doradox

After digging through a dozen tubs of parts I found these. The drivers side retractor still works and from outward appearances is pretty much rust free but the pass side is inoperable and, as can be seen, pretty rusty. I built a custom set of belts for my car so these are yours if you want them. Steve

I think I have some in parts only condition you could have. I have to do some work out in the barn today so I'll dig through my parts bins and see what I have. They'd be from a 72. Steve

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

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

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

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

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

If you don't have your heart set on OEM then here http://www.topyprecision.com/products/s_type_clips.aspx or here http://www.boltproducts.com/tinnerman/clips-s-fastener.html sell a similar product. Google "S Clip" and you will find many more. Steve

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

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 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

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

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

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

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

"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

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.

It's a Zorvette!

I don't have much of a singing voice so maybe an instrumental featuring the l24 running through the gears? Steve

You are mistaking honor with I'm not even sure what. A bunch of irrelevant IP protection and marketing stuff? Where did that come from? One does not have to intend to honor another for another to be honored by ones actions. I don't need to intend to honor the creator of the Z in order for him to feel honored by something I may have done with one of thousands of copies of his creation. A songs creator who licensed a song to someone for a cover version might well feel honored by the licensees work. The fact that someone even wanted to do a cover of his song could be considered an honor to him. Especially if the one wanting to do the cover is a well respected artist. But there is no need to intend to honor for this to happen. What you may be thinking is that even though one intends to honor another that does not make it an honor. Like your Peter Brock example. He, apparently, doesn't feel honored and so is not. Putting a V8 in a Z and thinking you are honoring the Z does not make it so. But since the Z does not think or feel I guess we'll never know how It feels about the whole thing. I don't expect the Mona Lisa will be chiming in anytime soon either. But I imagine there are plenty of folks willing to give their opinion on what is and is not an honor with respect to the Z. Steve

Seriously? A Z isn't a person it's an inanimate object. Ever heard of a recording artist state they were honored that another artist would produce a cover, often highly modified, of the one of his or her songs? That's ridiculous that an artist would never modify another's work to create a derivative work or that it would always be taken as an insult and not an honor by the original creator. Give me a break. Your definition of what honoring something can be is extremely narrow.

Choosing the Z to create a beautiful derivative work doesn't honor the original? Spending thousands of dollars and hundreds of hours doesn't honor the vehicle specifically chosen for the purpose when the "artist" could have chosen any other vehicle to commit their precious time and treasure to? Steve

You are very welcome!

There, I fixed it for you.

Front wheel drive? Why not just buy a Civic?