Crosspost...Powerhaus k27/6 for sale...
#46
what engineers? ok, but what i'm saying is that a stock car loses around 30 hp....and i'm making a rational assumption that dfast's car does not have a heavier drivetrain that stock and that it probably has synthetic oil, cv grease and wheel bearing grease...why wuold it take 50 more hp to move all of that than stock?
#47
Does someone has a dyno stand and a dynojet avaliable to test?
It would be pertty cool if it was possible to do that and answer this question once ond for all. Maybe if we get a engineer form the factory???
Regards,
AL
It would be pertty cool if it was possible to do that and answer this question once ond for all. Maybe if we get a engineer form the factory???
Regards,
AL
#50
times up. @ 205 hp flywheel, you'll lose just a hair over 30 hp. stock. That's 15%. at 564 - 86.4 hp you also get 15%. As I said, the ratio doesn't change. You are making more hp..YES, but you haven't changed the mechanical work that is systematically needed to transfer the power to the road. 15% is the accepted number across the board. It could be a hair higher or lower, but the ratio of the given vihicle can't change based on hp. It could change marginally based on a systematic change in the method of transfering hp to torque, but only marginally within the confines of a set system. You have to radically change the system to affect significant ratio change.
#51
I think 944CS poses an interesting question.....im not an engineer or know enough about engines to answer but I sure would like to know the answer. Perhaps if I rephrase his question so that there might not be any misunderstanding.
If it takes a stock car 30HP (dont know for sure but lets just say its 30) to move the stock drivetrain then why would another car (lets say DFastest's for instance) take more HP to move its stock drive train?
Or why isn't the amount of drivetrain loss a constant?
If it takes a stock car 30HP (dont know for sure but lets just say its 30) to move the stock drivetrain then why would another car (lets say DFastest's for instance) take more HP to move its stock drive train?
Or why isn't the amount of drivetrain loss a constant?
#52
parasitic loss is parasitic loss. If the common number were 50 % than a car with 500 hp would lose 250 hp at the wheels and a car with 200 would lose 100hp at the wheels. It doesn't matter. The accessory and crank pulleys require the same loss relationship at all power levels. The systems surrounding the engine, that allow for successful transfer of hp to motion, absorb a percentage of power.
Joe's 546 hp 951 has the same gear box, similar pulleys and accesories, similar capacity fluid pumps, similar crank rotational mass, similar or the same cv joints etc.
No engine transfers ALL of it's power to the ground. Even light loses energy.
It doesn't take MORE hp to move Joes stock drivetrain. That's litteraly putting the cart before the horse. The drivetrains steals the same ratio of power regardless of hp.
So the loss in question is a parasitic loss. His engine is capable of turning the accesory systems at 2.5 times a stock 951. At that level, the car's systems still require the same 15%rate to rotate at the additional power. By that I mean if one engine turns at 100 rpm, and an identical car (with an ehgine 2.5 times more efficient) turns at 250 rpm, they both are being drained of 15% of their overall power before it can get to the tire tread. It's not the difference in engines that's inportant here, but the similarity in accesory components.
So I have to go back to this. 15% is 15%. The faster you turn the engine, the more power you create. However, the corresponding systems now also have to be turned at that rate to transfer power. Since they are the same as they were if his car was at 200 hp then the ratio is the same.
Some dynos have shown marginally different loss ranges, but 15% is the accepted number. However if you went to a dyno that used say 12% than a car at 564hp would lose 67.68 hp and a car with 205 hp would lose 24.6. Either way, regardless of the vast differences in engine performance, the parasitic loss is the same.
In the end the other stuff saps hp before it can do its job. The job many of these things do (ie air conditioning, gears that turn cv joints, etc) are not direct links to creating torque. These things require energy in order to make the transfer of energy possible to the wheels.
Does that help?
Joe's 546 hp 951 has the same gear box, similar pulleys and accesories, similar capacity fluid pumps, similar crank rotational mass, similar or the same cv joints etc.
No engine transfers ALL of it's power to the ground. Even light loses energy.
It doesn't take MORE hp to move Joes stock drivetrain. That's litteraly putting the cart before the horse. The drivetrains steals the same ratio of power regardless of hp.
So the loss in question is a parasitic loss. His engine is capable of turning the accesory systems at 2.5 times a stock 951. At that level, the car's systems still require the same 15%rate to rotate at the additional power. By that I mean if one engine turns at 100 rpm, and an identical car (with an ehgine 2.5 times more efficient) turns at 250 rpm, they both are being drained of 15% of their overall power before it can get to the tire tread. It's not the difference in engines that's inportant here, but the similarity in accesory components.
So I have to go back to this. 15% is 15%. The faster you turn the engine, the more power you create. However, the corresponding systems now also have to be turned at that rate to transfer power. Since they are the same as they were if his car was at 200 hp then the ratio is the same.
Some dynos have shown marginally different loss ranges, but 15% is the accepted number. However if you went to a dyno that used say 12% than a car at 564hp would lose 67.68 hp and a car with 205 hp would lose 24.6. Either way, regardless of the vast differences in engine performance, the parasitic loss is the same.
In the end the other stuff saps hp before it can do its job. The job many of these things do (ie air conditioning, gears that turn cv joints, etc) are not direct links to creating torque. These things require energy in order to make the transfer of energy possible to the wheels.
Does that help?
#54
I don't understand peoples need to figure out a crank HP number when having dynoed a car, it is the power that reaches the ground that makes a difference. It is impossible to calculate a standard percentage for the driveline loss since it varies depending on how big the load is and the mechanical efficency in different gears.
As far as I know there is only one chassis dyno on the market that actually measue the torque, all other calculate it from two or more inputs. If you take a calculated value and add guesstimated percentage value you will drift pretty far from the real world power. It will get even more interesting if you have a automatic transmission since you do not only have more driveline loss you will also have slip in the torque converter. An example of that is a 928 GTS auto where we did a measurement of transmission slip and it turns out that it always slips at least 2.5% (even at 6000 rpm).
To simplify things I just say that we should leave it to how much power is being put down to the wheels.
As far as I know there is only one chassis dyno on the market that actually measue the torque, all other calculate it from two or more inputs. If you take a calculated value and add guesstimated percentage value you will drift pretty far from the real world power. It will get even more interesting if you have a automatic transmission since you do not only have more driveline loss you will also have slip in the torque converter. An example of that is a 928 GTS auto where we did a measurement of transmission slip and it turns out that it always slips at least 2.5% (even at 6000 rpm).
To simplify things I just say that we should leave it to how much power is being put down to the wheels.
#55
Horsepower = (Torque * 5252)/RPM
Torque = Ia = (mass moment of inertia of drivline)*(angular acceleration of driveline)
Hp = Ia*5252/RPM
Want double the HP, you have to double the angular acceleration of the driveline.
or looking at it another way:
The driveline requires "x" amount of torque for a given acceleration. To double the angular acceleration of the driveline will require double the torque.
205hp * .15% = 30.75
30.75 * 564/205 = 84.6
Make anymore sense?
Torque = Ia = (mass moment of inertia of drivline)*(angular acceleration of driveline)
Hp = Ia*5252/RPM
Want double the HP, you have to double the angular acceleration of the driveline.
or looking at it another way:
The driveline requires "x" amount of torque for a given acceleration. To double the angular acceleration of the driveline will require double the torque.
205hp * .15% = 30.75
30.75 * 564/205 = 84.6
Make anymore sense?
#56
Drivetrain losses are essentially all about friction. Friction is always expressed as a coefficient which is just a fancy way of saying it's a percentage.
Friction is energy being turned into heat. The faster you turn or accelerate something, the more heat you'll generate while you do it. If it didn't take more horsepower to turn the drivetrain, you would have a miracle of physics.
To put it in less technical terms, this is why high powered cars need things like transmission coolers, oil coolers, etc. They are turning things faster/harder and producting more heat.
Think of the simplest example. Sliding a block along a flat surface. At a slow speed you won't generate much heat doing that. But push it a lot faster and you'll be creating some heat. Push the block a lot faster and you'll have way more heat. That heat is a simple example of drivetrain loss. It takes more power to push that block at a faster speed but you're also losing more of that power to generating heat.
Now some people will say that a drivetrain isn't a block on a flat surface but the point of the example is friction and short of using electromagnets to suspend all the moving parts in a vacuum and push all the other parts, there will never be a friction free system of gears.
Friction is energy being turned into heat. The faster you turn or accelerate something, the more heat you'll generate while you do it. If it didn't take more horsepower to turn the drivetrain, you would have a miracle of physics.
To put it in less technical terms, this is why high powered cars need things like transmission coolers, oil coolers, etc. They are turning things faster/harder and producting more heat.
Think of the simplest example. Sliding a block along a flat surface. At a slow speed you won't generate much heat doing that. But push it a lot faster and you'll be creating some heat. Push the block a lot faster and you'll have way more heat. That heat is a simple example of drivetrain loss. It takes more power to push that block at a faster speed but you're also losing more of that power to generating heat.
Now some people will say that a drivetrain isn't a block on a flat surface but the point of the example is friction and short of using electromagnets to suspend all the moving parts in a vacuum and push all the other parts, there will never be a friction free system of gears.
#57
Originally Posted by 944CS
yess but all of those rotating parts have momentum once they are moving
I know of a 964 that lost 8 hp to the wheels. not a typo, 8....less than 5% of its power was lost through the drivetrain
I know of a 964 that lost 8 hp to the wheels. not a typo, 8....less than 5% of its power was lost through the drivetrain
was the engine pulled out of the car and dynoed on an engine dyno and then put back in and immediately dynoed on a chassis dyno? if not then how do you know how much hp was lost?
#60
Originally Posted by sweanders
IAs far as I know there is only one chassis dyno on the market that actually measue the torque, all other calculate it from two or more inputs.