[Halc]

So, how could it be that suddenly we don't have the technology to monitor the location of G stars at a sphere of 1,000LY or 4,000 LY around the solar system?

Technology is there up to a point. A type G star is quite difficult to see at 4000 LY.  It takes more resources to see than a bright star.  Resources, not technology, is what prevents a full mapping beyond a certain point.  I do not know where that point is. For all I know, they've very much done the survey you need. They don't write pop articles about each one, and your research seems confined to pop articles.

We could easily monitor the aria between the arms near our location.
So, why our scientists do not present this important verification?

They have. Nothing is hidden. Plenty of people have discovered new things by mining the public database. There is far more information in that data than there is manpower to process it.

[quoter=Halc]You don't have a theory.

That is incorrect
I have full valid theory. You just don't let me go on.[/quote]
Lying to yourself is fairly pathetic.  I've let you go one and all I get is endlessly repeated assertions that would be contradictory if they actually made some predictions, which they don't.

Therefore, by definition I offer an outcome which currently no one considers.

You've offered no outcome.  I can make no predictions of outcome without a theory.  You cannot predict the most simple model, let alone anything as complex as a galaxy.  Nobody considers the outcome because there isn't one proposed to consider.

We have to verify if my expectations are real. If so, my theory is real. If no - We will set it in the garbage.

You will do no such thing.  You will discard data that does not meet your expectations. That's how you've worked so far.

If you find one verification that contradicts any outcome from that theory - than this theory is incorrect by definition.

No theory is correct by that definition.  Your theory, were it to exist, would predict everything, and it would be more correct if it did a better job of predicting everything than any other theory, even if some measurements contradicted it.  One measurement will not do at all.  If it gets one thing right and totally fails everywhere else, it isn't a very good theory.

It isn't accepted to see any contradiction at any kind of theory - even if 100,001 scientists claim that this is the ultimate theory.

Who has claimed to have the ultimate theory?

If our scientists verify a contradiction between their expectations to the verified evidences, than - they must set this theory at the garbage of the history and start from point Zero - as we all expected for any theory.

No need to go to point zero.  What is known works very well.  All technology is based on theories that are not perfect. Newton might not have got everything right, but I still duck if a rock comes at my head, because his imperfect theory quantifies how much it will hurt. Going back to point zero, I'd have to let it hit me to test if still hurts as predicted by some totally new idea.

So run your idea past a few simple tests. Your refusal to do that demonstrates your lack of faith in the theory, and more importantly, the lack of the theory. You've made some amazing claims, many of which are far more trivial to verify than searching a petabyte database for a swath of deep space with no stars of a certain class.

I insist to get the data about G stars between the arms at the nearby aria.
You would find that my expectation is correct by 100%. (Not 90%, not even 99%).
This is real science! That is how our scientists should work.

Actually, it sort of is how it works. Go for it. The data exists. You just need to extract it.
 It will not verify the idea, but it will serve as a falsification test.  No evidence can verify any theory, only falsifications. A good theory is one that isn't easily falsified by any of multiple tests.

Set the expectation - and verify if your expectations are correct.

Not sure how that works here.  If you find a G-type star in a place you're looking, you figure you're looking in the wrong place and you go elsewhere. How common are these empty regions supposed to be that you'll declare failure if every region turns up with stars in it?  You have no predictions about what percentage of random volumes of space of size X will be empty and what percentage are these bridges.  Without that prediction, I don't see how any finding one way or the other is going to verify or sink the idea.

[Dave Lev (OP)]

So, how could it be that suddenly we don't have the technology to monitor the location of G stars at a sphere of 1,000LY or 4,000 LY around the solar system?

Technology is there up to a point. A type G star is quite difficult to see at 4000 LY.  It takes more resources to see than a bright star.  Resources, not technology, is what prevents a full mapping beyond a certain point.  I do not know where that point is.

Sorry - it shouldn't be so difficult to see them.
In the following article: http://www.solstation.com/stars3/100-gs.htm - it is stated:
"Yellow-orange "G" stars are among the most common stars seen with the naked eye in Earth's night sky due to their relative abundance. "
So, if we can see those nearby G Stars in naked eyes, why can't we see similar stars at 4000LY with simple technology?

Not sure how that works here.  If you find a G-type star in a place you're looking, you figure you're looking in the wrong place and you go elsewhere. How common are these empty regions supposed to be that you'll declare failure if every region turns up with stars in it?  You have no predictions about what percentage of random volumes of space of size X will be empty and what percentage are these bridges.  Without that prediction, I don't see how any finding one way or the other is going to verify or sink the idea.

I have already explained it and I'm ready to do it again:
The theory for Spiral arm is quite simple -
Spiral arm is an object. It is made out of stars. The stars are gravity bonded with each other at any given aria of the Arm. The density of stars is vital element to keep them all in a long arm.
At any distance from the galactic center, we might find different density. As we move inwards - the density should be higher. As we move outwards, the density should be lower.
Between the spiral arms there could gateways (bridges and branches) of stars. However, those getaways are not so common. Therefore, we might find here and there some gateways, but no more than that.
We are located at Orion arm at a radius of 28,000 LY from the center of the galaxy. We know that the density of stars in the nearby aria is 64 stars per 50LY sphere, or 512 per 100LY Sphere.
This fixed density is vital for the gravity bonding of the stars in the arm (at that 28,000 Ly from the center of the galaxy).
Therefore, if there are stars between the arms, those stars should have a similar density as we have in our location and they also must be connected at least to one of the arms or set a bridge between the arms.
The Sun is located at the edge of the Orion arm - facing Sagittarius arm.
As we start to move from our location to that nearby arm, we should see that as long as we are in the Orion arm the density of stars should be: 64 per 50 LY. However, at some point we have to cross the edge of the Orion Arm.
Once we cross that edge, we should get to a zone without stars. (Not even a single star). 
Therefore, the answer to your question is:
Technically it is expected to find wide arias between the arms without even a single star, while it must be more difficult to find those gateways and bridges between the arms.
So, simple and clear.
It is there. I have full confidence about it!

[Dave Lev (OP)]

The empty spaces should outnumber the occupied volumes, but no volume is specified.  If I select random spheres a light hour across within our solar system, I will find only one with a G class star in it.  Your prediction is met trivially right here in the dense arm, and it demonstrates nothing since nowhere are stars that dense.  If I do the same for random spheres of 10000 LY in diameter centerd between arms, I will find zero such volumes devoid of G type stars.  I would have to go between galaxies to find a space that big with nothing of that size in it, and it wouldn't falsifiy your idea because you don't posit empty volumes that large.  So at what size would I still expect to find more empty spaces than occupied ones if I look only between the arms?

Why do you take it to the extreme? (1 light hour or 10,000 LY)
I have specified a volume.
50 Ly or 100 Ly.
We have found 64 stars in a 50 LY sphere and 512 in 100 LY.
So this is the requested volume.
As I have stated, the chance to find a gateway/Bridge (of stars) between the arms is quite low.
If I understand it correctly, the distance between the Orion Arm to Sagittarius is about 1,000 LY - 2000LY.
Therefore, if we set a sphere of about 50 LY - 500 LY just at the center between the arms, (in front of our location) there is high chance that we won't find there even one star.
Why is it so difficult?

[Dave Lev (OP)]

It is difficult because you didn't really say that before.  I was pointing out the places that were lacking.  I'm satisfied now.

Thanks

So it is in your hands to prove us all wrong, and that only is evidence for your idea that spiral arms are objects, which was neatly refuted in the article I posted.

Yes, I agree

If they were objects like curved spokes on an alloy car wheel, their speed would be proportional to the distance from the center of rotation (a constant slope rotation curve).  The rotation curve only looks like that up to about 0.5 KPC and the arms don't even reach that far in.
Your unwillingness to face that falsification is a good deal of why nobody can take your ideas seriously.  All reasonable scientific proposals concentrate most on the parts where they fail predictions since those are the areas pointing out where improvement is needed.

Rotation Curve
Actually, our scientists try to give an explanation only for the spiral arms.
They don't even try to explain the activity at the Bulge, Bar and outside the arms.
I can explain it all.
Please look at the following diagram:
https://pages.uoregon.edu/jimbrau/BrauImNew/Chap23/6th/23_21Figure-F.jpg
Our scientists try to explain that rotation cure by dark matter as follow:
http://aerorocket.com/GalaxySpeed.html
"The MathCAD analysis presented below uses Newton's shell theorem to determine the combined effect of ordinary matter and dark matter on the rotation speed of stars as they orbit around the galaxy. Specifically the theorem states that when a star system is at radius r from the center of the galaxy, the portion of the galaxy that lies outside the shell of radius r does not produce a net gravitational force on the star system. In addition, only the portion of the galaxy that is inside a shell of radius r produces a net gravitational force on the star system and that mass is located at the center of the galaxy. Figure-3 presents the results of the analysis to determine the rotation curve of the Milky Way galaxy compared to the rotation curve using a two-body method. Finally, the rotation curve of the solar system is determined using the two-body method and the results are compared to actual planetary orbital speeds. These analyses clearly illustrate that a disk of matter that forms the Milky Way galaxy produces a flattened rotation curve that matches measured orbital speeds of stars that orbit the galaxy"
That is totally incorrect idea as as I will explain.
In order to understand that, we need to look at the following diagram which gives better visibility about the rotation cure:
https://www.researchgate.net/figure/Decomposition-of-the-rotation-curve-of-the-Milky-Way-into-the-components-bulge-stellar_fig4_45893184
We see that the rotaion cure is not an nice cure as we have seen in the first diagram.
We have the following sections:
1. Very close to the center, (at the bulge) we first see an increased velocity from about 200 Km/s to 250 Km/s.
2. Than it goes down to min level of 190 km/s at about 3Kpcs. (Bar)
during this phase, we see that the measurements fits almost perfectly the Blue line.
3. From 3KPC (Spiral arm) to 4KPC there is increase in the orbital velocity (from 190 Km/s to about 220 Km/s) and than  it starts to slow down at about 9KC to 190 Km/s. In that section we see some variations in the orbital velocities of different stars, but it is no so severe.
4. At 10KPC we see clearly significant variation in the orbital velocities of the stars. It gets from 140 Km/s to 230 Km/s. At 15 KPC we see even greater variations in the velocity (from less than 130 to over than 250 Km/s.
Dark Matter:
If the dark matter was real, than why do we see so significant dispersion in the orbital velocities of the stars at the same radius?
Our scientists need to explain that real rotation curve and not the one which they have used before (In Red).
That real dispersion of the orbital velocities proves that the dark matter in fantasy. It proves that there is no dark matter. Not in the galaxy and not outside the galaxy.
If you still believe in dark matter, would you kindly explain how could it be that we see at the same radius (for example - 10KPC) stars with orbital velocity of 140 Km/s while others at 230Km/s?
Would you kindly also explain why the velocity goes down from 250 Km/s at less than 1KPC to about 190 Km/s at 3KPC?

[Dave Lev (OP)]

Rotation Curve
In order to understand how the Rotation Curve works we need to look at the following diagram:
http://www.novacelestia.com/images/galaxy_hubble_classification.html
Let's start with Barred spiral galaxy SBb.
We see that it has a bar which is connected to two symmetrical spiral arms.
So, how does it works?
The Bar is used as a connection media between the Bulge to the Arm.
New matter & stars are formed in the Bulge. Those stars are drifted outwards cross the bar and get to the starting point of the Arm.
Let's assume that this starting point (the connection between the Bar to the Arm) is located at 3KPC from the center (as in the Milky way)
We also know that at this point the orbital velocity is minimal (Let's assume - 190 Km/s).
So, if we could force the star to stay at 3KPC, it's orbital velocity should be 190Km/s - that is quite clear.
However, although the arm is considered as an object, stars do not stay at the same point in the arm during any time interval. They are actually drifting outwards.
New stars emerge from the bar and replace them.
So, at any given moment we might see there stars. But all the stars in that point are new stars which had just joined the arm.
So, let's look again at the SBb diagram.
As the stars drift outwards (In the arm), they increase their orbital cycle and therefore they must also increase their orbital velocity.
If the arm was as a direct line which is going from the center, than as the stars goes outwards, they had to increase their orbital velocity. (higher and higher).
However, the arms are spirals. Therefore, as the stars drift outwards, they also drift backwards (due to the shape of the spiral arm).
Hence, as the stars increase their orbital radius, they also drift backwards and decrease the distance that they cross in a given time. The orbital velocity is a direct outcome of the distance per time interval.
Therefore, the balance between the orbital radiuses increased to the drifting outwards set the stars in almost a constant velocity.
In reality, it is never constant. In reality we might find two stars at two different arms at the same radius but with a slightly different orbital velocity.
This is quite clear as nothing is perfectly symmetric/ideal in the nature.
Let's look at the following diagram:
https://www.researchgate.net/figure/Decomposition-of-the-rotation-curve-of-the-Milky-Way-into-the-components-bulge-stellar_fig4_45893184
We see that at the bar the dispersion in the orbital velocity of the stars is quite minimal.
That is also correct at the early stage of the arm. However, as we move outwards, the dispersion in the velocity increases.
Stars that cross a bridge between the arms might have different velocity than other group in the same radius in the arm.
In this rotation-curve diagram we have no way to understand where each star is located and at what arm. Therefore, it is very confusing.
However, I'm quite sure that at any given location in any specific arm all the stars must orbit almost in the same orbital velocity.

With regards to mathematics:
Let's start with the assumption that the arm is rigid.
So, each star stay at the same location at the arm during all his life time.
In this case, it is clear that if a star 1 is located at a radius R1, than:
P1 = 2π R1
For a star 2 which is located at a radius 2R1, we get:
P2 = 2π R2 = 2π 2R1
So, it is clear that if the orbital velocity of star 1 is V1
Than the orbital velocity of a star 2 should be:
V2 = 2V1.
This of course is not realistic.
This proves that the arm is not rigid.
The star must drifts outwards and backwards (due to the spiral shape)
So, let's look again on a star 1 which is located at R1 in the arm.
As it orbits around the center it also drifts outwards and backwards.
Now we need to understand the impact of each activity.
Let's start with star 1 that in interval T drifts outwards a distance ΔS in the arm.
That ΔS represents two vectors as follow:
1. ΔR - the value of the increased distance in the radius.
2. ΔP - the distance that it moves backwards.
So, if the orbital velocity of star 1 is V1 than in order to keep the same velocity as it gets to R + ΔR it must drifts backwards at the same distance which is equivalent to the impact of the increasing radius.
Hence we first need to calculate the impact of ΔR.
It seems to me as a good estimation to assume that the average increase in the radius is ΔR/2 (for one full orbital cycle)
Therefore:
P1 = 2π (R + ΔR/2) = 2πR + 2πΔR/2 = 2πR + πΔR
However, we know that in order to keep the velocity, we need that the orbital distance per interval T will be (one full cycle):
P1 = 2πR
Hence, in order to keep the orbital velocity we must set that
ΔP = πΔR
In this case, we eliminate the impact of the radius increasing by the ΔP backwards drift.
This actually set the shape of the spiral arm.
It must meet that formula.
So simple and clear.
This actually also set the drifting outwards velocity.
As we are located closer to the center, the drifting velocity is lower. as we move outwards, the drifting velocity must be higher.
That also explains the density of the stars and the thickness of the arm.
Close to the center, the density and the thickness must be higher (about 3KLY at a radius of 3KPC)
As we move outwards, the density and the thickness of the arm get lower (about 1KLY at radius 28,000 LY).
At the edge of the arm the density is minimal while the thickness of the arm might be lower (I assume that it should be lower than 500LY).
So, the lowest thickness of the arm set the ending point of the arm.
However, as the density gets lower, the gravity bonding is lower.
At that point, due to the very low gravity force, the stars can't be hold in the arm any more.
Therefore they must be ejected from the arm.
As they are ejected from the arm they are also ejected from the disc. Therefore, it is expected to see that the disc gets thicker and thicker again as we move further away from the edge of the arm (it should be at about 10-15 KPC from the center)
At that aria we see high dispersion in the orbital velocities of the stars. (As stars are still connected in one arm, while other have already disconected from the other arm)
Some of them might still be connected to the arm, while other had already been disconnected.
All the stars, globular clusters, dwarf galaxies that we see around the Milky Way had been ejected from the galaxy.
So, the Milky Way is a giant star sprinkler.
She doesn't eat any star from outside as all of those stars had been born in the galaxy.
The Milky Way is the mother of all the stars that are still orbiting around the galaxy. As any good mother, she has no intention to eat any one of her children. She let them go and cross the space.
She also ejects the Nubile. If that Nubile has the ability to produce new matter as the Milky way, it will be converted over time to a giant spiral galaxy as her mother.
The Milky way should be very satisfy from all the stars and baby galaxies which she had produced so far.
As Andromeda galaxy should be satisfy from her baby - Triangulum Galaxy.
That new baby galaxies production gives perfect explanation for the Ultra high escape velocities of the far end galaxies which we see.
It will also set the Universe in the same matter density forever and ever while we see so many galaxies that are escaping from our observable Universe.
No need for Inflation or Space Expantion.
Spiral galaxies are the leading force in our Universe.
I will explain later on how it really works.
But first we need to understand how spiral galaxy works.
So let me highlight one more issue with regards to spiral galaxy:
It is very important to have a symmetrical shape in spiral galaxy (especially - at the center).
We would never ever see a spiral galaxy with only one arm.
We can compare it to propeller in an airplane.
If you set only one arm in the propeller is will break down the axis of the motor.
In the same token, there must be some sort of symmetrical shape in any real spiral galaxy in order to let the gravity works symmetrically in the center.
Is it clear?

[Dave Lev (OP)]

Based on the following formula we can trace a star as it drifts in the spiral arm:
ΔS↑ - Vector of the drifting outwards distance per time interval T
ΔR↑ - Vector of the value of the increased distance in the radius.
ΔP↑ - Vector of the distance that it moves backwards.

ΔS↑ =  ΔR↑ + ΔP↑

We have found that:
P = 2πR + πΔR
Hence, the following ratio set the change in the orbital velocity in the arm
ΔP ≈ πΔR
ΔR ≈ ΔP/π
If: ΔR ≥ ΔP/π
The impact due to the radius increasing is higher than the backwards drifting.
In this case the orbital velocity is increasing
That mainly represents the first segment in the arm (3KPC to 5KPC)
If: ΔR = ΔP/π
The impact due to the radius increasing is equivalent to the backwards drifting.
In this case the orbital velocity is constant
That mainly represents the second segment in the arm (5KPC to 10KPC)
If: ΔR ≤ ΔP/π
The impact due to the radius increasing is lower with regards to the backwards drifting.
In this case the orbital velocity is decreasing.
Once the stars is in the arm, we don't see a significant decreasing in the velocity. In any case, if there is any sort of decreasing - it is due to that ratio.

I have asked myself the following question:
How long the Sun might stay at the arm before it will be ejected from the arm?
My advice is as follow:
If we could eliminate the impact of the VHP1, we technically could measure the increasing in the radius ΔR per time interval T.
Assuming that we keep our current orbital velocity (around the galactic center), we could extract the drifting outwards vector as:
ΔR = ΔP/π
Once we have found the value of  ΔR and  ΔP we can calculate the value of the drifting distance in the arm per time interval T
ΔS↑ =  ΔR↑ + ΔP↑
As we know the length of the Orion arm, we can find how long it might take us to get to the ending point of the arm.
That doesn't mean that we will be automatically ejected.
We might be lucky and find a bridge to one of the nearby arms.
If not, we have to say By By to our protecting arm and the galactic disc and start a voyage to the unknown.
We might join to a Halo globular star (with other ejected stars).
But, that aria is very dangerous. Each star must obey to local gravity forces. So, as the stars orbit around each other, the chance for collision is quite high. Even without direct collision, the Sun might lose some of its planets.
Let's hope that we will survive and stay with our Sun as the globular star drifts away from the Milky Way.
Do you know that for any star in the galaxy there is at least one outside?
One day, our Sun will join those stars.

[Halc]

We see that it has a bar which is connected to two symmetrical spiral arms.

Responses are delayed.  Busy.  You need to state how long it takes the galaxy to make one rotation (for the bar and everything else around once) somewhat back to where it was before, even if the individual stars are elsewhere and replaced by different ones.  If you have a different figure, state it, but otherwise my responses assume something like 100 million years, a figure given by these scientists who you think don't know what they're doing.  It is a measured thing.  It fits in just a little with your idea of stars staying in their arms but drifting towards the ends of them, or for some reason jumping early on these bridges to more outward arms.  If the arm is moving faster than we are, we have no choice but to fall behind, which has us moving outward in the general direction of towards the end of the arm just like you say.  The galaxy form is moving at 480 km/sec at this radius and the solar system at a little under half that, so it would be moving quite quickly in that direction, not a slow 'drift'.

A rotating (not orbiting) galatic form (the bar and the arms and such) forms a straight rotation curve for an object that rotates once per 100 MY.  That line crosses the rotation curve of the material (the stars and gas and such) at only one point, which happens to be about where the bar and the arms meet, which really helps explain the transition that happens there.  If you posit a different rotation speed, it will intersect somewhere else and will be harder to explain.

Let's assume that this starting point (the connection between the Bar to the Arm) is located at 3KPC from the center (as in the Milky way)
We also know that at this point the orbital velocity is minimal (Let's assume - 190 Km/s).

It gets lower much further in, but yes, it is at a local minimum there.

New stars emerge from the bar and replace them.

Stars in the bar are moving faster than the bar itself since it is below the point where the two rotation curves meet.  That suggests that stars move quite quickly (up to 260 km/s) in a bar that is moving at say 40 km/s at that point.  It's why I say it isn't exactly a drift outward, but rather a sprint.  At that speed, it should take only a few 10's of millions of years to get to the end of the bar.

So, let's look again at the SBb diagram.
As the stars drift outwards (In the arm), they increase their orbital cycle and therefore they must also increase their orbital velocity.

They increase yes. This is why the motion is not considered Keplerian.  Orbiting planets move slower, not faster, with greater distance from the sun.
The speed of the stars is nowhere near the rate at which the arms increase their speed with radius.  They're moving at 190 km/s at the bar end but 480 out here at 8kpc and even faster further out.  The speed is cleanly 2πR/100 million years, which is the formula for any rotating object like the galaxy's form.  I call it 'form' because I am reluctant to call any of the form objects.  If they were objects, their motion would be an orbit, not a rotation, and you'd have to explain why one object orbits at a different speed than another at the same radius.  Somehow I suspect you will solve this problem by ignoring it.

If the arm was as a direct line which is going from the center, than as the stars goes outwards, they had to increase their orbital velocity. (higher and higher).

From whence comes the energy to accelerate them like that?  Somebody else posted that question not too far back and you ignored that post, which again is how you resolve inconsistencies in the idea.

However, the arms are spirals. Therefore, as the stars drift outwards, they also drift backwards (due to the shape of the spiral arm).

Very much so yes.  So much that in one rotation of the galaxy, the solar system, if it stays in its current arm, will have exited the end of the arm and be outside all the visible material in the galaxy.  That's a real problem for a view that we stay in the arms or worse, drift outward to exit even faster.  Find a picture and follow the arm.  It ends in half a rotation at best from where we are now.  Either the sun is only a couple hundred million years old, or the sun must regularly 'drift' to lower arms (not outward) in order to stay in the galaxy. No, we don't drift inward, but the arms move outward past us as we remain at a relatively fixed radius from the center, so we drift inward relative to the arms but not relative to the center. That puts the path of all stars in the spaces between the arms, making that space quite occupied, if not a quite the density of the arms.

Hence, as the stars increase their orbital radius, they also drift backwards and decrease the distance that they cross in a given time.

If their speed is going up as you suggest, that would increase the distance that they cross in a given time.  That's what higher speed means.

The orbital velocity is a direct outcome of the distance per time interval.

It's kind of the definition of speed, not the outcome of it.  Something with higher speed covers more distance in a given time, be it orbital motion or not.

Let's look at the following diagram:
https://www.researchgate.net/figure/Decomposition-of-the-rotation-curve-of-the-Milky-Way-into-the-components-bulge-stellar_fig4_45893184
We see that at the bar the dispersion in the orbital velocity of the stars is quite minimal.

Did they plot stars like S2 on there?  It's way off the top of the chart.

That is also correct at the early stage of the arm. However, as we move outwards, the dispersion in the velocity increases.

Yes it does, especially further out that us.  I notice you hunted down a chart that showed a lot more dispersion closer in than the ones you linked in earlier posts.  You must have a purpose in doing that.

Stars that cross a bridge between the arms might have different velocity than other group in the same radius in the arm.

The chart shows speed, not velocity and hence loses much measured information.  It would be instructive to get the actual velocity data and see if that was true.

In this rotation-curve diagram we have no way to understand where each star is located and at what arm. Therefore, it is very confusing.

It isn't a map of location and velocity.  It depicts what it means to and isn't confusing at all in meeting that purpose.  If you want position and velocity, don't look at a rotation curve plot.

However, I'm quite sure that at any given location in any specific arm all the stars must orbit almost in the same orbital velocity.

Discarding the odd ones out, yes, I agree with that.

With regards to mathematics:
Let's start with the assumption that the arm is rigid.

It does appear to be.  The galaxy makes a rotation without significant change in form, so in that sense it is rigid. It has a rotation 'curve' that is not curved at all, but a straight diagonal line from the origin on out. The form (appearance of arms and bar) can change if there is a significant collision with another galaxy, but the general stable picture is of the rotating barred galaxy going around once per 100 MY.
In a different sense it is not rigid any more than is my arm.  Stars don't stay put in the arm which is moving at very different speeds than the stars, sort of like the blood in my arm travels up and down my arm.

So, each star stay at the same location at the arm during all his life time.

Can't.  The arm moves at a massively different speed than the stars, at least around here.  It has to to get all the way around in 100 million years, less than half the time it takes the solar system to do it.  You even assert otherwise, saying stars drift outward from the center, not staying in one place all its life.

In this case, it is clear that if a star 1 is located at a radius R1, than:
P1 = 2π R1

That's the speed of the arm (which rotates), not the star (which orbits).  The difference is the speed at which the arm moves relative to the star.

For a star 2 which is located at a radius 2R1, we get:
P2 = 2π R2 = 2π 2R1
So, it is clear that if the orbital velocity of star 1 is V1
Than the orbital velocity of a star 2 should be:
V2 = 2V1.
This of course is not realistic.

For orbiting objects like stars, no. For the rotating arms (and the bar as well), this is exactly what is measured.  Anything else and the arms would quickly smear out and not be visible after a short time.

This proves that the arm is not rigid.

It shows that it doesn't move at the same speed as what is measured for the material within, yes.  If that's your definition of rigid (rather than 'holds form'), then the arm is not rigid, yes.

The star must drifts outwards and backwards (due to the spiral shape)

Backwards yes, but if they stay in the arm or move outward from them, they'd reach the ends of the arms in about a hundred million years.  Earth could not be the age it is even if there was something giving it the energy to move like you're suggesting.

Let's start with star 1 that in interval T drifts outwards a distance ΔS in the arm.
That ΔS represents two vectors as follow:
1. ΔR - the value of the increased distance in the radius.
2. ΔP - the distance that it moves backwards.

Yes.  Giving a figure to the rotation rate of the galaxy lets us put numbers to these symbols.  ΔP is trivial to compute.  The arm moves at roughly 480 km/s at this radius (per the 2πR/ω) and our star at say 220, so ΔP is 260 km/s. This has been measured.

So, if the orbital velocity of star 1 is V1 than in order to keep the same velocity as it gets to R + ΔR it must drifts backwards at the same distance which is equivalent to the impact of the increasing radius.

If it stays in its arm, yes.  So let's consider 50 million years.  We're at 8 kpc now, and in 50 million years the galaxy rotates about 180° and we move about 80° for a net change of 100°.  Follow our arm out on a map and see where 100° puts you.  We're now just short of twice the distance from the center of galaxy.  In 75 million years we'll have left the galaxy.  ΔR is about 7 kpc in 50 million years or 1/7 kpc per year or about 140 km/sec needed to stay in the Orion arm. That's not what I'd call a mere drift.
If ΔR is less than that (or zero), then we drift 'inward' towards the Carina-Sagitarius arm, or rather that arm moves outward towards us (at a ΔR of over 100 km/s) and we don't drift much at all.

Hence we first need to calculate the impact of ΔR.

Calculating ΔR first seems a good step to do before working out the impact of it.  I did that above and I don't see you doing it.

It seems to me as a good estimation to assume that the average increase in the radius is ΔR/2 (for one full orbital cycle)

You're guessing at something which is pretty trivial to work out?  ΔR (average) is ΔR/2?  That makes no sense.

As we are located closer to the center, the drifting velocity is lower. as we move outwards, the drifting velocity must be higher.

That it does.

So, the lowest thickness of the arm set the ending point of the arm.

What happens to all the young stars spinning of the ends of the arms?  Why hasn't that happened to us already?

However, as the density gets lower, the gravity bonding is lower.
At that point, due to the very low gravity force, the stars can't be hold in the arm any more.
Therefore they must be ejected from the arm.

OK, that's what happens.  Still, why hasn't this happened to us.  We're supposedly old enough to have gone around the galaxy about 20 times, but the galaxy has done about 45 full rotations in that time, a difference of 25.  The pattern of the galaxy does not have 25 windings.  1.5 at best for the arms.

She doesn't eat any star from outside as all of those stars had been born in the galaxy.

Tell that to the ones headed our way, or to the one coming our way but is large enough to eat us.

The Milky Way is the mother of all the stars that are still orbiting around the galaxy. As any good mother, she has no intention to eat any one of her children. She let them go and cross the space.

How very anthropomorphic.

In the same token, there must be some sort of symmetrical shape in any real spiral galaxy in order to let the gravity works symmetrically in the center.

There are asymmetrical galaxies, but most of them have seen recent violence (collisions) and have yet to reform.

[Dave Lev (OP)]

Responses are delayed.  Busy. 

Thanks, I was waiting for you!

Let's start with star 1 that in interval T drifts outwards distance ΔS in the arm.
That ΔS represents two vectors as follow:
1. ΔR - the value of the increased distance in the radius.
2. ΔP - the distance that it moves backwards.

Yes.  Giving a figure to the rotation rate of the galaxy lets us put numbers to these symbols.  ΔP is trivial to compute.  The arm moves at roughly 480 km/s at this radius (per the 2πR/ω) and our star at say 220, so ΔP is 260 km/s. This has been measured.

Wow!!!
Thanks for this great information/confirmation!
In my best dream I didn't expect that our scientists have measured the orbital velocity of the arm.
It fits perfectly with my theory and it proves that this theory is correct by 100%.
So, Yes - I fully agree. If the orbital velocity of the arm is 480K/s and the orbital velocity of the Sun is 220Km/s than by definition the backwards velocity of the Sun due (ΔP/T) is:
480 - 220 = 260 Km/sec.
Wow.
We are moving very fast backwards!!!

However, as the density gets lower, the gravity bonding is lower.
At that point, due to the very low gravity force, the stars can't be hold in the arm any more.
Therefore they must be ejected from the arm.

OK, that's what happens.  Still, why hasn't this happened to us.  We're supposedly old enough to have gone around the galaxy about 20 times, but the galaxy has done about 45 full rotations in that time, a difference of 25.  The pattern of the galaxy does not have 25 windings.  1.5 at best for the arms.

Thanks for the confirmation.
However, the Sun must be ejected from the arm and the disc. It is just matter of time.
At that high backwards velocity (260Km/s) we might be ejected from the arm even sooner than my expectations.
The Sun hasn't gone 20 times around the galaxy!
That outcome is based on the idea that we have born in the same radius as we are today.
That is a fatal mistake.
We have been born at the center of the galaxy from a gas cloud that was located very close to the SMBH.
That gas cloud was made out of the molecular jet stream that has been ejected from the SMBH excretion disc.
So, our real age is based on the time that it took the sun from day one to drift outwards from the center of the Bulge, cross the Bar and at 3KPC starts its long journey outwards In the arm to our current location.
I would assume that just at the Bulge we have orbit the galactic center by at least 1 Million times.
As I have stated, the bar is a media connection between the Bulge to the Arm.
It seems to me as a Rolette in the casino.
The Bar split the new born stars between the two symmetrical main arms.
https://phys.org/news/2011-06-distant-arm-milky-galaxy.html
Those two arms are: Perscus and Scutum-Crux
However, some of the stars miss the entrance of each main arm and they join the 3KPC arms.
The stars stay there for only half cycle just in order to join the other Arm.

Stars in the bar are moving faster than the bar itself since it is below the point where the two rotation curves meet.  That suggests that stars move quite quickly (up to 260 km/s) in a bar that is moving at say 40 km/s at that point.  It's why I say it isn't exactly a drift outward, but rather a sprint.  At that speed, it should take only a few 10's of millions of years to get to the end of the bar.

I wounded why do you claim that stars move at 260Km/sec (and up) at the bar.
Please look at the orbital velocity
https://www.researchgate.net/figure/Decomposition-of-the-rotation-curve-of-the-Milky-Way-into-the-components-bulge-stellar_fig4_45893184
The bar is located between the Bulge (about 1KPC) to the Arm 3KPC.
In that section we see clearly that the average stars orbital velocity get from 250 Km/s to almost 190Km/s at 3KPC.
That shows that the Bar reduce the orbital velocity.
So, why do you claim the oposite?
You even agree with the following:

Let's assume that this starting point (the connection between the Bar to the Arm) is located at 3KPC from the center (as in the Milky way)
We also know that at this point the orbital velocity is minimal (Let's assume - 190 Km/s).

It gets lower much further in, but yes, it is at a local minimum there.

It seems to me as a good estimation to assume that the average increase in the radius is ΔR/2 (for one full orbital cycle)

You're guessing at something which is pretty trivial to work out?  ΔR (average) is ΔR/2?  That makes no sense.

Let use the following examle:
if we are located at 8KPC and after one cycle we get to 10KPC.
It is clear that we set a spiral shape. However, what is the Length of that spiral shape?
Do you agree that we can assume that is is almost the average lengths of the cycles at 8KPC and at 10KPC?
If you do so, you would find that it is identical to the cycle lengths at 9KPC.
So, as ΔR = 10 - 8 = 2
ΔR/2 = 2/2 = 1
R + ΔR/2 = 8+1 = 9.
That exactly what I did.
Do you still see any problem with that?

Hence we first need to calculate the impact of ΔR.

Calculating ΔR first seems a good step to do before working out the impact of it.  I did that above and I don't see you doing it.

What do you mean?
If we know the value of ΔP/T we should extract the value of ΔR.
I had the impression that ΔP/T is more difficult to verify. Now we know that value - 260Km/s

So, if the orbital velocity of star 1 is V1 than in order to keep the same velocity as it gets to R + ΔR it must drifts backwards at the same distance which is equivalent to the impact of the increasing radius.

If it stays in its arm, yes.  So let's consider 50 million years.  We're at 8 kpc now, and in 50 million years the galaxy rotates about 180° and we move about 80° for a net change of 100°.  Follow our arm out on a map and see where 100° puts you.  We're now just short of twice the distance from the center of galaxy.  In 75 million years we'll have left the galaxy.  ΔR is about 7 kpc in 50 million years or 1/7 kpc per year or about 140 km/sec needed to stay in the Orion arm. That's not what I'd call a mere drift.
If ΔR is less than that (or zero), then we drift 'inward' towards the Carina-Sagitarius arm, or rather that arm moves outward towards us (at a ΔR of over 100 km/s) and we don't drift much at all.

I really don't understand your calculation. Please advice.
However, why do you calim that the sun stays at the arm for only 50 Million years. Based on what data?
Please if it is based on the density wave idea - than please forget it.
It is a fatal mistake.

Did they plot stars like S2 on there?  It's way off the top of the chart.

Yes I agree.
They also didn't plot the orbital velocity of the accretion disc (at about 0/.3 the speed of light)
But those orbital velocities are not so critical for our discussion abot the bar and the arm.

She doesn't eat any star from outside as all of those stars had been born in the galaxy.

Tell that to the ones headed our way, or to the one coming our way but is large enough to eat us.

Please, don't be in panic.
No one is going to eat us or the milky way. The Milky Way is not going to to eat any star or galaxy.
Actually the Milky way is like a giant Sweeper.
It's ultar high gravity force sweeps away any star or galaxy that cross our way. If I understand it correctly, the Milky way cross the space at a speed of 630 Km/s. The space is full with stars. However, all of them sweeped away. Therefore, we don't see even one star that gets into the disc from outside. Never - Ever!
We know that Andromeda is moving directly in our direction. However, I don't think that it will collide with the Milky way.
There are about 400 Billion spiral galaxies in the observable Universe.
How many collisions did we find? Could it be less than 10 or even 5?
so the chance for collisions between giant galaxies is almost 10 to 400 Billion. Similar to lottery win chance.
Therefore, I assume that those two giant spiral galaxies will probably clear the way to each other.

[Halc]

Somehow I suspect you will solve this problem by ignoring it.
...
From whence comes the energy to accelerate them like that?  Somebody else posted that question not too far back and you ignored that post, which again is how you resolve inconsistencies in the idea.

I see you have not responded to places that I predicted would be ignored. My predictions seem more reliable than your own. I point out many more contradictions below, which I also suspect will be ignored.

At that high backwards velocity (260Km/s) we might be ejected from the arm even sooner than my expectations.
The Sun hasn't gone 20 times around the galaxy!
That outcome is based on the idea that we have born in the same radius as we are today.

Yes it is.  If we were closer in and moving at the speeds shown in the rotation curve, it is more like 200 times around the galaxy in the age of the Earth, except not since Earth was created only a couple hundred million years ago (less than 1% of the accepted age).  The galaxy has not rotated that many times, so the sun cannot have stayed withing the bar/arm pattern.  It is either much younger than a billion years or it leaves the bar and arms.

We have been born at the center of the galaxy from a gas cloud that was located very close to the SMBH.

We must have sprung from the black hole with Dinosaurs and older fossils already in place. Very creationism.

That is a fatal mistake.

And yet the glaring mistake is ignored.

So, our real age is based on the time that it took the sun from day one to drift outwards from the center of the Bulge, cross the Bar and at 3KPC starts its long journey outwards In the arm to our current location.

Did you compute that at the speeds we're apparently moving along the arm and bar and such?  It would take only a few hundred million years, a brief journey.  Any other speed would not match the rotation curve, or would take us out of the arm, which you deny.  How do you explain Earth being 5 billion years old?  Just going to deny that as well?

I would assume that just at the Bulge we have orbit the galactic center by at least 1 Million times.

Then we go outside the bar, which rotates only once every 100 million years.  If the new stars lingered a long time down there and then spent only a brief moment (about 1% of its life) in the arms, then the arms would mass about 1% of the mass inside the 'galactic center' where you have us not staying inside the lines like you posit with the arms.  Why is there a bar if all the stars are whizzing around at much higher speeds?  Why is the center of a galaxy not 100 times brighter if it has 99% of the stars? If all the stars are from this excretion disk, how do they end up in the bulge, not part of the galactic disk?  How do they get out of that bulge and back into the original plane? Why do objects at 3kpc move so slow if 99% of the mass is inside their orbit?  Contradictions are piling up.  That was just 7 easy ones I mentioned.

I wounded why do you claim that stars move at 260Km/sec (and up) at the bar.

See any rotation curve at raidus 0.5 kpc.  The speed peaks at about 260 km/s.  The bar moves at 40 km/s there, which is what you get when plugging 0.5 kpc into R of the rotation speed 2πR/100 million years

Let use the following examle:
if we are located at 8KPC and after one cycle we get to 10KPC.

Not sure what a 'cycle' is, but one orbit about the galaxy is time for the arms to rotate more than twice, throwing us beyond 40 KPC, not 10.  Follow the arm out for one rotation.  It fades to nothing before that but you can still work out where it would continue.  We exit the galaxy in a very short time.  Remember the ΔR of 140 km/s (which is ever increasing)?  Plot that out over one 'cycle', whatever that is.

It is clear that we set a spiral shape. However, what is the Length of that spiral shape?
Do you agree that we can assume that is is almost the average lengths of the cycles at 8KPC and at 10KPC?

I just don't know what you think a cycle is.  It seems to be something different at every radius, so it probably isn't the rotation rate of the galaxy.  If we're spiraling out fast like that, we don't even orbit any more than does water spewing from a twirling sprinkler.  There doesn't seem to be anything that cycles.

So, as ΔR = 10 - 8 = 2

OK, you posit 2 KPC per 'cycle', so a cycle might be about 15 million years?  What is there that has a period of that length?  It takes 13 million years for the sun to move to 10 KPC out from 8 if it is moving down the arm at 260 km/s.  That's how far away the point is in our arm that is that distant from the center.

That exactly what I did.
Do you still see any problem with that?

You didn't define a 'cycle'.

If we know the value of ΔP/T we should extract the value of ΔR.

Yes.  All you need to know is the angle of the where we are.  Angle 0 would form a circle.  Angle 90 forms an asterisk shape.  Anything in between forms a spiral.  You can just look at a diagram of the galaxy and measure it.  It's about 20° here but it gets higher further out, and smaller further in.

However, why do you calim that the sun stays at the arm for only 50 Million years. Based on what data?

The arm, even if extended to the lengths of the larger arms, is only so long.  We exit the end of it in 50 million years if we move along it at 260 km/s and accelerating fast.  Further out (say at 17 kpc), the arm is moving at 1000 km/sec but the star only at 230 for a difference of 870, more than trice the current speed we move up the arm.

So I based it on actually working out the time to the end of the arm given your assertions.  I actually calculated it instead of making up a figure that makes me feel nice.

Please if it is based on the density wave idea - than please forget it.

Not based on that at all.  That idea doesn't have us traveling up the arm like that.

Please, don't be in panic.
No one is going to eat us or the milky way.

Not in either of our lives, no.  Denying what is coming is something else.  You don't need a telescope to see it, but a nice time exposure shows much better how close the thing is.
It is not a panic situation.  Collision happens, taking a lot of time to do so.  The solar system will likely not be much affected by the process.  Perhaps we are flung out, but more likely we find a new place in the combined thing.

The Milky Way is not going to to eat any star or galaxy.

It happens regularly.  You don't get big if you don't eat.

We know that Andromeda is moving directly in our direction. However, I don't think that it will collide with the Milky way.

I actually have little doubt that you hold such beliefs.

There are about 400 Billion spiral galaxies in the observable Universe.
How many collisions did we find? Could it be less than 10 or even 5?

Trying to find one that hasn't recently is more the challenge.

[Dave Lev (OP)]

At that high backwards velocity (260Km/s) we might be ejected from the arm even sooner than my expectations.
The Sun hasn't gone 20 times around the galaxy!
That outcome is based on the idea that we have born in the same radius as we are today.

Yes it is.  If we were closer in and moving at the speeds shown in the rotation curve, it is more like 200 times around the galaxy in the age of the Earth, except not since Earth was created only a couple hundred million years ago (less than 1% of the accepted age).  The galaxy has not rotated that many times, so the sun cannot have stayed within the bar/arm pattern.  It is either much younger than a billion years or it leaves the bar and arms.

Sorry
I don't understand why you claim that the sun cannot have stay within the arm pattern for long time.
Actually, the Sun had been born in the Center of the Bulge.
So, first we need to verify how long it should take it to drift all the way till the Bar.
Bulge -  what is the radius of the Bulge?
In the following article it is stated 6500LY.
http://astronomy.swin.edu.au/cosmos/B/Bulges
The bulge of the Milky Way appears to be fairly typical – a slightly flattened sphere of radius ~6,500 light years.
This value represents about 2KPC. I had the impression that the radius of the Bulge is only 1KPC.
However, the arms are already there at 3KPC.
So, based on this data, the bar is located between 2KPC to 3KPC.
If so, we need to find how long it took the Sun from its first day in the center of the Bulge to get to our current location at 28,000Ly from the center.
Bulge - We have already found the orbital velocity at the center of the bulge are quite high. (In the range of 260Km/s).
However, the drifting outwards is quite low.
Therefore, it seems to me that the sun could easily spent few Billion years only at the Bulge.
As the Sun gets to the main arm base at 3KPC, its orbital velocity had been decreased to 190Km/s.
At that aria the diameter of the arm is estimated to be 3000 LY.
I estimate that the density of stars there should be about four times than our current location (4 * 512 = about 2000 Stars per a sphere of 100LY).
Based on the following formula we can trace the Sun as it drifts in the spiral arm:
ΔS↑ - Vector of the drifting outwards distance per time interval T
ΔR↑ - Vector of the value of the increased distance in the radius per time interval T.
ΔP↑ - Vector of the distance that it moves backwards per time interval T.
ΔS↑ =  ΔR↑ + ΔP↑
We have found that:
P = 2πR + πΔR
Hence, the following ratio set the change in the orbital velocity in the arm
ΔP ≈ πΔR
ΔR ≈ ΔP/π
Hence, at the base of the arm it is clear that the drifting backwards ΔP is lower than the πΔR:
I estimate that if the ΔP/T at our location is 260 Km/s, the ΔP/T at 3KPC (at the base) should be less than 1Km/s.
We can also say that:
ΔR ≥ ΔP/π
Therefore, from 3KC to about 5KPC the orbital velocity is increasing.
Hence, it is clear to me that it took long time for the Sun to drift outwards while it is in that location.
I expect that it might take the sun few billion years just to drift from 3KPC to 4KPC.
As it gets to 4KPC, the orbital velocity had increased to about 200 Km/s.
That shows again that ΔR is higher than ΔP/π.
This is correct as long as the sun gets to 5KPC (with orbital velocity of about 220 Km/s)
It is clear to me that at 5KPC the diameter of the arm is lower than the diameter at the base (3KPC) and also the density should be lower. So, I expect that the density at that aria should be in the range of two times with regards to our location (2 * 512 = about 1000 stars per 100LY).
With regards to ΔP/T I expect that the value at 5KPC it should be less than 25Km/s
In any case, from 5KPC to about 10KPC, the orbital velocity is quite constant. (220Km/s).
In order to achieve it we have to set that:
ΔR = ΔP/π
The impact due to the radius increasing is equivalent to the backwards drifting.
The outcome is - constant orbital velocity.
There is a possibility that  ΔR ≤ ΔP/π
The outcome is that the orbital velocity is decreasing.
That can explain the dispersion in the orbital velocity of stars mainly from 5KPC to 10KPC.
So, it seems to me that it could easily take the Sun few more Billion years to get from the base of the arm to our current location.

  We exit the end of it in 50 million years if we move along it at 260 km/s and accelerating fast.  Further out (say at 17 kpc), the arm is moving at 1000 km/sec but the star only at 230 for a difference of 870, more than twice the current speed we move up the arm.
So I based it on actually working out the time to the end of the arm given your assertions.  I actually calculated it instead of making up a figure that makes me feel nice.

Yes, I agree with you that in that time frame we might be ejected from the Orion Arm.
I wonder what is the backwards velocity at the nearby arms?.
Somehow, it seems to me that the backwards velocity there should be lower than the 260 Km/s (even for the arm that is located further away from the center).

The arm, even if extended to the lengths of the larger arms, is only so long.  We exit the end of it in 50 million years if we move along it at 260 km/s and accelerating fast.  Further out (say at 17 kpc), the arm is moving at 1000 km/sec but the star only at 230 for a difference of 870, more than twice the current speed we move up the arm.
So I based it on actually working out the time to the end of the arm given your assertions.  I actually calculated it instead of making up a figure that makes me feel nice.

Yes, I also agree with this answer..

How do you explain Earth being 5 billion years old?  Just going to deny that as well?

I hope that by now you understand that there is no problem with the age of the earth.

Then we go outside the bar, which rotates only once every 100 million years.  If the new stars lingered a long time down there and then spent only a brief moment (about 1% of its life) in the arms, then the arms would mass about 1% of the mass inside the 'galactic center' where you have us not staying inside the lines like you posit with the arms.  Why is there a bar if all the stars are whizzing around at much higher speeds?  Why is the center of a galaxy not 100 times brighter if it has 99% of the stars? If all the stars are from this excretion disk, how do they end up in the bulge, not part of the galactic disk?  How do they get out of that bulge and back into the original plane? Why do objects at 3kpc move so slow if 99% of the mass is inside their orbit?  Contradictions are piling up.  That was just 7 easy ones I mentioned.

With regards to your questions:
1.  If the new stars lingered a long time down there and then spent only a brief moment (about 1% of its life) in the arms, then the arms would mass about 1% of the mass inside the 'galactic center' where you have us not staying inside the lines like you posit with the arms.
Answer - Stars stay long enough in the arms. So, in total - there is significant portion of the stars in the arms.
2.Why is there a bar if all the stars are whizzing around at much higher speeds? 
Answer - The bar is used as a media connection between the Bulge and the Arm base. It actually reduces the Orbital velocity of the Stars and delivers them from a free orbit in the Bulge to  a disc shape orbit in the arm.
3. Why is the center of a galaxy not 100 times brighter if it has 99% of the stars?
Answer - Please see answer 1. There is a significant portion of the stars in the arm. Therefore, it is not expected to see it brighter in the Bulge.
4. If all the stars are from this excretion disk, how do they end up in the bulge, not part of the galactic disk?
Answer - The stars had not been form in the excretion disc. However, their molecular/matter had been formed there. The stars had been formed in the gas cloud which is located in the Bulge near the SMBH.
5. How do they get out of that bulge and back into the original plane?
Answer - As I have stated before, all the orbital objects drift outwards. (even if it is just few Pico mm per cycle) Therefore, by definition - any orbital cycle is spiral.
The bar is the media which is taking care of setting the stars in the arms (and therefore in the galactic plane)
6. Why do objects at 3kpc move so slow if 99% of the mass is inside their orbit?
Answer -  First, it is incorrect. Significant portion of the stars are located at the arms. Second, the current  understanding about the impact of the mass/stars inside the orbital cycle is also incorrect.

Highlight -
I have introduced a simple idea why the stars can be in the arm while their orbital velocity is constant (or almost constant) at any radius. Therefore - the arm is an object. It is made out of stars, while those stars have a freedon to drift outwards. If any star will dare to move outside the arm (by itself) it will be ejected from the disc as a rocket.
This is the most important issue in this discussion.
Do you accept this idea?
If so, it's time for you to shift gear and understand the great impact of that theory.
If you have any more questions (especially if they are difficult), please do not hesitate to ask.
However, please base your questions on real evidences and not on outcome of wrong understanding from the unproved current theory.
I'm quite sure that one day Students will learn this theory in the University.
The question is - how long it might take? One year from now, or 100 years?
Sooner or later - our scientists will have to accept this theory as it is the only valid theory.

[Halc]

I don't understand why you claim that the sun cannot have stay within the arm pattern for long time.

Calculate the remaining length of the arm and the rate at which we're currently moving up it if we must stay within it.  Also take into account the higher ΔS rate the further out we go.  It is a simple calculation that we'll reach the end in under 100 million years.  Where do all the stars go that have done this before us?  There's a much lower density of stars out that far, and if the arms are continuously spewing their contents out into the cloud, it must be a pretty busy place.

Actually, the Sun had been born in the Center of the Bulge.

You've said it is excreted by the SMBH into the 'excretion disk', which is not the bulge.

the bar is located between 2KPC to 3KPC.

The bar goes all the way across, so it goes from -3KPC to +3KPC. 

We have found that:
P = 2πR + πΔR

P (circumference of a circle of radius R) is 2πR.  If R changes by ΔR, then ΔP = 2πΔR.

I estimate that if the ΔP at our location is 260 Km/s, the ΔP at 3KPC (at the base) should be less than 1Km/s.

Besides quoting me, how did you arrive at that 260 estimation and especially the 1km/s?  It didn't come from any of the formulas you gave.  I agree that at that point the rotating speed and orbital speed are the same.  They have to cross each other somewhere.  So the stars are making progress up the arm at 1/260th of the rate of here.  Why is the amount of material there not at least 260 times as much?  You say only 4x below.

If stars spend 10 billion years in the bulge and less than a 10th of that time in the arms, all the mass would be centralized and the rotation curve would look nothing like what we see.  Stars, like orbiting planets, would move slower the further out they are.  This has been brought up countless times and you don't begin to address any of it.  Your assertions make this problem worse since you put most of the mass at the center, and then you posit stars not being attracted by gravity but actually accelerating away from it over time. Where does the energy needed to do this come from?  Certainly not gravity, which causes objects to slow down as they move away from the gravity source.
The standard explanations don't have any of these problems, and yet you repeat what a fatal mistake they're all making.  The fatal mistake is this ignoring of all these contradictions.

We can also say that:
ΔR ≥ ΔP/π
Therefore, from 3KC to about 5KPC the orbital velocity is increasing.

Besides the fact that your formula there is false, how can the velocity conclusion possibly follow from the fact that a circle with a larger circumference happens to also have a larger radius?  It isn't true of Earth/Mars, which have a R of about a 3-5 ratio like that and corresponding larger P, and yet Mars's orbital velocity is not larger then Earth's at that radius.  An increasing ΔR has gravity resisting the motion since the gravity pulls in the opposite direction from the change in radius.  So the object slows down just as does a rock when I throw it upwards in a direction that gives it a positive ΔR.  So quite the opposite.  Any positive ΔR should decrease its orbital velocity.  Look at S2, which is at its fastest when its R is lowest.  When its ΔR is increasing, its speed in decreasing, and v-v.

As it gets to 4KPC, the orbital velocity had increased to about 200 Km/s.
That shows again that ΔR is higher than ΔP/π.

What has that speed have to do with your incorrect formula?  ΔR is in fact always exactly half that figure, so less than, not higher than. You can't just spout some random formula (especially if its wrong) and then make a random conclusion that does not follow from the thing you made up. Essentially, you are saying that circles are round, therefore skateboards increase speed when rolling up hill.

With regards to ΔP I expect that the value at 5KPC it should be less than 25Km/s

Is this a guess again?  The value is easily computed, and it isn't 25.

In any case, from 5KPC to about 10KPC, the orbital velocity is quite constant. (220Km/s).
In order to achieve it we have to set that:
ΔR = ΔP/π

This irrelevant thing again.  It's still wrong.  ΔR = ΔP/2π always.  All this says is that an increase of the radius of a circle will increase its circumference by 2π, which is simple geometry and has nothing to do with the speed of anything. Stating that it might be greater or less than this would mean that the circle does not increase its circumference by that amount, which would be simply wrong, at least in Euclidean geometry.

  We exit the end of it in 50 million years if we move along it at 260 km/s and accelerating fast.

Yes, I agree with you that in that time frame we might be ejected from the Orion Arm.

The Orion arm is really a short fragment and it will take only about 10 million years to exit that.  The 50 million figure was if the arm was as long as the main arms.

I wonder what is the backwards velocity at the nearby arms?.
Somehow, it seems to me that the backwards velocity there should be lower than the 260 Km/s (even for the arm that is located further away from the center).

So compute it instead of making up the numbers.  It's really easy. Both numbers have been measured, so all you have to do is subtract them.

I hope that by now you understand that there is no problem with the age of the earth.

I agree, because I don't buy your idea that we were excreted by Sgr-A.

With regards to your questions:
1.  If the new stars lingered a long time down there and then spent only a brief moment (about 1% of its life) in the arms, then the arms would mass about 1% of the mass inside the 'galactic center' where you have us not staying inside the lines like you posit with the arms.
Answer - Stars stay long enough in the arms. So, in total - there is significant portion of the stars in the arms.

Yet you agree that we're about to be flung off the end of the arm in about the time it takes to make ¼ orbit of the galaxy.  This sounds contradictory.

2.Why is there a bar if all the stars are whizzing around at much higher speeds? 
Answer - The bar is used as a media connection between the Bulge and the Arm base. It actually reduces the Orbital velocity of the Stars and delivers them from a free orbit in the Bulge to  a disc shape orbit in the arm.

The bar does not stop where the bulge starts.  It goes all the way across to the other side.

3. Why is the center of a galaxy not 100 times brighter if it has 99% of the stars?
Answer - Please see answer 1. There is a significant portion of the stars in the arm. Therefore, it is not expected to see it brighter in the Bulge.

You said above that 10 billion years are spent in the bulge and then this fast ride out through the arms. That contradicts what you just say here.

4. If all the stars are from this excretion disk, how do they end up in the bulge, not part of the galactic disk?
Answer - The stars had not been form in the excretion disc. However, their molecular/matter had been formed there. The stars had been formed in the gas cloud which is located in the Bulge near the SMBH.

This is also a change in story from the prior posts.  How do the stars get their velocity to end up in the bulge?  If the SMBH is just producing gas continuously, it should just gather around the thing without moving.  I think you mentioned magnetism, but gasses are not magnetic.  I cannot attract or repel a hydrogen or helium balloon with a magnet of any power.

5. How do they get out of that bulge and back into the original plane?
Answer - As I have stated before, all the orbital objects drift outwards. (even if it is just few Pico mm per cycle) Therefore, by definition - any orbital cycle is spiral.

Then they would move out in all directions, not in the plane of the disk.  Things in the bulge have random orbital planes.  If they drift outward, they retain those random planes and the bulge grows in all directions.  I was asking how the motion changes from one plane to the other.
Second comment:  Why do stars in the bulge drift by a few picometer per cycle (still unclear what a cycle is) while stars out by us do it at 140 km/sec?  We're maybe 1 order of magnitude further out, so why the 50 orders of magnitude drift difference?

6. Why do objects at 3kpc move so slow if 99% of the mass is inside their orbit?
Answer -  First, it is incorrect. Significant portion of the stars are located at the arms.

Your idea predicts otherwise.  Any star in the arms (especially from 3.5 KPC out) spends only 1% of its life in them since it travels so fast up the arm and is flung out as is about to happen to us. If 1% of stellar life is spent in the arms, the arms should mass 1% of the material in the galaxy.  You're exactly right with your answer.  It is incorrect.  We observe a significant portion of the stars located at the arms. This is a falsification of your idea. We observer stars that are very young.  We observe stars that are just forming now. All this completely falsifies what your idea predicts.

Second, the current  understanding about the impact of the mass/stars inside the orbital cycle is also incorrect.

You did deny Newton's law of gravity, but without a replacement for it, you don't have a theory.  Current law says that mass concentrated within one sphere will produce Keplerian orbital motion of objects outside that sphere.  The motion of stars outside 3.5 kps is anything but Keplerian, either observed or the strange motion that you assert.

Our measured ΔP (the difference between our orbital speed and the speed of the arm as it rotates by) has indeed been measured at 260 km/sec, but the ΔR of 140 has not.  A motion that fast would have been noticed.  The estimated ΔR may or may not be something more like a value that varies between a small positive and negative number (periodic) that might average perhaps 10 km/sec, mostly due to eccentricity and local perturbations.  That means that our current trajectory has us and the Orion arm parting company in a short time where we join the wealth of stars that fill the spaces between the arms.
The arms are brighter mostly because of the young bright stars that are born with every wave passing and die before they pass to the spaces between the arms.  The long lived stars like our own are nearly as dense between the arms as they are here. You asked about a survey of this. Here is a good place to look for your regions of empty space you're so sure exist: http://gea.esac.esa.int/archive/

Highlight -
I have introduced a simple idea why the stars can be in the arm while their orbital velocity is constant (or almost constant) at any radius. Therefore - the arm is an object.

A thing is an object if it has material moving at constant velocity through it? That makes no sense.
My definition was that it had inertia (mass).  If the arm moves here at 480 km/sec, there is some mass that has that sort of speed/momentum/kinetic-energy. None of it does, or we'd have to explain why it isn't flung from the galaxy due to moving faster than orbital speed.  The material is all moving at about 220.  With no mass moving at 480, the arm is not an object.  A pressure wave in the ocean moves at 1.5 km/sec even though no molecule of water ever travels anywhere near that speed.  Sound waves are thus not objects.  They have no momentum.

If the arm is an object moving at 480 km/sec here and the sun is an object moving at much slower 220 km/sec, why is it that the slow thing (sun) moves outward and the higher speed thing does not?  Why is the acceleration of the arm 3 times as much if it is in the exact same gravitational field as the sun? Answer: because it isn't an object, has no mass, and is thus not subject to gravity any more than one can measure the weight of a sound wave.

It is made out of stars, while those stars have a freedon to drift outwards. If any star will dare to move outside the arm (by itself) it will be ejected from the disc as a rocket.

If the star doesn't thus 'dare', how might it prevent this?

This is the most important issue in this discussion.
Do you accept this idea?

It's self-contradictory from one end to the other. Of course I don't accept it.

I'm quite sure that one day Students will learn this theory in the University.

Somehow this statement doesn't surprise me.

[Dave Lev (OP)]

Dear Halc

Unfortunately, it seems to me that instead of understanding my explanation you focus on highlight why this theory is incorrect.
Therefore, you don't understand my explanation.
Would you kindly set the effort to understand the theory and then claim why it is incorrect?

Let me start by mathematics:
 

P (circumference of a circle of radius R) is 2πR.  If R changes by ΔR, then ΔP = 2πΔR....
It's still wrong.  ΔR = ΔP/2π always.

I have already explained why we need to use ΔR/2 instead of ΔR. (in several answers)
The explanation was as follow:

Let use the following example:
if we are located at 8KPC and after one cycle we get to 10KPC.
It is clear that we set a spiral shape. However, what is the Length of that spiral shape?
Do you agree that we can assume that is almost the average lengths of the cycles at 8KPC and at 10KPC?
If you do so, you would find that it is identical to the cycle lengths at 9KPC.
So, as ΔR = 10 - 8 = 2
ΔR/2 = 2/2 = 1
R + ΔR/2 = 8+1 = 9.
That exactly what I did.
Do you still see any problem with that?

It seems to me as a good estimation to assume that the average increase in the radius is ΔR/2 (for one full orbital cycle)
Therefore:
P1 = 2π (R + ΔR/2) = 2πR + 2πΔR/2 = 2πR + πΔR
However, we know that in order to keep the velocity, we need that the orbital distance per interval T will be (one full cycle):
P1 = 2πR
Hence, in order to keep the orbital velocity we must set that
ΔP = πΔR

I really can't understand why do you insist on ΔR?
ΔR is valid only if at T=0 the star jump to the new radius (R+ΔR) and stay there for one full orbital cycle.
This is not our case.
The star is moving over time from R to R+ΔR.
Therefore, the average cycle is R+ΔR/2!!!
Hence, it is a severe mistake to use ΔR (as is) in the formula.
Is it clear to you by now?

[Halc]

Would you kindly set the effort to understand the theory and then claim why it is incorrect?

It is self-inconsistent, and something self-inconsistent isn't correct.  So work on the places where inconsistencies have been shown. That's how science works. You modify the idea until it is consistent with itself, and preferably consistent with empirical evidence.

Let use the following example:
if we are located at 8KPC and after one cycle we get to 10KPC.

You have yet to describe what a 'cycle' is, so I still have no idea what you're talking about with this statement.  This may or may not be consistent with observations, depending on what that cycle is.  How long is a 'cycle'?  Where do you figure 10 KPC?  Is that computed or just another figure you made up?

It is clear that we set a spiral shape. However, what is the Length of that spiral shape?
Do you agree that we can assume that is almost the average lengths of the cycles at 8KPC and at 10KPC?
If you do so, you would find that it is identical to the cycle lengths at 9KPC.
So, as ΔR = 10 - 8 = 2
ΔR/2 = 2/2 = 1
R + ΔR/2 = 8+1 = 9.
That exactly what I did.
Do you still see any problem with that?

Repeating unclear statements doesn't help.  I asked questions (what is a cycle?) which went unanswered.
ΔR is sometimes a speed and sometimes a distance.  That's inconsistent.  Distance seems more correct since Δ means change-in X and not 'rate of change of' X.
ΔR/2 = 2/2 = 1:  You divide the difference by two, but don't give any reason why.  OK, that's half the radius change.
R + ΔR/2 = 8+1 = 9 :  Yes, 9 is halfway between 8 and 10.  I still have no idea what this means.  We're presumably at radius 9 after half a 'cycle' instead of a whole cycle, whatever that is.  You've managed to find an obscure way to show that stars at radius 9 KPC orbit between stars orbiting at 8 and 10 respectively.  I cannot comment further because I have no idea what you are doing here.

It seems to me as a good estimation to assume that the average increase in the radius is ΔR/2 (for one full orbital cycle)

Here you call it an orbital cycle, so a cycle is presumably the time it takes something to orbit.  I am guessing the solar system orbiting the galaxy, except your idea has no such orbit since the solar system follows the arm in a spiral path and not a reasonably round orbital path.  Perhaps you mean the time it takes to orbit its VHP1, but no mention of that has been made. I'm not sure how long you envision a 'cycle' to be.  Perhaps you could use clear terms instead of assuming I can correctly guess at your meaning.  Clearly I'm failing to do that.

Therefore:
P1 = 2π (R + ΔR/2) = 2πR + 2πΔR/2 = 2πR + πΔR

If P1 is 2πR, then it is the circumference of a circle of radius R.  If you change R by ΔR, then that circumference goes up by 2πΔR or else the circumference at the new R will no longer be 2πR.  As such, the term on the right is just plain wrong.  If you mean totally different things by this equation above, then tell me what you mean by it, because you added no new explanation in this post.  It makes no sense to me as posted.

However, we know that in order to keep the velocity, we need that the orbital distance per interval T will be (one full cycle):

There seemed to be no mention of velocity or time in the above equation, but then again I seem to be misunderstanding the meaning completely.  As I said, sometimes R is a distance (radius) and sometimes it is a speed (like an outward drift speed or something).  If it is a speed, then ΔR would be a change in speed.

I really can't understand why do you insist on ΔR?

I'm insisting on anything. I'm not even clear what your ΔR is. I'm not telling you what it is.

ΔR is valid only if at T=0 the star jump to the new radius (R+ΔR) and stay there for one full orbital cycle.

Here you use R as a distance again, but there is time T involved, but there seem to be no temporal components to your equations.  At time 0 the star jumps to a different radius?  Like instantly?  Takes a left turn and start moving outward for a while?  Not sure what you're trying to convey with this.

The star is moving over time from R to R+ΔR.

Outward drift means an increase in radius, sure.  R is a distance here.

Therefore, the average cycle is R+ΔR/2!!!

A 'cycle' is a distance, not a time period?  So in our example of R=8, ΔR=2, the average cycle is 9KPC?  I really don't understand what a cycle is.  9KPC is an average radius between those two radii, like saying Mars sort of has an average orbital radius between Earth and the asteroid belt.  No idea how the word 'cycle' applies to that.  9 KPC is a distance and 'cycle' is usually one iteration of a repetitive process.

Is it clear to you by now?

I'm clearly baffled by your terminology.  I'm not saying any of it is wrong since I have no idea what you're trying to describe.

[Dave Lev (OP)]

Wow
You have totally got lost on a very simple exercise.
Please - just look at that from a mathematical point of view.
Let me start by asking:
What is the circumference of a circle if the radius is R?
You have already gave the answer:

P (circumference of a circle of radius R) is 2πR.

Therefore:
P = 2πR
So, if R1 = 8 cm, the formula is:
P1 = 2πR1 = 2π*8
If R2 = 10cm
P2 =2 πR2 = 2π*10
However, our mission is to verify the circumference of a circle while R increases from R1 (8cm) to R2 (10cm).
We actually try to calculate the circumference of a spiral shape as it moves from radius 8m to radius 10m in one full cycle.
So, please advice how do you calculate that circumference (P) of a spiral shape from 8cm to 10 cm (ΔR = 10-8=2)?
P is the circumference from R=8, t=0 to R=10 t=T
T = the time that it takes the object to move (in spiral shape) from 8 to 10 in one full cycle.
Somehow, you insist on the maximal radius:
P =2πR2 = 2 πR2 = 2 π (R1 + ΔR) = 2π*10
Is it realistic?
Why not the average radius (R1+R2)/2 or (R1+ΔR/2) or just (9)?

In order to help you, please see the following article:
https://sciencing.com/calculate-spiral-6544041.html
How to Calculate a Spiral
"Determine the inner diameter of the spiral. This is the diameter of the circle formed by the innermost ring of the spiral. Call this length "d."

Plug the numbers obtained in the first three steps into the following formula: L = 3.14 x R x (D+d) ÷ 2

For example, if you had a spiral with 10 rings, an outer diameter of 20 and an inner diameter of 5, you would plug these numbers into the formula to get: L = 3.14 x 10 x (20 + 5) ÷ 2.

Solve for "L." The result is the length of the spiral. Using the example from the previous step: L = 3.14 x 10 x (20 + 5) ÷ 2 L = 3.14 x 10 x 25 ÷ 2 L = 3.14 x 250 ÷ 2 L = 3.14 x 125 L = 392.5"

Please be aware that in this example it is stated:
"Determine the number of rings in the spiral. This is the number of times the spiral curve wraps around the center point. Call this number of rings "R."
We try to calculate only one ring!!!

[Halc]

However, our mission is to verify the circumference of a circle while R increases from R1 (8cm) to R2 (10cm).

OK, I see what you are doing.  You're computing the length of a spiral segment in a 360° arc which you are calling a cycle.  Your method is a close approximation for slow outward spiral just like sinΦ is a close approximation to tanΦ for a small Φ, but it gets pretty inaccurate for a larger Φ, and a glance at any picture of our galaxy shows that you get massively further out than 10 KPC  following any arm for one lap starting at 8 KPC.
So take the outer arm (a particularly long one) which is at radius 8 kpc about a quarter turn back.  It is at 11 kpc after 90° and 16 kpc at 180° after which it sort of ends.  If you extrapolate it all the way around, it is at something like 26 (not 10) kpc after one cycle.  By your approximation, the length of it between those two points is 2π17 or about 107 kpc.  I'm getting about 120 kpc myself, not a whole lot different. The arm ends after only about 50 kpc from that point where it is at the same radius as us.

We actually try to calculate the circumference of a spiral shape as it moves from radius 8m to radius 10m in one full cycle.

Here you talk about something moving again, so the definition of a cycle changes.  What exactly is moving for one full cycle? Countless times I've asked this and still I'm force to guess by your changing context.
Is it the time it takes the form (bar and arms and such) to make one rotation?  Is it the time it takes a star at a given radius to make one orbit (which can be worked out from the rotation curve) or is it the difference between the two (the time it takes for a star to make one trip around the pattern at a changing ΔP)?  Those are three very different values, and only the first one is any kind of regular cycle per your idea.  The orbit time is a cycle in standard theory, but an orbital motion does not keep a star in any arm, so that isn't a cycle in your view.  A trip with a finite end isn't a cycle, its just something that ends, especially if there's far less than one lap to go.

Anyway, you talk about going from 8 to 10 in this cycle.  What galaxy are you looking at?  The Orion arm reaches 10 kpc after only about a 15th of a cycle, after which that spur ends.

Not sure where we go after that since if there are no stars between the arms, all these stars shooting off the end of the Orion arm need to be destroyed by something.  A bridge to the larger arms on either side would be crowded enough to see.  How did we get in this arm in the first place?  The inner end is about 7kpc out and even closer than the far end.  We can't have been in this arm more than 10 million years at the rate we're traveling through it.

So, please advice how do you calculate that circumference (P) of a spiral shape from 8cm to 10 cm (ΔR = 10-8=2)?

The 8/10 figures are not representative for our (or any) galaxy, but for those numbers, the figure you're getting suffices as an approximation.  The way to do it proper is to integrate the cos of the spiral angle Φ for one cycle.  For a small angle like you get with 8/10, the difference is insignificant for our purposes.  Φ seems to be a around 20° for our galaxy.

P is the circumference from R=8, t=0 to R=10 t=T

I'll ask again: Where did you get 10 from?  It certainly isn't the radius of any arm in any barred galaxy after one cycle from some point at least twice the radius of where the arms start.  No galaxy is wound that tight.  Working with 10 means these calculations are meaningless.  The real figure for our galaxy is about 26.  Look at any depiction of it to see that.
This is part of why I didn't know what a cycle is.  It takes perhaps 10 million years for us to reach 10kpc if we stay in the arm, and I was trying to figure out something that had a period of 10 million years. 

T = the time that it takes the object to move (in spiral shape) from 8 to 10 in one full cycle.

I think you're talking about how long it takes a star to make one trip around the pattern.  Did you work this time out?  You have all the needed numbers.

Somehow, you insist on the maximal radius:
P =2πR2 = 2 πR2 = 2 π (R1 + ΔR) = 2π*10
Is it realistic?
Why not the average radius (R1+R2)/2 or (R1+ΔR/2) or just (9)?

I didn't know what you were trying to compute with these equations, which was the linear length of a spiral going from 8 to 10 in one lap, which didn't occur to me at all since our galaxy doesn't look like that.  Our arm goes from 8 to 10 in about a 15th of a lap.

"Determine the number of rings in the spiral. This is the number of times the spiral curve wraps around the center point. Call this number of rings "R."
We try to calculate only one ring!!!

OK, but where did 10 come from then?

[Dave Lev (OP)]

Would you kindly focus on my message?
I have stated:

Please - just look at that from a mathematical point of view

So, we ONLY discuss on a pure mathematical issue.
In that example I'm using a radius of 8 cm (centimeter) that increased in spiral way to 10 cm!!!
So we discuss about centimeters, not KPC.
I just want to show you the correct formula to calculate the length of a spiral segment in a 360° arc (which I'm calling a cycle).
I don't claim that in the Milky Way after one full cycle (or 360° arc) the spiral arm will get from 8KPC to 10KPC.
It is just a mathematical example.
Why is it so difficult?
So, If
R1 - the radius at starting point.
R2 - The radius after 360° arc.
ΔR = R2 - R1
Hence - the formula to calculate the length of a spiral segment in a 360° arc (from R1 to R2) is as follow:
P = 2π(R +ΔR)/2 = 2πR +πΔR
Do you agree with that?
Please!

Inner diameter of 5 means they start well away from the center.  Both types A and B can go all the way in, but the mathematics falls apart there for both.

If you think that the mathematics is incorrect, would you kindly advice how to calculate the length of a spiral segment in a 360° arc?

[Halc]

Would you kindly focus on my message?
I have stated:

Please - just look at that from a mathematical point of view

So, we ONLY discuss on a pure mathematical issue.

I did that in my prior post (294).  Did you read it before posting this?
The formula is only an approximation for a type-A spiral measured at a reasonable distance away from the center.  The site did not present the formula as inappropriate for other situations, nor did it identify it as an approximation.

In that example I'm using a radius of 8 cm (centimeter) that increased in spiral way to 10 cm!!!

18π cm is a reasonable approximation for that, yes.  It isn't exactly that.  The real figure is larger.

So we discuss about centimeters, not KPC.
I just want to show you the correct formula to calculate the length of a spiral segment in a 360° arc (which I'm calling a cycle).

It isn't correct, but close enough if that's all you need. Mathematically, close enough is the same as wrong.

So, If
R1 - the radius at starting point.
R2 - The radius after 360° arc.
ΔR = R2 - R1
Hence - the formula to calculate the length of a spiral segment in a 360° arc (from R1 to R2) is as follow:
P = 2π(R +ΔR)/2 = 2πR +πΔR
Do you agree with that?

I see what is being done, yes.  It is the sum of the tangential component of the line being measured, but discards the radial component, which isn't large for a mathematical spiral going up by 25% in one trip around.  The discarding of the radial component is what makes it wrong though.  It adds about a 28th of a cm to the length of your example.
The formula fails completely for type B (scale-invariant) spirals, and the article doesn't make the distinction, especially since they show a picture of a type B spiral.

If you think that the mathematics is incorrect, would you kindly advice how to calculate the length of a spiral segment in a 360° arc?

Integrate the distance along its entire length, just like I said in the post you are quoting.  I see you didn't quote that part. Translation: cut it up into little pieces and add up the length of the pieces.  The length is the limit of that process as the size of the pieces approaches zero.
For a type A spiral, a better approximation for one lap (still staying away from the center) is √((2πR)² + ΔR²) where R is the mean radius (9) and ΔR is 2.  This is still an approximation, but at least a better one.

I don't have at my fingertips a good formula for doing the same for a type B spiral.  It would have to be a scale-invariant formula, so there would be a fixed Φ input instead of a fixed ΔR per lap.  We'll leave it as an exercise to the reader.

[Dave Lev (OP)]

You insist that there are two spiral types -  Type A and Type B.
With regards to Type A you claim:

For a type A spiral, a better approximation for one lap (still staying away from the center) is √((2πR)² + ΔR²) where R is the mean radius (9) and ΔR is 2.  This is still an approximation, but at least a better one.

With regards to type B

The formula fails completely for type B (scale-invariant) spirals, and the article doesn't make the distinction, especially since they show a picture of a type B spiral.

Can you please direct me to the article which supports your explanation?
You also claim that:

It isn't correct, but close enough if that's all you need.

So, yes, that's all I need.
You have asked me to explain how we can get a constant velocity of stars in spiral galaxy while the arms are objects that based on stars.
In order to prove it, I need to use a simple formula to calculate the length of the spiral.
In all the articles which I have looked at the Web it is clearly show that the simple formula for that length of the spiral is:
P = 2πR +πΔR
I didn't find any relation in any article for Type A or Type B.
So, why do you try to make it so difficult?
Based on this formula I have proved my theory!!!
I have clearly proved that increased drifting velocity of stars in the arm can compensate the extra requested velocity due to increased radius.
If those types of spiral are real in the nature, why our scientists do not offer their solutions for those types of spirals?
Would you kindly show me the article about the modeling for each type of galaxy?
Why do you insist that I should offer a solution for those types of galaxies while our scientists have no obligation for that?
Why do you claim that:

Mathematically, close enough is the same as wrong.

All the modern science is based on "close enough"
In the idea of density wave the whole modeling is based on "close enough".
Our scientists do not offer any direct outcome from the BBT moment to the creation of spiral galaxy.
In their modeling they have used unrealistic idea of a very thin star disc in order to show that it could be evolved into spiral shape, but they don't show how we get this unrealistic thin star disc from the Big bang moment.
They also couldn't show why the disc get's so narrow as the radius increases.
We have solid evidence that at 3KPC the thickness of the disc is 3000LY.
At our location it is 1000LY.
I'm positively sure that if we move further away from the galactic center, the disc should be even less than 500LY. (If not than my theory is incorrect)
Our scientists could not explain this evidence by their theory.
Actually, based on their modeling they have found that the disc should be thicker as we move further away from the center.
But as usual, they have totally ignored this contradiction in the verification.
They also couldn't explain by the modeling the real functionality of the Bar.
Hence, the density wave is not even "close enough". It is just a fatal error.
So, would you kindly explain why "close enough" is not good enough for my theory, but it is perfectly OK for our scientists?

[Halc]

You insist that there are two spiral types -  Type A and Type B.
With regards to Type A you claim:

For a type A spiral, a better approximation for one lap (still staying away from the center) is √((2πR)² + ΔR²) where R is the mean radius (9) and ΔR is 2.  This is still an approximation, but at least a better one.

With regards to type B

The formula fails completely for type B (scale-invariant) spirals, and the article doesn't make the distinction, especially since they show a picture of a type B spiral.

Can you please direct me to the article which supports your explanation?

No article.  I worked it out.  There are probably real mathematical names given to those two types of spirals, which I might have learned had I bothered to hunt down an article.  So lacking that, I called them types A and B.  There are other types, some of them probably with names.

It isn't correct, but close enough if that's all you need.

So, yes, that's all I need.

So there's no problem.

You have asked me to explain how we can get a constant velocity of stars in spiral galaxy while the arms are objects that based on stars.

This isn't much of a question since you seem to realize that the arm is moving at one speed (480km/s locally) and the stars another (220).  The question was from a long way back when your personal definition of 'object' was unclear to me.  You seem to be talking about form as an object, so this question is answered so long as you don't start giving inertial properties to that form.

In order to prove it, I need to use a simple formula to calculate the length of the spiral.

You're going to prove a velocity question using a calculation of length?  I can't wait...

In all the articles which I have looked at the Web it is clearly show that the simple formula for that length of the spiral is:
P = 2πR +πΔR

Simple yes.  Mathematically correct, no, as I have pointed out.  But it works for our purposes here.  If you would like a demonstration of where the formula is wrong by a factor of 2 or more, I can show it.

I didn't find any relation in any article for Type A or Type B.

If you read my post on it, you'd know I made those terms up. They probably have official names, but I don't know them.

So, why do you try to make it so difficult?
Based on this formula I have proved my theory!!!

Your theory has yet to see light of day, and never will since it seems to require a rewrite of all physics.  A calculation of a length of a segment of a spiral is not going to prove or disprove 99% of the things you have proposed.

I have clearly proved that increased drifting velocity of stars in the arm can compensate the extra requested velocity due to increased radius.

Wait, did the proof go by?  What extra velocity requested?  You quoted a length calculation and suddenly you're saying this somehow 'proves' something concerning velocity of stars.  It doesn't mention velocity of anything.

If those types of spiral are real in the nature, why our scientists do not offer their solutions for those types of spirals?

What kind of solutions do you think are needed?  Their solutions are consistent at least, if not all correct.

Would you kindly show me the article about the modeling for each type of galaxy?

Not sure what you want.  Look them up yourself if you want to read about the various models.  I recall some good links going by a ways back in this thread.  I am not very familiar with the various models, or if 'articles' have been written about them.

Why do you insist that I should offer a solution for those types of galaxies while our scientists have no obligation for that?

Not sure what types of galaxies you're taking about.  Type A and B?  No galaxy is either.  I said that galactic spirals are type C in my post about that.  Type C is 'neither A nor B'.  The form is not describable with a 5 character formula.
Type A is: R=Xω where radius R is a linear function of the number of rotations ω.  X is some constant, and defines the constant separation of arms.
Type B is Φ=X where the X is a constant angle Φ of the spiral relative to the line tangential to a circle at any given point.
Those are really simple formulas.  No doubt there are others, but few that short.

I don't have at my fingertips a good formula for doing the same for a type B spiral.  It would have to be a scale-invariant formula, so there would be a fixed Φ input instead of a fixed ΔR per lap.  We'll leave it as an exercise to the reader.

This is clearly beyond the middle school mathematics skills of the reader, so I'll do it myself.
Formula for length (exact, not approximation) of type B spiral between two radius R values is ΔR/sinΦ. My initial difficultly was trying to make it a function of the number of laps (or cycles), but even that can be worked out from the above formula.

Mathematically, close enough is the same as wrong.

All the modern science is based on "close enough"

That's why I said mathematically, not scientifically, and usually science wants something more accurate than your crude formula for spiral length.  Your formula is wrong because if you integrate with that formula, the value approaches an incorrect figure (in fact the value doesn't change at all if you integrate yours).  If you integrate mine, it approaches the correct figure.  That's why even science would probably reject the P = 2πR +πΔR that you give above.  The fact that you got it from an article doesn't make it right.
I didn't know the correct formula, but I could see right away that the one you use can't be correct.  So I worked out a better one.  I didn't look it up.

They also couldn't show why the disc get's so narrow as the radius increases.

How so?  Where are these models that don't predict that?

Actually, based on their modeling they have found that the disc should be thicker as we move further away from the center.

The models do no such thing.

So, would you kindly explain why "close enough" is not good enough for my theory, but it is perfectly OK for our scientists?

I said 18π was close enough for an estimate of the length of a spiral doing one lap from R 8 to 10.  The formula fails in general, but it works fine for that example.  No idea what that example demonstrates since you seem incapable of getting to the 'therefore' part.

[Dave Lev (OP)]

After all your spiral types' formulas, I wonder which kind of formula our scientists have used in order to confirm the density wave theory.
As you claim that:

This is clearly beyond the middle school mathematics skills of the reader, so I'll do it myself.
Formula for length (exact, not approximation) of type B spiral between two radius R values is ΔR/sinΦ. My initial difficultly was trying to make it a function of the number of laps (or cycles), but even that can be worked out from the above formula.

So, would you kindly show me how our scientists have proved any sort of spiral formula by their density wave theory?
Not just nice hand wave and none realistic simulation - but please - based on real middle school mathematics.
This is vital.
Somehow, I have full confidence that they have never ever used any sort of spiral formulas or middle school mathematics which you are offering.
How can you explain that?