Jump to content

GR ch21 Mach?


Capiert

Recommended Posts

Rotational Velocity

 defies COM (conservation of momentum) & COE (conservation of energy)
 & is 1 of the strangest things to solve
 because of its non_linear acceleration.
Based on Pythagoras('s 90 degree triangle)
 coe & com don't add (up).
(But) Is that why we need relativity?
Einstein made a peculiar remark in ch21 (1920)
 commenting (that) Ernst Mach (had) recognized
 the rotational energy problem the most.(?)
 
 
Does anyone know what he was talking about there
 (about what Mach had recognized, exactly)?
 
I've been trying for years to make coe & com balance (linearly)
 but it just doesn't work.
I also know the Sine's acceleration
 peak's at 50 degrees.
E.g.
Which reminds me
 of the Tornado region
 for Oklahoma & Kansas latitude (angle).
 
Thanks in advance.
 

SIN_minus_4DA_max_50_degrees_2016_12_15_1313_PS_Wi.xlsx

Link to comment
Share on other sites

1 hour ago, Capiert said:

Rotational Velocity

 defies COM (conservation of momentum) & COE (conservation of energy)

No, neither are violated if you do the analysis correctly 

Quote

 

I also know the Sine's acceleration

 peak's at 50 degrees.

 

You'll need to explain why you think this is the case. Why not just take the derivative to find the maximum slope?

 

Link to comment
Share on other sites

On 18 May 2018 at 2:02 PM, Endy0816 said:

Talking about Mach's principle.

https://en.m.wikipedia.org/wiki/Mach's_principle

Thanks for the link.

 

It does seem to be (rather) involved

 just to describe a Foucault pendulum.

E.g. The tendency to continue moving

 in (apparently) a straight line

 wrt the stars (average position)

 approximated as (ruffly) fixed.

(We all know they are moving,

 & very fast, though.)

On 18 May 2018 at 2:15 PM, swansont said:

No, neither are violated if you do the analysis correctly 

You'll need to explain why you think this is the case.

As you can see from the excel table (given),

 a mark on a rotating disc

 would have a variable y speed.

Assuming the radius has 1 unit (e.g. 1 m).

The y distance travelled per degree

 varries.

The largest distance is between 49 to 50 degrees,

 & 50 to 51 degrees.

I've assumed the time duration is identical

 between each (single) degree

 (for constant rotational speed).

I've only (manually tracked=) compared the y displacement

 for the 1st quarter of a cycle

 assuming constant rotational speed.

The other parts of the cycle can also be done in the same way, if needed.

Quote

Why not just take the derivative to find the maximum slope?

I wanted a feel (ruff idea) for the real speed, component(s).

Does the derivative look like my graph's (asymmetric) curve?

No?

I've discussed that I don't trust your calculus (in other threads),

 concerning the +C error's definition (=my derivation).

Edited by Capiert
Correction, x vs y.
Link to comment
Share on other sites

1 hour ago, Capiert said:

 

As you can see from the excel table (given),

 a mark on a rotating disc

 would have a variable y speed.

Not surprising, for an object moving in a circle.

But to move in a circle it requires a force, which means that it's a situation where the momentum would not be conserved.

Quote

 I've discussed that I don't trust your calculus (in other threads),

Doesn't matter. You don't get to introduce a pet theory in a mainstream thread.

Max slope of sine is at 0 or pi radians. If you get a different answer, you've made a mistake.

 

Your calculations make no sense.

Link to comment
Share on other sites

2 hours ago, Capiert said:

Thanks for the link.

 

It does seem to be (rather) involved

 just to describe a Foucault pendulum.

E.g. The tendency to continue moving

 in (apparently) a straight line

 wrt the stars (average position)

 approximated as (ruffly) fixed.

(We all know they are moving,

 & very fast, though.)

...

He's talking about the motion of the stars that occurs as he 'spins' in place and the centrifugal force that is linked to that.

 

Edited by Endy0816
Link to comment
Share on other sites

16 hours ago, swansont said:

Not surprising, for an object moving in a circle.

But to move in a circle it requires a force, which means that it's a situation where the momentum would not be conserved.

Doesn't matter. You don't get to introduce a pet theory in a mainstream thread.

My intention

 was only to satisfy my curiosity

 if Relativity was necessary

 because of rotary motion

 because I don't get further on the idea.

I did not include the 50 degree problem

 except to indicate the type of answers

 I was looking for.

As you can assume non correlation

 unless Mach's ideas are still open.

Quote

Max slope of sine is at 0 or pi radians. If you get a different answer, you've made a mistake.

Your slopes may be that,

 but my calculations show

 y speed changes:

 (from increasing)

 to decreasing,

 at 50 degrees.

Quote

Your calculations make no sense.

Maybe you can suggest

 an appropriate (new) thread name

 (& where),

 to clarify the error.

Edited by Capiert
Link to comment
Share on other sites

7 hours ago, Capiert said:

 Maybe you can suggest

 an appropriate (new) thread name

 (& where),

 to clarify the error.

You reject valid math and introduce incorrect math. Even if you want to do this numerically, all you need to do is look at the difference in the values of sin(x) for small increments. No need to do the nonsense you did. 

Link to comment
Share on other sites

7 hours ago, swansont said:

You reject valid math and introduce incorrect math.

What's wrong?

I've used an alternative method

 to test the validity of your math.

I've eliminated the linear portion

 to examine (e.g. magnify) the curvature better.

Quote

Even if you want to do this numerically, all you need to do is look at the difference in the values of sin(x) for small increments.

That's what I did,

 (but) subtracting the linear part 1st.

I.e. 4 times the quarter cycle=1 (=100 %).

4 times zero cycle=0 (0 %).

From 0 to 100 % was a straight line,

 which I removed,

 to see the curve better.

Linear+curvature=total (y).

Quote

No need to do the nonsense you did. 

That (method of mine) leaves (=left, only) the curvature (=non_linear part) remaining

 (as fine tuning).

(Are you saying your own proposal is non_sense?)

Edited by Capiert
Link to comment
Share on other sites

40 minutes ago, Capiert said:

I've used an alternative method

 to test the validity of your math.

No. You have made up some meangless math of your own. 

Your results have no physical meaning or validity. 

Edited by Strange
Link to comment
Share on other sites

1 hour ago, Strange said:

No. You have made up some meangless math of your own. 

Your results have no physical meaning or validity. 

My results indicate the y acceleration is non_linear.

But that's obvious.

Link to comment
Share on other sites

21 minutes ago, Capiert said:

My results indicate the y acceleration is non_linear.

But that's obvious.

It is obvious but not really supported by your meaningless calculations.

A simple sine formula tells you this and correctly shows where the greatest acceleration is.

Link to comment
Share on other sites

3 hours ago, Capiert said:

What's wrong?

I've used an alternative method

 to test the validity of your math.

I've eliminated the linear portion

 to examine (e.g. magnify) the curvature better.

You get the wrong answer.

You want to know how much y varies, so you want to know how much sin(x) varies. sin(x) varies the fastest per degree near 0º (and 180º), not 50º

sin(1)-sin(0) = 0.0175

sin(50)-sin(49) = 0.01133

 

3 hours ago, Capiert said:

That's what I did,

 (but) subtracting the linear part 1st.

I.e. 4 times the quarter cycle=1 (=100 %).

4 times zero cycle=0 (0 %).

From 0 to 100 % was a straight line,

 which I removed,

 to see the curve better.

Linear+curvature=total (y).

That (method of mine) leaves (=left, only) the curvature (=non_linear part) remaining

 (as fine tuning).

(Are you saying your own proposal is non_sense?)

If your method is valid, why do we get different answers?  

1 hour ago, Capiert said:

My results indicate the y acceleration is non_linear.

But that's obvious.

If you want to get the acceleration, then you need to take an additional difference, because y is displacement. Derivative of displacement gives velocity (one difference in y). The derivative of velocity ( a second difference in y) gives acceleration.

Link to comment
Share on other sites

25 minutes ago, swansont said:

You get the wrong answer.

You want to know how much y varies, so you want to know how much sin(x) varies. sin(x) varies the fastest per degree near 0º (and 180º), not 50º

sin(1)-sin(0) = 0.0175

 0.011 Linear

+0.00634 non_linear

=0.01734 total

25 minutes ago, swansont said:

sin(50)-sin(49) = 0.01133

0.556

-0.544

=0.012 linear

 

0.21049

-0.21027

=0.00022 non_linear

 

0.012

+0.00022

=0.01222 total

25 minutes ago, swansont said:

If your method is valid, why do we get different answers?  

I suspect you are complaining about the inconsistent number of decimal places

 & rounding errors in excel.

25 minutes ago, swansont said:

If you want to get the acceleration, then you need to take an additional difference, because y is displacement. Derivative of displacement gives velocity (one difference in y). The derivative of velocity ( a second difference in y) gives acceleration.

How do you deal with non_linear acceleration? (Numerically).

Especially when your calculus is an approximation?

Link to comment
Share on other sites

9 minutes ago, Capiert said:

I suspect you are complaining about the inconsistent number of decimal places

No.

10 minutes ago, Capiert said:

Especially when your calculus is an approximation?

It isn't. Certainly not as much as just making stuff up.

Link to comment
Share on other sites

1 hour ago, Capiert said:

 0.011 Linear

+0.00634 non_linear

=0.01734 total

0.556

-0.544

=0.012 linear

 

0.21049

-0.21027

=0.00022 non_linear

 

0.012

+0.00022

=0.01222 total

How do you determine the "linear" part vs "nonlinear" part? How are you using those terms?

Why does linear vs nonlinear matter if you are interested in the change in y?

1 hour ago, Capiert said:

I suspect you are complaining about the inconsistent number of decimal places

 & rounding errors in excel.

I'm ignoring your excel spreadsheet, other than noting your unexplained and nonsensical equations.

1 hour ago, Capiert said:

How do you deal with non_linear acceleration? (Numerically).

Especially when your calculus is an approximation?

Calculus in not an approximation. And it gives instantaneous values (i.e. the differential is in the limit as the increment goes to zero), so nonlinearity isn't an issue.

Link to comment
Share on other sites

3 hours ago, swansont said:

How do you determine the "linear" part

The linear part is a straight line

 drawn from 0 degrees (value y=0)

 to 90 degrees (value y=1)

 using the fraction

 of a cycle

 (angle theta in degrees

 divided by 360 degrees)

 then multiplied by 4.

Let the decimal angle

 A=theta/(360 degrees).

That linear (straight) line('s y value)

 is 4*theta/(360 degrees).

Quote

vs "nonlinear" part?

The non_linear part (=Difference)

 dif=(Sin (theta) - (4*theta/(360 degrees))

 is simply

 the sin of the angle theta

 (then) minus the linear part.

Quote

How are you using those terms?

I've split them in 2

 so I can look at each separately.

Quote

Why does linear vs nonlinear matter if you are interested in the change in y?

I haven't seen non_linear (acceleration) physics delt with much.

I also have NOT seen a simple equation for angle, either.

Most are Taylor series.

You probably know I do NOT like endless series very much.

E.g. The sooner you start to solve them

 (by counting to infinity),

 the sooner you will be finished.

I don't like approximations (for precision) either

 unless I am lazy,

 &/or (when) things are too complicated.

 

The main reason is I do not (yet) have a (simple) continuum

 between linear & rotational motion.

So I have split up the structure (of motion)

 to interogate more.

 

Physics has split things into non_ & Eulclidian geometry

 to get around that big problem.

I.e. built a wall, not a bridge.

 

I see the problem

 similar to the Stern_Gerlach experiment.

An electron moving thru a (static) magnetic

 where the intensity changes wrt position,

 is acted on like a moving magnetic field exists

 to push the electron

 either up or down

 depending on which side of the (gradient) change (left or right)

 is an increasing or decreasing magnetic field.

Shoot the beam only into 1 half of the gradient (either left or right side)

 then the beam will go in only 1 direction

 either up or down;

 but it won't split into 2.

 

I attempted to break up the acceleration

 into its (linear & non_linear) components, also.

Quote

I'm ignoring your excel spreadsheet,

At least you're honest.

Quote

 other than noting your unexplained

The formulas are repeated in the cells,

 & produce trackable results

 to figure things out

 if needed,

 e.g. a working example

 if doubt arise.

Quote

and nonsensical equations.

The results speak for themselves,

 we get the same results with more (e.g. 6) decimal accuracy.

=So my values are NOT wrong,

 but instead confirm yours.

5 hours ago, swansont said:

sin(1)-sin(0) = 0.0175

 0.011111 Linear

+0.006341 non_linear

=0.017451 total

Quote

sin(50)-sin(49) = 0.01133

0.555556

-0.544444

=0.011112 linear

 

0.210489

-0.210265

=0.000224 non_linear

 

0.01222 linear

+0.000224 non_linear

=0.011336 total

Quote

Calculus in not an approximation.

I'm sorry, but I still can't trust it completely yet.

Quote

And it gives instantaneous values (i.e. the differential is in the limit as the increment goes to zero), so nonlinearity isn't an issue.

I can't say I can enjoy that statement for variable (non_constant) acceleration.

SIN_minus_4DA_max_50_degrees_2018_05_21_1552_PS_Wi_(6_Decimals).xlsx

 

 

Edited by Capiert
Link to comment
Share on other sites

8 minutes ago, Capiert said:

 The results speak for themselves,

 we get the same results with more (e.g. 6) decimal accuracy.

=So my values are NOT wrong,

Our answers do not agree if you think the maximum rate of change is at 50 degrees. Also, your results are not more precise or accurate than other results. 

It's hard to put this into context and see what your conceptual errors are. You say this is about rotational velocity, which has a fairly straightforward solution for e.g. an object moving in a circle, and does not have a nonlinear acceleration.  

What problem are you trying to solve?

 

 

Link to comment
Share on other sites

19 minutes ago, Capiert said:

The linear part is a straight line drawn from 0 degrees (value y=0) to 90 degrees (value y=1) using the fraction of a cycle (angle theta in degrees divided by 360 degrees) then multiplied by 4.Let the decimal angle A=theta/(360 degrees).That linear (straight) line('s y value) is 4*theta/(360 degrees).

You seem to be calculating the difference between the sine of an angle and a distance along a chord where that angle intersects. With some random multiplicative factors thrown in. This is clearly mumbo-jumbo. Not even numerology.

 

21 minutes ago, Capiert said:

Physicshassplitthingsintonon_&Eulclidiangeometrytogetaroundthatbigproblem.

Nonsense. This has nothing to do with non-Euclidean geometry. And the "big problem" only exists in your fevered imagination.

22 minutes ago, Capiert said:

I'm sorry, but I still can't trust it completely yet.

Maybe you should do a course in it.

22 minutes ago, Capiert said:

I can't say I can enjoy that statement

I doubt anyone cares about that.

Link to comment
Share on other sites

39 minutes ago, swansont said:

Our answers do not agree if you think the maximum rate of change is at 50 degrees.

If that's not clear,

 I mean the non_linear ("part of") acceleration is maximum at 50 degrees.

I.e. Excluding the linear (acceleration) portion.

It is a more dynamic part of acceleration.

I do NOT know its properties well enough yet.

I.e. Changes of changes (of changes..).

39 minutes ago, swansont said:

Also, your results are not more precise or accurate than other results. 

It's hard to put this into context and see what your conceptual errors are. You say this is about rotational velocity, which has a fairly straightforward solution for e.g. an object moving in a circle,

That's understandable (for me) til here.

Quote

and does not have a nonlinear acceleration.  

But the y "component" (of the mark on the rotating disc) has non_linear acceleration.

So that statement sounds ambiguous, & confusing.

Quote

What problem are you trying to solve?

(COM & COE Balance without trig.).

The y_component's speed(s) & acceleration(s).

A disc rotating at constant angular speed

 has x & y components which are variable (changing) speeds.

I'm curious how I can solve (what values) the angle, the speed & the accelerations (are)

 for a single component, e.g. y component.

Later they get applied to the mass in the disc.

E.g. a single atom

 on the circumference.

Both components are at 90° to each other,

 but you use the 3rd axis (virtually)

 for problems like that.

 

I want irrational exponent(ial) acceleration capability.

Edited by Capiert
Link to comment
Share on other sites

1 hour ago, Capiert said:

If that's not clear,

 I mean the non_linear ("part of") acceleration is maximum at 50 degrees.

I.e. Excluding the linear (acceleration) portion.

It is a more dynamic part of acceleration.

Your notion of what is linear and nonlinear seems arbitrary. There are an infinite number of linear functions you could choose, rather than theta/90º and there is no apparent physical relevance to your choice.

1 hour ago, Capiert said:

 But the y "component" (of the mark on the rotating disc) has non_linear acceleration.

 

So what? The x component would have it too. The total value is a constant.

And such systems are often analyzed in polar coordinates, rather than Cartesian.

1 hour ago, Capiert said:

 

So that statement sounds ambiguous, & confusing.

(COM & COE Balance without trig.).

The y_component's speed(s) & acceleration(s).

As i've pointed out before, you are not analyzing (such as it is) the speed. You are analyzing the velocity (distance/time)

 

Link to comment
Share on other sites

4 hours ago, swansont said:

Your notion of what is linear and nonlinear seems arbitrary.

I don't follow you there.

Quote

There are an infinite number of linear functions you could choose, rather than theta/90º

Would you like to state a few examples (please)?

Maybe it would occur to me then,

 why they would NOT be interesting to me

 & why I prefer my selection,

 in order to explain (to you) my preferences.

1 reason (=goal) is also to establish

 the (decimal) cycle (notation, syntax)

 for angle.

I suspect I might better interpret

 binary angle (for myself).

Quote

and there is no apparent physical relevance to your choice.

Said differently,

 looking at things

 from a different perspective

 (in a simpler syntax)

 I might get an aha (experience).

I'm not cean on the astronomical radian syntax,

 because it is redundant.

I.e. there is too much (garbage=luggage) there.

We live with many metric

 & decimal units.

But NOT everything

 is done=measured so to that=those standards.

I prefer to experiment

 with other( method)s, & simplifying.

Quote

So what? The x component would have it too.

Quite right.

If I can follow (=understand, or comprehend)

 1 component,

 then the 2nd component

 is (almost) guaranteed

 (to be similar, (but)

 just delayed by 90 degrees

 e.g. 1/4 of a cycle.

Thus the 1/4 selection

 (& the factor 4)

 is not random

 but instead

 with intention.

You simply had not recognized

 that it was

(so, for my choice).

Quote

The total value is a constant.

Yes, but I still can NOT follow only the parts

 to my satisfaction.

Quote

And such systems are often analyzed in polar coordinates, rather than Cartesian.

Yes, that's why I am disappointed

 & pursue my search

 to make it (=Cartesian x & y) work=fit

 as I would like to have=see it.

Quote

As I've pointed out before, you are not analyzing (such as it is) the speed.

Maybe you could explain what you mean there?

I assume you mean angular speed (there, e.g. polar coordinates);

Quote

You are analyzing the velocity (distance/time).

 & (there, you mean) the vectors (e.g. Cartesian x vs y components)

 of the ((polar) angular) speed.

 

I hope that helps (you, as an explaination)?

Does it?

Edited by Capiert
Link to comment
Share on other sites

16 minutes ago, Capiert said:

Would you like to state a few examples (please)?

  • theta
  • theta/10º
  • theta/45º
  • theta/180º
  • etc.
18 minutes ago, Capiert said:

Said differently, looking at things from a different perspective (in a simpler syntax) I might get an aha (experience).

I don't think you are going to get an 'aha moment' from looking at things wrongly, which is what you always do.

Quote

I'm not cean on the astronomical radian syntax, because it is redundant.I.e. there is too much (garbage=luggage) there.

Radians have nothing to do with astronomy. It is a natural unit, based on the value of pi. There is nothing redundant about it; it has exactly the same amount of information as using the arbitrary (man-made) units of degrees.

22 minutes ago, Capiert said:

I hope that helps (you, as an explaination)?Does it?

No. It is just more of your ignorant nonsense. Written in your usual infantile way.

Why not take a year or two out to do an introductory course in mathematics?  And then one on communication skills. Then you might be able to contribute something sensible to the forum.

Link to comment
Share on other sites

8 hours ago, Strange said:
  • theta

What do you mean there?

Quote
  • theta/10º

That 1 doesn't look useful (=interesting, for my intentions).

Quote
  • theta/45º
  • theta/180º

Those look interesting.

Quote
  • etc

Yes if you mean the binary relations.

Quote

I don't think you are going to get an 'aha moment' from looking at things wrongly, which is what you always do.

That's the interesting thing, orientation.

The do's & don't's.

I firmly believe,

 we learn from our mistakes (too);

 & if we have learned our lesson in life well,

 (then) we will NEVER make the same mistake again.

The do's & don't are both ends of the arrow.

The don't (=errors) reinforce

 & confirm which direction

 correct should be.

E.g. "Which direction is warmer", not necessarilly hot,

 (if you want to put it into child's language).

I generally stumble into every pothole that is possible.

But that's subjective testing.

At least I'm reassured

 I don't have to go that way again

 in the future.

I suppose (with a few clues) we (=I) can evaluate

 which direction is right.

Indicatores exist.

But I suppose your problem

 is you are impatient

 with such a method

 because it does NOT fit you timing scheme,

 & expectations.

We're all learning,

 & each at their own speed.

Each person has there own strengths & weaknesses.

You may be an ace in physics,

 & the fastest bunny on 2 feet;

 but an absolute looser in some other faculty,

 or quality of life.

That's all part of being human.

Quote

Radians have nothing to do with astronomy.

I thought they were a convenient unit for calculating stellar distances,

 (similar to the way wavenumbers are convenient notation).

Quote

It is a natural unit, based on the value of pi. There is nothing redundant about it; it has exactly the same amount of information as using the arbitrary (man-made) units of degrees.

The degree is also NOT my preference.

Decimal cycle is.

Quote

No. It is just more of your ignorant nonsense.

How can non_sense be ignorant? (when People are).

What purpose is info (to me) if I can NOT use it.

I can only digest info that "I" am capable of.

What is the purpose to learn things I will never use nor need?

Should education be infinite for a finite life?

I think it is meaningful to attack=tackle things from where we are.

Quote

Written in your usual infantile way.

Are you trying to play the parental role?

Quote

Why not take a year or two out to do an introductory course in mathematics?

Yes, that is an interesting proposal.

But who can afford it,

 where do I find it?

& will I be bored?

(Boredom is a diluted form of pain,

 & is (emotional) stress, & noticed as uncomfortable.

Boredom is NOT rewarding,

 & turns off the mind,

 so the message

 doe NOT sink in.

I doubt that I am a masochist;

 although being here sometimes I have my doubts.

Especially with the software.)

Quote

And then one on communication skills.

I wouldn't be this way if I didn't get bored

 or loose track (in all the confusion).

You guys blew it with a (scattered=) "non"_unified field.

Quote

Then you might be able to contribute something sensible to the forum.

Oh! Now we're getting subjective.

(I prefer my 6th.)

Edited by Capiert
Link to comment
Share on other sites

1 hour ago, Capiert said:

What do you mean there?

You are multiplying by an arbitrary factor. You could multiply by 1 or divide by a million. It would be equally meaningless.

1 hour ago, Capiert said:

The degree is also NOT my preference

Then why are you using degrees?

1 hour ago, Capiert said:

Decimal cycle is.

What is "decimal cycle"?

Do you mean gradian: https://en.wikipedia.org/wiki/Gradian

1 hour ago, Capiert said:

Are you trying to play the parental role?

It looks like someone needs to. And it isn't going to be you.

1 hour ago, Capiert said:

Yes, that is an interesting proposal.

But who can afford it,

 where do I find it?

There are lots of free online courses provided by good universities, etc.

1 hour ago, Capiert said:

will I be bored?

I doubt it. The courses are quite demanding. You might pretend to be bored as an excuse to avoid doing some hard work and learning, though.

You might also have to learn to write like a grown up.

 

Edited by Strange
Link to comment
Share on other sites

Guest
This topic is now closed to further replies.
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.