Crank to wheel power loss question.
01-03-06, 11:29 PM
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Crank to wheel power loss question.
OK, everybody, here's the question:
What is the crank to wheel power loss? Is it a fixed amount, or is it variable?
Here's what I think, I hope somebody has more than ideas about this.
First, power is "work", and it is torque x rpms (simplification).
Now, the difference between the power at the crank and the power at the wheels is caused by the fact that the engine has to spin things (and what I don't know is if the power at the crank is calculated before attaching power steering, alternator and so forth).
The spinning of "things" requires a force (torque) to produce a movement (spinning) -which in physics is called "work".
When you change the pulleys, for instance, you reduce the amount of weight of the pulleys, as well as the distance the pulleys need to be moved to complete a revolution, since they are smaller. This means that the engine has to use less torque to move the pulley and it has to move it for a lesser distance, therefore the power used to do that job is less, causing in return, an increase of rwhp. It goes without saying, that a lighter pulley of the same size, or a smaller pulley of the same weight still frees power to go to the wheels, although not as much as the reduction of both.
Now, here's the problem: let's say that the FD has 210 rwhp stock, which represents a 45 hp loss compared to crank (I am not sure about the numbers, but bear with me for a minute, as the numbers are anyway irrelevant).
Let's say that we do work on the car and now the rwhp has gone from 210 to 255, with a net gain at the wheels of 45 hp.
Let's say that the work we have done has not touched pulleys, flywheel, driveshaft, brake rotors, wheels and all that revolves, and that the power gains have come from bigger intercooler, boost, exhaust and what not. Let's also say, for the sake of the argument, that the engine's speed (rpms) hasn't been changed and is still exactly the way it was when the stock was put on the dyno and made 210 hp.
In my opinion, the power at the crank has also increased 45 hp, because nothing has been done in terms of reducing/increasing the amount of work necessary to spin all moving parts.
The conclusion of this first hypothesis is that if the FD loses 45 hp from crank to wheels, if you make 300 rwhp under the conditions I just mentioned, you must be making 345 hp @crank.
Now, let's say that we go back to our original stock Fd and we change pulleys, carbon fiber driveshaft, lighter brake rotors, wheels.
Now the car has gone from 210 rwhp to 225 (for example).
There has been no increase at the crank.
Lastly, let's say that we increase the power of the FD by increasing the rev limiter 1000 rpm. Let's say that our original stock FD went from 210 rwhp to 275.
In this case, the power at the crank has changed, but it's not 45 hp more (total of 315); it has actually become 330 (once again, just numbers, to explain the concept).
To conclude, the power loss at the wheels is not a percentage of the power at the crank.
Although we can always say that if the power at the crank is x and the power at the wheels i y the percentage loss is z%, that percentage does not stay the same when you increase the power in the car, but changes depending on what you do to increase the power.
That power loss will remain pretty much a fixed amount if your mods won't affect what rotates and how it rotates (weight of rotating masses and relative speed); it will increase if you raise the rev limiter without lightening the rotating masses; it will stay the same if there is an increase of rpms that is proportionally matched by the reduction of the weight of the rotating masses; it will decrease if the rpms stay the same and the weight of the rotating masses is reduced in any way.
Finally, it is my understanding that most people here tend to say that power loss on stock is 20%, so if they are now making 300 rwhp the car has 360 hp at the crank.
It is my belief that they are wrong and they are in most cases overestimating the power at the crank. This is because most people don't have significant rpm increases, while they do have lighter flywheels, driveshaft and pulleys while running much higher boost, so in general their power loss from crank to wheels will tend to be less (as a fixed number and as a percentage) than it was when the car was stock.
I apologize for the length, I am just trying to cause a constructive discussion, hopefully pulling in people that have a better factual knowledge of this matter.
...Just picking your brains...
Giovanni
What is the crank to wheel power loss? Is it a fixed amount, or is it variable?
Here's what I think, I hope somebody has more than ideas about this.
First, power is "work", and it is torque x rpms (simplification).
Now, the difference between the power at the crank and the power at the wheels is caused by the fact that the engine has to spin things (and what I don't know is if the power at the crank is calculated before attaching power steering, alternator and so forth).
The spinning of "things" requires a force (torque) to produce a movement (spinning) -which in physics is called "work".
When you change the pulleys, for instance, you reduce the amount of weight of the pulleys, as well as the distance the pulleys need to be moved to complete a revolution, since they are smaller. This means that the engine has to use less torque to move the pulley and it has to move it for a lesser distance, therefore the power used to do that job is less, causing in return, an increase of rwhp. It goes without saying, that a lighter pulley of the same size, or a smaller pulley of the same weight still frees power to go to the wheels, although not as much as the reduction of both.
Now, here's the problem: let's say that the FD has 210 rwhp stock, which represents a 45 hp loss compared to crank (I am not sure about the numbers, but bear with me for a minute, as the numbers are anyway irrelevant).
Let's say that we do work on the car and now the rwhp has gone from 210 to 255, with a net gain at the wheels of 45 hp.
Let's say that the work we have done has not touched pulleys, flywheel, driveshaft, brake rotors, wheels and all that revolves, and that the power gains have come from bigger intercooler, boost, exhaust and what not. Let's also say, for the sake of the argument, that the engine's speed (rpms) hasn't been changed and is still exactly the way it was when the stock was put on the dyno and made 210 hp.
In my opinion, the power at the crank has also increased 45 hp, because nothing has been done in terms of reducing/increasing the amount of work necessary to spin all moving parts.
The conclusion of this first hypothesis is that if the FD loses 45 hp from crank to wheels, if you make 300 rwhp under the conditions I just mentioned, you must be making 345 hp @crank.
Now, let's say that we go back to our original stock Fd and we change pulleys, carbon fiber driveshaft, lighter brake rotors, wheels.
Now the car has gone from 210 rwhp to 225 (for example).
There has been no increase at the crank.
Lastly, let's say that we increase the power of the FD by increasing the rev limiter 1000 rpm. Let's say that our original stock FD went from 210 rwhp to 275.
In this case, the power at the crank has changed, but it's not 45 hp more (total of 315); it has actually become 330 (once again, just numbers, to explain the concept).
To conclude, the power loss at the wheels is not a percentage of the power at the crank.
Although we can always say that if the power at the crank is x and the power at the wheels i y the percentage loss is z%, that percentage does not stay the same when you increase the power in the car, but changes depending on what you do to increase the power.
That power loss will remain pretty much a fixed amount if your mods won't affect what rotates and how it rotates (weight of rotating masses and relative speed); it will increase if you raise the rev limiter without lightening the rotating masses; it will stay the same if there is an increase of rpms that is proportionally matched by the reduction of the weight of the rotating masses; it will decrease if the rpms stay the same and the weight of the rotating masses is reduced in any way.
Finally, it is my understanding that most people here tend to say that power loss on stock is 20%, so if they are now making 300 rwhp the car has 360 hp at the crank.
It is my belief that they are wrong and they are in most cases overestimating the power at the crank. This is because most people don't have significant rpm increases, while they do have lighter flywheels, driveshaft and pulleys while running much higher boost, so in general their power loss from crank to wheels will tend to be less (as a fixed number and as a percentage) than it was when the car was stock.
I apologize for the length, I am just trying to cause a constructive discussion, hopefully pulling in people that have a better factual knowledge of this matter.
...Just picking your brains...
Giovanni
01-04-06, 10:22 AM
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Originally Posted by RETed
Fricken waste of time...
Put you car on a DynoJet (or any other dyno that can calculate drivetrain loss), and let the fricken computer figure it out.
-Ted
Put you car on a DynoJet (or any other dyno that can calculate drivetrain loss), and let the fricken computer figure it out.
-Ted
01-04-06, 10:34 AM
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Originally Posted by Mahjik
It's not a fixed amount. If it was, that would mean your drivetrain would become more efficient the more power you made. That's just not the case.
When you talk about efficiency, are you talking about volumetric efficiency? In that case, what's lost within the engine is lost before it hits the crank, so it is irrelevant in this discussion.
Furthermore, I asked for factual knowledge from somebody that has it, if I knew, or the subject was easily resolved, I wouldn't have put the post on the forum.
01-04-06, 11:36 AM
#6
Originally Posted by gio64
Well, I don't think I said it is a fixed amount. I said that there are changes in the car that will induce a higher power loss from crank to wheels, while there are others that won't. Once again, if I simply increase the boost in the car without changing anything else (assuming that nothing would brake), why would my power loss increase? What has come in to play to rob power from the engine that wasn't there before?
Originally Posted by gio64
Furthermore, I asked for factual knowledge from somebody that has it, if I knew, or the subject was easily resolved, I wouldn't have put the post on the forum.
01-04-06, 12:48 PM
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this is an example, if you are talking about losing say 20hp when you have 200hp at the crank, you will lose more than 20hp when you have 300hp at the crank, whether you turn the boost up or do other stuff. If anything you would lose as a percentage as hp increased, which probably is also not true, but then you would lose 30hp when you have 300hp at the crank, which in this case would be 10%.
But if you got down to it, and wanna do the mechanical forces on it, the more force you put on the drivetrain to make it accelerate quicker the more it is going to resist, which will cause more loss throught the drivetrain.
But if you got down to it, and wanna do the mechanical forces on it, the more force you put on the drivetrain to make it accelerate quicker the more it is going to resist, which will cause more loss throught the drivetrain.
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01-04-06, 01:22 PM
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OK.
I read the thread. I find that the only valuable (yet unclear) element in that entire thread is this:
"Good answer. 90% efficiency will always be 90% efficiency.
Power transmitted through a gear box (efficiency) is dictated by the geometry of the gears. There are three forces involved: Tangential force (or power transmitted) this is dependant on geometry, power, and rate. Axial force which is dependant on tangential force and gear geometry. And Radial force which, again, is dependant on tangential force and gear geometry.
Now, the gear box efficiency doesn’t give two ***** about how much power and the rate it’s being applied since the power losses are figured from axial and radial forces. Give it 1hp at 1 rpm and the formulas dictate there will be a certain amount of loss due to axial and radial forces. These forces, geometrically speaking , say there will be a certain amount of loss through friction caused by gear mesh and load transferred to the bearings. Put in 1000hp at 10,000rpm and the tangential force increases, but the geometry stays the same (provided the gearbox doesn’t explode ). Meaning there is more power being transmitted, but the axial and radial loads increase proportionately. Hence, the efficiency of the gear box stays the same."
This doesn't quite clear it to me. I'd like somebody to clarify.
As far as frictional force is concerned, those examples assume two things that I believe to not necessarily apply to this discussion:
1) Speed. Yes, the faster you rub your hands, the higher the heat generated; in a gear box, this only happens if the gears are spinning faster than before, which is always false, unless you have increased the engine rpms limit, so this factor is simply out of the discussion, as far as I am concerned, because I did say that I agree with the fact that if you increase the engine speed, you will consequently increase the amount of force needed to move things around; if you made 200 rwhp @ 6000 rpm before boosting and now you make 300 rwhp @6000 rpm, your velocity has not changed, so this part of the example goes out of the window.
2) Resistance to rotation between gears. Yes, rubbing the hands pressing harder will give you more friction. But to do that, you have to push harder with both hands, not with only one, which explains why your car will accelerate faster with more power. Pressing harder with both hands, equates to say that while the energy input from the engine increases, so does the resistance of the car to roll down the road. I still think this is an interesting and good one, but I have some doubts about this as well, and here's why: some where saying that when you apply more power, you are "pushing" on the gears harder than before, thus creating a higher resistance to rotation. In reality, however, most of that resitance to rotation is due to the resistance of the car itself to move, therefore it is depending on car's weight, stickyness of the tires and what not, which doesn't change when the car has more power (unless you accomplish that by increasing the weight). The increased momentary deformation induced by increased power is not proportional to the power increase, it cannot be; if it were, that would mean that the gearbox could deform limitlessly under stress and that's physically impossible (in fact, there is a certain amount of power that would deform it permanently, breaking it).
The fact that more power puts more stress on a gearbox is primarily depending on the fact that more power increases the loads on the gearbox under acceleration, because it causes the gearbox to accelerate more quickly than it did before you increased the power. In other words, if you were pushing your second gear for 2 seconds, that gear was being stressed by a certain amount of energy over that amount of time; now, with more power, you spend 1.5 seconds in that gear, thus concentrating the energy that causes the stress in less time, therefore stressing the transmission more.
In addition to this, the transmission is only one of the many elements that cause power loss. That doesn't include the rest of the things attached to the engine.
I am still not convinced.
I read the thread. I find that the only valuable (yet unclear) element in that entire thread is this:
"Good answer. 90% efficiency will always be 90% efficiency.
Power transmitted through a gear box (efficiency) is dictated by the geometry of the gears. There are three forces involved: Tangential force (or power transmitted) this is dependant on geometry, power, and rate. Axial force which is dependant on tangential force and gear geometry. And Radial force which, again, is dependant on tangential force and gear geometry.
Now, the gear box efficiency doesn’t give two ***** about how much power and the rate it’s being applied since the power losses are figured from axial and radial forces. Give it 1hp at 1 rpm and the formulas dictate there will be a certain amount of loss due to axial and radial forces. These forces, geometrically speaking , say there will be a certain amount of loss through friction caused by gear mesh and load transferred to the bearings. Put in 1000hp at 10,000rpm and the tangential force increases, but the geometry stays the same (provided the gearbox doesn’t explode ). Meaning there is more power being transmitted, but the axial and radial loads increase proportionately. Hence, the efficiency of the gear box stays the same."
This doesn't quite clear it to me. I'd like somebody to clarify.
As far as frictional force is concerned, those examples assume two things that I believe to not necessarily apply to this discussion:
1) Speed. Yes, the faster you rub your hands, the higher the heat generated; in a gear box, this only happens if the gears are spinning faster than before, which is always false, unless you have increased the engine rpms limit, so this factor is simply out of the discussion, as far as I am concerned, because I did say that I agree with the fact that if you increase the engine speed, you will consequently increase the amount of force needed to move things around; if you made 200 rwhp @ 6000 rpm before boosting and now you make 300 rwhp @6000 rpm, your velocity has not changed, so this part of the example goes out of the window.
2) Resistance to rotation between gears. Yes, rubbing the hands pressing harder will give you more friction. But to do that, you have to push harder with both hands, not with only one, which explains why your car will accelerate faster with more power. Pressing harder with both hands, equates to say that while the energy input from the engine increases, so does the resistance of the car to roll down the road. I still think this is an interesting and good one, but I have some doubts about this as well, and here's why: some where saying that when you apply more power, you are "pushing" on the gears harder than before, thus creating a higher resistance to rotation. In reality, however, most of that resitance to rotation is due to the resistance of the car itself to move, therefore it is depending on car's weight, stickyness of the tires and what not, which doesn't change when the car has more power (unless you accomplish that by increasing the weight). The increased momentary deformation induced by increased power is not proportional to the power increase, it cannot be; if it were, that would mean that the gearbox could deform limitlessly under stress and that's physically impossible (in fact, there is a certain amount of power that would deform it permanently, breaking it).
The fact that more power puts more stress on a gearbox is primarily depending on the fact that more power increases the loads on the gearbox under acceleration, because it causes the gearbox to accelerate more quickly than it did before you increased the power. In other words, if you were pushing your second gear for 2 seconds, that gear was being stressed by a certain amount of energy over that amount of time; now, with more power, you spend 1.5 seconds in that gear, thus concentrating the energy that causes the stress in less time, therefore stressing the transmission more.
In addition to this, the transmission is only one of the many elements that cause power loss. That doesn't include the rest of the things attached to the engine.
I am still not convinced.
01-04-06, 03:24 PM
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FYI guys
there are dyno's out there (or rolling roads as they call them in UK)
after u have done ur dyno pull
u leave the car in gear and just take ur foot off the gas
as the car decelerates on it's own the rollers "measure" the frictional loses as the car winds down.. (no wind friction and the rollers own friction is already factored out by computer)
apparently it can fairly accurately calculate your Flywheel HP this way
but also many ppl claim by changing engine and transmission fluids, and other drivetrain parts (such as the driveshaft or rear diff)
u can reduce losses as well
there are dyno's out there (or rolling roads as they call them in UK)
after u have done ur dyno pull
u leave the car in gear and just take ur foot off the gas
as the car decelerates on it's own the rollers "measure" the frictional loses as the car winds down.. (no wind friction and the rollers own friction is already factored out by computer)
apparently it can fairly accurately calculate your Flywheel HP this way
but also many ppl claim by changing engine and transmission fluids, and other drivetrain parts (such as the driveshaft or rear diff)
u can reduce losses as well
01-04-06, 04:12 PM
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Originally Posted by Ottoman
FYI guys
there are dyno's out there (or rolling roads as they call them in UK)
after u have done ur dyno pull
u leave the car in gear and just take ur foot off the gas
as the car decelerates on it's own the rollers "measure" the frictional loses as the car winds down.. (no wind friction and the rollers own friction is already factored out by computer)
apparently it can fairly accurately calculate your Flywheel HP this way
but also many ppl claim by changing engine and transmission fluids, and other drivetrain parts (such as the driveshaft or rear diff)
u can reduce losses as well
there are dyno's out there (or rolling roads as they call them in UK)
after u have done ur dyno pull
u leave the car in gear and just take ur foot off the gas
as the car decelerates on it's own the rollers "measure" the frictional loses as the car winds down.. (no wind friction and the rollers own friction is already factored out by computer)
apparently it can fairly accurately calculate your Flywheel HP this way
but also many ppl claim by changing engine and transmission fluids, and other drivetrain parts (such as the driveshaft or rear diff)
u can reduce losses as well
Thanks for the info.
This sounds like a good way to calculate the power loss.
In any case, I am still puzzled by how this entire system works in terms of physics, but this type of dyno would certainly be useful to evaluate that...
BTW, if your description of how this system works is correct, then you'd have to agree with me that many factors of power increase do not affect the power loss.
If I get it right, once you step off the gas and the car slows down, the power you were making has no influence in the process of calculating the drivetrain powerloss (unless you have increased compression on a naturally aspirated engine...)
What do you think?
01-04-06, 04:23 PM
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Correction:
increased compression should not influence that calculation because it happens before the flywheel and therefore its influence on the crank power would not be measurable.
My bad.
increased compression should not influence that calculation because it happens before the flywheel and therefore its influence on the crank power would not be measurable.
My bad.
01-04-06, 04:37 PM
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Originally Posted by gio64
If I get it right, once you step off the gas and the car slows down, the power you were making has no influence in the process of calculating the drivetrain powerloss (unless you have increased compression on a naturally aspirated engine...)
What do you think?
What do you think?
Also trying to explain the physics on an internet message board is not gonna be easy, not to mention if you really wanna explain all the physics with the whole drivetrain that would be a PIA. Good job explaining stuff by the way
01-04-06, 07:11 PM
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Originally Posted by gio64
I believe that the problem is not that simple. The computer has no way to establish what the loss is, unless you enter every single variable existing. If it were so simple, I don't see why I would see this subject appearing in magazines over and over.
You'd be surprised.
I know a DynoJet can calculate drivetrain friction through coast-down.
You can contact them for more info...or even argue with them if you want.
If this keeps getting brough up in magazines, it's because people have a hard time grasping it's concept - simple.
It's the same thing with hp versus torque - you ever seen the argument Paul Yaw has on his website?
Just mindless drivel unless you're into that sorta stuff...
The original poster is asking for an answer that takes college-level physics / ME over several hours to properly cover.
Personally, I like to use the example of dragging your arm through water...
Frictional losses are a percentage and not a fixed amount.
Your arm represents torque (horsepower).
Swing it slowly, and it moves through the water easily.
What happens when you try and slash through the water as fast and as hard as you can?
If it were a fixed amount, it should be *easier* to plow your hand through the water (higher power overcomes smaller, fixed friction)...but it doesn't.
This is a percentage increase of friction versus power.
The harder and faster you try to move your arm through water, the harder the resistance is...
-Ted
01-04-06, 08:31 PM
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Originally Posted by RETed
Have you seen what a modern dyno can do?
You'd be surprised.
I know a DynoJet can calculate drivetrain friction through coast-down.
You can contact them for more info...or even argue with them if you want.
If this keeps getting brough up in magazines, it's because people have a hard time grasping it's concept - simple.
It's the same thing with hp versus torque - you ever seen the argument Paul Yaw has on his website?
Just mindless drivel unless you're into that sorta stuff...
The original poster is asking for an answer that takes college-level physics / ME over several hours to properly cover.
Personally, I like to use the example of dragging your arm through water...
Frictional losses are a percentage and not a fixed amount.
Your arm represents torque (horsepower).
Swing it slowly, and it moves through the water easily.
What happens when you try and slash through the water as fast and as hard as you can?
If it were a fixed amount, it should be *easier* to plow your hand through the water (higher power overcomes smaller, fixed friction)...but it doesn't.
This is a percentage increase of friction versus power.
The harder and faster you try to move your arm through water, the harder the resistance is...
-Ted
You'd be surprised.
I know a DynoJet can calculate drivetrain friction through coast-down.
You can contact them for more info...or even argue with them if you want.
If this keeps getting brough up in magazines, it's because people have a hard time grasping it's concept - simple.
It's the same thing with hp versus torque - you ever seen the argument Paul Yaw has on his website?
Just mindless drivel unless you're into that sorta stuff...
The original poster is asking for an answer that takes college-level physics / ME over several hours to properly cover.
Personally, I like to use the example of dragging your arm through water...
Frictional losses are a percentage and not a fixed amount.
Your arm represents torque (horsepower).
Swing it slowly, and it moves through the water easily.
What happens when you try and slash through the water as fast and as hard as you can?
If it were a fixed amount, it should be *easier* to plow your hand through the water (higher power overcomes smaller, fixed friction)...but it doesn't.
This is a percentage increase of friction versus power.
The harder and faster you try to move your arm through water, the harder the resistance is...
-Ted
With the water example, you're once again bringing in a variable that is not part of the equation.
When you move water with your hand slowly, you are displacing an x amount of water/minute. This is "work" in physics and the "work" required to move that water I will call Robert.
Now, when you go and try again, but faster, you are trying to displace more water/minute than before (once again, you are increasing speed), you are doing more work, which I will call Vincent and Vincent is bigger than Robert.
That's the reason why you feel more resistance on your arm.
When you increase the power in a car without increasing the engine speed, you are increasing the force parameter of the force x distance product that gives power. You are doing more work, yes, but you are not involving more resitance in doing it.
Let me use this analogy, since we've been talking through metaphores:
Let's say I have a 1 foot wrench and it is placed on a 13 mm nut and the nut needs to be tightened.
Let's say that my arm force is such that I can push with the equivalent of 200 lbs. I put my might arm on the wrench and tighten the bolt; I encounter a certain amount of resistance.
Let's now say that I get my weak friend that has an arm that can push the wrench with only 100 lbs. Would you say that he's going to encounter any less resistance while tightening the bolt? Isn't that frictional resistance solely depending on the nature of the threads involved in the job together with the length of the wrench?
Now, if you were telling me that I am stronger and therefore I can rotate the wrench around the bolt faster, therefore I will produce more work in tightening the bolt that way, I agree, but that is to say that if I can accelerate the car faster because I have more power, than my engine can do more work than it could do before.
One thing is to say that a geabox takes so much to be turned, so if you're turning it in less time than you where turning it before, you'll have to do more work to do that; another is to say that it is now absorbing more power than it was before from the engine, because the engine can accelerate the car faster.
Just to give another example, imagine that I get a car and I don't touch anything on it.
Let's say that I accelerate using 1/2 of the throttle. is the crank to wheel power loss less than it was before? Does that therefore mean that there is no fixed power loss even when you are simply looking at a car which you are not going to modify and just drive differently?
Let's go back to the arms:
Let's say that I put 100 lbs on a bar and start benchpressing. Let's say that I can push 200 lbs max and my friend can push 100 (poor friend...). If I do 10 reps in 10 seconds and he does the same after me, would you say I encountered more resistance?
I'll give you this: 0-60 in 5 seconds vs 0-60 in 4 seconds.
In the first case, the gearbox has to be accelerated to that speed in 6 seconds, which will require less work than the second case, when it is accelerated to the same speed in less time. But this is not drive train power loss.
It would be like saying (keeping the 2 examples I just used) that when you do the 0-60 in 4 seconds, the car has become heavier...???
I think that what I have in mind here is the following: once you have reached peak power engine speed (say 6000 rpm) the amount of work necessary to keep everything spinning at that level is fixed. Whether your car is making 100 or 300 hp, the work needed to keep everything running at that speed is fixed.
A different thing is to say that you start accelerating the car after you have modified it; if your car is capable of accelerating faster now, it is obvious that the work needed to bring all rotating components (gearbox included) to that speed in less time is more than before. But I believe that is not drivetrain power loss, like I said before.
Still wondering...
01-04-06, 08:45 PM
#16
Originally Posted by gio64
it is obvious that the work needed to bring all rotating components (gearbox included) to that speed in less time is more than before
There are two main causes of power loss in transmissions.
1: The nice simple textbook loss due to sliding contact of the gears. Taken to be constant with speed, typically about 2% of power for spur gears and 3% for hypoid bevel.
2: The complex loss (never found in textbooks) due to oil fling and windage. As the gear starts to rotate it picks up oil, has a large wetted area and the loss follows a normal V^3 drag power law. As it picks up speed it tends to fling the oil and carve a groove in the oil bath reducing drag by entraining air. As it flings oil the oil depth reduces, again reducing drag. It moves to a loss approx proportional to speed regime. I don't know what happens at very high speed when the oil level has been reduced as low as it can go or a larger gear on the same shaft is still flinging oil and a smaller gear runs clear of the oil bath. Auto boxes main loss is due to pumping loss in the oil pump and hydraulic system as the gears are not dipped in oil.
Increased temperature will reduce viscosity and reduce drag.
My LSD diff has an oil cooler, not for the benefit of the gears but to maintain the oil at the working temperature of the Viscous LS unit. It has a warning light to tell me when it is possibly an open diff.
Transmission loss is a black art known only in the makers dyno house and consultants like Ricardo or SwRI, there is very little published information. Lots of people quote some % figure for loss but don't say what the power or speed was so the figure quoted is useless. Even assuming it's for peak power and speed doesn't tell you what it will be at normal cruise. In lots of engineering texts I have only ever found one graph of transmission loss against speed, it was in a book on dynometer testing of engines, 1936 reprinted in 1969!
1: The nice simple textbook loss due to sliding contact of the gears. Taken to be constant with speed, typically about 2% of power for spur gears and 3% for hypoid bevel.
2: The complex loss (never found in textbooks) due to oil fling and windage. As the gear starts to rotate it picks up oil, has a large wetted area and the loss follows a normal V^3 drag power law. As it picks up speed it tends to fling the oil and carve a groove in the oil bath reducing drag by entraining air. As it flings oil the oil depth reduces, again reducing drag. It moves to a loss approx proportional to speed regime. I don't know what happens at very high speed when the oil level has been reduced as low as it can go or a larger gear on the same shaft is still flinging oil and a smaller gear runs clear of the oil bath. Auto boxes main loss is due to pumping loss in the oil pump and hydraulic system as the gears are not dipped in oil.
Increased temperature will reduce viscosity and reduce drag.
My LSD diff has an oil cooler, not for the benefit of the gears but to maintain the oil at the working temperature of the Viscous LS unit. It has a warning light to tell me when it is possibly an open diff.
Transmission loss is a black art known only in the makers dyno house and consultants like Ricardo or SwRI, there is very little published information. Lots of people quote some % figure for loss but don't say what the power or speed was so the figure quoted is useless. Even assuming it's for peak power and speed doesn't tell you what it will be at normal cruise. In lots of engineering texts I have only ever found one graph of transmission loss against speed, it was in a book on dynometer testing of engines, 1936 reprinted in 1969!
01-04-06, 09:41 PM
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Regarding the fact that the new dynos can calculate power loss through coasting, doesn't that mean that the power the engine is making is not affecting the calculation?
When you coast, you're putting the foot off the gas, so...
And once again, I am not arguing with anyone. I am saying, this is what I know and this is what I believe happens. I asked for somebody to say yes, because blah blah, or no, because blah blah.
Using examples that use increased speed to justify increased work are in my opinion inappropriate, since you might even find that your modded car peaks power earlier in the rpm range than it did before.
I am simply looking for a precise answer for a confirmation or denial of what I think, I don't have a problem with the results. I have simply said what I said because I feel the examples and suggestions are incorrect, which doesn't necessarily mean that the matter itself is the way I think it is...
When you coast, you're putting the foot off the gas, so...
And once again, I am not arguing with anyone. I am saying, this is what I know and this is what I believe happens. I asked for somebody to say yes, because blah blah, or no, because blah blah.
Using examples that use increased speed to justify increased work are in my opinion inappropriate, since you might even find that your modded car peaks power earlier in the rpm range than it did before.
I am simply looking for a precise answer for a confirmation or denial of what I think, I don't have a problem with the results. I have simply said what I said because I feel the examples and suggestions are incorrect, which doesn't necessarily mean that the matter itself is the way I think it is...
01-04-06, 09:49 PM
#18
Originally Posted by gio64
Regarding the fact that the new dynos can calculate power loss through coasting, doesn't that mean that the power the engine is making is not affecting the calculation?
When you coast, you're putting the foot off the gas, so...
When you coast, you're putting the foot off the gas, so...
Nobody is saying that what the engine is putting out makes a difference. If it did, then your drivetrain loss percentage could grow or shrink depending on the amount of power/torque your motor is putting out (and we know that's not the case).
However you find the percentage for your car (comparing BHP dyno to RWHP dynos, or the coasting test), that percentage will remain constant as long as your driveline components remain the same. So if you are loosing say 40% to the drivetrain, then it will be 40% at 200hp and 40% at 1000hp; the percentage stays the same but the actually amounts are changed as expected.
But as I mentioned before, there is no forumla to just pull out and plug a few numbers into to find it.
01-04-06, 09:51 PM
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Originally Posted by Mahjik
No. You forgettting the fluid in the transmission. Just like the example posted, with fluid it takes more effort to move through it faster. Here's a quote for you:
Basically, there is no direct formula to calcuate it. You can get it for a specific car by testing it on a dyno and comparing it an engine dyno, but there are too many factors to just have a simple equation to calculate it.
Basically, there is no direct formula to calcuate it. You can get it for a specific car by testing it on a dyno and comparing it an engine dyno, but there are too many factors to just have a simple equation to calculate it.
Although, does that mean that the friction due to oil is less at higher rotational speeds?
If I read this right, it looks like increasing speed/constant speed have something to do with this entire problem as well...
And I agree, (since I was confused in the first place), when you say, it is complex and you cannot certainly say that there is a formula.
Would you however say that you'd have a power loss of 20% on a car that spits 350 hp at the wheels?
I feel it is unlikely...
01-04-06, 10:03 PM
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Originally Posted by Mahjik
Yes, that is correct. How much power doesn't matter what the drivetrain loss equals as you are just looking for the percentage difference (as that percentage will be applied for all power levels, of that particular car, using the same driveline components).
Nobody is saying that what the engine is putting out makes a difference. If it did, then your drivetrain loss percentage could grow or shrink depending on the amount of power/torque your motor is putting out (and we know that's not the case).
However you find the percentage for your car (comparing BHP dyno to RWHP dynos, or the coasting test), that percentage will remain constant as long as your driveline components remain the same. So if you are loosing say 40% to the drivetrain, then it will be 40% at 200hp and 40% at 1000hp; the percentage stays the same but the actually amounts are changed as expected.
But as I mentioned before, there is no forumla to just pull out and plug a few numbers into to find it.
Nobody is saying that what the engine is putting out makes a difference. If it did, then your drivetrain loss percentage could grow or shrink depending on the amount of power/torque your motor is putting out (and we know that's not the case).
However you find the percentage for your car (comparing BHP dyno to RWHP dynos, or the coasting test), that percentage will remain constant as long as your driveline components remain the same. So if you are loosing say 40% to the drivetrain, then it will be 40% at 200hp and 40% at 1000hp; the percentage stays the same but the actually amounts are changed as expected.
But as I mentioned before, there is no forumla to just pull out and plug a few numbers into to find it.
Allow me and forgive me.
Say I go with stock car and make 100 hp @ 5000 rpm; I coast, my drivetrain loss is 20 hp on the dyno. That can be worked into a percentage, that's fine, it is 20%.
Now let's say that I tune the car in, get another run and get 120 hp @ 5000 rpm; now, if what you say is what I understand, when I coast, my drivetrain loss has gone to 24 hp, still 20% of the power of the engine.
If this is correct, then something has changed in my drivetrain that is now robbing more power from the engine (and, if I get this right, my car would be slowing down the rollers more quickly than it was when I did my first run and scored 100 hp) is that correct?
01-04-06, 11:24 PM
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Originally Posted by Mahjik
Nope, you aren't losing "more" power. If you were making 80% before, you still are making 80% afterwards. It's just 80% of a larger total.
Or, otherwise, if your drivetrain was using 20 hp and is now using 24, you are not losing more power to the drivetrain than you were before.
Regardless of how we can word it, the fact is (based on my previous example), my drivetrain is using 4 more horse power after tuning than it was before tuning, can we agree on that?
That means that the work necessary to move my drivetrain has increased 4 hp, although the drivetrain is still the same and it is operating at the same speed, can we agree on that?
Now, I am not disagreeing nor agreeing, I just need to understand where that 4 hp has gone, what has sucked it in/up, given that, indeed, it is sucking that 4 more hp.
As far as the coasting on the dyno goes, what I was trying to say is this: if you coast on the dyno with a certain car before or after tuning, the results of the coasting won't change, but the results of the power will; so the dyno can only find that if the drivetrain absorbed 20 hp the first time on coasting, will also absorb 20 hp on coasting the second time around. Expressing that into a percentage is just an aftermath operation. Now, as I just said, if I am wrong, then it means that coasting after tuning becomes harder on the dyno and the dyno finds that the drivetrain is now taking 4 hp more. But that is what I don't understand. One last thing is that maybe the dyno does some other thing to figure this out, but I still don't understand how the data during coasting can change (increase that is) to maintain the power loss percentage costant.
01-05-06, 09:08 AM
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You are right that just by increasing power will not give you a greater hp loss reading when coasting, mostly because it should read out some kind of friction loss coefficent or something along those lines. And also the force required to keep the drivetrain at a constant speed is going to be the same, so the higher hp car will, for the most part, not work as hard(when comparing power use percentages) to keep moving at a constant speed, it will however require the same amount of hp.
So basically for a constant velocity it will not lose more power, but under accleration it will, but as a percentage.
here is a ME ex.
F=m*a
m=mass
a=accel
F=force
I hate standard so here it is in metric. These are just numbers off the top of my head, and is overly simplified.
1. say for the lower hp car your avg. accel is 10m/s^2 (0-60~3sec.)
2. higher hp car your avg. accel is 15m/s^2 (0-60~2sec.)
say the mass of the drivetrain is 10kg
1. The avg. force=100kg*m/s^2
2. =150kg*m/s^2
Therefore to accelerate faster it requires more force, which is where the extra loss comes from and is why it takes more power away from the crank hp.
So basically for a constant velocity it will not lose more power, but under accleration it will, but as a percentage.
here is a ME ex.
F=m*a
m=mass
a=accel
F=force
I hate standard so here it is in metric. These are just numbers off the top of my head, and is overly simplified.
1. say for the lower hp car your avg. accel is 10m/s^2 (0-60~3sec.)
2. higher hp car your avg. accel is 15m/s^2 (0-60~2sec.)
say the mass of the drivetrain is 10kg
1. The avg. force=100kg*m/s^2
2. =150kg*m/s^2
Therefore to accelerate faster it requires more force, which is where the extra loss comes from and is why it takes more power away from the crank hp.
01-05-06, 09:15 AM
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Originally Posted by gio64
Well, that would be like saying that if I make 10k a year and get taxed 10% and you make 100k a year and get taxed 10% we're paying the same taxes, but I think you'd disagree on that one.
Or, otherwise, if your drivetrain was using 20 hp and is now using 24, you are not losing more power to the drivetrain than you were before.
Or, otherwise, if your drivetrain was using 20 hp and is now using 24, you are not losing more power to the drivetrain than you were before.
01-05-06, 09:48 AM
#25
Originally Posted by SPICcnmGT
You are losing more but if you look at it in percentages you are not.
Yes. He's looking at the wrong number. He sees the 'total' loss as being greater and see that as a problem, rather than looking at the 'percentage' loss. He's looking at the loss number without comparing it to the beginning. It guess it's obvious to the rest of use that if the first number changes, then one of the other two HAS to change. Physics wise, it's not the middle number so it's the result that changes.
gio64, think about braking. Let's say it takes you 160 ft to stop when travelling 70mph. Now, nothing else has changed on the car (same force, etc), you only increase your speed to 100mph, will you still stop in 160 ft? Nope. Does that mean your braking is worse? No.