Results 1 to 71 of 71

Thread: Anyone try the MSA aluminum driveshaft?

  1. #1
    Registered User 71Nissan240Z's Avatar
    Member ID
    CZCC-19487
    Join Date
    Jun 2009
    Location
    Portland, Oregon
    Age
    37
    Posts
    146

    Default Anyone try the MSA aluminum driveshaft?

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

    Thanks!

  2. #2
    Registered User 71Nissan240Z's Avatar
    Member ID
    CZCC-19487
    Join Date
    Jun 2009
    Location
    Portland, Oregon
    Age
    37
    Posts
    146

    Default

    I guess I'll play guinea pig! ahah

  3. #3
    Registered User EuroDat's Avatar
    Member ID
    CZCC-26512
    Join Date
    Feb 2012
    Location
    The Netherlands
    Posts
    1,420

    Default

    Dont know anyone who has used one. Why do you want to go aluminium?
    Plenty of other ways to reduce weight and cheaper per ounce. Ligthening the flywhel etc.

  4. #4
    Registered User conedodger's Avatar
    Member ID
    CZCC-16545
    Join Date
    Apr 2008
    Location
    Reno, NV
    Posts
    2,053

    Default

    They list a carbon fiber model. Go with that if you want to spend that kind of money...
    Rob
    2000 BMW R1100 RT-SE (for sale)
    1999 Toyota 4Runner Supercharged
    1975 Porsche 914 stroker motor autoX car
    1973 Datsun 240Z Restoration project. New paint in original white. E31 head on 2.4 block. Nissan Motorsports header. R200 with Nissan motorsports LSD.

  5. #5
    Registered User
    Member ID
    CZCC-26864
    Join Date
    Apr 2012
    Location
    Durango, Colorado, United States
    Posts
    692

    Default

    Go to the junkyard and buy a used drive shaft for $30. Put your $ into something that would ACTUALLY boost performance, like a better exhaust system instead of something shiny on the bottom of the car.

  6. #6
    Registered User
    Member ID
    CZCC-23182
    Join Date
    Dec 2010
    Location
    Mid-Atlantic
    Age
    49
    Posts
    148

    Default

    Quote Originally Posted by 71Nissan240Z View Post
    Im considering an aluminum performance shaft since my slip yoke is no good.

    Thanks!
    Do you have any idea how much of a weight difference there is? If there is a significant difference then it might be worth it (in my mind). This is not a static component, it's rotating so there would be an added benefit. I have a feeling, though, that this is the type of component upgrade you make when you are going for a track car and have already tweaked everything else. If you do make the change please post some results. I'd be interested to know how it works out.

  7. #7
    Registered User Oiluj's Avatar
    Member ID
    CZCC-15388
    Join Date
    Nov 2007
    Location
    SF Bay Area
    Age
    67
    Posts
    1,746

    Default

    I would think the reduction in rotatonal inertia might have a bigger impact than the weight savings. Similar to a lightweight flywheel. Would be pretty easy to calculate. As mentioned above, it might not be the most bang for your buck.
    Julio
    1972 240Z (in-progress, 95% complete)
    CZC# 15388

  8. #8
    Registered User LeonV's Avatar
    Member ID
    CZCC-19146
    Join Date
    Apr 2009
    Location
    Bay Area, CA
    Posts
    1,961

    Default

    A driveshaft doesn't have much rotational inertia to begin with. Since it's long and skinny, the location of the mass doesn't stray far from the axis of rotation.
    2/74 260Z

  9. #9
    Registered User Oiluj's Avatar
    Member ID
    CZCC-15388
    Join Date
    Nov 2007
    Location
    SF Bay Area
    Age
    67
    Posts
    1,746

    Default

    I agree, the radius is relatively small, but they are also made of steel, and constantly accelerating / decelerating... My comment was simply that the weight savings probably has less impact than the rotational energy savings.

    Either way, it's not even worth doing the calculations. Like I said, there are other items / improvements that will provide a better bang for your buck.
    Julio
    1972 240Z (in-progress, 95% complete)
    CZC# 15388

  10. #10
    Registered User LeonV's Avatar
    Member ID
    CZCC-19146
    Join Date
    Apr 2009
    Location
    Bay Area, CA
    Posts
    1,961

    Default

    Quote Originally Posted by Oiluj View Post
    I agree, the radius is relatively small, but they are also made of steel, and constantly accelerating / decelerating... My comment was simply that the weight savings probably has less impact than the rotational energy savings.

    Either way, it's not even worth doing the calculations. Like I said, there are other items / improvements that will provide a better bang for your buck.
    Julio, I completely agree. Just adding some extra info...
    2/74 260Z

  11. #11
    Registered User MotoManMike's Avatar
    Member ID
    CZCC-26581
    Join Date
    Feb 2012
    Location
    Maryland, Eastern Shore
    Age
    40
    Posts
    173

    Default

    Let us know what you think of it. They look pretty nice. Pricy but hey, I don't know what it costs MSA to have them produced so I can't judge it.

  12. #12
    Registered User 71Nissan240Z's Avatar
    Member ID
    CZCC-19487
    Join Date
    Jun 2009
    Location
    Portland, Oregon
    Age
    37
    Posts
    146

    Default

    Lowering the rotating inertia weightl allow u to put a little more power to the wheels. So my slip yoke on my sdriveshaft is bad. Instead of getting a used part. Id rrather go with an upgrade. My car is far from stock at this point.

  13. #13
    Registered User Walter Moore's Avatar
    Member ID
    CZCC-3035
    Join Date
    Oct 2002
    Location
    Indianapolis
    Posts
    1,833

    Default

    And some of us just don't have junkyards nearby with 40+ year old cars in them...
    '71 240Z, Because any fool can drive fast in a straight line.

  14. #14
    Registered User conedodger's Avatar
    Member ID
    CZCC-16545
    Join Date
    Apr 2008
    Location
    Reno, NV
    Posts
    2,053

    Default

    The sad thing is that you really can't make an honest opinion without replacing a good stock one with the aluminum or carbon fiber one... All you can say is yup, now it works.
    Rob
    2000 BMW R1100 RT-SE (for sale)
    1999 Toyota 4Runner Supercharged
    1975 Porsche 914 stroker motor autoX car
    1973 Datsun 240Z Restoration project. New paint in original white. E31 head on 2.4 block. Nissan Motorsports header. R200 with Nissan motorsports LSD.

  15. #15
    Registered User LeonV's Avatar
    Member ID
    CZCC-19146
    Join Date
    Apr 2009
    Location
    Bay Area, CA
    Posts
    1,961

    Default

    Quote Originally Posted by 71Nissan240Z View Post
    Lowering the rotating inertia weightl allow u to put a little more power to the wheels. So my slip yoke on my sdriveshaft is bad. Instead of getting a used part. Id rrather go with an upgrade. My car is far from stock at this point.
    Lowering rotational inertia does not put any more power to the wheels.
    2/74 260Z

  16. #16
    Low Budget/High Value
    Member ID
    CZCC-20342
    Join Date
    Oct 2009
    Location
    Washington County, OR
    Posts
    3,629

    Default

    Quote Originally Posted by 71Nissan240Z View Post
    Lowering the rotating inertia weightl...
    Actually, I think it will allow you to send more shock load to the differential. Less mass to damp engine engagement through the clutch. Probably won't matter unless you're drag-racing, although I could see a higher modulus material transmitting higher frequency noise. Might be noisier, but I haven't seen any accounts. I have read stories about the aluminum shafts not liking harsh treatment through the clutch. Apparently the u-joint yokes can get damaged.

    You can buy a slip yoke and have a new steel shaft made for ~$250 bucks easy I would guess. I have found a local guy, up the Gorge (Portland talk...) who would shorten one for $90, which turned out to be essentially building it over again after cutting it. A good shop could probably source the yoke too. Just take your old one in and say "make me a copy of this, please". You might be spending more money than necessary for a relatively simple part.
    1976 280Z, with some minor modifications

  17. #17
    Registered User
    Member ID
    CZCC-17788
    Join Date
    Sep 2008
    Location
    So Cal
    Age
    35
    Posts
    8

    Default

    Iv'e done an aluminum driveshaft before (not on a Z) and it did make a noticable difference in performance. The throttle response felt quite a bit better and overall the car just felt lighter due to the quicker acceleration, especially in the corners. It's nothing like a light weight flywheel but worth doing if you have already done most of the major mods.

  18. #18
    Registered User doradox's Avatar
    Member ID
    CZCC-14809
    Join Date
    Aug 2007
    Location
    Colfax, IN
    Posts
    708

    Default

    Quote Originally Posted by LeonV View Post
    Lowering rotational inertia does not put any more power to the wheels.
    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.

  19. #19
    Registered User conedodger's Avatar
    Member ID
    CZCC-16545
    Join Date
    Apr 2008
    Location
    Reno, NV
    Posts
    2,053

    Default

    Quote Originally Posted by doradox View Post
    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.
    Leon is correct. There is no more power. There is just an increase, however large or small in efficiency. Somewhere, there is probably a trade though. Maybe they only have a service life of 5 years, there is always a trade off.
    Rob
    2000 BMW R1100 RT-SE (for sale)
    1999 Toyota 4Runner Supercharged
    1975 Porsche 914 stroker motor autoX car
    1973 Datsun 240Z Restoration project. New paint in original white. E31 head on 2.4 block. Nissan Motorsports header. R200 with Nissan motorsports LSD.

  20. #20
    Registered User doradox's Avatar
    Member ID
    CZCC-14809
    Join Date
    Aug 2007
    Location
    Colfax, IN
    Posts
    708

    Default

    Quote Originally Posted by conedodger View Post
    Leon is correct. There is no more power. There is just an increase, however large or small in efficiency. Somewhere, there is probably a trade though. Maybe they only have a service life of 5 years, there is always a trade off.

    "Lowering the rotating inertia weightl allow u to put a little more power to the wheels."

    Note the part about more power to the wheels. Less of the unchanged amount of power available from the engine is used to rotationally accelerate the lighter driveshaft leaving more available to be used at the wheels. No more total power. More available at the wheels during acceleration only. Just like reducing friction in a transmission doesn't give the engine more power but just leaves more available to do useful work.

    Steve

  21. #21
    Registered User conedodger's Avatar
    Member ID
    CZCC-16545
    Join Date
    Apr 2008
    Location
    Reno, NV
    Posts
    2,053

    Default

    Quote Originally Posted by doradox View Post
    "Lowering the rotating inertia weightl allow u to put a little more power to the wheels."

    Note the part about more power to the wheels. Less of the unchanged amount of power available from the engine is used to rotationally accelerate the lighter driveshaft leaving more available to be used at the wheels. No more total power. More available at the wheels during acceleration only. Just like reducing friction in a transmission doesn't give the engine more power but just leaves more available to do useful work.

    Steve
    Ok, Leon is still right, there is no more power being made, and you have restated what I said which is that "there is just an increase, however large or small in efficiency." You're correct. You'll probably want the last word but I agree, there is an increase in efficiency under acceleration. 8)
    Rob
    2000 BMW R1100 RT-SE (for sale)
    1999 Toyota 4Runner Supercharged
    1975 Porsche 914 stroker motor autoX car
    1973 Datsun 240Z Restoration project. New paint in original white. E31 head on 2.4 block. Nissan Motorsports header. R200 with Nissan motorsports LSD.

  22. #22
    Registered User doradox's Avatar
    Member ID
    CZCC-14809
    Join Date
    Aug 2007
    Location
    Colfax, IN
    Posts
    708

    Default

    Quote Originally Posted by LeonV View Post
    Lowering rotational inertia does not put any more power to the wheels.
    Quote Originally Posted by doradox View Post
    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.
    Quote Originally Posted by conedodger View Post
    Ok, Leon is still right, there is no more power being made, and you have restated what I said which is that "there is just an increase, however large or small in efficiency." You're correct. You'll probably want the last word but I agree, there is an increase in efficiency under acceleration. 8)


    Obviously Leon did not say "there is no more power being made" so I'm not sure how he can be "right" about something he didn't say. Maybe you didn't read what either Leon or I wrote.

    Your statement about efficiency is just another way of restating my original rhetorical question to Leon.

    So what exactly have you added to this?

    Steve

  23. #23
    Registered User conedodger's Avatar
    Member ID
    CZCC-16545
    Join Date
    Apr 2008
    Location
    Reno, NV
    Posts
    2,053

    Default

    And there you have it, 'the last word.'
    Rob
    2000 BMW R1100 RT-SE (for sale)
    1999 Toyota 4Runner Supercharged
    1975 Porsche 914 stroker motor autoX car
    1973 Datsun 240Z Restoration project. New paint in original white. E31 head on 2.4 block. Nissan Motorsports header. R200 with Nissan motorsports LSD.

  24. #24
    Registered User doradox's Avatar
    Member ID
    CZCC-14809
    Join Date
    Aug 2007
    Location
    Colfax, IN
    Posts
    708

    Default

    Quote Originally Posted by conedodger View Post
    And there you have it, 'the last word.'
    That's ironic. The guy who said I wanted the last word not being able to resist commenting one more time. This post gives you the opportunity to actually let me have the last word. If you think you can.

    Steve

  25. #25
    Registered User LeonV's Avatar
    Member ID
    CZCC-19146
    Join Date
    Apr 2009
    Location
    Bay Area, CA
    Posts
    1,961

    Default

    Quote Originally Posted by doradox View Post
    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."
    2/74 260Z

  26. #26
    Registered User doradox's Avatar
    Member ID
    CZCC-14809
    Join Date
    Aug 2007
    Location
    Colfax, IN
    Posts
    708

    Default

    Quote Originally Posted by LeonV View Post
    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

  27. #27
    Registered User conedodger's Avatar
    Member ID
    CZCC-16545
    Join Date
    Apr 2008
    Location
    Reno, NV
    Posts
    2,053

    Default

    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...
    Rob
    2000 BMW R1100 RT-SE (for sale)
    1999 Toyota 4Runner Supercharged
    1975 Porsche 914 stroker motor autoX car
    1973 Datsun 240Z Restoration project. New paint in original white. E31 head on 2.4 block. Nissan Motorsports header. R200 with Nissan motorsports LSD.

  28. #28
    Registered User 71Nissan240Z's Avatar
    Member ID
    CZCC-19487
    Join Date
    Jun 2009
    Location
    Portland, Oregon
    Age
    37
    Posts
    146

    Default

    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...lol...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.
    Last edited by 71Nissan240Z; 12-10-2012 at 12:43 PM.

  29. #29
    Registered User doradox's Avatar
    Member ID
    CZCC-14809
    Join Date
    Aug 2007
    Location
    Colfax, IN
    Posts
    708

    Default

    Quote Originally Posted by LeonV View Post
    Lowering rotational inertia does not put any more power to the wheels.
    Quote Originally Posted by conedodger View Post
    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...
    Since I'm an Engineer as well I think we'll both have a grand time.

    Steve
    I slipped into my jeans
    Lookin' hard and feelin' mean
    I took a spit at the moon

  30. #30
    Registered User Stanley's Avatar
    Member ID
    CZCC-24191
    Join Date
    Apr 2011
    Location
    Redondo beach
    Posts
    641

    Default

    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.

  31. #31
    Registered User conedodger's Avatar
    Member ID
    CZCC-16545
    Join Date
    Apr 2008
    Location
    Reno, NV
    Posts
    2,053

    Default

    I hate burnouts. It means the rest of the suspension isn't doing what I want.
    Rob
    2000 BMW R1100 RT-SE (for sale)
    1999 Toyota 4Runner Supercharged
    1975 Porsche 914 stroker motor autoX car
    1973 Datsun 240Z Restoration project. New paint in original white. E31 head on 2.4 block. Nissan Motorsports header. R200 with Nissan motorsports LSD.

  32. #32
    Registered User LeonV's Avatar
    Member ID
    CZCC-19146
    Join Date
    Apr 2009
    Location
    Bay Area, CA
    Posts
    1,961

    Default

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

    Quote Originally Posted by doradox View Post
    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!

    Quote Originally Posted by doradox View Post
    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.

    Quote Originally Posted by doradox View Post
    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...
    2/74 260Z

  33. #33
    Registered User LeonV's Avatar
    Member ID
    CZCC-19146
    Join Date
    Apr 2009
    Location
    Bay Area, CA
    Posts
    1,961

    Default

    Quote Originally Posted by Stanley View Post
    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.
    2/74 260Z

  34. #34
    Registered User LeonV's Avatar
    Member ID
    CZCC-19146
    Join Date
    Apr 2009
    Location
    Bay Area, CA
    Posts
    1,961

    Default

    Quote Originally Posted by doradox View Post
    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...

    Click image for larger version. 

Name:	tzun196l.jpg 
Views:	54 
Size:	68.1 KB 
ID:	58660
    2/74 260Z

  35. #35
    Registered User Stanley's Avatar
    Member ID
    CZCC-24191
    Join Date
    Apr 2011
    Location
    Redondo beach
    Posts
    641

    Default

    I really gotta go to work.

  36. #36
    Registered User conedodger's Avatar
    Member ID
    CZCC-16545
    Join Date
    Apr 2008
    Location
    Reno, NV
    Posts
    2,053

    Default

    Gentlemen, for the record, there will be no last word... Let the mud fly!
    Rob
    2000 BMW R1100 RT-SE (for sale)
    1999 Toyota 4Runner Supercharged
    1975 Porsche 914 stroker motor autoX car
    1973 Datsun 240Z Restoration project. New paint in original white. E31 head on 2.4 block. Nissan Motorsports header. R200 with Nissan motorsports LSD.

  37. #37
    Registered User doradox's Avatar
    Member ID
    CZCC-14809
    Join Date
    Aug 2007
    Location
    Colfax, IN
    Posts
    708

    Default

    Quote Originally Posted by LeonV View Post
    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

  38. #38
    Registered User
    Member ID
    CZCC-23182
    Join Date
    Dec 2010
    Location
    Mid-Atlantic
    Age
    49
    Posts
    148

    Default

    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.

  39. #39
    Crumudgeon
    Member ID
    CZCC-798
    Join Date
    Apr 2000
    Location
    La Habra, CA USA
    Age
    60
    Posts
    1,374

    Default

    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."

  40. #40
    Registered User conedodger's Avatar
    Member ID
    CZCC-16545
    Join Date
    Apr 2008
    Location
    Reno, NV
    Posts
    2,053

    Default

    Quote Originally Posted by John Coffey View Post
    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!
    Rob
    2000 BMW R1100 RT-SE (for sale)
    1999 Toyota 4Runner Supercharged
    1975 Porsche 914 stroker motor autoX car
    1973 Datsun 240Z Restoration project. New paint in original white. E31 head on 2.4 block. Nissan Motorsports header. R200 with Nissan motorsports LSD.

  41. #41
    Registered User LeonV's Avatar
    Member ID
    CZCC-19146
    Join Date
    Apr 2009
    Location
    Bay Area, CA
    Posts
    1,961

    Default

    Quote Originally Posted by ksechler View Post
    The problem with these types of discussion is that the terms power, torque, work, energy, etc. get misused.
    Quote Originally Posted by ksechler View Post
    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.

    Quote Originally Posted by doradox View Post
    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:

    Quote Originally Posted by Wikipedia, as seen in "Torque" under the "Relationship between torque, power, and energy" sub-section
    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.
    2/74 260Z

  42. #42
    Registered User LeonV's Avatar
    Member ID
    CZCC-19146
    Join Date
    Apr 2009
    Location
    Bay Area, CA
    Posts
    1,961

    Default

    Quote Originally Posted by John Coffey View Post
    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.
    2/74 260Z

  43. #43
    Registered User doradox's Avatar
    Member ID
    CZCC-14809
    Join Date
    Aug 2007
    Location
    Colfax, IN
    Posts
    708

    Default

    Quote Originally Posted by John Coffey View Post
    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."
    Quote Originally Posted by LeonV View Post
    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.
    Quote Originally Posted by LeonV View Post
    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

  44. #44
    Registered User Stanley's Avatar
    Member ID
    CZCC-24191
    Join Date
    Apr 2011
    Location
    Redondo beach
    Posts
    641

    Default

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

  45. #45
    Registered User LeonV's Avatar
    Member ID
    CZCC-19146
    Join Date
    Apr 2009
    Location
    Bay Area, CA
    Posts
    1,961

    Default

    Quote Originally Posted by doradox View Post
    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?
    2/74 260Z

  46. #46
    Registered User Stanley's Avatar
    Member ID
    CZCC-24191
    Join Date
    Apr 2011
    Location
    Redondo beach
    Posts
    641

    Default

    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?

  47. #47
    Registered User LeonV's Avatar
    Member ID
    CZCC-19146
    Join Date
    Apr 2009
    Location
    Bay Area, CA
    Posts
    1,961

    Default

    Quote Originally Posted by doradox View Post
    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:

    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
    Last edited by LeonV; 12-11-2012 at 03:01 PM.
    2/74 260Z

  48. #48
    Registered User LeonV's Avatar
    Member ID
    CZCC-19146
    Join Date
    Apr 2009
    Location
    Bay Area, CA
    Posts
    1,961

    Default

    Quote Originally Posted by Stanley View Post
    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.
    2/74 260Z

  49. #49
    Registered User doradox's Avatar
    Member ID
    CZCC-14809
    Join Date
    Aug 2007
    Location
    Colfax, IN
    Posts
    708

    Default

    Quote Originally Posted by LeonV View Post
    Lowering rotational inertia does not put any more power to the wheels.
    Quote Originally Posted by LeonV View Post
    Did you read post #41?

    I've already explained everything there. Simply put:

    Mechanical Power = Torque X RPM

    Are you denying that?
    Quote Originally Posted by LeonV View Post
    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
    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

  50. #50
    Registered User Stanley's Avatar
    Member ID
    CZCC-24191
    Join Date
    Apr 2011
    Location
    Redondo beach
    Posts
    641

    Default

    Quote Originally Posted by LeonV View Post
    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.

  51. #51
    Low Budget/High Value
    Member ID
    CZCC-20342
    Join Date
    Oct 2009
    Location
    Washington County, OR
    Posts
    3,629

    Default

    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.”
    1976 280Z, with some minor modifications

  52. #52
    Registered User
    Member ID
    CZCC-23182
    Join Date
    Dec 2010
    Location
    Mid-Atlantic
    Age
    49
    Posts
    148

    Default

    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.

  53. #53
    Registered User Stanley's Avatar
    Member ID
    CZCC-24191
    Join Date
    Apr 2011
    Location
    Redondo beach
    Posts
    641

    Default

    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.

  54. #54
    Registered User LeonV's Avatar
    Member ID
    CZCC-19146
    Join Date
    Apr 2009
    Location
    Bay Area, CA
    Posts
    1,961

    Default

    Quote Originally Posted by doradox View Post
    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.

    Quote Originally Posted by doradox View Post
    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?

    Quote Originally Posted by doradox View Post
    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.
    2/74 260Z

  55. #55
    Registered User LeonV's Avatar
    Member ID
    CZCC-19146
    Join Date
    Apr 2009
    Location
    Bay Area, CA
    Posts
    1,961

    Default

    Quote Originally Posted by Stanley View Post
    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.
    2/74 260Z

  56. #56
    Registered User doradox's Avatar
    Member ID
    CZCC-14809
    Join Date
    Aug 2007
    Location
    Colfax, IN
    Posts
    708

    Default

    Quote Originally Posted by LeonV View Post
    Lowering rotational inertia does not put any more power to the wheels.
    Remember, the above is what we are arguing.

    Quote Originally Posted by LeonV View Post
    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.


    Quote Originally Posted by LeonV View Post
    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.

    Quote Originally Posted by LeonV View Post
    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.


    Quote Originally Posted by LeonV View Post
    How about you do the math and prove me wrong?
    I just did.


    Quote Originally Posted by LeonV View Post
    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.
    I slipped into my jeans
    Lookin' hard and feelin' mean
    I took a spit at the moon

  57. #57
    Registered User LeonV's Avatar
    Member ID
    CZCC-19146
    Join Date
    Apr 2009
    Location
    Bay Area, CA
    Posts
    1,961

    Default

    Quote Originally Posted by doradox View Post
    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.

    Quote Originally Posted by doradox View Post
    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.

    Quote Originally Posted by doradox View Post
    Time is integral to power it can't be out of the picture.
    See above.


    Quote Originally Posted by doradox View Post
    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?

    Quote Originally Posted by doradox View Post
    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.

    Quote Originally Posted by doradox View Post
    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.

    Quote Originally Posted by doradox View Post
    I just did.
    Sort of...

    Quote Originally Posted by doradox View Post
    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.
    2/74 260Z

  58. #58
    Registered User doradox's Avatar
    Member ID
    CZCC-14809
    Join Date
    Aug 2007
    Location
    Colfax, IN
    Posts
    708

    Default

    Quote Originally Posted by LeonV View Post
    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

  59. #59
    Casey H Casey_z's Avatar
    Member ID
    CZCC-21264
    Join Date
    Mar 2010
    Location
    Smithville Ontario
    Age
    67
    Posts
    207

    Default

    Quote Originally Posted by 71Nissan240Z View Post
    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...............

  60. #60
    Registered User Oiluj's Avatar
    Member ID
    CZCC-15388
    Join Date
    Nov 2007
    Location
    SF Bay Area
    Age
    67
    Posts
    1,746

    Default

    Holy Cow!__Didn't think a little statement like "rotational inertia" could have generated so much discussion
    Julio
    1972 240Z (in-progress, 95% complete)
    CZC# 15388

  61. #61
    Registered User LeonV's Avatar
    Member ID
    CZCC-19146
    Join Date
    Apr 2009
    Location
    Bay Area, CA
    Posts
    1,961

    Default

    Quote Originally Posted by doradox View Post
    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.
    2/74 260Z

  62. #62
    Registered User LeonV's Avatar
    Member ID
    CZCC-19146
    Join Date
    Apr 2009
    Location
    Bay Area, CA
    Posts
    1,961

    Default

    Quote Originally Posted by Oiluj View Post
    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"...

    2/74 260Z

  63. #63
    Registered User conedodger's Avatar
    Member ID
    CZCC-16545
    Join Date
    Apr 2008
    Location
    Reno, NV
    Posts
    2,053

    Default

    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)
    Rob
    2000 BMW R1100 RT-SE (for sale)
    1999 Toyota 4Runner Supercharged
    1975 Porsche 914 stroker motor autoX car
    1973 Datsun 240Z Restoration project. New paint in original white. E31 head on 2.4 block. Nissan Motorsports header. R200 with Nissan motorsports LSD.

  64. #64
    Crumudgeon
    Member ID
    CZCC-798
    Join Date
    Apr 2000
    Location
    La Habra, CA USA
    Age
    60
    Posts
    1,374

    Default

    A prime example of Engineering Masturbation. :-)

  65. #65
    Registered User doradox's Avatar
    Member ID
    CZCC-14809
    Join Date
    Aug 2007
    Location
    Colfax, IN
    Posts
    708

    Default

    Quote Originally Posted by LeonV View Post
    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

  66. #66
    Registered User LeonV's Avatar
    Member ID
    CZCC-19146
    Join Date
    Apr 2009
    Location
    Bay Area, CA
    Posts
    1,961

    Default

    Quote Originally Posted by John Coffey View Post
    A prime example of Engineering Masturbation. :-)
    It helps to relieve the mind...

    Quote Originally Posted by doradox View Post
    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.
    2/74 260Z

  67. #67
    Registered User doradox's Avatar
    Member ID
    CZCC-14809
    Join Date
    Aug 2007
    Location
    Colfax, IN
    Posts
    708

    Default

    Quote Originally Posted by ksechler View Post
    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

  68. #68
    Registered User doradox's Avatar
    Member ID
    CZCC-14809
    Join Date
    Aug 2007
    Location
    Colfax, IN
    Posts
    708

    Default

    Quote Originally Posted by LeonV View Post
    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

  69. #69
    Registered User Oiluj's Avatar
    Member ID
    CZCC-15388
    Join Date
    Nov 2007
    Location
    SF Bay Area
    Age
    67
    Posts
    1,746

    Default

    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...
    Julio
    1972 240Z (in-progress, 95% complete)
    CZC# 15388

  70. #70
    Supporting Member pbarcher's Avatar
    Member ID
    CZCC-18713
    Join Date
    Feb 2009
    Location
    Fairfax, VA
    Age
    74
    Posts
    116

    Default

    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


    Peter Barcher
    Original Owner of HLS30-161241 - Mfg. 5/73
    Silver 1973 240Z

  71. #71
    Registered User
    Member ID
    CZCC-24680
    Join Date
    Jul 2011
    Location
    East Amherst, NY
    Age
    44
    Posts
    24

    Default

    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.

    More Information for MOOG/PRECISION 391

Thread Information

Users Browsing this Thread

There are currently 1 users browsing this thread. (0 members and 1 guests)

Similar Threads

  1. Cant remove driveshaft
    By tymarbry in forum Engine and Drivetrain (S30)
    Replies: 3
    Last Post: 07-19-2012, 01:06 PM
  2. Driveshaft came apart
    By KenshinX in forum Engine and Drivetrain (S30)
    Replies: 7
    Last Post: 05-15-2009, 02:37 PM
  3. I need a driveshaft
    By mmagnus in forum Help Me !!
    Replies: 5
    Last Post: 10-31-2005, 05:58 AM
  4. Dropped a driveshaft and need help
    By J.T. in forum Suspension and Steering (S30)
    Replies: 7
    Last Post: 04-11-2005, 07:46 PM
  5. driveshaft lengths
    By jasonparuta in forum Suspension and Steering (S30)
    Replies: 0
    Last Post: 04-15-2004, 03:07 PM

Bookmarks

Bookmarks

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •