Interesting Concept on Drivetrain Loss...
Teching In
Joined: Mar 2004
Posts: 29
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From: Independence, KY
Originally Posted by obZidian
I understood that you meant it as a bad exmple but i took your reasoning and enforced my own.
On a dyno, your not specifically measuring acceleration rate but power or effort force. Due to the nature of a dyno and since you are simulating a run, you are gradually raising your rpm to a specific point than let it rip, thus your initial force needed will not be as great as if you were at the line and hitting the throttle with all of your might. The power needed from a stop is greater than from a roll.
Also, you are taking my words out of context, please re-read and im sure yu will still back your statement after you have read it....
On a dyno, your not specifically measuring acceleration rate but power or effort force. Due to the nature of a dyno and since you are simulating a run, you are gradually raising your rpm to a specific point than let it rip, thus your initial force needed will not be as great as if you were at the line and hitting the throttle with all of your might. The power needed from a stop is greater than from a roll.
Also, you are taking my words out of context, please re-read and im sure yu will still back your statement after you have read it....
Apparently I didn't understand what you were saying if I took them out of context, so please expand on your point so I do not take it out of context. If you increase your power needed due to drivetrain losses from one reading to another, you are not using a fixed number for power loss.
Like solso posted, frictional losses are proportional to force applied.
TECH Junkie
Joined: Dec 2001
Posts: 3,242
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From: Lakeland, FL
I have both engine dyno'd and chassis dyno'd my engine at one point and it was real damn close to 15%
590 engine dyno, 500 to the wheels
Think of it like this when you have more power you are also turning the drivetrain at a faster rate which takes more power.
Go outside and spin a log, then spin it faster, it is easy to see just how much extra power can be used using this method. Its weird but but if you think about it makes sense.
A lower power car uses less power to spin the drivetrain but also cannot spin it as fast because it has less power.
A higher power motor could spin it at the same rate as the lesser power motor and use the same power but since it can't do this under WOT it will spin it faster which takes more power.
This can be seen with almost anything that you can spin, try spinning slowly and then really fast. Which one is harder?
590 engine dyno, 500 to the wheels
Think of it like this when you have more power you are also turning the drivetrain at a faster rate which takes more power.
Go outside and spin a log, then spin it faster, it is easy to see just how much extra power can be used using this method. Its weird but but if you think about it makes sense.
A lower power car uses less power to spin the drivetrain but also cannot spin it as fast because it has less power.
A higher power motor could spin it at the same rate as the lesser power motor and use the same power but since it can't do this under WOT it will spin it faster which takes more power.
This can be seen with almost anything that you can spin, try spinning slowly and then really fast. Which one is harder?
TECH Apprentice
Joined: Jan 2004
Posts: 337
Likes: 0
From: Frisco TX (Dallas Area)
Originally Posted by Mike K.
I have both engine dyno'd and chassis dyno'd my engine at one point and it was real damn close to 15%
590 engine dyno, 500 to the wheels
Think of it like this when you have more power you are also turning the drivetrain at a faster rate which takes more power.
Go outside and spin a log, then spin it faster, it is easy to see just how much extra power can be used using this method. Its weird but but if you think about it makes sense.
A lower power car uses less power to spin the drivetrain but also cannot spin it as fast because it has less power.
A higher power motor could spin it at the same rate as the lesser power motor and use the same power but since it can't do this under WOT it will spin it faster which takes more power.
This can be seen with almost anything that you can spin, try spinning slowly and then really fast. Which one is harder?
590 engine dyno, 500 to the wheels
Think of it like this when you have more power you are also turning the drivetrain at a faster rate which takes more power.
Go outside and spin a log, then spin it faster, it is easy to see just how much extra power can be used using this method. Its weird but but if you think about it makes sense.
A lower power car uses less power to spin the drivetrain but also cannot spin it as fast because it has less power.
A higher power motor could spin it at the same rate as the lesser power motor and use the same power but since it can't do this under WOT it will spin it faster which takes more power.
This can be seen with almost anything that you can spin, try spinning slowly and then really fast. Which one is harder?
TECH Apprentice
Joined: Jan 2004
Posts: 337
Likes: 0
From: Frisco TX (Dallas Area)
BTW, guys and gals, this thread has shown the maturity of the folks on the forum. This has always been a hotly debated issue with strong support in both camps, and with lots of opportunity for personalities and tempers to surface. And yet we have all tried to be logical, thoughtful and CIVIL. Props to you all, we get a lot smarter this way! 
Someone who claims the 15% rule applies everywhere explain to me why on the same dyno during the very same evening a "100 shot" of nitrous (only a .052 orifice of N2O) yeilded me 126 more HP at the tire. That would mean I really gained 150 at the flywheel (assuming the 15% rule to be true)....HIGHLY unlikely with that small a pill. I bet I made damn close to 126 more at the flywheel which found its way "net" to the back tires because the losses in my driveline were already accounted for. That is a fairly serious gain at the crank for such a small dose of juice.
Guys, an engine with more power doesn't mysteriously cause your driveline to become less efficient....The driveline (a fixed rotating mass) spins quicker due to the fact something more powerful is turning it. If you already have the losses in the driveline accounted for, you will see the bulk of any additional power made at the flywheel make its way to the rear tires. You might lose a small amount to additional heat, friction, and perhaps additional flexing of some of the components which would cause the driveline to rob a LITTLE more power (and become less efficient assuming the parts might be borderline for the application), but it won't be anything close to 15% IMHO. Also, modern synthetic fluids could help reduce the additional losses in the driveline due to their ability to withstand heat better and put up with higher shear forces and loads placed upon them from the higher output engine in this potential scenario.
Again...I will be doing more actual testing of this theory in the very near future.
Guys, an engine with more power doesn't mysteriously cause your driveline to become less efficient....The driveline (a fixed rotating mass) spins quicker due to the fact something more powerful is turning it. If you already have the losses in the driveline accounted for, you will see the bulk of any additional power made at the flywheel make its way to the rear tires. You might lose a small amount to additional heat, friction, and perhaps additional flexing of some of the components which would cause the driveline to rob a LITTLE more power (and become less efficient assuming the parts might be borderline for the application), but it won't be anything close to 15% IMHO. Also, modern synthetic fluids could help reduce the additional losses in the driveline due to their ability to withstand heat better and put up with higher shear forces and loads placed upon them from the higher output engine in this potential scenario.
Again...I will be doing more actual testing of this theory in the very near future.
Last edited by Tony Mamo @ AFR; May 9, 2005 at 01:26 AM.
TECH Junkie
Joined: Dec 2001
Posts: 3,242
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From: Lakeland, FL
Originally Posted by Tony Mamo @ AFR
Someone who claims the 15% rule applies everywhere explain to me why on the same dyno during the very same evening a "100 shot" of nitrous (only a .052 orifice of N2O) yeilded me 126 more HP at the tire. That would mean I really gained 150 at the flywheel (assuming the 15% rule to be true)....HIGHLY unlikely with that small a pill. I bet I made damn close to 126 more at the flywheel which found its way "net" to the back tires because the losses in my driveline were already accounted for. That is a fairly serious gain at the crank for such a small dose of juice.
Guys, an engine with more power doesn't mysteriously cause your driveline to become less efficient....The driveline (a fixed rotating mass) spins quicker due to the fact something more powerful is turning it. If you already have the losses in the driveline accounted for, you will see the bulk of any additional power made at the flywheel make its way to the rear tires. You might lose a small amount to additional heat, friction, and perhaps additional flexing of some of the components which would cause the driveline to rob a LITTLE more power (and become less efficient assuming the parts might be borderline for the application), but it won't be anything close to 15% IMHO. Also, modern synthetic fluids could help reduce the additional losses in the driveline due to their ability to withstand heat better and put up with higher shear forces and loads placed upon them from the higher output engine in this potential scenario.
Again...I will be doing more actual testing of this theory in the very near future.
Guys, an engine with more power doesn't mysteriously cause your driveline to become less efficient....The driveline (a fixed rotating mass) spins quicker due to the fact something more powerful is turning it. If you already have the losses in the driveline accounted for, you will see the bulk of any additional power made at the flywheel make its way to the rear tires. You might lose a small amount to additional heat, friction, and perhaps additional flexing of some of the components which would cause the driveline to rob a LITTLE more power (and become less efficient assuming the parts might be borderline for the application), but it won't be anything close to 15% IMHO. Also, modern synthetic fluids could help reduce the additional losses in the driveline due to their ability to withstand heat better and put up with higher shear forces and loads placed upon them from the higher output engine in this potential scenario.
Again...I will be doing more actual testing of this theory in the very near future.
I can't wait to hear your results, I had always thought like you do until I saw the 15% loss with my own eyes with my own engine. I really think it has to do with the more powerful engine turning it faster which takes more hp/energy.
My point is that the drivetrain doesnt have a fixed point at ALL power levels however i feel a percentage is again a generic way of measuring fwhp from a rwhp number.
Wats up mike? Nice cam!!!!!!!!!!!!!!! Gonna get my tuned soon, so watxh out now!!!!
Back to regularly scheduled programing....
A fixed number might not yeild the correct numbers we seek but a percentae allows for to much loss to be calculated out. i understand the proportionalities, its quit simple. However, i feel that though you DO lose more hp through a faster rate of rotation, frcition, load, etc; a variable, scalar number, and not a fixed percentage, is better able to secribe the amount of power lose on an engine dyno...
keep em coming folks, by far one of the best threads ever....hey, weres the originator of this post at?
Wats up mike? Nice cam!!!!!!!!!!!!!!! Gonna get my tuned soon, so watxh out now!!!!
Back to regularly scheduled programing....
A fixed number might not yeild the correct numbers we seek but a percentae allows for to much loss to be calculated out. i understand the proportionalities, its quit simple. However, i feel that though you DO lose more hp through a faster rate of rotation, frcition, load, etc; a variable, scalar number, and not a fixed percentage, is better able to secribe the amount of power lose on an engine dyno...
keep em coming folks, by far one of the best threads ever....hey, weres the originator of this post at?
Originally Posted by Tony Mamo @ AFR
Someone who claims the 15% rule applies everywhere explain to me why on the same dyno during the very same evening a "100 shot" of nitrous (only a .052 orifice of N2O) yeilded me 126 more HP at the tire. That would mean I really gained 150 at the flywheel (assuming the 15% rule to be true)....HIGHLY unlikely with that small a pill. I bet I made damn close to 126 more at the flywheel which found its way "net" to the back tires because the losses in my driveline were already accounted for. That is a fairly serious gain at the crank for such a small dose of juice.
Guys, an engine with more power doesn't mysteriously cause your driveline to become less efficient....The driveline (a fixed rotating mass) spins quicker due to the fact something more powerful is turning it. If you already have the losses in the driveline accounted for, you will see the bulk of any additional power made at the flywheel make its way to the rear tires. You might lose a small amount to additional heat, friction, and perhaps additional flexing of some of the components which would cause the driveline to rob a LITTLE more power (and become less efficient assuming the parts might be borderline for the application), but it won't be anything close to 15% IMHO. Also, modern synthetic fluids could help reduce the additional losses in the driveline due to their ability to withstand heat better and put up with higher shear forces and loads placed upon them from the higher output engine in this potential scenario.
Again...I will be doing more actual testing of this theory in the very near future.
Guys, an engine with more power doesn't mysteriously cause your driveline to become less efficient....The driveline (a fixed rotating mass) spins quicker due to the fact something more powerful is turning it. If you already have the losses in the driveline accounted for, you will see the bulk of any additional power made at the flywheel make its way to the rear tires. You might lose a small amount to additional heat, friction, and perhaps additional flexing of some of the components which would cause the driveline to rob a LITTLE more power (and become less efficient assuming the parts might be borderline for the application), but it won't be anything close to 15% IMHO. Also, modern synthetic fluids could help reduce the additional losses in the driveline due to their ability to withstand heat better and put up with higher shear forces and loads placed upon them from the higher output engine in this potential scenario.
Again...I will be doing more actual testing of this theory in the very near future.
Let's look at Mike's example. His motor made 590 FWHP and 500 RWHP. That is roughly a 15% loss from the drivetrain, and 90 HP. If his drivetrain required 90 HP to operate, a stock LS1 would have made around 250-260 RWHP going through that same drivetrain. I don't know the specifics of his drivetrain (maybe he can give some more insight) but I'd venture to say that a bone stock LS1 going through a TH400 and a 9" would still make at a minimum 275 rwhp, a ~ 65-75 HP loss. I just don't see, even a VERY strong (and power consuming) drivetrain like the aforementioned robbing 90 HP on a 340-350 HP motor. Assuming Mike was running a 6 speed and a 12 bolt, robbing 90 HP from a stock LS1 is still, IMO, way off.
Your comments suggest that a 70 HP engine couldn't turn his drivetrain, which is HIGHLY unlikely. The HP robbed is dependent upon the acceleration and terminal speed the engine/drivetrain is required to spin.
Again, I don't know any specifics about his drivetrain so these are just speculations.
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Verdad Gallardo TECH Apprentice
Joined: Jan 2004
Posts: 337
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From: Frisco TX (Dallas Area)
Original: ObZidian:
My point is that the drivetrain doesnt have a fixed point at ALL power levels however i feel a percentage is again a generic way of measuring fwhp from a rwhp number.
...
My point is that the drivetrain doesnt have a fixed point at ALL power levels however i feel a percentage is again a generic way of measuring fwhp from a rwhp number.
...
If the ME at 590 chp is 85%, then the loss is 90 hp, or 500 whp (what he said). Friction is by far the major contributor to loss, so the loss is the power required to overcome the drivetrain friction. If the loss is 90 hp with an 800 chp engine (the same loss as with 590 chp), then the ME has magically increased to 89%. I find that hard to envision because that says that there is no frictional increase as you add power. We all know that you can turn the darned thing virtually by hand with no load. If there IS no frictional increase, why would it take 90 hp to turn the drivetrain when you can turn it by hand in the first place?
(Hint: higher load increases the frictional loss in direct proportion to the power applied.)
TECH Junkie
Joined: Dec 2001
Posts: 3,242
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From: Lakeland, FL
Originally Posted by TeeKay
I am totally convinced from measured testing that the drivetrain does have a fixed point of reference, and that reference is the ME expressed as a percent.
If the ME at 590 chp is 85%, then the loss is 90 hp, or 500 whp (what he said). Friction is by far the major contributor to loss, so the loss is the power required to overcome the drivetrain friction. If the loss is 90 hp with an 800 chp engine (the same loss as with 590 chp), then the ME has magically increased to 89%. I find that hard to envision because that says that there is no frictional increase as you add power. We all know that you can turn the darned thing virtually by hand with no load. If there IS no frictional increase, why would it take 90 hp to turn the drivetrain when you can turn it by hand in the first place?
(Hint: higher load increases the frictional loss in direct proportion to the power applied.)
If the ME at 590 chp is 85%, then the loss is 90 hp, or 500 whp (what he said). Friction is by far the major contributor to loss, so the loss is the power required to overcome the drivetrain friction. If the loss is 90 hp with an 800 chp engine (the same loss as with 590 chp), then the ME has magically increased to 89%. I find that hard to envision because that says that there is no frictional increase as you add power. We all know that you can turn the darned thing virtually by hand with no load. If there IS no frictional increase, why would it take 90 hp to turn the drivetrain when you can turn it by hand in the first place?
(Hint: higher load increases the frictional loss in direct proportion to the power applied.)
yep my point exactly!
Originally Posted by TeeKay
I am totally convinced from measured testing that the drivetrain does have a fixed point of reference, and that reference is the ME expressed as a percent.
If the ME at 590 chp is 85%, then the loss is 90 hp, or 500 whp (what he said). Friction is by far the major contributor to loss, so the loss is the power required to overcome the drivetrain friction. If the loss is 90 hp with an 800 chp engine (the same loss as with 590 chp), then the ME has magically increased to 89%. I find that hard to envision because that says that there is no frictional increase as you add power. We all know that you can turn the darned thing virtually by hand with no load. If there IS no frictional increase, why would it take 90 hp to turn the drivetrain when you can turn it by hand in the first place?
(Hint: higher load increases the frictional loss in direct proportion to the power applied.)
If the ME at 590 chp is 85%, then the loss is 90 hp, or 500 whp (what he said). Friction is by far the major contributor to loss, so the loss is the power required to overcome the drivetrain friction. If the loss is 90 hp with an 800 chp engine (the same loss as with 590 chp), then the ME has magically increased to 89%. I find that hard to envision because that says that there is no frictional increase as you add power. We all know that you can turn the darned thing virtually by hand with no load. If there IS no frictional increase, why would it take 90 hp to turn the drivetrain when you can turn it by hand in the first place?
(Hint: higher load increases the frictional loss in direct proportion to the power applied.)
But, As power increase, yes, friction will also increase due to a greater amount of load but the machine is now also more efficient, due to increase of effort power and if reasonable percautions are meet to maintain that efficiency. Imo of course. though youa re losing more, you now have a more efficient system...Back to the original, a percentage, and this IS my point and not the fixed number theory, which is argueable, is a generic way to calculate drivetrain loss when solely based on types of transmissions and not on the actual components and materials used within that specific trans.
11 Second Club
Joined: Jul 2003
Posts: 2,311
Likes: 2
From: loudoun county,va
Most of the time i am in agreement with the things you post Tony(why i bought afr heads) but this time i disagree...
This is true... but what you are missing is that it does not make it any MORE efficient either. It remains for the most part the same. Perhaps not a true percentage persay, but parasitic loss does increase grow as the power applied to the drivetrain increases...
heres why,
i think what is wrapping most around the axle is that the weight of the drivetrain is rotating. but dont think about it that way, think about it as just what it is, weight...
if you do that then the rest becomes easy to understand...
i have a car that has a fixed weight...
it goes 0-60 in 5.0 seconds...
it has 300 rwhp
this is its RATE of acceleration , in relation to its weight and power.
now i want to make it go from 0-60 in 4.0 seconds
i can only do one of 2 things to make that happen, 1. lose weight,2 increase power.
now lets go back to the drivetrain, it works the same way.
the weight of the drivetrain is the fixed number.
in order to increase its RATE of acceleration, it will demand more power.
proof of this is simple.
lets take a auto set up with 20% stock drivtrain loss...
given this "stock" amount of loss. and the fact that most stock LS1 autos dyno about 280 rwhp then our motor power is 350 hp.
that means our loss was 70 hp... if your theory of loss is equal no matter the power applied, then you should HAVE to make greater then 70 hp in order to move that drivetrain, but if you put that same stock car on the dyno you can easly disprove this... simply give it only enough gas to get it moving at the same speed... i bet you could do that without breaking 70 hp... infact, with no resistance from the dyno, i am sure you could do it with less then half that.
hey, like anyone, i could be wronge... i would really like to see proof one way or the other...
Originally Posted by Tony Mamo @ AFR
Guys, an engine with more power doesn't mysteriously cause your driveline to become less efficient....
heres why,
i think what is wrapping most around the axle is that the weight of the drivetrain is rotating. but dont think about it that way, think about it as just what it is, weight...
if you do that then the rest becomes easy to understand...
i have a car that has a fixed weight...
it goes 0-60 in 5.0 seconds...
it has 300 rwhp
this is its RATE of acceleration , in relation to its weight and power.
now i want to make it go from 0-60 in 4.0 seconds
i can only do one of 2 things to make that happen, 1. lose weight,2 increase power.
now lets go back to the drivetrain, it works the same way.
the weight of the drivetrain is the fixed number.
in order to increase its RATE of acceleration, it will demand more power.
proof of this is simple.
lets take a auto set up with 20% stock drivtrain loss...
given this "stock" amount of loss. and the fact that most stock LS1 autos dyno about 280 rwhp then our motor power is 350 hp.
that means our loss was 70 hp... if your theory of loss is equal no matter the power applied, then you should HAVE to make greater then 70 hp in order to move that drivetrain, but if you put that same stock car on the dyno you can easly disprove this... simply give it only enough gas to get it moving at the same speed... i bet you could do that without breaking 70 hp... infact, with no resistance from the dyno, i am sure you could do it with less then half that.
hey, like anyone, i could be wronge... i would really like to see proof one way or the other...
11 Second Club
Joined: Jul 2003
Posts: 2,311
Likes: 2
From: loudoun county,va
jaberwaki, very interesting finds...indeed.
By far the car and driver article really caught my attention. I knew that air flow would also affect the shown hp/tq but that is a very intersting back to back comparison.
It seems that the main view or point is the fixed percentage, which after viewing most of the data here and on a few of your articles and others, seems to hold water. But ther is something about it that just doesnt site well with me. Maybe its something that is being over looked or not understood completely. I still feel to establish a static percentage is generic but i also understand that a static whole number is also false. Or off....a combination or not only frictional factororials but cooling, drag, load, temp, air temp and air flow rate, how the car feels like behaving in that particluar day..etc.
It all seems futile to attempt to calculate the actual, vehicle specific flywheel power rating in order to compare applications. The power to the ground is wat makes it all work. I feel that though the industry standard is flywheel rating becuase of the inconsistancies in rwhp, we should not worry so mucha bout the actual calculation of fwhp until we have an experiment were different drivetrain components have been disected and an acurate variable scalar number/percentage/fraction/watevr, is found per application.
I really like this thread, let see wat else i can find to stir up the pot!!
By far the car and driver article really caught my attention. I knew that air flow would also affect the shown hp/tq but that is a very intersting back to back comparison.
It seems that the main view or point is the fixed percentage, which after viewing most of the data here and on a few of your articles and others, seems to hold water. But ther is something about it that just doesnt site well with me. Maybe its something that is being over looked or not understood completely. I still feel to establish a static percentage is generic but i also understand that a static whole number is also false. Or off....a combination or not only frictional factororials but cooling, drag, load, temp, air temp and air flow rate, how the car feels like behaving in that particluar day..etc.
It all seems futile to attempt to calculate the actual, vehicle specific flywheel power rating in order to compare applications. The power to the ground is wat makes it all work. I feel that though the industry standard is flywheel rating becuase of the inconsistancies in rwhp, we should not worry so mucha bout the actual calculation of fwhp until we have an experiment were different drivetrain components have been disected and an acurate variable scalar number/percentage/fraction/watevr, is found per application.
I really like this thread, let see wat else i can find to stir up the pot!!
TECH Junkie
Joined: Dec 2001
Posts: 3,242
Likes: 0
From: Lakeland, FL
Originally Posted by obZidian
jaberwaki, very interesting finds...indeed.
By far the car and driver article really caught my attention. I knew that air flow would also affect the shown hp/tq but that is a very intersting back to back comparison.
It seems that the main view or point is the fixed percentage, which after viewing most of the data here and on a few of your articles and others, seems to hold water. But ther is something about it that just doesnt site well with me. Maybe its something that is being over looked or not understood completely. I still feel to establish a static percentage is generic but i also understand that a static whole number is also false. Or off....a combination or not only frictional factororials but cooling, drag, load, temp, air temp and air flow rate, how the car feels like behaving in that particluar day..etc.
It all seems futile to attempt to calculate the actual, vehicle specific flywheel power rating in order to compare applications. The power to the ground is wat makes it all work. I feel that though the industry standard is flywheel rating becuase of the inconsistancies in rwhp, we should not worry so mucha bout the actual calculation of fwhp until we have an experiment were different drivetrain components have been disected and an acurate variable scalar number/percentage/fraction/watevr, is found per application.
I really like this thread, let see wat else i can find to stir up the pot!!
By far the car and driver article really caught my attention. I knew that air flow would also affect the shown hp/tq but that is a very intersting back to back comparison.
It seems that the main view or point is the fixed percentage, which after viewing most of the data here and on a few of your articles and others, seems to hold water. But ther is something about it that just doesnt site well with me. Maybe its something that is being over looked or not understood completely. I still feel to establish a static percentage is generic but i also understand that a static whole number is also false. Or off....a combination or not only frictional factororials but cooling, drag, load, temp, air temp and air flow rate, how the car feels like behaving in that particluar day..etc.
It all seems futile to attempt to calculate the actual, vehicle specific flywheel power rating in order to compare applications. The power to the ground is wat makes it all work. I feel that though the industry standard is flywheel rating becuase of the inconsistancies in rwhp, we should not worry so mucha bout the actual calculation of fwhp until we have an experiment were different drivetrain components have been disected and an acurate variable scalar number/percentage/fraction/watevr, is found per application.
I really like this thread, let see wat else i can find to stir up the pot!!
Just look at the percentages as good estimates but not exact, its just a round about way to guess flywheel horsepower, only way to get exact is to put on engine dyno vs chassis.
11 Second Club
Joined: Jul 2003
Posts: 2,311
Likes: 2
From: loudoun county,va
I believe now, after reading all this, and also researching the subject because of this thread, that the real answer is that both are wronge...
there is no held true percentage...and there is no flat rate load...
instead i believe that it is one of those BIG math formulas that would be beyond most peoples ability to , one fathom,and 2 find all needed factors to work the equation....
somethings that would contribute to it would be,
1.friction, and as a byproduct the heat it generates.
2.weight and how rotating mass acts heavier the faster it is spun
3.the harmonics of each and every moving parts and the role it plays in adding resistance
4. all of these things would also effect the drum on the dyno
5. the reaction of the motor to a LOAD, it puts more stress on the parts and therefore may effect the power output.
these are the 5 i came up with....
any more?
there is no held true percentage...and there is no flat rate load...
instead i believe that it is one of those BIG math formulas that would be beyond most peoples ability to , one fathom,and 2 find all needed factors to work the equation....
somethings that would contribute to it would be,
1.friction, and as a byproduct the heat it generates.
2.weight and how rotating mass acts heavier the faster it is spun
3.the harmonics of each and every moving parts and the role it plays in adding resistance
4. all of these things would also effect the drum on the dyno
5. the reaction of the motor to a LOAD, it puts more stress on the parts and therefore may effect the power output.
these are the 5 i came up with....
any more?
https://ls1tech.com/forums/gears-axles/9518-finally-answer-driveline-loss.html
Here is the same thing from a few years ago. It may help, it may not.
Here is the same thing from a few years ago. It may help, it may not.
TECH Apprentice
Joined: Jan 2004
Posts: 337
Likes: 0
From: Frisco TX (Dallas Area)
Originally Posted by jaberwaki
I believe now, after reading all this, and also researching the subject because of this thread, that the real answer is that both are wronge...
there is no held true percentage...and there is no flat rate load...
instead i believe that it is one of those BIG math formulas that would be beyond most peoples ability to , one fathom,and 2 find all needed factors to work the equation....
somethings that would contribute to it would be,
1.friction, and as a byproduct the heat it generates.
2.weight and how rotating mass acts heavier the faster it is spun
3.the harmonics of each and every moving parts and the role it plays in adding resistance
4. all of these things would also effect the drum on the dyno
5. the reaction of the motor to a LOAD, it puts more stress on the parts and therefore may effect the power output.
these are the 5 i came up with....
any more?
there is no held true percentage...and there is no flat rate load...
instead i believe that it is one of those BIG math formulas that would be beyond most peoples ability to , one fathom,and 2 find all needed factors to work the equation....
somethings that would contribute to it would be,
1.friction, and as a byproduct the heat it generates.
2.weight and how rotating mass acts heavier the faster it is spun
3.the harmonics of each and every moving parts and the role it plays in adding resistance
4. all of these things would also effect the drum on the dyno
5. the reaction of the motor to a LOAD, it puts more stress on the parts and therefore may effect the power output.
these are the 5 i came up with....
any more?
Thomas Edison once hired a newly graduated engineer for his well established business (I believe in the '30's). The first morning, as an initial evaluation of the new guy's capabilities, he handed the young man an Edison light bulb. You know, the one with the funky tip on the top. He asked the guy to find the volume of the light bulb, then left the fellow alone.
Edison returned about lunch-time and asked the young man if he had the answer. The fellow replied that the shape was so complex that he had to make several assumptions and approximations, but that he had a good estimate. Edison replied that he wanted the EXACT number.
The poor fellow said that wasn't possible with the math of the day. Edison said nothing, but partially filled a lab beaker with water and then dunked the light bulb in it and measured the difference between the two water levels, giving the volume of the light bulb to 2 decimals. Edison said to the young engineer, "Now that's the exact volume!"
One last time, Mechanical Efficiency = Power Output/Power Input x 100% (by definition). All you have to do is measure the 2 values (power in and power out) for any given drivetrain and that ME is then true forever for that drive train and covers all variables. You CAN change the efficiency of the drive train (I posted earlier), but if you don't, then that's the ME you'll get whatever you put in, and it is the PERCENT of power input that is the loss you will experience. Like it or don't, believe it or don't, intuitive or not, that's the truth of the matter.
Good luck guys.