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Some people think that drivetrain losses are static, others think that it's percentage based. Others still, think that it's a combination of the two.

So if you have a 300hp car that puts 250whp down on a dyno, is it a static 50whp drivetrain loss, or is it a 17% drivetrain loss?

Does it take a fixed amount of power to turn the drivetrain, or does it consume a little more power as the output of the engine increases?

I personally think it's somewhat of a combination of things, but I lean towards a percentage loss. I googled for a bit, and here's what I came up with:


1 -
some guy on the internet said:
Some parts are constants (like seals, ring gear turning in oil etc.) and most are proportional to the load (gears etc).
If you where to have car with 500 flywheel/400 rear wheel hp, then it wouldn't be possible for a 100 hp 2 liter engine to move it an inch if it were a constant. A 302 from the seventies would have only 30 hp left at the wheels.... That is of course not the case.

2 - http://www.superstang.com/horsepower.htm#Proof1

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some other guy on the internet said:
A constant percentage should be what happens. That percentage is determined by every moving interface that the power is transmitted through. For example, if the bevel gears in your diff are 98% effecient than no matter if you send 100 or 1000 hp through them you're only getting 98% out(98hp or 980hp in this case). What you do is multiply all the efficiencies together to get the overall net.

Trans input shaft -> output shaft (could vary depending on gear selected)

output shaft -> driveshaft (universal joint eff)

driveshaft -> diff input (universal joint eff)

diff input -> axles (bevel gear eff)

And then there are all the losses associated with seals, bearings and friction, which are also proportional. I think the bulk of it is all the gear to gear interfaces. So just using complete guesses for numbers this might be 95x99x99x95x99 etc and you get (in this made up example) 87% eff, a 13% loss. There also might be some items which are more static than proportional also.

That's what the textbooks say at least, who knows how close real life is to this.

Eric
Feel free to discuss and debate your points. HOWEVER - As always, be respectful to one another.
 

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Well, in respose to robevo, the percentage loss theory doesn't work either because different dynos read different numbers. So one could read that an evo has a 15% loss, another 25%. This shows the inaccuracy of the % loss theory. Now, the static loss theory is not without faults, as it too will vary depending on the dyno, but tends to be a more consistent number to play with. My friend is a mechanical engineer for force protection, a military vehicle contractor, and this is what he had to say. "The percentage loss theory is correct, but so is the static loss theory. The % loss theory is only applicable to the stock motor/tranny configuration. The loss is commonly rated in a % to show the kind of force required to move the mechanical parts to get the power to the wheels, but this % is merely showing a fixed amount of power based off of what is being lost from the stock motor. The static loss theory is more correct in that the amount of force required to move a mechanical part is a constant, regardless of the force that is capable of being transmitted to that part. When more power is added over stock, the amount of power required to move the gears, axles, driveshaft, wheels, etc is not increased because the requirements of what they need to move have already been met. To give an analogy, let's say that a person picks up a tennis ball. Now, that same person takes a dose of steroids and then picks up that same tennis ball. He now has more available strength, so why would it require more effort to pick up the ball after the steroids?
 

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Well, in respose to robevo, the percentage loss theory doesn't work either because different dynos read different numbers. So one could read that an evo has a 15% loss, another 25%. This shows the inaccuracy of the % loss theory.
Not really. You can't compare apples to oranges. If a car has a 20% drivetrain loss, that would be according to one dyno. And as I'm sure you're about to point out, the static number is going to vary dyno to dyno as well...

Now, the static loss theory is not without faults, as it too will vary depending on the dyno, but tends to be a more consistent number to play with. My friend is a mechanical engineer for force protection, a military vehicle contractor, and this is what he had to say. "The percentage loss theory is correct, but so is the static loss theory. The % loss theory is only applicable to the stock motor/tranny configuration. The loss is commonly rated in a % to show the kind of force required to move the mechanical parts to get the power to the wheels, but this % is merely showing a fixed amount of power based off of what is being lost from the stock motor. The static loss theory is more correct in that the amount of force required to move a mechanical part is a constant, regardless of the force that is capable of being transmitted to that part. When more power is added over stock, the amount of power required to move the gears, axles, driveshaft, wheels, etc is not increased because the requirements of what they need to move have already been met. To give an analogy, let's say that a person picks up a tennis ball. Now, that same person takes a dose of steroids and then picks up that same tennis ball. He now has more available strength, so why would it require more effort to pick up the ball after the steroids?
The analogy doesn't hold true. With more power, the drivetrain will run hotter, have more friction, etc, and in turn be less efficient. How much less efficient is another debate altogether, but it's very hard to argue that when you're putting something under more stress and heat that it will not become less efficient. Make sure you check out the link I posted in my first post. The guy makes some good points.
 

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Well, I've read the link, and it is a very compelling argument, but has some holes in it. One thing worth mentioning is that when using the static loss theory to estimate the power gained at the crank, adding the initial loss recorded when the car was stock to the wheel hp number gives a much more conservative number that seems more reasonable. It's like saying, "well I'm making 800 at the wheels, so because of the % loss I'll tell people that I'm making 1000 at the motor instead of 850." Sounds like something a ricer would say..." yeah this intake gave me 100 hp, the exhaust added another 200, and that sticker there added 65." Yet again, without running on the wheel dyno and then pulling the motor out and running it on a motor dyno after each mod, then it cannot be proven. The theory of heat loss, and thus power loss, due to inreased friction when the drivetrain components are spun more quickly sounds plausible, but here's the problem. Heat creates friction, it doesn't dissipates it. By this logic, the more power at the motor, the more heat there woud be running through the tranny, which would increase friction and thus heat/power, which would put more power to the ground than at the motor, which is impossible. Friction creates resistance, and that's what creates an increased loss, but arguably if the gears are working poperly being spun faster shouldn't create that much more friction. And to attack another one of his points, when talking about radiators, the reason that bigger, more eficient radiators are needed is not because of more heat, and thus "power", being lost to the radiator, like he said. The radiator's job is to reduce heat, not pass it through to keep heat levels high and thus power levels high, by his logic. Heat is the enemy of horsepower. It's why when motors get too hot they lose power, it's why cold air makes more hp than hot air (and yes, a big part of that is because of the fact that the cold air has a higher oxygen content. But, it also cools the combustion chamber, which increases hp by REMOVING heat, not adding it), the list goes on and on. The reason for the bigger radiator being needed when power levels are high is because of the fact that the stock, small radiator gets heat soaked and cannot remove enough heat, thus passing it back to the motor and not cooling it off like it's supposed to. See what I mean?
 

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This is a topic covered in first year Physics with calculus. It is very easy to think of something requiring a certain amount of energy to move it. In this case, EVOXGSR, you are thinking of how much energy it takes to turn gears. However, this thought process is inaccurate. The reason is we arent calculating the amount of power it takes to move those gears, we are instead checking the output vs the input. Hench the tranny has an input and output shaft. We are applying some amount of power into the tranny in the input shaft. As that power is transferred from gear to gear, and finally along to the output shaft, we are losing some of that input power. You should remember the basic principle of work transfer, and that is what we are discussing here. You cannot have a 100% work transfer. Our current understanding of everything just doesnt allow for it. There is always some work loss through friction, mass issues, etc. If there wasnt, then we could easily make perpetual energy, and we would have a much different world.

There is yet another problem with your analysis. If all we had to worry about was how much power it took to get the gears moving, then we would also need to wonder about how much power it takes to keep those gears moving. Here we get to use another law that states "Objects that are in motion, prefer to stay in motion." This being the case, we would have a varying amount of power required to move those gears. When the gears arent moving, it would require more, and once the gears are moving, less. Therefore, it is impossible to have a static number across all power inputs. Does this make sense?

Loss isnt static, it is most definitely varying in this case. The best way to represent this loss is via a %, because the variance is highly Dependant on the input. Regardless of how much, or how little, power you apply, each gear is robbing a small percentage of each energy transfer. Combine those transfers, and you have your total loss percentage. Does this make sense? If not, MIT has an online course where I believe you can review this material. Ill see if I can find it.
 

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I agree that the number can vary depending on the speed at which the car is traveling, the rpms, load, etc., but couldn't all of these varying amounts of power loss be based off of a maximum that can be lost? I mean, to me these two theories are co-existing, to a degree atleast. I agree that there is a certain percentage loss through the transmission, but the point that I am making is that it's based off of a maximum loseable value. If this is the case, the number is static, or at least the varying losses that change with rpms, and thus gear speed, are based off of a maximum fixed amount of power that can be lost. My best friend is a mechanical engineer for a military vehicle manufacturer, and this is what he tells me. What are your qualifications to contest this?
 

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So you are saying you are trying to take the maximum possible loss and applying it across the board. That doesnt work. Ask your friend how he deals with the co-efficients for the input to output load functions. Its not quite a differential equation by any means, but a simple graph of the kinetic loss function will show that its not linear by any means.

My qualifications? Um, well, I know your buddy and I took the same math courses. Actually, for mechanical engineering Im not sure if he needs to take matrix, integrations, or Diff Eq. He most definitely took Vectors, but this doesnt mean anything. I know a ton of degree carrying people that dont know what they are talking about. Why dont you just go spend some time taking those free online MIT courses. Then, once you do the math work up on it, I believe you will see the problem with your buddies current thought process.

Remember, this is a transmission, and its entire goal is to change the ratio of the input to the output. So each gear has different loss values. Then the multiplier that is used for the output shaft is changing through the entire interval, so it cant be a constant variable. If it was, we would see some sort of asymptote as we approached that value.
 

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I hear ya, you make a compelling argument and it is well appointed with facts. I will read up and see if I can understand what the course is teaching. I'm just trying to make the point that there should be a maximum value that when reached, more loss is not possible. But, as much as I know about cars, and I do know quite a bit, I'm not overly affluent with advanced mathematics and physics principals, so you may well have me beat on this one.
:whipping:
 

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Have we ever thought of a deeper..connection then a little bit static and a little bit percentage....Seriously what if its a Static Percentage loss

The way I figure is the mechanical rotating parts are always going to weigh the same and the cars always going to rev and apply the same power to turn them...However The percentage part comes in when after all that is lossed through static also receives all the (What I'll call) Feed back resistance from the drive train and car as a whole. ... When the parts work together the percentage number affecting the static loss should be smaller making the overall parasitic loss smaller.
 

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There is a forumla for this that we can use. It will make more sense if I can find that formula. Im too lazy to search through my Physics books to find it, and I know google has this somewhere. The graph will make it very easy to understand.
 

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I have a 2012 EVO 10 and just put a turbo's X exhaust, turbo down pipe all the way back with freeflow cat. And an ETS cold air intake. Iv read on this forum that eveo 10's put's out 230-240 WHP stock. I did not get a base line with the car stock. My dinojet tuner got 280.21 WHP with the above modes. This seems wrong to me as the manufacturer of exaust says 9-10 engine HP max and the intake cant bee doing even that. So my question is this the math dos not seem to add up on the first run? Does anyone know why this is? Your help would be appreciated. Bruce
 

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My short answer is that, if you're looking for a single value or variable that determines drivetrain loss it is not a fixed amount nor a percent; instead, the amount of power lost to the drivetrain depends on the rotational speeds of the (pre-tranny) engine and the (post-tranny) shafts.
 

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I have a 2012 EVO 10 and just put a turbo's X exhaust, turbo down pipe all the way back with freeflow cat. And an ETS cold air intake. Iv read on this forum that eveo 10's put's out 230-240 WHP stock. I did not get a base line with the car stock. My dinojet tuner got 280.21 WHP with the above modes. This seems wrong to me as the manufacturer of exaust says 9-10 engine HP max and the intake cant bee doing even that. So my question is this the math dos not seem to add up on the first run? Does anyone know why this is? Your help would be appreciated. Bruce
Might wanna get a EBC.

Hard to tell with no baseline. But, I gained almost 100whp and 100wtq with just intake, uicp, ebc, hfc, and tune. My baseline was around 258whp and 261wtq IIRC.
 

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There is a forumla for this that we can use. It will make more sense if I can find that formula. Im too lazy to search through my Physics books to find it, and I know google has this somewhere. The graph will make it very easy to understand.
If we assume that there is 20 hp(I) of static loss to brake horsepower in the Evo X's drivetrain, and 1/6 percentage loss to it after that, we have:

bhp = whp × 1.2 + 20

Here is an Excel sheet calculation using this formula. Left is using a step of 100 and right is 25:

 
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