Engineering Applying Physics to Vehicle Dynamics: Do Wheel Number Matter?

[hutchphd]

`

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.

 

[Saalz]

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?

 

[erobz]

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.

 

[pbuk]

Firstly,



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

$2 \dfrac{T_{engine} - I * a}{r}$ [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?

 

[erobz]

Firstly,



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.

 

[erobz]

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.

 

[pbuk]

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.

 

[erobz]

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".

 

[Saalz]

@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} ##

 

[erobz]

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.

 

[erobz]

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: $\frac { 2 I \alpha}{r}$"

 

[jack action]

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).

If we know that Torque = I * a,

No, the torque is $I\alpha$ or $I\frac{a}{R}$.

 

[Saalz]

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?