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Temporal Divergence Meter


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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.

 

 

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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

 

 

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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.

 

 

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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

 

 

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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.

 

 

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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

 

 

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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

 

 

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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?

 

 

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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

 

 

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Are you willing to admit your error yet?

Not a chance for that happening. You can squiggle and squirm all you want. I don't care. I guess it's just in your nature.

 

In fact I would ask you to independently calculate the values for the hypotenuse yourself. Each successive value increases at a changing rate. That's a fact that I can not deny. And I'm not quite understanding why you refuse to accept this fact. The hypotenuse acceleration is along the same direction that the centrifugal force was on. Making it identical in behavior to gravity. The only difference is the direction of the force or acceleration is away from the center instead of toward the center as it is with gravity.You may recall quite a while ago I introduced the idea of negative rotation being responsible for gravity. That would be an equal but opposite force to inertial force. So it's really quite easy to extend this odd little mathematical jewel into a more understandable explanation of gravity.

 

Is this the mysterious acceleration causing the universe to accelerate its expansion? Right under our nose all the time.

 

 

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(sigh) Have it your way...I will just continue to shame you with facts. Actually, you will just shame yourself by ignoring the facts that I present.

 

In fact I would ask you to independently calculate the values for the hypotenuse yourself. Each successive value increases at a changing rate. That's a fact that I can not deny.

I am not questioning your calculations. They are correct. It is not that math that is wrong, it is your interpretation of what they are telling you that is wrong. I have explained (but perhaps you did not understand) that the changing rate is due to Pythagorean Theorem being based on the AREA of a triangle.

 

And I'm not quite understanding why you refuse to accept this fact.

Fundamentally, it is because the object you are trying to calculate the velocity and acceleration for (the mass) DID NOT TRAVEL ALONG THE LINES OF THOSE HYPOTENUSES. Kinematics of an object are based on the path the object traveled, and velocity and acceleration are time derivatives of that path. Your use of the hypotenuses is an incorrect method to calculate the acceleration of the object. Understand this: The hypotenuses do NOT describe the path of the object, hence they have nothing to say about the acceleration of the object.

 

The hypotenuse acceleration is along the same direction that the centrifugal force was on.

No it is not, because that force is no longer present. The force was imparted by the string. Once the string was cut, that force (which induced the curved path) ceased to act on the body. So the last time that force was present it was NOT AT ALL acting along any one of your hypotenuses. It was acting at a 90 degree angle to the tangent path that the mass ended up following.

 

Again....Your analysis is wrong and you should feel shame that you are not seeking to understand why what I am telling you is the actual truth (and the math backs me, not you).

 

Making it identical in behavior to gravity.

Wrong again. I will not even address any of the rest of your fallacious discussion beyond this statement of yours above. Because it is all conjecture based upon incorrect analysis, just like the analysis you gave above is incorrect. You are not doing vector dynamics correctly. Your refusal to understand how to do it correctly, and thus not admit your error, is what holds you back. But of course that same refusal is what feeds your belief that you have discovered some truth that no one else did. But it is not a truth. It is delusion, and nothing more.

 

RMT

 

 

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You like to conveniently ignore the factual points I make that destroy your ignorant analysis. But I will not let you continue to get away with that. I want you to either disagree or agree with these two facts about the "analysis" you have offered. And if you disagree, explain why and cite references:

 

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.

 

These are facts, according to how YOU drew the diagram. And these facts can do nothing but falsify your belief (unproven) that the mass is accelerating.

 

RMT

 

 

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Did I do something wrong?

 

Yes, you did, and I pointed it out. You just refuse to accept it. You incorrectly assumed that the mass traveled the path of the hypotenuses, when it did not. You analyzed the wrong thing. And you did not separate your x axis motion from your y axis motion. It is a sophomoric error which yields points lost on many undergrad kinematics exams. So you are less intelligent than those students, because at least those students learn from their mistakes.

 

RMT

 

 

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Fortunately, there are several different ways to explain to (willing) students how their analytic work is in error. This case is no exception. Another way to explain your error is that you are not using a consistent reference frame. You are mixing two reference frames with your hypotenuse calculations.

 

You introduce an X-Y grid via the means you set up the problem with an X-Y axis and constraining the motion to be zero along the Y axis and non-zero along the X axis. But then, you start to perform your analytic calculations in a completely different reference frame that is an "r-theta" (radius and angle) frame.

 

The most important rule of kinematics is: "Pick a reference frame and ensure all your calculations remain in that same reference frame." By calculating the hypotenuse, you are invoking the radial distance from the origin to the mass. That radius distance includes both X and Y measurements, wherease the X-Y axis system you used to setup the problem only contains motion in one of those two axes.

 

Once again, you give a good example of erroneous calculations that I can send my students to read to understand how they might commit errors, and hopefully avoid them. Classic mixing of reference frames. You fail Kinematics 101, Einstein.

 

RMT

 

 

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A. True

 

B. True

 

How convenient that your facts actually support mine.

 

But then your facts reference the inertial force and its ensuing direction. And my facts reference the centrifugal force and its ensuing direction.

 

Now I understand that I am challenging your religious beliefs. So naturally you are opposed. And you haven't really provided a valid argument on why I can't use real math to describe something that really happens. I mean if the math works, and it fits reality, it should be good to go.

 

Obviously I can see that you don't like what I did. Too late. It's done.

 

I'm off to bed. See you tomorrow.

 

 

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A. TrueB. True

 

How convenient that your facts actually support mine.

But actually, they do not. If you agree to the above, then there CAN BE NO ACCELERATION in either of the two directions.

 

And my facts reference the centrifugal force and its ensuing direction.

There is none. Anything "centrifugal" ceased to exist when the string was cut. Done. Over. Too late.

 

And you haven't really provided a valid argument on why I can't use real math to describe something that really happens. I mean if the math works, and it fits reality, it should be good to go.

And that is the problem. The math is correct, but does not work for the situation. Math is nothing devoid of context. And you are applying the wrong math to this context.

 

RMT

 

 

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You are right about one thing. It's too late to undo my exposing you to the truth. You can say there is no acceleration all you want. The math says otherwise.

 

You know this reminds me of a similar situation I remember when I was just a boy. Maybe 5 or 6 years old. It was just shortly after Christmas. There were several of us out playing, and a little girl had come over to join us. She was only 3 years old. We decided to tell her the truth about Santa Claus. She was devastated. Her happy little smile slowly changed to one of sadness. Then she started to pout. Which eventually led to tears. I can still see her rubbing the tears out of her eyes with her little hands. Then she got angry. She started to yell at us. She said her mother told her Santa Claus was real. Then she ran away. She didn't take the truth very well at all.

 

Kind of like you Rainman. Do you have a childhood memory of when you found out about Santa Claus? How long did you stay angry? Eventually you had to come to the realization that Santa was all a lie.

 

You're a grown man now. Eventually you'll come around and accept the reality of the math.

 

 

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You are right about one thing. It's too late to undo my exposing you to the truth. You can say there is no acceleration all you want. The math says otherwise.

(snip)

 

You're a grown man now. Eventually you'll come around and accept the reality of the math.

None of your gyrations or insults of me change the fact that your analysis is incorrect. The math does NOT show there is an acceleration, because your application of that math is in error with the facts. The fact is that the mass did NOT travel the path of those hypotenuses, did it? Of course, you will conveniently ignore this because it is the basis of you being wrong. But the derivative of an object can only be properly calculated by deriving along the path the object traveled. You are taking the derivative along paths (hypotenuses) which the mass did NOT travel. Hence, you are quite wrong. Stating that you are right and I am wrong does not change that. But I realize it makes your ego feel good. You still look like a fool.

 

I can bring professor after professor of engineering and mathematics to this board to tell you that you are wrong...and you will still not accept it. If that is not stubbornness, I don't know what is.

 

RMT

 

 

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You are correct. It's a weightless acceleration. Just like gravity. So there can be no mass.

 

There is no acceleration. You are not accounting for a deceleration. It is all because your coordinate system accounting is atrocious. Soon, very soon, you will be faced with math that proves you wrong. But yes, I know, you will ignore it.

 

RMT

 

 

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