[Professor Mega-Mind (OP)]

It's not high-level particle physics , nor does it require a mathematical treatise to broach the idea .  Any good tutorial on Perfectly Elastic/ Inelastic Collisions will break it down as I have .  The %2-%98 and %45-%55 ratios are quite adequate to calculate kinetic energy transfer.  These are the accepted figures for steel-ball on steel-wall collision , and sandbag on steel-wall collision .  Simple , clear , visible to all who wish to analyze the process .  I do so despise formulaic obfuscation .  There is a reason I said Erg instead of Joule !  Perhaps I should have just said Og instead ?
Anyhoo , I'm not going to jump through hoops to mystify and impress observers .  The concept is impressive enough , let's hear your brainstorm .  Nyuck , nyuck , nyuck !
P.M.

[Bored chemist]

It's not high-level particle physics

No. It isn't.
So, why don't you actually try to do the maths properly?
Then we can see where you are getting it wrong.
My best guess is that you think that this  sort of thing "

he %2-%98 and %45-%55 ratios are quite adequate to calculate kinetic energy transfer.

"
means that the momentum is also transferred in a 45:55 ratio.

It isn't.
Do you understand that?

The concept is impressive enough

It would be, if it wasn't wrong.

[Professor Mega-Mind (OP)]

For the sake of clarity , I will repeat what that means .  For steel-ball only %2 K.E. is transferred to the Massive Wall (box) , %98 remains with the ball (though direction is reversed) .  For sandbag , %55 K.E. is transferred to the Massive Wall , %45 is turned to thermal energy .  I would drop the momentum babble , it's just confusing you !  Make it a straightforward analysis of energy transfer .  No one will be bam-boozled then !
P.M.

[Kryptid]

I would drop the momentum babble , it's just confusing you !

Except that you are proposing that your device is reactionless (which, definition, means that it violates conservation of momentum). Since you claim to be able to violate conservation of momentum, you have to be able to express your claims in terms of that very momentum. You can't make a claim about momentum and then tell us to "drop the momentum babble". That's a fundamental contradiction. If someone tried to sell a device that they claimed could create water out of nothing and then said "drop the water babble" to their detractors who asked them to prove it, why should anyone take them seriously?

[Bored chemist]

I would drop the momentum babble , it's just confusing you !

Your problem is that you think the fundamentally important bit of the discussion is "babble".

Come back when you have learned enough to recognise how daft that view is (and, preferably, don't come back before you have done so)

[Professor Mega-Mind (OP)]

....Momentum = Mass × Velocity
"Look , I made an MTV !" .
Truthfully , you'd sound more credible without the fantasy .  Fact is , I prefer to "Keep It Simple , Simon !" .  That means simple math , and simple units . Sqwaking that I don't know the obvious , while I use it masterfully , sounds totally masochistic to me .  There is an audience out there , if my simple math was bad , at least some would save your pride . Well , I'm still listening !  No absurd put-downs please , come up with more than "they said so !" .
P.M.

[Bored chemist]

....Momentum = Mass × Velocity
"Look , I made an MTV !" .
Truthfully , you'd sound more credible without the fantasy .  Fact is , I prefer to "Keep It Simple , Simon !" .  That means simple math , and simple units . Sqwaking that I don't know the obvious , while I use it masterfully , sounds totally masochistic to me .  There is an audience out there , if my simple math was bad , at least some would save your pride . Well , I'm still listening !  No absurd put-downs please , come up with more than "they said so !" .
P.M.

Well done. You managed to find out how to calculate momentum.
Dow all you need to do is apply that to the problem (as you have been asked many times to do).
Then post your calculations and if you get the calculations right, they show that you are wrong about reactionless drive.
Or you will get the calculations wrong, and we will be able to explain your mistake to you.

Fact is , I prefer to "Keep It Simple , Simon !"

There is a lot to be said for keeping it simple. But your approach here is like saying that you plan to build a car but  to "keep it simple" you will not include wheels or an engine.

if my simple math was bad , at least some would save your pride

You have not done any maths.

That's the problem here. Please do the math- and then you will see that you are wrong.

[Professor Mega-Mind (OP)]

Speak for yourself Bubbatugs !  I learned that years ago , then learned to use calculators if I wanted specific values .  The percentage figures at the beginning of "Reactionless Drives" are all the number I need .  If moved to , I can plug any value I want in , fait accompli !  I have designed the first reactionless drive here , because I am a DESIGNER !  I let people who like crunching numbers get specific , I avoid it unless necessary , such as when I want to obtain operating capacities and parameters .  Now , show ME a quantitative analysis that disproves my design , and I'll show you a trick mirror !
P.M.

[Bored chemist]

show ME a quantitative analysis that disproves my design

I already did.
You were not able to understand it.

The percentage figures at the beginning of "Reactionless Drives" are all the number I need .

No.
The other number you need is the thrust produced (or the velocity if you like).
And the reason you need it is that it is zero.

[Professor Mega-Mind (OP)]

Just replace the % symbol in my breakdown with joules , and you've got it . 
Meanwhile , I'm busy at work right now .  Don't feel like unnecessary number crunching .
P.

[Bored chemist]

n my breakdown with joules

You didn't do any such breakdown.
If you had it would be irrelevant.
Energy does not work in the same way as momentum.

[Professor Mega-Mind (OP)]

KE = 1/2 mass × Velocity squared
Of course , I use a calculator !
A 100Kkgm ship is coasting at
100m per sec.
It represents .5 gigajoules of kinetic energy .
It crashes into a massive , steel-hard Tholian Web . 
%55 KE is transferred to the T.Web.
%45 KE is turned into heat .
In quantitative terms :
.275 gigajoules of kinetic energy is transferred to the T. Web .
.225 gj. becomes friction heat .
 There now , it's all calculated out .  Isn't that amazing ?  I could just stare at it for hours ! 
 Anyhow , I'm off to actually think .
......P.M.

[Bored chemist]

KE = 1/2 mass × Velocity squared
Of course , I use a calculator !
A 100Kkgm ship is coasting at
100m per sec.
It represents .5 gigajoules of kinetic energy .
It crashes into a massive , steel-hard Tholian Web . 
%55 KE is transferred to the T.Web.
%45 KE is turned into heat .
In quantitative terms :
.275 gigajoules of kinetic energy is transferred to the T. Web .
.225 gj. becomes friction heat .
 There now , it's all calculated out .  Isn't that amazing ?  I could just stare at it for hours ! 
 Anyhow , I'm off to actually think .
......P.M.

Now do the right maths.
The calculations of the momentum.

[Ophiolite]

It's not high-level particle physics , nor does it require a mathematical treatise to broach the idea .  Any good tutorial on Perfectly Elastic/ Inelastic Collisions will break it down as I have .  The %2-%98 and %45-%55 ratios are quite adequate to calculate kinetic energy transfer.  These are the accepted figures for steel-ball on steel-wall collision , and sandbag on steel-wall collision .  Simple , clear , visible to all who wish to analyze the process .  I do so despise formulaic obfuscation .  There is a reason I said Erg instead of Joule !  Perhaps I should have just said Og instead ?
Anyhoo , I'm not going to jump through hoops to mystify and impress observers .  The concept is impressive enough , let's hear your brainstorm .  Nyuck , nyuck , nyuck !
P.M.

Fair enough. You have opted for the response lacking integrity, or intellect, or both. Duly noted. Thank you for confirming you cannot do the math and can be safely ignored in future.

[Professor Mega-Mind (OP)]

Fairly said , fool yourself !
What I said withstands BASIC mathematical examination .  I don't  see any BASIC , TRANSPARENT , disproofs of the concept .  Whining that I'm not using opaque , overly complicated formulas does not discredit the design .  It does show that even the formula aces cannot undo this new effect .
Read it and weep , oh jealousy !
P.M.

[Bored chemist]

I don't  see any BASIC , TRANSPARENT , disproofs of the concept .

Yes you did.
https://www.thenakedscientists.com/forum/index.php?topic=75294.msg558852#msg558852

Now, can you let us know what there is "opaque , overly complicated formulas "?

[Professor Mega-Mind (OP)]

Ah , the incomplete counter-example .  It lacked a counter-throw , plus it had bouncy sandbag .  My sandbag would be loosely packed , and would yield a near-perfect inelastic collision .  This means the sandbag would hit the wall , transfer %55 of it's KE to it , and convert the rest to friction heat .  The bag would go "splat !" , then stop dead right near the wall .  No KE left in the bag . 
Well , how do you like the SigDif ?
P.M.

[Bored chemist]

It lacked a counter-throw ,

No.
It explained why a counter throw would make no difference.

So it doesn't matter if you throw the two things in opposite directions at the same time- neither of them "knows" about the other, so they can't change their behaviour because of the other throw.
The only difference it makes is we no longer need to fix your feet to the floor.

Feel free to do the maths (I showed you how) for both items.

Handy hint.
Stop focussing on energy (which can be lost from the ship) and look at the momentum which can't, but actually defines the reaction.

plus it had bouncy sandbag

Nothing has a zero coefficient of restitution. Everything is at least  slightly "bouncy".
But it doesn't matter, you can do the maths with the rebound equal to zero.
You get the same outcome; the ship does not move.


Incidentally, you keep complaining that you don't have time.
Well, all this dross you post takes longer than actually doing the maths, so that can't be your real reason.
It must be something else.
So, "professor", are you going to admit that you can't do basic school physics?
Or are you going to show that the maths  proves you wrong.

[Professor Mega-Mind (OP)]

First : The counter-throw makes all of the difference .  Without it , the thrower imparts KE to the ship , with it , he does not .
Second :  The "dead" sandbag makes a huge difference also .  It's powerful shove completely outweighs the weak shove of the metal ball .  Proper manipulation of this difference is the heart of the "Engine" .  I actually don't need ANY math to prove the concept , the words very big , very small , half & half would suffice .  It is the LOGIC involved that bulwarks the process , not numbers that mystify the readers .  I believe a third-grader could get this , if it was explained as I just illustrated . 
Third , I still don't see any "High-School math" that disproves my process .  Your prior attempt did not do too good !
Hooray for logic , thumbs down for mistaken math !
P.M.

[Bored chemist]

It's powerful shove completely outweighs the weak shove of the metal ball .  Proper manipulation of this difference is the heart of the "Engine" .

OK, here's a nice simple question for you.
Consider two collisions.
(1)
A small weight slides along a (frictionless) table and hits a second (bigger) weight.
The colliding surfaces are hard steel, the weight bounces off, but it imparts some momentum to the second one.
That secone weight slides away with velocity V(bouncy)
(2)
The same small weight slides along (at the same speed) and hits the same (big) weight. But this time, the two surfaces are velcro and the two weights stick together.
The weights slide off with a velocity V(sticky)


In which scenario does the bigger weight move away faster?
Is V(sticky) bigger or smaller than V(bouncy)?