Torque and power quiz 4

[yoshimitsuspeed]

I spend a lot of time in the automotive community and a lot of time on car forums and groups.
Ever since I started to learn the actual relationship of torque and power it drove me crazy how few others in the automotive world actually understood this basic formula. People capable of building motors that make 500 hp per liter and who still think that torque is low end power.
I have gotten into enough arguments with people to learn that most would rather argue relentlessly cause that's what theys daddy taught them than sit down and think about the simplicity of the formula long enough to understand the relationship of the two.
I have decided to try a different tact and make a little quiz that gets people thinking about this from a different angle and maybe hoping they will get the point that torque and power can't be compared, and that torque does not mean low end power.
I just started on this tonight and it's 3AM. I want to do more to improve it but I also would love some input from others on ways I could improve it.
I would like it to be as detailed and informative as possible while still being interesting and keeping the person engaged and interested.

Tell me what you think of what I have so far.
What could I do better?
What are other questions I could ask or ways I could put things to get people thinking about the relationship without getting too bogged down in the math to loose too many people?
On that note should I focus more on the math or stay more with the basic relationship and principles?

http://matrixgarage.com/content/think-you-understand-torque-and-power

 

[LionelHutz]

I don't really understand your point. Are you trying to claim HP doesn't matter???

Saying the gear ratio caused an increase in axle torque is no different than just saying it was because of a HP increase.

Calculate the new HP at the axle due to your torque change and then compare it to the new engine hp. They'd better be the same. The nice thing about HP is that there is no need to do all those torque and force calculations. If you are driving the same speed and increase the available HP then you will be able to do more work plowing or climb a bigger hill.

 

[rancherman]

If you'd read my first post.. I clearly state I was here for my information purposes. either right or wrong.
So, here we are:
I've been told that HP is all that is necessary to maintain speed when going up a hill.
I used the perfect vehicle to play with.. My Dodge cummins. (very flat torque curve between 1600 and 2600)
I proved that when going up hill, both engines degrade pretty much the same rate even though the faster engine was putting almost 2x the horsepower down.. The faster speed degraded quicker due to the higher drag on the pickup..

Simply shifting gears no doubt saves the day here. My 'point' is made from NOT shifting. I'm talking 'engine only'..

the 1600 rpm engine with 500 ft. lbs of torque hit the hill with 152 HP (about 50 MPH)
the 2600 rpm engine with 500 ft. lbs of torque hit the hill with 247 HP (about 80 MPH)

the faster engine lugged down much quicker (at the start) but I figured the drag on the truck was much higher as well.
Since I was told that HP maintains speed, why isn't the much higher HP helping out in this case? Is it because the torque applied to the rear tires is the same in both cases?
To me, 'maintaining speed' is just that: not losing rpm. Both failed in 5th gear at pretty much the same rate (after throwing out the drag thing) the vehicle was overloaded for the gear it was in.
I'm here to learn. NOT prove anyone wrong. I am here to see If I am on the right track. These are my 'findings', and what I have 'heard' on the streets in my recent lifetime.
Here it is, my jugular exposed. LOL. Someone explain it to me!

 

[140Airpower]

Rancherman, in my life I have only had one car that was an automatic. I love rowing through the gears and feeling the grunt from low rpms pull through right on up through the range. I really appreciate great low-end torque. But, when you say "...shifting gears no doubt saves the day here. My 'point' is made from NOT shifting. I'm talking 'engine only'..." realize that with a 9-speed auto you have no idea of what gear you are in or maybe not even that there are distinct gears. The flexibility and torque/power range of the engine is a non-issue. This is more and more the state and the future of engine operation. The engine you need for a Powerglide or a 5-speed manual is no longer needed. That is partly why a very small turbo engine can haul SUVs around. A lot of it is in the transmission.

 

[gruntguru]

Rancherman. Neglecting air and road resistance, maintaining speed on a given hill requires a particular value of torque at the wheels. If the same torque is available at double the road speed, the vehicle will also be able to maintain double the road speed on the hill. However doubling the roadspeed with the same rear wheel torque requires double the power. It is available power that ultimately determines the highest speed the vehicle can maintain up the hill.

Power = mgv
Where power = power consumed to gain altitude (increase potential energy), m = mass, g = gravitational acceleration constant, v = vertical component of vehicle velocity (similar to rate-of-climb for aircraft)

je suis charlie

 

[LionelHutz]

"I've been told that HP is all that is necessary to maintain speed when going up a hill."

To test what you're saying you want to prove, speeding up approaching the hill is not maintaining speed.


When your vehicle struggles to climb a hill in high gear:
You proved that increasing hp and speed at the same rate doesn't help to climb a hill.
You failed to prove that more HP doesn't help you climb a hill (not changing speed).

 

[yoshimitsuspeed]

I had someone pose a question on torque and power vs acceleration that managed to confuse me a bit. Maybe you guys can help me with this.
I have gotten so used to thinking in terms of maximizing acceleration through gearing that power and power curve should really be all that matters. Then they ask me something along the lines of
After thinking about this too much I believe the answer is yes but then I find it very confusing. This would mean that a motor that makes 100 lb ft at 1000 RPM and 50 lb ft at 2000 RPM would accelerate at half the rate at 2000 RPM as it would at 1000 RPM.
If you look purely at force applied to the mass this would make sense because at 2000 RPM the motor would be applying half the force to the road but the motor is applying half the torque at twice the rate so it should be able to do the same amount of work/time as it was at 1000 RPM.
Now if you shifted to a gear that halved your RPM it makes sense that your rate of acceleration should maintain the same because you are doubling torque at the same time so everything mostly makes sense. I'm just confused about why the rate of acceleration at 2000 would be less when the motor is doing the same amount of work/time.

 

[Compositepro]

"I'm just confused about why the rate of acceleration at 2000 would be less when the motor is doing the same amount of work/time"

Because, in the same gear the car is going twice as fast, so friction loses and wind resistance are much greater. Power is work/time. It is also force x distance. At twice the speed you cover twice as much distance per unit time. On a flat and level track you can plot the power required to drive the car as a function of speed. This curve will approximately be a squared function due to air resistance. So just going twice the speed will require four times the power. Of course, just to get the car to start rolling requires power so this overpowers the square relationship at low speeds.

 

[NormPeterson]

Because, in the same gear the car is going twice as fast, so friction loses and wind resistance are much greater. Power is work/time. It is also force x distance. At twice the speed you cover twice as much distance per unit time. On a flat and level track you can plot the power required to drive the car as a function of speed. This curve will approximately be a squared function due to air resistance. So just going twice the speed will require four times the power.

Nope. At double the speed, the force is quadrupled. But since you're now pushing that force at double the original speed, force x velocity is then 4 x 2 = 8 times the power requirement.


Norm

 

[yoshimitsuspeed]

I am talking about a situation with friction removed. Just talking about basic principle of mass and acceleration here.

Without friction would a motor accelerating a mass with 100 lb ft at 100 RPM accelerate the mass at twice the rate as the motor with the same gearing at 50 lb ft at 2000 RPM?

If so then how come when they do the same amount of work/time?

If I'm not mistaken work is a force/distance. Power is work/time.

A crankshaft spinning one rotation at 50 lb ft should to 50 ft lbs of work. A motor making 100 lb ft will do twice the work/distance but if it is spinning half the speed they will do the same amount of work per unit time.

 

[NormPeterson]

I had someone pose a question on torque and power vs acceleration that managed to confuse me a bit. Maybe you guys can help me with this.
I have gotten so used to thinking in terms of maximizing acceleration through gearing that power and power curve should really be all that matters. Then they ask me something along the lines of

Quote (In a given gear shouldn't the acceleration curve match the torque curve?)

After thinking about this too much I believe the answer is yes but then I find it very confusing.

This is part of what I was alluding to back on 15 March. Maximum acceleration at any given speed needs you to be geared to put the engine at peak power rpm at that speed. Once that gearing is fixed and you're looking at accelerations at different speeds, acceleration using that gearing must at least generally follow the torque curve.

It might help if the numbers used for illustration were not so closely related to one another (i.e. 2:1 for both torque and rpm), which actually implies that your motor is truly a constant power device. Say your 100 ft-lb @ 1000 rpm (~19 HP) motor makes peak power at 2000 rpm where 75 ft-lb is available (28.6 HP). You gear it for maximum acceleration at, say, 40 mph, so that's your 2000 rpm speed where you've got 75 ft-lbs times whatever gearing. At the peak torque point of 1000 rpm, you'll be doing 20 mph with 100 ft-lb available, times the same gearing . . . Ignoring aero and what-not, that sure sounds like a 33% improvement in acceleration to me.

Re-gearing the whole business to maximize acceleration at 20 mph only moves the max acceleration in gear point down to 10 mph, where it will be 33% better than the acceleration at 20. The new acceleration at 20 will be better than the old acceleration at 20 (as in the first arrangement), but that's not the question.


Norm

 

[yoshimitsuspeed]

NormPeterson that's a good way of explaining essentially the same conclusion I came to but I still don't feel like that accounts for all energy potential.

Acceleration really comes down to mass and force. The torque is the force so this makes sense but why does the motor need more power/energy at a higher RPM to apply the same amount of force?

I have been trying to think of an analogy or a different way to put it to try to clarify. I think I might have it.

If we hook your 19/28.6 motor up to a generator. To make it easier I'll convert that to 14/21 KW motor.
At 1000 RPM it will make 14 KW of power and at 2000 RPM it will make 21 KW of power.
Now if you took that power and applied it to an electric motor 21 KW of power would give you more energy for acceleration than 14 KW of power.
What am I missing to think that in the direct drive application there is lost energy when at 1000 RPM the motor is able to do more work than at 2000 RPM?

 

[JackAction]

I'm just confused about why the rate of acceleration at 2000 would be less when the motor is doing the same amount of work/time.

Because it only has half the time to do the work.

If you divide the time it takes to do the work by a factor of 2, then you must also divide the amount of work done by a factor of 2 ... if you want to keep the power constant, of course:

power = work1 / time1 = work2 / time2

if time2 = time1 / 2, then work2 = work1 / 2.

Assuming you travel the same distance (i.e. you make the same number of revolutions), at 2000 rpm you will loose half the torque over that same distance (50 lb.ft instead of 100 lb.ft) but you will do it in only half the time. That is the whole concept of power: How much work can I do in a given time frame.

Here's an analogy:

You have two workers. You need them to move a pile of bricks from one spot to another. You are paying them by the hour, so in it is important that this is done within 1 hour.

The first one is a big guy, so he takes the whole pile and moves it slowly to the new location, taking the full hour to do it;

The second one is not as strong, so he only takes one brick at a time, but runs to the new location. Making all the trips necessary to move the entire pile takes him an hour.

As a boss, you don't care how the work was done: The whole pile is moved (total force carried across a certain distance, a.k.a. work) in the same amount of time. If you didn't look at them work, you can't even tell which one took the whole pile in on trip and which one ran with one brick at a time. The same work was done in the same period of time. We say these two workers are equivalent.

But if you gave only ½ hour to these guys, the guy who run with one brick at a time will move only half the pile (like with your engine example) and the other guy would move the entire pile of brick half way. They would still be considered equivalent power-wise, because they had half the time, they did half the work. Give them another ½ hour and they will finish the job, even if they switch around (The big guy would lift half the payload and could now travel the full distance twice as fast; And the other guy would still moved all the bricks one by one at the same speed, but would only have to travel half the distance).

 

[NormPeterson]

Power doesn't define the force; it's kind of the other way around where a force over some distance per unit time (velocity) defines the power. So at a lower velocity than the velocity associated with peak power, the force can be greater than the force associated with the peak power velocity even as the amount of power available is less. Basically, the velocity term decreases faster than the power does as you drop from peak power rpm to peak torque rpm. At half the velocity (20 vs 40) you still have roughly 2/3 of the power (14 vs 21), so the force must be higher. F = Ma, where M is held constant . . .


Norm

 

[140Airpower]

Yoshimitsu, You said "...I'm just confused about why the rate of acceleration at 2000 would be less when the motor is doing the same amount of work/time...".

Don't forget that we are accelerating by applying thrust and power to the ground, ...and the ground is retreating at twice the speed at 2000rpm as at 1000rpm. Power is energy per unit time and energy is force x distance therefore power is force x velocity and for a constant power, when velocity doubles, force is halved. This is a simple fact that doesn't even account for friction and aero drag, etc.

 

[snoopnoon]

There are always people who will talk about how a vehicle without a lot of torque output will be slow, without any reference to horsepower; or, people will compare two vehicles of the same power output where one has more torque output, and assume that the one with more torque will be faster. Neither of these is true, as a lot of you know. I thought it would be fun to play with some numbers, to prove outright that horsepower is what matters, and that crank torque numbers are all but useless for comparing vehicles.

First, let's do a basic proof, using relations; none of the following are actual equalities, but I am going to use the equal sign to symbolize that they are proportional. In other words, I've dropped out constants to make the equations simpler and more general.

First, we start with three equations:

1.) Horsepower (HP) = Torque (TQ) * Engine Speed (RPM)
2.) Thrust (THR) = Torque (TQ) * Gear Ratio (GR)
3.) Speed = Engine Speed (RPM) / Gear Ratio (GR)

In this example, gear ratio is assumed to be a number describing the overall reduction from crank speed. Thrust is the force applied at the pavement, which allows you to find acceleration when given the weight of the vehicle.

Now, rearrange equation 1 to come up with the following:

4.) TQ = HP/RPM

Then substitute equation 4 into equation 2, for torque:

5.) THR = (HP*GR)/RPM

This can also be written as:

5a.) THR = HP * (GR/RPM)

We can rearrange the speed equation (eq. 3) above, to the form:

6.) 1/SPD = GR/RPM

And, through substituting this into equation 5a, we achieve the following:

THR = HP * (1/SPD) = HP/SPD

So, the result is that thrust is proportional only to horsepower and vehicle speed.

In a real-life situation, you don't have infinite gear ratios, and instead (assuming no gearchanges in the middle of our comparison) you have a constant in that place. The result actually remains the same, since then speed is only proportional to RPM, and you can drop it right into the bottom of the modified torque equation (eq. 4).

What does this prove? Well, this proves that at ANY given speed, the acceleration of a vehicle can be directly related to the horsepower production at that point, and the weight of the vehicle (assuming no wheelspin, of course).

So, if you have two identical vehicles traveling side by side, and one has an engine making 75 ft-lbs of torque at 14,000 rpm and the other has a sweet diesel making a cool 500 ft-lbs of torque at 2100 rpm, they'll accelerate at exactly the same rate. They're making the same power, and traveling at the same speed, so they have the same thrust.

On a side note, assuming they both have the same BSFC (which is a bit of a stretch, but it could be reasonably close), then they will both burn fuel at the same rate.

 

[yoshimitsuspeed]

Thanks guys I can see how that all works.
I guess my thought with seeing it in terms of energy such as using a generator to power an electric motor with higher wattage is kind of a roundabout way of achieving that more ideal gear ratio so to speak.
That makes sense.

 

[JackAction]

There are always people who will talk about how a vehicle without a lot of torque output will be slow, without any reference to horsepower; or, people will compare two vehicles of the same power output where one has more torque output, and assume that the one with more torque will be faster. Neither of these is true

The sentence in bold will be true in most cases. We might also reference to a previous quote in this thread:

Here we go with the science lessons. So when we put a V8 in a pick up we're wasting time, so put in a proper transmission instead? Torque and HP are related, give me something with torque then and forget the HP. Our 9 second Evo had 600 HP and 500 ft/lb, bu the 8 second dragster V8 had 600 HP and 900 ft/lb. Need we say more? Our Type R Honda had 650 HP and 450 ft/lb, it was not as fast as the Evo.

It is obvious that he's talking from experience, we cannot deny the fact that between 2 engines with the same peak power, the one with the highest peak torque will offer better performance. But it is not because of the peak torque per say.

People usually say: «Only the power counts, not torque». What they should say is:

«Only the average power in the selected rpm range counts, not the average torque in the selected rpm range.»

If the torque is less at a lower rpm, the power will also be lower and that will lower your average power available at the wheel since the peak power values are the same between the two compared engines.

Of course, if we have a lot more gear ratios in the car with the lower peak torque, such that we are using a smaller rpm range (closer to peak power), it is possible to maintain the same average power as the one from the engine with the highest peak torque.

 

[NormPeterson]

There are always people who will talk about how a vehicle without a lot of torque output will be slow, without any reference to horsepower; or, people will compare two vehicles of the same power output where one has more torque output, and assume that the one with more torque will be faster. Neither of these is true, as a lot of you know. I thought it would be fun to play with some numbers, to prove outright that horsepower is what matters, and that crank torque numbers are all but useless for comparing vehicles.

I hope that when your derivation reaches its conclusion that your HP variable is being taken from whatever RPM is consistent with both SPD and GR. Not max HP as might be inferred.

When you throw consideration of rotational accelerations of things like tires and flywheels into the mix, any approach working with HP should suddenly become a lot less attractive.


Norm

 

[JackAction]

It might help if the numbers used for illustration were not so closely related to one another (i.e. 2:1 for both torque and rpm), which actually implies that your motor is truly a constant power device. Say your 100 ft-lb @ 1000 rpm (~19 HP) motor makes peak power at 2000 rpm where 75 ft-lb is available (28.6 HP). You gear it for maximum acceleration at, say, 40 mph, so that's your 2000 rpm speed where you've got 75 ft-lbs times whatever gearing. At the peak torque point of 1000 rpm, you'll be doing 20 mph with 100 ft-lb available, times the same gearing . . . Ignoring aero and what-not, that sure sounds like a 33% improvement in acceleration to me.

Re-gearing the whole business to maximize acceleration at 20 mph only moves the max acceleration in gear point down to 10 mph, where it will be 33% better than the acceleration at 20. The new acceleration at 20 will be better than the old acceleration at 20 (as in the first arrangement), but that's not the question.

This is where the «torque» people get lost. You have to compare the torque output at the same vehicle speed, otherwise it is a useless comparison: At a higher speed, the traction force will always be smaller for a given power output.

When you «re-gear the whole business», of course the torque is still 33% at 1000 rpm than at 2000 rpm; It will always be the case no matter what gear ratio you use. The important thing is the torque output when the vehicle's speed is the same in both gears.

With your example, if the gear ratio was, say, 3:1 in the first case then the torque output was 300 lb.ft @ 333 rpm (20 mph) and 225 lb.ft @ 667 rpm (40 mph). If you switch to a gear ratio of 6:1, then you get 600 lb.ft @ 167 rpm (10 mph) and 450 lb.ft @ 333 rpm (20 mph). The important comparison is that at 20 mph, you get 450 lb.ft in low gear vs 300 lb.ft in high gear. Ignoring aero and what-not, that sure sounds like a 50% improvement in acceleration to me. Nobody cares about the fact that you get 600 lb.ft @ 10 mph, because you can get 900 lb.ft at the same speed with a 12:1 ratio (Which is still a 50% improvement). Only in first gear will the maximum torque of the engine will matter.

For the same output rpm, the output torque will always be greater in a gear ratio where the power is the greatest. That is the whole concept of power.

When you throw consideration of rotational accelerations of things like tires and flywheels into the mix, any approach working with HP should suddenly become a lot less attractive.

I never saw a transmission design that offered such an increase in inertia, that torque gain was overtaken. Even for the example, for a typical car, the effective mass of the car for a 3:1, 6:1 & 12:1 gear ratios would be 1.06, 1.13 & 1.40 times the mass of the car. So a 6.6% increase between 6:1 & 3:1 and 23.8% increase between 12:1 & 6:1. The 50% improvement in torque still makes a noticeable difference.

 

[NormPeterson]

This is where the «torque» people get lost. You have to compare the torque output at the same vehicle speed, otherwise it is a useless comparison: At a higher speed, the traction force will always be smaller for a given power output.

<sigh> . . . I understand that, and said as much.

Geared appropriately, running at max HP will provide maximum acceleration at any given speed <snip>

<double sigh> . . . and *power* people always manage to lose sight of what happens to acceleration when the engine cannot be operating at its peak power rpm; it's as if acceleration at any other rpm does not matter even if running at those rpms cannot be avoided. My point all along has been that (1) sometimes the reality of your road speed does not permit you to be at peak power rpm, (2) the acceleration at such other road speeds has nothing to do with either peak HP or its RPM, and (3) the maximum acceleration in any given gear occurs when the engine is operating (essentially) at its peak torque point.

But the maximum acceleration with that same gearing will still occur at max torque.

Sometimes it's nice to know what happens under conditions other than peak power operation. Enlightening, even, when you realize that acceleration curves droop toward the tops of each gear because of more than just the increases in the speed and drag.

I was hoping you'd look at your own derivation of rotational accelerations instead of blindly using somebody else's rules of thumb and all of their [unlisted] assumptions. You might have found that it's simpler to work with torque, and from there seen acceleration as a function of torque in a different light.

Have you taken yoshi's little quiz yet?


Norm