Drivetrain power loss question

[zraffz]

I was thinking about this for a while now, I know people usually say your driveline loss to the wheels is usually roughly done by X percent of your crank power. My question is why?

For instance, why does a 500 horse motor lose more power through a T56/10 bolt than let's say a 300 horse motor through the same drive train?

Why don't they both just lose "X" amount of power through the rest of the drive train?

[WS6_CAIT]

Thats how a percentage works...

[zraffz]

Originally Posted by WS6_CAIT

Thats how a percentage works...

But why does a 500 horse car lose more power than a 300 horse car? It makes no sense. Both are running through the same rotating mass throughout the rest of the motor, same gearing, etc...

Like if one car loses 60 horse from the engine to the tires, why doesn't the other one lose 60 horse as well?

[Blackbirdws6]

Harder you push, the more counteractive force you will experience.

[zraffz]

Originally Posted by Blackbirdws6

Harder you push, the more counteractive force you will experience.

I'm not understanding still. Can you explain it better?
I'm trying to relate to a pull cord on (as an example) a push mower. The more force you use the easier it cycles; does this mean I am wasting more energy?

[BonzoHansen]

I found some interesting answers on google.

http://www.google.com/webhp?hl=en&ta...w=1280&bih=858

might help you.

[zraffz]

The only thing that makes sense from what I've read so far is:

More power=more heat
More heat is usually due to more friction
The more friction the more resistance, right

I don't think I'll ever get a clear answer on it though. It seems like it's one of those things a very select few people understand and everybody else just follows that knowledge.

[Blackbirdws6]

Originally Posted by zraffz

I'm not understanding still. Can you explain it better?
I'm trying to relate to a pull cord on (as an example) a push mower. The more force you use the easier it cycles; does this mean I am wasting more energy?

That wouldn't be the right way to think about it. A lame example would be you stirring a glass of water with a spoon. The faster you accelerate the spoon, the harder it will be. While the minimum level of energy to stir the water at any given speed will be the same, the act of acceleration is what's affected. Keeping things simple, this is why you see a general % loss factor applied since getting into the fine details is not worth the effort (fluid choice, viscosity, mass of components, etc).

[zraffz]

Originally Posted by Blackbirdws6

That wouldn't be the right way to think about it. A lame example would be you stirring a glass of water with a spoon. The faster you accelerate the spoon, the harder it will be. While the minimum level of energy to stir the water at any given speed will be the same, the act of acceleration is what's affected. Keeping things simple, this is why you see a general % loss factor applied since getting into the fine details is not worth the effort (fluid choice, viscosity, mass of components, etc).

That makes sense, but when you switch to different fluids will you see the power loss decrease at all? I'm not interested in doing any of this but I want to understand it.

[V]

Originally Posted by zraffz

but when you switch to different fluids will you see the power loss decrease at all?

short answer, yes. but in order to still meet the physical aspect requirements and properties for a given fluid/additive, theres not TOO much of an advantage to be gained.

[WildBillyT]

Originally Posted by V

short answer, yes. but in order to still meet the physical aspect requirements and properties for a given fluid/additive, theres not TOO much of an advantage to be gained.

Yup. I forget the rule of thumb but you can gain maybe 5hp (+/- depending on overall power) going to a good quality synthetic package like Redline or RP has.

[Slow-V6]

Why is 15% off 100.00 more then 15% off of 50.00?? Its just the way it adds up.. 15% Drivetrain loss for a 300 hp is 45hp which means the car should dyno 255rwhp.. 15% drivetrain loss for a 500hp motor is 75hp which means the car should make 425rwhp ish..

[BullittSVT]

Originally Posted by Slow-V6

Why is 15% off 100.00 more then 15% off of 50.00?? Its just the way it adds up.. 15% Drivetrain loss for a 300 hp is 45hp which means the car should dyno 255rwhp.. 15% drivetrain loss for a 500hp motor is 75hp which means the car should make 425rwhp ish..

This.

[WildBillyT]

Originally Posted by Slow-V6

Why is 15% off 100.00 more then 15% off of 50.00?? Its just the way it adds up.. 15% Drivetrain loss for a 300 hp is 45hp which means the car should dyno 255rwhp.. 15% drivetrain loss for a 500hp motor is 75hp which means the car should make 425rwhp ish..

Jeff,

I think his question was more targeted at why it's a percentage vs. a flat deduction.

[zraffz]

Originally Posted by WildBillyT

Jeff,

I think his question was more targeted at why it's a percentage vs. a flat deduction.

Exactly my question... and answers like that make me stand stronger on what I said previously about the majority of people not understanding it. They read it is roughly a set percentage for each vehicle and just follow that knowledge.

I'm not hung up on the math aspect of this but the loss of power through the transfer of energy is confusing.

[sweetbmxrider]

Yes, I believe the OP understands the concept of percentages.

Its more or less just a general rule of thumb. To be completely accurate and say it only takes 40hp to turn a 4l60e wouldn't work. There's too many variables. If you really want to know your specific drivetrain loss, there's some expensive testing you can do. For now, just look at trap speed

[zraffz]

Originally Posted by sweetbmxrider

Yes, I believe the OP understands the concept of percentages.

Its more or less just a general rule of thumb. To be completely accurate and say it only takes 40hp to turn a 4l60e wouldn't work. There's too many variables. If you really want to know your specific drivetrain loss, there's some expensive testing you can do. For now, just look at trap speed

Yes you're 100% right but I don't care how much power I am making, knowing won't make it any faster. My curiosity just comes in from trying to understand how the loss is a percentage as opposed to a set number for an exact same setup; just with the motor making more/less power than originally.

[sweetbmxrider]

You are right in thinking how you are but for a manufacturer to rate given parts at a required amount of hp would cause chaos when owners had their car's dynoed and the numbers were low. See 99 cobras. Also, when rpm increases in the drivetrain, more heat/friction occurs so more power is required to move said parts. So you may need 20hp to turn a 4l60e at 1500rpms but you need 50hp to move the same 4l60e at 5000rpms. So, a general percentage keeps the world sane

[zraffz]

Originally Posted by sweetbmxrider

You are right in thinking how you are but for a manufacturer to rate given parts at a required amount of hp would cause chaos when owners had their car's dynoed and the numbers were low. See 99 cobras. Also, when rpm increases in the drivetrain, more heat/friction occurs so more power is required to move said parts. So you may need 20hp to turn a 4l60e at 1500rpms but you need 50hp to move the same 4l60e at 5000rpms. So, a general percentage keeps the world sane

But what about in a situation where you are not turning any more RPMs? As an example -- let's say the motor is supercharged. Now the motor is going from 285 to let's just say 385, would you expect the same loss to the wheels as you would from the car in stock form?

... or am I thinking too deeply into this now?

[sweetbmxrider]

No, there is probably a slight variance but I don't expect 12% power loss no matter the given power. Hell on stock cars manufacturers give a percentage window as to how much power will be used by the drivetrain. Like I originally said, its merely a general acceptance and not an accurate measurement.

[zraffz]

Originally Posted by sweetbmxrider

No, there is probably a slight variance but I don't expect 12% power loss no matter the given power. Hell on stock cars manufacturers give a percentage window as to how much power will be used by the drivetrain. Like I originally said, its merely a general acceptance and not an accurate measurement.

Alright I think I have a descent understanding of it now. Thanks everyone who was patient enough to explain it. I wish I had some fancy dynos to do some reserach but it's way too expensive to even rent one on an hourly rate