Time travel a reality.... what do you think?

[Einstein]

Well technically just one Tesla coil creates a toroidal field. The energy field created does fall off at one/r*squared. So that is an indication that it is a 3-D field. But I can't manipulate the energy field like I can with the Zero Time Generator. However because a Tesla coil produces a 3-D field, it is possible to wind a backwards coil. So I might be able to get two coils to interact with each other, rather than with just my physical proximity. It's something I'm going to try to see if a more stable frequency regulation can be achieved.

Go for it ;-)

How does a Zero Time Generator work exactly? I've read about them but haven't quite got my head round it. I would be very grateful if you could explain to me how they work.... thanks & much obliged

I started a thread on this device earlier this year. Check it out.

Tesla's Zero Time Generator | Paranormalis

 

[BeamMeUpScotty]

Well technically just one Tesla coil creates a toroidal field. The energy field created does fall off at one/r*squared. So that is an indication that it is a 3-D field. But I can't manipulate the energy field like I can with the Zero Time Generator. However because a Tesla coil produces a 3-D field, it is possible to wind a backwards coil. So I might be able to get two coils to interact with each other, rather than with just my physical proximity. It's something I'm going to try to see if a more stable frequency regulation can be achieved.

Go for it ;-)

How does a Zero Time Generator work exactly? I've read about them but haven't quite got my head round it. I would be very grateful if you could explain to me how they work.... thanks & much obliged

I started a thread on this device earlier this year. Check it out.

Tesla's Zero Time Generator | Paranormalis

Cool... you've got the grey matter going in my brain now! I'm going to do a lot of reading & research over the weekend, curiosity has got the better of me. Keep me posted with the toruses & Teslas, speak soon!

 

[Harte]

Earthmasque
Your statement is exactly what is taught in school. Yet I just clearly showed there is an acceleration going on between the center point and the moving weight after the string is cut.

Sorry I took so long to come back to this thread.

Your graph showing curvature is (I presume) what you base this "acceleration" on.
What you have done is graphed the function f(x)=SQRT(x^2 + 10), which is not a linear function.
In fact, it should (and does in your graph) closely resemble the function f(x)=SQRT(x)

Your graph is not a graph of postion using the normal method of such graphs (curvature in a position graph - aka a displacement vs. time graph - is what depicts acceleration.) Your argument should involve parametric equations, if you want to describe the path of an object with respect to some other point than the object's original position.

Using parametric equations, which, again, are used to describe the scenario in your papers (position change with respect to a point not on the obect's path,) the displacement vs. time graph is linear. In fact, it shows constant velocity, exactly like I said.


Harte

 

[Einstein]

Earthmasque
Your statement is exactly what is taught in school. Yet I just clearly showed there is an acceleration going on between the center point and the moving weight after the string is cut.

Sorry I took so long to come back to this thread.

Your graph showing curvature is (I presume) what you base this "acceleration" on.
What you have done is graphed the function f(x)=SQRT(x^2 + 10), which is not a linear function.
In fact, it should (and does in your graph) closely resemble the function f(x)=SQRT(x)

Your graph is not a graph of postion using the normal method of such graphs (curvature in a position graph - aka a displacement vs. time graph - is what depicts acceleration.) Your argument should involve parametric equations, if you want to describe the path of an object with respect to some other point than the object's original position.

Using parametric equations, which, again, are used to describe the scenario in your papers (position change with respect to a point not on the obect's path,) the displacement vs. time graph is linear. In fact, it shows constant velocity, exactly like I said.


Harte

I would ask you to do the length calculations of the hypotenuse yourself. After the string is cut. The path of the weight becomes one side of a right triangle. The path of the object is linear. But what I discovered is that the side of the right triangle that represents length against time being what we are taught in school, co-varies with the length of the hypotenuse. The hypotenuse does not change in length at a constant rate. The rate is changing. As it would if it was an acceleration. So from the perspective of someone standing at the center of the circle of previous rotation, the object is accelerating away. Now my main argument is that since the centrifugal force before the string was cut is directed away from the center of rotation, then after the string is cut, one should be measuring what is happening between the center of rotation and the object. Just using the Pythagorean theorem for this calculation does indeed show an acceleration to be present.

Another thing I want to point out is that before the string is cut the path of the object does have two forces acting on it. So that path requires those two forces to be present. So the resulting path even though it appears to be linear, is the result of the path of those two prior forces. So one path shows a velocity. And the second path shows an acceleration.

 

[Harte]

I would ask you to do the length calculations of the hypotenuse yourself. After the string is cut. The path of the weight becomes one side of a right triangle. The path of the object is linear. But what I discovered is that the side of the right triangle that represents length against time being what we are taught in school, co-varies with the length of the hypotenuse.

What you did was calculate a series of hypotenuses of right triangles, all of which have one side in common (your "y" axis length - 10, if I remember correctly) while varying the third side in step-wise fasion one unit at a time.

In other words, you found the value for f(x) for a series of x's for the function:
f(x)=SQRT(10+x^2)

That particular function is not linear. Nor does it represent the object's motion.

At least, not in the terms you used.

The hypotenuse does not change in length at a constant rate.

This is exactly correct. The length changes at the rate of SQRT(10+x^2).

The rate is changing. As it would if it was an acceleration.

It certainly would represent acceleration, if anything was following that path at that rate. But nothing is.

So from the perspective of someone standing at the center of the circle of previous rotation, the object is accelerating away.

Actually, no. If you put the x-axis along the line of the hypotenuse, you'll see that it is linear. But the function would be different.
The function for that path would be f(x) = o, which is linear.
The function for the velocity along that path (taking the x-axis [hypotenuse] as time) would be:
f(t)=V0 where V0 is the velocity of the weight when the string is cut.
Please understand, the motion of the weight is in a straight line. The path is that straight line. The variation you hope to see would have to be along that same straight line, not some other, unrelated line.

Another thing I want to point out is that before the string is cut the path of the object does have two forces acting on it. So that path requires those two forces to be present. So the resulting path even though it appears to be linear, is the result of the path of those two prior forces. So one path shows a velocity. And the second path shows an acceleration.

The only two forces on the weight are both along the string. One is the centripital force you are supplying from the center. The other is the reaction force to the force you are supplying. That is all the forces present, other than the small amount you have to keep adding to keep the weight moving (which, in a zero G situation, would be unnecessary.)

Harte

 

[IroncladMarshmallow]

Einstein, you're half right. To get the full picture, represent the system in polar coordinates. In this case, the velocity is split into two orthogonal components, u_r and u_theta. u_r is radially outward from the origin, and u_theta goes in a circle. The total velocity of the object is u = sqr(u_r^2 + u_theta^2).

Before the string is cut, u_r = 0, so the entire velocity is represented by the angular component of its velocity, u_theta. When the string is cut, both components of the velocity will change together, but, without any external forces, they will gradually approach an asymptotic value. At every moment, however, the total velocity will be constant.

Note that u_r should be the letter u with a subscript r. Same thing for u_theta.

 

[Einstein]

Harte

It certainly would represent acceleration, if anything was following that path at that rate. But nothing is.

I want to point out that what you are claiming is probably some technical detail that will have to be modified. All of our theory on force appears to be steered toward inertial force as being the only force of any concern. So there appears to be a very strong mathematical effort to make it so.

There is no sound mathematical framework for gravity at all. Yet right here in front of you is something obvious that could very easily become the factual foundation for an understanding of what gravity really is. I might point out that the acceleration computed is not associated with any current force present. There are two other physical observations in reality that have the same pattern. One is an object in orbit about the earth. The other is an object in gravitational free-fall within a vacuum. In both of those conditions the object is weightless.

Centrifugal force has its vector orientation away from the center. And the computed acceleration has its vector orientation directed away from the center. Gravitational weight and acceleration have vector orientations directed toward the center. It looks to me like we have equal behavior but opposite vector directions between gravity and centrifugal force. So for all intents and purposes this appears to be math based in a factual reality that does describe anti-gravity.

So I guess I'm off, sailing into the future. Just for clarification, which boat did you want to be on? The anti-gravity boat, or the curved space-time boat?

 

[Ayasano]

Although I believe I scared away Ayasano after he reviewed it.

That would be incorrect.

 

[Einstein]

Although I believe I scared away Ayasano after he reviewed it.

That would be incorrect.

Good. I'm glad you decided to return.

 

[walt willis]

In the past there were others that played with far less dangerous experiments in science and suffered greatly.

I hope you understand how dangerous it would be to try and manipulate space/time fabric.

Besides, our secret government has been there and all ready done that at the cost well beyond anything the average forum member can afford.

The Pegasus lab has been upgraded to what is now called the Atlas project with about five times the size capacitors.

Extreme high electrical energy is Very dangerous!

I lost a friend because he did not heed the warnings.

Please be careful and have an exit strategy...