Applying Physics to Vehicle Dynamics: Do Wheel Number Matter?

In summary, the conversation discusses the application of physics in modeling the dynamics of a vehicle. The individual is unsure about whether to consider the number of wheels in calculating the friction force and if the total mass of the vehicle should be used or if a single wheel's force is sufficient. The conversation also mentions the relationship between angular and linear acceleration and the need to consider the angular inertia of all wheels. The concept of combining forces from different wheels is also brought up for simplification. The conversation ends with a discussion about the work of frictional forces in this scenario.
  • #71
jack action said:
Where is the rolling resistance?
The model does not need "rolling resistance" the static friction is independent of rolling resistance. This is just basic physics...its not meant to examine tire deformations. It's a purely theoretical application of Work/Energy without resistive loss, or ##\sum F = ma## without resistive loss.
 
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  • #72
The front wheel has clearly a rolling resistance. There is a friction force that goes backward. Where is that force in your equation?
 
  • #73
jack action said:
The front wheel has clearly a rolling resistance. There is a friction force that goes backward. Where is that force in your equation?
Its not rolling resistance. Its static friction ##f_f##.

Front wheels:

$$ \circlearrowright^+ \sum \tau = R f_f = I \alpha = I \frac{a}{R} $$

The front wheels are not being driven by an active torque applied to the front axle, but they are none the less accelerating if the vehicle is accelerating ( angular ) and the no-slip condition applies. The static friction is responsible for the angular acceleration of the front wheels. Without it, they just slide without rotating.
 
  • #74
erobz said:
I want this to be over as much as anyone else. So I'm going to be as concise as possible about what I find confusing still:

$$\int F_{net} \, dx = \int 2 \left( f_r - f_f \right) dx = 0 $$

Will anyone explain this?
I see debate and understanding progress. :smile:
Could you explain that equation on your own words?
Sorry, I don't understand the function of fr and ff in it.
 
  • #75
erobz said:
Its not rolling resistance. Its static friction ##f_f#.
We have already shown that this static friction is a response to the rolling resistance.

erobz said:
The front wheels are not being driven by an active torque applied to the front axle
So you have a backward force applied to a wheel with no reaction torque? Nope. Something is reacting to that force. That front wheel torque wheel center force is ultimately coming from the rear wheel torque, passing through the body.

erobz said:
but they are none the less accelerating if the vehicle is accelerating ( angular ) and the no-slip condition applies.
Since the force is backward, it should contribute to decelerating, not only the front wheel itself but the whole vehicle, including the rear wheel.
 
  • #76
Lnewqban said:
I see debate and understanding progress. :smile:
Could you explain that equation on your own words?
Sorry, I don't understand the function of fr and ff in it.
##f_f## and ##f_r## are the static friction forces acting on the vehicle. They are the only forces acting on the vehicle externally in the direction of motion. Just like the example out of my old physics textbook that we examined earlier. Yet... if static friction cannot do work, by definition that integral must be ##0##.
 
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  • #77
jack action said:
We have already shown that this static friction is a response to the rolling resistance.
Its a steel wheeled car. virtually undeformed steel wheels on a steel track. rolling resistance is nill.

Since the force is backward, it should contribute to decelerating, not only the front wheel itself but the whole vehicle, including the rear wheel.

Your right. It does in the following ##\sum F## equation:
erobz said:
Total Vehicle:

$$ \rightarrow^+\sum F = 2 \left( f_r - f_f \right) = Ma $$
 
  • #78
erobz said:
##f_f## and ##f_r## are the static friction forces acting on the vehicle. They are the only forces acting on the vehicle externally in the direction of motion. Just like the example out of my old physics textbook that we examined earlier. Yet... if static friction cannot do work, by definition that integral must be ##0##.
Thank you very much, erobz. :smile:

Just like it happens in the picture shown in that book that you posted earlier, using the magnitudes of those forces at the contact patches, which are relatively easy to calculate, is a very convenient and traditional way to compute the work in or out a vehicle.

As you are an intelligent person trying to understand the apparent conflict of the equation related to that way, I am simply trying to reason that, by definition of work (force times distance), those static forces themselves are unable to perform any work (add to or subtract from the car any kinetic energy).

Nevertheless, the magnitudes of those forces are a direct reflection of the actual physical processes happening inside and outside that vehicle (heat, expansion of gases, transmission reduction, torque at shaft, pad friction on brake rotors, etc.).

Mathematically, there is no contradiction in that equation.
IMHO, physically, there is.
 
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  • #79
Lnewqban said:
Thank you very much, erobz. :smile:

Just like it happens in the picture shown in that book that you posted earlier, using the magnitudes of those forces at the contact patches, which are relatively easy to calculate, is a very convenient and traditional way to compute the work in or out a vehicle.

As you are an intelligent person trying to understand the apparent conflict of the equation related to that way, I am simply trying to reason that, by definition of work (force times distance), those static forces themselves are unable to perform any work (add to or subtract from the car any kinetic energy).

Nevertheless, the magnitudes of those forces are a direct reflection of the actual physical processes happening inside and outside that vehicle (heat, expansion of gases, transmission reduction, torque at shaft, pad friction on brake rotors, etc.).

Mathematically, there is no contradiction in that equation.
IMHO, physically, there is.
Thank you for being honest. Something is strange here in that model and we aren’t quite sure what. That is how I’m interpreting your opinion. I hope I'm not projecting.
 
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  • #80
erobz said:
Its a steel wheeled car. virtually undeformed steel wheels on a steel track. rolling resistance is nill.
Steel wheels do deform. There is no rolling possible without deformation. The simple fact that you have the static friction present means that it was created by some other opposing force or torque.
erobz said:
Your right. It does in the following equation:
Yes, but you are treating each ##a## in your equations as if they were independent of each other. They are not. ##a## is the acceleration of the vehicle but also the acceleration of each of your wheels. Furthermore, ##a =\alpha R## where ##\alpha## is the angular acceleration of your front and rear wheels. (They could have different angular acceleration if they had different radiuses.)

This means that if a wheel decelerates (both longitudinally and angularly), the other wheels must also decelerate, and the vehicle as a whole must decelerate. How is that represented in your equations in post #10?

Here is the problem with your 3 equations in post #10. There are 2 ways to look at it:
  1. Mine. ##f_f## is known. This means that the only unknows are ##a## and ##f_r##. Every other variable must be known. So you have 3 equations with 2 unknowns, so one of your equations is redundant. But we know that they all have pertinent information not available in the other equations. That cannot give a good result. You need more equations to solve this (post #28).
  2. Yours. ##f_f## is unknown. Then you have 3 equations with 3 unknowns, namely ##a##, ##f_r##, and ##f_f##. If you solve your system of equations in post #10, I end up:
$$f_r = \frac{T}{R}\left(1- \frac{I}{\frac{1}{2}MR^2 + 2I}\right)$$
$$f_f = \frac{T}{R}\frac{\frac{I}{R^2}}{\frac{1}{2}M + \frac{2I}{R^2}}$$
$$a= \frac{\frac{T}{R}}{\frac{1}{2}M + \frac{2I}{R^2}}$$

Doesn't that look weird that the front static force ##f_f## depends on the wheel inertia? What law of physics is implied here? If I assume ##I = 0##, then it disappears:
$$f_r = \frac{T}{R}$$
$$f_f = 0$$
$$a= \frac{2T}{MR}$$
And something is clearly wrong with the factor of ##2## with the acceleration.
 
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  • #81
jack action said:
Steel wheels do deform. There is no rolling possible without deformation. The simple fact that you have the static friction present means that it was created by some other opposing force or torque.

Yes, but you are treating each ##a## in your equations as if they were independent of each other. They are not. ##a## is the acceleration of the vehicle but also the acceleration of each of your wheels. Furthermore, ##a =\alpha R## where ##\alpha## is the angular acceleration of your front and rear wheels. (They could have different angular acceleration if they had different radiuses.)

This means that if a wheel decelerates (both longitudinally and angularly), the other wheels must also decelerate, and the vehicle as a whole must decelerate. How is that represented in your equations in post #10?

Here is the problem with your 3 equations in post #10. There are 2 ways to look at it:
  1. Mine. ##f_f## is known. This means that the only unknows are ##a## and ##f_r##. Every other variable must be known. So you have 3 equations with 2 unknowns, so one of your equations is redundant. But we know that they all have pertinent information not available in the other equations. That cannot give a good result. You need more equations to solve this (post #28).
  2. Yours. ##f_f## is unknown. Then you have 3 equations with 3 unknowns, namely ##a##, ##f_r##, and ##f_f##. If you solve your system of equations in post #10, I end up:
$$f_r = \frac{T}{R}\left(1- \frac{I}{\frac{1}{2}MR^2 + 2I}\right)$$
$$f_f = \frac{T}{R}\frac{\frac{I}{R^2}}{\frac{1}{2}M + \frac{2I}{R^2}}$$
$$a= \frac{\frac{T}{R}}{\frac{1}{2}M + \frac{2I}{R^2}}$$

Doesn't that look weird that the front static force ##f_f## depends on the wheel inertia? What law of physics is implied here? If I assume ##I = 0##, then it disappears:
$$f_r = \frac{T}{R}$$
$$f_f = 0$$
$$a= \frac{2T}{MR}$$
And something is clearly wrong with the factor of ##2## with the acceleration.
None of this is unexpected with the model.

##f_f## and ##f_r## are a reaction force to the acceleration. The underlying no-slip criterion here is simply ##f_f < \mu_s N_f## and ##f_r < \mu_s N_r##

If you let ##I \to 0## the wheels ( front or back ) provide no resistance to the vehicle's acceleration. A wheel without inertia takes no torque to accelerate.

The ##2## comes from the last equation. The torque acting on each wheel causes the reaction ##2 f_r##. e.g. there are ##2 f_r## forces accelerating the car forward.

That is all internally consistent.

What is not internally consistent is the work done by the static forces should equal zero, but...

$$ \int F_{net} dx = \int 2 \left( f_r - f_f \right) dx \neq 0 $$
 
  • #82
When you ( @erobz ) originally hijacked the post in #8 your question was whether the static friction forces did work on the car
erobz said:
Does static friction force do work in this case? It seems like the frictional forces must be doing work, but it often seems like this is "explained away" in some way I still don't fully understand.
and the answer was no. There is no net energy flowing from the contact. When there are contact forces between two bodies we only worry about net work done.
This is equivalent to asking how much work Elmer's glue does. The static friction "force" is merely a placeholder TBD by the rest of the problem. It is useful because it is usually accompanied by a limit at which the problem has a nonlinear response. It does no work itself because it is static by definition. It is unfortunate to call it a friction...I prefer stiction. In this problem essential work is done by the engine and dissipated by various rotational and linear dissipative friction. So there is a net force on the car which you can call static friction if you must but the Work is more complicated as @jack action has repeatedly noted.
In the canon of introductory physics we should ban "static friction" and revise the whole notion of "dissipative forces"
 
  • #83
hutchphd said:
In the canon of introductory physics we should ban "static friction" and revise the whole notion of "dissipative forces"
Yeah, it's not my fault I discovered the math doesn't jive the way it is typically presented. I'm taking a beating here defending my sanity.
 
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  • #84
I stopped worrying about the sanity part long ago...makes it easier
 
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  • #85
hutchphd said:
I stopped worrying about the sanity part long ago...makes it easier
Is it too soon to ask if I can be mentioned in a footnote somewhere when the theory is revised! :biggrin:

"Stubborn [insert expletive] challenges us to revise something no one cares about anymore" could be a good title of an insight?
 
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  • #86
Also sanity is grossly overrated !
When I was teaching it I thought about some revisions but nothing ever "jelled". Personally I just revert to more fundamental stuff when necessary. Energy and momentum are conserved although some people find further subdivision important. Carnot did do a few useful things after all
 
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  • #87
Rewinding back to post #8:

erobz said:
Does static friction force do work in this case? It seems like the frictional forces must be doing work, but it often seems like this is "explained away" in some way I still don't fully understand.

If I'm doing a force balance to solve the OP's problem, there are frictional forces acting as external forces on the vehicle. The rear frictional forces assisting acceleration and the front tires frictional forces opposing it. These forces must be doing the work...they are the only external forces (in the direction of motion). But I so often hear about static frictional forces not doing work because the point of contact between the tire and the road is instantaneously at rest w.r.t the road. Am I still confused, because I feel a bit unsure of my understanding?

...
But then again...in both cases (accelerating/not accelerating) the point of contact is instantaneously at rest w.r.t. the ground. So, I still am unsure what the actual resolution is! It all seems like a bit of a circular argument. It feels like whether or not the static frictional force does work or not is whether or not the body is accelerating, and whether or not the body is accelerating is whether or not the static frictional forces are doing work.

In the absence of external forces doing the work of accelerating the car, couldn't that energy come from an internal or onboard source, like accumulated potential energy in a spring, or compressed air, or steam, or battery?
 
  • #88
Lnewqban said:
Rewinding back to post #8:
In the absence of external forces doing the work of accelerating the car, couldn't that energy come from an internal or onboard source, like accumulated potential energy in a spring, or compressed air, or steam, or battery?
I don't think, because a body can't apply a net force to itself.
 
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  • #89
But that has been the principle of all automobiles, trucks, ships, locomotives, airplanes, old and new: to move at the expense of accumulated work or energy that is contained inside it.

That energy can be the chemical energy of fuel or batteries, converted to useful mechanical work via an engine or motor.
Is that energy (and its resulting internal output torque) the one doing the useful work.

These are the guts of a Tesla model S.
That grey plank is a battery capable of accumulating 100 Kw-h of energy.
The red motors can transfer enough of that energy onto the tires in 3.0 seconds as to accelerate the car's mass of 2,000 kg from 0 to 100 km/h.

2021-tesla-model-s_100777914_m.jpg
 
  • #90
Lnewqban said:
But that has been the principle of all automobiles, trucks, ships, locomotives, airplanes, old and new: to move at the expense of accumulated work or energy that is contained inside it.

That energy can be the chemical energy of fuel or batteries, converted to useful mechanical work via an engine or motor.
Is that energy (and its resulting internal output torque) the one doing the useful work.

These are the guts of a Tesla model S.
That grey plank is a battery capable of accumulating 100 Kw-h of energy.
The red motors can transfer enough of that energy onto the tires in 3.0 seconds as to accelerate the car's mass of 2,000 kg from 0 to 100 km/h.

View attachment 314662
That’s all fine. They just need something to “push off” (or mass to eject) to accelerate their own center of mass.
I just noticed an inconsistent model, which either is indication of a bad static force model, or we accept the fact that we can’t really do much if we have don’t have anything to interact with.

Perhaps just accepting that duality is the simplest resolution. Just treat this as looking at it from the universes perspective, instead of the “thing” in it. After all ## \Delta U_{universe}=0## . The static friction force does work on the universe. That has to be the case. And that’s actually what’s being calculated here. Maybe static friction is saved…from nothing. Because nothing was going so change anyhow.

Anyhow, I expect to be burned at the stake the more I talk! I’m going to bed.
 
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  • #91
`
erobz said:
Anyhow, I expect to be burned at the stake the more I talk! I’m going to bed.
Good choice. You are perseverating on a problem that is of your own making. In any object moving relative to me, by your your reckoning, each atom is doing work on its neighbor, and the neighbor is reflecting that work back. You can make a consistent argument (although this definition of work is not Galilean invariant) but it is not useful, nor is it the definition we've agreed upon.
 
  • #92
erobz said:
If there is no-slip between the tire and the road there is a relation between the angular acceleration of the wheel and the linear acceleration of the vehicle that must be held.

Then you get a system of equations using the following laws (one for a driven wheel, one for a non-driven wheel, and one for the entire vehicle):

$$ \sum \tau = I \alpha $$

$$ \sum F = m a $$

The last equation contains elements of the other two which you will use to solve for ##a##.

Hello mate.

I was out of town for a couple of days and just come back to continue with my work.

I I'd correctly understood everything, the first equation gives me the torque with which the inertia of the wheel is resisting the change in motion. That torque divided by the length of the arm (radius of the wheel), gives me the resistive force. The difference would be that in the driven wheels, there's a source providing a greater torque, the result of the difference being a force both greater and contrary in direction to the resistive force of the non-driven wheels.

Which leads to the third and last. I'd simply have to subtract the wheel/surface forces (one for each driven wheel) that propells the vehicle forward by all the forces contrary to the movement, this being the resistive force from the inertia of each non-driven wheel, and the drag forces acting on the vehicle (Aerodynamic resistance and rolling resistance)

Something like this:2 * ( ( T_engine - ( I * a) ) / r ) - 2 * ( (I * a) / r ) - ( μ * m_vehicle * g ) - ( ( c_d * A_frontal * ρ_air * v^2 ) / 2 ) = m_vehicle * a

( T_engine - ( I * a) ) / r -> propelling force from the traction wheels

(I * a) / r -> resistive force from the inertia of the non driven wheels

μ * m_vehicle * g -> would be the rolling resistance of the vehicle

( c_d * A_frontal * ρ_air * v^2 ) / 2 -> would be the aerodynamic drag force affecting the vehicleThat equation would give me the linear acceleration of the whole vehicle.

Would this be a valid resolution?
 
  • #93
hutchphd said:
Good choice. You are perseverating on a problem that is of your own making.
Yeah, in an act of remorse I was hoping a shift in perspective could save it. The conservation of momentum ideas brought up by lnewqban in the last couple posts just made it pop into my head.
 
  • #94
Firstly,

:welcome:

Perhaps you'd like to pop over to https://www.physicsforums.com/forums/new-member-introductions.240/ and say hello.

Secondly, I'm sorry your thread has been hijacked with a lot of distracting nonsense: PhysicsForums is usually better than this but, hey, its the internet.

Finally, in
Saalz said:
## 2 \dfrac{T_{engine} - I * a}{r} ## [edit: I made that look a bit prettier with ## \LaTeX ##]
You shouldn't be doubling up the torque from the engine because you have two driving wheels. You can see this easily from a thought experiment: bring the two wheels closer together until they touch, then glue them together so they are just one wheel. When would the force halve?
 
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  • #95
pbuk said:
Firstly,

:welcome:

Perhaps you'd like to pop over to https://www.physicsforums.com/forums/new-member-introductions.240/ and say hello.

Secondly, I'm sorry your thread has been hijacked with a lot of distracting nonsense: PhysicsForums is usually better than this but, hey, its the internet.

Finally, in

You shouldn't be doubling up the torque from the engine because you have two driving wheels. You can see this easily from a thought experiment: bring the two wheels closer together until they touch, then glue them together so they are just one wheel. When would the force halve?
Thats a good observation that was missed. As you can see, it's also a good place where we can come, challenge ourselves and learn from our mistakes.

That being said, I believe the correction is to bring that factor of ##2## inside and only apply it to the ##I \frac{a}{R}## term.

I think that first term should actually be:

$$\frac{ TR -2Ia}{R^2}$$

and likewise the next term should be:

$$ \frac{2Ia}{R^2}$$

@Saalz seems to be missing that factor of ##\frac{1}{R}## in each of the terms.
 
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  • #96
jack action said:
And something is clearly wrong with the factor of 2 with the acceleration.
Sorry to @jack action. That factor of ##2## was indeed a blunder in the original equations of motion.
 
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  • #97
erobz said:
That being said, I believe the correction is to bring that factor of ##2## inside and only apply it to the ##I \frac{a}{R}## term.
But the OP was not asking about losses due to rolling resistance or the moment of intertia of the wheels, the only question he was asking was "do I accelerate twice as fast with two wheels than with one". The answer to this is clearly "provided there is enough friction to prevent the wheels from slipping, no".

Once this had been resolved then perhaps the other complications can be introduced, but the 90 posts that it seems to have taken to discuss those should come AFTER the basic misunderstanding has been resolved.
 
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  • #98
pbuk said:
But the OP was not asking about losses due to rolling resistance or the moment of intertia of the wheels, the only question he was asking was "do I accelerate twice as fast with two wheels than with one". The answer to this is clearly "provided there is enough friction to prevent the wheels from slipping, no".

Once this had been resolved then perhaps the other complications can be introduced, but the 90 posts that it seems to have taken to discuss those should come AFTER the basic misunderstanding has been resolved.
Fair enough. I felt like we were going to get there eventually (based on past experience with this type of question), so I started that journey prematurely. Also, I'm pleading guilty of being greedy in the first degree, I want to know things too. And indeed, we did go down a rabbit hole into wonderland a bit. However, what we ended up discussing arose directly from this problem...and several of us did finally come to agreement that there appears to be an issue at its heart.

In the future I will refrain from "jumping the gun".
 
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  • #99
erobz said:
@Saalz seems to be missing that factor of in each of the terms.
I was indeed missing the 1 / R factor, but I can't figure out the meaning, tho.

If we know that Torque = I * a, and so, that Force = Torque / r, then the why woldn't it be correct for the first term to be ## \frac {T \m 2 I a} {r}, ##

And the second: ## \frac { 2 I a} {r} ##
 
  • #100
Saalz said:
I was indeed missing the 1 / R factor, but I can't figure out the meaning, tho.

If we know that Torque = I * a, and so, that Force = Torque / r, then the why woldn't it be correct for the first term to be $$ \frac {T \m 2 I a} {r] $$

And the second: $$ \frac { 2 I a} {r] $$
Don't use symbols in latex. I don't believe they parse. There is code ## \pi ## = \pi ## \alpha## = \alpha etc...

You have brackets and braces mixed together. Should be All braces there.

I think you actually have ## \alpha ## in your equations (looking more closely). If that is the case the factor of ##\frac{1}{R}## won't be in there yet.

That comes after substituting the no-slip condition ##a = R \alpha##

So, maybe not a problem. But that is why we like Latex here.
 
  • #101
I fixed up the Latex there. Just reply to this message to see what has changed.

"If we know that Torque = I * a, and so, that Force = Torque / r, then the why wouldn't it be correct for the first term to be ## \frac {T 2 I \alpha}{r} ##

And the second: ## \frac { 2 I \alpha}{r}##"
 
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  • #102
Saalz said:
2 * ( ( T_engine - ( I * a) ) / r ) - 2 * ( (I * a) / r ) - ( μ * m_vehicle * g ) - ( ( c_d * A_frontal * ρ_air * v^2 ) / 2 ) = m_vehicle * a
I strongly suggest using a simplified version with the equivalent mass (##m_e##) concept where:
$$F_{net} = m_e a$$
$$F_w - F_r - F_d = m_e a$$
$$F_w - C_{rr} mg - \frac{1}{2}\rho C_dAv^2 = m \left(1+\frac{4I}{mR^2}\right) a$$
$$a = \frac{\frac{F_w}{m} - C_{rr} g - \frac{1}{2}\rho\frac{C_dA}{m}v^2}{1+\frac{4I}{mR^2}}$$
Note that I use a coefficient of rolling resistance ##C_{rr}## instead of ##\mu## which usually represents the friction coefficient (not the same thing).

The force ##F_w## (sum from all driven wheels) is either the lesser between the maximum friction force available on all wheels (the normal force acting on each wheel is important here) OR the total power available over the speed of the vehicle (##F_w = \frac{P}{v}##), which is most likely. Using the total vehicle power available is less confusing than trying to identify how the torque is modified and split through the drivetrain.

The term in parenthesis for the equivalent mass represents the combined inertial effect of the vehicle mass ##m## and rotational inertia ##I## of 4 wheels. this term can be modified to include other rotational components from the brake system or drivetrain (more info here).

Saalz said:
If we know that Torque = I * a,
No, the torque is ##I\alpha## or ##I\frac{a}{R}##.
 
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  • #103
jack action said:
I strongly suggest using a simplified version with the equivalent mass (##m_e##) concept where:
$$F_{net} = m_e a$$
$$F_w - F_r - F_d = m_e a$$
$$F_w - C_{rr} mg - \frac{1}{2}\rho C_dAv^2 = m \left(1+\frac{4I}{mR^2}\right) a$$
$$a = \frac{\frac{F_w}{m} - C_{rr} g - \frac{1}{2}\rho\frac{C_dA}{m}v^2}{1+\frac{4I}{mR^2}}$$
Note that I use a coefficient of rolling resistance ##C_{rr}## instead of ##\mu## which usually represents the friction coefficient (not the same thing).

The force ##F_w## (sum from all driven wheels) is either the lesser between the maximum friction force available on all wheels (the normal force acting on each wheel is important here) OR the total power available over the speed of the vehicle (##F_w = \frac{P}{v}##), which is most likely. Using the total vehicle power available is less confusing than trying to identify how the torque is modified and split through the drivetrain.

The term in parenthesis for the equivalent mass represents the combined inertial effect of the vehicle mass ##m## and rotational inertia ##I## of 4 wheels. this term can be modified to include other rotational components from the brake system or drivetrain (more info here).No, the torque is ##I\alpha## or ##I\frac{a}{R}##.

Hi Jack,

I'm willing to use torque, to be fair. I've got the electric motor modeled. I have the torque curve across the whole RPM range of the electric motor that propels the vehicle and the drivetrain ratio, so I know the total torque available at the driven wheels.

In that case, I'll guess I could use the F_w in your expression as the total wheel torque available at a speed v, divided by the radius R of the wheels ( F_w = Torque / R ), right?
 
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