As you lighten the rotational weight--driveshaft, flywheel, crank, rods, pistons----horsepower increases. But you will lose torque....You will get to the higher rpms quicker however.

I have not checked yet, but the driveshaft is probably aluminum already....If it is not, you can have one made for under $300.00 I do believe.



<p>
hehehehehehehehehehe

 

I am sorry to burst your bubble but decreasing rotational weight will increase both torque and horsepower, i could explain this but i just dont want to take the time to.

It is always a good idea to reduce rotational weight, so let me know if you find a carbon fiber driveshaft.

Dan O

 

5/7/03 @12:51 per Dean Pleasanton Infiniti" the drive shaft is alum and the Z's won't fit" Well now we know. There are companies that make carbon drive shafts but I can't remember the site. I'll look to see what I can find.

03.5 Sedan/B. Silver/Prem/Winter/Aero/Tint/Rims
69 Chevelle Hard Top BB/4.11's/Posi

 

The reason rear wheel torque will go up is simply because the force (torque) that the engine puts out will be less into the rotaional weight and more of it will be left for the rear wheels.

Torque origninates by the pistons lever arm on the crankshaft which is pushed by the explosion in the cylinder, tell me how you can decrease that force by reducing the rotational moment of inertia of the rotating parts?

The conservation of energy shows that reducing the rotational moment of inertia will decrease the kinetic energy in that part but work stored is equal to the work put in minus frictional losses. You reduce the energy you have to put into the rotating parts, the more you have for the rear wheels.

Dan O

 

So.. crank torque remains constant, and lighter-weight parts reduce the drivetrain loss, so more torque makes it to the rear wheels. Yes?

2003.5 G35 Sedan Desert Platinum/Graphite Premium/Sport/Aero/Nav/Winter

 

DanO, I believe that power is "lost" to the drive shaft only when you are accelerating its rotational velocity. At a constant speed power (torque) is required only to overcome frictional losses. These frictional losses are independent of driveshaft weight. What do you think?

<font color=blue>Dan</font color=blue>

 

To me, it does not make sense....Let's say you yank out a steel flywheel that weighs 30 pounds--as in the one that my big block had on it at one time and replace it with a 15 pound one with steel inserts.....

Now with that 30 pound flywheel, the car would launch hard as hell. Downright violent. Those launches also tended to sheer all of the teeth off of first gear, and generally break all sorts of stuff....Mainly along the driveline....UJoints, pinion gears, etc.....With the 15 pounder, the car does not launch near as hard, but the RPM's get up high much quicker, it also destroys much less drivetrain parts. I guess my point is, in my limited professional opinion, the heavier the rotating mass is, the more inertia and energy is stored there in rotating form. The less weight would say that less horsepower is lost to turn the object. With the less weight being slung around, your horsepower would increase for sure. As far as the torque goes, I would be content with a loss, although I would much rather have more torque than horsepower.

My baby big block did the following on the dyno:

The High Torque Numbers

RPM CBTrqlb-ft CBDwrHP FHP HP ME% FA-PB lb/hr BSFC /lb/HPHr CAT
6600 405.3 509.3 166.5 73.6 219.8 .47 74

The High HorsePower Numbers


RPM CBTrqlb-ft CBDwrHP FHP HP ME% FA-PB lb/hr BSFC /lb/HPHr CAT
7000 401.5 535.1 192.9 71.6 235.4 .48 74

I thought that torgue was what broke things.....



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hehehehehehehehehehe

 

Horsepower is just a mathematical function of torque and rpm. Dyno's measure torque and calculate horsepower.

If you lose torque at a given rpm, you lose horsepower at that rpm, too.

2003.5 G35 Sedan Desert Platinum/Graphite Premium/Sport/Aero/Nav/Winter

 

You can have one custom made here:
http://www.pstds.com/driveshafts.cfm