[Professor Mega-Mind (OP)]

There are none so blind ...blah , blah , blah...!
By the way , does your math whiz agree with you on that !?
P.M.

[Bored chemist]

And I pointed out in your other, spurious, thread why this doesn't work

does your math whiz agree with you on that !?

What do you think that means?

[Professor Mega-Mind (OP)]

It means have the math person analyse the K. Benz flat engine , and see if there are any overall sideways forces from it .
P.

[Kryptid]

It means have the math person analyse the K. Benz flat engine , and see if there are any overall sideways forces from it .
P.

Who is "the math person"?

[Bored chemist]

any overall sideways forces from it .

Why would that matter anyway?

[Kryptid]

I admit that I had to think about this for a while, but here is my analysis of your most recent design. I'm going to make a couple of changes to make things easier to analyze while still retaining the essence of the machine.

We start off with a launcher mounted in the middle of the box with two guns. Each gun can shoot a rubber ball and is wired to fire both of them simultaneously. The right wall of the box is made of plain metal while the left wall is metal coated with a sticky substance. The mass of each rubber ball is equal to the mass of the rest of the assembly. Now we start the experiment.

Step 1: The launcher fires each rubber ball at the two opposing walls. As you say, the launcher (and therefore the box itself) does not move because each ball counteracts the force of the other.

Step 2: Each ball makes contact with their respective walls, causing each ball to compress and stretching the box itself slightly due to the force that each ball imparts on its own wall (every material has finite elasticity). The box still does not move because the force on each wall is exactly equal (each ball has the same mass and velocity upon impact).

Step 3: The box has finished stretching and now begins contracting, pushing each compressed rubber ball back towards the middle. The box as a whole still does not move as both balls are still in contact with the walls.

Step 4: The box finishes contracting, launching the ball from the right wall back towards the middle of the box. This is where a difference finally arises. The ball on the left wall tries to leave at the exact same time as the one on the right wall, but cannot as it is stuck to the wall. This causes this previously compressed ball to stretch out and away from the wall, pulling the wall along with it.

Now, if the ball on the right had also stuck to its wall, then it would pull on the right wall with equal force to that of the left wall and the box would not move. However, there is no ball on the right wall to counter the pulling force of the ball on the left. This, against your prediction that the box should move to the left, causes the box to move to the right instead. Since this ball is of equal mass and velocity to its partner, it moves the box to the right with the same amount of force that the free ball would have if it had been the only ball in the system.

Step 5: The box is drifting to the right and the free ball to the left. When the free ball makes contact with the sticky wall, it sticks and exactly cancels out the momentum that the other ball had imparted on the system. There is thus no net change in momentum for the system as a whole.

Bored Chemist, tell me if you see any problems with this as I may have made some mistakes.

[Bored chemist]

Bored Chemist, tell me if you see any problems with this

I predict one problem.
The OP will talk balderdash about a sufficiently rigid box not deflecting.
This will not actually make a difference, but that won't stop him.

[Professor Mega-Mind (OP)]

Wah-haha-haha-haha ! 
Good one , that's not my system though .  Massive Wall physics automatically implies heavy , extremely stiff walls .  I use steel because of this .  Your example involved entirely different material dynamics than mine .  I could , however , bastardise the two , just to make a point . 
Example : The thrower throws both balls at the opposite walls .  The RH ball hits the "sand-filled" wall , and imparts %50 of it's kinetic energy to it , then bounces back slowly .  The LH ball hits the steel wall .  It impart %5 of it's kinetic energy to the wall , then bounces back powerfully .  It continues across the room , then strikes the sand-filled wall , depositing %50 of it's remaining kinetic energy into the wall .  The sand-filled wall absorbed most of the energy that the thrower expended .  Half of that was converted to kinetic energy of the box , half to waste heat .
Any doubt as to which way the box drifts ?
Hold on... , I keep hearing the theme to "Mr. Ed" playing nearby !
Anyhoo , "Khan , I'm laughing at...".
P.M.

[Professor Mega-Mind (OP)]

For further clarification of this R.D., go to thread "Reactionless Drives Possible ?" .
P.M.

[Bored chemist]

Wah-haha-haha-haha ! 
Good one , that's not my system though .  Massive Wall physics automatically implies heavy , extremely stiff walls .  I use steel because of this .  Your example involved entirely different material dynamics than mine .  I could , however , bastardise the two , just to make a point . 
Example : The thrower throws both balls at the opposite walls .  The RH ball hits the "sand-filled" wall , and imparts %50 of it's kinetic energy to it , then bounces back slowly .  The LH ball hits the steel wall .  It impart %5 of it's kinetic energy to the wall , then bounces back powerfully .  It continues across the room , then strikes the sand-filled wall , depositing %50 of it's remaining kinetic energy into the wall .  The sand-filled wall absorbed most of the energy that the thrower expended .  Half of that was converted to kinetic energy of the box , half to waste heat .
Any doubt as to which way the box drifts ?
Hold on... , I keep hearing the theme to "Mr. Ed" playing nearby !
Anyhoo , "Khan , I'm laughing at...".
P.M.

You keep forgetting about the momentum transferred when you throw things.
Sure, you can cancel it by throwing wto things at once, but then you cancel the impacts on the walls too.
I explained that in the other thread you started for no good reason.

I keep asking you to show the maths.
I gave you an example of how to do it.
Are you scared to do it, or what?
Nobody on a science web page is going to take you seriously until you have calculations that back up your point.

[Professor Mega-Mind (OP)]

Math is just quantitative description , not concept .  I enjoy ideas , systems , and designs .  I'm not a big number cruncher by choice .  I may bother here , but you should have gotten this concept long ago .  For instance ; the opposing impacts are NOT the same , ergo , their effect on the box is NOT the same . Saying %2vs%55 is the same as saying 2,000 ergs vs 55,000 ergs .  As I said , I may bore myself later .  Busy now .
P.M.

[Bored chemist]

may bother here , but you should have gotten this concept long ago

We get the concept.
It's just that, when you do the maths, it does not work.
And that's why you are wrong to say this

Math is just quantitative description

Speaking of concepts, do you know the difference between energy and momentum?

[Professor Mega-Mind (OP)]

The problem you're having is that you are not seperating out the molecule momentum in the final product .  This "heat loss" reduces the amount left for the objects involved in the inelastic collisions .  So , yeah , momentum IS conserved , yet momentum imparted is different than momentum transferred .
Alright , still busy !
P.M.

[Bored chemist]

The problem you're having is that you are not seperating out the molecule momentum in the final product . 

That's because you just made up " molecule momentum ".

Now, would you like to come back when you are busy enough to do your working?

[Professor Mega-Mind (OP)]

Too critical man !  Molecules have molecular motion and vibration .  These , of course , can be categorized in terms of momentum .  While I'm doing math work , you need to do conceptual research . 
P.M.

[Bored chemist]

... These , of course , can be categorized in terms of momentum . 

Not really, no.
While you are doing the maths, you need to find out what the differences are among momentum, energy and vibration.

[Professor Mega-Mind (OP)]

Electrons gain momentum from friction , atoms & molecules vibrate more because of this .  This raises the temperature .  Electrons emit photons , atoms & molecules vibrate less , this decreases temp. .  Heat IS increased momentum .  In space the speed of free electrons is considered to be their temp.  .  The %45 figure I quote is moment-um transferred to electrons .
Basic research dude !
P.M.

[Kryptid]

I have yet to see any calculations showing that conservation of momentum is violated by your design. Calculating kinetic energy is not the same as calculating momentum. In the SI system, kinetic energy is measured in units of joules, whereas momentum is measured in units of kilogram meters per second: https://en.wikipedia.org/wiki/Momentum

[Bored chemist]

Electrons gain momentum from friction , atoms & molecules vibrate more because of this .  This raises the temperature .  Electrons emit photons , atoms & molecules vibrate less , this decreases temp. .  Heat IS increased momentum .  In space the speed of free electrons is considered to be their temp.  .  The %45 figure I quote is moment-um transferred to electrons .

you need to find out what the differences are among momentum, energy and vibration.

And you still need to find it out.

As you say.

Basic research dude !

[Ophiolite]

For further clarification of this R.D., go to thread "Reactionless Drives Possible ?" .
P.M.

Either provide a detailed mathematical treatment of your idea, or concede that you are way out of your depth. There are no other viable alternatives, unless we list those that lack integrity or intellect. So, maths, or concession?