# Calculating wheel torque from engine torque

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1. May 6, 2016

### Jack Duncan

Im currently writing a code to find the optimum rpm points to shift gears to maximize acceleration. Thus far I've found the shift points and the rpm's after an upshift, meaning I have my bounds on my rpm vs torque curve. For simplicity's sake I'm assuming and increase of 1000 rpm to take 1 second. My is, after I take my engine torque and multiply it by the gear and final gear ratios, that torque is presumably applied at the axel, so how would I find the rotational acceleration of the wheel? (And would I need to divide the engine torque by two assuming the power is going to two wheels?)
Thanks in advance

2. May 7, 2016

### jack action

The engine power (which is the same as the wheel power + the drivetrain losses) will split between the 2 wheels, and if the the 2 wheels are at the same rpm, then the axle torque will split between the 2 wheels.

But that doesn't really matter as you will need to sum them up for the next step. Knowing the total wheel torque and tire radius, you can find the traction force applied to the ground $F_T$. This force will go against the resistance forces $\sum F_{R\ @\ v}$ (aerodynamic drag, rolling resistance, slope of the road) at the given speed $v$. Then:
$$F_T - \sum F_{R\ @\ v} = ma$$
Once you found the acceleration of the vehicle $a$, you can find the new speed of the vehicle after the one second you assumed to be elapsed. Find the new engine rpm and torque at that new speed ($v = rpm_{wheel}\times r_{wheel}$), rinse and repeat.

3. May 7, 2016

### Jack Duncan

Thanks so much, exactly what I was looking for.

4. May 7, 2016

### cjl

There's a much simpler way to do this. Optimum shift point is where the power the engine will be making in the new gear is the same as the power will be making in the old gear, so you'll be shifting after the power peak, but in a place where the new gear will be before the power peak. This will maximize the average horsepower, and thus the acceleration.

5. May 7, 2016

### Jack Duncan

Well that makes sense, although for now I'm working with torque data, if I did use power data, wouldnt I need power output for ever gear? Or would the shift point be the same in each gear?

6. May 7, 2016

### cjl

It would be different for each gear depending on the ratio spread. Also, if you have torque data, power data is easy to get - it's just proportional to torque*rpm.

7. May 7, 2016

### Jack Duncan

Okay, well if I multiply rpm*torque, how would I go about finding the actual power if I wanted to use that to calculate acceleration, given that, that product would only be in proportion to, not equal to power?

8. May 7, 2016

### Baluncore

You do not need to compute power to get acceleration. Torque is sufficient since it will give you the force on the road.
Traction force = Torque / Wheel radius.
The mass of the vehicle is known, and F = m*a; So acceleration = torque / ( wheel radius * mass of vehicle).

If you need to convert RPM to angular frequency in radians per second, multiply RPM by 60 seconds * 2Pi radians.
Angular frequency = 277 * RPM.

9. May 8, 2016

### Jack Duncan

Right that's what I was trying, but because the torque (and therefore acceleration) is non constant, do I have to evaluate the acceleration at each time step dt or can I somehow calculate a net acceleration with net torque and total time?

10. May 8, 2016

### Baluncore

The start and stop points in different gears will be at slightly different RPMs.
You will need to have a table or piecewise function of torque against RPM.
The form of that data representation will decide the method used to get acceleration and speed.
Start with discrete computation and get some results.
If you later use a polynomial for the relationship, you might integrate the acceleration more quickly.

11. May 8, 2016

### Jack Duncan

Okay that makes sense, I've found the rpm shift points and the rpm post shift, so I have my bounds, I also have the area under the curve between these rpm's for each gear (the graph is rpm vs torque) I'll do it out on paper to check if it works, but I'm hoping that since I'm assuming it takes 1 second to increase the rpm by 1000, if I take these integrals, I can just say that the area under the curve is equal to net torque applied over the given rpm (time) interval. Thanks so much for all of the help

12. May 8, 2016

### cjl

True, but using the power curve and looking for shift points such that power in the new gear is the same as power in the old gear is one of the easiest ways to find optimal shift points. You can definitely achieve the same thing just by looking at tractive force too though (and very nearly as easily).

13. May 8, 2016

### Baluncore

I quite agree. But when selecting high and low RPM points with equal power, the constant of proportionality can be allowed to cancel.
RPM * torque is sufficient to find those points, no matter what data units are available.

14. May 13, 2016

### Rx7man

I learned you should aim to keep the engine between peak HP (at the top end) and peak Torque (at the low end)... It's quite rare to ever go above the engine's peak RPM for max acceleration, and usually the torque is declining very quickly at that point, and you're better off with the 'penalty' of a higher gear ratio but more engine torque.

15. May 13, 2016

### OldYat47

Using power is futile (more on that later). Just generate a set of output torque curves for each gear and overlay them. Your shift points may be determined by two factors: First, you run out of RPM so you have to shift, and second, output torque in the current gear falls below output torque in the next gear.

Power is not related to acceleration. Here's an example problem as illustration: You have a 3,220 pound car (mass of 100 slugs) in a vacuum (zero air resistance, keeping things simple). You observe that the net horsepower at the rear wheels is 100 HP. What is the rate of acceleration?

Here's another. That same car is accelerating at a rate of 10 ft/sec^2. How much horsepower does it take to do that?

16. May 14, 2016

### billy_joule

Power is the rate at which work is done and is directly related to acceleration:

P=Fv
F=ma
so
P=mav
and so
a = P / mv

P and m are constants so acceleration goes to zero as velocity increases.
At 100km/hr it'll accelerate at 1.8 m/s2, at 50km/hr it'll accelerate at twice that etc.

17. May 14, 2016

### Rx7man

Regardless of the torque the engine is making, if it's making peak power, and you're effectively applying it to the ground at whatever speed you're traveling, that's about as good as it's going to get... if you're going 100 kph and putting 100 hp, shifting into the next gear (where you're only making 60 hp prehaps) isn't going to make you accelerate any faster.. On the other hand, Peak power occurs when torque drops precipitously and going above that isn't going to do you any good either

18. May 14, 2016

### OldYat47

Look at that equation, a = P/mv. Note that power is force X velocity. So the equation can be written a = fv/mv, which reduces to a = f/m. Lastly, note that for a given constant acceleration and mass power changes whenever velocity does. Acceleration cannot be calculated without converting from power to force. So you can plug in an infinite number of values for velocity and calculate the acceleration from that, but the resulting acceleration curve will be some (multiple or fraction, depending on the values) of the torque curve.

In the second example problem, knowing the acceleration and mass you can't determine the power. It varies constantly with velocity. You can develop a power curve from that information, which is what chassis dynamometers do. They record the rate of rotation of a massive cylinder vs. engine RPM over time. From that a horsepower curve is calculated.

Rx7man, it often is better to run past the peak power curve for optimum acceleration. Remember that shifting to a higher gear reduces torque multiplication. Sometimes it's better to use (less torque X lower gear) than (more torque X next higher gear). You can't "see" this instinctively from the power curve. That's why it's better to map out rear wheel torque curves for each gear and overlay them. Peak power is useful if you are trying to design gear ratios, for example. The goal for optimum performance gearing would be for the highest gear to provide the highest top speed (balance between drag forces and propelling forces) using work and velocity. This can be done strictly using torque, too.

19. May 14, 2016

### Rx7man

OldYat, Every engine will have different characteristics.. In all cases though, peak power will occur after peak torque, and peak torque usually starts to fall off pretty quickly after peak power (It's been falling for a while, otherwise peak power would be later)... I should perhaps have reworded my last post a little.. going MUCH past peak power will usually not be beneficial because of how quickly torque drops at that point.

20. May 14, 2016

### jack action

When the output torque in the current gear equals the output torque in the next gear, then the wheel power is the same in both cases. And since engine power is equivalent to the wheel power, you don't need to calculate all the torque transformations done in the gearbox to find the shifting points.
Here's another: You have a 3,220 pound. You observe that the net maximum engine horsepower is 100 HP. What is the rate of acceleration?

With a simple equation, you can estimate the ¼-mile ET to be 18.51 s @ 73.64 mph.

Even with a more complete calculator, for a RWD, you can estimate the ¼-mile to be 18.21 s @ 75 mph, the 0-60 mph to be 11.18 s and the top speed to be 115 mph.

But try with this one instead: You have a 3,220 pound. You observe that the net maximum engine torque is 200 lb.ft. What is the rate of acceleration?

You will never find any relationship between engine torque alone and car acceleration. But you can with engine power and car acceleration.

Power is the most fundamental value you should know to estimate the acceleration of a car.

When you take a car at a given velocity and increase the wheel torque, you are right, you increase the car acceleration. But by doing so, you also have increased the wheel power as well, and that is the important variable. This means that you could have kept the same engine rpm and increase the engine torque, but you could just as well decrease the engine torque and increase the engine rpm and it would have had the same effect, if both cases produced the same power output.

When @billy_joule says:
It is a very important concept to understand.

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