Team-BHP - Understanding Power, Torque, Gearing and Top Speed
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This post continued a discussion from a transmission thread about gearing, specifically what overdrive means, as a response to a wikipedia definition of the overdrive concept. This developed into a discussion of power, torque, gearing, and the relevance of those for top speed and acceleration, which was a suitable to a thread of its own. - Support Team

EDIT: I have created some slides with charts to explain power, torque and gearing and their relevance to acceleration, top speed and efficiency (pdf attached to the bottom of this post). It is more comprehensive and thorough than the posts I made in this thread prior to creating that pdf. Be sure to check it out.

Quote:

The most fundamental meaning is that of an overall gear ratio between engine and wheels, such that the car is now over-geared and can no longer reach its potential top speed,
THis wikipedia definition is not completely straightforward. (wikipedia is a good source but not a definitive one.)

Firstly, The word "overdrive" itself gives us the real fundamental meaning: Any gear in any machine is said to be overdriven if it turns faster than the gear that is driving it.

In the context of automobiles, it refers to gear ratios that result in the output shaft turning faster than the input shaft. .

The wikipedia definition as quoted above needs clarification before it makes sense. and anyway, this definition has no relevance to most people. It is a meaningless simplification of a very complex technical concept. Allow me to attempt an explanation. You will find that many very high performance cars (such as the dodge viper) reach their top speed in 5th rather than in 6th. Let us take a hypothetical example. Let us assume a car has a top speed of 190 mph in 5th at 7000rpm but only 155mph at 3000rpm in 6th. What happened? Simple. At 3000rpm, the engine is not putting out enough power to overcome the aerodynamic drag at 155mph. So it can accelerate no further. If you down shift then to 5th, putting the engine at 4500rpm, you can accelerate because at 4500rpm the engine makes enough power to overcome the drag at 155. And it continues to increase in power as the revs rise, meaning the car can continue to accelerate till it reaches 190mph. At 190, it may be that the engine is at its power peak (the car's highest potential speed) and so the car can go no faster (if the engine were to rev higher, its power would drop off and the car would slow down while needing its wheels to turn faster. an impossibiity given adequate traction), or it may be that there is more power at 8000rpm but at 7000rpm there is not enough power to push the car past 190. If it is the latter scenario, some other gearing combination would allow the car to reach 8000rpm and thus a higher potential top speed.

In this case, "overdriven" means that in this ratio, the engine will not reach its power peak and thus the vehicle's potential top speed. It has little to do with the actual numerical ratio of the gear, and is a factor as much of the final drive (differential ratio) as it is of the gearbox gear ratio.

In the case of most cars, the top speed is reached in 5th because most cars are set up not to reach their maximum top speed but for real world drivability and efficiency. This applies even to the great McLaren F1. That car, fitted with standard gear ratios, runs out of power in 6th gear at 217mph and can accelerate no further. But that is not its true top speed. McLaren fitted a McLaren F1 with different gearing and no changes to the engine or aerodynamics and the car reached 240mph. By the wikipedia definition, the stock F1 was overdriven.

This is NOT the fundamental definition of "overdrive". And it is not a definition that is relevant to anybody except those who are seeking to set top speed records for their cars, or racers looking to optimize acceleration vs speed. It is not a definition relevant to most people.

For most people, overdrive means and ONLY means "higher than 1:1" ratio. Please note that taller or higher gears are numerically lower. I gave the example of the i10 and its 4th being a 0.71:1 overdrive. This means that for 0.71 revolution of the input shaft (driving gear), there will be 1 revolution of the output shaft (driven gear). Thus the output gear is turning faster than the input gear, the latter is overdriven, and this is an overdrive ratio.

There is no other definition of "overdrive" that is relevant to the modern motorist.

Harbir, thanks for taking the effort to type out such detailed/lucid/crisp explanations clap:. Makes for enjoyable reading :thumbs up !

Quote:

Originally Posted by Harbir (Post 2948258)
At 3000rpm, the engine is not putting out enough power to overcome the aerodynamic drag at 155mph. So it can accelerate no further. If you down shift then to 5th, putting the engine at 4500rpm, you can accelerate because at 4500rpm the engine makes enough power to overcome the drag at 155. And it continues to increase in power as the revs rise, meaning the car can continue to accelerate till it reaches 190mph.

In the above example, whenever you say power, you actually mean torque, right?

Quote:

Originally Posted by Samurai (Post 2948579)
In the above example, whenever you say power, you actually mean torque, right?

No, I mean power. You have to keep in mind that what matters is torque at the wheels, not at the crankshaft, which includes gearing. What limits top speed is aerodynamic drag. At a vehicle's maximum potential speed, you would have the force exerted by the tyres pushing the car forward exactly equalling the force of aerodynamic drag. It would seem that this ought to happen at the torque peak, but it doesn't because it is a complex equation involving speed, gearing, torque curve shape, and RPM.

THe physics of it is complicated but essentially involves the notion that speed and power are both rate units. THe rate of distance covered per unit time, and the rate of work done per unit time, respectively. When the kinematics formulas involving rotational velocity (rpm), rotational force (torque), aerodynamic resistance (drag coefficient and frontal area) are worked out we are left with an equation that relates power to maximum speed in terms of only the air resistance. If you know the frontal area and coefficient of drag of a car, then knowing its peak power, you can calculate the theoretical maximum speed.

This may not make sense unless you absorb the kinematics formulas yourself (too complex for this forum), but here is how to make sense of it. Top speed is not about how much force (linear torque) you can apply at any given instant, but how much force you can sustain over a period of time. To travel at any speed requires an ongoing expenditure of energy. How much energy you can expend per unit time determines how fast you go. Energy expenditure per unit time is power.

The peak torque will produce the maximum acceleration in any gear (acceleration = force/mass). But top speed will always occur at the power peak because ∆velocity = ∆power/∆force, where force is the force of wind resistance (plus the force to accelerate the car's mass, but at top speeds thats a minor factor compared to drag force so we'll ignore it here). This equation means that the change in velocity of a car is equal to the change in power divided by the change in the force acting on the car. As a car accelerates, the rising engine rpm increases power with speed. but the force of drag also increases with speed. As long as the increase in power with speed exceeds the increase in drag force, the car will continue to increase in velocity (accelerate). When the speed increases to a point where the drag force cannot be overcome by the available power, the engine can climb no higher and gain no more power. at that point ∆power and ∆velocity both become zero and the car ceases to accelerate.

It has reached its top speed in that gear.

So the trick of making any car reach its maximum potential top speed is to select its gearing from 1 through the top speed gear that will enable the engine to reach the power peak in each gear.

I have attached a scanned image from a 1989 issue of Car and Driver, a roadtest of a 1989 Corvette ZR1. It gives the manufacturer specified power and torque peaks, gear ratios, plus actual test top speeds in each gear along with RPM.

You notice that the top speed of 175mph is reached at the power peak in 5th gear (actually its 6300 vs the specified 6200 power peak, but this is an immaterial difference, considering that the both values are rounded off, that there are variations from unit to unit and error in the tach of the unit tested.)

We see that in gears 1 through 4, the car accelerated right up to redline. 5th we see tops out at 175, but 6th drops to 152 at 3650rpm. This is short of the torque peak of 4200rpm. What happened? At 3650rpm, the engine is not producing adequate power to overcome the drag force at 152mph. So the car stops accelerating. [the team-bhp logo has obscured the bottom of the image. the information for 6th gear is a ratio of 0.50, and mph/1000rpm of 41.6mph)

Note that the top speed is NOT achieved at the torque peak but at the Power peak. Why? The reason that the limitation is power and not torque is gearing. If the car stopped accelerating at 152 mph in 6th, its because the drag force at that speed equals the force the car is able to produce at that rpm in that gear. It can produce no more so it cannot accelerate any further.

But here is the KILLER observation: If you have reached 175 in 5th, and you upshift to 6th at that speed, what happens? Well, the car is geared so that 6th 41.6mph/1000rpm. So at 175mph, shifting into 6th, will drop the engine from 6300rpm to 175*1000/41.6 = 4207rpm. that is EXACTLY at the torque peak. So why shouldn't the car keep accelerating from 175mph when it is right at its torque peak? because it doesn't have enough power at that speed

lets figure this out. at 4200rpm, the car is making 370lb-ft of torque. Power = Torque x rpm/5250 = 370lbft * 4200rpm/5250= 296hp. At that speed, in 5th, the car was putting out 380hp. You've lost 84hp shifting from 5th to 6th! That is not enough power to hold the car at 175mph because the energy it takes to move the air out of the way at that speed exceeds the energy output rate (power) of the engine. The car will start to slow down as the drag force on it will exceed the force the engine is able to apply at the contact patches to push the car forward.

I hope this answers your question.

THe thing to keep in mind is that power is the application of torque over a period of time. acceleration is produced by torque and maximum acceleration is produced at the torque peak, but because top speed is a matter of energy dissipation over time, the controlling factor is power, because that is the application of torque over time.

To understand this, think of a single cylinder engine. At its torque peak, it produces the greatest expansive force in the combustion chamber, thus drives the piston down with the most force and thus producing the greatest torque at the crankshaft. But at its power peak, while it torque output is reduced, it has more combustion events per second than at the torque peak. This means that at the power peak, the engine is liberating more energy per second than it does at the torque peak. What determines the top speed of a vehicle is not the peak force the engine can apply, but the maximum energy that it can produce per second.

I hope that explains it.

Thanks Harbir, that was very enlightening. Especially the ∆velocity = ∆power/∆force formula, it just nailed the points home. Your last two posts by themselves deserve a separate thread, so it is here.:)

Quote:

Originally Posted by Harbir (Post 2948773)
If you know the frontal area and coefficient of drag of a car, then knowing its peak power, you can calculate the theoretical maximum speed.

Very true, link to an old thread & post in the same vein...

Quote:

Originally Posted by Harbir (Post 2948773)
I have attached a scanned image from a 1989 issue of Car and Driver, a roadtest of a 1989 Corvette ZR1. The team-bhp logo has obscured the bottom of the image. the information for 6th gear is a ratio of 0.50, and mph/1000rpm of 41.6mph. But here is the KILLER observation: If you have reached 175 in 5th, and you upshift to 6th at that speed, what happens ? Well, the car is geared so that 6th 41.6mph/1000rpm. So at 175mph, shifting into 6th, will drop the engine from 6300rpm to 175 * 1000/41.6 = 4207 rpm.

So in the Corvette, at its top speed of 175 mph in 5th, if you grab 6th, the RPM drops from the red-line to 4,207 RPM ? That's a pretty big drop isn't it ? Usually as you go up the gears, the ratios tend to close up, but for this Corvette the difference between the ratios is almost the same for the top 3 gears. Is this unique to the Corvette or do all (or most) performance cars share this trait ? Could it be that the ratio for the 6th gear has been chosen (solely) for fuel efficiency while cruising ?

Its a pretty big drop, but this typical with large capacity high horsepower engines. A high horsepower engine with lots of torque in the bottom end and midrange will in typical use not need short gearing to do anything expected of it. but when the demand is there, it has to perform. So the engineers will choose the gearing that will enable the car both to perform to its potential but also be relaxed, efficient and chilled out in normal driving. The gearing ratio that is best suited to help a car punch the atmosphere with 500hp of muscle at 175mph is not likely to be a gear appropriate for the most efficient and pleasant use of the engine at normal speeds. So cars like this will often end up with 4th being the top speed gear but tall and very tall overdrives on 5th and 6th.

There is no universal norm for gearing ratios, such as ratios tightening up as you go higher. Gearing ratios are selected based on the engine's output across the rev range, the purpose of the vehicle, and the expectations of performance from the vehicle. when you're talking about 7000 and 8400cc monsters in corvettes and vipers, the logic of gearing ratio selection is very different from what you'll find in 100 or 200hp cars.

In the context of top speed and acceleration, It occurs to me that there is not a lot of clarity on what power and torque are and how they are relevant to the car enthusiast. I thought I'd write out a small message about the physics of force, torque, acceleration, and power to help folks understand it better.

I'll use a simple example. Let us assume you have a 1000kg object sitting on the ground that you want to lift. You bring a crane. The crane will apply an upward force to lift the object. But how much force will it take to lift the object? Enough to fully counteract the force the object is exerting on the ground plus a tiny bit more. How much is that? Well, Force = Mass x Acceleration. The earth's gravity creates a downward acceleration of 10 m/s^2 on any object on its surface. (Its actually 9.8, but lets use 10 for simplicity). So the force the 1000kg object exerts on the ground:

F= m x a
=1000kg x 10m/s^2 =10,000kgm/s^2
=10,000 Newtons.

This is the force the object exerts on the ground. To lift this object, the crane will have to apply 10,000 newtons in the opposite direction.

Here is where it gets interesting. The electric motor that pulls up the cable to lift the object is rated in power not torque. Why is this? because we can get whatever torque we want through gearing. Lets say the cable of the crane is being pulled by a pulley of 1 meter diameter. The radius is then 0.5 meter. How much torque is applied to the pulley to lift this object?

Torque = force x distance from fulcrum of the point of application of force.

In this case, the force is 10,000 Newtons, the distance is the radius of the pulley, 0.5 meters.

Torque = 10,000N x 0.5 meters = 5000Nm

Lets say the motor produces 500Nm. We need 5000Nm. What do we do? Gearing. We install a 10:1 gearbox. This causes the pulley to turn at 1/10th the speed of the motor, but exert 10 times the torque.

Result? We lift a 10,000N object with a 500NM motor.

So far we have our motor torque, our drive wheel torque (the pulley), the force applied (the object being lifted). Where is the power? In how fast are we lifting the object.

Let us assume that it is a 1000rpm motor. We reduced the RPM by 10, to drive the pulley at 100rpm. *The circumference of the pulley is 2x pi x r

= 2 x 3.14 x 0.5meters = 3.14meters.

Tangential velocity of the pulley =3.14 meters/revolution x 100 revolutions per minute
= 314meters per minute
= 5.23 meters per second.

This means the cable and thus the object we are lifting is being pulled up at 5.2 meters per second

How much power are we expending? To understand this, you must understand what power is. It is a measure of how much energy has been transferred per second. When you are lifting the object above the ground, you are adding potential energy to it. The energy is coming from the electricity that drives the motor, which pulls the cable which lifts the object. So we have expended electrical energy to lift the object, transferring the energy from the electric cable to the object. How much energy is expended or transferred per second is power. (let us assume for simplicity that the process is 100% efficient with no waste, though its never like that in reality).

In mechanical engineering, energy is expressed by the term "work". If you appy a 1 Newton force to move an object a distance of 1 meter, you have done 1Nm of work. If you do 1 Nm or work in 1 second, you have got 1 Nm/s of power. 1Nm/s = 1 watt.

So in our case, we have applied 10,000 Newtons to move the object 5.2meters in one second.

Power = Force x distance/ time = 10,000 N x 5.2 m/ 1 sec = 52,000Watts = 52kilowatts = 69.7 Horsepower.

So we see that while torque determines how much force we apply on the object, power determines how fast we move it! Because power is the measure of how much energy we are expending per second!

So why are motors rated on HP and not torque? Because we are interested in how much work it can do in a given amount of time, not not much maximum force it can exert. We can always get more force/torque by gearing it down, but if we want heavier objects moved in the same amount of time, or if we want the same objects moved faster, we're going to need a more powerful motors.

Lets say we want to lift a 5 ton object. We have a 52kilowatt motor. 5,000kg will require a lifting force of 50,000Newton. We rearrange our power equation from above and get speed = power/force

Speed = 52,000watts/50,000Newtons = 1.04meters/second.

The guy making the decision can figure out the power he needs just from knowing the the force to be applied and speed he wants. He doesn't need to know the torque or the gearing. Knowing the power he needs, the torque and gearing can be selected.

Its the same way with a car and its top speed. TO know how fast a car will go, you only need to know its max power and the force of drag to be overcome. However, since drag is speed dependent, you get an exponential equation, but the consequence is the same. What determines the power requirement of a car is the total aerodynamic drag force on it, and the speed it has to go at. Alternatively the power of a car and its aerodynamic drag will determine the top speed. Top Max power determines the maximum amount of work that can be done per second. A car's top speed is reached at the power peak because that is the point where the car is able to produce the most energy per second to shove the air out of the way.

Now, we come to the difference between engines and electric motors. Electric motors, unlike combustion engines have a perfectly flat torque curve. the 500Nm motor in our example produces 500NM at 0 RPM, at its max RPM and at any RPM in between. Because power = torque x rpm/5250, and an electric motor's power output is constant, a motor's power output is directly proportional to the RPM. Double the RPM = double the power.

Combustion engines are not like that at all. Torque rises with RPM up to a point and then falls off. But power keeps rising because while the force produced by each power stroke is reduced, there are more power strokes per second, thus the energy density in time increases with RPM meaning the ability to do more work per second, push more air out of the way per second, drive the car faster per second.

Because the torque of a combustion engine is not linear, you get multi speed gearboxes. You will notice electric motors almost never have multispeed gearboxes. THey don't need them because they produce max torque at all times, and one gear ratio to match motor speed and torque to the application requirement is all thats needed.

Now here is another important point: What your body feels is the result of torque. Your body is able to perceive forces applied on it. It is able to perceive acceleration due to the sensitivity of the inner ear (your balancing mechanism). Both these are related to torque. The human body has no mechanisms for physically perceiving power. You perceive it only through mental reasoning.

A plane flying at 800km/h seems perfectly still to us as passengers when we cannot see it moving relative to something else. At that steady speed, the thrust of the engines (force) is equal to the drag force on the aircraft so our bodies experience no force, no acceleration, no speed, and we feel nothing. But when the aircraft is accelerating for take off or braking to slow down after landing, we feel the forces and the acceleration/deceleration. We do not feel the power that is being expended to accelerate or decelerate the aircraft.

What does all of this mean for you? You have to think about torque and power not as two separate entities that do different things for you, but as aspects of the larger phenomenon at play, which is the flow of energy. Mechanical energy is transferred in the form of a force being applied over a distance (or a rotation). The rate of transfer of energy is power. When you are looking for acceleration, you want to think about torque and gearing. When you are talking about speed, you are talking about power.

I hope that was useful and relevant.

@Harbir WOW! That was a good piece of information. Just got a feel of my School/Engineering College days.

One query, if Electric motor's torque is flat throughout the rpm, how it is utilized in a (pure) electric car? How the gearing system different from that of IC engine. How do acceleration happens in an electric car?

Cheers!

Vinu

OK, electric cars.

First a clarification, when I say that an electric motor has constant torque, I don't meant that it will produce the same torque at all times. I mean that the torque is independent of RPM.

THe torque of an electric motor is variable. It varies based on the current applied. There is a maximum current any given motor can handle so there is a limit to how much torque a motor can produce. So in an electric car, as you increase the throttle, that increases the current, which increases the torque, which results in acceleration.*

*Electric motors are inefficient at high torque levels because the high current heats up the coils, resulting in energy loss as the heat is radiated away. At low rpm, there is a mismatch between the input electrical power and the output mechanical power. The output power is low because the RPM is low, but the input electrical power is dependent on current, not RPM. So there is an optimal range of speeds that are appropriate for an electric motor.

You want the motor to spin fast enough to convert input electrical power to output mechanical power efficiently but not so fast as to start losing electrical power as heat. So, like a pison engine, electric cars would benefit from multispeed gearboxes. But most electric cars don't have one. They have a single speed gearbox that allows the motor to run at the most efficient RPM range.*

Why? well, Electric cars engineers are obsessed with weight, volume occupied by the system, and its cost. Present day electric cars, struggling to compete with combustion powered cars don't have the luxury of multi speed gearboxes. So the engineers choose one ratio that will work ok in the conditions the car is expected to be used in.*

This results in regular production electric cars having very limited top speed, even if they have the power to go much faster.*

Some electric cars do have 2 speed transmissions in an attempt to cover both the normal usage and also high speed usage, but they haven't excited anyone yet. Electric cars run out of charge at high speeds so quickly that nobody uses them like that, which leaves engineers thinking if the weight, cost and complexity burden was worth it.

What a brilliant thread!!! Reminded me the wonderful days of Resnick Halliday and I.E. Irodov. :D My friends keep asking me the difference b/w torque and BHP, I used to try my best in explaining it in layman terms. Finally, I have a thread I can recommend to them.

Rated a well deserved 5 star. :thumbs up

Quote:

Originally Posted by Harbir (Post 2948258)
Let us assume a car has a top speed of 190 mph in 5th at 7000rpm but only 155mph at 3000rpm in 6th. What happened? Simple. At 3000rpm, the engine is not putting out enough power to overcome the aerodynamic drag at 155mph. So it can accelerate no further.

I had some difficulty getting over this part, as I tend to think better spatially than in numbers. So I came up with an exaggerated thought to which I could associate more easily.

Assume that you are running a similar or same car above, with a badly designed spoiler that multiplies the drag drastically (OR with partially jammed brakes OR taking the car on a dirt track OR on a constant incline over a long distance), so that whenever you take your foot off the accelerator, the deceleration is too much. Now try taking the car to it's maximum possible speed. Just guess if you will ever be able to take her to 5th gear or even 4th. And it will definitely stall in higher gears. So on which gear top speed is achieved in this case? Definitely not the higher gears right?

Thanks Harbir.. This was very succinct. Had these questions in my mind & these explanations make it crystal clear. A very well deserved 5 stars.

@thoma - In these conditions the variable - in this case the drag on the vehicle changes, hence the top speed will not remain the same what you would get on a smooth road.

Holy pistons Batman, that was quite an explanation! clap:

(and to think it was my post on the AT thread that triggered it off!)

Harbir- can you explain in your trademark "Physics for Dummies" style why diesel engines generate more torque than petrols of the same displacement? The explanations I googled up (most leading to Team BHP) all spoke of bore and stroke and made my eyes glaze over.

What a brilliant thread Harbir! clap: I definitely recommend this one for a sticky in the Technical section. :thumbs up

Quote:

Originally Posted by Harbir (Post 2948258)
To understand this, think of a single cylinder engine. At its torque peak, it produces the greatest expansive force in the combustion chamber, thus drives the piston down with the most force and thus producing the greatest torque at the crankshaft. But at its power peak, while it torque output is reduced, it has more combustion events per second than at the torque peak. This means that at the power peak, the engine is liberating more energy per second than it does at the torque peak. What determines the top speed of a vehicle is not the peak force the engine can apply, but the maximum energy that it can produce per second.

Now here is another important point: What your body feels is the result of torque. Your body is able to perceive forces applied on it. It is able to perceive acceleration due to the sensitivity of the inner ear (your balancing mechanism). Both these are related to torque. The human body has no mechanisms for physically perceiving power. You perceive it only through mental reasoning.

I always had great difficulty in relating power and torque. Thank You Harbir. This is brilliant!


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