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

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.

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

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

For those who don't care to understand the math and the physics, the amusing thing is these two actually pretty much completely agree with each other. 8)

A prime example of Engineering Masturbation. :-)

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.

My last post was insulting and I apologize for that.

I was getting very frustrated since I do believe you have the knowledge to understand what I was trying to say yet seemingly did not. I have always found your posts to backed by a good understanding of basic principles. I thought this would be a short discussion.

I like using work/energy to look at a lot of different problems and sometimes people don't readily see the method in my madness.

Best Regards,

Steve

A prime example of Engineering Masturbation. :-)

It helps to relieve the mind...

My last post was insulting and I apologize for that.

I was getting very frustrated since I do believe you have the knowledge to understand what I was trying to say yet seemingly did not. I have always found your posts to backed by a good understanding of basic principles. I thought this would be a short discussion.

I like using work/energy to look at a lot of different problems and sometimes people don't readily see the method in my madness.

Best Regards,

Steve

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.

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.

ME. There seems to be a lot of engineers on this board.

Steve

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.

It was kind of fun. Right now I am crash testing some work I've done on a vehicle occupant safety system redesign.

Steve

There are a number of engineers on the board. Goes well with owning a 40 year old car running. One of us even works for NASA! (not me).

In my new job I'm developing & improving mfg processes/mfg automation in the medical device/pharma industry.

It's a nice change, but similar to Leon's comment, I kinda miss doing instrument design work. It involved not just mechanical design, but also applied physics. At least I still get to do some micro-fluidics...

To return to the original question, has anyone tried the MSA aluminum driveshaft? The only reason that I'm even considering this is because of cost, not performance.

I've been trying to track down some drive train vibration for some time, and when my new Michelin tires were being installed I asked the shop to inspect the entire drive train (again). They say that I need new driveshaft u-joints, which are $35 each from MSA for the heavy duty (assume Spicer) model, plus labor to install them (unknown tonight), plus balancing the assembly (also unknown at this point).

Given shop prices in Northern VA, there might not be a huge gap.

Bearings, bushings and all suspension components have been replaced, as I mentioned in other posts.

Thoughts?

Peter

RockAuto has Moog u joints for 11.78 each for the '73. Good place to get parts as long as you know exactly what you're looking for.

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