2000 rpm lugging...
Russ Smith
degrees advanced over stock. No changes made recently. Fresh plugs at
60K. Driven daily.
I recently noticed a lugging or frogging of the engine at 1900-2200 rpm.
Occurs when engine is fully warmed up (not sure if it happens when
cold) when I slow down and the rpm's drop to 2000 rpm or so the engine
bogs like you are lugging it in too low a gear. I know 2000 rpm is low
for the miata but this lugging does not occur below or above this rpm.
Engine does not hesitate when not under load--that is sitting in neutral
and doing any type of slow rev, no blips of any kind at any rpm noticed.
No error codes in computer, no warning lights ever came on. Good
performance and timing belt was changed at 60K and don't believe its a
timing belt slip due to overall performance is good. Plug wires are
aftermarket and not stock ones (unsure of age of wires). I am thinking
of swapping out the wires as a starting point.
Any thoughts?
Eric Lucas
> I am thinking
> of swapping out the wires as a starting point.
Good idea, it might help.
> Any thoughts?
Yep. Stop abusing your engine like that. The Miata should never be
driven below 2000 rpm, unless you're in 1st gear, coasting along at 5
mph in a traffic jam on a freeway. Even then it's hard on the engine.
Any accelerating should be above 2500 rpm, and any heavy acceleration
should be 3000 or above. Even the 2000-2500 regime you should reserve
for absolutely constant-speed, flat residential driving, where
minimizing noise is more important than performance.
Eric Lucas
Johannes Swartling
>
> Yep. Stop abusing your engine like that. The Miata should never be
> driven below 2000 rpm, unless you're in 1st gear, coasting along at 5
> mph in a traffic jam on a freeway. Even then it's hard on the engine.
> Any accelerating should be above 2500 rpm, and any heavy acceleration
> should be 3000 or above. Even the 2000-2500 regime you should reserve
> for absolutely constant-speed, flat residential driving, where
> minimizing noise is more important than performance.
I've heard this repeated so many times on this ng that it has become an
indisputable truth. But what is the mechanical hard evidence behind the
statement that you shouldn't drive the engine below 2000 rpm? OK, I know
that extreme lugging (the 1000 rpm regime) isn't good for any engine,
but most other engines can take down to around 1500 rpm without any
problem. What is so special about the Miata engine?
Johannes
Thomas Tran
position in relation to air intake and engine rpm). Other than that, and as
long as the motor is not knocking, there does not seem to be any adverse
effects.
Thomas
Russ Smith
original post. The reason I posted is that I never had this problem or
noticed it before. The reason for the low rpm's is just what you state,
its when I'm pulling into my devlopment and want to keep the car sounding
sensible. We know it should be revved but the neighbors don't. I'm more
concerned about a bigger issue of some problem coming to a head that I'm
not aware of. Does anyone else have this problem? If it is a problem at
all?
Eric Lucas wrote:
> Russ Smith wrote:
>
> > I am thinking
> > of swapping out the wires as a starting point.
>
> Good idea, it might help.
>
> > Any thoughts?
>
> Yep. Stop abusing your engine like that. The Miata should never be
> driven below 2000 rpm, unless you're in 1st gear, coasting along at 5
> mph in a traffic jam on a freeway. Even then it's hard on the engine.
> Any accelerating should be above 2500 rpm, and any heavy acceleration
> should be 3000 or above. Even the 2000-2500 regime you should reserve
> for absolutely constant-speed, flat residential driving, where
> minimizing noise is more important than performance.
>
> Eric Lucas
Jim Carr
>
>I recently noticed a lugging or frogging of the engine at 1900-2200 rpm.
A miss or hesitation in that region is typical of bad plug wires.
After that it gets into the expensive parts of the ignition system.
--
James A. Carr <j...@scri.fsu.edu> | Commercial e-mail is _NOT_
http://www.scri.fsu.edu/~jac/ | desired to this or any address
Supercomputer Computations Res. Inst. | that resolves to my account
Florida State, Tallahassee FL 32306 | for any reason at any time.
John Bobincheck
is above 2000 rpm. My VW GTI and BMW 1602 both liked to rev high as well.
I doubt you will be doing damage to the engine by running under the power
band, but the car will feel sluggish when you try to accelerate.
The mechanical evidence you are looking for would be to check a dyno of each
gear and see where your best shift points would be. Someone with a dyno
could do this, but I sure wouldn't (couldn't) pay for it.
My guess would be that somewhere around 3000 for the downshift and 6000 for
the upshift. At least thats what I usually do.
Johannes Swartling <Johannes....@fysik.lth.se> wrote in message
news:36E4D478...@fysik.lth.se...
>Eric Lucas wrote:
>>
>> Yep. Stop abusing your engine like that. The Miata should never be
>> driven below 2000 rpm, unless you're in 1st gear, coasting along at 5
>> mph in a traffic jam on a freeway. Even then it's hard on the engine.
>> Any accelerating should be above 2500 rpm, and any heavy acceleration
>> should be 3000 or above. Even the 2000-2500 regime you should reserve
>> for absolutely constant-speed, flat residential driving, where
>> minimizing noise is more important than performance.
>
John Bobincheck
It shows normally aspirated Miatas Having torque meeting HP at 5250 rpm.
Can someone translate that? I would assume you would need a graph like this
for each gear to determine the optimal shift points.
John Bobincheck <roc...@ccnet.com> wrote in message
news:7c55si$kl1$1...@news.ncal.verio.com...
BenChi1
>
>Can someone translate that? I would assume you would need a graph like this
>for each gear to determine the optimal shift points.
When the torque and HP curve intersect, it means torque=HP and therefore at
this intersection an equal number of HP and equal number of ft. lbs torque.
Since the powerband is the same regardless of what gear is used, you do not
need a graph for *each* gear.
Just remember that even though at 7000RPM the Miata is not generating peak
power, the momentum created by the crankshaft spinning at 7000rpm is more
important to maintain given the impending shift which because of gearing will
bring that figure own. Ideally, the next gear will bring you at or around peak
hp so that acceleration is done at peak power levels.
-bn
AkiraRdstr
>original post. The reason I posted is that I never had this problem or
>noticed it before. The reason for the low rpm's is just what you state,
>its when I'm pulling into my devlopment and want to keep the car sounding
>sensible. We know it should be revved but the neighbors don't. I'm more
>concerned about a bigger issue of some problem coming to a head that I'm
>not aware of. Does anyone else have this problem? If it is a problem at
>all?
>>
Umm, sorry to be the bearer of bad tidings, but...
When I had a similar problem on my early-prod '90, the igniters were going bad.
I was advised by my Mazda mechanic not to use platinum plugs (I had Bosch
platinums in for about 10K miles when I noticed the prob) as they can sometimes
mess up the igniters. I dunno why. I traded the car.
--
"Akira"
'96 Chaste A/T http://www.eunos.com/keith/stripes/akira.html
Leon van Dommelen
>
>
>John Bobincheck wrote:
>
>> I just checked out the link http://www.miata.net/misc/dynoday/normal.htm
>> It shows normally aspirated Miatas Having torque meeting HP at 5250 rpm.
>>
>> Can someone translate that? I would assume you would need a graph like this
>> for each gear to determine the optimal shift points.
>
>The intersection is not important, but the peaks of both curves are.
Right, the intersection is an artificial point that depends on what
your units of time and angle are, not on any real physics. You might
not have any intersection point if you choose your units right.
Note that maximum torque is always before maximum hp.
>I've read that...
>
>Torque = acceleration
>Horsepower = top speed
Well, the two are not unrelated. And gearing plays a role in whether
you can achieve the top speed allowed by your available hp. And in
how important maximum torque really is compared to maximum hp.
But in a rough sense there may be something in it.
>To drive fast, you want all you upshifts to land at the torque peak
>(5,500RPMs?). The gears are set out in such a way that to do this, you will find
>you are shifting at around 7,000-7,250RPMs (near the horsepower peak).
>
>My numbers might be off a little, but that's the idea.
>
>I have a little freeware program with *lots* of cars and their specs in it, and
>it will run little sequences to find the 'optimal' shift points and drag times
>and the like. What I thought was interesting is that as the speed increases,
>each gear's 'optimal' shift point gets lower and lower in terms of RPM. This
>didn't seem so much a matter of gearing, as a matter of utilizing torque to
>overcome wind resistance. Hmmmm.......
If you look at the ratios between gearing ratios, you see that gears
become closer together when you go up in gears. That directly implies
that the idealized operating rpms concentrate in an increasingly
narrow band around maximum hp. Which will move the shifting rpm
down.
Also, if you maximize available hp at a given speed, you also maximize
rear wheel torque (rather than engine torque), which is the torque
that really counts to overcome air resistance.
Leon
--
Leon van Dommelen, the man, and Bozo, the '96 white PEP Sebring Miata
REMOVE THE Z -> domm...@zmiata.net http://www.eng.fsu.edu/~dommelen
John Bobincheck
the taller gears would be a little bit shorter to account for wind
resistance and possible increased driveline losses at higher speeds.
BenChi1 <ben...@aol.com> wrote in message
news:19990310054623...@ng-ch1.aol.com...
>>It shows normally aspirated Miatas Having torque meeting HP at 5250 rpm.
>>
>>Can someone translate that? I would assume you would need a graph like
this
>>for each gear to determine the optimal shift points.
>
John Bobincheck
If peak HP in 1st is at 7000 RPM, when you shift into 2nd what RPM are you
in, and where are you in the power band for that gear?
My question is that it may be faster to shift a bit earlier to stay more
towards the middle of the band as opposed to the top (less torque), or
bottom (less HP) end.
What does your program say?
Brian and Denise <cla...@ihug.co.nz> wrote in message
news:36E656B9...@ihug.co.nz...
>
>
>John Bobincheck wrote:
>
>> I just checked out the link http://www.miata.net/misc/dynoday/normal.htm
>> It shows normally aspirated Miatas Having torque meeting HP at 5250 rpm.
>>
>> Can someone translate that? I would assume you would need a graph like
this
>> for each gear to determine the optimal shift points.
>
>The intersection is not important, but the peaks of both curves are.
>
>I've read that...
>
>Torque = acceleration
>Horsepower = top speed
>
>To drive fast, you want all you upshifts to land at the torque peak
>(5,500RPMs?). The gears are set out in such a way that to do this, you will
find
>you are shifting at around 7,000-7,250RPMs (near the horsepower peak).
>
>My numbers might be off a little, but that's the idea.
>
>I have a little freeware program with *lots* of cars and their specs in it,
and
>it will run little sequences to find the 'optimal' shift points and drag
times
>and the like. What I thought was interesting is that as the speed
increases,
>each gear's 'optimal' shift point gets lower and lower in terms of RPM.
This
>didn't seem so much a matter of gearing, as a matter of utilizing torque to
>overcome wind resistance. Hmmmm.......
>--
>Brian Clark
>'90 BRG (Spinach) http://homepages.ihug.co.nz/~clarkd
>Auckland, New Zealand
>
Jim Carr
}
} John Bobincheck wrote:
} > I just checked out the link http://www.miata.net/misc/dynoday/normal.htm
} > It shows normally aspirated Miatas Having torque meeting HP at 5250 rpm.
} >
} > Can someone translate that? I would assume you would need a graph like this
} > for each gear to determine the optimal shift points.
}
} The intersection is not important, but the peaks of both curves are.
Also the shape of the torque curve.
In article <36eab516....@206.214.99.7>
dommelen@delete_spam.SPAMeng.fsu.edu (Leon van Dommelen) writes:
>
>Right, the intersection is an artificial point that depends on what
>your units of time and angle are, not on any real physics. You might
>not have any intersection point if you choose your units right.
Time to teach some physics to the engineer. ;-) [I know Leon
knows this, he just spoke imprecisely up above.]
Since P = tau * omega * (units factor), there will always be a
cross-over point when omega = 1/(units factor). The only question
is whether it is at a value of omega that you care about.
For power in ft*lbf/sec and torque in ft*lbf (lbf meaning pounds force),
the cross over is at the uninteresting point omega = 1 rad/sec = 9.5 rpm.
For power in hp, the cross over is at 5252.113... rpm.
That number is 550*30/pi. See next article for details.
} I've read that...
}
} Torque = acceleration
} Horsepower = top speed
>Well, the two are not unrelated. And gearing plays a role in whether
>you can achieve the top speed allowed by your available hp.
What Leon left out here is that the top speed is obtained (eventually,
since it is approached asymptotically as one can see in NASCAR races
on a superspeedway) when the power put out by the engine at some
speed (which is related to rpm by the gear ratio) is equal to the
power required to push the air out of the way at that speed.
If a car reaches its top speed when the revs are right at the
hp peak, its top gear is just right to get the most speed out
of the motor for the given drag coefficient. More speed requires
more power or a 'cleaner' car -- and if you clean up the car, you
should change the gearing.
If you reach top speed above the peak power rpm point, that means
the car is 'clean' enough to go faster if you could use the peak
power at that higher speed. You need taller gears.
If you reach top speed below the peak power point, that means your
gear ratio is not allowing you to use all of the power you could.
The gears are too tall, probably to get higher mileage.
You will note that a stock Miata (at least for the pre-94 models
where I looked at this) reaches its top speed just over the power
peak in 5th gear. A taller gear (to reduce revs on the freeway
and probably improve mileage, unless the motor is optimally efficient
at 3500 revs) would lower the top speed. The designers did not
make this compromise.
>But in a rough sense there may be something in it.
Peak torque is where you get the largest force at the rear
wheels, so it does give the best acceleration. What you
really want is a flat torque curve across the working range
of the engine (its power band) so you never give up much
acceleration as the rpms vary.
Jim Carr
msc...@cs.utexas.edu (Michael Alan Schaeffer) writes:
>
> If I remember right, horsepower varies with torque as a function of
>this formula (if you're talking hp and lb/ft):
>
>hp = torque * (rpm/5250)
Yep. The conversion factor is required by the non-standard units.
When standard units are used,
Power (P) = torque (tau) * angular velocity (omega).
and there is no correction factor.
Indicating the units in square brackets, those relations are
P[W] = tau[N*m] * omega[radian/s]
P[ft*lbf/s] = tau[ft*lbf] * omega[radian/s]
for SI and 'british' units.
Since P[ft*lbf/s] = 550 * P[hp]
and omega[radian/s] = (pi/30) * omega[rpm]
we get P[hp] = tau[ft*lbf] * omega[rpm] / 5252.113....
where the denominator is 550*30/pi.
Rounding to 5250 is conventional because torque curves are often
plotted so that 250 rpm steps can be seen.
Jim Carr
>
>I would assume you would need a graph like this
>for each gear to determine the optimal shift points.
Yes and no. The engine performance only depends on revs so you
only need one power/torque curve for your motor. There are two
ways to go from there. One is to plot the set of straight lines
that relate mph (x axis) to rpm (y axis) for each gear and add
horizontal lines at the peak hp rpm, the peak torque rpm, and
maybe the lower rpm that gives the same torque as you have at the
peak power point. Vertical lines that take you from a high rpm
(around or above the power peak) with some torque to a low rpm
with the same torque make good shift points. This uses no more
than basic shop math and the ability to read a graph.
The other way is to calculate the rear-wheel output for each gear
as a function of speed and change gears to stay on the highest
curve. This will give the same answer, except you now get as many
significant figures as your computer can print out. ;-)
The advantage of the former method is you might keep the graph in
your head. It is also a method used to design gear combinations,
which is where I learned about it (SCCA article on doing that).
Eric Lucas
>
> I don't think the Miata engine is "special", It's just that the power band
> is above 2000 rpm. My VW GTI and BMW 1602 both liked to rev high as well.
>
> I doubt you will be doing damage to the engine by running under the power
> band, but the car will feel sluggish when you try to accelerate.
Well, for one thing, if you are in a region of poor power, it'll take
you longer to accelerate. This means that you'll be running the spark
advanced (this is part of the engine's mechanism for acclerating) for a
longer time, and it'll run the cylinders hotter. Pinging is just an
extreme form of this, and I've heard that the extra heat will damage the
surface of the pistons. Not sure how much of a problem this is, shy of
pinging, but given how much fun it is to keep the revs up, I wouldn't
chance it.
The original questioner pointed out that his reason for driving at 1900
rpm was indeed to minimize noise in residential areas. However, he
won't get a huge amount of extra noise from keeping the revs up around
2500 compared to 1900. Since it is likely easier on the engine than
allowing the revs to dip so much, and since he's having a lugging
problem at lower rpms, that might be the simplest solution. Someone
else's suggestion of checking the plug wires sounds like a good idea
too.
Eric Lucas
John Bobincheck
news:36ed4f5c....@206.214.99.7...
>"John Bobincheck" <roc...@ccnet.com> wrote:
>
>>Are you sure the powerband is the same for each gear? I would think that
>>the taller gears would be a little bit shorter to account for wind
>>resistance
>
>speed is somewhat favorable in that it tends to blow the hot air and
>heat out of the engine bay. And I could spin tall tales about Ram
>effects and the influence of the motion through the Earth's Magnetic
>Field.
I've been misunderstood. I'm trying to think of a clearer way to present
this:
As the car reaches higher speeds, more Torque and HP would be required to
maintain the curves set in the lower gears as a result of outside influences
(wind resistance).
This effect could be graphed as a ramp. Sort of a rising level of
diminishing returns.
Would this counteract the effective powerband (curve) of the higher gears
until at some point, RPM would be limited to:
Peak power minus operational losses (~20% internal+outside influences) = top
end of final gear powerband.
Since this is definately not occuring at 7200 RPM, what is the effect in 4th
gear? 3rd? While it is not as drastic (terminal) as the final gear, should
it be taken into account?
>> and possible increased driveline losses at higher speeds.
>
>Maybe, but the omnipotent God called We-Dont-Have-Data has decreed
>that power loss is 20% from the engine onwards. We will just have to
>learn to live with that. ;)
I can accept this answer as true.
As I think about it more, OG-WDHD seems correct as the machine does not
internally reconfigure itself at higher speed. Unless there is a change in
loss from different gear ratios.
Brian and Denise
John Bobincheck wrote:
> I just checked out the link http://www.miata.net/misc/dynoday/normal.htm
> It shows normally aspirated Miatas Having torque meeting HP at 5250 rpm.
>
> Can someone translate that? I would assume you would need a graph like this
> for each gear to determine the optimal shift points.
The intersection is not important, but the peaks of both curves are.
I've read that...
Torque = acceleration
Horsepower = top speed
To drive fast, you want all you upshifts to land at the torque peak
Brian and Denise
Russ Smith wrote:
> 1990 1.6L, 67K miles. excellent mechanicals. Timing at stock to 2
> degrees advanced over stock. No changes made recently. Fresh plugs at
> 60K. Driven daily.
>
> I recently noticed a lugging or frogging of the engine at 1900-2200 rpm.
Under acceleration? hard or slight? or just sitting there?
My '90 doesn't like low RPMs since I've bumped up the timing (14deg),
especially when cold. It hesitates a second at idle, but then revs freely.
This happens to me when starting from a stop. I know yours is a different
experience, but you could set your timing back to 10 (or even 8?) just to
make sure that it isn't that.
I would go with the plug wires too. I'm following you on this one. Then
maybe clean and gap the plugs (even though they're new). After that, I'm no
mechanic.
Good luck,
Eric Lucas
>
> >It shows normally aspirated Miatas Having torque meeting HP at 5250 rpm.
> >
> >Can someone translate that? I would assume you would need a graph like this
> >for each gear to determine the optimal shift points.
>
> this intersection an equal number of HP and equal number of ft. lbs torque.
The torque=HP intersection is meaningless, since they are not in any way
dimensionally equivalent. Torque is in units of force/distance, and HP
is in units of energy/time, or force*distance/time. You can change the
intersection point arbitrarily by simply changing the Y scale of
either. As someone else pointed out, the *peaks* of the two curves are
important, and I've heard that a good transmission is designed so that
the torque peak of one gear is at the HP peak of an adjacent gear (I
forget which is for the higher and lower gear--perhaps someone familiar
with the engineering can explain the rationale.) And I believe I've
heard that the optimum shift point is at this maximum.
Eric Lucas
Leon van Dommelen
>Are you sure the powerband is the same for each gear? I would think that
>the taller gears would be a little bit shorter to account for wind
>resistance
Wind resistance does not affect engine operations. In fact higher
speed is somewhat favorable in that it tends to blow the hot air and
heat out of the engine bay. And I could spin tall tales about Ram
effects and the influence of the motion through the Earth's Magnetic
Field.
> and possible increased driveline losses at higher speeds.
Maybe, but the omnipotent God called We-Dont-Have-Data has decreed
that power loss is 20% from the engine onwards. We will just have to
learn to live with that. ;)
Leon
Leon van Dommelen
>dommelen@delete_spam.SPAMeng.fsu.edu (Leon van Dommelen) writes:
>>
>>Right, the intersection is an artificial point that depends on what
>>your units of time and angle are, not on any real physics. You might
>>not have any intersection point if you choose your units right.
>
> Time to teach some physics to the engineer. ;-)
Fat chance. ;)
> [I know Leon
> knows this, he just spoke imprecisely up above.]
>
> Since P = tau * omega * (units factor), there will always be a
> cross-over point when omega = 1/(units factor). The only question
> is whether it is at a value of omega that you care about.
Well, not quite. ;) By choosing weird units, I can get one curve to
be above the other over the entire range. But then, I forgot that
typically these graphs have two different vertical axes, allowing
even more freedom in having the intersection point[s] {any|no}where.
> For power in ft*lbf/sec and torque in ft*lbf (lbf meaning pounds force),
I always express power as inch oz micro arcsec/h, as do many people.
[Teaching some International Engineering to the physicist here. ;) ]
Or should that have been eV/year? ?)
> Peak torque is where you get the largest force at the rear
> wheels, so it does give the best acceleration.
Not necessarily. Peak torque is where you get the largest force
at the rear wheels *for a given gear*. With a limited number of
gears this is likely to be the point of maximum force of any gear,
but that is not strictly needed. Furthermore, the advantage
becomes more likely to be nonexistent if you are near but not precisely
at maximum torque. In short, the higher *engine* torque is not
necessarily always the higher rear wheel torque.
> What you
> really want is a flat torque curve across the working range
> of the engine (its power band) so you never give up much
> acceleration as the rpms vary.
True. Unless you have a 100% effective Continuously Variable
Transmission. Then you don't care about anything but peak hp.
Eric Lucas
> > For power in ft*lbf/sec and torque in ft*lbf (lbf meaning pounds force),
>
> I always express power as inch oz micro arcsec/h, as do many people.
> [Teaching some International Engineering to the physicist here. ;) ]
>
> Or should that have been eV/year? ?)
I personally prefer stone-furlongs per fortnight.
And please excuse the brainfart in my previous post on the issue,
claiming torque in terms of force/distance, rather than force*distance.
Eric Lucas
Leon van Dommelen
>Leon van Dommelen <dommelen@delete_spam.SPAMeng.fsu.edu> wrote in message
>news:36ed4f5c....@206.214.99.7...
>>"John Bobincheck" <roc...@ccnet.com> wrote:
>>
>>>Are you sure the powerband is the same for each gear? I would think that
>>>the taller gears would be a little bit shorter to account for wind
>>>resistance
>>
>>Wind resistance does not affect engine operations. In fact higher
>>speed is somewhat favorable in that it tends to blow the hot air and
>>heat out of the engine bay. And I could spin tall tales about Ram
>>effects and the influence of the motion through the Earth's Magnetic
>>Field.
>
>I've been misunderstood. I'm trying to think of a clearer way to present
>this:
>
>As the car reaches higher speeds, more Torque and HP would be required to
>maintain the curves set in the lower gears as a result of outside influences
>(wind resistance).
>
>This effect could be graphed as a ramp. Sort of a rising level of
>diminishing returns.
>
>Would this counteract the effective powerband (curve) of the higher gears
>until at some point, RPM would be limited to:
>Peak power minus operational losses (~20% internal+outside influences) = top
>end of final gear powerband.
You are probably still misunderstood. You seem to say the powerband is
limited to what you can drive given an amount of driving resistance, or
to the excess power above that needed for resistance. I
thought the powerband is the region where the car puts out its
highest power, regardless of what the effect of that power is. But
maybe I am wrong. Anybody have an official definition?
In any case, if you define the powerband as the region where you
get the most power out of the engine, you can always continue to
accelerate immediately after shifting, but you might get stuck
subsequently in the gear when air resistance becomes equal to
the available power. (Assuming a steady windspeed.) Does that
mean you have reached the end of the powerband? What if the wind
shifts? Are you suddenly back in the powerband? Does not sound right.
>Since this is definately not occuring at 7200 RPM, what is the effect in 4th
>gear? 3rd? While it is not as drastic (terminal) as the final gear, should
>it be taken into account?
>
>>> and possible increased driveline losses at higher speeds.
>>
>>Maybe, but the omnipotent God called We-Dont-Have-Data has decreed
>>that power loss is 20% from the engine onwards. We will just have to
>>learn to live with that. ;)
>
>I can accept this answer as true.
>As I think about it more, OG-WDHD seems correct as the machine does not
>internally reconfigure itself at higher speed. Unless there is a change in
>loss from different gear ratios.
But of course. The forces on the various components and their speeds
are going to be different, and consequently their resistance. Think of
the gearbox. The differential is another example. I believe the 20%
figure is also supposed to include the rolling resistance of the wheels.
Anyway, the power loss is not even going to be a constant number
at low speeds. For example, various losses depend simply on the
speed of the components, regardless of how much power is being transferred.
Think of the dissipation of the oil. When would you expect a loss
proportional to power? Maybe if a rotating part is hitting a
stationary part with a force proportional to the transferred torque
and the friction is of the dry friction kind. That is not going
to describe much.
DLHagerman
Miata, the best acceleration is obtained by running the engine up to the cutout
in every gear. This does not say anything about how long the engine will last
if you do this, or how many tickets you will get.
Rexven
Wouldn't it have been cheaper to just change the plugs?? (grinning)
P. J. Remner
In a previous article, eal...@worldnet.att.net (Eric Lucas) says:
>
>I personally prefer stone-furlongs per fortnight.
>
I am going to calculate all my cars' horsepower in this standard,
just as soon as I find the standards for a stone (roughly 13lb?)
and a furlong (no clue, 8 feet? 16 yards? ??).
And, appropiately, I shall christen this measurement... the Lucas.
--
"Okay, how long has the band been together, as Metallica?"
"Well, since we started."
Pete '88 Subaru GL 4WD '72 Ford Thunderbird Schwinn s[9six].40 and more..
P. J. Remner
In a previous article, eal...@worldnet.att.net (Eric Lucas) says:
>
>The original questioner pointed out that his reason for driving at 1900
>rpm was indeed to minimize noise in residential areas. However, he
>won't get a huge amount of extra noise from keeping the revs up around
>2500 compared to 1900. Since it is likely easier on the engine than
>allowing the revs to dip so much, and since he's having a lugging
>problem at lower rpms, that might be the simplest solution. Someone
>else's suggestion of checking the plug wires sounds like a good idea
>too.
In many cars, if you keep the engine running in its powerband,
it will actually be quieter than if you lug it.
Eric Lucas
> I am going to calculate all my cars' horsepower in this standard,
> just as soon as I find the standards for a stone (roughly 13lb?)
> and a furlong (no clue, 8 feet? 16 yards? ??).
220 yds (1/8 mile). And that's 14 lb.
Eric Lucas
Jim Carr
}
} I am going to calculate all my cars' horsepower in this standard,
} just as soon as I find the standards for a stone (roughly 13lb?)
} and a furlong (no clue, 8 feet? 16 yards? ??).
In article <36E855C0...@worldnet.att.net>
eal...@worldnet.att.net writes:
>
>220 yds (1/8 mile). And that's 14 lb.
Just remember that an acre is a chain-furlong = 10 square chains
and it is easy to remember what a furlong is. ;-) It comes from
plowing long furrows. Seriously.
It is also a very "metric" system. 100 links to a chain (66'),
10 chains to a furlong, and 8 furlongs to a mile. OK, close.
That last 8 is there to make division by 2 easier.
Also, remember that the stone and pound are mass units.
Warren B. Focke
Quoth j...@ibms48.scri.fsu.edu (Jim Carr):
>
> Also, remember that the stone and pound are mass units.
The pound is a unit of force. The unit of mass for that system is
called a slug. A mass of one slug has a weight of 32.17 pounds (g) at
sea level.
Warren Focke
--
"Telopolies running the Internet would be almost exactly like foxes
guarding the henhouse, except that foxes are smart and agile."
Bob Metcalfe
Eric Lucas
> and 8 furlongs to a mile. OK, close.
No, exact. Multiply 220 yds by 8, and what do you get? I make it 1760
yds, which is 5280', which, last time I checked, is exactly 1 mile.
> Also, remember that the stone and pound are mass units.
Actually, my memory from college physics is that stone and pound are in
fact force units. I remember Doc Mills told us what the corresponding
mass units were called, but I can't remember now what they are.
Eric Lucas
Eric Lucas
>
> Quoth j...@ibms48.scri.fsu.edu (Jim Carr):
> >
> > Also, remember that the stone and pound are mass units.
>
> called a slug. A mass of one slug has a weight of 32.17 pounds (g) at
> sea level.
Thank you, that was the unit I was trying to remember in my previous
post.
It does raise the question, however....the usual conversion of lbs to g
is generally given as 1 lb = 454 g. The question then becomes, under
what conditions is that actually an accurate (exact) conversion--i.e.,
where (sea level, etc.) does a mass of 454 g weigh 1 lb.?
Eric Lucas
Leon van Dommelen
>
>Quoth j...@ibms48.scri.fsu.edu (Jim Carr):
>>
>> Also, remember that the stone and pound are mass units.
>
>The pound is a unit of force. The unit of mass for that system is
>called a slug. A mass of one slug has a weight of 32.17 pounds (g) at
>sea level.
I always thought a lbf was a pound force and a lbm was a pound mass.
But what do we SI people know? After I am done trying to guess where in
the British-units formulas to put the additional g's, I am off to the
supermarket to ask them where I can find my 32th of a slug of sugar.
"No, a pound will NOT do! And please lower the scale to sea level height.
I do not care if this is Denver."
Leon ;)
P.S. One of the formulas that has a "gEos ex imagina" is called "head loss".
I wonder why?
Matthew T. Russotto
Eric Lucas <eal...@worldnet.att.net> wrote:
}It does raise the question, however....the usual conversion of lbs to g
}is generally given as 1 lb = 454 g. The question then becomes, under
}what conditions is that actually an accurate (exact) conversion--i.e.,
}where (sea level, etc.) does a mass of 454 g weigh 1 lb.?
Whenever gravity is one standard g (9.80665 m/s^2). Or when "1 lb" refers to
pounds-mass, which it often does.
--
Matthew T. Russotto russ...@pond.com
"Extremism in defense of liberty is no vice, and moderation in pursuit
of justice is no virtue."
Jim Carr
|
| Also, remember that the stone and pound are mass units.
In article <7cbq9b$8...@baud.eng.umd.edu>
mo...@Glue.umd.edu (Warren B. Focke) writes:
>
>The pound is a unit of force. The unit of mass for that system is
>called a slug.
Only if you learned physics from US textbooks or did engineering
in Great Britain during a 10-20 year period. Consult the OED
if you do not believe me. Visit Gene Nygaard's web page on the
subject (and discussion with a troll in sci.physics) if you
need even more detail.
Otherwise, as in commerce in the US, the pound is a mass. That
has been true for centuries, with 19th century textbooks being
the best place to see the past use.
Force is measured in either poundals or pounds-force (lbf) in
the engineering system that has the pound as a mass.
Jim Carr
}
} "John Bobincheck" <roc...@ccnet.com> wrote:
} >Are you sure the powerband is the same for each gear? I would think that
} >the taller gears would be a little bit shorter to account for wind
} >resistance
}
} Wind resistance does not affect engine operations. In fact higher
} speed is somewhat favorable in that it tends to blow the hot air and
} heat out of the engine bay. And I could spin tall tales about Ram
} effects and the influence of the motion through the Earth's Magnetic
} Field.
In article <7c7qn8$662$2...@news.ncal.verio.com>
"John Bobincheck" <roc...@ccnet.com> writes:
>
>I've been misunderstood. I'm trying to think of a clearer way to present
>this:
>
>As the car reaches higher speeds, more Torque and HP would be required to
>maintain the curves set in the lower gears as a result of outside influences
>(wind resistance).
The "curve" is a property of the motor, whether measured at the
flywheel or at the rear wheels. It is what you get from the motor.
At higher speeds, the net effect of that motor is less because of
the effect of drag. Classic example: NASCAR. The motor is putting
out whatever it can (I don't know the "restrictor plate" numbers
since I only watch it for the crashes and 'love taps') and the car's
linear acceleration is zero. Why? The drag hp equals the motor's.
>Would this counteract the effective powerband (curve) of the higher gears
>until at some point, RPM would be limited to:
>Peak power minus operational losses (~20% internal+outside influences) = top
>end of final gear powerband.
What it does is narrow the part of the powerband you can use. For
example, the drag hp of a Miata is something like 12 or 15 at 60 mph
and grows rapidly with speed. If you are at a speed where the drag
is equivalent to 50 hp and shift into a higher gear where the motor
only makes 40 hp at the new revs, you will slow down no matter how
much gas you give it. Since this becomes a problem only when the
drag builds up, the gear ratios get narrower to keep you near the
power peak. Like Leon said, the only thing better is a continuously
varying transmission.
Most people are familiar with this effect when climbing mountain
grades in an underpowered car -- or a slight grade into a headwind
in something like a VW bus. You cannot maintain speed in top gear
because the car does not put out enough power at that speed/rpm
combination (rpm being the key factor) to do so.
Jim Carr
|
| and 8 furlongs to a mile. OK, close.
In article <36E9FD4B...@worldnet.att.net>
eal...@worldnet.att.net writes:
>
>No, exact. Multiply 220 yds by 8, and what do you get? I make it 1760
>yds, which is 5280', which, last time I checked, is exactly 1 mile.
Sure, but 8 is not 10. It would have been fully "metric" if
there were 10 furlongs to the mile, with 10 chains to the
furlong, etc, but the rod and furlong were already set. So
they worked backwards to build the decimal chain system.
| Also, remember that the stone and pound are mass units.
>Actually, my memory from college physics is that stone and pound are in
>fact force units. I remember Doc Mills told us what the corresponding
>mass units were called, but I can't remember now what they are.
That is a result of a silly attempt by some teachers of physics
in the US to ignore the actual definition of the pound in favor of
the colloquial one, and invoke a short-lived British aeronautical
unit as the analog of the kg. Since legally the pound is a mass,
and in 'weights and measures' a "net weight" is a mass so that
packages give the 'weight' (mass) in both pounds and kilograms,
this has only increased confusion.
More common system: mass in pounds, force in poundals.
Latecomers used mass in glb, force in pounds.
The best system is where mass is in pounds and force is in pounds-force.
Then you have (1 lbf) = (1 lb)*(32 ft/s^2). =8-0
Jim Carr
>
>It does raise the question, however....the usual conversion of lbs to g
>is generally given as 1 lb = 454 g. The question then becomes, under
>what conditions is that actually an accurate (exact) conversion--i.e.,
>where (sea level, etc.) does a mass of 454 g weigh 1 lb.?
It is always exact, because _that_ pound is a mass.
It is found by comparison of the pound mass to a specific fraction
of the kg mass on a balance, and the kg mass is referenced to the
"Grande Kilogramme".
Jim Carr
"John Bobincheck" <roc...@ccnet.com> writes:
>
>Say the total power band in 1st gear is from 3000 to 7000 RPM
Not in my Miata.
>If peak HP in 1st is at 7000 RPM, when you shift into 2nd what RPM are you
>in, and where are you in the power band for that gear?
Peak HP is at 6500 rpm. I have some notes that say 7000 in 1st takes
you to 4200 in 2nd, but the easiest way to check it for your car (I
have a '93) is to just do the experiment.
>My question is that it may be faster to shift a bit earlier to stay more
>towards the middle of the band as opposed to the top (less torque), or
>bottom (less HP) end.
Do people spend more money to get less power?
It is true that the torque starts to drop off very quickly after
6500 rpm (stock advance, etc), as does the power, and this can
favor an earlier shift (than redline) if you can go to the same
or higher torque when doing so. I am not looking at a dyno
chart so I can't say more. Maybe CSP will.
Leon van Dommelen
> The best system is where mass is in pounds and force is in pounds-force.
> Then you have (1 lbf) = (1 lb)*(32 ft/s^2). =8-0
This is the best system for tests for us professors. :) Some students
will always forget to insert the artificial "g" corrections to
go from consistent units (lbf + slug) to inconsistent ones (lbf + lbm).
Other students will remember them, but put them in the wrong place. ;)
Always good for lots of red ink. Great fun. :)
Especially funny for foreign students raised on SI units. "Newton said
Force was Mass times Acceleration divided by g?" "Kinetic energy
of a rocket out in space involves the value of gravity at sea level?"
"How do you mean forgetting a factor g is a major mistake for an
engineer since it could cause a bridge to collapse or a plane to
break up in the air? My basic method was correct!"
Leon :)
Ed Kleinhample
Not correct - the pound is a unit of force - 1 slug-foot/second.