Temporal Divergence Meter

[Designer2]

Here a video that shows multiple world theory is true thus event line can have a divergent point
since there are real quantum computers in existence. When a new event line is born gravity particles higgs boson enter into it which was formally just a probability and thus earth quake occurs since the line become real resulting in a new trajectory into the future or what we currently call a new time line. The Temporal Divergence Meter should detect this. I believe the gravity between timeline will be different. Please watch video.

 

[Robot]

"Centrifugal force is a real force, and not fictitious as I was taught."

By Newtonian Definition, "Centrifugal force" is not a force.

F=MA

"Centrifugal force" does not require an acceleration, only a continual vector change. An example would be a spinning ball on a string.

 

[Robot]

"If gravity changes in intensity, then so does the flow of time. And a change in intensity should produce a gravity wave."

Maybe it does. I spent time talking with Professor Gowdy about this. His was present at the testing of the "Weber Bar":

Weber bar - Wikipedia, the free encyclopedia

just because it hasn't been detected yet, doesn't mean it isn't there.

 

[Robot]

Ray,

You know the laws of physics. I know the laws of physics. But it appears that no one else even cares about the laws of physics else they would take some time and actually explore and learn them. .

I think it has been a long time since any Physics Models used the word "law".

One such reason is Ohms Law, which superconductors don't obey.

But yes, much ignorance out there..................

Finally figured out Reply is Quote, LOL

 

[Robot]

I don't believe the speed of light is constant in all reference frames. That one is easy to verify (hint: Radar guns). .

I would love to hear the Radar Gun logic that I have seen in a few posts

 

[Einstein]

"Centrifugal force is a real force, and not fictitious as I was taught."

By Newtonian Definition, "Centrifugal force" is not a force.

F=MA

"Centrifugal force" does not require an acceleration, only a continual vector change. An example would be a spinning ball on a string.

Don't believe everything you are taught in school, unless you can confirm it to be true and real. Here is a little paper I did on centrifugal force that does show an acceleration is present after the connecting string is cut. The only thing is the direction of centrifugal force is opposite to gravitational weight. As is the direction of the mathematically derived acceleration. I did it just for fun about 20 years ago.

Just copy this link into your browser address bar.

http://paranormalis.com/attachments/bogus-centrifugal-graph-jpg.1442/

 

[Robot]

Don't believe everything you are taught in school, unless you can confirm it to be true and real. Here is a little paper I did on centrifugal force that does show an acceleration is present after the connecting string is cut. The only thing is the direction of centrifugal force is opposite to gravitational weight. As is the direction of the mathematically derived acceleration. I did it just for fun about 20 years ago.

Just copy this link into your browser address bar.

http://paranormalis.com/attachments/bogus-centrifugal-graph-jpg.1442/

Must look like I must log in to view

How would an acceleration occur when the string is cut? If we go back to F=MA, we have a velocity at the time the string is cut, we would expect wind resistance etc, to slow that bull from that time. What would be causing an increase in velocity(acceleration)?

 

[Einstein]

You need to throw out F=MA. It is not applicable to non inertial reference frames. Centrifugal force is not an inertial force. It is the only force that opposes gravity with an effect of complete cancellation.

Try this link and download the image.

FileSnack | Easy file sharing

I used the Pythagorean theorem to calculate the distance between the moving object and the center of its previous rotation. The math doesn't lie.

 

[Robot]

You need to throw out F=MA. It is not applicable to non inertial reference frames. Centrifugal force is not an inertial force. It is the only force that opposes gravity with an effect of complete cancellation.

Try this link and download the image.

FileSnack | Easy file sharing

I used the Pythagorean theorem to calculate the distance between the moving object and the center of its previous rotation. The math doesn't lie.

DOH! "The link you are accessing has been blocked by the Barracuda Web Filter because it matches a blocked category. The name of the category is: "advertisement-pop-ups"

 

[Einstein]

DOH! "The link you are accessing has been blocked by the Barracuda Web Filter because it matches a blocked category. The name of the category is: "advertisement-pop-ups"

I'll find a way to get the image up eventually. Its really very hilarious to see something so simple totally debunk physics as we know it.

 

[Robot]

Centrifugal force is not an inertial force. It is the only force that opposes gravity with an effect of complete cancellation.

This is not true. I was taught in Classical Mechanics to usually view gravity as a vector, magnitude and direction.

Gravity here on earth for a bullet has that magnitude and direction. When I shoot a bullet straight up from a gun, that vector initially is greater than gravity.

As the bullet reaches the apex, you have two opposing vectors that cancel. It is "weightless" for that moment in time.

Gravity also cancels gravity. Gravity is strongest at the surface of the earth. It grows weaker the higher we go, or when we go down inside the earth.

If the earth were solid dirt, and I were able to make a tunnel to a cool core, I would weigh less all the way down. This is because the mass above me, influences the mass below me, until we have equilibrium and weightlessness at the earth's core.

 

[Einstein]

Here is my math to show centrifugal acceleration does exist.

If you graph the values for the hypotenuse you will see it is an acceleration curve. I suspect Newton's Principea had this information in it at one time. But the Principea has been heavily edited and modified.

 

[RainmanTime]

Here is my math to show centrifugal acceleration does exist.

If you graph the values for the hypotenuse you will see it is an acceleration curve. I suspect Newton's Principea had this information in it at one time. But the Principea has been heavily edited and modified.

I suspect you do not understand the calculus of position, velocity, and acceleration if you think this diagram shows an acceleration. Einstein, all you are showing here with the hypotenuse calculations are that the total distance of the object increases away from the origin. That is not acceleration, that is just the change in position. It appears you are ignoring gravity acting on the object, and that may be fine if you assume this experiment with the string and object is out in, essentially, gravity-free space. However, you have not even defined a velocity profile much less acceleration. So if you think what you have shown here displays an acceleration, I am afraid you are incorrect. But as I know you all too well, you will never admit it.

RMT

 

[Einstein]

I suspect you do not understand the calculus of position, velocity, and acceleration if you think this diagram shows an acceleration. Einstein, all you are showing here with the hypotenuse calculations are that the total distance of the object increases away from the origin. That is not acceleration, that is just the change in position. It appears you are ignoring gravity acting on the object, and that may be fine if you assume this experiment with the string and object is out in, essentially, gravity-free space. However, you have not even defined a velocity profile much less acceleration. So if you think what you have shown here displays an acceleration, I am afraid you are incorrect. But as I know you all too well, you will never admit it.

RMT

Graph the hypotenuse values against time. It's not that hard to do.

 

[RainmanTime]

Graph the hypotenuse values against time. It's not that hard to do.

I don't have to. I can do these things in my head. It still does not change your error that you are doing 2-dimensional calculations for what is 1-D rectilinear motion. The motion is only along one axis, so only the distances and times along that axis matter. The offset distance along the y-axis is constant and unchanging, but you are including it in your calculation of velocity erroneously.

Just like any vector problem, you must track your positions, velocities, and accelerations in orthogonal components separately (why vector math uses the <i,j,k> notation). If you were to do the math correctly, you would find there is no acceleration.

You are doing the math wrong. Plain and simple.
RMT

 

[Einstein]

But you are ignoring the fact that there are two forces present before the string is cut. It takes constant force along the circular path to make an object move in a circle. And then there is the additional centrifugal force which is always present in circular motion. That force is directed away from the central point of the circular motion. So that observation alone should tell you there should be two paths of motion. The math doesn't lie.

 

[RainmanTime]

And then there is the additional centrifugal force which is always present in circular motion. That force is directed away from the central point of the circular motion.

If that were true, then the object would not follow a tanget line when the string is cut. Newton's First Law of Motion applies to what we see when the string is cut:

"When viewed in an inertial reference frame, an object either remains at rest or continues to move at a constant velocity, unless acted upon by an external force."

When the string is cut, the external force imparted by the string ceases to act upon the body. This is why the body moves at a tangent to the circle. There is no longer an imbalanced force acting upon it.

So that observation alone should tell you there should be two paths of motion.

There is only one dimensional motion when the string is cut. It is along the line tangent to the circle at the point the string is cut. Period.

The math doesn't lie.

It doesn't lie when you do the math right. Unfortunately, you have done the math wrong. You can either understand why it is wrong, or keep trying to convince me you are right when more people than me know you are wrong. If you continue the latter, I will eventually shame you with the real math.

RMT

 

[RainmanTime]

Try thinking about it this way: Separate the X and Y directional motion in your diagram.

FACT: The Y coordinate of the mass DOES NOT CHANGE as the mass separates from the string. This is shown in your diagram. As such, since the Y position is not changing, there is ZERO y-component of velocity. All the velocity of the mass is along the x-dimension.

Because of this fact, ONLY the distance covered between each second (let's assume each point along the x-axis is 1 second) along the path of the mass which is parallel to the x-axis matters for the velocity calculation.

Velocity is approximated as a finite difference as:

Change in position / Change in Time.

The change of time is fixed at one second (each point where you draw a hypotenuse to the path of the mass)

The change along the axis of the path of the mass is the same length from one time point to the next.

Hence, (change in position/change in time) is going to be the same value for each point along the path of the mass.

If velocity is not changing, then there IS NO ACCELERATION.
RMT

 

[Einstein]

I totally agree that there is a velocity along the x axis. In fact I'm counting on it. And I never said there was an acceleration along the y axis. The apparent acceleration is along the hypotenuse. And are you saying a force has to be present for an acceleration? If it was an inertial force I would agree with you. In fact since there is no force present, I would have to say it is a weightless acceleration. Can you think of anything else that has the property of weightless acceleration? LOL...

The length of the hypotenuse changes as if there is an acceleration between the object and the previous center of rotation. The math says so. I got Pythagoras backing me up. Who you got?

 

[RainmanTime]

You are being obstinate again.

The length of the hypotenuse changes as if there is an acceleration between the object and the previous center of rotation. The math says so. I got Pythagoras backing me up. Who you got?

You don't have Pythagoras backing you up, because you are misappropriating his equation (one based on static geometry) towards a problem of dynamics. The equation of Pythagoras applies to an AREA (hence the squared terms) not a linear distance. There are only two things causing the length of the hypotenuse to change with each successive point:

1) The x-distance is changing at a constant rate (by what I have shown above, this is NOT causing any acceleration at all).
2) The AREA of the triangles increasing because of the constant Y distance and the changing X distance.

You only think there is an acceleration because you are being confused by item #2. This is because you are misapplying a formula based on area, and your mistaken acceleration is a remnant of the increasing area of the triangles. And this error of yours should show you why treating vectors by breaking them down into their orthogonal components is such an important technique to getting the proper answer.

A) There is no velocity, and hence can be no acceleration, in the Y direction.
B) There is only a constant velocity, and therefore no acceleration, in the X direction.
C) Since this is a planar motion problem these are the only two dimensions in which motion can take place in this problem. Hence there is NO sign of any acceleration.

Oddly enough, your formula is more akin to Kepler's Law (which speaks to a constant area swept of a body orbiting another on an elliptical path) than it does to rectilinear kinematics.

Are you willing to admit your error yet?
RMT