Calculate acceleration from power/torque graphs

[gen x]

Let say we want find out acceleration of two cars with same max power, comparing only power/torque graphs, both cars are 100% identical, same mass, aero drag,etc so difference is only in rpm and power/torque curve. Cars dont use CVT gearbox(allows to keep engine at max power) they both use gears, and both have optimal gear ratios to match optimal rpm window("power band") for ther power/torque curve for best acceleration.

So if I want find acceleration for lets say 10-14k rpm window, what I need to do? Find area under power curve in that limts 10-14k rpms(definite integral), does car that have more area under power curve accelerate faster, how to find optimal rpm window(gear ratios for my power/torque curve)?

 

[Baluncore]

Welcome to PF.

There is something wrong with that torque against RPM graph.
Power is the product of torque and RPM, so the power and torque lines should not cross as shown, they should be proportional.

 

[gen x]

Welcome to PF.

There is something wrong with that torque against RPM graph.
Power is the product of torque and RPM, so the power and torque lines should not cross as shown, they should be proportional.

Thanks.

I choose random photo from net.
We can look at this, just assume both have same max power.
Is there any advantge for engine that revs higher?

NA engine

turbo engine

 

[Baluncore]

You have a torque to RPM graph.
For each RPM being considered, multiply the torque by the gear ratio and by the differential ratio.
Knowing the radius of the wheel, convert that torque to a linear force, F, on the road at the tire contact patch.
You know the mass of the vehicle, m, and that F = m·a ;
Acceleration, a = F / m.

 

[gen x]

You have a torque to RPM graph.
For each RPM being considered, multiply the torque by the gear ratio and by the differential ratio.
Knowing the radius of the wheel, convert that torque to a linear force, F, on the road at the tire contact patch.
You know the mass of the vehicle, m, and that F = m·a ;
Acceleration, a = F / m.

Yes that is instatenuos acceleration for that specific rpm, but power and torque change in my operating rpm window, so called "power band".

There is two goal of my question.

1. Learn how to comparing acceleration/"performance" of two power/torque curves at their optimal rpm window.
2. Find does high revs engine have any advantage, two engines with same peak power, one revs 20 000rpm and one 6000rpm, what engine can produce higher acceleration or they will 100% the same? (excluding CVT)
Is key here width or power band, so engine can operate longer at higher average power?

 

[Baluncore]

1. Define "optimal RPM window".
2. High revs has an advantage if they have the same torque.
If they have the same power, they will have the same acceleration, with different gearing, but the faster engine will have a proportionally lower torque, because;
Power = torque * RPM.

 

[jbriggs444]

There is something wrong with that torque against RPM graph.
Power is the product of torque and RPM, so the power and torque lines should not cross as shown, they should be proportional.

The power and torque lines should cross at most once (discounting the edge cases end where both torque and power are zero). The choice of units and scale will mathematically dictate the rotation rate where the crossover will be depicted. In the graph as shown, I eyeball the crossing point at 5200 or 5300 RPM.

If we double that and look for 10400 to 10600 RPM, we should find that power is twice torque. Which is what we do see.

If we halve that and look for 2600 to 2650 RPM, we should find that power is half torque. Which is what we do see.

Let us attempt the calculation for the mathematically predicted crossover point using the units and scale from the graph. The graph equates 1 horsepower with 1 pound foot of torque.

1 horsepower is 550 pounds (force) times one foot per second. To get one horsepower from 1 pound foot of torque, we would need that torque to spin a load through 550 radians of rotation per second. That would be $\frac{550}{2 \pi} \approx 87$ rotations per second. Multiply by 60 seconds per minute and we get 5252 RPM.

 

[gen x]

1. Define "optimal RPM window".

Optimal rpm window is "power band" or "rpm power window" that engine use for max acceleration.

you can see this engine operate in rpm power window from 6500-9000rpm, as he shift in higher gears, rpm window become narrower.









@jbriggs444
If we have two engines with same peak power, but one revs 20 000rpm and one revs 4000rpm, does one of them can deliver better acceleration due to wider power band, using different gear ratios or both will always get same acceleration?

Engine that revs higher, can use shorter gears(bigger gear ratios number), which multiply torque at the wheels more effectively, so it has wider useful rpm range?
Goal is stay near peak power more often after every shift.

 

[Baluncore]

... you can see this engine operate in rpm power window from 6500-9000rpm, as he shift in higher gears, rpm window become narrower.

That is due to the supported gear ratio selection, decided by the manufacturer.

If you plot engine power against vehicle road speed, for each available gear ratio, you should change gear, when the power curves of adjacent gears cross, NOT when adjacent torque curves cross.

If you have a narrow power band, you will need more gears to cover the road speed range of the vehicle, or there will be a shortage of power during gear changes, which will make it difficult for the driver to accelerate the vehicle, when changing gears.

 

[gen x]

That is due to the supported gear ratio selection, decided by the manufacturer.

Yes manfacturer do that based on some calculation, they do this to get max acceleration.
But if they add few more gears and make them shorter, then can engine be more often close to peak power rpm. Shifting time today can be only 40miliseconds, so shifting is not a problem.

Has high revs engines wider power band compare to slow revs engines?

If you plot engine power against vehicle road speed, for each available gear ratio, you should change gear, when the power curves of adjacent gears cross, NOT when adjacent torque curves cross.

Ussualy that is graph torque at wheel vs car speed

 

[Baluncore]

But if they add few more gears and make them shorter, then can engine be more often close to peak power rpm.

Yes.

The valleys in the power, that must be crossed when stepping up, or down, a gear, become deeper if there are fewer gear ratios available.

More power, more of the time, requires more gear ratios, and quicker changes. That has led to dual clutch gearboxes, with two counter-shafts, and twice as many available gear ratios.

 

[Baluncore]

Usually that is graph torque at wheel vs car speed

Which is proportional to acceleration, but that is quite misleading, because it follows a reciprocal function on the graph, not a series of similar height humps.

If you are wanting to identify the optimum speed for gear changes, it is neater to plot engine power against road speed. Maximum torque at the road wheel is maximum acceleration, but it is also maximum power.

Consider using a log scale for the road speed, with a linear scale for engine power, you will then have only one shape of curve, that can be moved to overlap with copies of itself, to optimise gear ratio selection.

 

[gen x]

Maximum torque at the road wheel is maximum acceleration, but it is also maximum power.

In one gear (ex. in 3rd gear), max torque on wheel(max thrust=max acceleration) is not at engine max power, it is on engine max torque.
The highest instantaneous acceleration will be in 1st gear at max engine torque.

If you are wanting to identify the optimum speed for gear changes, it is neater to plot engine power against road speed.

I can't find these graphs.

 

[Baluncore]

In one gear (ex. in 3rd gear), max torque on wheel(max thrust=max acceleration) is not at engine max power, it is on engine max torque.

At the high RPM end of 3rd gear, where you would change to the low RPM end of 4th gear, the torque applied by the wheel to the road would be the same. Since the speed of the vehicle is the same in each case, the power will also be the same, because power is the product of torque and RPM.

I can't find these graphs.

Then you should draw them.

 

[gen x]

At the high RPM end of 3rd gear, where you would change to the low RPM end of 4th gear, the torque applied by the wheel to the road would be the same. Since the speed of the vehicle is the same in each case, the power will also be the same, because power is the product of torque and RPM.

Torque at wheel is smaller in 4th gear then in 3rd gear, acceleration is smaller in 4th gear compare to 3rd gear.

 

[Baluncore]

Torque at wheel is smaller in 4th gear then in 3rd gear, acceleration is smaller in 4th gear compare to 3rd gear.

But not at the instant before and after an optimum gear change, when road speed, road wheel torque, acceleration, and power are the same.

Look at the graph in post #10. That is a reciprocal curve. It is the reciprocal behaviour of wheel torque against road speed, that makes it a poor indicator when compared to graphing power.

 

[gen x]

But not at the instant before and after an optimum gear change, when road speed, road wheel torque, acceleration, and power are the same.

Look at the graph in post #10. That is a reciprocal curve. It is the reciprocal behaviour of wheel torque against road speed, that makes it a poor indicator when compared to graphing power.

You mean point where cruves from lower and higher gear intercept? yes here torque at wheel are the same

 

[Baluncore]

This is the force applied by the driving wheels to the road, plotted against road speed.
Because F = m · a ; it also represents vehicle acceleration.

This is the power that is accelerating the vehicle, plotted against road speed. Notice the depth of the valley that must be crossed when accelerating and changing through the gears.

Here, the power is plotted against Log(road speed), showing how each curve is now the same shape, only positioned by the variation in gear ratio steps.