[Bored chemist]

imagine 590 MILLION foot/pounds of work done .  Cram those into 1 second , and you have the force applied to the target CONTINUOUSLY ,

No.
You don't have a force, you have a power: specifically, you have 590 million foot pounds per second.
Or, in more sensible units 800 million watts.
That's a power, not a force.
It could refer to, for example,
a force of  590 pounds acting at a velocity of a million feet per second, or it could be
a force of a million pounds  acting at a velocity of 590 feet per second,
or any other combination which gives the same product (subject to a limit of the speed of light).

It could also refer to a large mass of water being heated rapidly, or a very loud noise, or the reduction of a lot of iron ore to the metal or lots of other things.

Why can't you learn simple physics and avoid making a fool of yourself in this way?

What happens to the track of the centre of the ball?
Are you unable to answer, or is it that you daren't because you realise that it undermines everything you have said?

[Professor Mega-Mind (OP)]

................Turd Flinging !
Once more , you "scientists" seem to forget what I laid out , a small mass of electrons impacting a target at relativistic velocities , not boiling coffee , or running an amp ! 
As for your silly exploding ball question , I did answer it , you did not confirm , I am tired of hearing it .  How's about you stop making up faults that don't exist , and take your misapplied formulas to a place that has a use for them .  I will take my 1.08 Mhp. engine to those who appreciate it . 
By the way , even if it has only turbojet levels of efficiency , that's still ~1 million pounds of thrust .
Rots of ruck , deceivers !
P.M.

[Bored chemist]

Turd Flinging !

Yes, you are, and we wish you would stop it.

Once more , you "scientists" seem to forget what I laid out

OK here's what you laid out

590 MILLION foot/pounds of work done .  Cram those into 1 second , and you have the force applied to the target CONTINUOUSLY

And I pointed out that it's wrong.
You don't have " the force applied to the target" you have the power  dissipated in it. I explained that could be a small force moving quickly or a large force moving slowly. Anyone who knows when to change gear in a car should understand that

As for your silly exploding ball question , I did answer it ,

No, you didn't .
You made stupid comments that are too ambiguous to be useful and, as far as it goes, are wrong.

I am tired of hearing it

Then answer it, plainly and clearly- or admit that you can't.

How's about you stop making up faults that don't exist ,

The faults plainly do exist.
And you made them up, not us.

take your misapplied formulas to a place that has a use for them .

they are not "misapplied"- that fundamental lack of understanding on your part is why this thread is so long and unproductive.
You refuse to learn.

I will take my 1.08 Mhp. engine

You don't have one.
I'd quite like to be a fly on the wall when you turn up at some university or NASA or whoever and try to "sell" them your idea.
Perhaps when they start laughing you will accept that you are wrong.

By the way , even if it has only turbojet levels of efficiency , that's still ~1 million pounds of thrust .

But it hasn't that efficiency.
Because it is a reactionless engine, we know exactly what the efficiency is, and it's zero.

Rots of ruck , deceivers !

The only deceiver here is you; you deceive yourself, rather than learning.

So, once again, here's the question. If you don't want to see it again, all you have to do is give a clear meaningful answer.
What happens to the track of the centre of the ball?
Are you unable to answer, or is it that you daren't because you realise that it undermines everything you have said?

[Kryptid]

Once more , you "scientists" seem to forget what I laid out , a small mass of electrons impacting a target at relativistic velocities , not boiling coffee , or running an amp ! 

I explicitly took that into consideration with my calculations.

take your misapplied formulas to a place that has a use for them

Misapplied? Which of my equations did not apply to your design? Arguably the most important one would be the equation governing impact force, which is obviously highly relevant.

[Professor Mega-Mind (OP)]

First , you need to elucidate your billiard ball question much better .
Second , the end of your post summarises the thing nicely ; this is ALL ABOUT impact force . Same as a firehose shooting a hard stream of water , or an artillery shell , striking a WALL .  Said wall being part of a ship .  Your calc.s continue to be off because you do not accept that there is no significant recoil on the launcher end , but major impact on the target end .  This imbalance in applied force is the gist of the engine . The trick is the conversion of internal photon energy (oscillation) into electron-kinetic energy , thus allowing for matter-style kinetic impacts .  There are a number of possible mechanisms to capture the electron-impacts , this is just the easiest for people to understand .
If you cannot fault the process , then don't try to confuse me with formulas I don't need .
P.M. 

[Bored chemist]

First , you need to elucidate your billiard ball question much better .

I already did, at the outset

You have a billiard ball flying through a big vacuum chamber at a few m/s - someone (in a space suit) threw it. Nothing complicated or relativistic.
It's the black ball and so it absorbs light really well.
I have a video camera pointed at it and hooked to a computer.
The centre of the ball is following a parabola and the computer is programmed to track the ball and work out the path  followed by the centre of the  ball so it can work out where the ball will go.

Now imagine that I fire a high power laser pulse at it and the light is absorbed by the ball.
The side of the ball nearest me  instantly boils.

 ( the computer now has a more difficult job because half the ball is now a cloud of hot gas + smoke).

What happens to the track of the centre of the ball?

Now, either clarify what information you think is missing, or answer it.

Your calc.s continue to be off because you do not accept that there is no significant recoil on the launcher end , but major impact on the target end .

No.
Our calculations are correct because we "do not accept that there is no significant recoil on the launcher end , but major impact on the target end".
Such an imbalance is impossible and you have not shown HOW it might happen, you have just repeatedly claimed THAT it happens.

[Bored chemist]

The trick is the conversion of internal photon energy (oscillation) into electron-kinetic energy ,

We will be able to get back to that once you have answered my  previous question.

[Professor Mega-Mind (OP)]

Firstly , I am not a flipping calculator !  I'll bother to answer for the sake of civility , is all . 
Obviously , whatever side boils will provide an explosive push in the opposite direction .  The arc of travel will be distorted by said push in a manner similar to shoving a swing which is already in motion .
There , I fail to see why that booger was worth picking !
P.M.
Note-It's NOT a perfect analogy !

[Bored chemist]

I fail to see why that booger was worth picking !

That's because you didn't read it properly.

The camera is tracking the centre of gravity of all the bits- even the soot  smoke + steam etc.

What happens to the path of the CoG of the ball?
(Hint; no maths is required)

[Professor Mega-Mind (OP)]

NOW I see your little point poking out !  You expect me to NOT see that the center of mass maintains it's initial trajectory , butt...  The initial burst of light would have pushed the ball an infinitesimal amount , before it ended .  This is analogous to a wave pushing a large ship a tiiiny bit , before the ship's magazine blows up .  There , isn't that all better now ?
Doctor P.M.

[Bored chemist]

You expect me to NOT see that the center of mass maintains it's initial trajectory ,

Well, It's not so much what I expect, as what you have done.
So far you have failed to see that, apart from the ting shove (less than a pound) from the laser, the ship will continue on its original trajectory.

Why do you think there's a difference between the ball- where you accept that it follows the laws of physics- and your ship where you claim it doesn't?

[Professor Mega-Mind (OP)]

Because YOU fail to see the full effects of the Compton Scattering Paradigm ; in otherwords , release of photon "heat" (oscillation) energy , to the electron as kinetic energy .
P.M.

[Bored chemist]

The fact that it's compton scattering rather than any other form doesn't make a difference.
The momentum transfer still follows the laws of physics.
You don't magically get more thrust.
So, six pages on,  you still don't have a reactionless drive.

[Professor Mega-Mind (OP)]

So sayeth the blind man .
P.

[Kryptid]

Second , the end of your post summarises the thing nicely ; this is ALL ABOUT impact force .

All right then, here is the equation for calculating impact force: Impact force in newtons = (2 x mass in kg x velocity in m/s)/ time of impact in seconds. One newton is 0.224809 pounds of force. That is a very simple equation. It can easily be solved. So feel free to do it yourself and show us how it produces a much larger force than we anticipate.

don't try to confuse me with formulas I don't need

These formulas are ones that anyone who graduated from high school should be capable of using because they're only algebra. If you haven't graduated from high school yet, then I can forgive that short-coming. If you have, then you really should know better...

[Bored chemist]

So sayeth the blind man .
P.

I presume that's self-referential.

[Kryptid]

 IN ONE SECOND ! Tell me you didn't realise that !  Even your Low-brow School classmates would have recognised that ! 

By the way, your calculations were off. The amount of force needed to give 600,000 pounds an acceleration of 1 foot per second per second isn't 18.2 million pounds. It's 18,688 pounds. Very big difference.

[opportunity]

Reactionless drives should be possible. Its one of the things we try to do as humans....the impossible. And yes, the math needs to be there. Anything is possible, the question is when and why.

[Professor Mega-Mind (OP)]

Whoops , rushed .  Still more than one pound though .
P.

[opportunity]

Yeah. Share that.