating weight? A car will go faster if you strip it down, but not touch the engine. The wheels accelerate faster, thus more energy is put into the wheels to accelerate the car. Should I also consider removing the passenger seat to "free up power to the wheels"? How about you do the math and prove me wrong? Here is my assertion: lowering the moment of inertia of a driveshaft, or any rotating driveline component that is powered by the engine, is effectively equivalent to removing weight from other parts of the car. How much effect it has is determined by where that rotating component is in relation to the differential and transmission, i.e. removing 1lb-ft2 of inertia from a tire is different than 1lb-ft2 from a driveshaft which is again different from removing 1lb-ft2 from the flywheel. The equivalent weight change of replacing the stock steel driveshaft with a super-duper lightweight Aluminum driveshaft is probably on the order of 1lb or less. About as effective as taking a dump right before hitting the track.

I don't get the point of predicting a dyno run? A dynamometer is a tuning tool. The numbers bear little significance, as you should be able to deduce from this thread - especially John Coffey's comment.

I will also point out that an inertial dyno inherently must calculate a time-averaged power, where: TAVG = Drum Inertia X ΔRPM/Δt and PAVG=TAVG*RPM which is really just another form of PAVG = ΔW/Δt Only at steady-state does PAVG = PINST

Did you read post #41? I've already explained everything there. Simply put: Mechanical Power = Torque X RPM Are you denying that?

I've removed that nut without unpeening as well, but with a cheater bar, not an impact. Threads were just fine.

Bingo! And I'll add that the power produced in the combustion chamber is the same as the (instantaneous) power seen at the wheels (forgetting about all friction, pumping, and heat losses) at any given moment in time. Average power consumption and amount of energy stored changes with inertia.

Precisely! I like where your head's at. I was thinking about this on my commute and I came to the conclusion that Steve and I are arguing two separate points. Steve is talking about energy, while I'm talking about (instantaneous) power, since the statement I challenged was that less inertia would "put more power to the wheels". In Steve's first reply, power and energy get muddled up. If you replace "power" with "energy" or "work" in the above quote, it would make more sense. Yes, something heavier takes more energy to accelerate and vice-versa, that's not being argued nor denied by me. However, with all this said, there are a couple of ways to look at power. What Steve is arguing is that AVERAGE power changes. Average power is a change in work (i.e. energy) over a given period of time, or ΔW/Δt. This is true and makes sense only if we are interested in just the average power consumed. I am arguing that the instantaneous power does not change. Instantaneous power represents mechanical power, and (instantaneous) Power = Torque X RPM. Notice there is no time nor weight/inertia dependency here! Please forgive me for quoting Wiki, but I thought they phrased it pretty well: Steve is right in that average power consumed decreases with less rotational inertia, and I am correct in that mechanical (instantaneous) power does not change no matter how much inertia you have. Let's get back to reality now and look on the practical side of things. An Aluminum driveshaft won't gain you performance but it sure will make your pockets lighter! If you're unsure, then try it: baseline a Z on a dyno, put in an Aluminum driveshaft (shouldn't take long, could even leave it strapped on the dyno if it's an elevated one) and run it again. I guarantee you won't see any appreciable results, let alone something perceivable in the seat-of-the-pants. In fact, you'd struggle to find any results even with something like a lightened flywheel (this goes back to my reply to Stanley about the effects of gearing on "effective" inertia). Dyno runs are typically done in 3rd or 4th gear, where the engine is brought up to speed slower than in a 1st or 2nd gear run, thus minimizing effects of slight changes in drivetrain inertia.

Same here, I noticed the sugar scoop almost right away and thought, "brilliant!" Looks snazzy.

Finish it and drive it. Worry about color later, once you're satiated with driving the car. Who knows, maybe you'll grow to like the color more, or at the very least you'll get to drive it for a year or two before making it undriveable again.

Oh god no, this is leading down a path of endless banter which will see no end...

Actually, that highly depends on what gear you're in. This is exactly what I was talking about with "effective weight" of the rotational components. The interesting thing is that this effective weight changes with gear ratio! In first gear, the engine has a mechanical advantage over the road. Because of this, rotational inertia has a much more prominent role. In overdrive, the opposite is true and rotational inertia has a comparatively small effect. What this means, is that the weight of rotating components will have a way bigger effect in lower gears than higher gears! If you think about this for a minute, it'll make sense.

We need to understand and identify the relationships between power, acceleration, and inertia. Here's what's going on: To start with your use of an inertial roller dyno as evidence: an inertial dyno approximates power by measuring the acceleration used to turn a drum of static weight. Therefore, decreasing rotational inertia will show an increase in power although there was no actual increase in power whether at the wheels or at the flywheel. This is the reason why inertial dynos are deemed less accurate than load-bearing dynos. A load bearing dyno has a brake that can hold the engine at a distinct speed, under a certain load and throttle position. The brake acts on a lever which then acts on a load cell, and approximates torque since it knows the force and lever arm length. This is a much better representation of how much power is actually making it to the wheels. Therefore, we must understand how we're quantifying power here. As you can now see, an inertial dyno will falsely give an increased power reading when decreasing rotational inertia. With that said, decreasing things like flywheel weight - let alone driveshaft weight - will have little (if any) effect on dyno power readings. Go ahead, try it! Now to your other point, about a super-heavy flywheel. That is incorrect, all the power produced by the engine is available at the wheels (forgetting about friction for the moment)! However, you will accelerate slower because you've got more weight to push (or turn). If we reach back to basic physics, this goes back to the good old Law of Conservation of Energy. Power = Torque X Speed Power is delivered from the engine to the wheels through a series of gear ratios. First gear reduces wheel speed when compared to engine speed, torque at the wheels is greater than torque made by the engine, through gear reduction. Fifth gear increases wheel speed compared to engine speed (overdrive) which results in less torque at the wheels. However, in all situations Torque X Speed (Power) remains constant whether you're accelerating quickly or slowly! Whether you're at steady-state or in a transient doesn't matter, the principles of physics stay the same. The amount of power required to move a mass is another topic. Whether that mass is attached to a rotating component or not, it'll take a certain amount of time in order to accelerate it because of inertia. This inertia may be linear or rotational, but that doesn't change the amount of power being produced at the wheels! It's completely relevant. It doesn't matter whether it's in the drivetrain or not, weight is weight! I've already explained how drivetrain weight/inertia can be deemed as an effective weight. That's all it is. It's not a magical power-robbing device. I think you're missing the main concept of the Conservation of Energy which ties this together. You're looking at only kinetic energy. There is also potential (stored) energy here. Kinetic + Potential = Total. Total energy is conserved in the system, therefore power is conserved (again, barring heat and friction losses). A driveshaft cannot simply "consume" energy. And we circle back once again, "lowering rotational inertia does not put any more power to the wheels." Hope this helps...