Calculating wheel torque from engine torque

[OldYat47]

The claim was that Fmax = Pmax/v. That premise is false. Look at Jack Action's 2nd gear torque curve. On that curve [Fmax (from about 20 KPH to 39 KPH)] > [Pmax/(v=40 KPH)]. From 20 to 39 KPH the torque curve is above the horizontal blue line. So Fmax doesn't equal Pmax/v. (Aside, if you plot torque vs. RPM instead of torque vs. velocity that constant power curve is a straight vertical line at max power. That's the way plots like this are generally done).

Power must be converted to force in order to calculate acceleration. If you have an example of calculating acceleration from mass and power please post it. Jack, I understand the 1/4 mile calculation but it's not a calculation of acceleration, it's a calculation of average acceleration. Here's a better sample problem: "A car is generating 1,000 Watts at the drive wheels. Its mass is 1,400 kg. How fast is it accelerating? Ignore friction and wind resistance".

Chris, you said, "The basic concept is almost insanely simple. Be as close to Fmax as possible at all times". In reference to a shift point chart it's better to say, "stay in gear until you run out of RPM or the rear wheel torque falls below what the rear wheel torque would be after the shift". In Jack Action's example, stay in 2nd gear until about 47 KPH before shifting to 3rd.

I said, "Sometimes it's better to use (less torque X lower gear) than (more torque X next higher gear)". Again, Jack Action's curves show this to be the case. The vehicle will accelerate faster in 2nd gear until about 47 KPH than it will after shifting to 3rd gear.

 

[cjl]

In reference to a shift point chart it's better to say, "stay in gear until you run out of RPM or the rear wheel torque falls below what the rear wheel torque would be after the shift".

And do you know when that is? It's when the power after the shift is the same as the power before the shift.

 

[jack action]

Damn it! I can't quit ...

@OldYat47, I'm sure you are in good faith, so let's go at it one more time. You have a tendency to choose and pick the statements you answer and ignore the other ones, so I indicated in my post where you should stop and think about what you read and give us your opinion about what I wrote. Don't ignore a single «check point» I inserted if you want to answer to this post. I just want to know where you agree with me and, if not, how you see it. If you don't understand what I said, please tell me, I'll try to find another way for you to picture it.

If you answer to argue this post in anyway, please tell me you understand and agree to the previous paragraph, got it? (answer here)

You seem to like numbers better than variables, so let's go back to the graph I made earlier:

First let's assume that the black dots on the «Constant power» curve are where the curve meets the other gear curves. Before & after is really close, but not quite on it. Let us assume that, such that we don't argue that the 2nd gear is tangent to «Constant power» curve at 40 km/h or 43 km/h. It is 40 km/h, because someone put the black dot there and he reads data better than he draws.

If you answer to argue this post in anyway, please tell me you understand and agree to the previous paragraph, got it? (answer here)

Now, you are saying that in 2nd gear there is more torque at 25 km/h than at at 40 km/h, and you are right. Don't think nobody else noticed that. Let us look at the 3rd gear instead. Let us look at just 40 km/h (The red line I drew). You can see that I am at maximum torque in 3rd gear, right? There is no way I can produce more torque into that gear ratio, right? But I can also shift it in 2nd gear and you can appreciate that - always at 40 km/h - I produce a larger tractive force in that gear. You must see this, since you said it yourself:

Again, Jack Action's curves show this to be the case. The vehicle will accelerate faster in 2nd gear until about 47 KPH than it will after shifting to 3rd gear.

If I shifted from 3rd to 2nd and increase my traction force - always being at 40 km/h - than someone could logically state:

«Let's shift in 1st gear, there is a lower gear ratio, therefore it must produce an even bigger force.»

But when we look at the graph, we see that it is not the case. The traction force in 1st gear is still slightly higher than in 3rd gear, but it is less than in 2nd gear. So that way of thinking is obviously wrong. What happened? What is so special about 2nd gear at 40 km/h?

If you answer to argue this post in anyway, please tell me if you agree with the previous observation first. (answer here)

You answer that question like there is some kind of voodoo going on, like nobody understand what is happening:

Sometimes it's better to use (less torque X lower gear) than (more torque X next higher gear).

In essence, you are telling us to go by trial and error: Try every possible gear ratio that you can think of, find the rpm the engine is in at 40 km/h, calculate the wheel torque for every case, and find out the gear ratio that gives out the larger wheel torque.

If you answer to argue this post in anyway, please tell me if you agree with the previous observation first. (answer here)

You even go further by saying:

You can't "see" this instinctively from the power curve.

But look at the graph once more. Where is situated the traction force on the 2nd gear at 40 km/h? It meets the «Constant power» curve, which is the tractive force produced with the maximum power of the engine. So you are wrong: We can «see» this instinctively from the power curve. For a given car speed (I can't stress this enough), you will always produce the maximum tractive force when the engine produces its maximum power (In math form: $F_{max} = \frac{P_{max}}{v}$, where $v$ is a constant for the purpose of comparison). So if you want to produce the maximum tractive force you can think of at 40 km/h, you need to find the gear ratio that will set the engine's rpm where it produces its maximum power at 40 km/h. No need for trial and error.

So what is the gear that produces the maximum tractive force at:

• 25 km/h? 1st gear. Why? The engine is at its maximum power;
• 40 km/h? 2nd gear. Why? The engine is at its maximum power;
• 60 km/h? 3rd gear. Why? The engine is at its maximum power;
• 85 km/h? Top gear. Why? The engine is at its maximum power.

If you answer to argue this post in anyway, please tell me if you agree with the previous observation first. (answer here)

If we had 1 million gear ratios, we could select a gear for every car speed. In any case, when the selected gear produces more tractive force that any other at a given car speed (again, I can't stress this enough), that is because the engine is producing its maximum power at this car speed.

SHIFT POINTS

What started this discussion? It is this statement:

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.

That was your quote, which was in direct opposition with a previous one made by @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.

Going back at the graph, where are the best shift points to maximize acceleration? There are where the different gear curves meet one another. That is:

• Go from 1st to 2nd at 35 km/h;
• Go from 2nd to 3rd at 47 km/h;
• Go from 3rd to top gear at 68 km/h.

If you answer to argue this post in anyway, please tell me if you agree with the previous observation first. (answer here)

Why is the tractive force the same in both gears at those particular speed? You will notice that - in all cases - the lower gear is slightly after the peak power and the higher gear is slightly before the peak power. If you would have the actual power curve, you would see that it would correspond to 2 points on that curve where the engine produces the exact same amount of power. It is easy to prove by stating that $P = Fv$ for the car. So if the tractive force is the same and the car speed is the same, therefore the power must also be the same. And the power at the wheel is the same as the power provided by the engine (not including losses).

If you answer to argue this post in anyway, please tell me if you agree with the previous observation first. (answer here)

If we add gears - say if we had 1 million gear ratios - The shift points would be closer and closer to the «Constant power» curve (Which correspond to keeping the engine's rpm at its maximum power at any car speed).

Why is your method working?

The «power» method is kind of cool, because we don't really need to know the engine's rpm; All we need is to know the actual power produced by the engine and we know the following will produce the best shift points:

Where the horizontal line's locations depends solely on the gear ratios themselves, i.e. where:
$$\frac{rpm_{lo}}{rpm_{hi}} = \frac{GR_{hi}}{GR_{lo}}$$

If you answer to argue this post in anyway, please tell me if you agree with the previous observation first. (answer here)

When you have access to the torque curve only, you will need to find the wheel torque & rpm from the engine torque & rpm and the gear ratio:
$$T_{w\ lo} = T_{e\ lo}GR_{lo}$$
$$rpm_{w\ lo} = \frac{rpm_{e\ lo}}{GR_{lo}}$$
$$T_{w\ hi} = T_{e\ hi}GR_{hi}$$
$$rpm_{w\ hi} = \frac{rpm_{e\ hi}}{GR_{hi}}$$

If you answer to argue this post in anyway, please tell me if you agree with the previous observation first. (answer here)

Then, you will visually inspect your data to find points where $T_{w\ hi} = T_{w\ lo}$ and $rpm_{w\ hi} = rpm_{w\ lo}$ or, in math form:
$$rpm_{w\ hi} = rpm_{w\ lo}$$
$$\frac{rpm_{e\ hi}}{GR_{hi}} = \frac{rpm_{e\ lo}}{GR_{lo}}$$
$$\frac{rpm_{e\ hi}}{rpm_{e\ lo}} = \frac{GR_{hi}}{GR_{lo}}$$
$$\frac{rpm_{e\ hi}}{rpm_{e\ lo}} = \frac{\frac{T_{w\ hi}}{T_{e\ hi}}}{\frac{T_{w\ lo}}{T_{e\ lo}}}$$
$$\frac{rpm_{e\ hi}}{rpm_{e\ lo}} = \frac{T_{w\ hi}}{T_{w\ lo}}\frac{T_{e\ lo}}{T_{e\ hi}}$$
By defintion:
$$\frac{rpm_{e\ hi}}{rpm_{e\ lo}} = (1)\frac{T_{e\ lo}}{T_{e\ hi}}$$
$$T_{e\ hi}rpm_{e\ hi} = T_{e\ lo}rpm_{e\ lo}$$
Or simply put:
$$P_{e\ hi} = P_{e\ lo}$$
We are again brought back to the power in low gear equals the power in high gear.

Whenever you need to know about torque and rpm, you will most likely have to multiply them at some point in order to find the power generated; whether you do it mathematically (like I just did) or that you do it intuitively by looking at a graph (like you are doing).

 

[OldYat47]

Let's do this in small steps. Let's look at two pieces of this discussion to start with.

Now, you are saying that in 2nd gear there is more torque at 25 km/h than at at 40 km/h, and you are right. Don't think nobody else noticed that. Let us look at the 3rd gear instead. Let us look at just 40 km/h (The red line I drew). You can see that I am at maximum torque in 3rd gear, right? There is no way I can produce more torque into that gear ratio, right?

Yes, at 40 KPH the engine is at maximum torque in 3rd gear. But you said that Fmax = Pmax/v. That is not the case. Fmax point will not coincide with the intersection of any gear ratio for this engine torque and the constant power curve.

Let's look at a 2nd to 3rd gear shift. From 40 KPH through about 48 KPH in 2nd gear there is more torque at the drive wheels than at any point on the 3rd gear curve. So shifting to 3rd at the maximum power point in 2nd will reduce acceleration, increasing the time to achieve any speed above 40 KPH.

Please start with these two points. We can, I am sure, reach agreement.

 

[OldYat47]

cjl, you may be able to find pairs of gear ratios where that would be true but it is not the general case.

 

[cjl]

cjl, you may be able to find pairs of gear ratios where that would be true but it is not the general case.

It is absolutely the general case, since that's the definition of power. Power is (at least in one definition) force times velocity. So, if the tractive force is the same in two gears at the same speed, the power must, by definition, be exactly equal (ignoring differing drivetrain losses for different gear ratios)

 

[OldYat47]

Yes, if the speed is the same for two different gear ratios at the same vehicle speed the power is the same. So if you shift when the torque is the same before and after the shift the power will be the same. But this is not always possible. One example is when peak torque is close to maximum RPM. The reduced torque at maximum RPM may be higher than available torque in the next gear at that speed. A similar variation of that case is when gear ratios are too far apart.

 

[xxChrisxx]

It's relatively clear you know where the shift points should be. Noone is disputing that. Everyone has all picked the same points.

The fact that you say you disagree with P=FV is slightly worrying from a conceptual point of view. Ignore engines, gearboxes wheels. Ignore all numbers.

Do you understand where P=FV comes from and why it is always valid? And by extension why its useful?

 

[OldYat47]

What I am pointing out that Jack Action's premise is incorrect. His premise is that maximum force (torque in this case) occurs at maximum power at some specific speed.

The reason it's "easier" to get shift point and acceleration charts with torque curves is that power is not related to acceleration or force. P=FV, so for any given value of power there is an infinite number of valid values for and force and acceleration. F=MA, so P=MAV. Again, there is an infinite number of values for M, A and V that will yield the same P.

 

[cjl]

What I am pointing out that Jack Action's premise is incorrect. His premise is that maximum force (torque in this case) occurs at maximum power at some specific speed.

The reason it's "easier" to get shift point and acceleration charts with torque curves is that power is not related to acceleration or force. P=FV, so for any given value of power there is an infinite number of valid values for and force and acceleration. F=MA, so P=MAV. Again, there is an infinite number of values for M, A and V that will yield the same P.

Maximum tractive force at any given speed does occur by maximizing power. As for your P = MAV, think about that for just a second. We can rearrange that to state that A = P/(M*V), and since for any given vehicle, M is not changing, we can see that max A for any given V occurs by maximizing P.

 

[jack action]

You obviously did not read what I wrote and/or think about what you've read at the check points I set in my last post. Time for more drawings. Note how I will answer EVERY statements you've made in your last post, just to show you what I understand about what you are saying (It is call listening).

Yes, at 40 KPH the engine is at maximum torque in 3rd gear. But you said that Fmax = Pmax/v. That is not the case. Fmax point will not coincide with the intersection of any gear ratio for this engine torque and the constant power curve.

This is what I'm taking about:

This is what you are taking about:

Nobody cares about the maximum tractive force in a particular gear. Why? Because in a lower gear, you can always have a higher tractive force for the same car speed. Why would you choose to be in 3rd gear at 40 km/h, if you can get more tractive force in 2nd gear? Remember, the goal is to have the highest tractive force to get the maximum acceleration.

Yes, in 2nd gear, I have more tractive force at 25 km/h. But when my car will be at that speed, I will choose 1st gear to get $F_{max}$.
$$F_{max} = \frac{P_{max}}{v}$$
See, there are no gear ratios in the equation. There is car speed $v$, though. You might be in 1st, 2nd, 3rd or top gear - You might even have an electric motor with no transmission at all - for any given speed (this should be in large bold red letters to emphasize the point, but I'm not sure other people can see it, tell me you are not seeing it), $F_{max} = \frac{P_{max}}{v}$.

Let's look at a 2nd to 3rd gear shift.

Okidoki.

Fro 40 KPH through about 48 KPH in 2nd gear there is more torque at the drive wheels than at any point on the 3rd gear curve.

That is what I've been saying all along. Actually there is more torque at the drive wheels from 0 to 48 km/h than at any point on the 3rd gear curve.

The real questions are:

1. Why is there more torque in 2nd gear than in 3rd gear before 48 km/h?
2. Why is there the same amount of torque in 2nd gear and in 3rd gear at 48 km/h?
3. Why is there less torque in 2nd gear than in 3rd gear after 48 km/h?

Please answer those. (HINT: It has to do with power.)

So shifting to 3rd at the maximum power point in 2nd will reduce acceleration, increasing the time to achieve any speed above 40 KPH.

Nobody is shifting from 2nd to 3rd at maximum power in this case. You obviously did not read the part of my last post entitled «SHIFT POINTS». A lot of people have told you that you are saying the same thing as I do shift-point-wise, but you don't seem to accept that. This next drawing should convince you. These are the shift points:

I'm sure we are on the same page now. So you shouldn't focus at all on this anymore. You should focus on why I (and all other people on this tread) say acceleration is directly related to engine power (for which, I don't think you are convinced).

Now, can you tell me why those are the shift points? (HINT: It has do to with power.)

What the heck - in case the scrolling feature is broken on your computer - let's make sure you will read my arguments thoroughly, so here they are again:

SHIFT POINTS

What started this discussion? It is this statement:

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.

That was your quote, which was in direct opposition with a previous one made by @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.

Going back at the graph, where are the best shift points to maximize acceleration? There are where the different gear curves meet one another. That is:

• Go from 1st to 2nd at 35 km/h;
• Go from 2nd to 3rd at 47 km/h;
• Go from 3rd to top gear at 68 km/h.

If you answer to argue this post in anyway, please tell me if you agree with the previous observation first. (answer here)

Why is the tractive force the same in both gears at those particular speed? You will notice that - in all cases - the lower gear is slightly after the peak power and the higher gear is slightly before the peak power. If you would have the actual power curve, you would see that it would correspond to 2 points on that curve where the engine produces the exact same amount of power. It is easy to prove by stating that $P = Fv$ for the car. So if the tractive force is the same and the car speed is the same, therefore the power must also be the same. And the power at the wheel is the same as the power provided by the engine (not including losses).

If you answer to argue this post in anyway, please tell me if you agree with the previous observation first. (answer here)

If we add gears - say if we had 1 million gear ratios - The shift points would be closer and closer to the «Constant power» curve (Which correspond to keeping the engine's rpm at its maximum power at any car speed).

Why is your method working?

The «power» method is kind of cool, because we don't really need to know the engine's rpm; All we need is to know the actual power produced by the engine and we know the following will produce the best shift points:

Where the horizontal line's locations depends solely on the gear ratios themselves, i.e. where:
$$\frac{rpm_{lo}}{rpm_{hi}} = \frac{GR_{hi}}{GR_{lo}}$$

If you answer to argue this post in anyway, please tell me if you agree with the previous observation first. (answer here)

When you have access to the torque curve only, you will need to find the wheel torque & rpm from the engine torque & rpm and the gear ratio:
$$T_{w\ lo} = T_{e\ lo}GR_{lo}$$
$$rpm_{w\ lo} = \frac{rpm_{e\ lo}}{GR_{lo}}$$
$$T_{w\ hi} = T_{e\ hi}GR_{hi}$$
$$rpm_{w\ hi} = \frac{rpm_{e\ hi}}{GR_{hi}}$$

If you answer to argue this post in anyway, please tell me if you agree with the previous observation first. (answer here)

Then, you will visually inspect your data to find points where $T_{w\ hi} = T_{w\ lo}$ and $rpm_{w\ hi} = rpm_{w\ lo}$ or, in math form:
$$rpm_{w\ hi} = rpm_{w\ lo}$$
$$\frac{rpm_{e\ hi}}{GR_{hi}} = \frac{rpm_{e\ lo}}{GR_{lo}}$$
$$\frac{rpm_{e\ hi}}{rpm_{e\ lo}} = \frac{GR_{hi}}{GR_{lo}}$$
$$\frac{rpm_{e\ hi}}{rpm_{e\ lo}} = \frac{\frac{T_{w\ hi}}{T_{e\ hi}}}{\frac{T_{w\ lo}}{T_{e\ lo}}}$$
$$\frac{rpm_{e\ hi}}{rpm_{e\ lo}} = \frac{T_{w\ hi}}{T_{w\ lo}}\frac{T_{e\ lo}}{T_{e\ hi}}$$
By defintion:
$$\frac{rpm_{e\ hi}}{rpm_{e\ lo}} = (1)\frac{T_{e\ lo}}{T_{e\ hi}}$$
$$T_{e\ hi}rpm_{e\ hi} = T_{e\ lo}rpm_{e\ lo}$$
Or simply put:
$$P_{e\ hi} = P_{e\ lo}$$
We are again brought back to the power in low gear equals the power in high gear.

Whenever you need to know about torque and rpm, you will most likely have to multiply them at some point in order to find the power generated; whether you do it mathematically (like I just did) or that you do it intuitively by looking at a graph (like you are doing).

 

[xxChrisxx]

What I am pointing out that Jack Action's premise is incorrect. His premise is that maximum force (torque in this case) occurs at maximum power at some specific speed.

This is where the disconnect is. You are thinking he means peak engine power output (as it sits on the engine curve) he doesn't. It's a boundary condition. Obviously this isn't possible with a real internal combustion engine and fixed gearing (where it would touch the curve once per gear). However the it's an important step conceptually, as it defines your envelope of performance.

You know your real power train and drive line combo must sit within this envelope. You will have heard of acceleration being 'traction limited' and 'power limited'. The 'power limited' is effectively the right hand curve, 'traction limit' would be a horizontal line.

The key difference here is how powerful the tool is in designing a system from a clean sheet vs. simply analysing a fixed known system.

For example. A theoretical maximum 0-60 is 5 seconds. The engine curve lines, sit with no gap to the Fmax line. If someone doesn't understand the implication of this they could spend the rest of time tinkering with clutches, gear ratios and final drives and never get any better. One look at that performance map would show instantly that there is no performance to be gained from drive line changes. The engine must produce more power.

 

[OldYat47]

Let's eliminate all the stuff where we already agree to simplify these posts.

- We all agree where the optimum shift points should be using either the torque curves or that latest power-vs-shift point curve Jack Action posted. No need to go over that again, let's eliminate that from the discussions. Agreed?
- We all agree that the rate of vehicle acceleration at any given speed in any given gear is proportional to the amount of torque available at that speed. We can eliminate that as well. Agreed?

Maybe we can come to agreement on two more items before proceeding further. The first is the formula Fmax=Pmax/v (this may be an issue of semantics). The second is the significance of the intersection of the power and ratio curves. The two issues are related, I think.

Some quantity of force is available at Pmax. In any gear that amount of force can be determined by dividing Pmax by the speed at which Pmax occurs. F=P/v. But that's not Fmax, is it? It is F at Pmax. To me, Fmax = (some value of P)/(some value of v), the "some value" part depends on the gear ratio.

What is significant about the intersection of a ratio curve and the power curve? It's not the optimum shift point. It's not where the vehicle is acceleration most quickly. It's not the upper limit of acceleration potential (the engine torque curve is that limit). So why is it important?

Let's stick with that first. One bite at at time? Please?

 

[jack action]

Some quantity of force is available at Pmax. In any gear that amount of force can be determined by dividing Pmax by the speed at which Pmax occurs. F=P/v. But that's not Fmax, is it?

It is Fmax @ speed v.

For another speed, say v₂, for the same Pmax, you will get another value, say Fmax₂.

If v₂ > than v the Fmax₂ < Fmax, and vice-versa.

This is shown in figure 1:

To me, Fmax = (some value of P)/(some value of v), the "some value" part depends on the gear ratio.

For any gear ratio, you have Fmax = GR / r * Tmax which is what you mean and what is shown by figure 2:

Can you see that Fmax at 40 km/h (figure 1) is greater than Fmax in 3rd gear (figure 2), which is also at 40 km/h? (Follow the vertical red line)

Therefore F = P / (40 km/h). If P is at its maximum value, then F must also be at its maximum value, don't you think?

It's not where the vehicle is acceleration most quickly.

Yes it is.

It's not the upper limit of acceleration potential

Yes it is.

 

[OldYat47]

OK, so part of it is semantics. I don't like using Fmax unless it is the highest value of F for that ratio. But I'm not going to argue semantics. I'll agree to use your terminology.

Let's suppose we are trying to accelerate. You shift from 1st to 2nd so you wind up at Fmax in 2nd gear. From that speed until the intersection of the 2nd gear ratio and the constant power line the vehicle is accelerating more quickly than it is at that intersection point (torque values higher, A=F/M, more F = more A). For that reason I make the statement that that intersection point is not the point of maximum acceleration. It is at 40 KPH in 2nd gear, but not overall in 2nd gear.

That intersection point is also not the upper limit of rate of acceleration in 2nd gear. The upper limit of acceleration in 2nd gear is around 25 KPH, Fmax in 2nd gear.

Can we agree on that?

 

[xxChrisxx]

OK, so part of it is semantics. I don't like using Fmax unless it is the highest value of F for that ratio. But I'm not going to argue semantics. I'll agree to use your terminology.

That intersection point is also not the upper limit of rate of acceleration in 2nd gear. The upper limit of acceleration in 2nd gear is around 25 KPH, Fmax in 2nd gear.
Can we agree on that?

You are both again talking at cross purposes, but it feels like a bit of progress. Noone has ever actually disagreed with your assertions about engines and gears. It's more your lack of appreciation of power as a tool for calculating performance and tuning.

Everyone agrees that 2nd gear the peak acceleration in gear is at the peak torque. If you'll look back Jack pointed out the peak force in gear at the peak of torque output with a big orange arrow.

Would it help if we rename the 'Fmax' line as 'power limited acceleration'?

The value of this line is that it is the boundary of performance for a given power output. Think of it this way; any gap between the ratio lines curves and this line is lost performance.

The goal of gear ratio selection is to minimise the area between the 'ratio curves' and 'power limit' curve. There will always be a compromise and some lost performance due to the nature of fixed gears and IC engines.

 

[OldYat47]

No, I don't want to start renaming things as long as I understand precisely what is meant by terms like Fmax. Jack Action is using Fmax as a function of v and the value of P at that v. My argument remains that all the information you need is present in the torque curves. I haven't seen any reason to say that the power curve is of any use in the context of vehicle performance. My secondary argument is that there is no way to calculate acceleration from power. In math terms, for any non-zero value of power there are infinite valid values of force (and by extension, acceleration).

I do feel like we're making progress, which is encouraging.

 

[jack action]

OK, so part of it is semantics. I don't like using Fmax unless it is the highest value of F for that ratio. But I'm not going to argue semantics. I'll agree to use your terminology.

OK.

So from now on, I will refer to «the highest value of F for that ratio» as Fmax.

From now on, I will refer to «the value of F under maximum engine power» as F@pmax.

Let's suppose we are trying to accelerate. You shift from 1st to 2nd so you wind up at Fmax in 2nd gear.

I'm stopping you right there. I'm not sure if you are talking about Fmax or F@pmax (I think it is Fmax), but in either case, nobody talked about shifting to «wind up at Fmax». I thought we already agreed that we shift when the wheel torque in first gear is the same as the wheel torque in 2nd gear (where 1st gear curve crosses 2nd gear curve)?

From that speed until the intersection of the 2nd gear ratio and the constant power line the vehicle is accelerating more quickly than it is at that intersection point (torque values higher, A=F/M, more F = more A).

Agreed.

For that reason I make the statement that that intersection point is not the point of maximum acceleration. It is at 40 KPH in 2nd gear, but not overall in 2nd gear.

Agreed.

That intersection point is also not the upper limit of rate of acceleration in 2nd gear.

Agreed.

The upper limit of acceleration in 2nd gear is around 25 KPH, Fmax in 2nd gear.

Agreed.

My turn.

Can we agree that when you are at Fmax in 3rd gear at 40 km/h, at that same car speed, you can also be at F@pmax in 2nd gear?

Can we agree that F@pmax in 2nd gear is greater than Fmax in 3rd gear and therefore will produce a larger acceleration (torque values higher, A=F/M, more F = more A)?

For that reason I make the statement that the intersection point is the point of maximum acceleration at 40 km/h, no matter what gear you choose (but you have to be in 2nd gear to reach that point).

Can we agree that this same intersection point is the upper limit of acceleration at 40 km/h? Meaning that if the car is set in any other gear (1st, 3rd, top) at 40 km/h, it will not have a higher acceleration than in 2nd gear.

Can we agree that when you are at Fmax in 2nd gear at 25 km/h, at that same car speed, you can also be at F@pmax in 1st gear?

Can we agree that F@pmax in 1st gear is greater than Fmax in 2nd gear and therefore will produce a larger acceleration (torque values higher, A=F/M, more F = more A)?

For that reason I make the statement that the intersection point is the point of maximum acceleration at 25 km/h, no matter what gear you choose (but you have to be in 1st gear to reach that point).

Can we agree that the intersection point at 25 km/h is the upper limit of acceleration at 25 km/h? Meaning that if the car is set in any other gear (2nd, 3rd, top) at 25 km/h, it will not have a higher acceleration than in 1st gear.

 

[xxChrisxx]

My secondary argument is that there is no way to calculate acceleration from power. In math terms, for any non-zero value of power there are infinite valid values of force (and by extension, acceleration)

You could make the exact same argument about the general case of torque:
Traction force = Engine torque * gear ratio * FD ratio / Rolling radius
For any non zero value of engine torque, there are infinite valid values of gear ratio, FD and wheel size to give a traction force. and by extension.

What happens is you then define the variables to give a specific case.

F = P/v
You have a power @ an engine speed, and the engine speed relation to road speed, is the overall ratio. Ie the exact same thing you have to define above to give the specific case.

It's all one and the same.

 

[cjl]

OK, so part of it is semantics. I don't like using Fmax unless it is the highest value of F for that ratio. But I'm not going to argue semantics. I'll agree to use your terminology.

But let's say for a second that we don't care about the ratio at all. Let's say we can pick the ratio arbitrarily - we have a CVT, for example. If we're going 25mph and want to maximize our force, what gear ratio do we pick? Do we pick a ratio such that the engine is at peak torque? No, because that won't actually give us the maximum tractive force available at 25mph. To obtain the maximum possible tractive force at that speed, we need to pick a ratio such that the engine is at peak power, hence why jack is calling it fmax - given that engine, it is the highest possible tractive force the vehicle could generate at that speed (no matter what gear ratio you pick).

 

[OldYat47]

Jack Action, we are talking the same language finally and I agree with all your points. Whoopee! It was all semantics. And yes, I meant Fmax, not F@Pmax.

You could make the exact same argument about the general case of torque:
Traction force = Engine torque * gear ratio * FD ratio / Rolling radius
For any non zero value of engine torque, there are infinite valid values of gear ratio, FD and wheel size to give a traction force. and by extension.

Not the same thing at all. The bottom line is if you know the drive wheel torque you can directly calculate acceleration.

Ignoring friction, power at the drive wheels is always the same as engine power regardless of gearing. But acceleration does change with gearing. The values of power remain constant as the values for acceleration change depending on the reduction ratio. Knowing the drive wheel power, you cannot calculate acceleration. You have to know velocity, which brings you back to A=[(F*v)/v]/M, or A=F/M.

cjl, that's not correct. Look at a any torque and power vs. RPM set of curves. You can choose to program your CVT so as to maintain any engine speed. Your goal is maximum acceleration. If the torque is lower at (maximum power RPM) than at (maximum torque RPM) then acceleration will be less at maximum power than at maximum torque.

Just in time. I'm "unretiring" for a few days (or more).

 

[jack action]

@OldYat47, I'm glad we agree. Now let me share a thought experiment with you.

Let's say that we are cruising at the constant speed of 35 km/h in 3rd gear and we happen to have the throttle wide open. It would be the yellow dot on the red line in the figure below:

• The green dots represent where Fmax is for each gear;
• The black dots represent where F@pmax is for each gear;
• The blue dots represent the traction force at 35 km/h in each gear.

We now want to accelerate to 36 km/h as fast as we can. What do we do?

1. Any point in top gear provides a smaller traction force than the one we already have, so they're eliminated;
2. Fmax in 3rd gear offers a larger traction force, but we need to be at 37-38 km/h to get it. We can't instantaneously be at 38 km/h to get the traction force to help us reach 36 km/h; That makes no sense;
3. We could downshift in 2nd gear and get the largest traction force possible at 35 km/h (even larger than in 1st gear at 35 km/h);
4. But there is an even greater traction force at Fmax in 2nd gear (or even the largest traction force possible, i.e. Fmax in 1st gear). The problem is that we need to decelerate to 25 km/h (or even 20 km/h in 1st gear) to get it. That is not really smart, especially knowing that once we'll be back again at 35 km/h (remember our goal is to reach 36 km/h), we'll have the traction force that we would have had by staying at 35 km/h and simply downshift into 2nd gear (point #3).

Can we agree that point #3 is the smartest move one can do to get the greatest acceleration?

But can I get a greater traction force at 35 km/h, just by modifying the gear ratio of my 2nd gear? Let's see.

Let's set a 2nd gear with a lower gear ratio. Let's call it gear 2a (orange in the next figure, which is an enlargement of the previous figure):

Hurray! We have more traction force at 35 km/h! Note also that F@pmax moved to the left (to a lower car speed) on the «Constant power» curve.

Let's try an even lower gear ratio. Let's called it gear 2b:

It provides an even larger traction force at 35 km/h! Note that the traction force at 35 km/h in gear 2b is now exactly F@pmax.

It goes so well, let's try a gear ratio even lower. Let's call it gear 2c:

Oh no! Now we got less traction force than with gear 2b. Actually, we have the same amount of force than with gear 2a (because we have the same engine power in both cases, i.e. slightly less than the maximum engine power). Note that F@pmax is now at an even lower car speed, still moving left on the «Constant power» curve.

So, can we agree that when you are at a given car speed - say 35 km/h - to get the maximum traction force (thus, the maximum acceleration), the engine rpm must be where it produces its maximum power?

Can we agree that this will be true at any speed?

Therefore, can we agree that the maximum acceleration at any given speed depends on the engine maximum power?

 

[OldYat47]

Yes, #3 is the best shift option.

And, you can manipulate gear ratios to make maximum power and maximum torque coincide (sort of like horsepower and pound feet of torque are always equal at 5,252 RPM). That may or may not be the best option for optimum vehicle acceleration. It's an artifact. You could also set the 2nd gear ratio such that maximum torque occurred at 35 KPH. That case would give the best acceleration from 35 KPH to 36 KPH after a 3rd to 2nd downshift.

 

[cjl]

cjl, that's not correct. Look at a any torque and power vs. RPM set of curves. You can choose to program your CVT so as to maintain any engine speed. Your goal is maximum acceleration. If the torque is lower at (maximum power RPM) than at (maximum torque RPM) then acceleration will be less at maximum power than at maximum torque.

This is not true. Wheel torque will be highest if the CVT is programmed to maintain engine speed at max power, not at max (engine) torque. Thus, for maximum acceleration, you always want to maintain the engine speed for maximum power, even though the engine torque is not maximized at this value. This is because max power RPM will always be above max torque RPM, which allows you to use a lower gear ratio for better mechanical advantage, providing higher wheel torque despite the lower engine torque.

 

[cjl]

Yes, #3 is the best shift option.

And, you can manipulate gear ratios to make maximum power and maximum torque coincide (sort of like horsepower and pound feet of torque are always equal at 5,252 RPM). That may or may not be the best option for optimum vehicle acceleration. It's an artifact. You could also set the 2nd gear ratio such that maximum torque occurred at 35 KPH. That case would give the best acceleration from 35 KPH to 36 KPH after a 3rd to 2nd downshift.

He's not manipulating gear ratios to make max power and max torque coincide. He's showing you that maximum wheel torque at a given speed always coincides with the gear ratio that allows the engine to be at max power.

 

[jack action]

Yes, #3 is the best shift option.

I'm glad we agree.

You could also set the 2nd gear ratio such that maximum torque occurred at 35 KPH. That case would give the best acceleration from 35 KPH to 36 KPH after a 3rd to 2nd downshift.

I like that. You are so tenacious! Let's see what happens when we redo the same thought experiment as I did in my previous post, but by changing the 2nd gear ratio to a higher gear instead of a lower gear. Let's call it gear 2d:

Can we agree that by setting a higher gear ratio, the whole traction force curve will go down and cover a greater speed range?

Can we agree that Fmax in gear 2d will be smaller and at a faster car speed than in 2nd gear?

Let's go to an even higher gear ratio. Let's call it gear 2e:

Look at that, Fmax is now precisely at 35 km/h! And it is a lot less than the traction force in 2nd gear. Note also that if I just set it to a slightly higher gear ratio, I will be in 3rd gear.

Now you should also be able to appreciate that Fmax in 2nd gear is not the greatest force you can achieve at 25 km/h; F@pmax in 1st gear is much greater at that same speed.

No matter what gear ratio you're in when at Fmax, there is a lower gear ratio that will create a greater force at the same speed with F@pmax.

If you want to move a point from left to right, it will also have to move down. That is because engine power is always the same, which is equal to the wheel power, and wheel power is F X v, so if v increases and power is the same, then F must go down.

Can we agree that Fmax will never offer the greatest possible acceleration at a given speed?

Can we agree that the maximum acceleration at any given speed depends on the engine maximum power?

You may now appreciate more my answer to one of your previous posts:

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).

And I answered (I added comments in red to relate to examples from my previous post):

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 [--> corresponds to gear 2a]
242.8 * 7.25 = 1760 unit torque [--> corresponds to gear 2b]
233.3 * 7.5 = 1750 units torque [--> corresponds to gear 2c]

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).

 

[OldYat47]

Tenacious perhaps, but also concerned that I missed the significance of cjl's comment about maximum power and the effects of multiplying torque. He or she is, of course, correct, which means I am missing stuff again. That's distressing. I'll sign off for a while, looks like more time at the neurologist for me. Ya gets old, you gets feeble.

Thank you all for a refresher and a refreshing debate.

 

[Emre]

What about accelerating in first gear from a dead stop?It makes sense that the max achievable acceleration comes when we choose the gear corresponding to max horsepower RPM at a given speed.But what can we say when the speed is 0?I mean,why don't we choose the max torque RPM for a dead stop for max acceleration since the wheel torque is engine torque x gear ratio x axle ratio?Am i wrong?Is this the reason why diesel cars accelerate as if they are too quick for their power?

 

[CWatters]

In an ideal engine the power would be high/maximum at all rpm right down to zero. Now since power = torque * angular velocity this would mean the torque should increase to infinity at zero rpm.

Unfortunately we can't make an ideal gas/petrol or diesel engine, they have a maximum torque that limits power at low rpm. However diesel are "more ideal" than gas/petrol. Diesel have greater torque at low rpm so they develop more power at lower rpm.

Electric motors are even "more ideal" than diesel engines. They can develop max torque at very low rpm. Take a look at some of the Teslas vs super car videos. Typically the Tesla wins over short distances but super car has higher top speed..



Some think that one day an electric dragster should be able to take the top fuel drag records..

https://newatlas.com/top-ev-electric-drag-racing-top-fuel/50741/

 

[Randy Beikmann]

What about accelerating in first gear from a dead stop?It makes sense that the max achievable acceleration comes when we choose the gear corresponding to max horsepower RPM at a given speed.But what can we say when the speed is 0?I mean,why don't we choose the max torque RPM for a dead stop for max acceleration since the wheel torque is engine torque x gear ratio x axle ratio?Am i wrong?Is this the reason why diesel cars accelerate as if they are too quick for their power?

Remember that power = torque*rotation speed (P=T*omega), so when omega is zero, power is zero. Axle torque might be huge at zero speed (limited by traction), but except for the fact the tires will slip a bit acceleration, you can't use much power right at launch.

 

[jack action]

I mean,why don't we choose the max torque RPM for a dead stop for max acceleration since the wheel torque is engine torque x gear ratio x axle ratio?Am i wrong?

You are not wrong. From a dead stop, you should launch at a rpm where the torque is maximum to get the maximum acceleration (given you have the traction to support it), and that is what racers do. Then you increase the rpm until maximum horsepower and change the gear ratio as often as necessary to keep the rpm constant as much as possible.

 

[miltos]

A better idea is to start launching at max rpm so you can make use of the motor rotational inertia. Using the clutch you should take care to transfer to the wheels the maximum torque, accordind to the max available friction with the road (while you keep accelaration pedal to the floor)

With a low horsepower car, the engine rpm will drop a lot. When they tend to drop below max torque rpm you should idealy stabilize motor rotation speed to that of max torque, by deppresing the clutch pedal.

With a race car, it is likely that the maximum friction will correspond to a motor torgue less than the maximum available. Then, there are more than one way to achieve max accelaration (even without full throttle) and probably other factors are crucial for the starting rpm (eg turbo lag, clutch wear).

 

[Emre]

You are not wrong. From a dead stop, you should launch at a rpm where the torque is maximum to get the maximum acceleration (given you have the traction to support it), and that is what racers do. Then you increase the rpm until maximum horsepower and change the gear ratio as often as necessary to keep the rpm constant as much as possible.

Thanks for the answer.But for many cars when you are close to peak hp RPM at launch,tire spins become too much in comparison to peak torque RPM.Then why do this happen?

 

[miltos]

The answer is at my post above

 

[jack action]

But for many cars when you are close to peak hp RPM at launch,tire spins become too much in comparison to peak torque RPM.Then why do this happen?

RPM plays a role too. If an engine revs at 8000 rpm and is suddenly connected to a drivetrain at 0 rpm, one or more of three things must happen:

• The engine must stall;
• The clutch must slip;
• The tires must slip.

When slipping occurs, the greater the rpm difference, the greater the chance to be dealing with the lower kinetic friction coefficient instead of the higher static friction coefficient (Hence, promoting even more spinning).

If you are at a higher engine torque, and lower rpm, not only the difference in rpm is less and a higher friction coefficient might result, but if the car accelerate fast, that rpm difference will decrease fast. Furthermore, if the torque is higher, the clutch might also slip more, meaning the wheel rpm is even less, thus resulting in an even smaller rpm difference, that will fade even faster.