i understand you completely!!

but maybe the idea isnt in how fast you can crush the can( i think the speed is only a bi-product of the over all "left-over"force acting on the can) but how much force is need to crush it.

basically, in our case, if the transmisson is kept "identical" (every factor included) then it should still require a fixed amount of force to "power it" therefore you are now displacing more power to the ground, with the new add hp from your new setup, and accelerating quicker because of the added power of the motor, not the rate at which the trans spins.

i hope im not missing anything and in clearly puttng my opinoin out ther cause i do tend to ramble on and lose my train of thought!

great thread, i cant wait to see the kind of info that can be generated from this kind of discussion.

 

I am no expert on this thing, but is the 15-20% rule of thumb used because when large amounts of power are being added, typically the drivetrain is also beefed up accordingly (because our stock drivetrain is not the best as we all know). Possibly even that percentage needs to be reached in-order for it not to bust (assuming that the tougher drivetrain is to turn, the beefier drivetrain)? If two different powered engines are being rotated to the same speed on the same drivetrain then it seems as if the drivetrain loss would be a fixed number not a percent. This might be wrong but Ill throw it out there...if the more powerful motor was spun to a higher speed then the higher the speed the more loss there will be exponentially?

 

It WILL take more power to increase the acceleration rate of a given combination's drivetrain mass. That is WHY it accelerates faster in the first place. If comparing two different power plants in the same vehicle, the engine that will accelerate the vehicle quicker is doing so because it does generate more power. Understand that "power" is the rate at which TQ is applied. An engine that produces more power WILL accelerate a given mass quicker....thats exactly what the term "power" is all about and thats why an engine with more power will spin an inertia dyno to "redline" in a much shorter period of time. The energy required to spin the rollers at a given speed (or propel your car at a given speed) remains the same. How quickly you get it there is determined by the power of the engine accelerating it.....a driveline spinning 2000 RPM's - 6000 RPM"s will use the same energy to spin it at EVERY RPM point between 2 and 6 K (just using these #'s as examples). An engine with more power will simply accelerate it faster from one point to the other. The same driveline will not get less "efficient" if tested within the same parameters of RPM....it goes against the laws of physics.

Tony M.

(97 Bowtie....not looking to single you out, but I think the statement you made reflects the "general" thinking....that is why I quoted it.)

 

Tony, I hate to be the bearer of bad news, but you might look at the definition of "mechanical efficiency". It is expressed in terms of percentage of power applied. That is the DEFINITION! A drivetrain's percent efficiency is defined because of mass, moment of inertia, bearing friction from thrust force, pressure angle of the gears and some other factors. Just like you lose acceleration or power because you add larger, heavier wheels, the faster you try to accelerate a given drivetrain, the more power it takes to achieve the same speed in the shorter time. Those ARE the laws of physics. No flame, just scientific fact.

Now I'm not saying your Miata friend is wrong, I would just like to see how he figured out how much power he was losing to the drivetrain on both cases. For instance, did he dyno both engine and RWHP on the same setup? I'm sorry, basic physics says it just ain't so.

 

if there is anything that the "laws" of physics has shown me, is that with time, nothing is absolute. That is, with time, newer ideas, fundamental facts, the views in which we analysis a problem. Also the reasoning wich we gather information and more importantly, our interpretaion of the gathered information, not to mention the motivatives in accepting theory into law will change with time. We as a "whole" viewed the world as the center of the universe. As a "whole" couldn't fathom the idea that a round planet, that there was a large gate of water surrounding the planet, which is why the sky was blue. We as a "whole" couldnt determine, presently, the importance of of einteins theories until now, and still he is bein rejected by many. Just a few years ago the plate tectonic theory was established and no one thought it to be true...until now.

get my point? I would love to just rest on the set laws of now and not ponder on the beyond. However, though it would be easier, and i am not trying to start a flame war; i know that with time, again, nothing is absolute and that these laws that we, and understand that we still have not even reached the haf way mark on "absolute knowledge" ("in essence, absolute knowledge is the absence of the motivation to know all and the evocation of not knowing but being..." well, my take on it anyways. hehehe!!) or even the half way to the half way mark, and so on; see now are but a reflection of our current level of understanding and that they WILL be changed...either for the good of man kind (cough! bullshit cough! ) of scientific/politcal gains or "advancements in science".

i thing that we are all looking at this in different points of views but my opinion, is that the trans in any system, acts more like a kink in a water hose if you will. The more pressure or force you inact before the kink(opening the valve full instead of half), the faster the rate at which the water will travel after the kink because of the pressure building up behind it.

no flames intended, just a reasonable (i feel ) counter arguement...

 

You guys seem to assume that the power lost in the drivetrain is due to acceleration of same. That is simply not true. By far and away the major power loss is due to friction and such. Besides, if we use a Mustang dyno, we don't have acceleration

 

How does this apply to the topic we are discussing?....I agree with your last sentence....It takes more power to accelerate a given drivetrain faster. What I am suggesting is once a given amount of power is determined to accelerate that SAME drivetrain from one given RPM point to another, any additional power found will be added to the bottom line...that being the rear wheels.

I will prove or disprove that statement in the next few months as I spend considerable time both on the engine dyno and the chassis dyno working with a few 225 based engine configurations.

Should prove to be interesting....

Tony M.

EDIT....

To clarify...if "X" amount of power is gained at the flywheel and then tested at the rearwheels thru the use of a chassis dyno on a known baselined vehicle, I feel you will see a one to one relationship of the power gained at the flywheel. Take for instance the night I sprayed my car on the same chassis dyno I just did a pull normally aspirated but minutes before....I picked up 126 HP at the wheels with a Nitrous Express "Shark Nozzle" jetted for a 100 shot (.052 No2 jet). Looks like I picked up every pony that little 100 shot could have given me at the flywheel right to the tire....With the 15% theory, that would indicate I would have actually made approaching 150 additional HP at the flywheel with a small .052 oriface to disperse nitrous thru....highly unlikely....I was quite pleased to see a 126 HP increase at the tire with that small a whack on it....

 

lets see wat you can find tony. I know its hard to accomplish a muliple engine test like this, but could you do the same with a diferent engine, trans setups?

good luck!!

 

I agree with what obZidian said, since Hp is a calculation of torque vs. RPM the increase in acceleration of a constant mass will cause a percentage of loss(try rolling a heavy ball slow then fast, see how much more energy it takes to move it faster).

Brad

 

Of course accelerating the driveline takes power. But is that really fair to call it power loss in the drivetrain? You put heavier wheels on your car, maybe ZR1 replicas with big Nittos, and lose power on a Dynojet. Well, where did the power go? Did you really lose power? You used more of the available power to accelerate the heavier wheels, but how does putting heavier wheels on a car change the mechanical efficiency of the drive line? The power loss is simply because you are using a DynoJet to measure the power. Since we all use chassis dynos to get numbers for bench racing, what is the point of knowing what the drivetrain loss is??? Because we want to convert our RWHP numbers to FWHP. Well, those heavier wheels did NOT affect what an engine dyno would read for your engine, so why do you want to take it into account?

Once again, you have the mechanical efficiency of the drivetrain, which is losses do to friction, churning oil, etc. Then you have an apparent power loss when you try to measure power with an inertial dyno. Changing the weight of anything in the driveline will change the apparent power loss. So will using a lower gear that accelerates faster. So will making more power which produces faster acceleration. But none of those change FWHP. Heavier rods/pistons would, but we aren't talking about them.

 

yes, in a way it is fair and it isnt. The power "lost" is not really lost but redirected or used, changed form; to maintain a certan balance or work consistency. personally, these numbers, the seen on a chassis dyno are not only used to bench race but to measure the changes done to your setup. Furthermore, yes, you do "lose" or redirect power when you put on heavier wheels just like you would see a difference is effeciency when you install lighter rods... The mechanical efficiency you speak of is thus altered because the numbers seen on a chassis syno dont just stop at the trans but are read though the revolutions and resistance from the tires. If you add a heavier set of wheels of lighter than you are changing the rotational mass thus alteruing the necesary energy requirement to spin those wheels at the same speed as before.

the dynojet isnt the only thing taking away or converting power from your net.

you second paragraph seems to contradict you intial arguement, imo. Also, i havea question. When you say FWHP, you mean flywheel hp correct? ok, than why not take a change that affects your rwhp if it doesnt directly affect your fwhp? It is much easier to put your car on a dyno and see wat you have accomplished instead of tearing the motor out everytime. Therefore, if you want as accurate a reading as you can get, you must take all the variables into consideration that affect your system and overall net power seen on a chassis dyno.

 

Once more, by definition, percentage is the measure of mechanical efficiency: Power out divided by Power in x 100. The position you hold says that as power is increased at the input, the efficiency of the powertrain improves. As attractive as that notion is, does that sound reasonable to you?

Just as an example of one of the factors in a gear set, the pressure angle of a helical gear produces a frictional force as the teeth mesh and unmesh related directly to the cosine of the gear helical angle. That angle never changes. What does change is the pressure (or force) applied to the gear contact area.

Frictional force is defined as normal force times the coefficient of friction. The normal force is the force perpendicular to the contact point, that's determined by the cosine of the helical angle. The coefficient of friction does not change for a given gearset and lubrication. By adding power to the input of the gear set (from the crankshaft) we get a larger normal force at the gear mesh contact point. Since the cosine of the gear angle doesn't change, the frictional loss will go up in direct proportion to the increased normal force caused by the added input power.

We haven't discussed additional bearing side/thrust loads and the higher frictional loss from that. There are other factors at work too, like oil viscosity.

As I said, this is a nice notion. It just ain't so.

 

understood, however, i feel that you might have misunderstood my point. I never stated that efficency will improve as power is increased. MY point is that fixing a percentage on a drivetrain loss might is not accurate enough to show actual power (hp/trq.) lost when seen on a chassis dyno.

I will not dispute any of the arguements or points that were brought up but i feel that assigning a fixed percentage is just not correct. Your inducing more power on the same drivetrain so why lose more power? yes, you need to displace more power when trying to achieve faster rates of speed or revolutions but your not (typially) measuring any of that on a chassis dyno. Your just measuring engine power at the wheels. (When you trace a before and after dyno graph, do you not measure the power increased at the same rpm's? well, unless you raised you rpm limiter due to inproved valvtrain, etc.)

When adding power to a motor with the same drivetrain (hyperthetically of course), the trans, for example needs a certain amount of force to drive, like you accesories: a/c. As soon as you have all of these values, that are absorbed by the multiple elements of the drivetrain, you measure the left over power at the wheels. When measuring after an agumentation of the engine power, you are now measuring the new amount of "left-over" power throught the same drivetrain..

these ar my views of course and i see that wat we learn in engineering courses for example, is all subjectible to change.... theories that were once converted to law have been discredited due to further experiementation....Thought they do make sence now, im sure there might be a few that wont in the years to come.

 

What I keep coming back to in my head (not sure if it's right or wrong) is...

If you attached say, a 10-15 HP engine to a stock f-body driveline, you could most likely turn the drivetrain, correct? I'm sure you couldn't turn it at a very high rate of speed, but I think it could be turned. Given this, you might see at most a few HP at the rear tires. Let's say you see 3 rwhp from the 10 hp engine. You lost 70% of your HP and it took 7 hp to turn the stock f-body drivetrain. Now, you attach a 400 hp engine to the same drivetrain, it's not safe to say you will see a 7 hp or 70% loss of power at the rear tires. Thus, we are dealing with a much more dynamic percentage or constant loss of power.

Let's use the 300 and 600 flywheel HP example. Even though you have twice the HP at the crank with the 600 hp engine, which will accelerate at a much higher rate, you still have to accelerate the drivetrain at a faster rate which will require more power (torsional losses in the transmission, rear end, etc.)...the higher HP at the crank has an adverse affect on the HP seen at the wheels as well.

To use a real extreme example, I don't think you could attach a 400 hp SBC up to a stock f-body drivetrain and see the same static loss at the rear tires as you would see with a 7,000 hp top fuel engine (this is an extreme example only...I understand the fop fuel engine would tear the stock f-body to pieces). I think you would see a much higher (relative) final HP loss at the rear tires with the 7,000 hp engine.

Just thinking out loud.

 

Power out / Power in x 100% = ME. Empirical fact. Tested many times since the beginning of the Industrial Revolution in the early 1800's.

3 out / 10 in x 100%. ME = 30%. Not likely. If the ME of the Drivetrain is 85% (determined by testing), then it would be 8.5 out/10 in x 100% = 85%. You can turn an M6 drivetrain with a breaker on the trans input shaft with the drivetrain in the air. You can't generate even 1 hp that way! Even the best athletes only generate about 1/2 hp with their entire body. So it obviously doesn't take 20, 30 or 50 hp to turn the drivetrain. So the power it takes to turn the drivetrain VARIES, and it varies with power transmitted. Step up to 600 in and you will get 600 x 85% out = 510 out. In both cases the remainder of the hp will be used to heat up the drivetrain, mainly from internal friction.

If that doesn't seem right, think of what happens to drivetrain temp when you increase hp from, say, 300 to 600. You start to need additional coolers to stay in the safe range for the trans and diff. What can be causing the extra heat, except the extra power? You don't have an extra heater in there, do you?

The fallacy of your position is that you are assuming numbers, not quoting measured numbers. You're right that many scientific notions have changed over the years. They have changed because measurements did not bear out the theory. So they had to re-think the theory. In this case, detailed, careful measurements DO bear out the theory, over 200 years of measurements! ...that's just "The Way It Works". HTH.

 

OK, new thinkers, here are a couple of ways to change the overall ME of the total drivetrain! That includes all the components between the flywheel and the ground. Use slicker lubricant like a synthetic (less friction). Lighten the unsprung weight (wheels, tires, and brakes) for less rotational mass. Lighten the other components (flywheel, driveshaft) for the same effect. Note that a higher stall TC lowers the ME, because it slips more (viscous fluid loss). But it does increase acceleration because it puts the engine in its best power band sooner/longer. You can also use shorter gears (4.10 vs. 3.42) with the same results as the looser TC, dyno less, but accelerate faster.

If you don't do anything to change the ME, then the percentage is still good to use for the calculations of whp to chp. That's all it is, BTW, a calculation, unless you dyno the engine as well as the chassis hp for each change you make. If you're into more exotic things, go to straight gears instead of helicals for better ME. If you can put up with the extra howl.

 

I don't see the contradiction.

FWHP is taken on this board to be the TLA for FlyWheel HorsePower as RWHP is take to be the TLA for Rear Wheel HorsePower. (BTW, I really hate all the use of TLAs here (Three Letter Acronyms)) - they may make typing easier, but they tend to interfere with the communication of information.

The problem here is that people are trying to prove or disprove a theory (drivetrain loss is not a fixed percentage, but a constant) proposed in the first post of this thread with anecdotal evidence (the Miata has a 26HP loss from engine to chassis dyno regardless of the power level; Injuneer has 20% loss regardless of the power level) and guesses. There is no magic here, all the normal rules of nature apply. TeeKay understands this. For lack of a better term, it is an engineering problem. You identify the causes of loss and quantify them, and you have your answer for a particular test with a particular vehicle. But the point is that the "drivetrain loss" (implicitly defined in the first message as the difference between engine dyno and chassis dyno power measurements) varies with a number of things as both TeeKay and I have pointed out. These are real differences. Changing transmission oil or rear end oil can make a difference in "drivetrain loss" that is measureable and repeatable. What I am arguing is that measuring "drivetrain loss" with a DynoJet dynomometer is meaningless. A conventional (or maybe classical is a better term) dynomometer measures torque with an absorption device and RPM. From these you compute HP = Torque * RPM / 5252. Note that torque is a measured variable. The DnyoJet measures the RPM of the drum instead of torque. It then tries to infer torque but taking the rate of change of the RPM of the drum, or the acceleration of the drum, and convert that to force at the surface of the drum by the formula force = mass * acceleration using the mass of the drum. Then, using the radius of the tire, they convert that force at the drum surface to torque at the axle. Now with torque at the axle and RPM of the wheel, you can use that same formula to compute HP at the wheel. Note though that they use torque derived by computation, not measurement. The problem is that you are accelerating other mass than the drum - tires, wheels, axles, driveshaft, flywheel, crank, etc. The power used to accelerate these masses shows up in the result, but the amount is unknown, so they take a guess and add it in so that the HP number is not less than it would be on an absorption dynomometer. So any change you make in those unknown masses shows up as a power difference (ie "drivetrain loss") on an inertial dynomometer (like the DynoJet) but does not on an absorption dynomometer (like a Mustang). The power consumed in accelerating the drivetrain mass is real and can make a noticable difference in dragstrip times. I doubt very much that it makes as significant a difference in cup car lap times simply because the rate of acceleration of the driveline components is much less. Now, I say that measuring "drivetrain loss" on a dynojet is meaningless because it is a function of variables other than those that control the mechanical efficiency of the drivetrain, like HP. The more HP you have, the more you will "lose" on a DynoJet because of accelerating the mass of the drivetrain, but you will not "lose" that power on a Mustang because it does not accelerate the mass. (Well, it can, but then it will have the same problem as the DynoJet). You can't say that a T-56 absorbs X HP or even Y% of HP when the apparent absorption varies with the weight of the tires.

 

I agree with everything you said except for this:

Although you can't generate 1hp with a breaker bar, you can generate hundreds of lb. ft. of torque (which HP is a function of) which is more than enough to turn the drivetrain, then some...granted you would generate this torque at a VERY low rate of speed so you wouldn't see even 1 hp.

 

Torque = rotational moment of inertia x angular acceleration (the rotational accleration of whatever is spinning). The rotational moment of inertia depends on the weight and shape of whatever is spinning (in this case, all the components of the drivetrain). The moment of inertia is not changing so if the angular acceleration is to increase, then the torque consumption must increase as well

 

Simple, you first explain that heavier wheels have nothing to do and should not be acounted when measuring rwhp on a inertial dyno.

yet, here you say that that it will....

maybe im misunderstanding you but that does sound like a contradiction to me. I wish not to discredit but just bring up an observation.