# Re: [Vo]:Successful Mechanical OU

```LOL the quote you're referring to is expressly a calculation of the energy
efficiency of a hypothetical fully-asymmetric distribution of momentum, ie.
an effective N3 symmetry break, in the context of Bessler's 'toys page'.
The purpose is to illustrate the decoupling of input to output energies as
they evolve as a function of accumulating this notionally 'reactionless
momentum'.  Like i say, this EXAMPLE sequence rises in 25% steps, reaching
unity at 4, and 125% OU at five.  Thus, items 'A' and 'B' on the toys page
are consistent with an allusion to this form of symmetry break.. if not its
means.```
```

Had you actually read what i'm saying more carefully, you'd note that i'm
actually claiming a much simpler process for achieving the same end,
WITHOUT having to perform 5 discrete reactionless accelerations to reach
125% of unity.  Instead, i'm claiming 190% in a single interaction, in a
single second.  So, even more worserer, bashically...

I absolutely encourage you to keep having fun working out why it's not
possible, and just how confounding any exception would be... but do keep in
mind that i'm going to show you something that shatters such certainties..

On Mon, Jun 4, 2018 at 5:25 PM, John Shop <quack...@outlook.com> wrote:

> On 4/06/2018 11:19 PM, Vibrator ! wrote:
>
> .  .  .
> The only precondition there is that we can apply a force between two
> inertias, which nonetheless only accelerates one of them.
>
> This I suggest is your problem.  If you apply a force between two masses
> or inertias, then one must accelerate in the opposite direction to the
> other (Newton's first law).  If one of them is massive enough (eg make it
> the earth), then only the light one is accelerated by any measurable amount
> (but the tiny acceleration of the heavy one ensures that momentum is
> conserved).
>
> You could apply a force between two equal inertias so that one accelerates
> forward and the other accelerates backwards, and then bounce one of them
> off a wall fixed to the earth say.  Now you would have them both moving in
> the same direction and with the same speed.  But their total kinetic energy
> would be equal to that put in during the acceleration phase (the bounce
> being elastic and conservative).  So each would contain say 0.5 joules of
> energy for a total of one joule put in by the initial acceleration
> impulse.  Let's call this square one.
>
> At this stage you could then apply the same accelerating impulse as the
> first time between the two inertias (which are now both travelling along
> together) and the speed of one would double, while the other would become
> stationary.  Here the kinetic energy of one has gone up by a factor of 4
> (due to v^2) to become 2 joules while the energy of the other has gone down
> to zero - the total being the 2 joules that have been put in by the two
> accelerations (so no gain).  Call this square two.
>
> Then we inelastically collide them (as by a length of string being pulled
> taut), equalising their velocity, and keep repeating that process, whilst
> monitoring input / output efficiency (how much energy we've spent vs how
> much we have).
>
> As you note, inelastic collisions waste kinetic energy by turning it into
> heat.  So joining the stationary mass to the travelling mass inelastically
> with a piece of string will produce a combined speed which is just the same
> as the speed of both masses before applying the second impulse (from
> conservation of momentum).  So the entire effect of the second impulse will
> have been undone taking us back to square one.
>
> I see no way to progress beyond square two that does not simply take us
> back to square one?
>
```