Anyone try the MSA aluminum driveshaft?

[LeonV]

Really? So the power used to accelerate the driveshaft is not decreased by lowering the rotational inertia and therefore more of the engine's power is not available to accelerate everything else?

At steady state you are correct though.

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."

[doradox]

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."

I've already said that at steady state you are correct. When accelerating you are not.

I am not saying we are "making more power" only that more of what there is makes it to the rear wheels but only while accelerating. That's why an inertia dyno will register more power if you lighten a flywheel. The dyno doesn't see the power used to accelerate the flywheel, only what makes it to the wheels. The more power used to accelerate a heavy flywheel or any other part of the drive train, the less available at the wheels.

Imagine attaching a gigantic flywheel to the engine and you might see that all the power able to be produced by the engine would be consumed by trying to rotationally accelerate the flywheel and almost none would make it to the wheels.

The spare tire analogy is irrelevant because the tire isn't in the drivetrain like the driveshaft is.

Rotationally accelerating a driveshaft takes power. Work/Time.

During acceleration the increasing rotational kinetic energy stored in the driveshaft comes from the engine. Where else could it come from?

Therefore some of the engine's fixed amount of power is consumed rotationally accelerating (doing work on) the driveshaft.

Reducing the amount of power that the driveshaft consumes during acceleration means more power makes it to the rear wheels during acceleration.

At steady state the rotational kinetic energy of the driveshaft does not change and therefore it has no effect on the power available at the wheels. As I already said.

Steve

[conedodger]

Warning! Steve, you are arguing with an engineer. A famous Z guy once pointed out that arguing with an engineer is like mud-wrestling with a pig. After a while you're covered with mud and you realize the pig loves it!

I told you Leon said there is no more power being made...:classic:

[71Nissan240Z]

Well I've seen a dyno on a friends car gain a few ponies and some torque with a driveshaft from driveshaft specialists.

Ive decided to just install the new ujoint on my bad slip yoke for now. I need to fix part of my passangerside floor and do some other repairs like my valve seals and some otherthings.

On a side note...Im an electrical engineer student. You dont see me trying to buff people about rebuilding engines......just because I know how to do it doesnt make me an expert. And whether or not you agree on spending $400 on an aluminum driveshaft is my decision not yours. Do what you want with your car right?

Okay thanks.

Edited December 10, 2012 by 71Nissan240Z

[doradox]

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

Warning! Steve, you are arguing with an engineer. A famous Z guy once pointed out that arguing with an engineer is like mud-wrestling with a pig. After a while you're covered with mud and you realize the pig loves it!

I told you Leon said there is no more power being made...:classic:

Since I'm an Engineer as well I think we'll both have a grand time.

Steve

[Stanley]

Sometimes if a problem has unknown variables or is complex and you need a quick solution you can ballpark it with boundry conditions. Suppose your rear wheels were off the ground then you might notice faster revving with a lighter driveshaft. This corresponds to a burnout with a strong motor, street tires & a wet or greasy track. You might get a better burnout (disregarding the question of how much burnout is good - once bet a new Mustang GT in the 1/8 due to his excessive burnout, he creamed me in the quarter) with the lighter shaft. Once you're going, weight & rotational enertia of the driveshaft won't make much difference compared to the weight of the car. On the plus side, the MSA comes with heavy-duty u-joints.

Let em argue- makes a good discussion. One thing that's kept the engineers that know it from making a lot of wrong decisions (and caused the ones who don't know it a lot of wrong decisions) is those three little words - "I don't know".

Maybe I should get one - can't do much of a burnout with the automatic, ha ha.

[conedodger]

I hate burnouts. It means the rest of the suspension isn't doing what I want.

[LeonV]

We need to understand and identify the relationships between power, acceleration, and inertia. Here's what's going on:

I've already said that at steady state you are correct. When accelerating you are not.

I am not saying we are "making more power" only that more of what there is makes it to the rear wheels but only while accelerating. That's why an inertia dyno will register more power if you lighten a flywheel. The dyno doesn't see the power used to accelerate the flywheel, only what makes it to the wheels. The more power used to accelerate a heavy flywheel or any other part of the drive train, the less available at the wheels.

Imagine attaching a gigantic flywheel to the engine and you might see that all the power able to be produced by the engine would be consumed by trying to rotationally accelerate the flywheel and almost none would make it to the wheels.

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!

The spare tire analogy is irrelevant because the tire isn't in the drivetrain like the driveshaft is.

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.

Rotationally accelerating a driveshaft takes power. Work/Time.

During acceleration the increasing rotational kinetic energy stored in the driveshaft comes from the engine. Where else could it come from?

Therefore some of the engine's fixed amount of power is consumed rotationally accelerating (doing work on) the driveshaft.

Reducing the amount of power that the driveshaft consumes during acceleration means more power makes it to the rear wheels during acceleration.

At steady state the rotational kinetic energy of the driveshaft does not change and therefore it has no effect on the power available at the wheels. As I already said.

Steve

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...

[LeonV]

Once you're going, weight & rotational enertia of the driveshaft won't make much difference compared to the weight of the car.

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.

[LeonV]

Since I'm an Engineer as well I think we'll both have a grand time.

Steve

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

[Stanley]

I really gotta go to work.

[conedodger]

Gentlemen, for the record, there will be no last word... Let the mud fly!