If you are going to the trouble of doing a cam swap do a REAL cam swap
a sleeper cam for a 346 that would be good for spray with such a huge split?
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Last edited by chpmnsws6; 01-18-2006 at 02:57 PM..
Ls7 cam has already been installed in 346. It lost 10 HP.
The 62% I/E flow should dictate cam specs accordingly.
vs a stock ls1 cam? OUCH
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Was that LS7 in the 347 tuned properly as well? What mods? Any more details than "it lost 10 hp"?
Drivetrain loss is a percentage.
Physics dictate that when you increase the acceleration, the power involved to increase that should also go up.
HP is a measure of torque and acceleration since you want to increase the amount of acceleration of the mass of the drivetrain, more force is required to provide that acceleration. Simple F=ma applies here, or rather a=F/m.
The perspective you are missing is that you are not looking at the power needed to overcome friction within bearings and gears, you are looking at increasing the amount of rotational acceleration of the mass of those parts.
So a 505 HP car will have it's drivetrain sap MORE HP and TORQUE than a 305HP car in order to achieve increased acceleration for the same mass.
__________________ 1998 Camaro Z28
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mine's stock too...
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F=ma and a=f/m are the exact same equation, same as 6=2x3 and 3=6/2 is the same equation. I tend to believe that drivetrain loss is a constant, delta. The mass of the rotating objects in your drivetrain does not change, so neither does the force required to turn them, but they should accelerate a whole hell of a lot faster.
The reason that the 10-15% percent loss stays the same is that when you have a 500hp car you need a beefier (heavier, or more massive) drivetrain so that you can actually put that power to the ground. That's why the percentages are used, for guesstimation purposes.
F=ma and a=f/m are the exact same equation, same as 6=2x3 and 3=6/2 is the same equation. I tend to believe that drivetrain loss is a constant, delta. The mass of the rotating objects in your drivetrain does not change, so neither does the force required to turn them, but they should accelerate a whole hell of a lot faster.
The reason that the 10-15% percent loss stays the same is that when you have a 500hp car you need a beefier (heavier, or more massive) drivetrain so that you can actually put that power to the ground. That's why the percentages are used, for guesstimation purposes.
AJ
There is a lot of confusion on this topic. You are correct about 2 things:
F=ma and the mass of the rotating objects does not change.
But that is not the whole story. The drivetrain losses are not due to acceleration alone. There are drivetrain lossses even at a constant speed - i.e zero acceleration. If drivetrain losses were entirely due to acceleration, then the max amount of power (at a fixed rpm, say 6000) put down at the rear wheels would be the same whether you had one transmission behind the engine or 1000 transmissions in series behind the engine. Granted it would take a lot longer to accelerate all the moving parts of 1000 transmissions up to speed than it would take to accelerate one transmission's parts up to speed, but once up to a contatnt speed there is no acceleration and therefore not force due to acceleration (f = ma)
What you are overlooking is friction.
Friction is proportional to the normal force between two rubbing surfaces and the coefficient of friction ( a constant for a given material / lube combination)
As power increases even at a constant speed, the normal forces between all rubbing surfaces in the engine / drive train (bearings, gears, etc.) increases therefore the total friction forces increase. This may be non intuitive at first. As people have said, you could easily turn the drive train by hand so it seems that it doesn't take much force. This is equivalent to saying I could easily tie a rope to a 2x2 sheet of plywood and drag it across a concrete parking lot, but if I put a 1000 lb block of lead on top of it will be just as easy to drag at a constant speed. It won't! Given F=ma, it will require a lot more force to accelerate the plywood with the block of lead on it which you would expect, but it will also require a lot more force just to drag it at a constant speed (zero acceleration), once it has been accelerated up to a given speed, due to the increase in the normal force.
I do have a degree in mechanical engineering - not saying this to brag in any way, but rather to show I didn't just make this stuff up...
There is a lot of confusion on this topic. You are correct about 2 things:
F=ma and the mass of the rotating objects does not change.
But that is not the whole story. The drivetrain losses are not due to acceleration alone. There are drivetrain lossses even at a constant speed - i.e zero acceleration. If drivetrain losses were entirely due to acceleration, then the max amount of power (at a fixed rpm, say 6000) put down at the rear wheels would be the same whether you had one transmission behind the engine or 1000 transmissions in series behind the engine. Granted it would take a lot longer to accelerate all the moving parts of 1000 transmissions up to speed than it would take to accelerate one transmission's parts up to speed, but once up to a contatnt speed there is no acceleration and therefore not force due to acceleration (f = ma)
What you are overlooking is friction.
Friction is proportional to the normal force between two rubbing surfaces and the coefficient of friction ( a constant for a given material / lube combination)
As power increases even at a constant speed, the normal forces between all rubbing surfaces in the engine / drive train (bearings, gears, etc.) increases therefore the total friction forces increase. This may be non intuitive at first. As people have said, you could easily turn the drive train by hand so it seems that it doesn't take much force. This is equivalent to saying I could easily tie a rope to a 2x2 sheet of plywood and drag it across a concrete parking lot, but if I put a 1000 lb block of lead on top of it will be just as easy to drag at a constant speed. It won't! Given F=ma, it will require a lot more force to accelerate the plywood with the block of lead on it which you would expect, but it will also require a lot more force just to drag it at a constant speed (zero acceleration), once it has been accelerated up to a given speed, due to the increase in the normal force.
I do have a degree in mechanical engineering - not saying this to brag in any way, but rather to show I didn't just make this stuff up...
Great explanation
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I know this is 2 years too late, but the drivetrain loss thing is a pet peeve of mine. 15% drivetrain loss is a rule of thumb estimate, true. But it is based on sound physics. a=f/m is correct.
If you spin a wheel by hand, it takes effort to spin it. To accelerate the wheel more quickly, you have to apply more force. The wheel could be floating on a magic superconducting magnetic frictionless bearing, and you would still have to use more force to accelerate the wheel faster. It is the law of conservation of energy.
If you double the horsepower of a car, it will accelerate twice as fast, yes? Turn that around, if a car is accelerating twice as fast, it must have twice the horsepower, assuming mass is constant.
If you are accelerating the car twice as fast, then you are also accelerating the drivetrain twice as fast. In this case, it is not the forward motion of the drivetrain we are accelerating, but the rpm of the transmission all the way back to the rear wheels. So to accelerate the drivetrain twice as fast takes twice as much torque.
On a chassis dyno, the drivetrain is BETWEEN the engine and the dyno, so the dyno has no way to measure the horsepower being absorbed by the drivetrain. However, by looking at the difference between an engine dyno and a chassis dyno, you can calculate how much power is being absorbed.
the exact amount of power is dependent on the rotational moment of inertia of a particular drivetrain combined with the moment of inertia of the particular dyno the car is on. In the case of a standard GM transmission, driveshaft, rear end, and wheels, they absorb about 15% of the torque and the dyno absorbs the other 85% (or 20% and 80% respectively, in the case of an automatic). That's why Mustang dynos generally read a little lower - the moment of inertia for a Mustang dyno is lower, and the drivetrain of the car absorbs a greater percentage of the horsepower. It would be more correct to use the 15% figure for Dynotech dynos and another figure (say, 19% just as an example) for the Mustang dynos.
__________________ 1996 Impala SS
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thats all well and good. But we have empirical data which indicates that the loss is a constant rather than a set percentage. I think Tony Mamo will have another data point to add soon. I think 3 or 4 data points which show a constant rate of loss over a wide range of HP will help settle this debate.
[quote=SladeX;6742541]Was that LS7 in the 347 tuned properly as well? What mods? Any more details than "it lost 10 hp"?
[quote]
Damian did the swap of an Ls7 cam into a 2002 Z06 and according to him it was a complete pain (relative to other cam swaps). If memory serves me correctly the car picked up about 20 peak rwhp but lost a bunch of TQ in the low-mid areas. According to him it was definitely not worth it. He said he would never do it again.
I think overall it picked up a little peak power, and then lost so much across the board that there was more loss than there was gain.
To the thorny discussion of drivetrain losses on a chassis dyno, allow me to contribute a recent (10 days ago) data point:
The SB2.2 (Nascar cast-off bits) 358 in our SCCA Trans Am Corvette had engine problems at the Atlanta race. After repairs, we took the car to a Mustang chassis dyno to see if all was well.
Last fall it made 830 HP and 560 lb-ft on the builder's dyno in North Carolina. The Mustang unit registered 701 and 490, a loss of 15.5% and 12.5 % respectively.
Notes:
1.The Mustang unit appears to be quite low inertia, as it took only about 4.0 sec. to go from 4500 to 8800 (~140 MPH) in 4th gear.
2. However, the drivetrain inertia is also lower than most, with lightweight wheels and slicks, a CF driveshaft and a 4" triple disc CF clutch.
3. Engine-driven accessories included water pump, power steering, alternator and dual driveshaft -driven pumps for transmission and rear axle lube.
4. Engine, tranny and diff oil remaind 'off-scale' cold throughout.
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