engineers, i have a physics question. if an engine in a car with peak power of 250hp at the flywheel and 200hp at the wheels experiences 20% d/t loss, will an engine of 500hp at the flywheel in that same car experience a 100hp d/t loss, all other variables remaining the same? or are there factors which make that equation not linear?

 

The powerloss can be modeled as linear or can be linearized. So yes normally it takes a percentage of flywheel hp to drive the powertrain. It is not normally a given number like 50hp.

 

Part of the loss is fixed, such as certain bearing frictions, fluid pumping losses, and so forth; and part is proportional to load, such as gear teeth friction. Additionally, there are things that are totally variable depending on how things are built; for instance, the tighter you stack up the clutches in an auto trans, the better it will shift; but the higher the losses will be in other gears, from those clutches rubbing against each other.

So there's no universal "conversion factor" that says if you're getting X HP at the wheels, there's Y HP at the crank.

 

i was talking to a certain engineering student at u maine orono regarding this topic and this is the response i got. thought it was worth posting:

ok here is a basic primer on power and its relation to loss. Power is of the units force x distance / time, and any combination of values with these units will yield a specific power consumption. Your example of dragging a rock down the street, you drag it the same distance, but you do it in half the time, as a result you get say 5lbs force x 10ft distance / 2 seconds = 25 (forget units since the english system is useless here), now say you do the same thing in 1 second because you are going twice as fast, 5 x 10 /1 = 50, you have doubled the power required to pull the rock by halving the time you pull it the same distance in. Now this unit _expression can be applied to anything else as well, you are driving at 100mph (distance / time), the wind resistance on your car causes a net resisting force of 400lbs (force), you have to have 400(force)x100(distance/time)=4000 (force*distance/time) (units again are meaningless without conversion factors but you get the point), this is also why it is so difficult to increase top speed, because velocity acts against you twice, once to increase the force required, and twice by increasing the actual velocity and thus power consumption. so in the case of a differential you have fluid that is being beaten to **** by a bunch of gears, when you increase the force on them you are doing several things, you are increasing acceleration of the fluid, which causes internal force as fluid friction; you are accelerating rotating mass of the gears/driveshaft/etc...; and you are also increasing the velocities of the spinning things inside of the fluid. We have seen that simply increasing velocity will increase power consumption, so when you drive at a higher velocity you are blowing off the same amount of heat from friction per mile as a lower velocity, but you are doing it more rapidly, ie: less time. Since time is on the denomenator of the equation it will make the power consumption go up; if you did anything in an infinitely small amount of time it would take an infinite amount of power because you would have an infinite velocity (remind you of the theory of relativity at all?). So we have determined that velocity is a factor but it really doesnt apply to acceleration, only to constant cruise. What happens when you accelerate? You are applying a force to the internal fluids by applying a force from the engine to the wheels; when you double this force you are beating the fluids twice as much and as a result heating them twice as much. So now we are seeing that we are doubling the resistive forces in the fluids by doubling the power output of the engine. This is confirmed by the force part of the power equation (F*D/T) causing double the resulting P. Consequently if you do the power calculations to find out exactly how much power is lost through accelerating a wheel, driveshaft, internal gear, etc... the sum of all the parts is surprisingly low in power consumption. It obviously will vary with acceleration here, because the faster you accelerate them the more power you will lose, but ultimately it appears by calculation that the total losses from accelerating parts are around 1% on modern cars with lightweight wheels, flywheels, and driveshafts, insignificant in comparison to the power lost through heat of friction it seems. Incidentally powerloss in lower gears seems to be larger than in higher gears, this is counter intuitive except for the fact that gear efficiencies improve when you get closer to 1:1 and get worse the farther out you get, at a 100:1 ratio a lot of gears are only 65% efficient! So the extra losses from the differential and after-gears in the transmission being at higher velocity turns out to be less of a factor than the total gearing efficiencies and as a result you get a net gain when going closer to a 1:1 ratio. One last note, like i said any combination of values which will yield a F*D/T with units will yield a power P. Torque is force*distance, and rpm is rotations (unitless) per time. so F*D/T=P here as well. When you go uphill gravity applies a force to drag you back down, the faster you oppose this force the more power you need, so this is why you need to downshift when going uphill, even when you start out going fast. Mechanical advantage, force applied to the wheels, torque output at the engine, distance, time, velocity, acceleration, every single aspect is related by power through the conservation of energy. There is a mathematical _expression to account for all of it and to calculate everything you could possibly do with power, electrical, mechanical, anything (except perhaps quantum, since we as an existing set of beings dont have a proven firm grasp on that one yet). Hope it helped, talk more about it if you have questions. Sorry for the novel =P

 

Somebody really likes to type just to hear the sound of their own keyboard..... that's a whole bunch of technical-sounding drivel that contains no information.

I'm all done with being an engineering student, myself. Although actually my major was math and physics. But I've spent a number of decades since then out in the real world BEING an engineer, electrical, electronic, and manufacturing, at various times; and one of the things that matters out here in meatspace (in addition to having one's facts straight of course) is the ability to communicate. That guy could use a bit of that. If of course he had something to communicate in the first place.

 

I agree. He needs a few more years in school before it finally clicks.

RB nailed it in his first post.

A cliff notes version could also be summed up by the following.

Viscous friction (friction due to fluids, ie air, tranny fluid, rear end lube) increases as velocity increases. So the faster it goes the more friction there is. Engines that accelerate faster will experience more viscous friction.

Coulum or dry friction stays the same no matter how fast it is going. Assuming you don't start from a stop the friction factor of that draging rock never changes no matter how much faster you move it. It is a certian fixed amount. If you dragged it through water or through mid air, the fluid forces will increase as its speed increases.

 

A percentage loss is only a ballpark figure and only for an unmodified, relatively stock drivetrain.

Take a 300 hp engine with a calculated 20% drivetrain loss. That means 240 hp is getting through the wheels since the drivetrain eats up 60 hp. Now swap out that engine for one that puts out 500 hp. Theoretically, 440 hp should be getting through the wheels but since you want to use a percentage of 20% then only 400 hp is getting through. How can you lose 40 more hp through a static drivetrain when only the engine has been changed?

Stop thinking of drivetrain losses as percentages. Each component loses a specific amount of power. Different transmission use more power to operate than others. Same goes for differentials. It would be like saying all cars with 500 hp can run 11 second 1/4 miles. It takes more hp to move a 3800 pound car than it does a 2800 pound car so guessing numbers doesn't mean much. My 454 truck runs a 1/4 mile in 15 seconds. 305 third gens can do that but my truck is also over 1000 pounds heavier.

Small numbers to play with. A powerglide tranny only uses up 18 hp. A TH350 uses 36 hp while a TH400 uses 44 hp. Each one can use more or less depending on how they've been set up or modified so even those numbers are not static.

The only way to know exactly how much hp you're drivetrain is eating up is to put the engine on a dyno then put the whole car on a dyno and compare the 2 numbers.

 

because the portion of that equation that you are missing is time . if you have more power, then you are putting it down in a shorter amount of time and that is what is sucking up the extra power, force over time. so if you do the math, then give or take a slight amount for other variables, the equation is indeed pretty much linear and therefore the percentage rule applies no matter what the power output level of your engine is. true you will never know exactly until you dyno it, but id bet a dyno would back up the laws of physics.

 

I disagree, if you compare lots of engine and chassis dyno numbers you will see that the loss is best modeled as a percentage loss.l

 

I agree with ME Leigh and the calculations support his statement.

 

and i disagree with this statement. i didn't find it that difficult to understand and the vast majority of it is factual and makes sense.

 

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I dont think its real easy to just asign it a neat compact value. There are far too many factors involved. Theres also the rotational inertia playing against you as well. I did some basic math using datalogs and found that there was about a 30% loss in peak power in first gear vs. second since some of the work was going into spinning up the engine/TC/trans internals rather then accelerating the car.