# Calculating wheel torque from engine torque

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

## Answers and Replies

jack action
Science Advisor
Gold Member
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.

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.
Thanks so much, exactly what I was looking for.

cjl
Science Advisor
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.

• CWatters
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.
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?

cjl
Science Advisor
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.

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.
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?

Baluncore
Science Advisor
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.

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.
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?

Baluncore
Science Advisor
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.

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.
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

cjl
Science Advisor
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.

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).

Baluncore
Science Advisor
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.
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.

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.
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.

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?

billy_joule
Science Advisor
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?
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.

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

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.

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.

jack action
Science Advisor
Gold Member
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.
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.
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?
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:
Power is the rate at which work is done
It is a very important concept to understand.

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

You won't with power, you will with torque. Rear axle torque is engine torque times reduction ratio. Knowing the radius of the drive wheels gives you drive wheel torque. Now you have force and mass so you can directly calculate acceleration.

You can estimate 1/4 mile times, but that's just an estimate. And restating, you must convert power to force in order to calculate acceleration. I'd be interested in any equations which get from power and mass to acceleration directly. Please post them if you have one or more.

Lastly, power is an "invented" concept, not a fundamental property. James Watt came up with the concept in order to compare water lift capacities of different types of pumps. You can do any and all engineering tasks without ever using power.

You won't with power, you will with torque. Rear axle torque is engine torque times reduction ratio. Knowing the radius of the drive wheels gives you drive wheel torque. Now you have force and mass so you can directly calculate acceleration.

You can estimate 1/4 mile times, but that's just an estimate. And restating, you must convert power to force in order to calculate acceleration. I'd be interested in any equations which get from power and mass to acceleration directly. Please post them if you have one or more.

Lastly, power is an "invented" concept, not a fundamental property. James Watt came up with the concept in order to compare water lift capacities of different types of pumps. You can do any and all engineering tasks without ever using power.
Really? Pretty bold statement... Seems to me that any equation that has time involved requires power to be known...
Power and force are *related* but not the same thing.

Baluncore
Science Advisor
Lastly, power is an "invented" concept, not a fundamental property.
Power is the rate of flow of energy. It is the number of joules of energy being converted per second.
Seems to me that any equation that has time involved requires power to be known...
Power and force are *related* but not the same thing.
Power or the rate work is done, or energy is converted, are related by force multiplied by the distance moved.

• Rx7man
jack action
Science Advisor
Gold Member
You won't with power, you will with torque. Rear axle torque is engine torque times reduction ratio. Knowing the radius of the drive wheels gives you drive wheel torque. Now you have force and mass so you can directly calculate acceleration.

You can estimate 1/4 mile times, but that's just an estimate. And restating, you must convert power to force in order to calculate acceleration. I'd be interested in any equations which get from power and mass to acceleration directly. Please post them if you have one or more.

Lastly, power is an "invented" concept, not a fundamental property. James Watt came up with the concept in order to compare water lift capacities of different types of pumps. You can do any and all engineering tasks without ever using power.

Let's take a vehicle powered with an engine (any type) that is transmitting its power through the wheels. What is the maximum force transmitted by the wheels, assuming no friction limit?

First, what do we know about this vehicle? We know it has an engine that produces power and that power cannot be unlimited: It has a maximum power (All engine/motor are defined by their power). We also know that the velocity of the car will change. Now, thanks to the concept of power (Thank you, Mr. Watt!), we can find the maximum force the engine can produce at the wheels:
$$F_{max} = \frac{P_{max}}{v}$$
Which gives in graph form: As you can see, we now know the maximum tractive force the car can produce at any speed, no matter the radius of the wheels, the gear ratios (if any), the engine torque or RPM, even the engine's type (piston, turbine, electric motor, etc.).

As you already mentioned, this available force can be translated into acceleration, if you know the vehicle mass (which is also a fundamental value of your vehicle).

The only thing left to the vehicle designer is to select the appropriate method to make sure that the maximum power will be available at any speed (For example, using a gearbox with a piston engine like in the figure below. Note that - even though each curve correspond to the torque curve of the engine - the «Constant power» curve meets the maximum power in each gear). From that simple concept of power, I can determine what type of engine will be needed to achieve the acceleration curve I desired, and the fundamental value of the engine will be its power.

So I can choose a diesel engine, do a first draft, find out it pollutes too much, and then decide to change to four electric motors. If I expect the same performance, what will be the common point between the diesel engine and the four electric motors? The sum of the power produced by the four electric motors will be equal to the power produced by the diesel engine.

As for the equations to estimate the ¼-mile, it is an estimate based on a real physics base, not just a statistical curve fitting. The energy gain by the vehicle from 0 to $v$ must be equal to the one delivered by the engine at constant power $P$ during the time $t$ or:
$$\frac{1}{2}mv^2 = \int_0^t Pdt = Pt$$
or (equation 1):
$$v = \left(\frac{2Pt}{m}\right)^{\frac{1}{2}}$$
Also, the distance $d$ traveled by the car during time $t$ is easily found by integration:
$$d = \int_0^t vdt = \int_0^t \left(\frac{2Pt}{m}\right)^{\frac{1}{2}}dt = \frac{2}{3}\left(\frac{2P}{m}\right)^{\frac{1}{2}}t^{\frac{3}{2}}$$
or (equation 2):
$$t=\left(\frac{3}{2}\left(\frac{m}{2P}\right)^{\frac{1}{2}}d\right)^{\frac{2}{3}}=\left(\frac{3}{2}d\right)^{\frac{2}{3}}\left(\frac{m}{2P}\right)^{\frac{1}{3}}$$
Since we know that for a ¼-mile, $d$ = 402.336 m (SI unit), then:
$$t=56.68\left(\frac{m}{P}\right)^{\frac{1}{3}}$$
or, converting from SI unit to hp and lb:
$$t_{\lbrack s]}=4.802\left(\frac{m_{\lbrack lb]}}{P_{\lbrack hp]}}\right)^{\frac{1}{3}}$$
And combining equation 2 with equation 1:
$$v = \left(3d\frac{P}{m}\right)^{\frac{1}{3}}$$
Since we know that for a ¼-mile, $d$ = 402.336 m (SI unit), then:
$$v = 10.65\left(\frac{P}{m}\right)^{\frac{1}{3}}$$
Converting from SI unit to mph, hp and lb, you get:
$$v_{\lbrack mph]} = 281.2\left(\frac{P_{\lbrack hp]}}{m_{\lbrack lb]}}\right)^{\frac{1}{3}}$$

The only thing that vary slightly from the actual equations used across the web are the constants, such that it takes into account some other variables of minor importance. (ref.: http://stealth316.com/2-calc-hp-et-mph.htm)

• Dale, cjl and billy_joule
The equation you posted says force = (force X velocity) / velocity. But does maximum force (torque in this case) occur at maximum power? Look at some torque vs. power curves for internal combustion engines. Peak torque occurs somewhat before peak power, not at the same RPM. Remember that power increases linearly with speed. At constant torque, or if torque falls off in a less than 1:1 ratio with RPM the power will continue to rise. If torque falls linearly with RPM the value for power remains the same.

And all those shift point curves could be generated directly from torque curves and overall reduction ratios.

I should add that the last part of your post assumes constant power. Assuming you had a power source that supplied a constant power, then the force falls off linearly with velocity so acceleration decreases throughout the 1/4 mile.