unfortunately, the way it works seems to me to be a bit redundant when viewed in the eyes of history. but that is a discussion that will get us no where. Also, this is not a whos's smarter rivalry or a showing of "Empirical knowledge". (i love hearing this and thinking that when in the renaissance the "new thinkers" were discredited though they had great ideas of machines helping our trade and goods production. You can never see it even if its right there inbetween your very eyes!") Also, basing theories and proving them on ithout exact measurements isnt anything new. Columbus had no way of calculation his theory but just follow the guiding stars. Primative civilizations had no way of knowning, for example, which material will make for better pottery, arrow heads, weapons, etc. Things were daelt with uncertainty until proven. I, in a way, am using a similar method of reasoning. Its not that im in complete denial of what science has acheived, im also an engineeer major, but i also like to think outside of the box. Reason wat has been denied for the majority of the community and see where it leads me. Its that quintessentially the basis of science?

Anyways,

If im not mistaken, mechanical efficiency is formulated by dividing the "ideal" mechanical efficiency of the machine and the "actual" mechanical eficiency of the machine. (ME= IMA or De / Dr / AMA or R / Ea) Inessence, an ideal machine has 100% efficiency which unfortunately, we are no dealing with. Our system or machine is a complex machine with multiple "anchor points" and measured values of resistance, friction, distances traveled, and the single force work that drives them. ( these anchor points, gears for example, are mulitpliers of power and increase the efficiency of the machine.) AMA is calculated by dividing the resistance by the effort force. The point that is taken here is that the resistance or friction, which contributes to power loss, is a percentage that is fixed. Thus when increasing the effort force on the object, the amount of friction will increase and therefore will result in more power loss. A power loss that is dictated by a fixed percentage.

I feel, that a fixed or a slightly variable coifficient, unlike a percantage but more like a scalable whole number or variable scalar number should be used instead (e.g. 25-26hp lost through the trans.) The problem i see here is that the internals have not been changed (just the trans for example) thus resistant force or amount of friction produced is the same. When enacting more effort force on the system or trans, you running it through the same amount of friction as before you increased the power. However, since you are now increasing power, yes you will have the gears, for example, spin at a faster rate and displace more heat, which is a bi-product of the energy being enacted by your power source, and not enact any more resistance on the power see through the rest or your drivetrain. The coefficient of friction is found through measurements that dictate the force or work (energy needed) to overcome friction and produce the necesary work. (in this case, spinning the trans.)

Thus, imo, when increasing the amount of power enacted on a system (trans) the frictional force stays relatively (and i say relative because since parts do wear, there strength is being jeopardized and the amount of efficiency produced will gradual decrease, though not from the increase of power but more of the factors that we are not dealing with a perfect system with an efficiency of 100%, which is why im proposing a variable fixed scalar number and not a percentage which would enact more energy lossed than necesary.) the same, though larger amounts of heat is increased and created from the extra work being enacted, which subsiquently isnt subtracted from the power enacted since it is a bi-product of the work unless the molecular sturture of the material used have reached dangerous levels and their bonds bein to deteriate.

simply, back to my first example, you have two EXACT cans being crushed by two EXACT seperate press'. One is exhibiting 100lbs of pressue and the other is at 200lbs. The can needs approximately, and im making this up, about 50lbs of initial force to overcome its molecular structure and be crushed COMPETELY. The first will have 50lbs of "left-over" or mechanical efficiency aka ME, while the last will have 150lbs of ME. The rate of the crushing of the can is proportional to this "left-over" or ME while the heat generated is proportional to the crushing rate, frictional coefficient of the can and the "left-over" power, yet it was created by the work being enacted by the press on the can, not subtracted by the power exerted on the can (power loss). IF the heat is untolerable and will somehow jeopardize the operation, than again a cooler of some sort could be added to maintain that balance....(cough* trans cooler! )

One of the basic principles of energy was that it couldnt not be created but transfered from one form to another. However, though energy cannot be destroyed, (though im starting to feel that you can in fact destroy energy..i know, crazy, but just another of my ideas that im working on) it can surely be created in many forms due to operations, releasing of potential energies, and numerous other activities that we see everyday. (the heat created by your engine, it is a product of the engine being created by the operaton and not subtracted by the net. though again our engines are cooled to maintain a certain level molecular support and longitivity.)

 

Tony you are a smart guy but this has to be one of the most NAIVE statements I have ever seen on this forum.

With this post you are ASSUMING that the "rated" 100 shot is an actual 100 shot when in reality it probably isnt. Manufacturing tolerances, manufacters marketing etc all play a factor in what those jets actually produce. In reality you probably were making more like 150 FWHP with the "100 shot jets".

I think the driveline loss should remain fairly constant no matter how much power is add but the nitrous statement is a bad argument.

 

I dropped some of the quote in the hope of making all this discussion clearer, not to squelch anything. Where I did, I added ellipses marks (...). The full post is above.

Paragraph 1. My sincere congratulations! An inquiring mind is always welcome to and admired in any scientific pursuit.

Paragraph 2. I abbreviated it some because you're exactly correct and we are both saying the same thing. Put all that you said together and it comes out to: P out/P in x 100% = ME%.

Paragraph 3. Although I hear what you're saying, I don't agree. Here's why: you aren't just "spinning the trans" like a top, you're getting the trans to DO WORK. In that case the ME comes into play.

Paragraph 4. The frictional force CANNOT remain the same! The loads on the components have increased and that WILL increase the friction and the force required to overcome it.

Paragraph 5. ... Sorry, I don't believe this is a good analogy.

Paragraph 6. Well, all I can say to that is I believe you're well on the way to a perpetual motion machine! Maybe you're the one to do it after all these years. Good luck!

Finally, if what you say has merit, then I'm sure that you could propose some practical experiment to prove your theory with measured data. Maybe with a small electric motor and a miniature drivetrain of some sort? You sound like you have the skill and training to make it happen. Skeptical experimentation is the scientific method after all, isn't it?

 

Homer Simpson to Lisa Simpson after she invents a perpetual motion machine: "Young Lady, we obey the laws of thermodynamics in this household!!! GO TO YOUR ROOM!!!"

hehe, figured I'd lighten the mood

 

LOL, thanks, that's a good one!

I really believe ObZidian is sincere. I don't want to discourage anyone who might have something! I consider this a worthwhile thread. To a lot of intelligent folks, the idea of increasing loss just doesn't seem to make sense!

 

You have to also remember on a chassis dyno that all the accessories need to be hooked up as well. Such as water pump, alternator, and power steering. All of these affect hp on the engine. So whoever is doing these tests need to realize that it takes power to turn all of these things. If the engine dyno has all of this and then you transfer all of this into the car and dyno we should have an accurate account.

 

yes, and i thank you all for keeping this most interesting thread alive and fruitful. I believe if we put all of our heads together we could, after much and possible heated arguements, which the laws of thermodynamics are obeyed respectively, we could result in an understanding and possible solution to our current problem.

back to the topic though, mabe i didnt word myself correctly but my point is that fixing a percentage when concerning power loss is like a predator tune, its generic. There tons of factors or variables that contribute to not only power loss but to how we achieve or try to achieve a perfect, simple machine with 100% efficiency. I only wish that we had the proper lab and funding of course to do such an experiement ut unfortunately, we mostly spend watevet money we have left on our cars.....which isn to bad but when looked at as a net worth value, not to promising..well, not yet!!!!

A percentage to me seems like a value given to estimate the frictional coifficient of our drivetrains. I understand that it isnt a fixed value either, i had stated that im proposing a scalar value or whole number that relative to a gradual decrease of increase instead of a precentage. I feel that though you do need more power, thus more frictional force applied, to spin or create or transfer more power through the trans, for example, a fixed percentage that isnt vehicle specific isnt correct but a false value.

the only way to truly have a variable number system of whole numbers to dictate the exact power loss through our drivetrains is to have each vehicle striped down and each anchor point, gear, disc, clutch, nut, bolt, the viscosity of the fluids used, all of the thermodyanamic values for each part, all of the frictional cooiffient of each part and catagorized for its specific grouping...etc. you get my point.

again, i dont think a fixed percentage is the way to go because saying that all auto trans loss this muh (17-20 percent) and all manual trans loss this much (15-17 percent) seems wrong to me since all of this trans are purposely built by different manufactures, with differnt materials, fluids, etc. So how can one that is better equipped to handle high temp power (materials used, coolers, frictional design) loss be catagorized with another trans that is a generic "performance" built unit? It just doesnt seem right.

The analogy i used was to desrcibe the different is mechanical efficency not through friction or resistance but so the significance of wat is mechanicl eficiency; the "lft-over" power is exactly that. THe more you have left over the more efficient your system will be....

tag, you it!!!!!

 

Unfortunately your can crusher is not a good example of ME. You only took into account the amount of force that the crusher can generate- you didn't take into account the power going INTO the can crusher to generate the 100 and 200 pounds of force. Just because the 200 lb system has 150 lbs of force available after crushing that can does not make it the more efficient system in terms of ME. That system could require (for example) a driving motor with 600 lb of force to operate for a ME of 33%. The 100 lb system may only require a motor with 200 lbs of force to operate for a ME of 50%. The units aren't correct, but you they work for the example. If you would relate this to a car drivetrain, you would be measuring the amount of power required to turn the drivetrain at a specific rpm, and not accelerating the drivetrain. The engine can obviously produce more power, but it's not required. While this is efficiency (power out/power in), it's not in the overall efficiency of the system.

 

I don't think a fixed percentage is going to be accurate all the time, but it will get you closer than a fixed value. For example, I'm pretty sure a 10 hp (say, lawnmower) engine could turn a stock f-body drivetrain. Given this, how is it logical to conclude that the drivetrain requires a fixed amount of hp to turn it? In order to support this, a 300 rwhp car would have ~ 310 fwhp, a 600 hp car would be making ~ 610 fwhp, etc...if nothing else, I think we can all agree that these flywheel HP figures aren't anywhere near reality. This isn't logical and it supports the idea that the faster you turn the drivetrain and the more power you generate at the crank, the more power the drivetrain requires to do its required work.

I would like to hear someones argument to this from the 'fixed hp figure' camp.

 

I dont think it is either a fixed number or a percentage. It is more likely a combination of both. I think a lot of the loss comes from friction and rotating weight at a given rpm. The reason a 12 bolt is less efficient than a 7.6 is from weight. The more you move that weight (rpm) the more power it takes. Since in most n/a applications more power is made from more rpm (heads,cam,etc) it would automatically take more power. Although at the same rpm I think both would see a close to equal percentage. Also, a 9" is less efficient than a 12 bolt, both due to rotating weight and a less efficient pinion location. Through testing thay have figured 3% more loss than a 12 bolt.
I'm sure if your were to take a trans and put it on a trans dyno it would take "x" amount of power to turn it at a steady 5,000 engine rpm, but it would probably take more power to acclerate it through that rpm.
I think the specific number idea has some merit to it, but it doesnt account for acceleration and actual speed. If you add a cam to an engine and increase its peak rpm by 10% (from 5500 to just past 6000) then there is no way you would see the same # loss at peak, altough you may see it at the same rpm (both at 5500).
I guess the easiest way to figure it out would be to look at a V6 and LS1 Camaro. They have nearly identical drivelines for automatics. If the percentage thing was true the V6 would have 164 rwhp, if the # thing was true it would have the same total loss as a LS1. What would those numbers look like? Anybody have any general numbers?

Like I said, I dont think either is exactly right, but I think the percentage is far more accurate than the "given #" theory.

 

my point exactly, the example was given to demonstrate, though i did leave other factors out for simplicity as you stated and much more on the table, it was just a showing of the ME of that, though simplie operation.

Again, that is the point im trying to acheive. The measuring of power increased/decreased on a dyno when mods are introduced and its apparent loss of power through different mediums.

When on a dyno, and you have swapped out you stock lid and exhaust for and aftermarket gmmg and slp lid. Prior you had done a dyno and ran your stock numbers. e.g. 292hp @ 6k. after the bolt ons, you now made e.g. 312p @ 6k. Your ME of your system or motor has now changed and is now measured at the same specific RPM as the dyno before......now, if you take it to the track, THAN you are now measuring its acceleration properties but on a dyno, you are only measuring its ME at specific rpm ranges, compared to previous pulls done.

 

Simple, the lawnmower has applied enough power to overcome the frictional coifficient of the trans or drivtrain thus spining it. Not very fast mind you, but enough to get it going.

You see, the percentage, like you said, gets you close but im a person of "exacts". That is the reason i dont have a hhp3 or predator and i own a hp tuner. I like to make sure everything is precision tuned. Also, the percentage is going to yield way to much power lose, imo. Thus imo not being close at all but further away from the truth.

From the fixed number camp, which i frequent, it simple when measuring ME at specific rpm ranges e.g. 2k-6k. Now if you increase your rpm range, you will require more power but not much. Thus the power loss through your drivetrains, will be a fixed number or very close to one. Not a percentage that will yeild more power loss as you increase engine power.

The problem is when measuring acceleration. As you spin the drivetrain faster, e.g. lower et's and higher mph in the 1320; you now require more power to spin the drivetrain becuase of its increase of fricitonal load, well, among other factors. A percentage at this point isnt correct either becuase it will yield more power loss than necesary because of the comparisons of different drivetrains into one category and the generic nature of its calculations. If anything, a formula could be introduced that would take that initial power loss than multiplied by the increase in load with the frictional coifficient at a specific speed or rate of acceleration.

just a guess....anyone have any ideas?

 

I should have said your can crusher is not an example of efficiency, instead of just saying it wasn't a good example. On the dyno, you are accelerating the drivetrain (just like at the strip) - the power reading you are getting is an instantaneous reading at a specific rpm taken during the whole run. You aren't holding the rpms at 6000 (or whatever it is) and taking a reading- I'm sure you realize that.

You did say "Now if you increase your rpm range, you will require more power but not much."
This is a contradiction to your next statement of "Thus the power loss through your drivetrains, will be a fixed number or very close to one". You said that the power loss through the drivetrain is not a fixed number by stating "you will require more power but not much". If it were a fixed number you would not "require more power".

 

The fact that the 10 hp lawnmower engine will turn the drivetrain throws out the fixed number theory. As you said, it WILL turn the drivetrain, but not very fast...but it is enough power to turn the drivetrain. Given this, the faster you wish to turn/accelerate the drivetrain, the more power it requires. The amount of power the drivetrain requires is dependent upon the speed you wish to accelerate/turn the drivetrain.

I understand where you're coming from, I too like things exact. The problem in this case is, there doesn't seem to be an exact answer. The rate of acceleration of the drivetrain, viscosity losses in the tranny/rearend, torsional losses in the tranny/rearend, different weight wheels/driveshaft, etc. all play a factor in the amount of HP required to operate the drivetrain. It's a dynamic function that can't simply be calculated by adding a number to RWHP dyno numbers or adding a certain percentage to chassis dyno numbers. However, I still believe the percentage will get us closer than a fixed number. I go back to the lawnmower engine example. If you can turn/accelerate the drivetrain at a very slow rate with a 10 hp engine, but can achieve a must faster speed/rate of acceleration (of the drivetrain) with a more powerful engine, but lose more than say 9 hp to the rear wheels, it's obvious we are dealing with a much more dynamic number than saying it takes 9 HP to turn the drivetrain. A 400 hp engine is going to lose much more than 9 hp through the drivetrain.


So, if I spin my 10 hp lawnmower engine from 2k-6k, I will see the same total HP loss at the rear wheels as I would with say a 500 hp engine spinning from 2k-6k? It's pretty clear the HP loss at the rear wheels is a dynamic number. A percentage loss to the rear wheels gives you a dynamic number that changes relative to the factors that increase the HP required to operate the drivetrain, a fixed number doesn't.

 

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....

 

again, i understand your point but please re-read my post and think out side of the box with a different set of perception glasses.

initailly, no, the amount of power to spin or cause a work function on any objuect is the surpassing of its frictional coifficient. If you wish to spin the object faster or work faster, than you are now increasing the load thus lossing more power. A percentage is a generic form of doing this since there hasnt been a conrete experimentaltion with concrete results.

A fixed number isnt my point but a percentage, imo, again, isnt either. IT will yield a closer reading because it is a variant figure but not close enough because of its genericity.

 

I understand 100% what you are saying. Believe me, confirmation bias isn't an issue here.

Agreed.

This idea is probably as close as we could get to a concrete formula. It would take A LOT of experimentation and some sort of regression to come up with a decent formula. However, we'd still be dealing with a good sized margin of error, but it would get us closer than a straight percentage.

I agree, it is generic. However, I think it at least gets us in the ballpark. Until someone invests thousands of dollars to conduct the aforementioned experiment, I'm afraid it's the closest we will come to estimating crank HP with any kind of consistency.

 

I am digg'n this thread....I have been thinking about this for the last 2 weeks and decided to see what was out there on the topic.

building on the previous, my input:
......formula .... that would take that initial power loss, multiplied by the increase in load with the frictional coefficient (of resistance) at a specific RPM. The coefficient would be different for a static environment (dyno test) vs dynamic environment (ie drag strip -- wind resistance or drag).

I do not agree that this has not been done yet. I am sure that one or more of the major car companies have had a team of physicists and mathmaticians figure this out (probably before many of us were born) -- we just need to hunt around and find the results.

 

I posted this same thing over 2 years ago here:
http://www.ls1.com/forums/showthread.php?t=289550

First let me apologize if this has been addressed - I didn't read the whole thread.

 

im sure, the idea of turbocharging has been around since the begining of motor vehicles. IF im not mistaken, i think a basic i dea was made in 1887 or cloe to that a used later in a steam engine int he early 1900'z

the formula should be accomidated fo specific rom and rate of acceleration induced by power output.