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Why do you take it to the extreme? (1 light hour or 10,000 LY)I have specified a volume.

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?

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

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.

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.

Quote from: HalcIf 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

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

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).

New stars emerge from the bar and replace them.

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.

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_45893184We 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 2R1So, it is clear that if the orbital velocity of star 1 is V1Than 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)

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)

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

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.

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.

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.

Responses are delayed. Busy.

QuoteLet'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.

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.

QuoteHowever, 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.

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.

QuoteLet'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.

QuoteIt 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.

QuoteHence 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.

QuoteSo, 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.

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

QuoteShe 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.

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.

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.

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

That is a fatal mistake.

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.

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

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?

So, as ΔR = 10 - 8 = 2

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

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

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.

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.

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?

Quote 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.

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.

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.

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

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 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.

the bar is located between 2KPC to 3KPC.

We have found that:P = 2πR + πΔR

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.

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

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.

With regards to ΔP 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

Quote from: Halc 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.

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

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).

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

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.

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?

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

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.

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 = 2ΔR/2 = 2/2 = 1R + 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 + πΔRHowever, we know that in order to keep the velocity, we need that the orbital distance per interval T will be (one full cycle):P1 = 2πRHence, in order to keep the orbital velocity we must set thatΔP = πΔR

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

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 = 2ΔR/2 = 2/2 = 1R + ΔR/2 = 8+1 = 9.That exactly what I did.Do you still see any problem with that?

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):

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.

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

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

Is it clear to you by now?

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

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*10Is it realistic?Why not the average radius (R1+R2)/2 or (R1+ΔR/2) or just (9)?

"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!!!

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) ÷ 2For 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.

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

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.

Would you kindly focus on my message?I have stated:Quote from: Dave Lev on 05/04/2019 04:57:13Please - just look at that from a mathematical point of viewSo, 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).

So, If R1 - the radius at starting point.R2 - The radius after 360° arc.ΔR = R2 - R1Hence - 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 +πΔRDo you agree with that?

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?

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.

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.

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

Mathematically, close enough is the same as wrong.

You insist that there are two spiral types - Type A and Type B.With regards to Type A you claim:Quote from: HalcFor 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 BQuote from: HalcThe 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?

Quote from: Halc on 05/04/2019 17:55:24It 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?

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.

Quote from: HalcMathematically, close enough is the same as wrong.All the modern science is based on "close enough"

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

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

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

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.