Anyone try the MSA aluminum driveshaft?

[doradox]

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

Did you read post #41?

I've already explained everything there. Simply put:

Mechanical Power = Torque X RPM

Are you denying that?

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

You have not explained anything. Average power converges to instantaneous as Δt approaches 0 not just at steady state. So what?

Mechanical power need not be torque x rpm. Look up drawbar power.

To an engineer, in basic units mechanical power = (force * distance) / time.

force*distance happens to be the units for both energy and work.

So when you do an energy balance over a fixed amount of time, even dt, as I previously suggested and use conservation of energy you can see that I am right.

The power from the engine is, during acceleration in a fixed interval of time, partly consumed and stored by the rotating masses in the drivetrain as when their rotational speed increases they have an increase in rotational kinetic energy. This leaves less power at the wheels to increase the linear kinetic energy of the entire car because the sum of all those energy changes MUST be equal to the energy the engine was able to convert in that fixed interval of time.

Do the math.

My assertion from the beginning has been that lowering the moment of inertia of a driveshaft allows more power, however small, to be transmitted to the wheels during acceleration ONLY. I have never asserted that engine power is changed or any other thing.

Steve

[Stanley]

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.

Just trying to find a convenient way to record performance changes after little tuning tweeks and mods.

[Zed Head]

This won't help the resolution of the discussion, but it has the word "engineer" in it -

“An optimist will tell you the glass is half-full; the pessimist, half-empty; and the engineer will tell you the glass is twice the size it needs to be.”

[ksechler]

Stanley:

Without a dyno you're fighting an uphill battle. Maybe someone here can help.

Steve:

I still concur. What type of engineer are you? ME? I'm an AE.

[Stanley]

No biggie for me, just trying to get from a 15.7 et to the high 14's without spending any money. Guess I take those tenths of a second pretty seriously. I'm a civil/structural PE so I know a lot about boards, no help with this stuff. Glad to be learning something. Hope it helped the original poster. He said his car was far from stock, so every little bit helps, maybe.

[LeonV]

You have not explained anything. Average power converges to instantaneous as Δt approaches 0 not just at steady state. So what?

Mechanical power need not be torque x rpm. Look up drawbar power.

To an engineer, in basic units mechanical power = (force * distance) / time.

force*distance happens to be the units for both energy and work.

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.

So when you do an energy balance over a fixed amount of time, even dt, as I previously suggested and use conservation of energy you can see that I am right.

The power from the engine is, during acceleration in a fixed interval of time, partly consumed and stored by the rotating masses in the drivetrain as when their rotational speed increases they have an increase in rotational kinetic energy. This leaves less power at the wheels to increase the linear kinetic energy of the entire car because the sum of all those energy changes MUST be equal to the energy the engine was able to convert in that fixed interval of time.

Do the math.

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?

My assertion from the beginning has been that lowering the moment of inertia of a driveshaft allows more power, however small, to be transmitted to the wheels during acceleration ONLY. I have never asserted that engine power is changed or any other thing.

Steve

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.

[LeonV]

No biggie for me, just trying to get from a 15.7 et to the high 14's without spending any money. Guess I take those tenths of a second pretty seriously. I'm a civil/structural PE so I know a lot about boards, no help with this stuff. Glad to be learning something. Hope it helped the original poster. He said his car was far from stock, so every little bit helps, maybe.

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.

[doradox]

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

Remember, the above is what we are arguing.

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.

At it's most basic power is the time rate of change of energy. So it does need to be something and that is why I have chosen to use energy to make my point.

One could also do an analysis by using force analysis with free body diagrams of all the affected components. F=MA so it will become clear than as we move away from the engine the force available to rotationally accelerate each component decreases exactly by the amount that was needed to accelerate the previous component until we get all the way back to the wheel. we can then convert what torque is left to a linear force and apply that to the vehicle mass.

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.

Time is integral to power it can't be out of the picture.

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"?

Easy, F=MA. Since the mass of the car is lower the same amount force , your precious torque, applied to the pavement causes A to be bigger. Simple physics.

Now explain why increasing the rotational inertia of the driveshaft but not it's, weight causes the same car to accelerate slower.

Perhaps since M in unchanged and A is less there is less F at the wheel. That's exactly what the math says. And if there is less torque then there must be less power because, as you have said..." Power = Torque * Angular Displacement / Time." Remember the F at the wheel, or more precisely at the contact patch, is what affects the acceleration of the whole vehicle.

How about you do the math and prove me wrong?

I just did.

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 agree. BUT that isn't what your post, quoted at the top of this post, is saying. If it did we wouldn't be having this discussion.

[LeonV]

Remember, the above is what we are arguing.

At it's most basic power is the time rate of change of energy. So it does need to be something and that is why I have chosen to use energy to make my point

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.

One could also do an analysis by using force analysis with free body diagrams of all the affected components. F=MA so it will become clear than as we move away from the engine the force available to rotationally accelerate each component decreases exactly by the amount that was needed to accelerate the previous component until we get all the way back to the wheel. we can then convert what torque is left to a linear force and apply that to the vehicle mass.

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.

Time is integral to power it can't be out of the picture.

See above.

Easy, F=MA. Since the mass of the car is lower the same amount force , your precious torque, applied to the pavement causes A to be bigger. Simple physics.

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?

Now explain why increasing the rotational inertia of the driveshaft but not it's, weight causes the same car to accelerate slower.

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.

Perhaps since M in unchanged and A is less there is less F at the wheel. That's exactly what the math says. And if there is less torque then there must be less power because, as you have said..." Power = Torque * Angular Displacement / Time." Remember the F at the wheel, or more precisely at the contact patch, is what affects the acceleration of the whole vehicle.

Nope. See above.

I just did.

Sort of...

I agree. BUT that isn't what your post, quoted at the top of this post, is saying. If it did we wouldn't be having this discussion.

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.

[doradox]

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.

Your arguements are really light on math and physics and heavy on " because two things seem the same to me then they must be the same".

This is why I think you can't comprehend that decreasing rotational inertia of the drivetrain will cause an increase in power at the rear wheels during acceleration over the non reduced inertia driveline.

Since reducing rotational inertia and reducing any fixed mass both result in increased acceleration you think the REASON why they do that is the same. It's just a reduction in "apparent" mass. That's rediculous and the kind of logic someone who has no grasp of math or physics believes.

Reducing the rotational inertia of the driveline results in an increase in the force available to the rear wheels to apply to the road to accelerate the car.

F=MA . Increased force applied to the road results in increased acceleration. Total actual mass being accelerated by the force has not increased.

There is an ACTUAL increase in the force applied to the road as will be explained below.

Reducing the actual mass of the car results in the increased acceleration because..once again..

F=MA. Same force because we did NOTHING to change that and less mass MUST result in increased acceleration. No sort of increase in a rediculous "available" force caused the acceleration.

Let me say it like this.

While accelerating at one instant in time, at the engine crank, we have 100 lb*ft of torque and 1000 rpm.

If the torque needed to rotationally accelerate the flywheel at the current rate is 10 lb*ft then the next component, say the transmission in 4th gear, in the drive train sees 90 lb*ft and 1000rpm.

Say the tranny requires another 10 lb*ft to rotaionally accelerate it's rotating components. That leaves 80 lb*ft@1000 rpm,

Then next the driveshaft requires 10 lb*ft of torque to rotationally accelerate. Now we're down to 70 lb*ft@1000rpm.

Let's then say the differential has a gear ratio of 1:1. So now the diff and axles ect. use 10 lb*ft of torque to rotationally accelerate those rotating parts.

We are left with 60 lb*ft @1000 rpm to the wheels.

Horsepower at the crank = 100*1000/5252

Horsepower at the wheels = 60*1000/5252

All this is while accelerating, as I have stated over and over, because if there was no acceleration then the rotating parts in the drivetrain would not require any of the engines torque output. Because F=MA ( T=I*alpha) if you prefer. Alpha is 0 so T is 0.

If we reduce the driveshaft's moment of inertia so it only takes 5 lb*ft of torque to rotationally accelerate it at the current rate then....

Horsepower at the wheels = 65*1000/5252

That is an ACTUAL increase in the amount of horsepower at the rear wheels which is counter to your assertion.

Steve

[Casey_z]

Im considering an aluminum performance shaft since my slip yoke is no good.

Thanks!

Original question was has anyone used one:)

So far the simple answer is no. Just sayin...............

[Oiluj]

Holy Cow!__Didn't think a little statement like "rotational inertia" could have generated so much discussion