Low-end Torque
Joined: Feb 2001
Posts: 4,061
Likes: 363
From: Redondo Beach
Tim,
The short answer is that F = m*a is for linear (i.e. non-rotational) systems. This is, of course, the system in which the car operates, but the engine does not! The twisting force of the engine's torque must be translated into the *linear* acceleration that we actually feel in the car.
The confusion, again, arises from the fact that torque is NOT a true physical force; it's an *angular* force applied to a moment (lever) arm, and its units are N*m, which in turn are just Joules, the unit of energy! But power is energy divided by time, i.e., the rate of expending energy (or rate of doing mechanical work).
Here's the long answer:
Let's go back to the CVT experiment. We again set the CVT to operate at either the power peak or the torque peak (at which power is assumed to be lower), and we do 1/4-mile runs. We find that one of the two is indeed faster, but unfortunately, the guy who recorded the times forgot which was which, and so now we have to figure it out!
I claim that the faster run is the power-peak run.
Proof:
I'll prove it by contradiction, i.e. I'll assume that the torque-peak run was fastest, and look for a reason that can't be true. Let the peak engine power be Pp, and the engine power at the torque peak be Pt. By definition of the torque peak and power peak, we have
(1) Pp > Pt.
Now, first we observe that we did mechanical work during the runs, i.e. we exerted a (linear) force through a distance. The work done during the torque-peak run is:
Wt = Ft*D (Joules)
where Ft is the force (in Newtons), and D is the constant distance (400 meters). The corresponding work rate is the power done during the torque-peak run. If we assume a lossless CVT, perfect traction, and negligible air resistance (all fine for our purposes), then this power is equal to the engine power produced during the run. So,
Pt = Wt/Tt = Ft*D/Tt (Watts)
where Tt is the time of the torque-peak run, in seconds. Similarly, during the power-peak runs, we have
Pp = Fp*D/Tp.
Now, our assumption that the torque-peak wins means
(2) Tp > Tt,
so we can write
(3) Pt > Ft*D/Tp.
Further, in order for the torque-peak run to be fastest, it must have had greater linear acceleration than the power-peak run, which in turn implies a greater linear force. So again by assumption,
(4) Ft > Fp.
We may then preserve the inequality of (3) by substituting Fp for Ft:
(5) Pt > Fp*D/Tp = Pp.
But this contradicts definition (1), so the assumption (2) is invalid, and the desired result is proved.
[Edit: minor changes for clarity.]
The short answer is that F = m*a is for linear (i.e. non-rotational) systems. This is, of course, the system in which the car operates, but the engine does not! The twisting force of the engine's torque must be translated into the *linear* acceleration that we actually feel in the car.
The confusion, again, arises from the fact that torque is NOT a true physical force; it's an *angular* force applied to a moment (lever) arm, and its units are N*m, which in turn are just Joules, the unit of energy! But power is energy divided by time, i.e., the rate of expending energy (or rate of doing mechanical work).
Here's the long answer:
Let's go back to the CVT experiment. We again set the CVT to operate at either the power peak or the torque peak (at which power is assumed to be lower), and we do 1/4-mile runs. We find that one of the two is indeed faster, but unfortunately, the guy who recorded the times forgot which was which, and so now we have to figure it out!
I claim that the faster run is the power-peak run.
Proof:
I'll prove it by contradiction, i.e. I'll assume that the torque-peak run was fastest, and look for a reason that can't be true. Let the peak engine power be Pp, and the engine power at the torque peak be Pt. By definition of the torque peak and power peak, we have
(1) Pp > Pt.
Now, first we observe that we did mechanical work during the runs, i.e. we exerted a (linear) force through a distance. The work done during the torque-peak run is:
Wt = Ft*D (Joules)
where Ft is the force (in Newtons), and D is the constant distance (400 meters). The corresponding work rate is the power done during the torque-peak run. If we assume a lossless CVT, perfect traction, and negligible air resistance (all fine for our purposes), then this power is equal to the engine power produced during the run. So,
Pt = Wt/Tt = Ft*D/Tt (Watts)
where Tt is the time of the torque-peak run, in seconds. Similarly, during the power-peak runs, we have
Pp = Fp*D/Tp.
Now, our assumption that the torque-peak wins means
(2) Tp > Tt,
so we can write
(3) Pt > Ft*D/Tp.
Further, in order for the torque-peak run to be fastest, it must have had greater linear acceleration than the power-peak run, which in turn implies a greater linear force. So again by assumption,
(4) Ft > Fp.
We may then preserve the inequality of (3) by substituting Fp for Ft:
(5) Pt > Fp*D/Tp = Pp.
But this contradicts definition (1), so the assumption (2) is invalid, and the desired result is proved.
[Edit: minor changes for clarity.]
Registered User
Joined: Jan 2001
Posts: 6,455
Likes: 0
From: Glendale/Burbank/LA
While we're pumping out *supposedly correct* numbers 
, let's get some satisfaction against those Mustang GT's -- which one has the greater peak acceleration, that or the Honda S2000? (Using F = P / RPM --> F = Torque x RPM x Gearing / RPM ---> F = Torque x Gearing, and then finally, Acceleration = Torque x Gearing / Mass)
From CarPoint:
2003 Mustang GT
Torque: 302 @ 4000 RPM
Curb Weight: 3477 lbs
Gear Ratios: 3.37, 1.99, 1.33, 1.00, 0.67, and 3.27 Final.
2002 Honda S2000
Torque: 153 @ 7500 RPM
Curb Weight: 2810 lbs
Gear Ratios: 3.133, 2.045, 1.481, 1.161, 0.971, 4.1 Final. 1.16 Reduction Gear.
1st Gear:
GT: 302 x (3.37 x 3.27) / 3477 = 0.957
S2K: 153 x (3.133 x 4.1 x 1.16) / 2810 = 0.811 DOH!
2nd Gear:
GT: 302 x (1.99 x 3.27) / 3477 = 0.565
S2K: 153 x (2.045 x 4.1 x 1.16) / 2810 = 0.530 DOH!
3nd Gear:
GT: 302 x (1.33 x 3.27) / 3477 = 0.378
S2K: 153 x (1.481 x 4.1 x 1.16) / 2810 = .384 WOOHOO!
4th Gear:
GT: 302 x (1.00 x 3.27) / 3477 = 0.284
S2K: 153 x (1.161 x 4.1 x 1.16) / 2810 = 0.300 WOOHOO!
5th Gear:
GT: 302 x (0.67 x 3.27) / 3477 = 0.190
S2K: 153 x (0.971 x 4.1 x 1.16) / 2810 = 0.252 WOO HOO
6th Gear:
WOO HOO!
P.S. I'm bored.
P.P.S. Does anyone know for sure whether the reduction gear is multiplied in with the rest of the gearing, or did I not do that right?
, let's get some satisfaction against those Mustang GT's -- which one has the greater peak acceleration, that or the Honda S2000? (Using F = P / RPM --> F = Torque x RPM x Gearing / RPM ---> F = Torque x Gearing, and then finally, Acceleration = Torque x Gearing / Mass)From CarPoint:
2003 Mustang GT
Torque: 302 @ 4000 RPM
Curb Weight: 3477 lbs
Gear Ratios: 3.37, 1.99, 1.33, 1.00, 0.67, and 3.27 Final.
2002 Honda S2000
Torque: 153 @ 7500 RPM
Curb Weight: 2810 lbs
Gear Ratios: 3.133, 2.045, 1.481, 1.161, 0.971, 4.1 Final. 1.16 Reduction Gear.
1st Gear:
GT: 302 x (3.37 x 3.27) / 3477 = 0.957
S2K: 153 x (3.133 x 4.1 x 1.16) / 2810 = 0.811 DOH!
2nd Gear:
GT: 302 x (1.99 x 3.27) / 3477 = 0.565
S2K: 153 x (2.045 x 4.1 x 1.16) / 2810 = 0.530 DOH!
3nd Gear:
GT: 302 x (1.33 x 3.27) / 3477 = 0.378
S2K: 153 x (1.481 x 4.1 x 1.16) / 2810 = .384 WOOHOO!
4th Gear:
GT: 302 x (1.00 x 3.27) / 3477 = 0.284
S2K: 153 x (1.161 x 4.1 x 1.16) / 2810 = 0.300 WOOHOO!
5th Gear:
GT: 302 x (0.67 x 3.27) / 3477 = 0.190
S2K: 153 x (0.971 x 4.1 x 1.16) / 2810 = 0.252 WOO HOO
6th Gear:
WOO HOO!
P.S. I'm bored.
P.P.S. Does anyone know for sure whether the reduction gear is multiplied in with the rest of the gearing, or did I not do that right?
Joined: Nov 2001
Posts: 432
Likes: 0
From: Louisville
twohoos, torque is measured in foot-pounds or newton-meters to account for the distance of the force from the axis of rotation. It's not because it represents a unit of energy, or force over a distance.
Ignore the transmission and CVT effects completely. If the car's in first gear, it will be accelerating faster at the torque peak than it does at the HP peak, even though it's doing "more useful work" at the HP peak (force expended over distance in a unit time).
Yes, ultimately what moves the car is the tires spinning, so at some point, you have to account for the gearing and the transmission. However, now you're talking about torque peak at the engine vs torque peak at the transmission, and suggesting that my simple statement of F = M/a (or F x M = a, for those of you who perfer to use "Tim's new math" <g>), is too simple.
I KNOW IT IS!
However, in any given gear, the torque at the rear wheels will be proportional to the torque at the engine. Appropriately, the acceleration--within that gear--will be greatest at the torque peak. Changing the force from angular to linear doesn't change a single thing! By definition, torque is a measurement of "equivalent linear force" at a specific distance from the axis. All the gears do is change the distance from the axis (and hence the effective torque at the wheels).
Now, in retrospect, I ignored the CVT, which was a foolish mistake by me, exceeded only by my even more foolish physics comment and poor math. The CVT changes the effective toque axis length as we go.
However, my ignorance notwithstanding, I still maintain that the acceleration (rate of change of velocity) is greater at the torque peak than it is at the HP peak. Eliminate the CVT, use a single gear, and show me I'm wrong. We go through this argument about twice a year, and I don't remember anything changing about the argument since the last time we went through it.
Tim
P.S. Why is it that I never see this discussed on other car forums? Is it because we're always so busy defending our car's lack of low-end torque, or because the effective RPM range is great enough for it to be a noticable issue?
P.P.S. The torque units are no different from a linear force. The units may LOOK like energy units, but the terms have much different meanings. Torque is NOT watts.
Ignore the transmission and CVT effects completely. If the car's in first gear, it will be accelerating faster at the torque peak than it does at the HP peak, even though it's doing "more useful work" at the HP peak (force expended over distance in a unit time).
Yes, ultimately what moves the car is the tires spinning, so at some point, you have to account for the gearing and the transmission. However, now you're talking about torque peak at the engine vs torque peak at the transmission, and suggesting that my simple statement of F = M/a (or F x M = a, for those of you who perfer to use "Tim's new math" <g>), is too simple.
I KNOW IT IS!
However, in any given gear, the torque at the rear wheels will be proportional to the torque at the engine. Appropriately, the acceleration--within that gear--will be greatest at the torque peak. Changing the force from angular to linear doesn't change a single thing! By definition, torque is a measurement of "equivalent linear force" at a specific distance from the axis. All the gears do is change the distance from the axis (and hence the effective torque at the wheels).
Now, in retrospect, I ignored the CVT, which was a foolish mistake by me, exceeded only by my even more foolish physics comment and poor math. The CVT changes the effective toque axis length as we go.
However, my ignorance notwithstanding, I still maintain that the acceleration (rate of change of velocity) is greater at the torque peak than it is at the HP peak. Eliminate the CVT, use a single gear, and show me I'm wrong. We go through this argument about twice a year, and I don't remember anything changing about the argument since the last time we went through it.
Tim
P.S. Why is it that I never see this discussed on other car forums? Is it because we're always so busy defending our car's lack of low-end torque, or because the effective RPM range is great enough for it to be a noticable issue?
P.P.S. The torque units are no different from a linear force. The units may LOOK like energy units, but the terms have much different meanings. Torque is NOT watts.
Registered User
Joined: Feb 2001
Posts: 2,895
Likes: 1
From: You wish
Thank you Luis.
John, that's why your CVT example is causing you so much consternation. At the power peak, the engine is producing the best overall output (torque and rpm). If your torque peak is at 6000 rpm and the power peak is at 9000 rpm, you'll be able to run 50% tighter gearing if you're accelerating at 9000 rpm. Which means that as long as you're making at least 2/3 the torque at 9000 that you are 6000, you're all good.
Luis put things very succinctly and clearly.
Steve, on the topic of acceleration and Mustangs, couple things to consider. First, peak acceleration doesn't tell you too much unless the torque curve is flat. I think you'll find that the S2K torque curve is flatter than the Mustang's over the normal acceleration range (6k-9k for the S2K vs. 4k-6k for the Stang). If you take an average torque value over that range (just do a piecewise integration every 500 or 1000 rpm for simplicity) I think you'll find that the S2K's deficits are smaller and advantages grow bigger. The other thing to note is that on street tires, you'll be hard pressed to pull more than about 0.8 g in longitudinal acceleration (I don't know why that is, since we can brake at over 1 g). This is why a Mustang can light the tires on a roll in 1st gear while an S2K can't (calculate g's using wheel torque and add in the weight of a driver for accurate results).
UL
John, that's why your CVT example is causing you so much consternation. At the power peak, the engine is producing the best overall output (torque and rpm). If your torque peak is at 6000 rpm and the power peak is at 9000 rpm, you'll be able to run 50% tighter gearing if you're accelerating at 9000 rpm. Which means that as long as you're making at least 2/3 the torque at 9000 that you are 6000, you're all good.
Luis put things very succinctly and clearly.
Steve, on the topic of acceleration and Mustangs, couple things to consider. First, peak acceleration doesn't tell you too much unless the torque curve is flat. I think you'll find that the S2K torque curve is flatter than the Mustang's over the normal acceleration range (6k-9k for the S2K vs. 4k-6k for the Stang). If you take an average torque value over that range (just do a piecewise integration every 500 or 1000 rpm for simplicity) I think you'll find that the S2K's deficits are smaller and advantages grow bigger. The other thing to note is that on street tires, you'll be hard pressed to pull more than about 0.8 g in longitudinal acceleration (I don't know why that is, since we can brake at over 1 g). This is why a Mustang can light the tires on a roll in 1st gear while an S2K can't (calculate g's using wheel torque and add in the weight of a driver for accurate results).
UL
Administrator
Joined: Oct 2000
Posts: 20,274
Likes: 4
From: Toronto, Canada
Originally posted by Luis
...actually what pushes you back in your seat is force and what you feel is acceleration. ... an you're just a piece of mass, C3
...actually what pushes you back in your seat is force and what you feel is acceleration. ... an you're just a piece of mass, C3
Some guy said that once Acceleration is the measure of the change of speed of a mass upon which a force is applied, enough force to counteract the other forces we experience as a normal part of life; resistance, gravity, wind blowing etc.
When force is applied to the car in enough quantity to counteract the other forces acting upon it, it changes speed. That change is measured as acceleration. We, the passengers in the car are forced to accelerate at the same rate by the force being applied to us by the back of our seats.
In summary:
The measure of torque is meaningless in terms of acceleration. Consider a wrench. When you put a wrench on a bolt head and apply pressure to the wrench to turn the bolt you are applying force. You apply the same force regardless of how fast you turn it. How fast the bolt turns is related to the force counteracting the force you are applying. If the bolt is rusted and seized then 100 ft/lbs of force may not move the bolt at all even though you are applying force to it.
Power is the measure of the force applied in relation to how far the bolt actually turned. It is the measure of work. If you apply 100 ft/lbs or torque to a rusty bolt and it doesn't turn then you haven't actually done any work; the bolt didn't move. However, if you applied a constant 100 ft/lbs of torque to the bolt and it turned one revolution (360*) then you have done work; the bolt moved. The torque applied in both cases was the same.
So...
If a car is travelling along a straight and level road at sea level with no wind and no force is applied to it, it will slow down and stop. To keep it moving at a constant rate of speed, enough force needs to be applied to it to exactly counteract those forces which are acting upon it to make it stop. The faster the car is moving, the stronger the forces acting on it to stop and thus the more force must be applied to keep it moving at a constant rate of speed. If you reduce the forces acting on it to stop -- make it lighter and more aerodynamic -- then you need less force to keep it moving. Make sense?
In conclusion:
A Mustang GT is a big fat hunk of poo and thus requires a lot of force to accelerate its bulbous profile through the atmosphere. The S2000 is a sleek and light weight gem of engineering and thus requires less force to do the same work. A 125cc shifter cart will out accelerate both of them with only a fraction of the force being applied. A locomotive is massive and will never accelerate anywhere close to as quickly regardless of how much torque it has.
Torque is NOT a measure of acceleration, it's a measure of force. The rate of acceleration is determined by what that force is applied to. Requiring 1/3 more torque to accomplish the same amount of work is not necessarily something to brag about.
peace
Joined: Feb 2001
Posts: 4,061
Likes: 363
From: Redondo Beach
Sigh. One more try -- it's tough being a lone voice.
Luis seems to have followed my CVT proof (thanks, Luis!
), but remains unconvinced because the CVT changes the gearing. So let's look at another example that removes gearing from the issue -- top speed.
As most of us know, a vehicle's top speed is basically a function of two things: its horsepower, and its drag coefficient. This says that max speed is achieved at peak engine power, not peak torque (remember, we can gear the car any way we want). Now, at that max speed, clearly the aero force resisting the car is maximum; therefore, the net *linear* force propelling the car must be maximum. But as we've agreed, it occurs at the engine's power peak, not at its torque peak! (Tim: I propose this as your requested non-CVT, single-gear example.)
Anyway, my point all along has been that "gearing" is a red herring. Bruce originally said that power is important because it "takes advantage of gearing." This is just not correct. It's wheel *torque* that is multiplied by gearing. Wheel *power* is not -- it's the only constant, and that's why it's important!
Summary: I believe I've proposed two examples in which the net linear vehicle acceleration (i.e. what we FEEL) is constant, and I've compared that acceleration at the engine torque peak vs the power peak. Both results have favored the power peak. What more can I do?
Final thought: One thing I've learned in addressing tricky physical problems like this is that you *must* start by following the ENERGY. Energy is what nature conserves, not force. In our case, looking at engine power is the way to understand where the energy is going.
Luis seems to have followed my CVT proof (thanks, Luis!
), but remains unconvinced because the CVT changes the gearing. So let's look at another example that removes gearing from the issue -- top speed. As most of us know, a vehicle's top speed is basically a function of two things: its horsepower, and its drag coefficient. This says that max speed is achieved at peak engine power, not peak torque (remember, we can gear the car any way we want). Now, at that max speed, clearly the aero force resisting the car is maximum; therefore, the net *linear* force propelling the car must be maximum. But as we've agreed, it occurs at the engine's power peak, not at its torque peak! (Tim: I propose this as your requested non-CVT, single-gear example.)
Anyway, my point all along has been that "gearing" is a red herring. Bruce originally said that power is important because it "takes advantage of gearing." This is just not correct. It's wheel *torque* that is multiplied by gearing. Wheel *power* is not -- it's the only constant, and that's why it's important!
Summary: I believe I've proposed two examples in which the net linear vehicle acceleration (i.e. what we FEEL) is constant, and I've compared that acceleration at the engine torque peak vs the power peak. Both results have favored the power peak. What more can I do?
Final thought: One thing I've learned in addressing tricky physical problems like this is that you *must* start by following the ENERGY. Energy is what nature conserves, not force. In our case, looking at engine power is the way to understand where the energy is going.
Joined: Feb 2001
Posts: 4,061
Likes: 363
From: Redondo Beach
Yet a third example -- this one we should all be familiar with: Rolling-start acceleration in a given gear.
The S2K's torque curve is, as UL noted, pretty flat. If acceleration is greatest at the torque peak, then it stands to reason that acceleration is directly proportional to the torque value at any rpm, as long as the gearing is constant.
Now, at 3K rpm, UL's dyno said my car's wheel torque was 130 lb-ft, compared to its peak of 143 lb-ft at 6700 rpm. So my car's net linear acceleration at 3K rpm should be 91% (=130/143) of its linear acceleration at 6.7K. I suspect most S2Ks are similar. Now -- do you guys really think you're getting about 90% of your car's peak acceleration at 3K rpm? If not, then what's causing the difference? I'll tell you for sure: it isn't gearing.
The S2K's torque curve is, as UL noted, pretty flat. If acceleration is greatest at the torque peak, then it stands to reason that acceleration is directly proportional to the torque value at any rpm, as long as the gearing is constant.
Now, at 3K rpm, UL's dyno said my car's wheel torque was 130 lb-ft, compared to its peak of 143 lb-ft at 6700 rpm. So my car's net linear acceleration at 3K rpm should be 91% (=130/143) of its linear acceleration at 6.7K. I suspect most S2Ks are similar. Now -- do you guys really think you're getting about 90% of your car's peak acceleration at 3K rpm? If not, then what's causing the difference? I'll tell you for sure: it isn't gearing.