heya all just wandering if anyone could tell me or give me a web site that can convert or calculate the difference from fly wheel to rear wheel in kilowatts....


thanks
shaun

there isnt any real coverter...

although not completly accurate people generally say 30% drivetrain loss for an auto and 20-25% for a manual.

as i said this isnt really accurate but gives you an idea.

eg standard |6.. 157kw * .7 = 110kw.. people generally report between 105-115rwkw standard i believe

what he said.....

its impossible to give an exact figure for drivetrain loss as every car will have a different amount, plus different components can affect it too.... like, a 9" diff sucks a bit more power than a borg-warner

There is a big thread arguing about it somewhere, do a search and you might get something useful out of it.

hi

here is a few threads u might wanna read

http://www.fordmods.com/forums/viewtopic.php?t=17981

and these ones to

http://www.fordmods.com/forums/viewtopic.php?t=19941
( very interesting result in this thread )

http://www.fordmods.com/forums/viewtopic.php?t=14282

cheers

Last edited by FPV_GTp on Tue Feb 07, 2006 10:10 am, edited 1 time in total.

Lol - dont hold your breath though!

30% give or take... should get you within the ballpark... but prob best to do yourself a favour and talk in rwkw....

it depends on your driveline and each is different, the only way to measure engine output is on an engine dyno and that requires you to pull the engine and spend mucho penny's.
Having said that take a given value for example the standard 290KW boss cars are turning between 220 to 225 rear wheel KW on a chassi dyno which just happens to be what my AU is churning at the wheels (224rwkw) therefore I can safely say that it would be making some where around the same flywheel KW (except mines lighter and faster he he) (just kidding guys.)
Put it on a chassis dyno and re-post your figures and we all should be able to give you a ball park figure from known given values.

have a read of this...
link to the site http://www.pumaracing.co.uk/trans.htm


Puma Race Engines - Transmission Losses - The Last Piece Of The Jigsaw

When I started out writing these technical articles there was some sort of overall plan that the pieces would fit together in the end to form a picture of how maths and physics operate to determine how engines and vehicles work. The aim was to try and dispel at least some of the myth and s**t that pervades the tuning industry in the attempt to make people buy tuning parts that aren't properly designed and don't work. Fundamental principles are very powerful tools to help us decide what is and what is not possible. Unfortunately most people don't have the maths or education to apply these tools. Hopefully this website will have helped those who want to learn to appreciate some of those fundamental principles. One of my biggest hobby horses is the supposed flywheel power figures that tuning firms and magazines attempt to derive from wheel bhp figures taken from rolling road tests. In the main these are vastly overinflated. Many people seem to think that as much as 30% or more of the flywheel power is lost in the transmission. A simple reposte to that is that gearboxes would melt if they soaked up so much energy but there are more powerful arguments to use. Copyright David Baker and Puma Race Engines

My own figures for transmission losses have been stated in previous articles but to recap they are usually no more than 15% of the flywheel power for FWD vehicles and no more than 17% for RWD ones. A good guide is to deduct 10% of the flywheel figure plus another 10 bhp for FWD and 12% plus 10 bhp for RWD. If you are starting from a wheel bhp figure then you have to apply those equations in reverse - add 10 bhp then divide by 0.9 for FWD or 0.88 for RWD to get back to a flywheel figure. Those figures were not just plucked out of thin air. They represent the culmination of many years of research and testing combined with the views of reputable companies like Bosch and VW.

There is however a final way of estimating what transmission losses really are based on the most accurate dyno in the world - the car itself. Physics tells us that it takes a certain amount of energy to push a given shape through the air at a given speed. The maths behind this is not open to debate. It's part of the basic physics that determines how the universe operates. The article on how top speed and engine power are related goes into this in some detail. The power available to force a car to its top speed is obviously the net power at the wheels after all tyre and transmission losses. If we can work out this net power and we also know the flywheel bhp figure then the transmission losses must be the difference between the two. To work out the power requirements of a car based on its top speed we need to know its drag coefficient and its rolling resistance. For light vehicles, like passenger cars, the rolling resistance is about 0.013 x the vehicle mass. Drag coefficients are measured and published by the manufacturers. If we take a couple of examples we can see how this required power ties in with the engine's flywheel bhp. The equations are explained in more detail in the previous article so read that first if you haven't already done so.

The 1.8 Vauxhall Astra GTE (new shape from 1985 on) has a frontal area of 20.5 sq feet and a Cd of 0.31. With two people and some test equipment on board (which is how most reputable magazines do their tests) the car weighs about 2460 lbs. The engine is rated at 115PS (about 113 bhp) and the tested top speed is about 123 mph. let's see how much net power is required to achieve that speed.

Rolling resistance power is 0.013 x 2460 x 123 / 375 = 10.5 bhp

Air resistance power = 20.5 x 0.31 x 0.00256 x 123 cubed / 375 = 80.7 bhp

Total bhp at the wheels must be about 91.2 bhp to achieve that speed. If we apply my formula for FWD cars to the quoted flywheel power we get (113 x 0.9) - 10 = 91.7 bhp at the wheels. Hmmm - so you gonna step outside and fight me about 0.5 bhp or is this starting to make some sort of sense? Copyright David Baker and Puma Race Engines

Let's try a more powerful car.

The 2WD Sierra Cosworth was rated at 205 PS (about 202 bhp). Top speed was in the 145 mpg region according to most magazines. Test weight with 2 people and 50 lbs of equipment on board is around 3060 lbs. Frontal area is 21 sq feet and Cd is 0.35.

Rolling resistance power is 0.013 x 3060 x 145 / 375 = 15.4 bhp

Air resistance power is 21 x 0.35 x 0.00256 x 145 cubed / 375 = 153 bhp

Total net power required is 168.4 bhp. Apply the RWD formula to 202 bhp and we get (202 x 0.88) - 10 = 167.8 bhp.

The conclusion here is pretty obvious. If transmission losses were as high as 30% then there just wouldn't be enough power left at the wheels to achieve the top speeds that the cars actually show. The Cosworth would only have 141 bhp at the wheels if this were the case and its top speed would therefore only be around 136 mph. You can work that out for yourself by applying the formulae above. The Vauxhall would only have 79 bhp at the wheels and be capable of around 117 mph. Copyright David Baker and Puma Race Engines

Of course every magazine test achieves a slightly different top speed and published drag coefficients vary a bit depending on the source. If you carefully select your data depending on what you are trying to prove you can show just about anything. But if you run enough of these calculations and also factor in the acceleration from computer simulations you start over the years to build up a picture of how things really work. The conclusion is that transmission losses are much lower than commonly quoted. The only reason to apply big transmission loss percentages is to flatter the supposed flywheel power outputs from poor tuning work.

Ask any tuning firm which applies these big transmission losses why they do so and about the best you are going to get in reply is "that's what we've always done" or "we read it in a book somewhere once". Hopefully a proper scientific argument like the above will be a bit more convincing. You don't even have to take my own word for it. All the equations I use are part of fundamental physics and commonly quoted. The book "Internal Combustion Engine Fundamentals" by J.B.Heywood is a good source and he is the professor of automotive engineering at MIT so dispute him at your peril.

I was thinking the same thing...

Interesting read there tickford_6!
Pretty much blows the 30% figure out of the water!
I think it's best to think of dynos as a tuning device, and quarter mile times as bragging rights.

i agree with you.. and a good tunning device too. but there is still a need for finnal tunning to be done real time on the street...

most 1/4 mile are crap too and have not a great deal to do with power.
to many other factors on the strip to trust 1/4 mile times.
by the power my car makes on a dyno and the calculators to work out 1/4 time form power and weight, my car should run 14.5seconds. best i can do is 15.00



i've posted that befor. and i'm going to post it in every thread like this.
because it's the only scientific argument i've seen, and it make more sence then %30 uleh

hi guys

The amount of crap people post and then there all experts on dynos and drag racing ,engine building and quote mathematical equations and some of the people haven't even driven a car on a chassis dyno nor engine dyno and for that matter run a car down the quarter mile drag strip makes me laugh

Do i enter the argument here na i give up LOL

again i will put anyones car on my dyno and lets see the figures

cheers

it dosnt have a flywheel one but there is a bunch of others fun to have a look at any how http://www.surefloexhaust.com.au/frame1024x768.htm

well put it to the test. you find an engine dyno, you already have a chassis dyno. i'll donate my engine.

well run it as it is now. then when i built by new one we'll run that too,

we'll see if your much beleived in %30 s**t is right or not..

i will bet my car on it not not having doulbe the driveline loss when it has double the power.

by the %30 rule my bosses 180sx has 160kw drive line loss. and it's power output passes the safe limmit for a stock crank........

am i understanding this correct ( example this is testing totally on chassis dyno at this stage please note this

let me pose this to u , if my engine produces 300 rwkwatts using the 30% method i should have 200 flywheel kwatts YES do we agree here ???


now i make more power out of my motor just say for arguement sake i ( bigger cam , more head work , up the compression ratio and so on ) now have the same engine now producing 600 rwkwatts im using the same car same gearbox same diff same tyres and rims just producing more power

u tell me what the drivetrain losses are now this the new engine in the same test vehicle ? and WHY ?

just think about ur answer carefully ????

i can speak from experince on dynos and race track thats why i laugh some times LOL

cheers

hey, tickford 6
have you taken wheelspin & gear changes into consideration?
that can cost 1/2 second
I dont believe the max HP and vehicle weight is accurate enough to calculate 1/4 times
for example the last car I built for street/strip racing had 581hp at the flywheel, but this power had a very short band 4500 to 6800rpm, if my car wasn't geared to achive max speed by the finnish line, it would not get its most efficient ET ,the more time a car spends in its peak power the quicker it will be.
another factor is how the suspension is set up, eg how much power is wasted by pulling the a** down to the tar, thats why drag cars use quite firm rear settings and use low pressure in the tyres
play around with ur tyre pressures and see how it helps/costs