Universal UP or DOWN (split from Fields and ether)

[steveupson]

1 hour ago, Strange said:

The sign of what?

And why does introducing distance make a difference? Direction has a sign (is that what you are referring to?) but distance doesn't.

And this seems unbelievably complicated. You are starting with a very complex equation, then you think you need to add octonions (another very complex mathematical structure) to address its shortcomings. But you can't provide any rational reason for that, it is just a belief.

And for what? You have not yet shown that this equation has any advantages. You claim it will solve all sorts of problems, but have presented no reason for anyone to believe that.

You can't even use this equation to solve problems that are trivial using existing geometry. 

While we are are on that, can you go from this equation:

to an equation for v or lambda?

Go on then.

A couple of questions (which have been asked before).

1. You asked people to provide an example circle and position and the equation would calculate alpha, the direction they were facing. When I do that for the most recent example, I get the value of alpha = 0°. Can you explain what that means? 0° with respect to what?

2. For some value of v and lambda, the resulting value of alpha is imaginary; what does that mean?

 

I think that the sign will represent chirality in the completed geometry (when distance is derived.)  And as far as it being very complicated, I can't really offer any apologies for that.  The fact that vector manipulation won't readily solve the problem should provide ample illustration of its worth.  (Actually, we've had excellent results using vector calculus in order to solve this type of problem.  The difference is that instead of having to integrate many discrete rotations or projections, here we have a smooth function that performs the same task.  Mathematically, for some reason, the two give the same result for this problem, but I don't think that this is the case with relativistic problems.  The two methods should definitely give different results because they use completely different methods.)

I'm not really that good with algebra, but the equation is the resulting function from combining two simultaneous equations that relate alpha and lambda to another variable phi.  It's in the proof.  Do the math.  How many times do I have to tell you that as far as I can tell it is completely impossible to understand any of this without doing the math.

1.  The alpha = 0° result means that you are facing normal to the meridian plane.  Or, the tangent to the circle where you are standing is parallel to the surface normal to the meridian plane.  There is no angle between them.  The two planes that set up this relationship are the one that is tangent to the cone having the small circle as its base, and the meridian plane, where both are intersecting at your feet.  

2.  I'm not sure what that means, or what you are referring to.  If we use negative values for some of the inputs then the outputs are also negative.  That could be what you're talking about.  Maybe?

There is a complicated issue of how to keep track of the sign for alpha.  I may have gotten that particular calculation crossed up so that I am actually calculating the complementary angle for alpha instead of alpha, but it really is the best that I can do considering my very limited abilities with algebra.  I wasn't able to come up with the actual formula that is being used.  It was someone (Did) from mathstackexchange who helped with that.   The simultaneous equations that give the parameterization of alpha are:

As a check, we were able to compare results of the formula to the results that we already had from the original mathematica model.  This gave us a high degree of confidence that the formula is correct.  I don't know if you recall, but there was quite a bit of discussion over the concept that mathematica was able to create the model without knowing the equation.  I'm sure that this is related to the issue stated above about how we were able to solve these problems long before we had this formula.  

 

Also, in addition to alpha=f(lambda), these same equations can be combined in order to produce the tridentity:

 

Edited January 4, 2019 by steveupson

[Strange]

6 hours ago, steveupson said:

I think that the sign will represent chirality in the completed geometry (when distance is derived.) 

I have only asked "the sign of what". But I assume, from later comments, that you mean the sign of alpha, is that right?

You said earlier that you couldn't tell N from S or E from W. Isn't that the meaning of the sign of alpha? So +alpha would be "forward" out of the plane and -alpha would be "backward" relative to the plane. No? (That is the usual sense of the sign of a surface normal.)

I'm not even going to ask what you mean by chirality in this context. It just looks like another rabbit hole!

6 hours ago, steveupson said:

The fact that vector manipulation won't readily solve the problem should provide ample illustration of its worth.  

Won't solve what problem?

Can you give an example of a problem that can't be solved using vectors, and show us how this system solves it?

6 hours ago, steveupson said:

2.  I'm not sure what that means, or what you are referring to.  If we use negative values for some of the inputs then the outputs are also negative.  That could be what you're talking about.  Maybe?

So if we set v to 0.1 (radians) and lambda to pi/2 then the value of alpha is 2.98986 i

In other words an imaginary number. I can't understand what that means. Maybe I have calculated wrong, see here: https://www.wolframalpha.com/input/?i=cot^-1(cos(v)+tan(sin^-1(sin(L%2F2)%2Fsin(v)))),+v%3D0.1,+L%3Dpi%2F2

 

[steveupson]

Oh, ok.   I think you might be getting this result because you’re no longer on the circle.  The extent of the circle is from zero at the north pole, to 2V at its southernmost point, so the range of L is 0,2V.  This is because whatever the latitude of the southernmost point of the circle is, the midpoint of this latitude will be the latitude where we would find the circle center, if we were to dig deep enough.  I want to look into this further, but that might take a little while.

 

Here’s a brief history of the project that led us to where we are.  We were trying to perform a dynamic analysis of the behavior of a small steel ball that is trapped in the intersection of two grooves on the surface of two spheres.  As the two spheres are rotated relative to one another, the steel ball trapped in the intersection will be forced to travel along the grooves in a prescribed manner.  In order to do the analysis, we needed to determine the kinematics of the system.

In the 3D illustration of the model, only the part of the sphere that represents the groove is shown for the outer groove, and this groove is able to be tilted in a manner such that its relationship the other groove is variable.  For the analysis, we held the inner sphere stationary (which we call the driven groove Dn) while we turned the outer groove (which we call the driving groove Dg.)   At different inclinations of the driving groove, we get different reactions from the ball trapped in the intersection.

This next illustration shows how we went about resolving the forces on the ball.  During a complete rotation, the ball would travel up the driven groove, stop, then travel back down the driven groove to its original position, experiencing different rates of acceleration, or in other words, storing and returning energy.

To model the forces, we chose the “shear plane” for our reference frame.  The forces in the shear plane are modeled in the diagram below.  The shear plane is tangent to the spheres at the intersection of the grooves. 

Everything is rather mundane up to this point.   Then, in order to determine the path of the ball in the intersection, it is necessary to know the slope of the grooves as they move past one another, and to relate this slope back to the rotation of the driving groove.

Initially we tried to resolve these slopes using spherical trigonometry, without success.   We gave up in frustration with that idea, and instead we used finite rotations in order to relate the two slopes back to the axis of rotation.   We would perform a series of three rotations (projections) such that we were in the shear plane, move the ball according to the two slopes, then reverse the three discrete rotations back to the original frame, advance the model a tiny bit, and repeat.  The process called for a series of successive approximations in order to find the slopes each time the model was advanced.  This is because of the fact that the two small circles that parameterize the two different slopes were different radii.  There was no other solution to the problem. 

These notes show the transforms that we used.

pg transform.pdf

These notes show the bookkeeping that was required in order to do the calculus.

Theory of Operation.pdf

Here are typical results we obtained from our model in the Win 96 version of Excel.

Many years later, while looking through our notes, I came across a cryptic illustration that was based on how we originally approached the problem.  In our naiveite we had stumbled across a completely different approach.  That’s when I started posting online to try to solicit help with the project.  You’re one of the people who has been along for the whole ride, since those early days.

In any event, I think that when we try to solve this type of problem using lines with arrowheads for our symbols we don’t have as much power as we have when we replace those symbols with a plane.  That change allows us to do this type of calculation without having to result to the successive approximations that we were using in the first case.  I’m pretty sure that within the next few years someone will prove that this type of problem cannot be solved using vector manipulation.  But that’s simply a prediction based on what I know about the math, and how the two are substantially different than one another.

I'm not quite sure what you mean about chirality being a rabbit hole.  Don't tell me that you don't believe in chirality...

Edited January 4, 2019 by steveupson

[Strange]

On 1/4/2019 at 8:19 PM, steveupson said:

Oh, ok.   I think you might be getting this result because you’re no longer on the circle.  The extent of the circle is from zero at the north pole, to 2V at its southernmost point, so the range of L is 0,2V.  This is because whatever the latitude of the southernmost point of the circle is, the midpoint of this latitude will be the latitude where we would find the circle center, if we were to dig deep enough.  I want to look into this further, but that might take a little while.

If that is the case, then you need to do a better job of defining the domain of the function.

On 1/4/2019 at 8:19 PM, steveupson said:

I'm not quite sure what you mean about chirality being a rabbit hole.  Don't tell me that you don't believe in chirality..

Of course I believe in chirality (after all, I wear left and right socks!). It is a fundamental property of asymmetrical systems.

However, it has no obvious place in your equation which is why I think it would be (yet another) unnecessary distraction at this point.

So, again: Can you give an example of a problem that can't be solved using vectors, and show us how your equation solves it?

[Strange]

On 1/4/2019 at 8:19 PM, steveupson said:

n any event, I think that when we try to solve this type of problem using lines with arrowheads for our symbols we don’t have as much power as we have when we replace those symbols with a plane. 

In computer graphics, there are certainly problems that are easier to solve using the equation of a plane than vectors; this includes clipping, shading and some hidden surface algorithms. This is not the same thing you are talking about (it is very simple mathematics) but it does confirm that transforming from one representation to another can simplify some problems.

Which brings us to another important question: how does one transform a "traditional" definition of direction (an x,y,z unit vector) to your system? And then how does one reverse the process?

 

23 hours ago, Strange said:

Of course I believe in chirality (after all, I wear left and right socks!)

[studiot]

7 minutes ago, Strange said:

?

  23 hours ago, Strange said:

Of course I believe in chirality (after all, I wear left and right socks!)  

 



I laughed my socks off when I read this first time round.

It's still funny now.

+1

[steveupson]

22 minutes ago, Strange said:

Which brings us to another important question: how does one transform a "traditional" definition of direction (an x,y,z unit vector) to your system? And then how does one reverse the process?

The two systems are not in communication with one another at this time.  There is someone working on trying to relate them to one another through creation of physical models for force or energy.  Until we can add distance to the geometry, I think these two things will not transform from one to the other.  There's a lot of work to be done yet, in order to answer this question.

20 minutes ago, studiot said:

I laughed my socks off when I read this first time round.

It's still funny now.

+1

[Strange]

On 1/4/2019 at 10:43 PM, Strange said:

So, again: Can you give an example of a problem that can't be solved using vectors, and show us how your equation solves it?

Can I assume the answer to this is "no", then?

[steveupson]

55 minutes ago, Strange said:

Can I assume the answer to this is "no", then?

I don't know why you make that assumption.  You're definitely not understanding the mathematical issues involved in this type of problem.  I've provided two such examples in this thread already, together with the following statement:

"In any event, I think that when we try to solve this type of problem using lines with arrowheads for our symbols we don’t have as much power as we have when we replace those symbols with a plane.  That change allows us to do this type of calculation without having to result to the successive approximations that we were using in the first case.  I’m pretty sure that within the next few years someone will prove that this type of problem cannot be solved using vector manipulation.  But that’s simply a prediction based on what I know about the math, and how the two are substantially different than one another."

If you think that I'm mistaken somehow, then show me.  Where are the existing equations?

[Strange]

2 minutes ago, steveupson said:

I don't know why you make that assumption.  

Because you didn’t answer the question several times.

But I see you have one example - I can see that this specific problem might be easier to handle in terms of the angles of the forces on the surface of the sphere. (Although without the ability to transform the results back to a more general coordinate system, that may be of limited use.)

I can’t see it easily generalising to different cases, though.

What was the other example? I must have missed it. 

[studiot]

2 hours ago, Strange said:

What was the other example? I must have missed it. 

I seem to have missed both of them.

Can someone please identify them.

Thanks.

[Strange]

Just now, studiot said:

I seem to have missed both of them.

Can someone please identify them.

Thanks.

One was this, I think:

 

 

[steveupson]

.Again, I don't think that you understand the nature of the mathematical issues involved.  The forces are not complicated at all, and can be calculated in a very straightforward manner.

The mathematical question is about how to determine the kinematics.  Or, how can we use vectors to find which direction we're facing from knowing our latitude and the diameter of the circle we're going around.

 

 

[studiot]

Thank you Strange.

 

Quote

Strange

So if we set v to 0.1 (radians) and lambda to pi/2 then the value of alpha is 2.98986 i

In other words an imaginary number. I can't understand what that means. Maybe I have calculated wrong, see here: https://www.wolframalpha.com/input/?i=cot^-1(cos(v)+tan(sin^-1(sin(L%2F2)%2Fsin(v)))),+v%3D0.1,+L%3Dpi%2F2

 

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steveupson

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Posted Friday at 07:19 PM (edited)

Oh, ok.   I think you might be getting this result because you’re no longer on the circle.  The extent of the circle is from zero at the north pole, to 2V at its southernmost point, so the range of L is 0,2V.  This is because whatever the latitude of the southernmost point of the circle is, the midpoint of this latitude will be the latitude where we would find the circle center, if we were to dig deep enough.  I want to look into this further, but that might take a little while.

 

What is 0.2V in Steve Upson's reply?

Is capital V a new variable or what?

And is this a reply to Strnages question in the preceeding post?

[steveupson]

4 minutes ago, studiot said:

Thank you Strange.

 

What is 0.2V in Steve Upson's reply?

Is capital V a new variable or what?

And is this a reply to Strnages question in the preceeding post?

Strange used V to replace upsilon, and L for lambda.  Lambda must be within twice upsilon since upsilon is the center of the circle.  

[studiot]

49 minutes ago, steveupson said:

Strange used V to replace upsilon, and L for lambda.  Lambda must be within twice upsilon since upsilon is the center of the circle.  

Thank you for that explanation.

 

It means that your defining expression is not an identity (for your purpose) and therefore cannot be used as a definition.
This is because one or more variables are limited in extent and you have not shown this.

[steveupson]

II don't think that you understand the mathematical issues that are involved.

There is an equation that expresses a family of functions.  Strange modeled this in mathematica.

The tridentity is a different equation.

Someone asked what the tridentity is good for, and it allows you to make constructions like one that tells you what direction you're facing when on a small circle on a sphere.

Edited January 7, 2019 by steveupson

[steveupson]

4 hours ago, studiot said:

Thank you for that explanation.

 

It means that your defining expression is not an identity (for your purpose) and therefore cannot be used as a definition.
This is because one or more variables are limited in extent and you have not shown this.

It is currently not known by modern mathematics that the three angles have a relationship with one another.  

1) Diameter of the small circle, expressed as an arc length of the sphere

2) Latitude of your feet

3) Direction that you are facing

[Strange]

8 hours ago, steveupson said:

Strange modeled this in mathematica.

No I didn't. I have never used Mathematica.