Anyone try the MSA aluminum driveshaft?

[doradox]

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...

I don't even know where to start.

The inertia dyno measures the rate of change of the rotational speed of the drum. The more quickly the rotation increases the more power was being applied to it. A simple fact of physics. Less power being used to rotationally accelerate a lightened drivetrain means more available at THE WHEELS TO BE APPLIED TO THE DRUM. Since that power is transmitted by the tires to the drum there MUST be a change in the power available at the tire if the drum's rotation increases more quickly with a lighter flywheel, driveshaft, whatever. The engine never produces more power than it ever has but there simply is not as much consumed by increasing the rotational kinetic energy of the rotating masses in the drivetrain.

The inertia dyno doesn't "falsely" report how much power is being APPLIED to the drum. It's simple physics. A drum with moment of inertia X was accelerated from 0 to Y rad/s in Z amount of time. The power to do that comes from one place.The power at the wheels. Drum turns faster in less time = more power at the wheels.

The inertia dyno doesn't know the exact inertia of the driveline and therefore doesn't report ENGINE power accurately. Put a huge flywheel on it and the dyno says you have less engine power. Put a light flywheel and now suddenly you have more engine power. We all know how that works. But the result is the same. If you can turn the drum faster you can also accelerate he car faster.

Energy is conserved. The energy in the fuel is converted in the engine and the some of that energy gets stored as rotational kinetic energy in the driveshaft. In order for that to happen the engine did work on the driveshaft. Work that wasn't used to accelerate the car. So the driveshaft most certainly consumes energy. It does not DESTROY energy. That consumed, or stored if you prefer, energy stays there until it is converted into some other form. Likely brake heat when it comes time to slow down.

I agree that at steady state there is no more power at the wheels or that somehow the engine has more power because the driveshaft is lighter. But during acceleration the power used to rotationally accelerate a heavy driveshaft is no longer available to accelerate the rest of the vehicle. Reducing the amount of power required to rotationally accelerate the driveshaft frees up power that can now be used to accelerate the rest of the car. Since that driveshaft is between the engine and the wheels it necessarily reduces the amount of power available at the wheels for the period of acceleration.

Steve

[ksechler]

Steve's right. The problem with these types of discussion is that the terms power, torque, work, energy, etc. get misused. I'm also an engineer and pulled one of my texts on Engineering mechanics, and quickly got bored and put it away. I will say this. The argument could probably be better made by using energy not power. Moment of inertia figures into that which means (over simplified) that if the moment of inertia is smaller (i.e. lighter flywheel/driveshaft, whatever) there will be more energy available to do work at the wheels. The car will accelerate faster.

[John Coffey]

Power is generated during the combustion process.. That power is used to accelerate mass (pistons, rods, crank, flywheel, valve train, clutch, transmission, driveshaft, differential, axles, wheels, tires...) Reducing the mass and/or MOI in everything that is being acted upon by the combustion process increases the rate of acceleration. The measuring system (inertia dyno) directly measures rate of acceleration of ALL the mass being acted upon by the combustion process and uses a formula to calculate an artificial number called "horsepower." Reducing the mass anywhere in the range of measurement increases the rate of acceleration but has no effect on power produced in the combustion process. It does affect a proprietary calculation provided by inertia dyno manufacturers that generates an artificial number called "horsepower."

[conedodger]

Power is generated during the combustion process.. That power is used to accelerate mass (pistons, rods, crank, flywheel, valve train, clutch, transmission, driveshaft, differential, axles, wheels, tires...) Reducing the mass and/or MOI in everything that is being acted upon by the combustion process increases the rate of acceleration. The measuring system (inertia dyno) directly measures rate of acceleration of ALL the mass being acted upon by the combustion process and uses a formula to calculate an artificial number called "horsepower." Reducing the mass anywhere in the range of measurement increases the rate of acceleration but has no effect on power produced in the combustion process. It does affect a proprietary calculation provided by inertia dyno manufacturers that generates an artificial number called "horsepower."

Word!

[LeonV]

The problem with these types of discussion is that the terms power, torque, work, energy, etc. get misused.

The argument could probably be better made by using energy not power. Moment of inertia figures into that which means (over simplified) that if the moment of inertia is smaller (i.e. lighter flywheel/driveshaft, whatever) there will be more energy available to do work at the wheels. The car will accelerate faster.

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.

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.

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:

Note that the power injected by the torque depends only on the instantaneous angular speed – not on whether the angular speed increases, decreases, or remains constant while the torque is being applied

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.

[LeonV]

Power is generated during the combustion process.. That power is used to accelerate mass (pistons, rods, crank, flywheel, valve train, clutch, transmission, driveshaft, differential, axles, wheels, tires...) Reducing the mass and/or MOI in everything that is being acted upon by the combustion process increases the rate of acceleration. The measuring system (inertia dyno) directly measures rate of acceleration of ALL the mass being acted upon by the combustion process and uses a formula to calculate an artificial number called "horsepower." Reducing the mass anywhere in the range of measurement increases the rate of acceleration but has no effect on power produced in the combustion process. It does affect a proprietary calculation provided by inertia dyno manufacturers that generates an artificial number called "horsepower."

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.

[doradox]

Power is generated during the combustion process.. That power is used to accelerate mass (pistons, rods, crank, flywheel, valve train, clutch, transmission, driveshaft, differential, axles, wheels, tires...) Reducing the mass and/or MOI in everything that is being acted upon by the combustion process increases the rate of acceleration. The measuring system (inertia dyno) directly measures rate of acceleration of ALL the mass being acted upon by the combustion process and uses a formula to calculate an artificial number called "horsepower." Reducing the mass anywhere in the range of measurement increases the rate of acceleration but has no effect on power produced in the combustion process. It does affect a proprietary calculation provided by inertia dyno manufacturers that generates an artificial number called "horsepower."

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.

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

John is right and no one is argueing that engine power changes in any way due to drivetrain inertia. Nor that interia dynos aren't

notoriously innacurate.

If the drum spins faster due to a reduction of drivetrain inertia there is more power at the wheels. The wheels are the only thing that touch

the drum so power can come from no other place. That is a simple fact. This instantaneous vs. average power is irrelevant.

1. Plot the rotational kinetic energy of the drum against time and the slope at any point is the instantaneous power being applied to it.

Applied by the power available at the rear wheels.

2.Do the same for all the drivetrain and engine component energies, again forgetting about friction etc. The dyno tries to estimate this,

including friction drag, which is why it is inacurate.

The total of 1 and 2 is the total instantaneous power. That sum is equal to the instantaneous power generated during the combustion

process by the engine.

If it's not some of our energy or time has gone missing. Simple energy balance.

In time interval (even an inifinitely small one) X energy changes by Y amount. That's power. Energy is conserved so all the energy

changes during time interval X must balance. If any drivetrain component's energy increases during time interval X then it has consumed

engine power during that time interval. That power is now unavailable at the rear wheels and cannot be used to accelerate the car.

Leon, if the instantaneous power of the engine is the same as that at the rear wheels, the dyno would accurately calculate engine HP

because it calculates instantaneous power, applied to the drum.

Steve

[Stanley]

So don't go to sleep in math class, street racers.

[LeonV]

John is right and no one is argueing that engine power changes in any way due to drivetrain inertia. Nor that interia dynos aren't

notoriously innacurate.

If the drum spins faster due to a reduction of drivetrain inertia there is more power at the wheels. The wheels are the only thing that touch

the drum so power can come from no other place. That is a simple fact. This instantaneous vs. average power is irrelevant.

1. Plot the rotational kinetic energy of the drum against time and the slope at any point is the instantaneous power being applied to it.

Applied by the power available at the rear wheels.

2.Do the same for all the drivetrain and engine component energies, again forgetting about friction etc. The dyno tries to estimate this,

including friction drag, which is why it is inacurate.

The total of 1 and 2 is the total instantaneous power. That sum is equal to the instantaneous power generated during the combustion

process by the engine.

If it's not some of our energy or time has gone missing. Simple energy balance.

In time interval (even an inifinitely small one) X energy changes by Y amount. That's power. Energy is conserved so all the energy

changes during time interval X must balance. If any drivetrain component's energy increases during time interval X then it has consumed

engine power during that time interval. That power is now unavailable at the rear wheels and cannot be used to accelerate the car.

Leon, if the instantaneous power of the engine is the same as that at the rear wheels, the dyno would accurately calculate engine HP

because it calculates instantaneous power, applied to the drum.

Steve

Did you read post #41?

I've already explained everything there. Simply put:

Mechanical Power = Torque X RPM

Are you denying that?

[Stanley]

Sorry if this is off topic but it seems like my best chance with engineers on the thread. With the dragstrip a long trip away, and the dyno a once a year thing due to recession, been doing 0 to 60's and mapping steady state revs at various mph to check the tune.

Is there some way to turn this information into torque numbers to predict next dyno run? How to account for wind resistance? Does (tire) size matter?

[LeonV]

John is right and no one is argueing that engine power changes in any way due to drivetrain inertia. Nor that interia dynos aren't

notoriously innacurate.

If the drum spins faster due to a reduction of drivetrain inertia there is more power at the wheels. The wheels are the only thing that touch

the drum so power can come from no other place. That is a simple fact. This instantaneous vs. average power is irrelevant.

1. Plot the rotational kinetic energy of the drum against time and the slope at any point is the instantaneous power being applied to it.

Applied by the power available at the rear wheels.

2.Do the same for all the drivetrain and engine component energies, again forgetting about friction etc. The dyno tries to estimate this,

including friction drag, which is why it is inacurate.

The total of 1 and 2 is the total instantaneous power. That sum is equal to the instantaneous power generated during the combustion

process by the engine.

If it's not some of our energy or time has gone missing. Simple energy balance.

In time interval (even an inifinitely small one) X energy changes by Y amount. That's power. Energy is conserved so all the energy

changes during time interval X must balance. If any drivetrain component's energy increases during time interval X then it has consumed

engine power during that time interval. That power is now unavailable at the rear wheels and cannot be used to accelerate the car.

Leon, if the instantaneous power of the engine is the same as that at the rear wheels, the dyno would accurately calculate engine HP

because it calculates instantaneous power, applied to the drum.

Steve

I will also point out that an inertial dyno inherently must calculate a time-averaged power, where:

T_AVG = Drum Inertia X ΔRPM/Δt

and

P_AVG=T_AVG*RPM

which is really just another form of

P_AVG = ΔW/Δt

Only at steady-state does P_AVG = P_INST

Edited December 11, 2012 by LeonV

[LeonV]

Sorry if this is off topic but it seems like my best chance with engineers on the thread. With the dragstrip a long trip away, and the dyno a once a year thing due to recession, been doing 0 to 60's and mapping steady state revs at various mph to check the tune.

Is there some way to turn this information into torque numbers to predict next dyno run? How to account for wind resistance? Does (tire) size matter?

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.