someone told me that through a 9 inch you lose 50 rwhp? how is this possible?

 

Originally Posted by Injuneer
Numbers published by Evan Smith in National Dragster show that because of the reduced offset between the pinion shaft axis and the axle centerline, the Dana 60 has LOWER drivetrain losses than either the 12-bolt or the 9-inch:

Dana 60: 5%
GM 12-Bolt: 7%
Ford 9-inch: 10%

According to this information, the Ford 9" has the greatest loss of the three rear-ends. 10% of 500 HP = 50, therefore you are correct.

WD

 

You have to be careful with the numbers..... because the loss is made up of two major components. 1) Friction losses are variable, and proportional to the torque being transmitted. 2) Inertial losses are relatively fixed, given a rate of acceleration. But it could be argued that the more powerful the engine, the greater the rate of rotational acceleration of the components, making inertia a somewhat variable part of the equation.

 

Thanks guys!

 

Of course you need to consider the fact that whatever rear you use will also have loses, therefore even if you went for the more efficient Dana 60, it would loose 25 hp (in the above example), so the difference between the 2 loses is 25 hp (50 hp - 25 hp = 25 hp).

 

As mentioned above, there will be lose factors with whatever rear-end you select (no perfect situation). While the Dana will have less drive-train loss then the 12-bolt, the Dana is also about 40 lbs heavier that the 12-bolt.

WD

 

I agree with your point. 90 degree energy transfer via gears correctly calculated would be based on non-linear progression. My math was an over-simplification.

WD

 

Resurrecting this old post.
What is the percentage loss for the 10 bolt?

 

yeah, I want to know the percent loss through the 10 bolt as well.

 

If your losing up to 10% from the rear end alone how much are you losing from the trans?

 

The numbers published in National Dragster, in a tech article by Evan Smith show:

Dana 60 - 5%
GM 12-bolt - 7%
Ford 9" - 10%

Since the losses are very heavily dependant on the offset of the pinion shaft centerline from the ring gear centerline, the 10-bolt and 12-bolt would likely have similar losses, since they have similar geometry.

With a drivetrain consisting of a steel FW Street Twin, stock T56, 3" chrome moly DS and a Strange 12-bolt w/ 3.73 gears, and 17" wheels with 275/40-17 street tires, we measured total drivetrain losses in the range of 12.2 - 12.5% for power levels ranging from 490 -760 flywheel HP (measured in 1:1 4th gear). Subtracting the nominal 7% value for the rear axle assembly would leave only 5% or so for the rest of the drivetrain. The tranny is not going to have a lot of losses in the direct drive gear, since there is little or no power transmitted through the gear teeth. Its mainly some friction losses in the bearings. The rear eats up power because of the fact the gears are transmitting the power through a 90-deg bend, and involve a lot of friction losses as the gear teeth slide on each other. The trend now is to reduce those frictional losses with the Mikronite treatment of the gear teeth faces.

 

I don't understand how you would lose more the more power you have.....

So 500hp loses 50 hp in a 9" but 1,000 hp loses 100 hp? Thats not right.
It may lose more but not that much......

What are some ways to get more out of a 9" star flanges, gun drilling, back cut gear and microniting? how much of a gain do you get from that stuff?

 

I just got a 9" for my car, no regrets yet. Going by this information I will be eating around 18 RWHP or so with this choice. I hope its not something that can actually be elt when driving around. I understand I can launch the hell out of it now to offset those losses with a better 60 foot time.

But the fact that the Dana is the lowest is a good selling point.