Calculating wheel torque from engine torque

  • #1
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
 

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  • #2
jack action
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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 [itex]F_T[/itex]. This force will go against the resistance forces [itex]\sum F_{R\ @\ v}[/itex] (aerodynamic drag, rolling resistance, slope of the road) at the given speed [itex]v[/itex]. Then:
[tex]F_T - \sum F_{R\ @\ v} = ma[/tex]
Once you found the acceleration of the vehicle [itex]a[/itex], 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 ([itex]v = rpm_{wheel}\times r_{wheel}[/itex]), rinse and repeat.
 
  • #3
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 [itex]F_T[/itex]. This force will go against the resistance forces [itex]\sum F_{R\ @\ v}[/itex] (aerodynamic drag, rolling resistance, slope of the road) at the given speed [itex]v[/itex]. Then:
[tex]F_T - \sum F_{R\ @\ v} = ma[/tex]
Once you found the acceleration of the vehicle [itex]a[/itex], 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 ([itex]v = rpm_{wheel}\times r_{wheel}[/itex]), rinse and repeat.
Thanks so much, exactly what I was looking for.
 
  • #4
cjl
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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
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?
 
  • #6
cjl
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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
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?
 
  • #8
Baluncore
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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
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?
 
  • #10
Baluncore
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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
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
 
  • #12
cjl
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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).
 
  • #13
Baluncore
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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.
 
  • #14
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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.
 
  • #15
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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
billy_joule
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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.
 
  • #17
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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
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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
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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
jack action
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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.
 
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  • #21
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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.
 
  • #22
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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.
 
  • #23
Baluncore
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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.
 
  • #24
jack action
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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:
[tex]F_{max} = \frac{P_{max}}{v}[/tex]
Which gives in graph form:

fmax-vs-v.jpg


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).

image003.jpg


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 [itex]v[/itex] must be equal to the one delivered by the engine at constant power [itex]P[/itex] during the time [itex]t[/itex] or:
[tex]\frac{1}{2}mv^2 = \int_0^t Pdt = Pt [/tex]
or (equation 1):
[tex]v = \left(\frac{2Pt}{m}\right)^{\frac{1}{2}}[/tex]
Also, the distance [itex]d[/itex] traveled by the car during time [itex]t[/itex] is easily found by integration:
[tex]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}}[/tex]
or (equation 2):
[tex]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}}[/tex]
Since we know that for a ¼-mile, [itex]d[/itex] = 402.336 m (SI unit), then:
[tex]t=56.68\left(\frac{m}{P}\right)^{\frac{1}{3}}[/tex]
or, converting from SI unit to hp and lb:
[tex]t_{\lbrack s]}=4.802\left(\frac{m_{\lbrack lb]}}{P_{\lbrack hp]}}\right)^{\frac{1}{3}}[/tex]
And combining equation 2 with equation 1:
[tex]v = \left(3d\frac{P}{m}\right)^{\frac{1}{3}}[/tex]
Since we know that for a ¼-mile, [itex]d[/itex] = 402.336 m (SI unit), then:
[tex]v = 10.65\left(\frac{P}{m}\right)^{\frac{1}{3}}[/tex]
Converting from SI unit to mph, hp and lb, you get:
[tex]v_{\lbrack mph]} = 281.2\left(\frac{P_{\lbrack hp]}}{m_{\lbrack lb]}}\right)^{\frac{1}{3}}[/tex]

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)
 
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  • #25
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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.
 
  • #26
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In reply to the two earlier posts, since power is force X velocity any calculation that involves power can be done using force X velocity. The reverse is not true. There are many cases where you cannot substitute power for force X velocity. The example problem of calculating acceleration of a mass knowing the mass and the power is a good example. To solve that you have to know the velocity, converting power to force.
 
  • #27
jack action
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But does maximum force (torque in this case) occur at maximum power?
For a given car speed, yes, maximum tractive force always happens at maximum power, that is what [itex]F_{max} = \frac{P_{max}}{v}[/itex] means.
I should add that the last part of your post assumes constant power.
Of course, I assume I'm at maximum power, all the time (like with a CVT for example).
Assuming you had a power source that supplied a constant power,
That is the purpose of a transmission in a race car.
then the force falls off linearly with velocity so acceleration decreases throughout the 1/4 mile.
That is why I put the graphs, I wanted you to see that.

You should re-read my post with an open mind, because you are just repeating my arguments.
 
  • #28
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For a given car speed, yes, maximum tractive force always happens at maximum power, that is what Fmax=PmaxvF_{max} = \frac{P_{max}}{v} means.
and
Of course, I assume I'm at maximum power, all the time (like with a CVT for example).

In the first quote, that is not true. As I showed, torque (and so force) can decline while power increases. So maximum force may not (and most probably doesn't) coincide with Pmax.

In the second quote, that also is not generally true. Similarly, it depends on the shape of the torque curve. For example, suppose you have an engine that generates 250 units torque at 7,000 RPM and 233.3 units of torque at 7,500 RPM. The power in both cases is the same (1.75X10^6 units of power). Suppose maximum power occurs at 7,250 RPM. Then it's easy to generate a power curve that would be maximum at that point and yields 242.8 units of torque (1.76X10^6 units of power).

The purpose of any transmission is to change input torque and speed vs.output torque and speed. In cars their purpose is to multiply torque. Drive wheel power is always net engine power less drive train frictional losses. Drive wheel torque is not, it is (net engine torque less drive train losses) X reduction ratio (and modified by rear wheel diameter). And restating, acceleration is related to force, not power. That's also why, for maximum acceleration, it may be better to delay shifting into the next higher gear until past the peak power point. The next higher gear reduces torque across the RPM range.

You can try it yourself. Take a set of engine torque curves from an actual vehicle, use the transmission ratios and lay out rear axle torque curves vs. RPM.
 
  • #29
jack action
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In the first quote, that is not true. As I showed, torque (and so force) can decline while power increases. So maximum force may not (and most probably doesn't) coincide with Pmax.
Yes, torque declines when power increases. That is because the rpm increased, and with it the speed of the car too. At a higher speed, the maximum tractive force will always be smaller for a given power (P = F X v). So that is why I said:
For a given car speed, yes, maximum tractive force always happens at maximum power, that is what [itex]F_{max} = \frac{P_{max}}{v}[/itex] means.
In the second quote, that also is not generally true. Similarly, it depends on the shape of the torque curve. For example, suppose you have an engine that generates 250 units torque at 7,000 RPM and 233.3 units of torque at 7,500 RPM. The power in both cases is the same (1.75X10^6 units of power). Suppose maximum power occurs at 7,250 RPM. Then it's easy to generate a power curve that would be maximum at that point and yields 242.8 units of torque (1.76X10^6 units of power).
But the car doesn't go at the same speed at 7000, 7250 or 7500 rpm.

Let's say I want the wheel to turn at 1000 rpm. The gear ratio needed for having the engine at 7000 rpm will be 7:1. For 7250 rpm, you will need a 7.25:1 gear ratio and a 7.5:1 for 7500 rpm. What are the wheel torque then, knowing the wheel rpm is the same in all cases?

250 * 7 = 1750 units torque
242.8 * 7.25 = 1760 unit torque
233.3 * 7.5 = 1750 units torque

Note how the wheel torque is the same when at 7000 or 7500 rpm; That is because the power is the same in both cases. Yet, the wheel torque is greater at 7250 rpm; That is because the power is greater. And the wheel torque increase is directly proportional to the power increase (1.76X10^6 / 1.75X10^6 = 1760 / 1750).
You can try it yourself. Take a set of engine torque curves from an actual vehicle, use the transmission ratios and lay out rear axle torque curves vs. RPM.
I did, I put a graph in my previous post especially for that. Here's that same graph with added info:

force-vs-speed.jpg


All black dots on the «Constant power» curve correspond to the maximum power in each gear.

At a speed of 40 km/h, where is the maximum tractive force? In 2nd gear. Why? Because I'm at maximum engine power. In 1st gear (I extended the 1st gear torque curve in green), I'm passed the maximum power rpm, so I produce less wheel torque (even though I'm in a lower gear ratio) because there is less power available. In 3rd gear, I'm at maximum engine torque, but because I have a higher gear ratio, I produce less wheel torque than in 2nd gear. How could I've known that in a simpler manner? Well, I also knew that I produce less power at maximum engine torque, and the wheel torque decrease is proportional to the engine power decrease. If the power would have been the same, it would have produce the same traction force as in 2nd gear. In top gear, there's even less power because of the higher gear ratio that set my engine rpm even lower.

See, at any car speed, the maximum tractive force is always at maximum power.

That is why you can draw the curve [itex]F_{max} = \frac{P_{max}}{v}[/itex] and you will know for sure that you will never be able to set the tractive force outside the green envelop (see below); The upper (white) region is off limit, no matter the gear ratio, tire diameter or if you produce that power with an electric motor, a diesel engine or a gas turbine.
force-vs-velocity.jpg
 
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  • #30
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OH NO. NOT THIS THREAD AGAIN.
 
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  • #31
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Yeah, xxChrisxx, me too. But there are some very basic and important issues here.

Remember this thread started with the question of optimum shift points for maximum acceleration.

You've selected 40 KPH, but I don't understand why. Where is the maximum force in 2nd gear? Not at 40 KMH, but at about 25 KMH, below the peak power curve (it's important to understand why). Where is the maximum force at 45 KPH? In 2nd gear, below the constant power curve. So shifting to 3rd at the maximum power point in 2nd is not the way to accelerate most quickly. Look at your 2nd and 3rd gear curves. Past 40 KPH until about 50 KPH 2nd gear is delivering more force than 3rd gear. Shifting at the "dot" in 2nd gear would reduce vehicle acceleration between 40 and about 50 KPH. Same with the other gears.
 
  • #33
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Yeah, xxChrisxx, me too. But there are some very basic and important issues here.

Remember this thread started with the question of optimum shift points for maximum acceleration.

You've selected 40 KPH, but I don't understand why. Where is the maximum force in 2nd gear? Not at 40 KMH, but at about 25 KMH, below the peak power curve (it's important to understand why). Where is the maximum force at 45 KPH? In 2nd gear, below the constant power curve. So shifting to 3rd at the maximum power point in 2nd is not the way to accelerate most quickly. Look at your 2nd and 3rd gear curves. Past 40 KPH until about 50 KPH 2nd gear is delivering more force than 3rd gear. Shifting at the "dot" in 2nd gear would reduce vehicle acceleration between 40 and about 50 KPH. Same with the other gears.
Right. I've had my Saturday night beers. So it's time for some abrasive posting.

Take the time to go back and actually READ AND COMPREHEND JACK'S POST AND THE VERY SPECIFIC POINT IT WAS ADRESSING. (hint it wasn't talking about shift points). Capped for effect.

Then once you've done that. You will probably realise that if asked to go away calculate the optimum shift point. You'll both give the same answer.

The basic concept is almost insanely simple. Be as close to Fmax as possible at all times.

You can calcuate it using torque or power. If you do your sums correctly, you'll both arrive at the exact same answer.

Job. Done.
 
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  • #34
jack action
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Sorry @xxChrisxx for still arguing on this issue, but when I read statements like these:
Using power is futile
Power is not related to acceleration.
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.
Lastly, power is an "invented" concept, not a fundamental property.
I feel an obligation to spread the good words. Though, I'm learning to quit when I began to repeat myself to the same crowd.

But I still do it as I even learn something new sometimes; like I never went in depth before to see where those ¼-mile equations came from. Somehow, it gives me a feeling of accomplishment :partytime:.
 
  • #35
Baluncore
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But I still do it as I even learn something new sometimes;
There are many ways to explain the theory and practice. We learn to explain it better every time.

The different answers from different members will hopefully provide an answer that can be understood by questioners who come from different educational paths and diverse backgrounds.
 
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