Altima SE V-6 vs. Accord LX V-6

Actually torque is much more important at the lower rpm than you seem to
understand. In fact, that low end power you are thinking of is actually
more torque than anything else. Torque *is* a specific measurement of the
vehicle's ability to push/pull an object/weight from a stop or lower speed.
You will reach the peak torque of an engine in the lower rpm range, exactly
where most of your acceleration occurs, while your peak horsepower occurs in
the upper rpm range, where you will accelerate slower. Don't beleive me?
Look at Horsepower and Torque charts and you will see where the peaks occur.

Why does this occur? You want to drop your rpms to a point where you can
take the most advantage of torque and horsepower, usually around 5200rpm, to
accelerate. Not saying that horsepower does nothing, in fact it does help
sustain and increase speed at the upper rpm when the vehicle is already
moving. Think of horsepower as related to the top speed of the vehicle.
BTW, just because an engine is bigger, doesn't mean it has more power.

Actually, just because and engine is twice the size of another doesn't mean
that it is twice as powerful. Much of an engines abilitly is based on it's
design. If your idea held true, all 2 liter engines would put out the same
power. BTW, what do you think torque is twisting? I would say it's the
axles which is connected to the wheels which moves a car forward. How does
horsepower move the car?

I highly doubt you could produce more torque than the VQ35DE on your
bicycle.

If you want to get a little more technical:
http://www.datsuns.com/Tech/tech_dept.htm

 

Sean is correct and you are incorrect. Looking at a chart to find the torque
and hp peaks doesn't tell you where maximum acceleration occurs.
Easy to show from Newton's second law (F=ma) that maximum acceleration occurs at
the POWER peak in any gear. Sorry, but you just have to do the physics here.

Again, Sean is basically correct. The physics of combustion is a constant for
all engines so maximum PRESSURE within the cylinders without detonation is
basically a constant (all 2 liter engines don't put out the same power because
not everyone achieves the same volumetric efficiency, designs for premium fuel
etc) Torque is pressure x displacement - so yes, an engine twice the size puts
out twice the torque AND hp (hence acceleration) at a given rpm - provided
intake and exhaust systems give the same volumetric efficiency at that rpm.
These are all basic concepts for Engine Design 101.

Frank

 

Sean is correct and you are incorrect. Looking at a chart to find the torque
and hp peaks doesn't tell you where maximum acceleration occurs.
Easy to show from Newton's second law (F=ma) that maximum acceleration occurs at
the POWER peak in any gear. Sorry, but you just have to do the physics here.

Again, Sean is basically correct. The physics of combustion is a constant for
all engines so maximum PRESSURE within the cylinders without detonation is
basically a constant (all 2 liter engines don't put out the same power because
not everyone achieves the same volumetric efficiency, designs for premium fuel
etc) Torque is pressure x displacement - so yes, an engine twice the size puts
out twice the torque AND hp (hence acceleration) at a given rpm - provided
intake and exhaust systems give the same volumetric efficiency at that rpm.
These are all basic concepts for Engine Design 101.

Frank

 

1. Torque, by definition, is just a term to describe the strength of the
twisting motion. It has nothing to do with acceleration. Just because there are
a lot of idiots, including magazine editors, miss understand the term, that does
not make it right. You should replace your "torque" with "horsepower".

2. At low rpm, like 1k-3k rpm, most engine put out the same horsepower/volume.
Hence, double the engine, you double the power. What differentiate a good engine
from an average engine is how much power it could produce at high rpm, like
above 5200. This region of rpm is where VTEC and Vanos and such enhance engine
power.

3. 100 lb person on a 3' crank on a bicycle produce 300 ft*lb of torque. A 300
lb person standing on a 1' crank produce 300 ft*lb of torque. I, standing on a
4' breaker bar on the wheel axle produce more torque than a 3.5 L Altima.
However, the car doesn't move much because I make little horsepower. There is no
doubt that I could produce 1000 ft*lb. All I need is to extend the crank to 10'.

 

1. Torque, by definition, is just a term to describe the strength of the
twisting motion. It has nothing to do with acceleration. Just because there are
a lot of idiots, including magazine editors, miss understand the term, that does
not make it right. You should replace your "torque" with "horsepower".

2. At low rpm, like 1k-3k rpm, most engine put out the same horsepower/volume.
Hence, double the engine, you double the power. What differentiate a good engine
from an average engine is how much power it could produce at high rpm, like
above 5200. This region of rpm is where VTEC and Vanos and such enhance engine
power.

3. 100 lb person on a 3' crank on a bicycle produce 300 ft*lb of torque. A 300
lb person standing on a 1' crank produce 300 ft*lb of torque. I, standing on a
4' breaker bar on the wheel axle produce more torque than a 3.5 L Altima.
However, the car doesn't move much because I make little horsepower. There is no
doubt that I could produce 1000 ft*lb. All I need is to extend the crank to 10'.

 

Horespower does not get you off the line or out of the corners. It is
torque.

 

Horespower does not get you off the line or out of the corners. It is
torque.

 

Raybender <> floridly penned in

Also do some reading:
http://tinyurl.com/5hgqe

 

Raybender <> floridly penned in

Also do some reading:
http://tinyurl.com/5hgqe

 

I've wondered about torque and HP too and didn't know the answer but
would like to.

As you state, F = ma (force equals mass times acceleration)

So, a = F/m (acceleration equals force divided by mass).

To accelerate the car, we need F (force).

Torque is force times distance. T = Fd

Solving for F and plugging that into our acceleration equation, we get

a = T/d/m (acceleration equals torque divided by distance divided by
mass).

For your particular car, the distance and the mass don't change. Hence
acceleration depends on torque only.

As the car goes faster and faster, you need horsepower to maintain
that torque at the higher RPMs.

There's some interesting background at
http://science.howstuffworks.com/fpte.htm.

 

I've wondered about torque and HP too and didn't know the answer but
would like to.

As you state, F = ma (force equals mass times acceleration)

So, a = F/m (acceleration equals force divided by mass).

To accelerate the car, we need F (force).

Torque is force times distance. T = Fd

Solving for F and plugging that into our acceleration equation, we get

a = T/d/m (acceleration equals torque divided by distance divided by
mass).

For your particular car, the distance and the mass don't change. Hence
acceleration depends on torque only.

As the car goes faster and faster, you need horsepower to maintain
that torque at the higher RPMs.

There's some interesting background at
http://science.howstuffworks.com/fpte.htm.

 

Here's the way you need to work it out. Then to prove it, go test the acceleration
of your car in second gear somewhere past the torque peak and up on the hp curve, and
compare to acceleration AT THE SAME SPEED in third gear down near the peak of the
torque curve.

Start from

F= ma Newtons second law

Now the car is moving as you accelerate so.....

F x velocity = mav
But this is just

Power = mav,
so that for the moving car we have,

acceleration = Power / (mass x speed)

This is the equation you need to work with. Now to make all the units work out
correctly in the English pound - foot system, you need to express

Power = Horsepower x 550 ft-lb/sec
Mass = Weight (in pounds) / (32 ft / sec^2)
speed is in feet / sec
acceleration will then be in feet / sec^2

Hope this helps.

Frank

 

Here's the way you need to work it out. Then to prove it, go test the acceleration
of your car in second gear somewhere past the torque peak and up on the hp curve, and
compare to acceleration AT THE SAME SPEED in third gear down near the peak of the
torque curve.

Start from

F= ma Newtons second law

Now the car is moving as you accelerate so.....

F x velocity = mav
But this is just

Power = mav,
so that for the moving car we have,

acceleration = Power / (mass x speed)

This is the equation you need to work with. Now to make all the units work out
correctly in the English pound - foot system, you need to express

Power = Horsepower x 550 ft-lb/sec
Mass = Weight (in pounds) / (32 ft / sec^2)
speed is in feet / sec
acceleration will then be in feet / sec^2

Hope this helps.

Frank

 

You're getting things mess up.

T = Fd. F is perpendicular to d. This is a twisting motion, as in angular force.
F = ma. a is linear acceleration. You can't just plug those 2 equations together and
solve them. You need to deal T equation with angular acceleration, not linear.
Basically, you need more equations to translate those 2.

That site sucks. First it says the Mustang engine is not good for pulling a truck
because it lacks torque. Then it contradict its self by proving that the Mustang's
engine could put out as much torque as the diesel engine, through the use of gear.
And then it says that the Mustang's engine won't last pulling a truck.

If the author had a bit of brain cell, he would forget his argument about torque and
just conclude that the Mustang's engine could pull as well as the diesel engine, but
that it won't last as long operating full time at full load at 5k rpm. It has nothing
to do with torque, just pure power.

As I said before, massive recital of myth does not make it a fact.

 

You're getting things mess up.

T = Fd. F is perpendicular to d. This is a twisting motion, as in angular force.
F = ma. a is linear acceleration. You can't just plug those 2 equations together and
solve them. You need to deal T equation with angular acceleration, not linear.
Basically, you need more equations to translate those 2.

That site sucks. First it says the Mustang engine is not good for pulling a truck
because it lacks torque. Then it contradict its self by proving that the Mustang's
engine could put out as much torque as the diesel engine, through the use of gear.
And then it says that the Mustang's engine won't last pulling a truck.

If the author had a bit of brain cell, he would forget his argument about torque and
just conclude that the Mustang's engine could pull as well as the diesel engine, but
that it won't last as long operating full time at full load at 5k rpm. It has nothing
to do with torque, just pure power.

As I said before, massive recital of myth does not make it a fact.

 

More than the Honda anyway. What was that, 214 lb/ft? ( what's a "tq" ?).
Assuming the pedal on the bicycle is one foot long, which it might not be,
and that I weigh 214 pounds, all I have to do is stand on the pedal.
214 lbs/ft of torque. I could get more by pulling up on the handlebar as I
pressed down on the pedal.
How many horespower is that? If I could pedal the bicycle at that rate for
one full second it would be 214/550 = .38 hp.

 

More than the Honda anyway. What was that, 214 lb/ft? ( what's a "tq" ?).
Assuming the pedal on the bicycle is one foot long, which it might not be,
and that I weigh 214 pounds, all I have to do is stand on the pedal.
214 lbs/ft of torque. I could get more by pulling up on the handlebar as I
pressed down on the pedal.
How many horespower is that? If I could pedal the bicycle at that rate for
one full second it would be 214/550 = .38 hp.

 

(ravelation) wrote in :

i you buy a 3.0 v6 accord don't talk about the milage/gallon
otherwise you would drive a 2.0 i-vtec accord (the european one)

marcel

 

Fianlly, someone who understands the difference between torque and
horsepower!

 

Fianlly, someone who understands the difference between torque and
horsepower!