Twin paradox without same clocks switching frames In the following picture I am trying to eliminate the role of acceleration in twin-paradox like scenario.

*

*I is moving from right to left at constant speed .8c represented by a
small orange bar on top line on right.

*S is stationary represented by a small red bar on middle line

*O is moving from left to right at constant speed .8c represented by a
small blue bar on bottom line on left.

*I and O are moving at SAME speed of .8c.

When the scenario starts then the bottom left black arrow happens first, then after some time the upward pointing right black arrow and then at last the down pointing top left black arrow.
I am also trying to avoid the case of any of the clocks jumping their own inertial frames.
Given that, will S find that duration SWs > SWi + SWo ? Why?
Of course S, O and I are using light clocks for counting the duration on their stop watches.

Comparing Videos
Let us assume that S, O and I are identical triplet so we can sort of visualize comparison of their aging. Let us assume that video cameras were recording the triplets aging process. After the experiment was over we wanted to put the videos side by side. Obviously the length of the video recorded by S will be longer than the sum of the length of the video recorded by O and I. Obviously we must align the start of videos of O and S. Obviously we should align the end of the video of I and S. I have slightly modified the diagram by @Cleonis.


*

*Will the S video section AB align O video section AM ?

*Will the S video section TZ align I video section MZ ?

*the S video section BT will not have any corresponding part to O and I videos? If so why don't all the explanations of Twin paradox not state this clearly?

*At the end of S video section AB and O video section AM, will S and O look equally old?

*At the start of S video section TZ and I video section MZ, S will look older that I by BT time? However the age difference between S and I will seem to remain the same (BT) over the duration of TZ and MZ ?

So basically S video section BT accounts for S aging more right?
If we stretch the videos AM and MZ by equal amount to make it same length as AZ, will it give a visualization of how the aging difference happens? Or the aging of S should not be amortized but should be left as a sudden jump in the aging?
I hope it makes sense.
 A: I went ahead and uploaded an image that I had uploaded to wikipedia many years ago.

I'm not sure what your intended scenario is, but you state that you want to explore a scenario in which none of the clocks involved undergoes actual physical accelation, so I will proceed with a scenario that implements that.
Let's assume the setup uses atomic clocks that keep time with such a level of accuracy that the setup does not require the clocks to move at a very large velocity.
Clocks that move past each other with a constant relative velocity can be synchronized at the instant that they are at their closest proximity. I will refer to that (whimsically) as drive-by-synchronization.
The setup requires three drive-by-synchronisations.
1 At the origin. The clock that will count as the outward traveling clock has started moving earlier, the drive-by synchronization occors at the origin.
2 At the "turnaround" point. A third clock has started moving earlier, such that second and third clock pass each other at the pre-planned "turnaround" point. At that point a drive-by-synchronization is performed.
3 As the third clock moves past the first clock the amount of proper time that has elapsed is compared.

Under those conditions it will be seen that for the stationary clock (the first clock) a larger amount of proper time has elapsed than the amount of proper time as counted by the traveling clocks.
The amount of difference in elapsed proper time is given by the Minkowski metric.
Using the convention of represending spatial distance in terms of the speed of light: the amount of proper time of the traveling clocks, expressed in comparison with the proper time of the stationary clock is as follows:
$$ \tau^2 = t^2 - x^2  \qquad (1) $$
In (1) the quantity $t$ is the amount of proper time that has elapsed for the stationary clock, and $\tau$ is the amount of proper time that has elapsed for the traveling time keeping.
Note especially that (1) does not contain a term that represents velocity, nor one that represents acceleration.
To assess difference in amount of elapsed proper time it is sufficient to evaluate (1).
Of course, if the setup does involve physical acceleration then you have to set up an integration. What that integration does is to evaluate a concatenation of short spatial distances.
The point is: whatever integration that is done is a mathematical operation to accommodate the specific circumstances of a setup; in terms of the physics taking place expression (1) is exhaustive.

In structure the Minkowski metric is analogous to Pythagoras' theorem. Both in the case of Pythagoras's theorem and in the case of the Minkowski metric the units that go into the expression are squared. The one difference of course is that minus sign.
In Euclidean space, if you travel from point A to point B along a path that is not a straight line you cover a spatial distance that is longer than that of the straight line.
Pythagoras' theorem expresses the metric of Euclidean space.
The Minkowski metric expresses the metric of Minkowski spacetime. In Minkowski spacetime if you travel from point A to point B along a path that is not a straight line the amount of proper time that elapses for you is less than when moving along a straight line.

Special relativity does not explain why the Minkowski metric has the form that it has. The Minkowski metric is something that you have to grant in order to formulate special relativity at all. With the Minkowski metric granted all aspects of special relativity follow logically.
A: UPDATE: Since the OP updated the question, I have updated this answer (see below).
UPDATE2: In order to further address the comments of the OP, I have added two more spacetime diagrams.

It might be helpful to visualize the spacetime-diagram of the light-clocks involved in the Clock Effect.
It provides a mechanism that can be used to operationally define how time is measured.
On rotated graph paper, it can be seen that the "light-clock diamonds" (the causal diamond between consecutive ticks of an inertial clock) have the same area [as required by lorentz invariance (since det L=1)].
Below are clocks for inertial observers RED, BLUE, and GREEN.
BLUE has velocity 0.8c(= (4/5)c ) with respect to RED, and
GREEN has velocity of -0.8c(= (-4/5)c ) with respect to RED.
These inertial light-clocks measure the proper time along these inertial paths.

*

*Of course, the "traveler" who left RED and later reunites with RED
undergoes acceleration since she must use a non-inertial path BLUE-THEN-GREEN (which is piecewise-inertial, but nevertheless non-inertial).

*One can imagine the "traveler" releasing the BLUE clock and grabbing onto [or merely traveling along with] the GREEN clock, or merely using the clocks along those inertial worldlines. These clocks are inertial.

*One can also imagine the "traveler" carrying her own clock along the piecewise-inertial worldline, which must match the intervals with the corresponding inertial clock alongside on that inertial portion of the trip.(*) In this case, the "traveler"-clock also undergoes acceleration.

*(*) From Geroch's General Relativity from A to B, p. 80:


The upshot of our property of clocks is this. Given any world-
line of a particle, one acquires an assignment of times to points
of that world-line. (Physically, the assignment is obtained by
carrying a clock alongside the particle, and using the readings
of the clock to obtain the "times.") This assignment of "times"
to the points of the world-line of any particle is unique as far as
time differences are concerned. (Here is where we are using our
property of clocks. When we say "carry a clock alongside the
particle" we do not specify the past history of the clock. We
are thus assuming implicitly that ticking rates are independent
of past history.) To summarize, world-lines of particles now
acquire time-functions.

UPDATED
https://www.desmos.com/calculator/8kr9uc9zwu 
(v.9e of my spacetime diagrammer [for the Twin Paradox / Clock Effect] ) 


Update
To answer your questions, you can read off information from my diagram.
This approach is a "visualization of proper time"
by visualizing the spacetime-diagram of a light-clock,.
My interactive Desmos application has some features that can be turned on:
"simultaneity" and "time dilation" and "doppler".
"Simultaneity" is identified by the spacelike diagonal of the observer diamonds.
I'll address your added questions below this diagram.
Note that the traveler in the diagram has, by counting diamonds,

*

*speed = (4/5)c = 0.8c, there and back.


*the time-dilation factor $\gamma =\ 5/3 $.
Hence, (5)=(5/3)(3) along RED's simultaneity (your "S")
and (3)=(5/3)(9/5)=(5/3)(1.8) along BLUE's simultaneity (your "outbound O").
--Refer to the diagram above.--


*Doppler factor $k=3$, which is associated with the stretching of the diamonds, by a factor 3 in forward direction and by 1/3 in the other direction.Follow the worldlines of light-rays along the light-cones.


There is an important subtlety involving such "videos":
one needs to distinguish

*

*"simultaneity of distant events" (the "assignment of t-coordinates" done by each observer using radar-measurements or a line of distant clocks associated with the observer) following the lines parallel to the clock's spacelike diamond-diagonals. This involves "time-dilation". This is essentially "map making".

*"visual appearance" (the "seeing" done by each observer using light-signals that are received on that observer's worldline) following lines along the light-cones (along 45-degree lines, the diamond-edges). This involves the "Doppler effect". This is essentially "viewing a movie [of light-rays striking your eye or film]".

Onward to your questions...

*

*Will the S video section AB align O video section AM ?

NO.
According to S

*

*event B (on S @ 1.8) occurs at t=1.8 according to S


*event M (on O @ 3) occurs at t=5 according to S (see the first diagram)


*If S makes an animated chart (an animated spacetime diagram) that develops as S's clock evolves, when B appears at t=1.8, the event Bsim (on O @ 1.08) on O's worldline appears. S assigns both events t=1.8 and says that B and Bsim are simultaneous. (Why 1.08? That's 1.8/(5/3), from Time Dilation. Think similar triangles, comparing with first diagram.)


*If S makes a movie with a camera (receiving light from outbound observer O),
when B appears at t=1.8, the light-flash-signal from the event Bpas (on O @ 0.6 ) appears in the movie. Imagine that each clock emits the light on its display telling the time.
All S can say is that when S's clock reads (at B) 1.8, S received light from O with the reading of 0.6. (Why 0.6? That's 1.8/(3), from Doppler.)
According to O

*

*event B (on S @ 1.8) occurs at t=3 according to O


*event M (on O @ 3) occurs at t=3 according to O


*If O makes an animated chart (an animated spacetime diagram) that develops as O's clock evolves, when B appears at t=3, the event M (on O @ 3) on O's worldline appears. S assigns both events t=3 and says that B and M are simultaneous.


*If O makes a movie with a camera (receiving light from observer S),
when B appears at t=3, the light-flash-signal from event B (on S @ 1.8 ) appears in the movie. Imagine that each clock emits the light on its display telling the time.
All O can say is that when O's clock reads (at M) 3, O received light from S with the reading of 1.8.


*What would O say when O's clock reads 1.8? 
O obtains the same results when O's clock reads 1.8
that S obtained when S's clock reads 1.8
...in accordance with the principle of relativity since O and S have been inertial observers since their separation event.

I think this gives you enough of a headstart
(and an interactive tool) to answer the rest of your questions.

UPDATE 2
My Desmos application has a "boost" feature to view the situation
from different frames. (In this version, the rotated graph paper grid
for the incoming leg hasn't been drawn.)
Here is the incoming leg, followed by the outgoing leg
in the same spirit of my answer to another question (linked at the end)
to attempt to construct a spacetime diagram for a non-inertial observer.
Stitching together the diagrams for outgoing (AM) and incoming (MZ) legs
leads to a single diagram which is not a real spacetime diagram because of some peculiar features discussed in that earlier answer (linked below).
For example, the inertial observer (AZ) has a discontinuous worldline in this diagram (jumps in t-assignments, irregular clock-readings, and jump in position [when the incoming leg has a different speed from the outgoing leg]).
This Frankensteined diagram is clearly not equivalent to the spacetime diagram of the inertial traveler from A to Z.
By following light-rays along the light-cones (along the diamond-edges), one can track
the transmission and receptions of the "clock-face images",
if one want to understand what each traveler "sees" or "captures on film".



For more details [including a reference to my paper], and for a discussion of how the non-inertial "traveler"
is not equivalent to the inertial stay-at-home RED observer,
visit my answer
https://physics.stackexchange.com/a/507592/148184

for this question
What is the proper way to explain the twin paradox? .
A: Acceleration has virtually no effect on the different aging experienced by the twins. That depends only on their relative velocities and how long they are apart, i.e. their relative paths through spacetime. And indeed it is possible to construct scenarios (such as you've done here) in which there is no acceleration.
The reason that acceleration is often pointed to as the explanation for the twin paradox is that it's an easy way to show that the situation is not symmetric: in the typical setup only one twin experiences acceleration, and that shows that the two twins are not at all the same. If you eliminate acceleration but have two traveling clocks instead, there's still no symmetry: one time period is experienced by a single clock (the "stay at home" clock), the other is experienced by two clocks (the outbound and inbound clocks). Again, the paths through spacetime are different.
Think of a clock as being like an odometer, but measuring time instead of space. There's no mystery if two twins leave point A and arrive at point B with different amounts showing on their odometer -- it just means that they've taken different paths from A to B.
A: David Morin, Introduction to Classical Mechanics, Chapter 11, p. 14: "Twin A stays on the earth, while twin B flies quickly to a distant star and back. [...] For the entire outward and return parts of the trip, B does observe A's clock running slow, but ENOUGH STRANGENESS occurs during the turning-around period to make A end up older." http://www.people.fas.harvard.edu/~djmorin/chap11.pdf
So, all along, the traveler sees himself aging FASTER than stationary people, but, during the short turning-around period, "enough strangeness" occurs and stationary people suddenly get very old:
"At the same time, the twin in the spaceship considers himself to be the stationary twin, and therefore as he looks back towards Earth he sees his brother ageing more slowly than himself. [...] Ah, but in order to return to Earth, the spaceship must slow down, stop moving, turn around and go back the other way. During those periods of deceleration and deceleration, it is not an inertial frame and therefore the normal rules of special relativity don't apply. When the twin in the spaceship turns around to make his journey home, the shift in his frame of reference causes his perception of his brother's age to change rapidly: he sees his brother getting suddenly older. This means that when the twins are finally reunited, the stay-at-home twin is the older of the two." http://topquark.hubpages.com/hub/Twin-Paradox
Physics Girl (4:30): "One last question. What's happening to the clocks during the period of acceleration? We still get time dilation, but we have to use a different set of rules from the general relativity. General relativity states that clocks runs slower in accelerated reference frames. So while your twin is turning around, her clock runs slower, and she sees the same thing. She sees your clock running faster than hers, so you're aging quicker. IT'S DURING THIS PERIOD OF ACCELERATION THAT YOU BECOME THE OLDER TWIN." https://www.youtube.com/watch?v=ERgwVm9qWKA
Can you imagine anything more absurd than the sudden ageing of stationary people while the traveling twin turns around? Actually, the "enough strangeness" is a euphemism for "homogeneous gravitational field" - an incredible absurdity Einstein produced in 1918. This homogeneous gravitational field is due to the turning-around acceleration of the traveling twin and affects stationary clocks - makes them run very fast (which means that stationary people suddenly get very old):
Albert Einstein 1918: "A homogeneous gravitational field appears, that is directed towards the positive x-axis. Clock U1 is accelerated in the direction of the positive x-axis until it has reached the velocity v, then the gravitational field disappears again. An external force, acting upon U2 in the negative direction of the x-axis prevents U2 from being set in motion by the gravitational field. [...] According to the general theory of relativity, a clock will go faster the higher the gravitational potential of the location where it is located, and during partial process 3 U2 happens to be located at a higher gravitational potential than U1. The calculation shows that this speeding ahead constitutes exactly twice as much as the lagging behind during the partial processes 2 and 4." http://sciliterature.50webs.com/Dialog.htm
