Drive Train Loss Explained
Team Owner
Joined: Jun 2001
Posts: 26,714
Likes: 344
From: Houston, TX
Re: Drive Train Loss Explained (AP)
Actually, I got to thinking about this some more, and I have a question that frankly leads me to wonder about the validity of this premise. The dyno is just a contrived measuring device. The computer takes the data from the machinery, and converts it to a RWHP number. Now, if all the dyno mfrs. know of this issue, why don't they simply change the factors to correct for the losses in the dyno itself?
Team Owner
Joined: Aug 2000
Posts: 24,091
Likes: 1,325
From: Spring Texas
Re: Drive Train Loss Explained (jschindler)
All I can say is that it is a good thing that Horsepower is a relative measure and is actually calculated based upon torque and rpm. In other words, it doesn't really matter what the number is, it only matters that when comparing, you are using the same measuring device.
I feel better now. :crazy:
I feel better now. :crazy:
Torque is a measure of how far the wall moves.
Thread Starter
Burning Brakes
Joined: Sep 2001
Posts: 1,173
Likes: 1
From: Aguila AZ
Re: Drive Train Loss Explained (alc2001)
alc2001, Kewlbrz and others,
Your equations and comments are correct, the problem is that the dyno only knows the inertia of its drum. It ignores the inertia of the engine and all the rotating parts of the drive train... And it is too cumbersome to try to put a value on each and every car that drives up on the roller.
Let me illustrate with an example... Consider the car and dyno to be a 'system'... When the dyno pull is made, the algorithms inside the dyno's computer only consider the inertia of the drum which, on a DynoJet, is 3000 Lbs. The other part of the 'system' is the car and it has an inertia equivelent of 300 Lbs of crankshaft, flywheel, clutch, gears, driveshaft, tires and wheels which must also be accelerated. So the 'system' has a total inertial equivelent of 3300 lbs which must be accelerated, but the dyno only recognizes it's portion of the inertia which is 3000 lb... That's a 10% error. Because of this, the car will accelerate the roller 10% slower and the power read out will be 10% lower as a result... Couple this 10% absorbtion error with the 5% frictional losses in the drive train and Wah Lah... 15% perceived power loss...
So the problem is not in the 'math', the problem is in what the math inside the dyno computer ignores... Hope this makes it clearer...
Another way to look at the dyno results is to consider the dyno RWHP to be the HP available to accelerate the car down the quarter mile... Your engine power at steady state would be higher and would ALL be available in a top speed run where maximum available HP and wind resistance reach equivelency and nothing is being accelerated...
Shirl
Your equations and comments are correct, the problem is that the dyno only knows the inertia of its drum. It ignores the inertia of the engine and all the rotating parts of the drive train... And it is too cumbersome to try to put a value on each and every car that drives up on the roller.
Let me illustrate with an example... Consider the car and dyno to be a 'system'... When the dyno pull is made, the algorithms inside the dyno's computer only consider the inertia of the drum which, on a DynoJet, is 3000 Lbs. The other part of the 'system' is the car and it has an inertia equivelent of 300 Lbs of crankshaft, flywheel, clutch, gears, driveshaft, tires and wheels which must also be accelerated. So the 'system' has a total inertial equivelent of 3300 lbs which must be accelerated, but the dyno only recognizes it's portion of the inertia which is 3000 lb... That's a 10% error. Because of this, the car will accelerate the roller 10% slower and the power read out will be 10% lower as a result... Couple this 10% absorbtion error with the 5% frictional losses in the drive train and Wah Lah... 15% perceived power loss...
So the problem is not in the 'math', the problem is in what the math inside the dyno computer ignores... Hope this makes it clearer...
Another way to look at the dyno results is to consider the dyno RWHP to be the HP available to accelerate the car down the quarter mile... Your engine power at steady state would be higher and would ALL be available in a top speed run where maximum available HP and wind resistance reach equivelency and nothing is being accelerated...
Shirl
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