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Aluminum driveshaft - Prove me right- please!

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Old 12-27-2009, 01:07 PM
  #21  
ryan4096
 
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Originally Posted by Norm Peterson
"They" probably are. The math is a bit more complicated than that. Even a lb at the wheels is less than about 1.5 lbs deadweight in most cases. I'd guess that a "gain" at the wheels due to reduced driveshaft rotational inertia might be closer to half of one percent (measured in 4th gear). Different U-joint operating angles could account for more than that.


Norm

So ... I was wanting to look into how to make my ride quicker and with all the 'opinions' on both sides, I found this writeup. Which makes the absolute most sense:

http://www.w8ji.com/rotating_mass_acceleration.htm

Basically, drive shaft weight means close to nothing in terms of rwhp. Same with quite a few other 'rotational mass' changes ... except wheels, apparently wheels with less weight on the outer edge, either by tire or rim weight a benefit.

Nice.

so, Norm, so it seems you're near dead-on, it's hard to argue with physics. :-)

Last edited by ryan4096; 12-27-2009 at 01:13 PM.
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Old 12-27-2009, 01:19 PM
  #22  
S197steve
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Originally Posted by ryan4096
so, Norm, so it seems you're near dead-on, it's hard to argue with physics. :-)
Its hard to argue against Norm, he always seems to be right... :-)
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Old 12-27-2009, 01:26 PM
  #23  
ejanderson05gt
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It's the same exact concept as underdrive pulleys. You're reducing the parasitic HP loss from some of the OEM equipment, but you're not actually gaining any power.

Eric
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Old 12-27-2009, 03:07 PM
  #24  
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Seems a little odd that people state benefits in ET from the driveshaft and yet that link makes it seem like a complete waste of money. I've heard of nearly 2/10ths off ET from the driveshaft.
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Old 12-27-2009, 03:07 PM
  #25  
Norm Peterson
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Originally Posted by Nuke
Is that really true? Since you're reducing a bit of the driveline loss and thus adding even a few RWHP, isn't top end going to improve, even if by a mere 1 MPH due to those extra 2 or 4 HP's overcoming a tad of the aerodynamics?
Offhand, I can think of three possible effects from installing a lighter (lower MOI) driveshaft.

Less MOI = less torque lost rotationally accelerating it. BUT . . . at top speed, both forward vehicle acceleration and powertrain rotational acceleration are by definition zero. At zero rotational accelerations, everything that rotates is now at constant speed, so the rotational MOI effect also drops out. IOW, the MOI effects will not affect top speed.

Less weight = minutely less rolling drag = a theoretical gain in top speed. But it'll be tiny. 10 lbs weight difference times a rolling drag coefficient of something like 0.02 is 0.2 lbs less drag (out of several hundred lbs or so total drag @ 140+).

Friction loss effects from U-joints that are either more or less efficient and/or operate at either slightly greater or slightly smaller angularities. I'm betting on this being potentially the greatest effect, and in the absence of reliable U-joint data this could work either way.


Norm

Last edited by Norm Peterson; 12-27-2009 at 03:11 PM.
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Old 12-27-2009, 04:52 PM
  #26  
Nuke
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Reasonable explanations as usual, Norm. Thanks.
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Old 12-28-2009, 08:04 AM
  #27  
07VistaChicago
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Where are the specifications for this driveshaft?
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Old 12-28-2009, 08:58 AM
  #28  
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So yes or no. Do you get any performance improvements with the driveshaft or not? That article says pretty clearly you will NOT.
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Old 12-28-2009, 09:40 AM
  #29  
Norm Peterson
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That article deals only with the MOI effects, which are finite but quite small as far as acceleration is concerned (and depend somewhat on axle gearing and tire size, actually) and are nonexistent for top speed.


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Old 12-28-2009, 04:17 PM
  #30  
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I am going to recall my physics class, but that rule of thumb of about one pound lost from rotational mass is 3 rwhp is a complete lie.

If you ever sat in a spinny chair and spun with your legs and arms out, then brought them in your angular velocity will increase. The closer you bring that weight (of your arms and legs) towards the center axis you are spinning around the faster you will spin.

In other words to build the most successful drive shaft, locating all the mass of the shaft closer to the center axis (of which the shaft spins about) will reduce rotational inertia.

This means that the shaft could weigh as much as the stock shaft, but if it was denser it would inevitably give you more power to the wheels.

In case anyone cares, Force = mass x acceleration which is equivalent to Torque = inertia x angular acceleration. Inertia is complicated to calculate but basically it is a function of mass and radius. If radius is zero then the inertia zero, in other words the closer the weight is to the center axis the less the rotational inertia.

also why a crank shaft has such a goofy shape.

I hope this statement to end all debates but I doubt it will
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