Which has more parasitic loss?

[Buck531]

FWD auto, or a RWD auto? or are they the same..

 

Trying to prove a point to my friend. I'm pretty sure it's FWD....

[Tractor]

I'm thinking its FWD also.

 

Evan

[controler]

I would think it'd be the rwd simply because of the added weight of unsprung weight.

[BiG BeN]

i was kinda thinkin rwd too, because of the extra weight and the 90degree change in direction at the differential. hard to say for sure though.

[street pilot]

Originally posted by BiG BeN:

i was kinda thinkin rwd too, because of the extra weight and the 90degree change in direction at the differential. hard to say for sure though.

Exactly what I would guess.

[CobraKevin]

Put it on a dyno and find out, lazy bastard. smile.gif

 

I don't know, I always thought the more drivetrain the power had to be transferred through, the more power that would be lost in transition. Just a guess, though.

[OGRE]

FWD is typically less efficient than RWD. A good example is that honda Manual trannies typically lose 20%!!!!

[Mr.COS]

You're spinning a driveshaft and a rear diff with a RWD. A RWD has more loss.

[Conesmasher]

Originally posted by Shrek:

FWD is typically less efficient than RWD. A good example is that honda Manual trannies typically lose 20%!!!!

I've seen mainly in the range of 14-15%

[Buck531]

Right now I'm at 22% loss. What are the RWD auto's loss?

[Mowgli1647545497]

The rule of thumb loss for the T56 and IRS equipped Cobras has been 12-16%

[BiG BeN]

Originally posted by B1ackout:

Put it on a dyno and find out, lazy bastard. smile.gif

 

I don't know, I always thought the more drivetrain the power had to be transferred through, the more power that would be lost in transition. Just a guess, though.

the only good way to do that though is to have identical engines configured for FWD,and RWD then dyno them on an engine dyno, and again on a chassis dyno.

[Drunkendubber]

Originally posted by BiG BeN:

the only good way to do that though is to have identical engines configured for FWD,and RWD then dyno them on an engine dyno, and again on a chassis dyno.

sounds like we need ourselves a DSM we can do it the hillbilly way and just unhook drive shafts

[1Quik7]

i would think that RWD is less, the loss relies on the amount of parasitic drag produced by friction, one way to overcome friction would be to increase the torque applied...in the instance of RWD you have the long shaft which increases the mechanical advantage and thus rotational torque.

 

i might be a genious or just plain retarded.

[Tinman]

Originally posted by 1Quik7:

i would think that RWD is less, the loss relies on the amount of parasitic drag produced by friction, one way to overcome friction would be to increase the torque applied...in the instance of RWD you have the long shaft which increases the mechanical advantage and thus rotational torque.

 

i might be a genious or just plain retarded.

Are you fucking simple?

[Guest Tony_K]

RWD (assuming it has front engine) most often has more moving parts, therefore more loss due to friction.

 

The added weight of the driveshaft and other components add to the inertia of the assembly, which affects it in a way similar to having a *heavier* flywheel.

 

Generally, you want the fewest and lightest parts possible from crankshaft to wheels.

[nathan]

RWD will have more loss, but the weight transfer under acceleration will be what makes the RWD more preferable. The less distance for the power/torque to be lost will always be the more efficient. In most cases you just cannot get the power to the ground with a front wheel drive car even though the weight is mostly over the drive wheels.

[Awfulcanawful]

I would guess FWD because it is indepent suspension and not a solid axle and a lot of the losses come in the cv joints. A RWD only has one U joint and the gear power transfer is highly efficient.

[Guest Spyder550]

Originally posted by nathan96:

RWD will have more loss, but the weight transfer under acceleration will be what makes the RWD more preferable. The less distance for the power/torque to be lost will always be the more efficient. In most cases you just cannot get the power to the ground with a front wheel drive car even though the weight is mostly over the drive wheels.

+1

 

Assuming the same auto trans, power loss through this item will be roughly 50HP.

 

Spinning a driveshaft will cause torque to be lost in simply moving that mass.

 

Assuming suspension characteristics are roughly equal, i.e. independent suspension, I don't believe that will be a factor.

 

Therefore, more loss in RWD, generally.

 

If you want to argue real world performance, weight transfer will help a RWD car hook up and work the same way against a front driver. Also, in extreme handling situations, breaking up the work done by the wheels will help car go faster around a road course because all the fronts have to do is steer, not also deal with the acceleration forces.

 

So... Given equal engines, transmissions, and suspension, a RWD should always be faster in the real world. Driver skill could cancel these advantages, though.

[Guest Spyder550]

Originally posted by Awfulcanawful:

I would guess FWD because it is indepent suspension and not a solid axle and a lot of the losses come in the cv joints. A RWD only has one U joint and the gear power transfer is highly efficient.

A rear driver with independent rear suspension will not have a solid axle. CV joints will connect the halfshafts or transmission to the rear axles.

[Dave1647545494]

the difference between the two is negligeble

 

figure 20% minimum either way for autos

 

figure 15% minimum for stick

 

the real loser in powertrain loss is awd and 4wd. it can be 25% or higher auto or stick

[Doug1647545489]

rwd in gerneral will have more loss. Mainly because of the driveshaft.