Physics Discussion

[2015ghost]

When considering a dyno run everyone says you average 20% parasitic loss for the drive train. Cool, I agree. 300 crank HP motor makes 240whp. So then for a 600 crank HP motor is should only take the original 60hp to spin the trans drive shaft and related components right?

But this is where it begin to defy the laws of Physics. People still argue it takes 20% so a 600 CHP motor would lose 120hp and only show 580hp?

How is this possible? Nothing changes, where is the mystical HP gremlin eating all this HP?

Now, consider this! If an alternator takes 15hp to run and you take it off the 300hp motor and put it on the 600hp motor why does it not take 30hp to run? this argument could be made with anything you attach to the motor right?

[Mr_Draco]

Law of motion. The faster something moves or spins the more friction that is built up. The more friction, the more power needed to overcome said friction.

When it's all said and done, a 600hp engine will create twice as much friction and rotational forces among all moving parts and tires as a 300hp engine. Twice as much friction means you need twice as much power to overcome said friction.

[Stingr69]

(Hip shot explanation) The 20% plus or minus something is a "wide brush". Over the span of different torque outputs measured on a typical car engine pull, the loss empirically observed is in the area of 20%. You can not extrapolate the torque numbers by multiples and expect any accuracy.

[pnwdan]

And 600 - 120 is 480

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[pnwdan]

Originally Posted by Mr_Draco

Law of motion. The faster something moves or spins the more friction that is built up. The more friction, the more power needed to overcome said friction.

When it's all said and done, a 600hp engine will create twice as much friction and rotational forces among all moving parts and tires as a 300hp engine. Twice as much friction means you need twice as much power to overcome said friction.

I don't necessarily agree with this in terms of drivetrain. If the same engine is modified to create more power and it remains coupled to the same drive train, this would not apply.

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[Camaro_QC]

They call it the rule of thumb. Might be less than 20% but it is a good general estimate.

[Mr_Draco]

Originally Posted by pnwdan

I don't necessarily agree with this in terms of drivetrain. If the same engine is modified to create more power and it remains coupled to the same drive train, this would not apply.

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Doesn't matter if you agree or not, fact of the matter is that's the reason why. There are a lot of articles and research done on the subject.

Here's one I found with a quick google search that goes in depth about the whole "rule of thumb" thing and what actually causes the power loss. http://www.superstreetonline.com/how...in-power-loss/

They even state the same thing I did above.

It's also worth noting that the more powerful you make your engine, the greater the thrust force and angular acceleration it's able to exert on the drivetrain, generating even more friction and heat in the process.

[DGthe3]

Originally Posted by pnwdan

I don't necessarily agree with this in terms of drivetrain. If the same engine is modified to create more power and it remains coupled to the same drive train, this would not apply.

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You are thinking of inertial losses, not frictional.

Consider what is actually happening in the system, at a moment frozen in time. The power may as well be torque (since power at a given moment equates to torque). More than that, torque is effectively a rotating force. What this means is that the drivetrain components are transmitting force from one part to the next. Follow so far?

Alright, so that force transmission happens because of friction. And friction is proportional to the forces applied. Its a touch more complicated that that, as it involves calculating the normal forces at the tangent angles between gears and such. But the end result is that as you increase power, you increase the friction within the system.

[CamaroSkooter]

So, theoretically you should be able to take you whp numbers from the dyno, divide by .8-.85 and get a decent estimate of your crank hp numbers?

Cool...