960mm bore, 2800mm stroke. Three injectors per cylinder, inlet ports at the bottom, air operated exhaust valve at the top, and four superchargers. Started using 30 bar compressed air, blown directly into the cylinder
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220 + BHP from a 172 (Cheap)
- Thread starter Captain Slarty
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960mm bore, 2800mm stroke. Three injectors per cylinder, inlet ports at the bottom, air operated exhaust valve at the top, and four superchargers. Started using 30 bar compressed air, blown directly into the cylinder
it would be a bit hot, down side of summink this big made of metal is you have to keep it warm all the time it is in service, its standstill temp is about 85 degrees.
half of clio sport would fit in the crankcase, the other half in the exhaust. they dont do them in fast fit centres either.............
nah, they dont give you the points for company purchases!
See the difference? To get 172BHP at the wheels from a standard motor aint going to happen. Final Specific output depends on your final drive ratio, if it was 1:1 then you would get 172BHP at the wheels, but its not, it is probably something like 1.15:1 so your loss is what ever this works out at.
BUT..
.that is NOT the transmission loss, that is a mathematical relationship of the gearing to the torque and rate of change. it is not lost.. what you are referring to is the power scaling.the loss comes from absorbtion of power by the drivetrain friction and the accessories. at 1:1 you would never get 172 at the wheels.
your bhp remains a constant, its the torque that varies with drive train dividers or multiplyers, acceleration is the rate of work (ie - rate indicated a time elelment).. you can never have more than the flywheel figure of bhp, but the rate of acceleration of the mass can be varied by gearing to multiply torque.
Yes you can multiply torque in gearing, but this will absorb power. You have missed the point.
Final drive ratio is this simple, you have a flywheel creating 172 BHP output power. If you drive the output at the same revolution as the engine, i.e one rev of engine equals one rev of wheel, the output power will be the same, less frictional losses. But you mention torque multiplying, this is were its lost, and thats why ALL manufacturers must quote the final drive ratio, i think this is about 1.125:1 for the 172, then the maximum available at the wheel is only ever going to be 152 Bhp, 20 bhp down, and still within standards as acceptable loss.
Bhp is not constant, it changes with road/engine speed and full BHP is only available at maximum engine rated speed, thats why you have to go flat out on the rollers. A good transmission will not scale the power, it will maitain it and multiply its torque, making this power more usable, something which the 172 does really well in gear.
If this is not good enough then you can sue Renault for not quoting the specific output, selling under false pretense i think, thats why the euro boffins changed the law, so there is no grey area. 172 specific or 172 rated output? you decide.............
but, the power is measured on a rolling road for the purposes of this discussion. And, power does not change with road speed, it changes with engine rpm............
the losses from torque multiplication/division are frictional.
the reason you will never get the flywheel power at the wheels is nothing to do with final drive ratio as you elude to..
if you have an overall ratio of 2:1, the torque is divided by 2, Forgetting the frictional losses for the moment, the power remains the same at the same rpm with the same gross weight.. it doesnt change with road speed. it is the acceleration that changes.
in your example calculation, you are confusing torque with bhp.. a fairly common problem.. with a gearing of 10:1, the power is still the same. the ability t o cange the speed is determined by wieght of the car, and torque applied, if you reduce the torque by devision, the power remains constant at the same engine rpm, it is the acceleration of the overall weight that changes...
again, the ratio has nothing to do with the power loss apart from frictional losses.
I am not confusing anything, torque is a turning moment which determines the accelaration (pulling power) of the engine.
If youre into splitting hairs, Bhp is actually an illegal terminology in the EU, it is non specific. It is kilowatts, as this is an accurate measure of the engines output ENERGY, and it is this output ENERGY we seek to measure on the rolling road. If you are old fashioned you multiply the kw figure by 0.742 and a bit to give you Bhp. Some of this ENERGY is absorbed by the gearing by creating that turning moment necessary to drive the road wheels.
to back this up, my motor has a shop test certificate (like all handmade engines) it was dyno tested at 268kw (360 Bhp) before being mated to the gearbox. This Brake test also determines the turning moment imparted on the brake in newton metres, you know, how much force to rotate it through 2 pi rads if you want to get heavy, also known as torque.
Any way I collected it, drove round a bit, witnessed the roller test, and an indicated 252kw at the wheels (343 bhp), hence this is quoted as specific output, it is actually whats delivered to the road.
Literature for my motor quotes a final drive ratio for the SMG drivelogic box of 1.064:1. So if you divide dyno Kw output (rated power) by roller Kw output (specific power) you get the same ratio, plus a bit. That is how it works.
If you have created perpetual motion then let me know how, you have solved my biggest problem to date, I could half the size of my engine for the same geared output speed.
sorry Sir, but you are again wrong in your assumtions.
the power delivered to the road is only reduced by frictional losses in the transmission. Kw or bhp are of no relevance as the result is determined by the factors used and still identical.
the point here is that the bhp is not reduced at the wheels by some mathematical formula dictated by ratio... it is the frictional losses in the drive train - including road contact if you wish to be pedantic - that reduces the figure.. NOTHING else.
the frictional losses are not directly proportional to the ratio hence your argument holds no context.
the LOSS by friction is the only factor occuring. To use ratio to determine POWER loss is not appropriate.
I see a lot of this type of confusion. With no loss due to friction, the power at the wheels would be identical. in any gear - but the torque would NOT BE.!!!!.
thats what you need to comprehend. the torque available to turn the road wheels would be altered in proportion to the gear ratio which effects the ability to accelerate the vehicle mass. The acceleration is proportional to the reduction or otherwise of the turning force applied AFTER the gearing.. however, the ability to ACCELERATE this turning moment is the POWER, and that Sir, remains the same at a given engine speed. The final result of losses are frictional only.
So, loss dictates power figures at the wheels, NOT gear ratios. There is NO Mathematical link, the only factor is that the transmission losses due to friction are increased with more cogs in the equation.
If you take your math as a starting point.. (again assuming zero frictional loss)
with a ratio of 1.064:1, the turning force at the road wheel is turning force at the flywheel / 1.064.
The power does not change by this ratio range - the torque does, however.....
and the ABILITY TO ACCELERATE a given mass (AFTER the gearing) does....
the SAME power is applied to the SAME MASS (the vehicles weight in this case) so the RATE of change is different (again OF THE VEHICLE MASS)
The fundamental mistake you are making is to presume that power multiplication occurs with torque multiplication, the ability to reduce the amount of WORK is provided by gearing, the RATE (POWER) of doing that work remains the same. But, that is applied to a FIXED weight (the vehicle)
By confusing the fixed weight with the reduction in FORCE available from a geared moment, you are crossing the math. As the torque is multiplied, the ability to accelerate increases with the same power.. According to your math, the POWER also alters, it doesnt. its simply the work that can be done on the final FIXED mass (the car)..
Let me pose a question to you...
if the power loss is proportional to the gearing, and at 1.064 you lose 17bhp due to the gearing.. what happens if the gearing is .8:1// do you GAIN 90 bhp at the wheels ??.. ie 360/.8 ??
no m8.. you need to measure the coast down resistance on the rollers to calculate that approximately.. there is no set figure for transmission loss.. and no relationship of transmission loss to gearing.
or better still, put it on a dyno..
as discussed above, the rolling road is not interested in the power in a gear.. it is interested in the figure that most closely equates to the real world use and attempts to relate this to the ENGINE.. not the road wheel,,, the term AT THE WHEELS is the original measurement the road takes, not the result. In RR terms this means AT THE FLYWHEEL MINUS TRANSMISSION LOSSES DUE TO FRICTION.. its the terminology thats confusing.... use of any gear does not alter the result as the output is calculated based on the inputs AND added known factors (such as weight AND MPH !!!) specifically to eliminate any differential.
You appear not to have read the post, the dyno figure was 268 kw, the at the wheels figure was 252kw, the calculated final drive ratio, reciprocal of all the gear ratios, 1/2/3/4/5/6 is equal to 1.064. When you divide dyno by roller you should achieve the same figure as the calculated one (1.064), proved in my example by the fact the engine put out less power when mated to a transmission. This power difference (or loss) is quantified by the calculation. You appear to be unable to accept this justification, this simple calculation has been applied for years, and is known as an Industry standard, all manufacturers of engines sold mated with tranmissions are required to quote this. its called a Benchmark. As earlier stated torque measurement is done entirely differently, and we are not concerned by that.
It is that simple, you can recite Reeds Mechanical theory books all you want, You cannot simply say that these rule apply to all power transmissions because in practice it doesnt. Power is not lost, it is absorbed, rubber tyres, clutches, differentials etc.
Again, no one gives a sh*t about torque, moment what ever, basically all they want to know why the 172 doesnt do what it says on tin, and this is not a theory of mine, its an explanation automotive fact. Like i said, 150 bhp at the wheels is damn good, and is within an ISO tolerance, you cant argue against that.
aha.. now I see where you are going awray, and please dont take this as personal, but you do not appear to understand the way a rolling road works, and this is what we are discussing.
Some of the aspects you discuss are related to the measured ACTUAL torque at the wheels... this is NOT how the rr works directly.. but I can see that not understanding or appreciating this could cause the confusion in your argument.
A rr does not give you the ACTUAL torque reading as such, as said above.. If you apply standard theory, as you are doing, you will come to the wrong conclusion !.
A RR works by taking this ACTUAL reading in terms of torque and wheel distance travelled. It also looks at rpm.
From rpm and speed (mph) the RR calculates the transmission RATIO. This ratio (which I presume is what you are referring to, is only relevant to the final figure that is produced (the one on the paper print out or screen - you DO NOT SEE THE ACTUAL READING -).
If you read the posts again, then you will find I am discussing the rolling road measurement of power and tx loss, you are not.. you are discussing standard power theory and applying that without the factors that are used to produce a valid RR result.
the rr may well measure a different figure of power and hence bhp from the engine rating - based on the ability to spin the rollers, but - .. this is NOT the figure you get, and not how your output was given or calculated.
230 at the wheels or any figure is mathematically calculated from the actual torque, rpm of the engine, AND distance covered in a given time by the rollers.
So, dont mean to piss on yer fire m8.. but the figures you quote are ALREADY factored for in the result of the RR... please note that. !! this is again, what you appear to be overlooking.
Now, you are left with a difference in power from the calculations.. that is LOSS BY FRICTION, nothing else.
I will stress again, the rr result is NOT the figure in bhp torque DIRECTLY at the wheels.. it is a calculation based on this and the other factors that fully allow for ratio calculations.
A said, you cant factor them in TWICE lol...
I hope that makes sense.
thanks for the interesting discussion too.
Joe.
lol Dash, thats a concept... !!
naw.. dont like Diesels.. bloody great chug buckets m8...
Joe...
Are you able to sum up in a few words then and tell paddymph what his actual power at the fly wheel is and how you arrive at that figure?
Nothing is factored into a rolling road test, it is a simple indication of the output power being transmitted through the wheels. We are talking about where the 20 to 30 Bhp loss comes from that 172 owners appear to be experiencing on the test. Remember, the car has passed a type approval, it is within an international tolerance, i have merely explained how this tolerance is calculated.
Where does this loss come from then, you need to enlighten me, you obviously did a different masters degree than the one I did.
I design engines for a living, and I am NOT ALLOWED to claim my engine has a certain output if it doesnt, i get sued. You need to explain why it is lower and how Renault are allowed to get away with it being lower, instead of bringing torque and other irrelavent points to the argument. You said all this was factored in right?
So why does my engine produce more power WITHOUT the gearbox attached? and when it is fitted, why DOES the power reduction at the output shaft correspond to the final drive ratio?
QUOTE !!!
Nothing is factored into a rolling road test, it is a simple indication of the output power being transmitted through the wheels
this is becoming increasingly difficult to get through to you...
IT IS NOT a simple indication of power at the wheels !!!. with nothing factored.
It is HEAVILY factored.. .
thats how it works..
instead of bringing torque and other irrelavent points to the argument
my learned friend
.. you HAVE to bring torque into it....your xyz kw engine did NOT lose a percentage due to gear ratios on the rr.. it lost a percentage due to friction... the RR allowed for other apsects becasue they are known or can be EASILY calculated..
I am sorry if you cannot comprehend that, but please feel free to contact your local rolling road distributer or SUN themselves.. and ASK them....
and as for...
So why does my engine produce more power WITHOUT the gearbox attached -
it doesnt, it produces exactly the same... but, the output of the gearbox is of a lesser reading (from 2 factors.. friction and a multiplying / divider effect.... )this, is WHERE you are going wrong...
the power at the OUTPUT SHAFT varies, not of or at the engine.. I trust you agree ??
And again... you are not measuring the output shaft or wheels directly in the final result of a rolling road !!!. only the INITIAL data used to CALCULATE the result you, as the customer... sees....
it simply uses that to calculate appropriate figures.
The result is.... THE FIGURE AT THE FLYWHEEL MINUS A LOSS DUE TO FRICTION... but, you cannot accurately assess the amount of loss... but, you CAN say WHERE it is lost... presuming the starting figure was the same...
5 psi on tyre pressure can make 10 bhp difference on a RR.... simply due to contact area * 2 as opposed to * 1 on the road.. the tyre is not designed to be deflected in two areas in such a short space of a rev... to compare accurately you need to know all variables....
again, I am sorry, but you DO NOT seem to appreciate the way a rolling road works.
Joe.
Im happy that you understand what you are on about, I and probably others do not. I have yet to hear a valid explanation of where the 172 loses this 20-25 bhp.
I certainly wont be losing any sleep over it, tell me when someone achieves 172 Bhp at the rollers from a standard car.
but m8.. you will never achieve 172 at the rollers.. due to friction..
the variation is probably more down to the following, in no paticular order ..
Frictional losses and oil drag.
tyre pressures.. with a double contact point the rolling resistance is increased relative to wheel rpm.. the pressure in the tyre is critical to reading repeatability.. also, the design of tyre wall which is operating outside of normal parameters.
Temps.. not air specifically.. although this can vary from car to car.. mainly engine temps.... for an accurate comparison each needs to be at the same temp.
Fuel octane and history.. use of superplus or regular.. even a sligh fill of regular will take a few tanks of super to change the flash memory in the eeprom related to knock.. ie - you will not be getting the same ignition curve on an adaptive ecu unless the recent history is the same.
Weight.. of fuel, crap in the boot etc..
these are a few... there are others..
I hope that is a valid explantion for ya m8...
Thanks for the input...
Joe.
WOW!!!!
what a fab talk!
technical man, i couldnt imagine what sort of combustion goes on inside you gargantuan motor. Does it have a combustion chaber as such (i.e designed). ANd no offence, but a 2 stroke diesel must sound bloody awful......and geez, the fuel consumption, must carry half of the shipping tonnage in fuel alone haha.
But reading the posts you were both guilty of arguing totally different but confusingly similar topics, which i think was great! nothing like being pissed off to come up with an answer.....
TOPS!
Yes it was good. Agreed you dont and wont get the power at the wheels. I say there is a calculation for this, slarty says different. Both trying to prove a point and getting lost, need me white board.
No, Ben, two stroke sounds good, smooth, it only does 98 rpm flat out and its drowned out by the turbos screaming. 12000 tons of fuel in this particular ship, owner wants to get 8000 nautical miles on a fill, having bother with that at the minute, it just wont do.
lol m8, yep, when we meet up, we will take a pad n paper.. (a PUB Whiteboard
)Joe.
or you coudl always use the univerasl technical talk sign language which looks like you have epilepsy.........but at least you understand right ...haha
and agree..i love strokers...i was weened on semiburnt fuel from karts.....but doin it with diesel!...pong!
and how do you ignite it? (initially, assuming it still used compression combustion?), or does it only stop running for short periods, sice you asy its standing temp is 85 deg!