Drivetrain Losses
06-06-2007, 12:02 AM
#21
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If a stock 2002 Z06 with 405rwhp made 350rwhp on a DynoJet then it has a 13.5% drivetrain loss from the Crank to the rear wheels.
Same car made only 340rwhp on a Mustang Dyno, and shows a 17.1% loss from Crank to rear wheels.
If the drivetrain stayed the same, then which is the correct drivetrain loss?
As far as dyno loads, DynoJets have a constant load, basically the rotating mass of the drums, where Mustang dyno's are loaded by a computer controlled magnetic brake.
Same car made only 340rwhp on a Mustang Dyno, and shows a 17.1% loss from Crank to rear wheels.
If the drivetrain stayed the same, then which is the correct drivetrain loss?
As far as dyno loads, DynoJets have a constant load, basically the rotating mass of the drums, where Mustang dyno's are loaded by a computer controlled magnetic brake.
because each dyno measures it a bit differently.
That's my point, you can not accurately express it as a percent.
For any given drivetrain, on a particular car, you will have a certain amount of loss, which will remain fairly FIXED, unless of course, there are changes made to the drivetrain itself, i.e. rear diff gear, shorter wheel/tire combo, etc., which will affect the amount of power applied to the ground.
I agree with your point about mustang vs dynojet, but it goes back to my point about each dyno simply measuring the results a bit differently.
That's why it's important for people to stick to one dyno for their measuring instrument when modding their car.
If the proper weather correction factors are used each time the car is dynoed, then it is possible to see accurate representaions of the gains made from said mods.
If someone simply wants a big dyno sheet, then go ahead and shop around until you get the numbers you're looking for.
06-06-2007, 01:11 AM
#22
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Thread Starter
Neither is correct if you attempt to express it as a percentage,
because each dyno measures it a bit differently.
That's my point, you can not accurately express it as a percent.
For any given drivetrain, on a particular car, you will have a certain amount of loss, which will remain fairly FIXED, unless of course, there are changes made to the drivetrain itself, i.e. rear diff gear, shorter wheel/tire combo, etc., which will affect the amount of power applied to the ground.
I agree with your point about mustang vs dynojet, but it goes back to my point about each dyno simply measuring the results a bit differently.
That's why it's important for people to stick to one dyno for their measuring instrument when modding their car.
If the proper weather correction factors are used each time the car is dynoed, then it is possible to see accurate representaions of the gains made from said mods.
If someone simply wants a big dyno sheet, then go ahead and shop around until you get the numbers you're looking for.
because each dyno measures it a bit differently.
That's my point, you can not accurately express it as a percent.
For any given drivetrain, on a particular car, you will have a certain amount of loss, which will remain fairly FIXED, unless of course, there are changes made to the drivetrain itself, i.e. rear diff gear, shorter wheel/tire combo, etc., which will affect the amount of power applied to the ground.
I agree with your point about mustang vs dynojet, but it goes back to my point about each dyno simply measuring the results a bit differently.
That's why it's important for people to stick to one dyno for their measuring instrument when modding their car.
If the proper weather correction factors are used each time the car is dynoed, then it is possible to see accurate representaions of the gains made from said mods.
If someone simply wants a big dyno sheet, then go ahead and shop around until you get the numbers you're looking for.
So then we're in agreement that there is no commonly accepted drive train loss amount. Unless you know what motor makes at the crank during the pull you're measuring at the wheels you can never test for this, and because differenent dynos measure power differently, you can get a million different results. It doesn't matter if you base it on a fixed amount or percentage. It's kind of like the Heisenburg Uncertainty Principle. This was the exact point of my original post, questioning the "accepted" value associated with drive train loss. It all started with being able to compare one measuring device (our dyno) and another measureing device (calculated torque from the scanner) on the same pull.
As for taking it to a the track, I'm all for that. I've never been one to proclaim the highest horsepower numbers. If I wanted to get into that game, we would have bought a DynoJet for tuning. I'm far more concerned with performance, fueling and timing, not benchmarks.
Last edited by Tuner@Straightline; 06-06-2007 at 01:18 AM.
06-06-2007, 01:39 AM
#23
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So then we're in agreement that there is no commonly accepted drive train loss amount. Unless you know what motor makes at the crank during the pull you're measuring at the wheels you can never test for this, and because differenent dynos measure power differently, you can get a million different results. It doesn't matter if you base it on a fixed amount or percentage. It's kind of like the Heisenburg Uncertainty Principle. This was the exact point of my original post, questioning the "accepted" value associated with drive train loss. It all started with being able to compare one measuring device (our dyno) and another measureing device (calculated torque from the scanner) on the same pull.
As for taking it to a the track, I'm all for that. I've never been one to proclaim the highest horsepower numbers. If I wanted to get into that game, we would have bought a DynoJet for tuning. I'm far more concerned with performance, fueling and timing, not benchmarks.
As for taking it to a the track, I'm all for that. I've never been one to proclaim the highest horsepower numbers. If I wanted to get into that game, we would have bought a DynoJet for tuning. I'm far more concerned with performance, fueling and timing, not benchmarks.
One more example.
I think we've all seen a person mod, for example, their 2002 z06 which in stock spec dynos 355rwhp, for a suggested % loss of 50/405= 12.3%.
Then they do TT's, heads, cam, etc, and end up with 750 rwhp.
Following the % loss method, you would end up with
750/ .877 = 855 flywheel hp.
855-750= 105 hp loss, which can't be true, if the drivetrain remained the same.
In my theory, actual flywheel hp is closer to 750+50 = 800.
My suggestion is that any power gained at the rear wheels is straight up gained at the flywheel, which flies in the face of many aftermarket parts manufacturers, who claim both rear wheel and flywheel numbers for thier "chips", "intakes", etc.
Again, I agree that a % loss as commonly applied is a fallacy.
06-06-2007, 10:30 AM
#24
Le Mans Master
A stock c6 z06 rated at 505 flywheel hp, dynoes around 450 rwhp, for a drivetrain loss of 50 or so hp.
A stock c5 z06 rated at 405 flywheel hp, dynoes around 350rwhp, for a drivetrain loss of 50 or so hp.
Therefore, my estimate for a manual equipped vette is a loss of 50 or so hp thru the drivetrain.
Even if you increase the engine hp by turbos or nitrous, for example, the drivetrain loss remains the same.
A stock c5 z06 rated at 405 flywheel hp, dynoes around 350rwhp, for a drivetrain loss of 50 or so hp.
Therefore, my estimate for a manual equipped vette is a loss of 50 or so hp thru the drivetrain.
Even if you increase the engine hp by turbos or nitrous, for example, the drivetrain loss remains the same.
For example, you can take the drive train out of the car and crank it by hand to some very slow speeds and obviously it doesn't take 50hp to do so. This is where the percentage rule comes in. An additional 100hp from a C6 over a C5 will definitely cause the drivetrain to heat up more. This afterall, is where the extra energy goes. Heat.
So don't rule out the percentage rule. It probably is still more accurate than a fixed loss. The difference comes from how many rpms you have to crank it as well as the forces applied.
06-06-2007, 01:35 PM
#25
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I agree that a set percentage of loss is probably not accurate but I think the true answer is somewhere in between. Your example above seems sensible at first glance but you are leaving out the fact that the drive train absorbs more power when you apply more force on the gear faces, accelerate all the rotating pieces faster and to higher rpms.
For example, you can take the drive train out of the car and crank it by hand to some very slow speeds and obviously it doesn't take 50hp to do so. This is where the percentage rule comes in. An additional 100hp from a C6 over a C5 will definitely cause the drivetrain to heat up more. This afterall, is where the extra energy goes. Heat.
So don't rule out the percentage rule. It probably is still more accurate than a fixed loss. The difference comes from how many rpms you have to crank it as well as the forces applied.
For example, you can take the drive train out of the car and crank it by hand to some very slow speeds and obviously it doesn't take 50hp to do so. This is where the percentage rule comes in. An additional 100hp from a C6 over a C5 will definitely cause the drivetrain to heat up more. This afterall, is where the extra energy goes. Heat.
So don't rule out the percentage rule. It probably is still more accurate than a fixed loss. The difference comes from how many rpms you have to crank it as well as the forces applied.
Good food for thought.
One more example.
I think we've all seen a person mod, for example, their 2002 z06 which in stock spec dynos 355rwhp, for a suggested % loss of 50/405= 12.3%.
Then they do TT's, heads, cam, etc, and end up with 750 rwhp.
Following the % loss method, you would end up with
750/ .877 = 855 flywheel hp.
855-750= 105 hp loss, which can't be true, if the drivetrain remained the same.
In my theory, actual flywheel hp is closer to 750+50 = 800.
Look again at this example from my previous post.
Please explain to me how drivetrain loss doubles in this example?
Nothing in the drivetrain has changed.
The same parts are being used.
In your reply, I agree there may be additional parasitic losses from heat, but in my estimate they are very small.
Remember, the dyno pull lasts only 5-10 seconds, depending on engine power.
Heat is minimized.
RPM's are generally the same for most tests.
The reason these pieces accelerate faster is because of the greater torque generated by the engine.
The parts themselves don't generate any additional resistance.
The reason we see additional hp and torque on the dyno chart is because the engine is able to accelerate a given mass or load quicker.
The percentage rule can't be more accurate than a fixed loss, because if it was, under wide open throttle, which is where these parts are tested, your suggestion is that all of a sudden parts that required about 50 or so hp to turn them, now require over 100 hp to turn the same parts.
You're correct, we can always assign a percentage to the loss involved thru the drivetrain, but once the engine is making more power, all bets are off as to what that percentage is.
Maybe the % loss is a rising rate, such that as the engine begins to turn the dyno at low speeds, less hp is lost, and at higher rpms there is greater loss associated.
That's why I suggest it is easier and probably more accurate to assign a general amount of loss to a given drivetrain, rather than a %.
Then again, who really cares what the flywheel hp is, because we'll never really know, and it really only matters what is hitting the ground.
Personally, I wish someone with the ability to derive the actual mathematical truth to this question, thru mechanical engineering knowledge, would come on here and prove it one way or the other.
If I am wrong, then I'll gladly admit it and learn from it.
06-06-2007, 02:53 PM
#26
Le Mans Master
Good food for thought.
855-750= 105 hp loss, which can't be true, if the drivetrain remained the same.
In my theory, actual flywheel hp is closer to 750+50 = 800.[/B]
Look again at this example from my previous post.
Please explain to me how drivetrain loss doubles in this example?
Nothing in the drivetrain has changed.
The same parts are being used.
855-750= 105 hp loss, which can't be true, if the drivetrain remained the same.
In my theory, actual flywheel hp is closer to 750+50 = 800.[/B]
Look again at this example from my previous post.
Please explain to me how drivetrain loss doubles in this example?
Nothing in the drivetrain has changed.
The same parts are being used.
The reason these pieces accelerate faster is because of the greater torque generated by the engine.
The parts themselves don't generate any additional resistance.
The reason we see additional hp and torque on the dyno chart is because the engine is able to accelerate a given mass or load quicker.
The parts themselves don't generate any additional resistance.
The reason we see additional hp and torque on the dyno chart is because the engine is able to accelerate a given mass or load quicker.
The percentage rule can't be more accurate than a fixed loss, because if it was, under wide open throttle, which is where these parts are tested, your suggestion is that all of a sudden parts that required about 50 or so hp to turn them, now require over 100 hp to turn the same parts.
You're correct, we can always assign a percentage to the loss involved thru the drivetrain, but once the engine is making more power, all bets are off as to what that percentage is.
Maybe the % loss is a rising rate, such that as the engine begins to turn the dyno at low speeds, less hp is lost, and at higher rpms there is greater loss associated.
Maybe the % loss is a rising rate, such that as the engine begins to turn the dyno at low speeds, less hp is lost, and at higher rpms there is greater loss associated.
That's why I suggest it is easier and probably more accurate to assign a general amount of loss to a given drivetrain, rather than a %.
Then again, who really cares what the flywheel hp is, because we'll never really know, and it really only matters what is hitting the ground.
Then again, who really cares what the flywheel hp is, because we'll never really know, and it really only matters what is hitting the ground.
06-06-2007, 04:23 PM
#27
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Here's something I found while poking around on a few mechanical engineering sites.
RobHughes (Automotive)
3 Jun 04 23:58
I have been wondering about tranny power losses....wouldn't the loss be better defined as an actual number as opposed to a percentage.
Lets say motor X makes 100hp (at the crank) and looses 15hp with an auto and only 10hp with a manual. Now through the magic of the aftermarket the same engine now makes 200hp at the crank...will it loose 15 or 30hp with the auto and 10 or 20 with the manual? Oh yea...the transmission is the same.
Thanks guys!
NormPeterson (Structural)
4 Jun 04 6:55
I'd think more in terms of a fairly constant portion (for the parasitic losses) plus a portion that's a percentage of the amount of power being transmitted. With the above numbers, I'd guess a little less than 20 vs slightly under 25.
Norm
Now, that appears to be what you are saying.
Unfortunately, what I get from this is that, while it appears that loss is expressed in percentage, as power increases greatly, you still can not affix an accurate % to it unless you do the math!
So, back to square one, let's forget about accurate flywheel estimates and focus only on what a given dyno reads at the wheels!
It seems we both have some basis for our individual conclusions in this regard.
RobHughes (Automotive)
3 Jun 04 23:58
I have been wondering about tranny power losses....wouldn't the loss be better defined as an actual number as opposed to a percentage.
Lets say motor X makes 100hp (at the crank) and looses 15hp with an auto and only 10hp with a manual. Now through the magic of the aftermarket the same engine now makes 200hp at the crank...will it loose 15 or 30hp with the auto and 10 or 20 with the manual? Oh yea...the transmission is the same.
Thanks guys!
NormPeterson (Structural)
4 Jun 04 6:55
I'd think more in terms of a fairly constant portion (for the parasitic losses) plus a portion that's a percentage of the amount of power being transmitted. With the above numbers, I'd guess a little less than 20 vs slightly under 25.
Norm
Now, that appears to be what you are saying.
Unfortunately, what I get from this is that, while it appears that loss is expressed in percentage, as power increases greatly, you still can not affix an accurate % to it unless you do the math!
So, back to square one, let's forget about accurate flywheel estimates and focus only on what a given dyno reads at the wheels!
It seems we both have some basis for our individual conclusions in this regard.
Last edited by vrybad; 06-06-2007 at 04:27 PM.
06-07-2007, 08:53 PM
#28
Here's something I found while poking around on a few mechanical engineering sites.
RobHughes (Automotive)
3 Jun 04 23:58
I have been wondering about tranny power losses....wouldn't the loss be better defined as an actual number as opposed to a percentage.
Lets say motor X makes 100hp (at the crank) and looses 15hp with an auto and only 10hp with a manual. Now through the magic of the aftermarket the same engine now makes 200hp at the crank...will it loose 15 or 30hp with the auto and 10 or 20 with the manual? Oh yea...the transmission is the same.
Thanks guys!
NormPeterson (Structural)
4 Jun 04 6:55
I'd think more in terms of a fairly constant portion (for the parasitic losses) plus a portion that's a percentage of the amount of power being transmitted. With the above numbers, I'd guess a little less than 20 vs slightly under 25.
Norm
Now, that appears to be what you are saying.
Unfortunately, what I get from this is that, while it appears that loss is expressed in percentage, as power increases greatly, you still can not affix an accurate % to it unless you do the math!
So, back to square one, let's forget about accurate flywheel estimates and focus only on what a given dyno reads at the wheels!
It seems we both have some basis for our individual conclusions in this regard.
RobHughes (Automotive)
3 Jun 04 23:58
I have been wondering about tranny power losses....wouldn't the loss be better defined as an actual number as opposed to a percentage.
Lets say motor X makes 100hp (at the crank) and looses 15hp with an auto and only 10hp with a manual. Now through the magic of the aftermarket the same engine now makes 200hp at the crank...will it loose 15 or 30hp with the auto and 10 or 20 with the manual? Oh yea...the transmission is the same.
Thanks guys!
NormPeterson (Structural)
4 Jun 04 6:55
I'd think more in terms of a fairly constant portion (for the parasitic losses) plus a portion that's a percentage of the amount of power being transmitted. With the above numbers, I'd guess a little less than 20 vs slightly under 25.
Norm
Now, that appears to be what you are saying.
Unfortunately, what I get from this is that, while it appears that loss is expressed in percentage, as power increases greatly, you still can not affix an accurate % to it unless you do the math!
So, back to square one, let's forget about accurate flywheel estimates and focus only on what a given dyno reads at the wheels!
It seems we both have some basis for our individual conclusions in this regard.
Drivetrain HP loss on a chassis dyno is composed of the following:
1) second order or exponential losses like friction
2) first order or linear losses like inertia and gears
3) constants like windage of gears through the oil in the transmission/differential, shafts rotating through seals, and sidewall flex all at a particular RPM. In other words, those losses are independent of HP transmitted.
That leaves us with the equation HP (loss)= ax^2+bx+c where a, b, and c are constants and x is FWHP. "c" is the predominate term meaning most of the HP loss is a constant.
Other variables to consider is when a dyno run is done in a lower gear, the tire, transmission shafts/gears, and differential shafts/gears RPM is less reducing the HP loss to sidewall flex/windage and giving more RWHP. On a Dynojet, this is countered by a shorter dyno run increasing loss to inertia, not a factor on a Mustang dyno though since it keeps the time of the dyno run constant. Conversely, a dyno run in 5th or 6th on a Dynojet will have less loss to inertia (longer dyno run time) but more loss to windage/tires...but there's also a speed limitation on chassis dynos which prevents a run in 6th (unless you have some 4.56s in the rear). On the road, quicker acceleration in the lower gears results in more loss (as a percentage) to inertia, the same effect as a Dynojet.
As far as the 1:1 ratio of 4th gear being the best to dyno in, there is fact as a basis to that statement. Quality involute gears have a HP loss of ~1%, so when we dyno in 1st, 2nd, 3rd, 5th, or 6th, part of the drivetrain loss is ~2% of the HP transmitted through any of those gears (it's 2% because power flows through one set of gears from the input shaft to the countershaft then through another set of gears from the countershaft to the output shaft). You may notice the distinct absence of 4th gear in the list above...that's because there are no "gears" for 4th gear, something a lot of people don't know. When you make the shift to 4th gear, the synchro mechanism mechanically couples/locks the input shaft to the output shaft and power flows straight through the transmission making 4th gear the most efficient "gear" to dyno a car in as far as the transmission is concerned. It doesn't have anything to do with the fact the "ratio" is 1:1 since we could also put the power through gears at a 1:1 ratio and achieve the same output RPM.
Hopefully this shines a little light on the subject. You guys are on the right track, it's neither a percentage or a constant. But if you had to pick one, you'll be closer to FWHP if you use a constant. I.E. if you dyno a stock Z06 and come up with 450 RWHP for a 55 HP/10.9% loss, then mod it to 600 RWHP on the same dyno, you'll be much closer to the true FWHP if you add the 55 HP loss from stock rather than back calculate using the percentage number. And the guys mod their cars to 600+RWHP that just pull a percentage number out of their *** (like 15%) to back calculate their FWHP are just looking to stroke their ego with a big number and impress people. The only way to know the percentage is to remove the engine and put it on an engine dyno to get the FWHP. But then the first time you mod the engine, that number is not valid anymore.
And this statement you made is absolutely correct:
"So, back to square one, let's forget about accurate flywheel estimates and focus only on what a given dyno reads at the wheels!"
06-07-2007, 09:23 PM
#29
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If we're talking about RWHP on a dyno, it depends on the type of dyno used. A Mustang dyno can put a load on a car and control the rate of engine acceleration (250 RPM/sec, 500 RPM/sec or whatever) or it can hold the RPM constant. That way, HP loss due to the inertia of engine components, drivetrain components and wheels/tires/rotors are held constant. On a Dynojet dyno, there is only the inertia of the drum to resist the torque at the wheels so when we add more HP or dyno in a lower gear, the time for the dyno run is less which means more HP loss to inertia.
Drivetrain HP loss on a chassis dyno is composed of the following:
1) second order or exponential losses like friction
2) first order or linear losses like inertia and gears
3) constants like windage of gears through the oil in the transmission/differential, shafts rotating through seals, and sidewall flex all at a particular RPM. In other words, those losses are independent of HP transmitted.
That leaves us with the equation HP (loss)= ax^2+bx+c where a, b, and c are constants and x is FWHP. "c" is the predominate term meaning most of the HP loss is a constant.
Other variables to consider is when a dyno run is done in a lower gear, the tire, transmission shafts/gears, and differential shafts/gears RPM is less reducing the HP loss to sidewall flex/windage and giving more RWHP. On a Dynojet, this is countered by a shorter dyno run increasing loss to inertia, not a factor on a Mustang dyno though since it keeps the time of the dyno run constant. Conversely, a dyno run in 5th or 6th on a Dynojet will have less loss to inertia (longer dyno run time) but more loss to windage/tires...but there's also a speed limitation on chassis dynos which prevents a run in 6th (unless you have some 4.56s in the rear). On the road, quicker acceleration in the lower gears results in more loss (as a percentage) to inertia, the same effect as a Dynojet.
As far as the 1:1 ratio of 4th gear being the best to dyno in, there is fact as a basis to that statement. Quality involute gears have a HP loss of ~1%, so when we dyno in 1st, 2nd, 3rd, 5th, or 6th, part of the drivetrain loss is ~2% of the HP transmitted through any of those gears (it's 2% because power flows through one set of gears from the input shaft to the countershaft then through another set of gears from the countershaft to the output shaft). You may notice the distinct absence of 4th gear in the list above...that's because there are no "gears" for 4th gear, something a lot of people don't know. When you make the shift to 4th gear, the synchro mechanism mechanically couples/locks the input shaft to the output shaft and power flows straight through the transmission making 4th gear the most efficient "gear" to dyno a car in as far as the transmission is concerned. It doesn't have anything to do with the fact the "ratio" is 1:1 since we could also put the power through gears at a 1:1 ratio and achieve the same output RPM.
Hopefully this shines a little light on the subject. You guys are on the right track, it's neither a percentage or a constant. But if you had to pick one, you'll be closer to FWHP if you use a constant. I.E. if you dyno a stock Z06 and come up with 450 RWHP for a 55 HP/10.9% loss, then mod it to 600 RWHP on the same dyno, you'll be much closer to the true FWHP if you add the 55 HP loss from stock rather than back calculate using the percentage number. And the guys mod their cars to 600+RWHP that just pull a percentage number out of their *** (like 15%) to back calculate their FWHP are just looking to stroke their ego with a big number and impress people. The only way to know the percentage is to remove the engine and put it on an engine dyno to get the FWHP. But then the first time you mod the engine, that number is not valid anymore.
And this statement you made is absolutely correct:
"So, back to square one, let's forget about accurate flywheel estimates and focus only on what a given dyno reads at the wheels!"
Drivetrain HP loss on a chassis dyno is composed of the following:
1) second order or exponential losses like friction
2) first order or linear losses like inertia and gears
3) constants like windage of gears through the oil in the transmission/differential, shafts rotating through seals, and sidewall flex all at a particular RPM. In other words, those losses are independent of HP transmitted.
That leaves us with the equation HP (loss)= ax^2+bx+c where a, b, and c are constants and x is FWHP. "c" is the predominate term meaning most of the HP loss is a constant.
Other variables to consider is when a dyno run is done in a lower gear, the tire, transmission shafts/gears, and differential shafts/gears RPM is less reducing the HP loss to sidewall flex/windage and giving more RWHP. On a Dynojet, this is countered by a shorter dyno run increasing loss to inertia, not a factor on a Mustang dyno though since it keeps the time of the dyno run constant. Conversely, a dyno run in 5th or 6th on a Dynojet will have less loss to inertia (longer dyno run time) but more loss to windage/tires...but there's also a speed limitation on chassis dynos which prevents a run in 6th (unless you have some 4.56s in the rear). On the road, quicker acceleration in the lower gears results in more loss (as a percentage) to inertia, the same effect as a Dynojet.
As far as the 1:1 ratio of 4th gear being the best to dyno in, there is fact as a basis to that statement. Quality involute gears have a HP loss of ~1%, so when we dyno in 1st, 2nd, 3rd, 5th, or 6th, part of the drivetrain loss is ~2% of the HP transmitted through any of those gears (it's 2% because power flows through one set of gears from the input shaft to the countershaft then through another set of gears from the countershaft to the output shaft). You may notice the distinct absence of 4th gear in the list above...that's because there are no "gears" for 4th gear, something a lot of people don't know. When you make the shift to 4th gear, the synchro mechanism mechanically couples/locks the input shaft to the output shaft and power flows straight through the transmission making 4th gear the most efficient "gear" to dyno a car in as far as the transmission is concerned. It doesn't have anything to do with the fact the "ratio" is 1:1 since we could also put the power through gears at a 1:1 ratio and achieve the same output RPM.
Hopefully this shines a little light on the subject. You guys are on the right track, it's neither a percentage or a constant. But if you had to pick one, you'll be closer to FWHP if you use a constant. I.E. if you dyno a stock Z06 and come up with 450 RWHP for a 55 HP/10.9% loss, then mod it to 600 RWHP on the same dyno, you'll be much closer to the true FWHP if you add the 55 HP loss from stock rather than back calculate using the percentage number. And the guys mod their cars to 600+RWHP that just pull a percentage number out of their *** (like 15%) to back calculate their FWHP are just looking to stroke their ego with a big number and impress people. The only way to know the percentage is to remove the engine and put it on an engine dyno to get the FWHP. But then the first time you mod the engine, that number is not valid anymore.
And this statement you made is absolutely correct:
"So, back to square one, let's forget about accurate flywheel estimates and focus only on what a given dyno reads at the wheels!"
I appreciate your response!
You seem to have a pretty darn good grasp of this subject, and I personally enjoy the whole discussion.
What I've highlighted in bold in your response is part of my initial point about this whole subject.
Everyone wants to have the most power that they paid for, so the numbers eventually get very inflated when a fixed percentage is used to estimate flywheel hp, especially as rwhp numbers increase by a large margin.
Thanks again for getting involved in this thread!
Last edited by vrybad; 06-07-2007 at 09:25 PM. Reason: spelling
06-08-2007, 12:28 AM
#30
Melting Slicks
You are 100%correct on this. Don't forget "wear" as this makes a much more inefficient power transmission.
Rule of thumb: @ new,perfectly mated straight cut gears will reduce torque between 2 aligned shafts by 7%. Now, 2 helical cut gears changing direction (such as a rearend) will reduce about 11%. This is for perfectly mated, "machined as a set" gears. Worn gears will only escalate the reduction. You are also correct in oil being critical to efficiency. Ok, now I'll get off of my tribologist & mechanical engineering soapbox.
Rule of thumb: @ new,perfectly mated straight cut gears will reduce torque between 2 aligned shafts by 7%. Now, 2 helical cut gears changing direction (such as a rearend) will reduce about 11%. This is for perfectly mated, "machined as a set" gears. Worn gears will only escalate the reduction. You are also correct in oil being critical to efficiency. Ok, now I'll get off of my tribologist & mechanical engineering soapbox.
Most rear ends, including our Corvettes don't use helical gears, they use hypoid gears. The loss is in the nighborhood of 5% for a set of hypoid gears.
You also over rate the effects of reasonable wear. Reasonably well cared for automotive gears at 100,000 miles aren't much lossier than just broken in gears.
