skinny z

Reading a few threads here and I was reminded that its always struck me as odd that the HP loss through a given drivetrain is expressed as a percentage. At some point in the engine output vs parasitic loss graph, x% would hold true but for a given chassis, doubling the power will not double the effort to turn the transmission or rear gear.
Does that not make sense?
Now, further to that, it would be interesting to know how much HP a certain transmission requires to turn. Same for the differential.
Case in point, one of the earliest vehicles I had on a Dynojet produced about 300 rwhp. The question here is what is the engine's actual output? You can't say add 15 or 20% because if I cranked up the boost or sprayed it, I'm not suddenly increasing driveline friction.
Thoughts? Opinions?

Fast355

iTrader: (2)

A transmission will have a fixed loss but it will also have a variable loss. Meaning the fluid pump takes X amount of power to turn. The pump loss is very low however, probably only a HP or two at full line pressure. If it were higher than that your engine would never idle in reverse. However the gear train, u-joints, differential gears and bearings will eat up power at a very fixed percentage. A 700r4/4L60E and 10-bolt eat right at 18% HP regardless of the crank HP. 300 HP at the tires through a 700r4 and 10-bolt is right at 360-370 crank HP.

skinny z

Think about those last two sentences for a minute. Sure, 18% on a 360 HP engine happens to work out that way. But if I take that same driveline and double the engine output, I'm not now losing double the power. It would be about 65 HP regardless. Otherwise you'd never go faster with additional power.
At least that's the way I see it (and keeping upper RPM limits the same).

red rock

iTrader: (2)

I totally get what your say'n. 300hp @ 20% loss = 60hp, 600hp @ 20% loss = 120hp, 1200hp @ 20% loss = 240hp. It shouldn't lose that much hp the more hp you get. Correct? At some point the loss should go down I would think.

skinny z

That about sums it up. For the most part (and there may be a few variables that I'm not aware of), the losses are fixed and not percentage based.
As for the 18% loss example on 300 rwhp, (that Fast posted) that falls in line with that particular cars MPH at the strip. But rather than saying it's 18%, it's really more like 60 HP. If I sprayed that car on the dyno, it would still lose 60 HP regardless of the power increase.

red rock

iTrader: (2)

I'm thinking the only way to know, is to dyno the engine first, then a chassis dyno to figure the loss at a given hp? But then you get into the differences between the two dyno's.

skinny z

It wouldn't be a loss at a given HP but a loss for a given driveline. Take that same chassis, plunk in an engine with twice the power and the driveline losses should be the same (all else being equal).

red rock

iTrader: (2)

Somebody on here that knows physics should respond because now it's over my head. Type of transmission, length and diameter of driveshaft, and type of rear end with whatever gears are in there, tire size = hp loss.
Maybe sort through these and make some sense of it. http://www.wallaceracing.com/Calculators.htm

sofakingdom

iTrader: (1)

Drive shaft length and diameter, and tire size, have no effect on loss.

The comments that there are some fixed, and some variable, components of loss, are true. No matter how much HP the parts are transmitting, there are some things that just take x amount of power to work. Auto trans pumps for example. There are others that extract a loss just from the relative motion of the parts; pinion bearings for example. But then there are others that are more a proportion of the force placed on them; the friction of rear end gears for example. Too much variation in the variables to generalize from any one car to any other.

I don't know of any good way to "calculate" or "predict" drive train loss. Even putting an engine on a dyno, then putting the car it gets put into on a chassis dyno, has too many sources of error. Exhaust systems, intakes, cooling systems, all would affect the numbers. Then, worst of all, there's the way a chassis dyno works... most of them, DynoJet in particular, are "inertial". Which means, it is set up to apply however much INERTIAL load (mass that has to accelerate) to the wheels, to limit the engine's acceleration rate to some # of RPMs per second; usually around 200 - 250 for a street or stock-car application, maybe up to 500 - 600 for a drag racer. Which is why a "pull" on a dyno always seems to take about the same amount of time for every car, no matter whether it's a 150 HP 6-cyl, or a 1000 HP twin-turbo monster or something... the operator starts the pull at 3000 or whatever, and lets it run out to 6500 or whatever, and at 200 RPM per second, that's gonna take about 18 - 20 seconds. Period. They're designed that way to simulate the natural load of accelerating the CAR, wherein the wheels, drive shaft, etc. all have to start at some very slow speed and accelerate to 100 mph or something, and so reflect the losses from accelerating all the car's moving parts. An "electrical" chassis dyno, like a "water brake" engine one, is more "steady state", and measures the engine output at a fixed RPM level. So right there you are generating numbers in 2 very different ways, and it stands to reason they're not going to correlate in a one-to-one fashion.

That said, applying a 25% reduction for an auto trans and 20% for a manual, gets close enough to be of value, when applied to a water brake on the engine and an inertial for the chassis. Alternatively, the trap mph is EXTREMELY accurate for drive wheel HP, if 2 assumptions can be made: that the engine stays at or near its max HP for nearly all of the run, and that the aerodynamic and rolling losses are low enough to be ignored.

red rock

iTrader: (2)

[ Drive shaft length and diameter, and tire size, have no effect on loss.] I just threw that in for effect. But at what point does HP override the 20% loss and lessen the percentage? If that makes sense. Or as Skinny z said, is 20% a constant 60 hp?

sofakingdom

iTrader: (1)

Nothing really ever "overrides" anything.

The constant losses stay constant, and the ones that are a fraction of HP remain fractional.

If you were to plot a curve of loss HP vs engine output HP, it would be almost flat near 0 engine HP, at some distance above 0 loss; then as HP increases, loss would also increase. At very low HP levels, for any given drive train, loss would then be near 100%; then as HP increases it would decrease as a %, levelling off at maybe 20%ish or so for the average car. Lower for synthetic fluids, efficient drive train setup, low brake drag, etc. etc. etc; higher for auto trans, weird U-joint angles, and so forth.

skinny z

But that puts us backs to the generalization of a fixed percentage for any power output doesn't it? I can see that holding true if we keep engine output to a certain range. At some point, the 20% is exactly what the losses are.
Take the same chassis and plunk two different engines in it, where HP levels are doubled, the parasitic losses aren't doubled. Those losses remain the same. Which is kind of the point of the discussion.
That said, I do understand the thinking behind that. Then again, as perhaps might be the case with my own personal projects (as in disarray as they are), having measured output on a Dynojet and then hoping to return with about a 50% increase in HP, I certainly don't expect the trans and differential (et al) to consume another 30 or so HP.
I appreciate the replies and am interested in further comments too.

AlkyIROC

All things being equal and that may be true however the driveline of a 300HP engine is not the same as a 1200 HP engine. A 300 HP engine may use a converter that stalls around 2400 rpm. A 1200 HP engine probably won't see a converter stall less than 5000 rpm and a 300 HP engine couldn't use a 5000 stall converter. It's unlikely they would use the same type of transmission either. For these engines to produce best power to the wheels, they won't have identical drivelines to do an apples to apples comparison.

As far as I'm concerned, each transmission requires a certain amount of HP to spin the internals. The more power going into the transmission, the more power coming out. Rear gears are torque multipliers but the same gear ratio in different diffs will use a different amount of power to turn them although the difference is very low.

As for how much drivetrain loss your specific vehicle has, you would need to put the engine on a dyno to see how much flywheel HP it produces then put the car on a chassis dyno to see how much is getting to the wheels. The difference is lost in the driveline. Now you can play with driveline changes to see what makes the vehicle faster or slower without changing anything on the engine. You can make a big power change to the rear wheels just by changing the converter stall. That doesn't change the engine HP. That doesn't change the driveline loss, it just changes where the torque from the engine is used in the powerband which can translate into more power to the rear wheels.

red rock

iTrader: (2)

While what you said was true as far as more HP will require different drivelines, What i meant by stating the HP increases, was all things staying the same, @ the standard 20% loss, why would the loss double as HP doubles? I too think there must be a set HP loss for different drivetrains, vs. a set percentage loss, or maybe there is a set percentage loss for different drivetrains? AlkyROC, what would your drivetrain percentage be for your set up? Would it still be the 20%, or more? Or maybe as HP increases, bigger drivetrain increases and still stays at 20%? Now I'm trying to wrap my head into this and getting confused.

GeneralDisorder

iTrader: (1)

It's relatively easy to calculate engine HP on MAF based vehicles. From lots of dyno time, and general experience.... take MAF grams per second and divide by .82

In practice this is quite close to actual crank HP.

You can test it out on the dyno. Just hold the MAF g/s at half its peak value. Does it scale linearly? Probably not. Is it a flat number like 12 HP? Definitely not! Some various sample data and curve fitting would be in order. There's an appropriate polynomial for it can tell you that much. LoL.

GD

red rock

iTrader: (2)

Unless I'm missing something here, we were discussing drivetrain loss percentage and how it equates to HP loss. I'm not sure what your meaning here is?

sofakingdom

iTrader: (1)

Because for things in the drive train that RUB against each other, such as the R&P, the loss is caused by sliding friction, which within a certain range, is directly proportional (IOW a fraction or percentage of) the force mashing them together.

red rock

iTrader: (2)

I guess that makes sense too. So then as hp increases the extra force applied to the driveline requires more hp to turn it.

GeneralDisorder

iTrader: (1)

My meaning is it's easy enough to measure power at the wheels and calculate power at the crank from MAF readings (possible with speed density also but the calculations are more involved and less accurate). Then you simply calculate drivetrain loss.

GD

sofakingdom

iTrader: (1)

Correct, for some parts of the losses... high-school physics.

"Calculating" it however, while "theoretically" possible, is totally impractical; too many variables, too many that vary from car to car, or even moment to moment (lube temp for example, or the degree to which the rear end housing flexes and allows gear misalignment at any given power level).

When I was a math & physics major, one of the favorite exercises in certain classes (differential equations especially) was to give us a list of boundary conditions and/or observations, then ask us if this was enough info to solve the problem. Needless to say, this was often harder to answer than actually deriving the solution, if there even was one. Proving that there isn't enough info is HARD sometimes.

red rock

iTrader: (2)

Well said. And High school physics was 43+ years ago for me. I didn't retain that info very well.

AlkyIROC

Lots of examples like that. A quick google search and this pops up.

If a ship has 26 sheep and 10 goats on board, how old is the captain?

Although there is no real answer based on the data, there actually are a couple of common answers given by the majority of people who are asked the question.

Fast355

iTrader: (2)

It may not be quite double the loss with 2x the HP but it would be close. It is definately not a constant loss. You go faster with increased HP because you are putting more power to the ground even with the increased loss.

skinny z

It's still a situation where as power output and frictional losses are not linked by a constant. I can see how increased torque may result in increased friction due to a number of factors. But to put 500 HP through a 9" rear housing and then 1000 HP through the same gear set can't possibly mean twice the effort required to turn it. Same for any transmission, transfer case, u-joint, etc.
That's really what I'm getting at here. That and some kind of intellectual discussion.
I would expect that the example I first posted has a reasonable and expected 15-20% loss. This is verified by data collected from the dyno and the dragstrip.
Dynojet results. 282 RWHP (corrected). Track performance 12.78 @ 105.1 / 8.088 @ 83.94 /1.71 60’ . Weight: 3750 lbs.
Some on-line race calculator suggest that about 350 CHP is about right based on the MPH/ET/weight. Others are out to lunch and state that 350 RWHP is required for that kind of performance.
Now, if theoretically I sprayed a 300 shot of nitrous through the same engine, I might see frictional losses doubled? I doubt it.
But as has been suggested here, to prove that would be difficult and time consuming. Back to back dyno tests. Then back to drag strip testing. And even then a bunch of environmental factors come into play.
On post claims CHP can be calculated via chassis dyno results and MAF readings. That might offer an easy calculation. however for this OP, FI is not an option.

Kingtal0n

When we model an engine there is some inertial component times the acceleration of the engine, so the J term would be larger the faster the engine is accelerating.
There is also a friction component times the velocity of the engine (first derivative of angle) which remains constant at steady state.

If you increase engine output without changing the friction component (using all the same drivetrain parts that turn and provide counter torque) then there will not be any additional parasitic loss due to friction.
In theory if we increase engine acceleration there will be a larger J term (moment of inertial counter torque will increase) but this is easily realized in 1st gear vs 4th gear (it is negligible for our purposes) without increasing power, and each dynometer will have it's own rate of change domain during measurement which obscures the link between actual drivetrain losses experienced on the road and dyno drivetrain losses while it was on the dyno.


when increasing power, the major 'power parasites' are more likely to be:
-any attempts to cool the air/fuel mixture, i.e. richer a/f, water injection, intercooling all reduce power
-inadequate flame propagation (wrong plug for the application, wrong combustion chamber design for the pressure used, poor timing, wrong combination of parts)

basically engine setup, and tuning have a far more dramatic effect on losses (expected vs actual)
This can be 'seen' a number of subtle ways. For example many compressor maps will give airflow in terms of lb/min, where typically 1lb/min is worth 10 horsepower.
However if you pick up a thermodynamics text you will find that questions concerning power and mileage never consult the airflow quantity, but instead only use the fuel mass to determine how far or how much power something will produce.
This is because fuel provides the power, and air is just a substrate for the chemical reaction. So how can turbo manufacturers claim that 1lb/min of airflow is worth 10 horsepower?
It is because they anticipate specific engine setup/tuning behavior. Furthermore it should be obvious that engine setup parameters such as compression ratio will also influence output.
So really what they are giving us is a generic, general idea of what power is possible to make given a specific air mass and typical fuels. If you use a fuel that contains a large percentage of oxygen for example it would throw the whole general equation out the window. However if we know exactly what the fuel is we could determine exactly how much power is possible given a specific fuel mass, regardless of how much air is being pumped through the engine.

red rock

iTrader: (2)

Then we are at what i posted back on #6. https://ls1tech.com/forums/general-l...-amount-2.html On post 32 is a graph of drivetrain loss. This is from 2005, https://www.thirdgen.org/forums/engi...tion-20-a.html. There are a lot of discussions about this subject on the net. Here, Hot Rod magazine actually did engine and chassis dyno's. https://www.hotrod.com/articles/ccrp...in-power-loss/

skinny z

If I take the preceding and succeeding statements around the quote above and call it all else being equal, then that quote represents exactly what I'm getting at.
Not to discount the other statements however keeping tuning and setup out of the equation is more in line with the intent of the thread. Much like making any comparisons as equal as possible and only changing one component and that's engine torque output.



A quick read over those articles and I see the arguments are the same as those reflected here. And, some the comparisons are often not apples to apples. Different chassis combinations in particular.
Some of the discussions have acceleration included in the equation. Again I'm only trying to consider the rear wheel output of two engines, in a fixed chassis, at a fixed RPM where each engine is making maximum torque and one engine has twice the power output of the other.
The end result can't purely be percentage based.

red rock

iTrader: (2)

I realize the HOT ROD test didn't show adding HP to the test engine, but you can see the percentage go up as HP is applied. I guess that's what i getting at. (in the 455 test)

Kingtal0n

I didn't read it but i would happily go out on a limb and say that is it highly unlikely when doubling the output of an engine that all else is held equal
i.e.
-it doesn't run at the same a/f ratio (more fuel 'wasted' to achieve optimal performance is seen as 'power lost' since fuel was basically poured on the ground just to keep the engine happy, and fuel = power)
-it doesn't use the same amount of ignition timing (less timing means less degrees of crankshaft rotation over which to extract torque)
-it will put out more heat given identical constraints (changes exhaust gas volume which affects a slew of variables such as scavenging and turbine behavior if there is one)
-add your favorite

The real answer is a set of differential equations utilized via simulation environment would yield the best data for a model. and this is what engineers do when designing the engine to begin with. Modern computers after 1998 include variables, graphs, relating airflow friction, engine friction, accessory torque, all of that is accounted for within the model of modern engines.

skinny z

Finding answers to this question for let's say, a 4L60, and looking for varying tests with different input levels might provide some interesting insights. That is to say, pushing 300 ft/lbs into that transmission vs putting 600 ft/lbs might reveal what?
I'll agree on one thing . The intraweb is full of this kind of discussion. Prior to posing this in the first place, I hadn't thought to see if the question had been asked before. For as long as I can remember, the accepted 15-20% has been the standard (when referenced to our 3rd gens and what the typical engine/driveline configurations are likely to be). And for an average 400-500 HP engine package (again saying that's typical of our platform) that percentage is close enough to count.
Go outside that norm, in particular with the same chassis/driveline but greater engine output, then the percentage loses it's accuracy.
Sound reasonable?

Fast355

iTrader: (2)

I actually have a very good comparsion of what the same driveline will do V6 vs V8 in front of it. Thinking back a couple of years ago I had a friend dyno his bone stock 2012 Infiniti G37 Sedan the same day I had my 2011 Infiniti M56 on the dyno. I messaged him last night to see if he still knew what his V6 car put down when it was stock and he responded with 297 rwhp and 252 rwtq. Why is this a valid comparison you say. They use the same 7 spd transmission (V8 uses more clutches) and a very similar driveline. The ring and pinon ratios are different (2.62:1 for the V8 and 3.36:1 for the V6) but the housings, rear hubs, rear CV shafts, driveshaft, etc are for all purposes the same. The V8 car was crank rated at 420 HP and 418 TQ. The V6 car was crank rated at 330 HP and 270 TQ. My V8 did 378 HP and 390 TQ to the tires. The V6 did 297 HP and 252 TQ to the tires. Because neither car could be run in 1:1 to peak HP without overspeeding the dyno, they were both run in 4th gear manual mode which locks up the torque converter even at WOT on both cars. Having the numbers in hand, I started calculating percentage differences. When you actually crunch the math, the drivelines that are for all purposes the same, lose 6.9% at peak torque and 10.5% at peak HP. It was shocking to me that the numbers were the same down to the 10ths digit. I would put money that the driveline losses are a in actuality a bit bigher and both engines just outperform their ratings however by a substantial margin. I also pulled up another dyno of a G37 stick shift coupe. Run in 1:1 the stick car produced 303 hp and 256 tq to the tires. My V8 car weighing ~4,600 lbs with driver went 13.​​​​​2 @ 106 mph. His 3,800 lbs V6 car went 13.5 @ 103. Both cars were on factory size tires and had could have run better ETs if they could have left the line without wheelspin and traction control pulling power in response. If you use a 1/4 mile MPH/Weight/ET calculation both cars are dead on when you compare dyno numbers, crank HP rating to MPH.

Abubaca

iTrader: (3)

It is....here's how I've always understood it. The variable I think is time.

Think about it this way. If you double the horsepower, then wouldn't you cut your quarter mile time in half? Same friction on the track? Same size tires, right? But of course that's not how it works at all ....because the SAME weight is moving faster and requires more "work"....ie using force to move an object.

100 Horsepower engine, take 20 HP to move the transmission, car goes 15 seconds in the quarter mile.
200 Horsepower engine still takes only 20 HP to move the transmission and make the car run 15 seconds, if your ran part throttle....but it takes 40 Horsepower to make the transmission move to get the car down the track in 12 seconds.

Orr89RocZ

iTrader: (20)

skinny z

.

red rock

iTrader: (2)

Other than a period, I missed your post. Good conversation going on here, we need your input.

QwkTrip

iTrader: (22)

Been far too many years for me to remember exactly, but I think gear efficiency is a function of the force applied to the gear and velocity. So yes, if engine torque is increased then there are more losses through the gear train. Increase velocity and there are more losses through the gear train. The losses produce heat, noise, ect... That's why you need larger trans and axle coolers when engines make more power.

An automatic transmission has an efficiency curve too and it is not a straight line. I don't remember any of the calculations so can't go much beyond that. But clearly losses do increase at higher power levels or you wouldn't need larger oil coolers.

It wouldn't surprise me at all if drivetrain losses do follow the "Percentage" estimate at power level of most street cars.

Fast355

iTrader: (2)

Its not ludicrous. Its physics. For every force applied their is an opposite force being applied to things like your pinon bearing and carrier bearings as well as your automatic transmissions thrust bearings and bushings or manual transmissions main shaft bearings. The more power you put through the driveline the more force it exerts on the bearings and the higher the loss. I hope you really do not think if you were to put a 100 hp shot of nitrous on an engine you would see 100 hp to the tires. You will see maybe 75-80 hp at the tires even though you gained 100 hp at the engine. I am sure the loss is not exactly linear but it is pretty close. Take a Mercruiser bravo drive fro example. If you put an engine on an engine dyno then bolt the exact same engine to a prop dyno you will lose roughly 10% through the drive. Does not matter if it is a 200 hp 4.3 or a 420 hp 8.1 it is right at 10%.
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QwkTrip

iTrader: (22)

Already off topic with your first sentence as usual. This isn't about engine efficiency. The OP's question was about losses from crankshaft to tire.

Fast355

iTrader: (2)

It IS a percentage and the more power you put through the driveline the more HP it eats.

QwkTrip

iTrader: (22)

All horsepower is taxed just like your money. None of it gets through the drivetrain magically unaffected by friction. The only question at hand is whether the relationship has a constant slope or if the slope changes at higher power levels.

QwkTrip

iTrader: (22)

I don't know what else to tell ya, bud. I tried to explain it as simply as I could to help you. Some people just don't have the gift.

Fast355

iTrader: (2)

Actually you are wrong. As mentioned above drivetrain loss absolutely will increase in near percentage to the power flowing through it. It is not a fixed loss because friction losses increase as power increases.

Orr89RocZ

iTrader: (20)

skinny z

Lets put in the simplest terms and avoid some of the confusion.
We're not talking about acceleration.
For the purposes of this discussion I want to strap a 500 HP engine to a simple driveline and measure output on a chassis dyno. This engine makes peak torque at 4500 RPM. That RPM is held steady and the dyno output recorded.
Now take a 1000 HP engine, install it in the same simple chassis. This engine also makes peak torque at 4500 RPM. That RPM is held steady and the dyno output recorded.
What are you going to see as result?
Some here suggest, through some math process or by empirical data, that the outputs will be significantly different. Falling back on a 20% loss, the first engine will produce 400 rwhp and the 2nd will produce 800 rwhp. From that then the more powerful engine will cause the driveline to lose an additional 100 HP? This happens despite the RPM being the same.
I'd like to see the math behind that.
I might have to go back to my college studies and dig it up myself.
And as always, thanks for the input from everybody.

skinny z

Doing a little internet digging (rather than digging up my old papers) and I came across this article.

https://x-engineer.org/automotive-en...es-efficiency/

In summary, it characterizes drivetrain efficiency as a percentage.
I guess that answers that.
Still can't see it but the math and the data don't lie I guess.

Orr89RocZ

iTrader: (20)

skinny z

I was mentioned yes but I think only in passing and not part of the overall explanation I was looking for. At that point, there was kind of hybrid theory that appeared to be emerging.
Still I find it interesting that any given gearbox, differential, u-joint etc. will consume varying amounts of power. Intuitively, one would think that x transmission requires 100 HP to drive it and that's it. Put that transmission behind triple the input power and it will take three times the amount to drive it.
At any rate, I got what I wanted and that's some discussion and some debate.
And I guess now I have to consider that the additional 100 HP I've added to my engine's output isn't going to be realized at the rear wheels. No 11's for me.

Abubaca

iTrader: (3)

TL, I think what you're missing in the equations isn't the transmission or engine, but the work being done. It's not just the engine side of the equation that's changing. The DYNO side it changing too, in that more power is being transferred to the dyno. THAT is what's pulling the extra power, expressed in a ball park percentage. What your quote above should say is:

If you want less power loss through the drivetrain, do less work. -in which case you'll only NEED a less powerfull engine.

Or you could just trust 100 years of engineers hot rodding cars. I'm not one for blind faith, but sometimes you can simply trust smarter people. Not being snotty, just sayin' you don't have to trust me.

Orr89RocZ

iTrader: (20)

skinny z

I understand the varying efficiencies but what I was getting at relates to the first post.
I was postulating that the drivetrain loss is fixed and that 100 HP added is seen at the rear wheels too. Example: Transmission A takes 50 HP, rear gear B takes 20 HP and so on. All cumulative but still fixed. Instead, only a portion will be transmitted as the rest is eaten up by the accumulating losses. Output power equals input power times the drivetrain efficiency.
Certainly more to it than the obvious.