free book - Impulse Drive.

[Kirk]

http://www.amazon.com/dp/B00BXWNG96

Free Sunday

Drawings and description of inertial thruster.

I didn't go into free energy but it is capable of that as well.

My birthday present to you.

edit - free is tomorrow, March 25, Monday

[Ed]

Thanks for the link Kirk.

Your Amazon page has an impressive collection of products. Well done!

[Kirk]

free is tomorrow, March 25
sorry about that.

[Kirk]

no comments?

[eccentrically1]

What does the cue ball represent?

[Kirk]

the cue ball carries the momentum to the 3 ball collision.

We have a black box and we power it by using electricity to impart momentum to ball 1 or "the cue ball".

[eccentrically1]

Then it hits the other two balls, and then what?

[Kirk]

I suspect you didn't download the book.

[Kirk]

If you plot the vectors of the forces of the balls you see they form a
Y
Imagine if you will a billiard table, and upon this table you place a pair of balls somewhere near the middle. You then strike them with a third ball, the cue ball. The resultant collision should see the momentum of the cue ball transferred to the pair. They should separate with an angle of 90° relative to each other. This means that the original vector has been modified. Of the original vector only half remains. The new vector is at right angles to the original vector and of equivalent magnitude. The momentum has been split.
The momentum (mv) of ball 1 transmits into the 2 balls halving velocity but doubling the mass (mv is preserved). It also rotates the vectors to form a new path, the 2 arms of the Y. Half of the forward vectored thrust has disappeared to reappear as sideward force. The sideways thrust of the partner ball has a vector 180 degrees different from its partner so they cancel. The net sum is only half the energy originally imparted by the cue ball solenoid appears as symmetric reaction force.

[eccentrically1]

I did, it's ten pages. I didn't look for the 1935 magazine.
I don't see from your book and pictures how this works in a vehicle/vessel to drive it forward.
It depends on how much electricity you provide to "m" , the cue ball?
It doesn't look very efficient if half of the thrust is going sideward instead of forward.
Your book is sort of confusing, sorry.

[Kirk]

follow the step by step states of the machine and the vectors of the momentum. The sideways momentum is how you make the vectors asymmetric. No thrust if it is symmetric.

You get to keep 40 - 50% of the momentum of the cue ball as acceleration to the black box which is fastened to what you wish to apply a force to.

This would of course be proportional to the electrical energy and the efficiency of your transducer. Think solenoid or loudspeaker for styles of transducer.

[honza]

Trying to imagine how to obtain benefit of this vector splitting method and get rid of the unpleasant impulse form of the productive force I came to an analogical setup using water flow.

Imagine a bathtub on wheels, having a submersible pump in one end and a nozzle above the water level pointing towards the other end of the bathtub (to cancel the pump torque opposing momentum 2 opposing rotation pumps would actually be needed).
If we start the pump the reaction force of the jet at the nozzle would equal the impact force at the location of jet impact at the other end of the bathtub. The bathtub would not move anywhere.

Now - if we place at the jet impact location a suitably shaped object designed to split the water into 2 equal parts and deflect these in opposite directions (one 90 deg CW other half 90 deg CCW).
Would we achieve a reduction in the reaction force at the splitter / deflector?
Would the two water jets after splitting and deflecting retain ~ 1/2 of the original kinetic energy?

If the reaction force at the nozzle in this setup would be near double of the reaction force at the splitter/deflector that would get the bathtub rolling !!!

Or is it possible that nearly all kinetic energy would be retained in the two deflected water jets and this aspect has nothing to do with the reaction force of the splitter / deflector?

Is there something else in play I don't understand?

What do you think?

[eccentrically1]

follow the step by step states of the machine and the vectors of the momentum. The sideways momentum is how you make the vectors asymmetric. No thrust if it is symmetric.

You get to keep 40 - 50% of the momentum of the cue ball as acceleration to the black box which is fastened to what you wish to apply a force to.

This would of course be proportional to the electrical energy and the efficiency of your transducer. Think solenoid or loudspeaker for styles of transducer.

No thrust if the vector is symmetric? Why would that be?
So you're saying if the other two balls weren't there, the cue ball isn't going to provide thrust to the black box, which is fastened to the the vehicle?
That just doesn't make sense to me.
Wouldn't it be more efficient to directly drive the vehicle with the black box and not lose momentum from the balls' collisions?

[Kirk]

Trying to imagine how to obtain benefit of this vector splitting method and get rid of the unpleasant impulse form of the productive force I came to an analogical setup using water flow.

Imagine a bathtub on wheels, having a submersible pump in one end and a nozzle above the water level pointing towards the other end of the bathtub (to cancel the pump torque opposing momentum 2 opposing rotation pumps would actually be needed).
If we start the pump the reaction force of the jet at the nozzle would equal the impact force at the location of jet impact at the other end of the bathtub. The bathtub would not move anywhere.

Now - if we place at the jet impact location a suitably shaped object designed to split the water into 2 equal parts and deflect these in opposite directions (one 90 deg CW other half 90 deg CCW).
Would we achieve a reduction in the reaction force at the splitter / deflector?
Would the two water jets after splitting and deflecting retain ~ 1/2 of the original kinetic energy?

If the reaction force at the nozzle in this setup would be near double of the reaction force at the splitter/deflector that would get the bathtub rolling !!!

Or is it possible that nearly all kinetic energy would be retained in the two deflected water jets and this aspect has nothing to do with the reaction force of the splitter / deflector?

Is there something else in play I don't understand?

What do you think?

would be nice if it worked with water but divergence is not supplied by a collision, rather it is supplied by a knife or wedge which transmits the force to the enclose. As a result it won't work. you need 3 elastic objects.

[Kirk]

follow the step by step states of the machine and the vectors of the momentum. The sideways momentum is how you make the vectors asymmetric. No thrust if it is symmetric.

You get to keep 40 - 50% of the momentum of the cue ball as acceleration to the black box which is fastened to what you wish to apply a force to.

This would of course be proportional to the electrical energy and the efficiency of your transducer. Think solenoid or loudspeaker for styles of transducer.

No thrust if the vector is symmetric? Why would that be?
So you're saying if the other two balls weren't there, the cue ball isn't going to provide thrust to the black box, which is fastened to the the vehicle?
That just doesn't make sense to me.
Wouldn't it be more efficient to directly drive the vehicle with the black box and not lose momentum from the balls' collisions?

Don't be confused, be methodical. step by step pencil the forces. Symmetric would be net sum zero. The asymmetry is where you get the useful product