LeonV

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...

Give me a shout (send a PM) if you're coming through SF and need to get your Z fix. I'm in San Bruno.

Yes, "really". In fact, you already know why. As you stated, steady state power is unchanged. Steady state power is power to the wheels. This does not change whether you're accelerating or not, you don't magically make more power under acceleration than at steady state! Your statement is akin to saying, "I took the spare tire out of my Z which allows my engine to deliver more power to the wheels." Is that true? Rotational inertia is an "effective" weight, meaning decreasing 5lb from your flywheel does not alter performance the same as taking that 5lb out of the body, but it has a similar effect which can be quantified with an "effective" weight (e.g. 5lb off the flywheel is like taking 20lb off the car - not real numbers, I made those up). Power is a function of torque and RPM (not the time-rate of change of RPM), i.e. it is independent of acceleration. To sum this up, just because the car accelerates faster does not mean there is any more power available at the wheels. We now circle back to my original statement, "lowering rotational inertia does not put any more power to the wheels."

Had an awesome freeway encounter today! I drove my 260Z to work for the first time in a few weeks since it's been pretty wet lately. As I was heading up 101 North in San Rafael (Northern Bay Area), I saw a very original-looking brown 240Z, down to the wheel caps. Behind the wheel was an older gentleman and a co-pilot, who I can only assume is his wife. Gave them a honk, smile and a thumbs up and got the same (minus the honk) in return. Made my day!

More like Porsche 935...

Rob, look up "Primadonna Z". It actually started life as an apartment parking lot project in San Bruno!

Exhaust systems don't just get "plugged". TPS is my first thought, vac leaks seem unlikely from his description, a stock EFI Z won't idle smoothly at all if there's a leak large enough to hinder WOT operation that much. RT, have you washed the engine bay lately or been in a downpour? Water can screw with the TPS, even if it's adjusted correctly. Follow Captain's advice and download the FSM and EFI guide.

Nice! It's definitely not a job for the faint of hear but perseverance will get it done.

Lowering rotational inertia does not put any more power to the wheels.

What's the reason for swapping the 260Z unit out?

http://atlanticz.ca/zclub/techtips/suspension/rearend/index.html Bookmark this site: http://atlanticz.ca/index.php/tech-tips.html Then thank Blue for his "Tech Tips".

You've already got one?