GofasterFirebird

iTrader: (3)

I am wondering why rear wheel power ratings are expressed as a percent of crank hp? It is my theory that driveline friction is essentially constant for similar rpm combinations at different power levels. For example, a T-56 takes 15 hp to turn at 6000 rpm. I would take my crank hp and subtract 15 but many people would subtract say 5 percent. Thoughts?

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1991 Firebird
350 L98 (was a 305 TBI),T-5,Edelbrock TES and cat back,Accel manifold
NOS,subframes,jegster torque arm,MSD Digital 6
AFPR,Lakewood lcas
Hurst linelock,SLP cam (206 212 .480 .486),relocated battery,cold air,Hypertech chip,centerforce df,clutch
poly bushings and mounts
AFR 190s
Harland sharp 1.5 rockers
autopower rollbar

12.33 @ 114.83 juiced uncorrected

13.510 @ 102 non juiced uncorrected

johnsjj2

I have pondered this also. Why does the driveline take more power to turn, if you have more power sent back? Say you have a 500hp engine buzzing to 7500rpm. Lets say it takes 100hp to turn everything. You should see 400 at the wheels. Now if you have 300hp, why does it not take the same 100hp to turn everything at 7500rpm. This is all hypothetical. RPM is rpm. I fail to see why you would see more of a loss to driveline friction if you produce more power. And BTW, I have heard anywhere from 15-30% loss.



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Joshua Johnston
1991 Z-28
350, T-5, K&N, Ported Vortec heads, Edelbrock RPM, Holley 750 D.P., HEI, 11.07:1 CR, Comp Cams Roller-.510"/.520"-282*/288* dur., Shorty Headers, Dual 2.5 Exhaust, Dynomax Bullet Mufflers, T&R Motorsports custom air intake, Bald tires

RB83L69

Consider what happens with a gear... the harder you push those 2 sets of teeth against each other, the more friction there will be. That is, if it's unloaded and you spin it at 6000 driveshaft RPM, there will be certain losses in the bearings, and in slinging fluid around, etc. Now if you add load, the losses will go up, because of the added forces causing things to slide harder against each other.

So, there's a mix of constant losses that don't change with HP, and of ones proportional to the power being transmitted. There is no one exact calculation method that will produce accurate results. It's just like the "rule of thumb" about 100 lbs = .1 sec; that only works right for 3500 lb. cars with 350 HP (or other 10 lb / HP combinations). Everything else is just a guess.

Then, figure in the differences in how you measure crank HP compared to RW HP. You've got the engine on a stand, who knows what fuel, nice cool room-temp air, maybe some other pump driving the cooling system, no alternator, an exhaust that (whatever else it may be) is totally different from the one in the car. So you come up with this supposed crank number that's not representative of what the engine itself is going to do once it's in the chassis. And then you're going to try to come up with a "precise" equivalence between that random number and the RW number... I don't think so. It's all a guess. The only one that really matters is the RW number: the real world. Most of us don't drive dyno stands around much.

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"So many Mustangs, so little time..."
ICON Motorsports

Tas

Force= Mass time acceleration. Acceleration varies with RPM and wheel speeds, different forces are produced. Overall its an averaged percentage, its not constant(like saying 15hp).

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-Tas
'89 Formula WS.6

JPrevost

RB83L69 nailed it on the head. It's all about friction!
Yes F=ma applies but you have to understand that as F increases the coefficient of friction increases, it isn't a constant. This increase in coefficient of friction formula is; (Force of friction) = (coefficient of friction) x (Normal force or in this case, the engines crank force)
Now if the coefficient of friction is variable and the Normal force is constant, you notice that the force of friction will be directly proportionate to the Normal force. Hence the percent of power loss and not a constant! It's completely variable depending on how much power your engine is pumping at the crank.

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, Jon (350 TBI!)
91 Red My website

Tas

I was trying to keep it simple

johnsjj2

Well, now that you put it into a math formula, I understand. Thanks guys.



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Joshua Johnston
1991 Z-28
350, T-5, K&N, Ported Vortec heads, Edelbrock RPM, Holley 750 D.P., HEI, 11.07:1 CR, Comp Cams Roller-.510"/.520"-282*/288* dur., Shorty Headers, Dual 2.5 Exhaust, Dynomax Bullet Mufflers, T&R Motorsports custom air intake, Bald tires

GofasterFirebird

iTrader: (3)

Honestly Jprevost, what a poor, elementary explanation. I just wanted to get the point across that the percent rule is just a guess and is no way intended to be more than an approximation.