Max Horsepower vs. Max Torque
Mike Christensen
also
Max Horsepower = Max Force at the rear wheel
However
Max Torque at the engine -not equal to - Max Force at the rear wheel
Power = F x v
For any given velocity, the maximum force at the rear wheel (and therefore
the maximum acceleration) will be achieved when the engine is delivering maximum
horsepower (obvious from the above equation). At any other power level, such
as the maximum torque operating point, which always occurs at a power level that
is lower than the maximum, the force will be less as will the acceleration.
This argument can be made for every velocity encountered. Therefore to achieve
maximum acceleration, operate your engine at its maximum horsepower point, not
its maximum torque point.
Mike Christensen
Darren Lyon
What if you have a samll bore/long stroek engine. The acceleration off
the line would be good, but top end speed would be horrible. This is
because the engine has an incredible amount of torque compared to its
horsepower, so it can get off the line much more quickly. By having
a big bore/small stroke engine of the same capacity, are you saying
it would accelerate better than the other engine ??
I think the cars today have proven this wrong, why would they have
1.6 V6 engines then?? or for that matter why do F1 cars have V10 1.6
or V12 litre engines ???
--
============================================================================
Darren Lyon, Griffith University, Australia.
GUYNN, RICHARD CARL
>1.6 V6 engines then?? or for that matter why do F1 cars have V10 1.6
>or V12 litre engines ???
The two major reasons for increasing the number of cylinders for a given
displacement (which I don't see how this relates to bore stroke relationship) is
to: 1) Make the power delivery smoother; the more cylinders the more power
strokes per engine cycle, regardless of displacement and more importantly 2) to
decrease dynamic load on the mechanical parts. F1 cars in particular would use
a larger number of cylinders so that they can rev higher. I think they generally
rev in the 10,000 - 12,000 rpm range. Doing this with, say, 4 cylinders
displacing the same volume would either destroy the engine instantly, or cause
it to wear out in a real hurry.
I'll have to check my books on the bore/stroke relationship, it's been
awhile since I read that part.
***************************************************************************
*Rick Guynn -MG driver soontobe. * MGA 1600 MkII *
*RCG...@zeus.tamu.edu * Rebuild (complete) to be finished ?? *
*Texas A&M University * *
*Keeper of the eternal octagon * `69 MGB Roadster *
* a.k.a. The marque symbol that * I'll have it registered soon, *
* refuses to die. * honest! *
***************************************************************************
Peter Tong
hp rating! I assume you mean shifting at the max hp rpm point. This is
untrue. For maximum acceleration you want to shift at that point where the
torque loss of the next higher gear will be minimized. You want to maximize
the area under the torque curve in each gear. To do this means that you
usually rev past max power rpm. Torque is what accelerates the car, power
is what holds it at that speed. So to figure the best rpm to shift at you
must have a torque curve for your engine as well as the gear ratios, and
then figure out where to shift to maximize the area under the torque curves
when shifting using the available gear ratios.
Peter Tong
Peter Tong
>mmm interesting, but could someone answer this question...
>What if you have a samll bore/long stroek engine. The acceleration off
>the line would be good, but top end speed would be horrible. This is
>because the engine has an incredible amount of torque compared to its
>horsepower, so it can get off the line much more quickly. By having
>a big bore/small stroke engine of the same capacity, are you saying
>it would accelerate better than the other engine ??
>I think the cars today have proven this wrong, why would they have
>1.6 V6 engines then?? or for that matter why do F1 cars have V10 1.6
>or V12 litre engines ???
>
acceleration depends. By acceleration do you mean 0-60, 50-70, or 60-100?
Or do you mean 1/4 mile times? Engines having low rod length to stroke
length ratios (eg. 1.5)have pistons that don't accelerate and decelerate to TDC
nearly as quickly as motors having high rod/stroke ratios (eg 1.9). high
rod/stroke ratios mean the pistons approach and leave TDC much faster and
generate their "sucking" action in a smaller time. These engines tend to
have most of their torque concentrated in the upper ranges of the power band.
Engines that have lower ratios generate a broader span of torque at lower
RPM but usually with less peak torque than the high ratio motors. Acceleration
characteristics also very much depend on the final ratio of the tranny. You
can compensate lack of low end torque on a high rpm high rod ratio motor
by increasing the final ratio - multiply the torque such that the motor
reaches its powerband quicker (to mask lack of low end torque off the line)
and applies its torque over a smaller mph interval. For eg. on my
Rabbit GTI (3.45 1st gear, 3.91 final ratio) 1st gear did 2.95s 0-30 and at
redline I was doing about 33mph in 1st. I could be even faster off the line
by increasing the final ratio but then 1st gear would only be going 27 mph
at redline. There is a limit to how much you can mask the lack of low
end torque on high ratio engines. You can mask it somewhat by adding
displacement (eg. running a wild cam and compensating for low end torque loss
by adding displacement). This all applies mostly to first gear since you
must run the engine in its lower rpm band from a stop (unless you like
dropping clutches and burning them out all the time). In second gear
acceleration really depends more on the higher rpm torque characteristics
of the engine and where the gearbox drops you rpm wise when you do the 1st
-2nd shift. On a highrpm power motor, a wide ratio gearbox wouldn't be optimal
since you would be losing too much torque when you shifted = too much of an
rpm drop. You would want closer spaced gears to keep you where the engine
really pulls. Concerning 1.6 V6s , F1 V-12s you have to realize that
racing engines don't really need to generate low end oomph as the run optimally
at extremely high rpm 10000-11,000 rpm in some cases. They are aiming for
maximum hp = maximum cylinder breathing /perstroke at the rpm range the car
will need to sustain very fast racing speeds. The reason they run with V-12s
is simply to rev higher to generate more power. A v-12 has lighter rotating
rods, valve train, pistons, and can therefore rev higher. If you could get
a 4 cylinder of the same displacement to rev as high without flying apart
it would make more hp, simply because it would have less motoring friction
than the v-12. But you won't find an automotive 4 that will rev that high
and hold together. So going back to your original question, generally
speaking given 2 cars with the same mass, aerodynamics, gearing, the
low ratio engine would beat the high ratio engine in the lower rpm ranges
(0-30 mph) but the high ratio engine once it winded up would accelerate
faster in the higher mph range. Who would wind really depends therefore
on what mph range (0-30, 40-60) you are comparing. The longer the
comparing distance the more honest the comparison. That is why the quarter
mile is a much better indicator of overall engine transmission performance
than 0-30 or even 0-60, because it measures not only off the line squirt but
high end power as well.
Peter Tong (from Rec.autos.vw)
Patrick Taylor, The Sounding Board
>What if you have a samll bore/long stroek engine. The acceleration off
>the line would be good, but top end speed would be horrible. This is
>because the engine has an incredible amount of torque compared to its
>horsepower, so it can get off the line much more quickly. By having
>a big bore/small stroke engine of the same capacity, are you saying
>it would accelerate better than the other engine ??
>I think the cars today have proven this wrong, why would they have
>1.6 V6 engines then?? or for that matter why do F1 cars have V10 1.6
>or V12 litre engines ???
I don't think the points are incompatible. All he's saying is that you
should gear the car so that the engine operates in its highest bhp range,
for acceleration. This range will be lower for the type of engine you
describe. To get the best out of such an engine (like mine), gear it low,
so that torque gets converted to motion.
---------Visit the SOUNDING BOARD BBS +1 214 596 2915, a Wildcat! BBS-------
Psychotherapist ... Psycho-the-rapist ... coincidence? You decide.
Patrick Taylor, Ericsson Network Systems
exu...@exu.ericsson.se "Don't let the .se fool you"
Erik van Bronkhorst Code C02313 Phone 939-1421
I agree with Peter. However there is one semantic discrepency amongst the
participants of this discussion regarding acceleration.
We are constantly accelerated by 1g toward the surface of the earth. The
fact that we do not "speed up" is because there is a constraint, or
balancing force of equal value holding us up (our feet against the floor
for example). When speaking of acceleration, it is confusing to say
A = F / M (from f=ma) since it assumes quite a lot. In the real world,
there are many forces an engine must work against in order to increase
the velocity of the car. Saying that a given torque produces a given
force and *therefore* a given acceleration is all well and good, but
the car will not actually SPEED UP by that amount of acceleration because
some of the engine's force is used up elsewhere. I will only mention
wind resistance since from 50 mph it is important and at 100 mph it is
quite significant. One common flaw when considering torque as the unique
driving force is that it assumes the car is at rest and the rpm is not
changing.
Ideally, a CVT holding the engine at its torque peak would produce the
greatest actual change in speed of the vehicle from zero to 50 mph. IMHO
Without constantly variable gearing, in order to achieve max change
in speed from zero to 50, the area under the torque peak must be maximized
by selecting shiftpoints and/or ratios.
It is easy to believe the above two paragraphs if one looks at high
performance torque converters for auto trannys. They all have a
"stall speed" around 2500-4500 rpm. This allows the engine to wind up
to just below its TORQUE PEAK before engaging the wheels. If peak HP
were the panacea, the stall speed of drag converters would be around
5000-7000.
However, at higher speeds, the force generated at the tires is not *all*
accelerating the car, some of it is fighting frictional forces!!!
I.E. Torque / Mass is NOT Proprotional to acceleration anymore!
Torque determines the force applied to the ground by the wheels. This
force in a frictionless vacuum would produce A equal to F/M. However,
in the real world, there is friction. Friction which goes up non-linearly
with respect to velocity. This is where the confusion over power enters in
since the frictional force (at a given velocity) may be though of as a power.
Basically all I'm saying here is that
(1) Fe = T * DR Engine Force (at the tires) = Torque times Final Drive Ratio
(2) A = (Fe - Fd)/M Acceleration = Engine force minus Friction force / Mass
Note however
(3) Fd = Cd * V**2 Friction equals coefficient of drag * Velocity squared
Substituting 3 into 2
(4) A = (Fe - Cd * V**2)/M
This is the acceleration of the vehicle. It is dependent on velocity,
and therefore seems to bring the consideration of power into play (but
not really).
Given P = F * V
(5) Pe = Fe * V Engine power = Engine Force times Velocity
(6) Pd = Fd * V Drag Power (or frictional power) = Drag Force times Velocity
(7) Fe = Pe / V
Substituting 7 into 4
(8) A = (Pe / V - Cd * V**2)/M
For clarity assume Cd = 1
(9) A = (Pe / V - V**2)/M
From this, one can see that at low velocities, V**2 will be smaller than
Pe/V and that at a certain Velocity Pe/V will equal V**2. That point is the
maximum speed of the vehicle.
Of course, 9 can be expanded into
A = (Fe * V / V - V**2)/M
A = (Te * DR - V**2)/M
But this Te is at some particular engine rpm, that is, it is in effect
a function of V and DR. This is where the confusion begins.
If you start with
Pa = Pe - Pd Available power = Engine power - Drag power
In order to use
F = Pa / V
A = F / M
And go through all the substitutions the other way, you will wind up
with exactly equation 4 again and, hopefully realize that the what
determines how quickly your car gains speed is the *shape* of the
torque vs rpm curve and the gearing. Both are equally important.
Peak torque is not since it if the car is in fact accelerating (non
CVT), it is only at the rpm where peak torque occurs for an instant.
Engine Horsepower is a simple function of Torque. It has no other
special significance but in some circles it is more comfortable to
use.
Feel free to correct me where I have screwed up.
;-)
--
Erik van Bronkhorst KC6UUT DoD#4342585443 AMA#[classified]
"Truth is false and logic lost, now the fourth dimension is crossed..."
"You slice Orc child's head"
Craig Bradford
Phone 939-1421) writes:
>>>>> LOTS DELETED >>>>>
>
>It is easy to believe the above two paragraphs if one looks at high
>performance torque converters for auto trannys. They all have a
>"stall speed" around 2500-4500 rpm. This allows the engine to wind up
>to just below its TORQUE PEAK before engaging the wheels. If peak HP
>were the panacea, the stall speed of drag converters would be around
>5000-7000.
>
>However, at higher speeds, the force generated at the tires is not *all*
>accelerating the car, some of it is fighting frictional forces!!!
>I.E. Torque / Mass is NOT Proprotional to acceleration anymore!
>
>
>;-)
>--
>Erik van Bronkhorst KC6UUT DoD#4342585443 AMA#[classified]
>"Truth is false and logic lost, now the fourth dimension is crossed..."
>"You slice Orc child's head"
Erik, the determination of optimum stall speed for drag converters is based on
a number of variables. Probably the most important is available traction to
transfer the power into motion. Next is the amount of heat generated in the
converter during stall. Finally, it is a means of storing energy much like a
flywheel. Numerous racers regularly use stall speeds well above 5,000rpm.
In addition, most pro stock and alcohol racers launch in excess of 8,000 rpm.
I would suggest that energy storage is the largest contributor to launch
acceleration as long as there is sufficient traction.
Fuel cars, on the other hand, have more power than can be transfered to the track
for most of the 1,320 feet of the race. This is why they launch at relatively
low rpm, and use a sliding clutch mechanism for at least 1,000 feet to transfer
the maximum power that traction can handle.
Regards,
Craig
===============================================================================
A drag racer's perspective of life -----
"How much horsepower can I have and still go to heaven?????"
===============================================================================
Mark Shaw
|> >>For any given velocity, the maximum force at the rear wheel (and therefore
|> >>the maximum acceleration) will be achieved when the engine is delivering maximum
|> >>horsepower (obvious from the above equation).
|> >>
|> >>
|> ..................................Torque is what accelerates the car, power
|> >is what holds it at that speed. So to figure the best rpm to shift at you
|> >must have a torque curve for your engine as well as the gear ratios, and
|> >then figure out where to shift to maximize the area under the torque curves
|> >when shifting using the available gear ratios.
|> >
|> >Peter Tong
|>
|> I agree with Peter. However there is one semantic discrepency amongst the
|> participants of this discussion regarding acceleration.
|>
|> there are many forces an engine must work against in order to increase
|> the velocity of the car. Saying that a given torque produces a given
|> force and *therefore* a given acceleration is all well and good, but
|> the car will not actually SPEED UP by that amount of acceleration because
|> some of the engine's force is used up elsewhere.
Mike, Peter and Eric are all right at different points. But Mike's recurring
claim is only that max acceleration occurs at maximum horsepower. This can
be empirically shown to be true by selecting a given vehicle speed and allowing
different engine speeds for max torque and max hp points. The end result says
that you will get maximum force at the wheels at that speed only if:
5252 x Max Hp > Max Torque x (rpm at max torque)
which is true for practically all production IC engines. But Mike's argument
is of little value in the real world were you must operate with fixed gear
ratios over a range of speeds. For a single gear the maximum force on the
wheel will occur when:
Max Torque x (rpm at max hp) > Max Hp x 5252
which is also true for most production IC engines!
So if you have an infinitely variable gear box so that you can maintain your
engine at its max hp point that would be great, but in the real world your have
to choose a gear! This is where Peter and Eric (and some others) show that they
really have a grasp of the problem. You simply cannot use either argument to
predict acceleration alone.
If and only if you take the time to run an analysis of the available torque at
each speed in each gear and offset it with the vehicle loads for those speeds; then
and only then can you predict which setup or shift points will work best. So you
not only need the engine performance curves and gear ratios, but also the wind load
vs. speed, rolling resistance vs speed, grade (if any) and inertia loading of
drivetrain (which is worse at higher rotational speeds). But with my own past
history in doing these types of analyses, when the dust settles and the smoke
clears you will usually (not always, but usually) find that most vehicles are
geared for maximum performance when shifting up shortly past the HP peak and
shifting down at the torque peak. Running to redline or lugging below torque
peak does little good.
Optimization beyond that takes a lot more source data and analytical work than
most of the respondents on this net seem willing to do.
So, Mike's point may be correct in a limited case; but lacks real value in
determining any real world performance. Now, can we get on to something else?
Mark
Unknown
derived an equation to approximate forward thrust of an engine operated
with a 'real' world fixed gear tranny.
If we recall that expressing HP as a function of force and velocity
given the definition of power : P=W/t then HP=W/t/33,000ftlbs/min. W (work done)
can be expressed as : W=s*F where s=distance(movement) & F=lbs of force applied.
(s is defined by revolutions*PI*Diameter)
Substitute s*F for W in the HP equation : HP=s*F/t/33,000. Seeing that s/t is
in reality distance / time (velocity in feet/minute) we substitute v=s/t
Thus... HP=Fv/33,000. To get to MPH we divide by 1mph = 88 feet/min.
HP= Fv/33,000/88 ----> HP=Fv/375 ----> F=375*HP/v
Using the final force-velocity equation as a basis, the fixed gear foward thrust
(and potential G acceleration) can be approximated. Additional data is required
1. Torque curve or Horse power curve throughout the rpm range (or estimation)
2. axle ratio
3. gear ratio of each transmission gear
4. tire diameter
5. Vehicle weight
mph = (RPM*tiredia) / (axle_ratio * trans_gear_ratio * 336)
hp = (torque * rpm) / 5252
F = (hp * 375) / mph
A = (F - road_load - friction) / weight {assume friction and road load = 0}
{for hypothetical comparisons }
Calculate F for as many points on the torque curve as you want. Do this for as
many gears as you have. Where F(gear n) =< F(gear n+1) that's your shift point.
If you plot F on a graph against mph you will generate several curves. One for
each gear. The peak values for F will occur where the peak torque occurs.
EX: *=2nd gear curve
o=3rd gear curve
+=theoretical optimal thrust curve if engine is operated at max power
(if you print this and connect the dots its a lot clearer)
note that where 2nd gear & 3rd gear curves intersect the upshift should
occur. The actual thrust curves become tangent to the optimal thrust
curve at max hp, however max acceleration occurs at the peak of each
actual gear curve.
Lbs
Thrust
4000|
|+
3500|+
|
3000| +
|
2500| +
| +
2000| +
| ***+
1500| * *+
| * ooooo* +
1000|* oo * oo+
|o * o +
500| * o +
| * o
0|
+---+---+---+---+---+---+---+---+---+---+---+---+---+--
20 30 40 50 60 70 80 90 100 110 120 130 140 150
mph
By looking at the graph it is obvious that the more times you can bring the
actual thrust curve up to the optimal curve the better you will accelerate.
so. A six speed close ratio tranny achieve optimal thrust six times wheras
a three speed only three times.
Okay I've bored you all enough,
$0.02
Ericy
*---------------------------------+---------------------------*
| Eric Youngblood |
| Bell-Northern Research _ |
| Richardson, Texas 75082 _| ~- |
| \, _} |
| \( +---------------------------|
| | Peon w/o Email privs |
*---------------------------------+---------------------------*
Clive Apps
>In article <C36pA...@gucis.cit.gu.edu.au> dl...@gucis.cit.gu.edu.au (Darren Lyon) writes:
>>mmm interesting, but could someone answer this question...
>>What if you have a samll bore/long stroek engine. The acceleration off
>>the line would be good, but top end speed would be horrible. This is
>>because the engine has an incredible amount of torque compared to its
>>horsepower, so it can get off the line much more quickly. By having
>>a big bore/small stroke engine of the same capacity, are you saying
>>it would accelerate better than the other engine ??
>>I think the cars today have proven this wrong, why would they have
>>1.6 V6 engines then?? or for that matter why do F1 cars have V10 1.6
>>or V12 litre engines ???
>>
>> Darren Lyon, Griffith University, Australia.
>Basically, whether an long stroke or shortstroke motor would win in
>acceleration depends. By acceleration do you mean 0-60, 50-70, or 60-100?
>Or do you mean 1/4 mile times? Engines having low rod length to stroke
>length ratios (eg. 1.5)have pistons that don't accelerate and decelerate to TDC
>nearly as quickly as motors having high rod/stroke ratios (eg 1.9). high
>rod/stroke ratios mean the pistons approach and leave TDC much faster and
>generate their "sucking" action in a smaller time. These engines tend to
>have most of their torque concentrated in the upper ranges of the power band.
>Engines that have lower ratios generate a broader span of torque at lower
>RPM but usually with less peak torque than the high ratio motors. Acceleration
RANGE>characteristics also very much depend on the final ratio of the tranny. You
>by increasing the final ratio - multiply the torque such that the motor
>reaches its powerband quicker (to mask lack of low end torque off the line)
>and applies its torque over a smaller mph interval. For eg. on my
>Rabbit GTI (3.45 1st gear, 3.91 final ratio) 1st gear did 2.95s 0-30 and at
>redline I was doing about 33mph in 1st. I could be even faster off the line
>by increasing the final ratio but then 1st gear would only be going 27 mph
>at redline. There is a limit to how much you can mask the lack of low
>end torque on high ratio engines. You can mask it somewhat by adding
>displacement (eg. running a wild cam and compensating for low end torque loss
>by adding displacement). This all applies mostly to first gear since you
>must run the engine in its lower rpm band from a stop (unless you like
>dropping clutches and burning them out all the time). In second gear
>acceleration really depends more on the higher rpm torque characteristics
>of the engine and where the gearbox drops you rpm wise when you do the 1st
>-2nd shift. On a highrpm power motor, a wide ratio gearbox wouldn't be optimal
>since you would be losing too much torque when you shifted = too much of an
>rpm drop. You would want closer spaced gears to keep you where the engine
>really pulls. Concerning 1.6 V6s , F1 V-12s you have to realize that
>racing engines don't really need to generate low end oomph as the run optimally
>at extremely high rpm 10000-11,000 rpm in some cases. They are aiming for
>maximum hp = maximum cylinder breathing /perstroke at the rpm range the car
RATIOS
>will need to sustain very fast racing speeds. The reason they run with V-12s
>is simply to rev higher to generate more power. A v-12 has lighter rotating
>rods, valve train, pistons, and can therefore rev higher. If you could get
>a 4 cylinder of the same displacement to rev as high without flying apart
>it would make more hp, simply because it would have less motoring friction
EXPOSE TO THE VALVES AND INCREASE VOLUMETRIC EFFICIENCY
EXPOSE TO THE VALVES ;IE: BETTER OLUMETRIC EFFICIENCY
>than the v-12. But you won't find an automotive 4 that will rev that high
>and hold together. So going back to your original question, generally
>speaking given 2 cars with the same mass, aerodynamics, gearing, the
>low ratio engine would beat the high ratio engine in the lower rpm ranges
>(0-30 mph) but the high ratio engine once it winded up would accelerate
AGAIN THESE ARE REVERSED
Mark Shaw
|> I know everyone is basically sick of this subject, but as I promised I
Well, yes.
|> A = (F - road_load - friction) / weight {assume friction and road load = 0}
|> {for hypothetical comparisons }
Ignoring road load can have a big effect on the available torque, especially
over the wide speed range given. The net effect will be to turn the acceleration
curves downward much steeper and make it harder to find optimal shift points.
But, I have to give you an "A" for effort and understanding of the problem.
It is far more complex than just some simple rule about max torque or max hp.
BTW, this month's Road & Track had an interesting discussion in the Technical
Correspondence section about two cars with the same engine, but different
transmissions. The importance of a good launch and quick shifts actually
made the automatic quicker than the manual in a quarter mile. This is
despite the fact that the automatic tends to be less efficient.
Mark
George Pachulski
The car has about 75K 's on it and is in good condition. The roblem
stems from the fact that after starting , it cannot idle smoothly. It has
had a tune up and runs fine. If you switch from park to drive too quickly
it will die. if you drive it it runs fine until you stop then usually idles
very rooughly and dies ocationally. Since it has so few miles I wonder if
someone could help diagnose the problem.
Also when the crate is hot it floods easily---- a manufacturing problem?
the saleman told my wife the day after she bought it this was normal
for this beast !!! is this true ? or can this be fixed?
One last request Does anyone know the baud rate at which the Chrysler car
computers run . I may attempt to talk to it , unless Diacom dosn't temp
me first?
Thank You
George
Unknown
|> In article <C3L8C...@news.rich.bnr.ca>, (Eric Youngblood) writes:
|> |> I know everyone is basically sick of this subject, but as I promised I
|>
|> Well, yes.
|>
|> |> A = (F - road_load - friction) / weight {assume friction and road load = 0}
|> |> {for hypothetical comparisons }
|>
|> Ignoring road load can have a big effect on the available torque, especially
|> over the wide speed range given. The net effect will be to turn the acceleration
|> curves downward much steeper and make it harder to find optimal shift points.
|>
I almost omitted the "theoretical" acceleration equation for this very reason,
but its fun to play what if?
The force equation is still valid as F will be the same whether the vehicle
is going up hill or being pushed by a tail wind. The vehicle will just "feel"
heavier or lighter to the engine/drive train.
|> But, I have to give you an "A" for effort and understanding of the problem.
|> It is far more complex than just some simple rule about max torque or max hp.
|>
|> BTW, this month's Road & Track had an interesting discussion in the Technical
|> Correspondence section about two cars with the same engine, but different
|> transmissions. The importance of a good launch and quick shifts actually
|> made the automatic quicker than the manual in a quarter mile. This is
|> despite the fact that the automatic tends to be less efficient.
|>
|> Mark
Hard to believe isn't it?
Patrick Taylor, The Sounding Board
>Correspondence section about two cars with the same engine, but different
>transmissions. The importance of a good launch and quick shifts actually
>made the automatic quicker than the manual in a quarter mile. This is
>despite the fact that the automatic tends to be less efficient.
That having been said, anyone know where I can get a good 4 speed auto for
my chevy?
>Mark
----------------------------------------------------------------------------
---------Visit the SOUNDING BOARD BBS +1 214 596 2915, a Wildcat! BBS-------
Patrick Taylor, Ericsson Network Systems
P. J. Remner
In a previous article, cl...@problem.uucp (Clive Apps) says:
>In article <1993Mar3.1...@calvin.edu> pto...@calvin.edu (Peter Tong) writes:
>>at redline. There is a limit to how much you can mask the lack of low
>>end torque on high ratio engines. You can mask it somewhat by adding
>>displacement (eg. running a wild cam and compensating for low end torque loss
>RUNNING A WILD CAM WILL TOTALY DESTROY ANY CHANCE YOU HAD OF LOW END TORQUE
>>by adding displacement). This all applies mostly to first gear since you
Clive, you're right. So was he. What he was saying was that you can
take a motor, pop in a really wild cam, then add displacement (preferably
in the stroke) to get some semblance of low-end torque. Why do you think
Ford/Yamaha stroked the SHO to 3.2 for the automatic? Did you notice that
the automatic has 15 lb-ft more torque?
Also, another case in point: Chrysler Slant-6. Tons of stroke. Also,
and not coincidentally, great big gobs of torque in relation to HP.
>NOT NECCESSARILY AS THE MORE CYLINDERS YOU HAVE THE MORE AREA YOU CAN
>EXPOSE TO THE VALVES AND INCREASE VOLUMETRIC EFFICIENCY
>
>EXPOSE TO THE VALVES ;IE: BETTER OLUMETRIC EFFICIENCY
Oh? Since when? The number of cylinders has no effect on volumetric
efficiency. Valves, yes, but not cylinders.
Please, delete unneccessary (sp?) info, and STOP SHOUTING.
--
/* aj...@cleveland.Freenet.Edu (Windows Freak)
* I want my... I want my... I want my Win/NT......
* You are not expected to understand this.
*/
Phil Houseago DNM22
in a lot of previous articles, much has been said.
the original question is now lost, but relating acceleration to engine
performance is wrong anyway. -
the torque at the back wheels is what you makes the car go - either
accelerate or mantain a continuous speed.
all this chatter about what the engine does is important , but only in
the context of gear ratios and how often you change gear.
one thing no-one that i've seen has commented on is how wide the useful
rev-range of the engine is. this will determine how often you change
gear.
in terms of overall performance two engine with the same power will not
have the same torque. they are bound peak at different (engine) speeds.
if two engines produce the same power, but one does it at twice the
revs of the other, then at that point, the two will produce the same
acceleration if they are geared to give the same back axle speed.
(assuming identical tyres).
the gearboxes will give differing mechanical advantages, but the power
is the same. therefore if the revs (axle) are the same so is the
torque.. if that is true, then if the cars weigh the same, they will
accelerate the same. until its time to change gear.
what happen to the *axle* torque when you do change is a function of
the engine characteristics, the gearing, and of course, how fast you
actually change gear.
usually, lower revving engines produce more torque over a wider rev
range than high revving engines. if the power output is similar, or the
same, you'd likely find that the low revving lump would give better
performance. (flatter characteristics) there would not be much in it.
the problem is that the higher revving engine produces more power
(thats a characteristic of the engine) so for a given *axle* speed,
more power = more torque = more acceration.
phew.
imho
James W. Adams
>
> [...]
>
>the original question is now lost, but relating acceleration to engine
>performance is wrong anyway. -
Ridiculous. The engine is the source of motive effort, so obviously
acceleration *must* be related to engine performance.
>the torque at the back wheels is what you makes the car go - either
>accelerate or mantain a continuous speed.
Technically, the percentage of that torque transmitted successfully to
the pavement, hence all the talk about tires, etc. Also, the percentage
of that force available to accelerate the car is reduced as aerodynamic
drag and frictional forces build with increased road speed.
> [...]
>one thing no-one that i've seen has commented on is how wide the useful
>rev-range of the engine is. this will determine how often you change
>gear.
Quite true.
> [...]
The source of confusion seems to be that everyone is arguing from
different frames of reference.
At any given drive ratio, maximum *instantaneous* output torque will
occur when input *torque* is maximum. Thus, the maximum *instantaneous*
acceleration will be available at peak *instantaneous* engine torque.
At any given output angular velocity, the maximum *average* output
torque will be obtained at that drive ratio which operates the engine at
its *power* peak. Thus, we would want to operate at peak power if we
had a continuously variable transmission.
In fact, neither of these conditions is fully sufficient to describe the
average acceleration of an automobile with several fixed gear ratios
over finite but relatively noninstantaneous times and distances.
Given the behavior of internal combustion engines, the range between the
measured torque and power peaks gives some indication of the width of
the useful power band. The reason is that as RPM rises, volumetric
charging efficiency falls off due to air inertia and intake/valve
restrictions. As RPM rises from idle, the engine produces more average
torque due to more power strokes/minute. Once the engine passes the
"torque peak," the strength of each power stroke falls off due to
volumetric restrictions, but slowly enough at first that peak power is
reached at a higher RPM where enough power events are produced to more
than offset the reduced strength of each one. Past the power peak, the
BHP falls off rapidly due to a rapid drop in volumetric charging
efficiency. There simply isn't enough time during the intake phase to
allow much air to flow into the cylinder.
Cylinder volume and, in particular, piston area is a determinant of the
force from each power stroke, so large bore engines tend to produce more
torque at lower RPM.
Empirically, one wants to keep the engine above the measured torque peak
and close to the BHP peak when accelerating.
--
James W. Adams -- NIH Scientific Computing Resource Center
Building 12A, Room 1050 phone: (301) 402-4154
National Institutes of Health uucp: uunet!nih-csl!jwa
Bethesda, MD 20892 Internet: j...@alw.nih.gov
Mark Walker
...
> The source of confusion seems to be that everyone is arguing from
> different frames of reference.
Yes, and by now we are all painfully aware of this, and the fact that
the original case was extremely constrained and idealized to make a
point (which was subsequently ignored to the point we were all so sick
of this thread we considered unsubscribing from rec.autos.tech).
Can we just quit now, while we're behind?
Thanks.
--
Mark Walker mwa...@chama.eece.unm.edu
