Quantifying the Physical Property of Direction.

[steveupson]

Hans Milton seems to have been posting questions on http://community.wolfram.com/ 9 days ago. Maybe you can contact him through that forum.

 

As he had made some computer graphics, I imagine that he has used some parameterisations to construct the circle on the sphere and then derive the tangent space.

 

 

Do you understand that the tangent space in this case represents the quantification (sum or magnitude of all the existing directions in the space surrounding the origin of the vector) rather than simply the surface vector?

 

Or, more importantly, if you do understand, can you see how these two would be different and have different mathematical properties?

 

Hans used some animations posted to youtube in order to create the animation. I don't know if he can derive the formula, but I will ask, thanks for the suggestion.

Edited May 19, 2016 by steveupson

[ajb]

Do you understand that the tangent space in this case represents the quantification (sum or magnitude of all the existing directions in the space surrounding the origin of the vector) rather than simply the surface vector?

The tangent space at a point is the collection of all vectors at that point tangent to the manifold in question. In the graphics one presents, in this case, two bases vectors and the 'fill this in' to give you part of that tangent plane.

 

The tangent space is always of the same dimension as the manifold, more carefully the tangent bundle is of dimension twice the manifold. In this sense any basis of the tangent space at a point give you some 'local directions'. Once you have a coordinate system on the manifold you could of course employ a holonomic or coordinate bases.

 

Or, more importantly, if you do understand, can you see how these two would be different and have different mathematical properties?

I do not understand what you are getting at... of course what you are asking could be very special to spheres and so I am missing some point.

 

I don't know if he can derive the formula, but I will ask, thanks for the suggestion.

He either derived it, or took it from a book or something. Either way he should be able to help you more than we can by trying to 'reverse engineer' some animations.

[steveupson]

The tangent space at a point is the collection of all vectors at that point tangent to the manifold in question.

 

Correct, but here we have the collection of all vectors in the space that is surrounding that point.

 

 

 

Edited for clarification> By "space that is surrounding" I mean points adjacent to the origin that are not in the surface of the manifold.

 

We are quantifying the direction of what's in the manifold by quantifying the direction of everything else that isn't in the manifold.

Edited May 19, 2016 by steveupson

[ajb]

We are quantifying the direction of what's in the manifold by quantifying the direction of everything else that isn't in the manifold.

Not that I quite follow what you are saying, but the problem now is that this seems to rely heavily on the embedding of the 2-sphere into 3 dimensional Euclidean space. This suggests that your notions are not intrinsic to the 2-sphere.

[steveupson]

Either way he should be able to help you more than we can by trying to 'reverse engineer' some animations.

 

Here's the model in Mathematica. He constructed the mathematical model using another set of visual animations.

 

Trust me on this one thing, no one, not even me, has ever derived this function. It seems that the folks with the know-how to do it have some sort of bizarre math phobia that I have trouble understanding.

 

Those who have argued countless times "you're wrong, do the math and you'll see" are now arguing "you are wrong and we don't need to be concerned with math at all."

 

 

This is a .cdf so you have to change the extension after downloading it.

Not that I quite follow what you are saying, but the problem now is that this seems to rely heavily on the embedding of the 2-sphere into 3 dimensional Euclidean space. This suggests that your notions are not intrinsic to the 2-sphere.

 

 

That was my biggest, most childish mistake.

 

There are no spheres. There are only the normal structure that exist with any surface.

 

What people seem to be missing, what is difficult to comprehend without actually doing the math, is that 3space has a structure that survives when it is stretched.

NewSphericalTrigFunction, Nr 2, v9.txt

Edited May 19, 2016 by steveupson

[Strange]

Trust me on this one thing, no one, not even me, has ever derived this function. It seems that the folks with the know-how to do it have some sort of bizarre math phobia that I have trouble understanding.

 

Those who have argued countless times "you're wrong, do the math and you'll see" are now arguing "you are wrong and we don't need to be concerned with math at all."

 

You keep making it sound like it is other people's fault that you are not able to work out what this function is. (Rather than a "phobia" I assume no one else cares about this function that bugs you so much. Why would anyone else waste any time on it?)

 

What people seem to be missing, what is difficult to comprehend without actually doing the math ...

 

Interesting statement from someone who can't do the math. Given that, how do you known anyone is missing anything?

[steveupson]

 

Interesting statement from someone who can't do the math. Given that, how do you known anyone is missing anything?

 

 

Your participation is very much appreciated. I'm not being facetious, either. Without your help, and many others like you, this discussion would never have come this far.

 

But to answer your question, I've always been a failure at algebra. I don't know why. Everyone tells me it is lack of effort, but that isn't the case. I can read and understand algebraic expressions fairly adequately, but I can't conjugate the formulas myself.

 

This has nothing at all to do with any ability to do the math. If you honestly think that I don't have the ability, how do you think that the model was created?

 

I don't know what your particular vision of math is, but mine doesn't include algebra, except in the superficial sense of being the language that we use to communicate with each other.

 

Why would anyone else waste any time on it?

 

 

If you think that proving or disproving things using math is a waste of time, then I'm afraid you'll not be able to follow any explanation that I could give.

Edited May 19, 2016 by steveupson

[Mordred]

I find I'm a little confused here.

 

The article is excellent for describing the principles of axis for you.

However you are still looking for one single function that would control those three principles of axis.

 

The problem is the only value shown in the video is e. Which is a common connection point on the animation.

Part of the problem here is the animation only shows one number for e. E being the small connection circle on the animation which follows a tilted circular axis. However Hans doesn't show the coordinate changes of e, and only shows one the degree changes made as e moves. So I assume he's showing the degree change of e as it follows its tilted circular path

 

 

To add complication all three rotating planes have three different relation changes to (let's call it reference e). One plane undergoes an additional tilting compared to the other two planes, and e doesn't maintain the same location on another plane but cuts through that plane in a circular arc.

 

Yet your asking everyone to build one function that describes this?

Especially when there is far more going on in the program than what is shown in the animation.

 

Looking at this the clue is in the plane that e cuts through the plane.

This is the plane that involves both degree changes simultaneously ( though all three planes do ) the point where e resides on that particular plane coincides with e along the tilted circular path.

 

Degrees between the y to x axis and degrees to z to x axis. The change to each is equal and simultaneous.

Edited May 19, 2016 by Mordred

[Strange]

 

 

Your participation is very much appreciated. I'm not being facetious, either. Without your help, and many others like you, this discussion would never have come this far.

 

But to answer your question, I've always been a failure at algebra. I don't know why. Everyone tells me it is lack of effort, but that isn't the case. I can read and understand algebraic expressions fairly adequately, but I can't conjugate the formulas myself.

 

This has nothing at all to do with any ability to do the math. If you honestly think that I don't have the ability, how do you think that the model was created?

 

I don't know what your particular vision of math is, but mine doesn't include algebra, except in the superficial sense of being the language that we use to communicate with each other.

 

 

So, how do you know something is being missed?

 

 

 

If you think that proving or disproving things using math is a waste of time, then I'm afraid you'll not be able to follow any explanation that I could give.

 

That isn't what I said.

[Mordred]

If you plot the degree changes on each plane then the rate of degree change is identical, the only difference is the two simultaneous direction changes for each plane.

[studiot]

 

I find I'm a little confused here.

 

 

 

Only a little confused? +1

 

I don't see the point of the planes at all.

They only serve to confuse things.

 

You have a pole which wanders halfway round a small loop (ie 180) on a sphere as the slider is moved from 0 to 90.

 

This is like a demonstration of half a Chandler wobble.

So?

Edited May 19, 2016 by studiot

[Mordred]

Plot movement of e, then plot each planes relation to e. However here is the thing as e moves two planes are getting another rotation.

 

The plane that e follows undergoes

 

Two rotations, (your reference plane) this is the plane that e dissects. however the other two planes which follows the coordinate change in e under go a rotation as e moves. (Which is the degree change)

 

Lol it looks fancy but really it's not. The solution is surprisingly easy.

 

Referring to the graphic Steve wanted to reverse engineer. Programming that is surprisingly easy.

 

 

Here's a hint because of how easy a math problem it is I don't want to give away the solution. If you remove the reference plane. All quadrants of the x,y,z graph undergo the identical movement.

Edited May 19, 2016 by Mordred

[studiot]

But what does this have to do with direction?

[Mordred]

Nothing not in the manner the OP thinks it does. Quite frankly what's shown in the animation is basic trig.

Edited May 19, 2016 by Mordred

[studiot]

The OP originally pedelled this stuff as

 

"A new function in Spherical Trigonometry", coupled with the claim that spherical trigonometry only deals with great circles, not small circles.

 

I don't think it bears any relation to that either

Edited May 19, 2016 by studiot

[Strange]

The OP originally pedelled this stuff as

 

"A new function in Spherical Trigonometry", coupled with the claim that spherical trigonometry only deals with great circles, not small circles.

 

 

That is obviously nonsense. F`or example, it is quite straightforward to calculate the ratio of radius to circumference of a circle on the surface of a sphere. Or the angles of a triangle.

 

The OP seems to have missed out on several centuries of non-Euclidean geometry. ("No, I understand all that. I just can't do anything with it.")

[Mordred]

Not that I can see from the rotation plot on each quadrant. All points on the rotations are easily definable via standard trig. The graphic representation merely plots an identical set of relation change between axis.

Edited May 19, 2016 by Mordred

[studiot]

 

The OP seems to have missed out on several centuries of non-Euclidean geometry. ("No, I understand all that. I just can't do anything with it.")

 

 

We have only just reached the eighth page so we have a long way yet to go here.

There are twelve pages on the last website ( a mathematics one) and there could well be others on other sites I haven't seen.

Edited May 19, 2016 by studiot

[Mordred]

Roflmao

[pzkpfw]

(And it's all our (in the general sense) fault: http://www.thephysicsforum.com/mathematics/9252-defining-new-function-merged.html see post #11. Sigh. )

[studiot]

In one sense Steve has reached the late 20th / early 21st centuries.

 

Few realise the enormous benefit that ensued by moving from a malleable geodetic manifold to a defined XYZ coordinate system that satellite positioning has permitted.

Prior to this calculations on the Geoid were exhausting.

 

@pzkpfw I thought post#23 on your link was more fun +1 for that.

Edited May 19, 2016 by studiot

[Mordred]

Lol there is a significants at pi/4 that does make doing trig relations easier to understand. However it's well known lol.

[Strange]

I must have been really bored ... the attached graph shows the value of the angle from the animation as a function of E (blue line). The red line is a sine curve over the same range for comparison (I thought the result was simply going to be a sine, at first).

[Mordred]

You need to apply the same trig function to two axis. Y to x and z to x. Which I think you've already applied.

 

Lol this reminds me of the problem set I had learning 3d graphics back in the trash 486 days. I asked myself how do you program a 2d screen to have a third dimension.

 

The trig functions bogged down that poor PC lol. Granted I was recalculating each pixel along a line.

Edited May 20, 2016 by Mordred

[steveupson]

Thank you Strange. You have no idea how envious I am.

 

http://mymathforum.com/math/331919-i.html

Edited May 22, 2016 by steveupson