Solve the Flywheel-IVT Problem with This Simple Solution in 15 Seconds

[jimgram]

I’ve seen this problem phrased several ways and posted several times – a couple of times by me. But I have not seen an explanation or a solution that can be applied in a real-life situation.

The problem is simple: Given two flywheels coupled together by an infinitely-variable-transmission, with one flywheel at rest and the other at an initial angular velocity, calculate the ending velocities for both wheels when the transmission varies through 0 (geared neutral or infinity) to n in t(time) seconds.

The only known parameters are:
A. Flywheel_1 inertia = 1000*ft²*lb
B. Flywheel_2 inertia = 4000*ft²*lb
C. Initial angular velocity of FW1 = 10,000*rpm
D. Initial angular velocity of FW2 = 0
E. The transmission ratio (n) = 0...0.3 (I.E. 1/n = inf…..3.33)
F. The period change ratio: t = 15 sec

You can set n (ratio) as a function of t (time) this way: n(t)=t*n_e/t_e
Where n_e=ending ratio (.3) and t_e=ending time (15*s)

My problem has been that solving for ending velocities (or velocity[n]) based on the conservation of momentum the subsequent torque calculation will not be correct; when calculating based on the correct torque relationship (which cannot be anything except fw1_tor=fw2_tor*n ) then momentum is not conserved.

I know that momentum is a vector quantity and that torque is applied to the gear reducer, but even when doing the comparison based on small grounding mass (e.g. a space station) I still cannot make velocity, torque, and momentum to all come out correctly.

I suspect (hope) that I’m missing something relatively simple. Any insight will be greatly appreciated

 

[rcgldr]

Seems like conservation of momentum and energy should be enough to provide an answer.

 

[jimgram]

rcqldr: This is one of the first fundamental issues (questions?): Energy is a function of angular velocity squared. Momentum is a function of angular velocity not squared. So, if velocity is calculated to conserve momentum, then energy cannot be conserved. In fact, I've been to web sites of people who believe this must be a way to "create free energy".

But as I said, an even more fundamental issue is that when you calculate the change in velocities based on the obvious need to maintain equilibrium of torque, even momentum is not conserved.

 

[rcgldr]

After looking into this a bit more, an IVT model doesn't work. Instead consider the situation similar to a collision, except that the two flywheels share a common axis, and that a massless and elastic (zero loss) spring suddenly connects the two axles. During the transition, the spring gains energy, then returns the energy. Given the initial state (state #1), there is a second state where the spring has zero energy. I end up with this result for the 2 states:

State #1:
flywheel 1 = +10,000 rpm, flywheel 2 = 0 rpm.

State #2:
flywheel 1 = -6,000 rpm, flywheel 2 = +4,000 rpm.

It would be similar to an elastic collision where ball #1 collides with stationary ball #2 with 4x the mass. Ball #1 ends up moving backwards and ball #2 forwards at a slower speed.

 

[jimgram]

I think you have highlighted the problem. But there IS a transmission and there will be a result from changing the ratio over a period of time. And certainly, at least initially, it will not need to turn to a negative ratio. Also, I think your analogy was based on the requirement that the ending momentum equals the initial momentum. But couldn't you size the spring any way you wanted and choose a time period such that flywheel 1 slowed from its initial velocity to say 1/2 its initial velocity, and then apply the energy in the spring to accelerate fw 2?

Keep in mind that in the automotive industry, this exact system has been applied for a number of years, and so I'm certain that the modeling has been accomplished, however, I've been unable to locate any references.

 

[RonL]

I think you have highlighted the problem. But there IS a transmission and there will be a result from changing the ratio over a period of time. And certainly, at least initially, it will not need to turn to a negative ratio. Also, I think your analogy was based on the requirement that the ending momentum equals the initial momentum. But couldn't you size the spring any way you wanted and choose a time period such that flywheel 1 slowed from its initial velocity to say 1/2 its initial velocity, and then apply the energy in the spring to accelerate fw 2?

Keep in mind that in the automotive industry, this exact system has been applied for a number of years, and so I'm certain that the modeling has been accomplished, however, I've been unable to locate any references.

You might look at air compression and expansion as the transfer mechanism, it's has the ability to take in heat energy on the expanded end, that might offset much or all of the losses to bearing and windage friction.
Protect against heat loss on the high side, but encourage spontaneous (almost free) heat flow on the cold side.

Ron

 

[jimgram]

A very interesting idea Ron. I know of a compressed air car developed by a consortium in Luxenbourg and being mass-produced in India and also I know a bit about CAES - Compressed Air Energy Storage, but neither of these applications take advantage of the thermal dynamics of compressing and de-compressing. I don't understand how that concept would play into the transfer of rotational momentum?

 

[RonL]

A very interesting idea Ron. I know of a compressed air car developed by a consortium in Luxenbourg and being mass-produced in India and also I know a bit about CAES - Compressed Air Energy Storage, but neither of these applications take advantage of the thermal dynamics of compressing and de-compressing. I don't understand how that concept would play into the transfer of rotational momentum?

I'm poor with words, but I'll try. As for the air car, I think ultra high pressure is not needed and that action and reaction between high and low pressure can be almost as fast as generation and use of electricity between generator and motor. A closed loop of air, can give a high and low pressure if controlled by one or more air motors or turbines. This is where linear gas flow is transformed into rotational motion.
The push of high pressure and the pull of low pressure through the motors turning flywheels, is a bounce effect that I view the same as the bounce in a basket ball, be it a 6" dribble, a 4 foot high, or slam against the floor and the energy reacts as high as 20'.

If that can be clear in ones mind, then heat loss control and spontaneous flow absorption will become the defining energy value.

Insulate the hot side, encourage the cold side. Action/reaction between flywheels controls the pressure/heat balance.

Ron

 

[rcgldr]

Couldn't you size the spring any way you wanted and choose a time period such that flywheel 1 slowed from its initial velocity to say 1/2 its initial velocity, and then apply the energy in the spring to accelerate fw 2?

You'd need some sort of mechanism to allow either end of the spring to be connected or disconnected from either flywheel, and held in place when disconnected. In this case energy would be conserved, but not angular momentum, unless you take into account that the mechanism used to hold the spring in place would ultimately transfer that torque onto the earth, and this closed system composed of 2 flywheels, the spring, and the earth, angular momentum would be conserved, as well as energy.

If you were simply observing an always connected spring in action, then half way through the process, angular momemntum would be conserved, energy would be conserved, but much of that energy would be potential energy within the spring.

Another option would be to use a third flywheel, which then provides an option for "storing" excess energy or momentum during the transitions.

Getting back to the case of the IVT, wouldn't it generate a net torque force on what ever it was mounted to (eventually the earth) during transitions, which would mean that angular momentum of the 2 flywheels would not be preserved (since some torque would go into angular momentum of the earth)? The other option would be to mount the 2 flywheels offset and not sharing a common axis and IVT onto a third frictionless axle, allowing the entire system to rotate during transitions in order to conserve angular momentum.

 

[jimgram]

RonL - I think I get your idea and it would be very interesting to investigate further, but it doesn't help with the current path which is pretty fixed. rcgldr - I suspect you are correct regarding the lost momentum.

 

[RonL]

RonL - I think I get your idea and it would be very interesting to investigate further, but it doesn't help with the current path which is pretty fixed. rcgldr - I suspect you are correct regarding the lost momentum.

Pardon I guess is your fixed with a spring, just work in a couple of clutch bearings.

Ron

 

[Cleonis]

After looking into this a bit more, an IVT model doesn't work. Instead consider the situation similar to a collision, except that the two flywheels share a common axis, and that a massless and elastic (zero loss) spring suddenly connects the two axles.

I agree with jimgram that this highlights the problem.

Another version would be to posit a clutch mechanism. A perfect clutch would give instantaneous mechanical connection of the two flywheels. In the real world any clutch mechanism will involve slip. If a clutch mechanism is all you have then you must allow slip, otherwise the device will tear itself apart.


With an IVT and an infinite amount of time you can avoid slip altogether.

But of course in the case presented by jimgram available time is limited. I think that rcgldr has indeed pointed out that if available time is finite then the transfer mechanism must transiently store energy. (That would give rise to a torque mismatch.)

It seems to me that in the real world not all energy is recoverable, some of the energy will dissipate.

 

[jimgram]

I have found a root of omega fw1 that yields correct torque transfer and CoM, but cannot paste the text so have attached a pdf.

Can anyone confirm (or deny) that this may be a correct solution?

 

[Cleonis]

I have found a root of omega fw1 that yields correct torque transfer and CoM, but cannot paste the text so have attached a pdf.

Can anyone confirm (or deny) that this may be a correct solution?

What I can is reason in terms of general considerations.

I believe it is physically unrealistic to use a model in which the mechanical device (the variable transmission) produces a lossless transfer of torque. There is non-negligable friction, resulting in non-negligable loss of efficiency. I think it is physically impossible to transfer all of the torque.

(More specifically, I think transferring all of the torque is possible only if an infinite amount of time is available.)

 

[rcgldr]

If available time is finite then the transfer mechanism must transiently store energy. (That would give rise to a torque mismatch.)

Using the spring concept I mentioned before, there's no torque mismatch ... updating the IVT case (gearing and unequal torque issue) in my next post

 

[Cleonis]

Using the spring concept I mentioned before, there's no torque mismatch,

Yes, you are correct.

I erred; it's not about torque mismatch, it's whether absorbed energy can be recovered.

 

[rcgldr]

spring based device

Assuming a massless spring and instant transfer of torque impulses through the spring, the spring exerts equal and opposing torques to the 2 flywheels at all times, but has stored potential energy whenever the magnitude of those torques is non-zero. The torques are zero only at initial state #1, and the other state #2:

State #1: flywheel 1 = +10,000 rpm, flywheel 2 = 0 rpm.

State #2: flywheel 1 = -6,000 rpm, flywheel 2 = +4,000 rpm.

During the transition from state #1 to state #2, flywheel 1's momentum changes by -16000 rpm x 1000 ft² lb = -16000000 rpm ft² lb, while flywheel 2's momentum changes by +4000 rpm x 4000 ft² lb = +16000000 rpm ft² lb, so momentum is conserved. The energy is also conserved, = 5 x 10¹⁰ rpm² ft² lb.

If the transition occurs over a finite amount of time, and since torque = dL / dt (where L = angular momentum), and since angular momentum is conservred at all times, the spring is generating equal and opposing torques on both flywheels, and gaining or losing energy during this transitions corresponding to the magnitude of the torques involved at any moment in time. It's potential energy and the torques are zero only at state #1 or state #2.

IVT based device

The effective gearing of the IVT means that the opposing torques on the flywheels are not equal in magnitude, so there needs to be a third component of angular momentum in order to preserve angular momentum. Consider the case where the two flywheels are mounted at the ends of a massless rod in space. During the transition, the rod and the two connected flywheels will rotate, and angular momentum and energy of the system willl be conserved because there are no external torques, and no energy losses. The issue here is I don't see how energy related to the rotation of the rod can be extracted to end up at state #2.

 

[jimgram]

I would not disgree that there are losses do deal with (from several sources, in fact). However, IVT's do exist and they do provide ratio variations from 0:n wherein at zero no torque is transferred (I.E. torque = 1/n)(see attached pdf).

Nevertheless, one must find the root of the basic mechanical system in order to then apply the relevant losses.

I don't understand the comment that time must be infinite: I've found that time can be almost any desired period (e.g. n=0...n_mx in 15 seconds or 30, etc.). The issue is maximum torque which increases as the period decreases. Could you elaborate please?

 

[rcgldr]

... find the root of the basic mechanical system in order to then apply the relevant losses. I don't understand the comment that time must be infinite. The issue is maximum torque which increases as the period decreases.

I don't see a reason for the time to be infinite either. As far as torque goes with the idealized IVT, update - I'm not sure.

As I mentioned in my previous post, because the torques are not equal (except for one instantaneous moment during the transition), angular momentum between the two flywheels is not maintained, so there is a net torque that affects some third component of angular momentum. In my example, the two flywheels are mounted at the ends of a massless rod in space, and the rotation of the rod and flywheel system provides the third compenent of angular momentum, and it also consumes some of the angular energy.

 

[jimgram]

I have to take issue with the notion that either torque must be infinitely large: Initially, thanks to the IVT, flywheel 2 is de-coupled from flywheel 1, so there can be no torque generated at all. As the coupling ratio changes from zero to n, momentum transfers from flywheel 1 to flywheel 2, producing a torque that increases angular velocity of flywheel 2 and decreases angular velocity of flywheel 1 as a function of deltaL/t. At all times torque produced by flywheel 2 is reflected to flywheel 1 as tor_fw2*n=tor_fw1. Flywheel 1 does not need to stop if n varies only enough to transfer some amount of momentum from fw1 to fw2.

The torque applied to the gear reducer housing is equal to the torque of fw1 plus the torque of fw2 (generally in opposing directions dependent on the gearing - reversed direction or not). I have found that when this is done, all original momentum of fw 1 is conserved. It is not an objective or requirement that all of the original momentum is transferred.

 

[Cleonis]

The effective gearing of the IVT means that the opposing torques on the flywheels are not equal in magnitude, so there needs to be a third component of angular momentum in order to preserve angular momentum.

That leads to the following setup: an IVT (such as the torotrak IVT system) is mounted in a car.

During energy transfer between the flywheel and the driveshaft the car as a whole experiences a torque. Because of the weight of the car its four wheels stay on the ground. The angular momentum of the Earth/car/flywheel system is conserved. The angular momentum of the car/flywheel system is not conserved.

Now the following setup: two flywheels, the two of them mounted on a single bench. The two flywheels are connected by an IVT. The IVT allows you to transfer kinetic energy from one flywheel to another. You can transfer at will.

In addition: the bench is equipped with a sensor that registers whether a torque is exerted upon the bench. What happens if you add the constraint that at all times the torque-sensor of the bench must report zero? That imposes a serious limitation.

 

[jimgram]

In addition: the bench is equipped with a sensor that registers whether a torque is exerted upon the bench. What happens if you add the constraint that at all times the torque-sensor of the bench must report zero? That imposes a serious limitation.

I believe that would only be a constraint in a closed system such as a spacecraft - and then I see no way around it. But for automotive applications I don't see a problem.

I understand that there is torque on the housing and dL=tq/t and so momentum is conserved. This works in the math models. It's difficult to grasp the concept that since L=I*\omega there must be some value of I (the Earth?) and some value of omega but Ek is a function of \omega²*I the amount of energy is negligible and does not result in conservation of energy

 

[Cleonis]

But for automotive applications I don't see a problem.

OK.

This was your original question:

My problem has been that solving for ending velocities (or velocity[n]) based on the conservation of momentum the subsequent torque calculation will not be correct; when calculating based on the correct torque relationship (which cannot be anything except fw1_tor=fw2_tor*n ) then momentum is not conserved.

I believe rcgldr has completely clarified the problem.

 

[Cleonis]

I believe rcgldr has completely clarified the problem.

On second thoughts: I've been rereading your original post, and I realized I don't know your question with sufficient precision. You may have been too terse.

What has been happening in this thread is that two different kinds of setup have been discussed.
A. Two flywheels, free floating, connected through an IVT. (The mass of the IVT is negligable compared to the masses of the flywheels.)
B. A car fitted with a flywheel and an IVT, with kinetic energy transferred to and from the flywheel.

Those two cases are very different. So different that I think there is no merit in comparing them. One is not a stepping stone to the other.I surmise that your ultimate goal is to better understand what engineers must model if they want to equip a car with an IVT + flywheel system.
What you seek, I surmise, is a mathematical model for an IVT + flywheel system in a car.

In your original post you ask about a two-flywheels setup. In my opinion analysing a two-flywheels setup does not bring you closer to analysing a car-with-IVT-and-flywheel setup.

 

[rcgldr]

I updated my previous post to inidicate I'm not sure of the torques involved. In the IVT you'd have tension on the "driven" side of the belt and zero tension on the "return" side. The main issue is flywheel #2's initial state is 0 rpm, which in real life would require some type of clutch and/or spring mechanism. If the initial state had flywheel #2 rotating, even 1 rpm, then the initial state wouldn't be an issue.

The tension in the belt would be correspond to the torque / radius of IVT belt at each flywheel, and would be the equal and opposing at each end of the belt, a constraint on the torque versus radius at each end of the belt. The torque at each flywheel would correspond to the rate of angular acceleration (or deceleration) times angular inertia. Assuming that the IVT belt doesn't stretch or slip, then the rate of acceleration would be related to the rate of radius increase on flywheel #1 and rate of radius decrease on flywheel #2, with the constraint that tension in the IVT belt is equal and opposing at each end. Given the constraint that tension is the same at both ends which affects the rate of radius change at each end, I'm not sure if the belt length is naturally constant or if some type of tensioner on the return side would be needed to take up slack to allow belt path to vary.

The main point of my previous posts remains, since the torques on each flywheel aren't equal and opposing (except for the instant where the belt radius is the same on each end), angular momentum between the two flywheels is not conserved, and requires some third factor for angular momentum to be conserved, such as mounting the two flywheels at the end of a massless rod in space.

 

[Cleonis]

I updated my previous post to inidicate I'm not sure of the torques involved. In the IVT you'd have tension on the "driven" side of the belt and zero tension on the "return" side.

As an example of what is possible with CVT check out the http://www.torotrak.com/content/54/torotraks-variator.aspx" .

For simplicity I think it's best to assume that the IVT device has zero elasticity. In the particular case of Torotrak CVT the moving parts can slip relative to each other if the force is too big.

 

[rcgldr]

As an example of what is possible with CVT.

The issue I was trying to address was the flywheel and conservation of angular momentum abstract problem from the first post in this thread. In the case of a real CVT, the engine's torque is modulated by a combination of throttle input and reaction by the CVT to allow the engine to rev to it's peak power rpm at full throttle application. Conserving angular momentum isn't a design issue, only trying to minimize energy losses in the drive train.

 

[Cleonis]

What has been happening in this thread is that two different kinds of setup have been discussed.

Let me discuss differences between two particular setups.

1. Two free-floating flywheels, mechanically connected via an IVT. No (transient) storage of potential energy.

Take the moste symmetrical case: initially the two flywheels are counter-rotating, with the same angular velocity. The IVT can accommodate that. Next adjust the IVT to gradually decrease the velocity difference between the two flywheels. This depletes the kinetic energy stored in the system. The location of energy dissipation is the IVT.

Once the velocity difference between the two flywheels has been reduced to zero all the kinetic energy that could be dissipated is dissipated.

To calculate the transition from initial state to final state you have two options:
- Use conservation of angular momentum
- The two flywheels are doing work upon each other. Given the timeline of IVT ratio you can integrate the amount of work done, which should give you the end state of the process.

Notable points:
In this setup all you can do is decrease the relative angular velocity of the two flywheels, dissipating kinetic energy in the process



2. An IVT device is mounted on a bench, such that the IVT will withstand any torque; the IVT is immovable. To either side of the IVT a flywheel is mounted. No (transient) storage of potential energy.

In that setup you can transfer kinetic energy between the two flywheels. However, you can transfer kinetic energy only if the two torques that the flywheels exert upon the IVT reinforce each other.
If you make the mistake of setting up the IVT ratio in such a way that the torques exerted upon the IVT cancel each other then kinetic energy will be dissipated, not transferred.

To calculate the transition from initial state to final state you must integrate the work done.

 

[jimgram]

Those two cases are very different. So different that I think there is no merit in comparing them. One is not a stepping stone to the other.

I surmise that your ultimate goal is to better understand what engineers must model if they want to equip a car with an IVT + flywheel system.
What you seek, I surmise, is a mathematical model for an IVT + flywheel system in a car.

In your original post you ask about a two-flywheels setup. In my opinion analysing a two-flywheels setup does not bring you closer to analysing a car-with-IVT-and-flywheel setup.

I believe the transition from the driven "flywheel" (in other words, a driving tire) is thus:

Linear momentum p = vehicle mass ms times vehicle linear velocity v: (p=ms*v)
Linear momentum p can be converted to angular momentum (tire) as L_tr=p*r_tr, where r_tr is the radius of the tire.
Then the equivalent moment of inertia at the tire (ignoring all drive line components including the tire) is: I_tr = ms * r_tr².

Therefore beginning with the reflected momentum at the tire as being analogous to a "flywheel" should be correct

 

[jimgram]

rcgldr:cleonis: You have both been very helpful. Through this process I believe I now have an understanding of not only the factors that must be considered when evaluating based on conservation of momentum, but also the mechanics associated with the conservation of energy.

I did not intend to conceal the ultimate objective that this is a straight-forward application of use of a flywheel in the capture and restoration of braking kinetic energy in an automobile - rather I thought that the question was more easily understood by removing that conversion.

While this should be a rather simple application, the ultimate objective is to fully model a concept that is patent pending utilizing an epicyclic differential with two variable-inertia counter-rotating flywheels (real flywheels) supplying the two inputs to the differential: the inertia of one flywheel is increased (slowing it) and the other inertia is decreased (speeding it) thereby affecting an output. This function is analogous to the IVT discussed. Thank you again.