drivetrain loss in relation to rwhp?
03-18-2006, 09:28 PM
#1
drivetrain loss in relation to rwhp? this seems to be an dvanced question so ill ask it here. ok everyone says that drivetrain loss is measured in percentage........but im failing to see how this could be true. say 1 car with the identical drivetrain. with the one motor it makes 400 hp. using the 15% drivetrain loss number u get 340 rwhp. a loss of 60 hp. now lets say this same car gets a heart transplant and now makes 650 hp. u get 552.5 rwhp. thats a loss of 97.5 hp. im failing to see how identical drivetrains could cause such a greater loss because the car makes more power. please discuss............... 
03-18-2006, 09:57 PM
#2
I have always wondered about that myself. The only thing my feeble brain could come up with is that the higher horsepower motor puts more "stress" on the drive train componets which in turn creates more friction and the resultant power loss. Whether it remains say at 15% in a linear fashion with the horsepower is another question.
03-19-2006, 06:24 AM
#3
drivetrain loss in normaly a percentage! for a noramal manual RWD aplication you are looking at about 10-15%. on a mid rear you are looking at 15+%. on a WFWD set up your looking at 15-20% and for 4WD you are gunning at 25% ish!
as for why its a percentage rather than a fixed number? well thats obvious. the more you run though a drive systems the more you loose!
Chris.
as for why its a percentage rather than a fixed number? well thats obvious. the more you run though a drive systems the more you loose!
Chris.
03-19-2006, 07:27 AM
#4
Originally Posted by chuntington101
drivetrain loss in normaly a percentage! for a noramal manual RWD aplication you are looking at about 10-15%. on a mid rear you are looking at 15+%. on a WFWD set up your looking at 15-20% and for 4WD you are gunning at 25% ish!
as for why its a percentage rather than a fixed number? well thats obvious. the more you run though a drive systems the more you loose!
Chris.
as for why its a percentage rather than a fixed number? well thats obvious. the more you run though a drive systems the more you loose!
Chris.
03-19-2006, 08:44 AM
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Originally Posted by ls2 bait
this seems to be an dvanced question so ill ask it here. ok everyone says that drivetrain loss is measured in percentage........but im failing to see how this could be true. say 1 car with the identical drivetrain. with the one motor it makes 400 hp. using the 15% drivetrain loss number u get 340 rwhp. a loss of 60 hp. now lets say this same car gets a heart transplant and now makes 650 hp. u get 552.5 rwhp. thats a loss of 97.5 hp. im failing to see how identical drivetrains could cause such a greater loss because the car makes more power. please discuss...............
03-19-2006, 12:57 PM
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I can see a fixed number loss for a given drivetrain- EXEPT for the torque converter.
Say the stock torque converter is 95% efficent. That means more power through it will be more of a loss plus the fixed number of the rest of the drive train.
For example:
350 crank hp---95% of 350 is 332.5. minus drivetrain loss (30hp lets say) = 302.5
a total loss of 47.5
500 crank hp---95% of 500 is 475 minus drivetrain loss (30hp) = 445
a total loss of 55
A M/T would be a fixed number. This is the way i see it, i could be wrong.
Say the stock torque converter is 95% efficent. That means more power through it will be more of a loss plus the fixed number of the rest of the drive train.
For example:
350 crank hp---95% of 350 is 332.5. minus drivetrain loss (30hp lets say) = 302.5
a total loss of 47.5
500 crank hp---95% of 500 is 475 minus drivetrain loss (30hp) = 445
a total loss of 55
A M/T would be a fixed number. This is the way i see it, i could be wrong.
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03-19-2006, 03:05 PM
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Drivetrain losses are vastly overstated in modern dyno testing, in many cases. It is far more common than one would often think. But I will leave this out of the arguement for now.
Drivetrain (specifically gearset; for now do not consider a torque converter or AT hydraulic pumps) energy consumption is a function of friction, angularity, and velocity. While it is true that at least on a simple observational analysis, 'the more power you put through a drivetrain, the more power you lose to it', it is far from a 1:1 direct proportionality, as it is often assumed.
It is obvious that the difference between power in and power out is equal to power lost, which is lost chiefly as two forms of energy: heat, and vibration. We can use this knowledge in conjunction with general observation to realize some significant physical phenomena for what they really are.
If we take velocity out of the equation for a second, we are left with force and energy*distance, instead of power and speed. In other words, instead of talking about the 'power' loss, let's look at it (from an equally correct viewpoint) as torque dissipation as heat and vibration, per revolution, without regard to the time it takes for that revolution to occur. This perspective simplifies the relationships significantly.
If a gearset is less than 100% mechanically efficient, as all gearsets inherently are, then by definition, some of the energy put into them will be converted to heat and vibration; they will 'consume' this energy in order to operate. This is largely due to sliding friction of the mechanical parts and fluid pumping losses due to viscosity of the lubricants. This is ignoring system-specific variables such as pumps, filters (which create pressure differentials that demand energy flow across them that must be accounted for) and other similar items.
If one were to have a system with an input of 100 lb-ft at 100rpms, with an associated loss of 10 lb-ft measured at the output, this would of course represent a 10% drivetrain torque loss. If one raised the input torque to 150 lb-ft at 100rpms, one might expect the 10% loss to result in 15 lb-ft reduction at the output. In nearly all cases, this is not the result. Less than the expected 15 lb-ft will be dissipated as additional heat. This would represent an increase in rated mechanical efficiency. Why would this be observed?
It is observed because the energy dissipated, while strongly influenced by sliding friction (which itself is not a purely linear relationship to imposed loading anyway), is not solely dependent upon gear friction; which is all that has changed by adding 50 lb-ft more to the input.
If we account for your losses by their source pathways, we find that the resultant force loss accountability is represented as sliding friction + viscosity + vibration + system specific variables. The point is that the major variable that was changed, sliding friction, accounts for less than 100% of the losses, and therefore less than 100% of the consumed torque is amplified.
Even if we were to evaluate sliding friction as the sole pathway for torque dissipation/loss, we would likely not see a linear relationship between increases in input, and increases in dissipation. This is because of the extensive use of friction modifiers (lubricants) and tooth profiles that do not respond in a purely linear fashion to increased gear loading.
One of the issues when attempting to seriously evaluate drivetrain efficiency is the need to account for pathways of power loss. Power loss is highly dependent upon rpm, which poses a problem for ratings due to the need to represent different information (peak loss or peak efficiency for example) when the peaks may not be very representative of the average in the operational speed range, anyway.
As velocity increases, so does additional friction. But lubricants complicate this relationship immensely, as does geometry that is influenced by thermal expansion, varying elasticity under load, angle, temperature, etc. Often times when either input torque or rpm is changed, some variables will contribute to more loss, while others will contribute to increases in efficiency instead.
The result can be very difficult to model with great accuracy. The good thing is that most of the time, this kind of accuracy means little to most of us other than an educational exercise. 'Good enough' measurements and correlations can be drawn from simpler models. More often these models get reduced to:
1.As input torque at a specified rpm increases, so does the associated torque dissipation, but in a less than purely linear fashion, so long as the system materials are not loaded to the point of deforming beyond geometrically efficient limits. This is a good thing.
2.As rpm increases, so does torque dissipation. This is a bad thing. Torque is measured per revolution, so if torque/rev increases with increasing rpm, this means a greater than linear power loss with increasing rpm.
3.Torque dissipation most often is greater influenced by lubricant properties than gear loading or rpm. Often times gear loading is less responsible for increased power loss compared to increased rpm (which makes sense due to the nature of the measurement.)
What can you take from all this? In short: If you put more power through a drivetrain, YES, it will dissipate more power than it currently does. HOWEVER, it will almost always GAIN in terms of 'rated mechanical efficiency' simply because the gains in dissipation will be less than linear. Such is the nature of the beast.
Does this mean that the less power you put through a drivetrain, the less efficient it inherently is? Absolutely not. It means that if you stop looking at output torque as your product, and measure heat instead, that you get diminishing returns of ADDED heat output with increased power in.
If you have 100 lb-ft input and 10 lb-ft lost, and double the input to 200 lb-ft, you will likely not see 20 lb-ft lost, but less. Of that 200 input lb-ft, 100lb-ft of it is still 'operating at it's 10% loss' just as before. The difference in the system is that the ADDITIONAL 100 lb-ft fails to contribute an additional 10 lb-ft loss to heat due to the variables in the system.
The faults are the method of measurement, and the method of rating the system, not of the system itself, in my opinion.
Drivetrain (specifically gearset; for now do not consider a torque converter or AT hydraulic pumps) energy consumption is a function of friction, angularity, and velocity. While it is true that at least on a simple observational analysis, 'the more power you put through a drivetrain, the more power you lose to it', it is far from a 1:1 direct proportionality, as it is often assumed.
It is obvious that the difference between power in and power out is equal to power lost, which is lost chiefly as two forms of energy: heat, and vibration. We can use this knowledge in conjunction with general observation to realize some significant physical phenomena for what they really are.
If we take velocity out of the equation for a second, we are left with force and energy*distance, instead of power and speed. In other words, instead of talking about the 'power' loss, let's look at it (from an equally correct viewpoint) as torque dissipation as heat and vibration, per revolution, without regard to the time it takes for that revolution to occur. This perspective simplifies the relationships significantly.
If a gearset is less than 100% mechanically efficient, as all gearsets inherently are, then by definition, some of the energy put into them will be converted to heat and vibration; they will 'consume' this energy in order to operate. This is largely due to sliding friction of the mechanical parts and fluid pumping losses due to viscosity of the lubricants. This is ignoring system-specific variables such as pumps, filters (which create pressure differentials that demand energy flow across them that must be accounted for) and other similar items.
If one were to have a system with an input of 100 lb-ft at 100rpms, with an associated loss of 10 lb-ft measured at the output, this would of course represent a 10% drivetrain torque loss. If one raised the input torque to 150 lb-ft at 100rpms, one might expect the 10% loss to result in 15 lb-ft reduction at the output. In nearly all cases, this is not the result. Less than the expected 15 lb-ft will be dissipated as additional heat. This would represent an increase in rated mechanical efficiency. Why would this be observed?
It is observed because the energy dissipated, while strongly influenced by sliding friction (which itself is not a purely linear relationship to imposed loading anyway), is not solely dependent upon gear friction; which is all that has changed by adding 50 lb-ft more to the input.
If we account for your losses by their source pathways, we find that the resultant force loss accountability is represented as sliding friction + viscosity + vibration + system specific variables. The point is that the major variable that was changed, sliding friction, accounts for less than 100% of the losses, and therefore less than 100% of the consumed torque is amplified.
Even if we were to evaluate sliding friction as the sole pathway for torque dissipation/loss, we would likely not see a linear relationship between increases in input, and increases in dissipation. This is because of the extensive use of friction modifiers (lubricants) and tooth profiles that do not respond in a purely linear fashion to increased gear loading.
One of the issues when attempting to seriously evaluate drivetrain efficiency is the need to account for pathways of power loss. Power loss is highly dependent upon rpm, which poses a problem for ratings due to the need to represent different information (peak loss or peak efficiency for example) when the peaks may not be very representative of the average in the operational speed range, anyway.
As velocity increases, so does additional friction. But lubricants complicate this relationship immensely, as does geometry that is influenced by thermal expansion, varying elasticity under load, angle, temperature, etc. Often times when either input torque or rpm is changed, some variables will contribute to more loss, while others will contribute to increases in efficiency instead.
The result can be very difficult to model with great accuracy. The good thing is that most of the time, this kind of accuracy means little to most of us other than an educational exercise. 'Good enough' measurements and correlations can be drawn from simpler models. More often these models get reduced to:
1.As input torque at a specified rpm increases, so does the associated torque dissipation, but in a less than purely linear fashion, so long as the system materials are not loaded to the point of deforming beyond geometrically efficient limits. This is a good thing.
2.As rpm increases, so does torque dissipation. This is a bad thing. Torque is measured per revolution, so if torque/rev increases with increasing rpm, this means a greater than linear power loss with increasing rpm.
3.Torque dissipation most often is greater influenced by lubricant properties than gear loading or rpm. Often times gear loading is less responsible for increased power loss compared to increased rpm (which makes sense due to the nature of the measurement.)
What can you take from all this? In short: If you put more power through a drivetrain, YES, it will dissipate more power than it currently does. HOWEVER, it will almost always GAIN in terms of 'rated mechanical efficiency' simply because the gains in dissipation will be less than linear. Such is the nature of the beast.
Does this mean that the less power you put through a drivetrain, the less efficient it inherently is? Absolutely not. It means that if you stop looking at output torque as your product, and measure heat instead, that you get diminishing returns of ADDED heat output with increased power in.
If you have 100 lb-ft input and 10 lb-ft lost, and double the input to 200 lb-ft, you will likely not see 20 lb-ft lost, but less. Of that 200 input lb-ft, 100lb-ft of it is still 'operating at it's 10% loss' just as before. The difference in the system is that the ADDITIONAL 100 lb-ft fails to contribute an additional 10 lb-ft loss to heat due to the variables in the system.
The faults are the method of measurement, and the method of rating the system, not of the system itself, in my opinion.
03-20-2006, 04:43 PM
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^^^Very very nice. I agree totally, but couldnt articulate it nearly this well.
Think about it, why would a driveline be the same efficiency over varying speeds, loads, etc.
On Vorshlag.com I also read a great analysis of this. The basic point was that expressing the power lost as a fixed number and a smaller percentage gain depending on power levels is not perfect, but much more accurate than the current trend of saing "12%".
Makes sense. These guys running 1000WHP, if they were really burning up 15% through the drivetrain, thats over 150hp just disappaited as heat and vibration. I dont think there are any transmissions capable of disapating that kind of heat and taking that kind of stress. The things would simply melt and fall apart.
Think about it, why would a driveline be the same efficiency over varying speeds, loads, etc.
On Vorshlag.com I also read a great analysis of this. The basic point was that expressing the power lost as a fixed number and a smaller percentage gain depending on power levels is not perfect, but much more accurate than the current trend of saing "12%".
Makes sense. These guys running 1000WHP, if they were really burning up 15% through the drivetrain, thats over 150hp just disappaited as heat and vibration. I dont think there are any transmissions capable of disapating that kind of heat and taking that kind of stress. The things would simply melt and fall apart.
Last edited by Sparetire; 03-20-2006 at 04:45 PM. Reason: Grammar LOL
03-20-2006, 05:34 PM
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This possibly started back in the '80s' when chassis dynos started to become popular. We simply needed a multiplication factor to compare what we measured at the drive wheels to compare to the flywheel numbers that an engine dyno or the vehicle manufactures provided. We simply called it "drive train loss" which is only part of the equation. One other significant part of the equation is "proceedural loss". Engine dynos and vehicle manufacures normally use the "RPM step" proceedure versus a chassis dyno's "RPM sweep" proceedure, meaning that the engine dyno holds the engine at given static RPM increments then measures the HP and TQ. This proceedure in itself provides significantly higher readings, sometimes more then 50% of what we call "drive train" losses, because there are no engine/drivetrain acceleration (inertia) losses. Yet, when we compute our standard "drive train" loss factors, we seem to get consistanly close to, for example, what the vehicle manufacurers rate thier engines, low or high HP.
03-28-2006, 09:14 AM
#11
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We had this discussion over on Corvetteforum.
http://forums.corvetteforum.com/show...052&forum_id=1
I'll throw in my $.02.
Do a couple of simple experiments. Take a stock Z06, pull the rear brakes off the car, and put a superlight wheel and tire combo on the car and dyno it. My guess is the car will register about 5-8 RWHP just from the removal of the brakes, and depending on wheel/tire weight as much as 20 RWHP (but probably closer to 10-15 RWHP). Now, where has all that deflection and frictional loss everyone keeps talking about gone to?
See here: heavy wheels vs lighter wheels (+/-20RWHP)
http://forums.corvetteforum.com/show....php?t=1328054
-20RWHP by adding 33 lbs of rotating mass.
Most of your driveline losses are simply due to weight (mass) that you have to accelerate. Component mass is a necessary evil. Reduce the mass, and the "percentage of loss" gets smaller. That 12-15% is a simple number based on the HP levels we are dealing with, and not based on an across the board rule.
Let look at a class where small incremetal gains are costly (NASCAR). Tell me, do you honestly think if on an unrestricted cup motor that if you had to throw away 15-20% of the usable power to "losses" that it would not have NASCAR guys going over every inch of the driveline looking for how to get "free" HP. I mean C'mon they still use Ford 9" rears which are notorious HP robbers. Why is that? Its because its not a percentage, its pretty much a simple "we loose 50 HP (or whatever it is) in the driveline" equation.
I do agree that some of the point that have been brought up about firction losses, deflection, etc... are all completely valid points. But, I think where everyone differs is how much of a percentage of the total loss that is. In other words, they are part of the overall equation, but I don't think they are the majority of the loss.
Lets look at that loss as heat energy. On a stock 405 HP Z06. 15% of 405 is 60 HP. 60HP is 152801.9 BTU/Hr 152K BTU's is a lot of energy to get rid of. Lets go to a 505HP C6 Z06 75 HP - 191002.4 BTU/hr. Now, you increase that to say 600 Crank HP that 90 HP loss, so 229202.9 BTU/hr. Even if only a fraction of that is converted to heat energy, you're talking about tons of heat. Then the question becomes where does the rest of that energy go?
Let look from the other side. ~50 HP loss in the Z06 driveline
405-50 = 355
505-50 = 455
Hmmmm... seems to come out pretty close.
I've seen this plenty in drag racing. We replace a lighter component with a heavier component, we take a basic HP loss. I.E. you swap from a TH350 to a TH400 you see a loss. You swap from a 10 bolt to a 12 bolt you see a loss. But, the motors are run on an engine dyno, so we know the HP we make there. We know what to expect, and it pretty much jives up with what we see.
We have a pretty good feel what kind of drivline loss we expect to see when swapping tranmsissions, etc...
I guess what I'm saying is that if the losses you are talking about were real world, much more effort would be placed on drivline improvement. There is simply too much power on the table if its a percentage vs a fixed constant.
Also, I'd like to point out what Tony was talking about on N2O orifice size. You can pretty much figure based on orifice size how much fuel you'll need to keep up with the mixture, and from that you can pretty well predict the ammount of power you can make with an engine.
I consider Tony's A to B testing to be good data which shows at least in my mind that most of the losses are driveline losses, and can be expressed as a simple ammount, and not as a percentage IMHO.
http://forums.corvetteforum.com/show...052&forum_id=1
I'll throw in my $.02.
Do a couple of simple experiments. Take a stock Z06, pull the rear brakes off the car, and put a superlight wheel and tire combo on the car and dyno it. My guess is the car will register about 5-8 RWHP just from the removal of the brakes, and depending on wheel/tire weight as much as 20 RWHP (but probably closer to 10-15 RWHP). Now, where has all that deflection and frictional loss everyone keeps talking about gone to?
See here: heavy wheels vs lighter wheels (+/-20RWHP)
http://forums.corvetteforum.com/show....php?t=1328054
-20RWHP by adding 33 lbs of rotating mass.
Most of your driveline losses are simply due to weight (mass) that you have to accelerate. Component mass is a necessary evil. Reduce the mass, and the "percentage of loss" gets smaller. That 12-15% is a simple number based on the HP levels we are dealing with, and not based on an across the board rule.
Let look at a class where small incremetal gains are costly (NASCAR). Tell me, do you honestly think if on an unrestricted cup motor that if you had to throw away 15-20% of the usable power to "losses" that it would not have NASCAR guys going over every inch of the driveline looking for how to get "free" HP. I mean C'mon they still use Ford 9" rears which are notorious HP robbers. Why is that? Its because its not a percentage, its pretty much a simple "we loose 50 HP (or whatever it is) in the driveline" equation.
I do agree that some of the point that have been brought up about firction losses, deflection, etc... are all completely valid points. But, I think where everyone differs is how much of a percentage of the total loss that is. In other words, they are part of the overall equation, but I don't think they are the majority of the loss.
Lets look at that loss as heat energy. On a stock 405 HP Z06. 15% of 405 is 60 HP. 60HP is 152801.9 BTU/Hr 152K BTU's is a lot of energy to get rid of. Lets go to a 505HP C6 Z06 75 HP - 191002.4 BTU/hr. Now, you increase that to say 600 Crank HP that 90 HP loss, so 229202.9 BTU/hr. Even if only a fraction of that is converted to heat energy, you're talking about tons of heat. Then the question becomes where does the rest of that energy go?
Let look from the other side. ~50 HP loss in the Z06 driveline
405-50 = 355
505-50 = 455
Hmmmm... seems to come out pretty close.
I've seen this plenty in drag racing. We replace a lighter component with a heavier component, we take a basic HP loss. I.E. you swap from a TH350 to a TH400 you see a loss. You swap from a 10 bolt to a 12 bolt you see a loss. But, the motors are run on an engine dyno, so we know the HP we make there. We know what to expect, and it pretty much jives up with what we see.
We have a pretty good feel what kind of drivline loss we expect to see when swapping tranmsissions, etc...
I guess what I'm saying is that if the losses you are talking about were real world, much more effort would be placed on drivline improvement. There is simply too much power on the table if its a percentage vs a fixed constant.
Also, I'd like to point out what Tony was talking about on N2O orifice size. You can pretty much figure based on orifice size how much fuel you'll need to keep up with the mixture, and from that you can pretty well predict the ammount of power you can make with an engine.
I consider Tony's A to B testing to be good data which shows at least in my mind that most of the losses are driveline losses, and can be expressed as a simple ammount, and not as a percentage IMHO.
03-29-2006, 04:16 AM
#13
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There's some good info here.
I think drivetrain loss is a very confusing area and combine this with poorly operated dyno's and dyno's such as Dynojets which are non load bearing dyno's, these are the major factors why so many people beleive their cars are underatted from the factory.
IMO drivetrain loss can not be a static value nor a fixed percentage either. I beleive there is probably a fixed loss and a decreasing percentage loss.
I say decreasing because a small percentage of a large number can be bigger than a large percentage of a small number, example:
10% of 600bhp = 60bhp
15% of 350bhp = 52.5bhp
The reason I beleive this, is because there are some components in the driveline which remain static. An object of X mass requires Y force to rotate it at Z speed. The mass does not change with increased power/torque and neither should the balance so friction and vibration and thus heat should also remain static in this respect.
But then there is the problem of Newton's Laws of Physics:
"every action has an opposite and equal reaction"
This simply means if you push on something, that something pushes back at you. But in order to move the object some of that energy must be dissipated else where.
In estimating drivetrain loss I use several calculations. For a manual rwd vehicle I use the 'classic' 15% rule, along with the 12% + 10bhp rule (as this loads different HP levels differently allowing for any static increase).
Then if possible try and work out what the drivetrain loss is from a stock vehicle so you have a base level of what the loss is likely to be.
Example:
A car is factory rated at 345bhp SAE Net, assuming this is accurate (and in many to most cases it is).
If the same car makes 295rwhp on a load bearing dyno (comparable to the engine dyno that would have been used by the manufacturer), with SAE conversion and all other correction factors taken into account. This would then be as close as we could get to a real world comparable number.
So 345 - 295 = 50bhp drivetrain loss.
we can assume this will be the minimal that will be lost regardless of how much extra power is produced.
If that same car has engine mods performed (drivetrain remains stock) and re-dyno's 400rwhp then we know the minimal it will have lost via the drivetrain is 50bhp.
So combining the other rules:
15% Rule: 400 / 0.85 = 471bhp (loss of 71bhp)
12% + 10bhp Rule: (400 / 0.88) + 10 = 465bhp (65bhp loss)
We can then take a mean average (50 + 65 + 71) / 3 = 62bhp
Therefore we can make an educated guess that the engine is producing 462bhp SAE Net equiverlent +- 10bhp
I don't beleive you can ever be anymore accurate than that.
One way of helping to reduce the inaccuracy of measuring the torque from the rear wheels is to use a chassis dyno like the Rototest or similar. This type of dyno mounts directly to the hubs of the driven wheels, thus removing any inaccuracy of the wheels and it also avoids the possibilty of tyre slip on the rollers.
I think drivetrain loss is a very confusing area and combine this with poorly operated dyno's and dyno's such as Dynojets which are non load bearing dyno's, these are the major factors why so many people beleive their cars are underatted from the factory.
IMO drivetrain loss can not be a static value nor a fixed percentage either. I beleive there is probably a fixed loss and a decreasing percentage loss.
I say decreasing because a small percentage of a large number can be bigger than a large percentage of a small number, example:
10% of 600bhp = 60bhp
15% of 350bhp = 52.5bhp
The reason I beleive this, is because there are some components in the driveline which remain static. An object of X mass requires Y force to rotate it at Z speed. The mass does not change with increased power/torque and neither should the balance so friction and vibration and thus heat should also remain static in this respect.
But then there is the problem of Newton's Laws of Physics:
"every action has an opposite and equal reaction"
This simply means if you push on something, that something pushes back at you. But in order to move the object some of that energy must be dissipated else where.
In estimating drivetrain loss I use several calculations. For a manual rwd vehicle I use the 'classic' 15% rule, along with the 12% + 10bhp rule (as this loads different HP levels differently allowing for any static increase).
Then if possible try and work out what the drivetrain loss is from a stock vehicle so you have a base level of what the loss is likely to be.
Example:
A car is factory rated at 345bhp SAE Net, assuming this is accurate (and in many to most cases it is).
If the same car makes 295rwhp on a load bearing dyno (comparable to the engine dyno that would have been used by the manufacturer), with SAE conversion and all other correction factors taken into account. This would then be as close as we could get to a real world comparable number.
So 345 - 295 = 50bhp drivetrain loss.
we can assume this will be the minimal that will be lost regardless of how much extra power is produced.
If that same car has engine mods performed (drivetrain remains stock) and re-dyno's 400rwhp then we know the minimal it will have lost via the drivetrain is 50bhp.
So combining the other rules:
15% Rule: 400 / 0.85 = 471bhp (loss of 71bhp)
12% + 10bhp Rule: (400 / 0.88) + 10 = 465bhp (65bhp loss)
We can then take a mean average (50 + 65 + 71) / 3 = 62bhp
Therefore we can make an educated guess that the engine is producing 462bhp SAE Net equiverlent +- 10bhp
I don't beleive you can ever be anymore accurate than that.
One way of helping to reduce the inaccuracy of measuring the torque from the rear wheels is to use a chassis dyno like the Rototest or similar. This type of dyno mounts directly to the hubs of the driven wheels, thus removing any inaccuracy of the wheels and it also avoids the possibilty of tyre slip on the rollers.
03-29-2006, 09:01 AM
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Quote from 300bhp/ton, "One way of helping to reduce the inaccuracy of measuring the torque from the rear wheels is to use a chassis dyno like the Rototest or similar. This type of dyno mounts directly to the hubs of the driven wheels, thus removing any inaccuracy of the wheels and it also avoids the possibilty of tyre slip on the rollers."
We want to know how much HP/TQ is being produced at the roller, not the hub. Most professional chassis dyno shops, like ours, use our dynos mainly for tuning purposes, which includes speedo calibrations and tuning everything from very small SCCA formula cars to commercial dually pickups with heavy loads. Many cars, especially drag cars, want to tune their tire/gear combos for MPH Vs HP curve Vs RPM shift points. Granted, I can see some advantages to such a non roller dyno and may be a good choice for some, but it would be of limited value at our shop.
We want to know how much HP/TQ is being produced at the roller, not the hub. Most professional chassis dyno shops, like ours, use our dynos mainly for tuning purposes, which includes speedo calibrations and tuning everything from very small SCCA formula cars to commercial dually pickups with heavy loads. Many cars, especially drag cars, want to tune their tire/gear combos for MPH Vs HP curve Vs RPM shift points. Granted, I can see some advantages to such a non roller dyno and may be a good choice for some, but it would be of limited value at our shop.
03-29-2006, 09:18 AM
#16
TECH Addict
Well I'll lean on your experience but I do know this type of dyno is very popular in motorsport outside the US (where this type of dyno is more common anyhow), in such things as rallying and circuit racing.
03-29-2006, 11:31 AM
#17
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The 15% or whatever is a good rule of thumb for across the board estimates, the real loss is something that must be measured, and it will depend on alot of things like engine temp, oil type, power/load, rpm, etc.
In engineering, efficiencies and losses all generally scale as percentages. The variables which effect losses are formulated as proportional to power, because that's what we're interested in.
If you want to get technical, for a given tranny, higher power/load will reduce oil film thickness, and power loss is highly sensitive to oil film thickness.
Look at Petrov's equation for bearing power loss to get an idea: Power dissipated (power loss=heat) = pi*mu*w^2*L*D/er
where er is the film thickness, w is the rotational speed. mu is oil viscosity, L and D are bearing sizes.
In engineering, efficiencies and losses all generally scale as percentages. The variables which effect losses are formulated as proportional to power, because that's what we're interested in.
If you want to get technical, for a given tranny, higher power/load will reduce oil film thickness, and power loss is highly sensitive to oil film thickness.
Look at Petrov's equation for bearing power loss to get an idea: Power dissipated (power loss=heat) = pi*mu*w^2*L*D/er
where er is the film thickness, w is the rotational speed. mu is oil viscosity, L and D are bearing sizes.
03-29-2006, 11:42 AM
#18
6600 rpm clutch dump of death Administrator
In Australia, they use a load dyno (Eddy Current) rather than an intertia dyno. In fact many of the V8 Super Car teams will do an entire "event" on the dyno to ensure how the car is going to perform.
Again, in testing on an engine load dyno vs a chassis dyno with multiple combinations, the delta remained a constant 110 RWHP loss. It was not a percentage.
Again, in testing on an engine load dyno vs a chassis dyno with multiple combinations, the delta remained a constant 110 RWHP loss. It was not a percentage.
03-29-2006, 04:48 PM
#20
TECH Enthusiast
Good information here.
Has anyone ere every played with coating any hard parts of the trans? I know in higher classes racing this is the norm to reduce friction and find power, but has anyone done this for a street car?
Could be an interesting side by side comparision, before and after with a 4L60e
Has anyone ere every played with coating any hard parts of the trans? I know in higher classes racing this is the norm to reduce friction and find power, but has anyone done this for a street car?
Could be an interesting side by side comparision, before and after with a 4L60e