power and torque curve, and maximum acceleration
Discussion
Obviously the engine torque means nothing, it is the torque to the wheel that takes precedence. but suddenly if we neglect the constraints of bearing and air resistance, for the same gearbox speed (the 2nd) the highest force in g is found to be at the level of the maximum torque. In general, this is what it feels like when driving.
Murci.sv said:
Obviously the engine torque means nothing, it is the torque to the wheel that takes precedence. but suddenly if we neglect the constraints of bearing and air resistance, for the same gearbox speed (the 2nd) the highest force in g is found to be at the level of the maximum torque. In general, this is what it feels like when driving.
Yes, for the Golf GTI it will be at 20 mph.For clarity, rolling resistance is not bearing resistance, it's the friction between tyres and the road.
Modern tyres have a coefficient of rolling resistance of around 0.01.
The weight of the car (i.e. the mass multiplied by earth's gravity, g) bearing down on the tyre creates a friction force as the tyre rolls over the road.
The horizontal friction force is about 1% of the cars weight, expressed (correctly) as a force.
A car with a mass of 1000 kg has a weight of 9.81 kN, the rolling resistance force would be 98 N.
This friction force remains quasi-constant regardless of speed.
However, as described, power = force x velocity.
The rule of thumb for rolling resistance is 6 bhp per 100 mph per ton, i.e. just 1% of the power required to accelerate the car at 1 g.
The linearly increasing power with speed for rolling resistance can be seen in the BMW iX plot I posted.
Drag force is more complex and requires information about the car's CdA, it's drag coefficient multiplied by the frontal area.
A very low drag car will get below 0.5 for CdA, heffalumps are closer to 1.
A Tesla model 3 is 0.5 for example, an FFRR is just under 1.
I'm only using Tesla as examples because their data is well established, not for any other reason....
Drag also depends on the outside air temp, the cooler it is, the more dense the air, and a higher drag force results.
Given that drag force increases with the square of velocity, and thus the power increases with a cube law, the future of low energy consumption cars is very much related to improving aero drag, especially if we want efficient cars that can travel fast.
There isn't really a rule of thumb for calculating drag power, but these three reference points for a 15 deg C temperature are useful I guess:
9 bhp per unit of CdA at 50 mph
73 bhp per unit of CdA at 100 mph
583 bhp per unit of CdA at 200 mph
The cube law effect is very apparent.....
Even if you were travelling in a vacuum with zero loss tyres, there are other power losses in the car such as cooling, heating, ventilation, bearings, gearboxes, differentials, fans, pumps, etc. which will all reduce the torque at the driven wheels available for acceleration. I've ignored these however as they won't change the outcome as fas as max attack acceleration is concerned.
Travelling up or down an incline would also change things of course.
GT9 said:
You are conflating force as being 'the same as' power, it very much is not.
You need to go back to engineering first principles.
Power = force x velocity
What that means is that is the power required depends on what speed you are travelling at, generally, the force does not.
The only force acting on the car that has a speed dependency is drag force.
The force to accelerate the car at a given rate of acceleration is independent of speed, as is the force to overcome rolling resistance.
All of the torque values you have been referring to are at the engine, NOT at the wheels.
The multi-ratio gearboxes on ICE engine vehicles confuse the understanding of the fundamental principles of motion for two reasons.
Firstly, because they give access to the engine's full power output at low road speed, which is totally unnecessary, it's simply a function of having high ratios between the engine and the driven wheels to provide sufficient acceleration at low speed from an imperfect torque device, i.e. an engine. By imperfect, I mean that the engine cannot produce the same torque at all rpm.
Secondly, the torque figure that manufacturers and everyone quotes are the torque at the engine output. The gearbox then multiplies that torque by different amount, depending on what gear you are in.
The same torque in first gear provides a higher accelerative force than it will in second gear, but the engine runs out of revs to go any faster, so second gear must be selected.
Nobody, except me it seems, has every quoted or referred to driven wheel torques, yet that's the actual torque turning the driven wheels and providing the acceleration.
It's easier to understand all of this by taking the multi-ratio gearbox out of the equation, as is the case with a single speed powertrain, such as that in an EV.
The EV also has a perfect torque device driving it, i.e. an electric motor that can provide the same torque at all rpms.
I concede that statement is not entirely true, because most EVs have a motor that operates in two regimes, fixed torque (up to say 60 mph) and the fixed power (over 60 mph). The reason for this is related to engineering optimisation to do with battery voltages, and typical drive cycles, amongst other thing , so rather than confuse the issue, let's just examine the fixed torque regime.
Here is a plot for a typical large EV.
It shows the individual power demands required to overcome drag, rolling resistance and the power required to accelerate the car at different g values. The acceleration power needs to be read of the righthand vertical axis, the permanent consumers of the left hand vertical axis.
None of the power demands are high at low road speeds, they all start from zero.
It is totally unnecessary to have high power output at low road speed.
Here is proof, this is the power output plot of one of the fastest accelerating production cars ever, one that is also a very heavy car by most measures. It also has no multi-ratio gearbox.
The reason it can accelerate so quickly to 60 mph is because it is supplying a fixed value of driven wheel torque at all road speeds up to 60 mph. This results in the power output rising linearly from zero to full power over the 0-60 run.
The torque at the driven wheels is what is providing the ability for the speed to increase, the power is simply a measure of how fast you are travelling.
Torque matters, it is what is providing the instantaneous force.
Power and the resulting energy consumption that is defined by power multiplied by time, is essentially a measure of how long this force can be sustained for, either to keep the car moving or to accelerate it.
To accelerate, the car must be able to increase the instantaneous torque at the driven wheels, otherwise nothing will happen.
If you want to understand the role the gearbox plays, here is a plot showing the comparison between two similar cars, a Golf GTI and a Cupra Born. Despite the fact that the EV cannot muster full power until 60 mph, it accelerates at the around same rate as the Golf which gives access to full power at 30 mph. Look at the driven wheel torque curves for the two cars. The ICE drivetrain is trying to mimic the fixed driven wheel torque of the EV, and it does so by supplying excessive torque to make up for the sections where it has less torque. The gear change in the Golf is at 35 mph.
The data supplied above for the BMW, the Golf and the Cupra is my own, obviously the Tesla plot comes straight from their website.
Note that EVs do have a fixed final drive ratio between the motor and the wheels, so even the quoted torque values for these you always read about, e.g. 350 Nm only tell you what the motor is producing, they DO NOT tell you what is being supple to the driven wheels.
You can only determine this if you know the final drive ratio, in the case of the Born, its about 8:1, which is why the driven wheel torque is just over 3000 Nm.
A small number of EVs also have second gear ratio to optimise high speed driving, the Taycan for example.
And finally, just to complicate things a bit more, converting driven wheel torque into the actual accelerative force or force to overcome drag/friction depends on the working outer radius of the tyres, which can alter slightly with speed/wear, as well as obviously differing between cars.
All of which is why people try to define acceleration by power-to-weight ratios, as it takes the engine's instantaneous torque value, the gearing being used at a given speed and tyre diameters out of the discussion.
The key thing to remember though is that the power required to accelerate is not a fixed value, it depends entirely on speed.
Hence the rule of thumb I posted at the beginning of the thread:
6 bhp per ton per g per mph.
You're quoting gear ratios but the initial question said:You need to go back to engineering first principles.
Power = force x velocity
What that means is that is the power required depends on what speed you are travelling at, generally, the force does not.
The only force acting on the car that has a speed dependency is drag force.
The force to accelerate the car at a given rate of acceleration is independent of speed, as is the force to overcome rolling resistance.
All of the torque values you have been referring to are at the engine, NOT at the wheels.
The multi-ratio gearboxes on ICE engine vehicles confuse the understanding of the fundamental principles of motion for two reasons.
Firstly, because they give access to the engine's full power output at low road speed, which is totally unnecessary, it's simply a function of having high ratios between the engine and the driven wheels to provide sufficient acceleration at low speed from an imperfect torque device, i.e. an engine. By imperfect, I mean that the engine cannot produce the same torque at all rpm.
Secondly, the torque figure that manufacturers and everyone quotes are the torque at the engine output. The gearbox then multiplies that torque by different amount, depending on what gear you are in.
The same torque in first gear provides a higher accelerative force than it will in second gear, but the engine runs out of revs to go any faster, so second gear must be selected.
Nobody, except me it seems, has every quoted or referred to driven wheel torques, yet that's the actual torque turning the driven wheels and providing the acceleration.
It's easier to understand all of this by taking the multi-ratio gearbox out of the equation, as is the case with a single speed powertrain, such as that in an EV.
The EV also has a perfect torque device driving it, i.e. an electric motor that can provide the same torque at all rpms.
I concede that statement is not entirely true, because most EVs have a motor that operates in two regimes, fixed torque (up to say 60 mph) and the fixed power (over 60 mph). The reason for this is related to engineering optimisation to do with battery voltages, and typical drive cycles, amongst other thing , so rather than confuse the issue, let's just examine the fixed torque regime.
Here is a plot for a typical large EV.
It shows the individual power demands required to overcome drag, rolling resistance and the power required to accelerate the car at different g values. The acceleration power needs to be read of the righthand vertical axis, the permanent consumers of the left hand vertical axis.
None of the power demands are high at low road speeds, they all start from zero.
It is totally unnecessary to have high power output at low road speed.
Here is proof, this is the power output plot of one of the fastest accelerating production cars ever, one that is also a very heavy car by most measures. It also has no multi-ratio gearbox.
The reason it can accelerate so quickly to 60 mph is because it is supplying a fixed value of driven wheel torque at all road speeds up to 60 mph. This results in the power output rising linearly from zero to full power over the 0-60 run.
The torque at the driven wheels is what is providing the ability for the speed to increase, the power is simply a measure of how fast you are travelling.
Torque matters, it is what is providing the instantaneous force.
Power and the resulting energy consumption that is defined by power multiplied by time, is essentially a measure of how long this force can be sustained for, either to keep the car moving or to accelerate it.
To accelerate, the car must be able to increase the instantaneous torque at the driven wheels, otherwise nothing will happen.
If you want to understand the role the gearbox plays, here is a plot showing the comparison between two similar cars, a Golf GTI and a Cupra Born. Despite the fact that the EV cannot muster full power until 60 mph, it accelerates at the around same rate as the Golf which gives access to full power at 30 mph. Look at the driven wheel torque curves for the two cars. The ICE drivetrain is trying to mimic the fixed driven wheel torque of the EV, and it does so by supplying excessive torque to make up for the sections where it has less torque. The gear change in the Golf is at 35 mph.
The data supplied above for the BMW, the Golf and the Cupra is my own, obviously the Tesla plot comes straight from their website.
Note that EVs do have a fixed final drive ratio between the motor and the wheels, so even the quoted torque values for these you always read about, e.g. 350 Nm only tell you what the motor is producing, they DO NOT tell you what is being supple to the driven wheels.
You can only determine this if you know the final drive ratio, in the case of the Born, its about 8:1, which is why the driven wheel torque is just over 3000 Nm.
A small number of EVs also have second gear ratio to optimise high speed driving, the Taycan for example.
And finally, just to complicate things a bit more, converting driven wheel torque into the actual accelerative force or force to overcome drag/friction depends on the working outer radius of the tyres, which can alter slightly with speed/wear, as well as obviously differing between cars.
All of which is why people try to define acceleration by power-to-weight ratios, as it takes the engine's instantaneous torque value, the gearing being used at a given speed and tyre diameters out of the discussion.
The key thing to remember though is that the power required to accelerate is not a fixed value, it depends entirely on speed.
Hence the rule of thumb I posted at the beginning of the thread:
6 bhp per ton per g per mph.
Edited by GT9 on Tuesday 12th December 17:11
for the same gearbox ratio, at what moment is the acceleration felt (G force) the strongest? at maximum torque rpm or at maximum power rpm?
I assume this to mean essentially a 1speed manual gearbox, at which rpm will you accelerate fastest, at the point of peak hp or point of peak torque from idle to red line.
GravelBen said:
I *think* that on a simplified level it will be felt the most where the power curve is steepest, but the actual maximum acceleration will be at peak power.
But I could be wrong.
I think you are absolutely correct. A prize also for the best answer you don't need an Engineering Degree to understand.But I could be wrong.
popeyewhite said:
GravelBen said:
I *think* that on a simplified level it will be felt the most where the power curve is steepest, but the actual maximum acceleration will be at peak power.
But I could be wrong.
I think you are absolutely correct. A prize also for the best answer you don't need an Engineering Degree to understand.But I could be wrong.
popeyewhite said:
GravelBen said:
I *think* that on a simplified level it will be felt the most where the power curve is steepest, but the actual maximum acceleration will be at peak power.
But I could be wrong.
I think you are absolutely correct. A prize also for the best answer you don't need an Engineering Degree to understand.But I could be wrong.
And yet....
Ben is correct in his first statement and wrong in his second.
Nobody is ever going to get their head around this until they understand that the relationship between power and rate of acceleration depends on speed.
Simply saying, it's where power is highest ignores that, and is why it is wrong.
Look at the Golf power and wheel torque curve I posted, the blue curves.
The torque curve peaks where the power curve is steepest, not where it is highest.
This is where the maximum rate of acceleration occurs.
It doesn't occur where the peak power is reached because at that point you are now travelling at a higher speed and therefore require more power to achieve the same acceleration.
The reason you require more power to achieve the same acceleration is because kinetic energy = 0.5 mv^2.
As the speed increases the amount of energy being added is rapidly increasing such that for fixed g acceleration, the power required increases proportionally with speed.
I'll put it another way using the same curves, in first gear.
At 20 mph the Golf produces 110 kW.
At 35 mph the Golf produces 160 kW.
110 divided by 20 is 5.5.
160 divided by 35 is 4.6.
5.5 is bigger than 4.6.
110 kW at 20 mph will accelerate the car faster than 160 kW at 35 mph.
That's the crux of it.
The maximum acceleration occurs where the power delivered to the road divided by the road speed is highest, assuming loss of traction has not occurred.
For the EV, the power divided by the speed is constant to 60 mph, so it accelerates completely linearly and there is no peak felt.
That is achieved by delivering a constant wheel torque throughout the run....
Edited by GT9 on Tuesday 12th December 23:57
popeyewhite said:
GT9 said:
Fair enough, you think I'm overcomplicating it.
I just gave up halfway through, that's all. Not only your post BTW, and sorry if you find my response a bit rude.You can basically ignore everything I said if you just use the rule of thumb I started out with.
1g acceleration requires 6 bhp per ton, per mph.
The highest acceleration occurs where the ratio of power available to vehicle speed is the highest.
That occurs when the engine hits peak torque in first gear.
If the car can sustain traction at that point...
popeyewhite said:
GravelBen said:
I *think* that on a simplified level it will be felt the most where the power curve is steepest, but the actual maximum acceleration will be at peak power.
But I could be wrong.
I think you are absolutely correct. A prize also for the best answer you don't need an Engineering Degree to understand.But I could be wrong.
trevalvole said:
popeyewhite said:
GravelBen said:
I *think* that on a simplified level it will be felt the most where the power curve is steepest, but the actual maximum acceleration will be at peak power.
But I could be wrong.
I think you are absolutely correct. A prize also for the best answer you don't need an Engineering Degree to understand.But I could be wrong.
torque increases from 600nm+ to 700nm from 3500rpm to 4500rpm. At this time the power curve is steepest at around 1000 rpm. if we stick to the previous statement of the maximum acceleration in g will be at the maximum torque 6800rpm, but the most important rate of change takes place between 3500-4500. when will the acceleration (feel) the fastest?
Murci.sv said:
trevalvole said:
popeyewhite said:
GravelBen said:
I *think* that on a simplified level it will be felt the most where the power curve is steepest, but the actual maximum acceleration will be at peak power.
But I could be wrong.
I think you are absolutely correct. A prize also for the best answer you don't need an Engineering Degree to understand.But I could be wrong.
torque increases from 600nm+ to 700nm from 3500rpm to 4500rpm. At this time the power curve is steepest at around 1000 rpm. if we stick to the previous statement of the maximum acceleration in g will be at the maximum torque 6800rpm, but the most important rate of change takes place between 3500-4500. when will the acceleration (feel) the fastest?
Edited by trevalvole on Wednesday 13th December 15:36
Murci.sv said:
when is the real acceleration of the vehicle the strongest?
When power at the road divided by road speed minus resistance is greatest. How that relates to gearing, revs, torque and engine power will depend entirely on the vehicle and there is no one simple answer,Murci.sv said:
at what moment is the acceleration felt (G force) the strongest?
If you're talking about perception rather than actual acceleration then that's highly subjective and there is no simple answer. It muight be when the rate of increase of acceleration is greatest, but it might not be. If you believe it is then you can find the answer by plotting acceleration against time and finding where the slope is greatest. Again, the relationship between this point and gearing, revs, torque and engine power will depend entirely on the vehicle and there is no one simple answer.Why do you care about any of this? I mean, why do you care whether the hypothetical thrill of driving some arbitrary vehicle is best at 4000 rpm or 5000 rpm?
Murci.sv said:
the green between 3500_4500 rpm is the steepest where the torque goes from 600+ to ~720nm over 1000 rpm. more steep than the blue line 
Using the blue torque curve:Whilst the rate of change of acceleration is highest at 5000 rpm, the absolute value of acceleration, i.e. the g number, is higher at just under 7000 rpm.
In other words, the force pushing you into the seat will be increasing more quickly at 5000 rpm but the highest absolute force will be felt approaching 7000 rpm.
Referring to the steepest slope of the power curve as the point of 'highest acceleration' is misleading (or wrong....) if the torque curve is lumpy like the one you posted.
The maximum force in any given gear is felt at peak torque, unless there has been some loss of traction that results in some of that torque not making it to the road.
Edited by GT9 on Wednesday 13th December 18:12
It is not rocket science. All someone needs to do is post their HP and torque curves and the results of data logging acceleration in g's. You will see that the acceleration curve closely matches the torque curve.
Now to get your gears turning. If you have a CVT will you accelerate quick locking the RPM at peak torque or peak HP?
Stan
Now to get your gears turning. If you have a CVT will you accelerate quick locking the RPM at peak torque or peak HP?
Stan
Stan Weiss said:
You will see that the acceleration curve closely matches the torque curve.
They will very much NOT match if there are any gear changes. I know you know that, but your simplification is glossing over that.Once the relationship between torque, power, speed, drag and acceleration are understood, the answer is obvious. But without that understanding, these simplisitic explainations will be either misleading or flat out wrong.
Stan Weiss said:
It is not rocket science. All someone needs to do is post their HP and torque curves and the results of data logging acceleration in g's. You will see that the acceleration curve closely matches the torque curve.
Now to get your gears turning. If you have a CVT will you accelerate quick locking the RPM at peak torque or peak HP?
Stan
The acceleration curve will match the torque curve if traction is available to deliver it to the road.Now to get your gears turning. If you have a CVT will you accelerate quick locking the RPM at peak torque or peak HP?
Stan
However, the absolute value of acceleration depends entirely what gear you are in.
It should go without saying that the same engine torque in first gear will accelerate the car faster than that same engine torque in 4th gear.
Reason: for the same engine torque, the driven wheel torque is much higher in first than fourth.
For the CVT question, the answer is, it depends.
In theory, you would keep the engine at maximum power to maximise the ratio of power available to road speed.
In practice though this will most likely break traction at low road speed and you would probably accelerate faster by limiting the power available until it is needed at a higher road speed.
Once again, I will quote the rule of thumb, 6bhp per ton per mph per g.
This is an unbreakable and non-negotiable law of physics.
It come back to the fact that acceleration is the addition of kinetic energy, and that the instantaneous kinetic energy the car carries is a direct function of the speed squared.
Adding kinetic energy at 100 mph requires twice as much effort as it does at 50 mph.
Once you understand this unbreakable rule, understanding the role torque and power play is far easier.
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