Flywheel-IVT Problem

  1. 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*ne/te
    Where ne=ending ratio (.3) and te=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 fw1tor=fw2tor*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
  2. jcsd
  3. rcgldr

    rcgldr 7,691
    Homework Helper

    Seems like conservation of momentum and energy should be enough to provide an answer.
  4. 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.
  5. rcgldr

    rcgldr 7,691
    Homework Helper

    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.
    Last edited: Jan 4, 2011
  6. 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.
  7. RonL

    RonL 901
    Gold Member

    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.

  8. 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?
  9. RonL

    RonL 901
    Gold Member

    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 absorbtion will become the defining energy value.

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

  10. rcgldr

    rcgldr 7,691
    Homework Helper

    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.
    Last edited: Jan 4, 2011
  11. 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.
  12. RonL

    RonL 901
    Gold Member

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

  13. Cleonis

    Cleonis 691
    Gold Member

    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.
  14. 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?

    Attached Files:

  15. Cleonis

    Cleonis 691
    Gold Member

    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.)
  16. rcgldr

    rcgldr 7,691
    Homework Helper

    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
    Last edited: Jan 5, 2011
  17. Cleonis

    Cleonis 691
    Gold Member

    Yes, you are correct.

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

    rcgldr 7,691
    Homework Helper

    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.
    Last edited: Jan 5, 2011
  19. 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...nmx in 15 seconds or 30, etc.). The issue is maximum torque which increases as the period decreases. Could you elaborate please?

    Attached Files:

  20. rcgldr

    rcgldr 7,691
    Homework Helper

    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.
    Last edited: Jan 6, 2011
  21. 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 torfw2*n=torfw1. 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.
Know someone interested in this topic? Share this thead via email, Google+, Twitter, or Facebook

Have something to add?