Author Topic: SpaceX Starship IFT-2 : Starbase TX : 18 Nov 2023 DISCUSSION  (Read 438413 times)

Offline InterestedEngineer

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Re: SpaceX Starship IFT-2 : Starbase TX : 18 Nov 2023 DISCUSSION
« Reply #820 on: 11/29/2023 11:28 pm »
Interesting video on CFD simulations of prop slosh in the booster:



I'm not seeing any gravity effects, so I'm suspicious this CFD has proper inputs.

at 30 degrees slope from horizontal, there should be a noticeable angle to the top of the liquid.
I don't think you should be seeing any.

The booster is ballistic, except for the inputs from the engines (and from Starship)

To me, this slosh pattern does indicate the booster sees negative acceleration for a quick moment.

MeekGee is correct. The pendulum fallacy is as devilishly deceptive as ever, I see!

I can only plead I was groggy that day.   It was also about a week back, why does this keep getting quote posted?  I was hoping it would die a timely death, like it would in most threads.   <shrug>

Thanks for the gentle correction, unlike <cough> someone who PM'd me.

The accelerations are happening all over the place, rotational, net force changing quickly from Starship engine impingment, changes in Booster throttle, and probably something else I'm not thinking of.  (I'd guess the 0.5m/sec centripetal acceleration acts the same as gravity... right?)

Online meekGee

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Re: SpaceX Starship IFT-2 : Starbase TX : 18 Nov 2023 DISCUSSION
« Reply #821 on: 11/29/2023 11:30 pm »
Interesting video on CFD simulations of prop slosh in the booster:



I'm not seeing any gravity effects, so I'm suspicious this CFD has proper inputs.

at 30 degrees slope from horizontal, there should be a noticeable angle to the top of the liquid.
I don't think you should be seeing any.

The booster is ballistic, except for the inputs from the engines (and from Starship)

To me, this slosh pattern does indicate the booster sees negative acceleration for a quick moment.

MeekGee is correct. The pendulum fallacy is as devilishly deceptive as ever, I see!

I can only plead I was groggy that day.   It was also about a week back, why does this keep getting quote posted?  I was hoping it would die a timely death, like it would in most threads.   &lt;shrug&gt;

Thanks for the gentle correction, unlike &lt;cough&gt; someone who PM'd me.

The accelerations are happening all over the place, rotational, net force changing quickly from Starship engine impingment, changes in Booster throttle, and probably something else I'm not thinking of.  (I'd guess the 0.5m/sec centripetal acceleration acts the same as gravity... right?)
Sorry.   I'm a simple creature.  Tapatalk pings me, I respond, repeat...

Topic killed. :)
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Offline CorvusCorax

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Re: SpaceX Starship IFT-2 : Starbase TX : 18 Nov 2023 DISCUSSION
« Reply #822 on: 11/29/2023 11:41 pm »
Look at the bright side, noone who follows this thread is gonna fall for that pit again, anytime soon :)

Offline clongton

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Re: SpaceX Starship IFT-2 : Starbase TX : 18 Nov 2023 DISCUSSION
« Reply #823 on: 11/30/2023 02:14 am »
The F9 has header tanks?

It has sumps

Sumps is the essence of what I was suggesting several pages back for the SH when I was mentioning porous baffles. I was envisioning all the engine intakes being within the sump with a porous baffle on the top. Nominal operations would draw all propellants from the tanks - thru the sumps - and during zero-g moments or flipping, the baffles would secure enough propellant within them to give a good engine start sequence. The baffles on the sump tops would prevent the loss of propellant from the engine intakes. As soon as the engines ignite, all the remaining propellant will drop immediately to the bottom of the tank.
« Last Edit: 11/30/2023 02:19 am by clongton »
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Online TheRadicalModerate

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Re: SpaceX Starship IFT-2 : Starbase TX : 18 Nov 2023 DISCUSSION
« Reply #824 on: 11/30/2023 12:11 pm »
Look at the bright side, noone who follows this thread is gonna fall for that pit again, anytime soon :)

You do realize that if you fall victim to Poe's Law and your user handle is Raven, the irony police will fast-rope into your bedroom, Tom Cruise-like, and haul you away?

Offline CorvusCorax

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Re: SpaceX Starship IFT-2 : Starbase TX : 18 Nov 2023 DISCUSSION
« Reply #825 on: 11/30/2023 06:57 pm »

Nevermore

Offline InterestedEngineer

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Re: SpaceX Starship IFT-2 : Starbase TX : 18 Nov 2023 DISCUSSION
« Reply #826 on: 12/01/2023 12:52 am »
Someone extracted raw data from the videos for trajectory, velocity, etc.

https://osemplacyc146.owncloud.online/s/uGkK2mFVqqb39qA?path=%2F2023-11-18%20IFT-2

Online catdlr

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Re: SpaceX Starship IFT-2 : Starbase TX : 18 Nov 2023 DISCUSSION
« Reply #827 on: 12/01/2023 01:02 am »
Someone extracted raw data from the videos for trajectory, velocity, etc.

https://osemplacyc146.owncloud.online/s/uGkK2mFVqqb39qA?path=%2F2023-11-18%20IFT-2

Be careful that the site (not the data) is coming up with a Malware Security warning.
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Offline InterestedEngineer

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Re: SpaceX Starship IFT-2 : Starbase TX : 18 Nov 2023 DISCUSSION
« Reply #828 on: 12/01/2023 01:21 am »
Someone extracted raw data from the videos for trajectory, velocity, etc.

https://osemplacyc146.owncloud.online/s/uGkK2mFVqqb39qA?path=%2F2023-11-18%20IFT-2

Be careful that the site (not the data) is coming up with a Malware Security warning.

Not sure what you are using that's warning you, but I've got the latest Chrome.

Your security software probably doesn't like old school file share sites or XLS sheets.  I've loaded most of the files with no ill effects or warnings. (OSX)

Offline InterestedEngineer

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Re: SpaceX Starship IFT-2 : Starbase TX : 18 Nov 2023 DISCUSSION
« Reply #829 on: 12/01/2023 01:27 am »
Someone extracted raw data from the videos for trajectory, velocity, etc.

https://osemplacyc146.owncloud.online/s/uGkK2mFVqqb39qA?path=%2F2023-11-18%20IFT-2

I took the vertical and horizontal velocities from the trajectory sheet and came up with an angle of 33 degrees near MECO (T+151s).

About what is expected.

sin(33) = 0.54
cos(33) = 0.84

So given an acceleration of about 16.6 m/sec2 that's pretty close to break even with gravity and the centrifugal acceleration that occurs at 1.5km/sec at 60km altitude. (maybe 0.5m/sec2 shy)

For the X component it's 14m/s2.   

Offline InterestedEngineer

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Re: SpaceX Starship IFT-2 : Starbase TX : 18 Nov 2023 DISCUSSION
« Reply #830 on: 12/01/2023 01:57 am »
Someone extracted raw data from the videos for trajectory, velocity, etc.

https://osemplacyc146.owncloud.online/s/uGkK2mFVqqb39qA?path=%2F2023-11-18%20IFT-2

There's negative acceleration at MECO before separation, which is weird.  It continues to be negative, gets a big negative glitch at Starship ignition, until the booster clears the plume and the 3 remaining engines kick in a positive acceleration just before the flip fully starts.

This is of course in the reference frame of the ground station.  Since we know the angle I'm wondering if we can now figure out what the accelerations look like in the Booster's reference frame.  (For example gravity takes over which might be reflected in the velocity magnitude we are presented with)

Offline Lee Jay

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Re: SpaceX Starship IFT-2 : Starbase TX : 18 Nov 2023 DISCUSSION
« Reply #831 on: 12/01/2023 02:31 am »
Someone extracted raw data from the videos for trajectory, velocity, etc.

https://osemplacyc146.owncloud.online/s/uGkK2mFVqqb39qA?path=%2F2023-11-18%20IFT-2

There's negative acceleration at MECO before separation, which is weird.  It continues to be negative, gets a big negative glitch at Starship ignition, until the booster clears the plume and the 3 remaining engines kick in a positive acceleration just before the flip fully starts.

This is of course in the reference frame of the ground station.  Since we know the angle I'm wondering if we can now figure out what the accelerations look like in the Booster's reference frame.  (For example gravity takes over which might be reflected in the velocity magnitude we are presented with)

It's inertial guidance, right? So it's in the reference frame of the launch site where accelerations were zeroed?

Offline Twark_Main

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Re: SpaceX Starship IFT-2 : Starbase TX : 18 Nov 2023 DISCUSSION
« Reply #832 on: 12/01/2023 02:50 am »
Since we know the angle I'm wondering if we can now figure out what the accelerations look like in the Booster's reference frame.

Hint: decompose the velocity into horizontal and vertical components.  :)

This is essentially what I did in my "napkin" analysis.  I was lazy and just estimated the angle at staging, but naturally if you curve fit the angle you can get more precision.
« Last Edit: 12/01/2023 02:52 am by Twark_Main »
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Offline InterestedEngineer

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Re: SpaceX Starship IFT-2 : Starbase TX : 18 Nov 2023 DISCUSSION
« Reply #833 on: 12/01/2023 03:08 am »
Since we know the angle I'm wondering if we can now figure out what the accelerations look like in the Booster's reference frame.

Hint: decompose the velocity into horizontal and vertical components.  :)

This is essentially what I did in my "napkin" analysis.  I was lazy and just estimated the angle at staging, but naturally if you curve fit the angle you can get more precision.

Have to decompose the accelerations into horizontal and vertical components.   The vertical acceleration goes negative because thrust is reduced, which *may* make the video speed readout go down, but relative to the fuel that doesn't change. 

Making my head spin.

When I do that it's almost all gravity, magnitude 8.8m/s2, sign negative, assuming the thrust drops to 1/11 of what it was prior.

But the readouts right after MECO are showing -3.3m2.   

If I assume the thrust dropped to 1/11/.4 or .223 of before MECO, I get +3.2m/s2 horizontal and -7.5m/s2 vertical, or still magnitude -8.1.

But I don't think I have the frame of reference correct to compare to the -3.3m2 on the readout.

so my head is still spinning.
« Last Edit: 12/01/2023 03:23 am by InterestedEngineer »

Offline Brigantine

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Re: SpaceX Starship IFT-2 : Starbase TX : 18 Nov 2023 DISCUSSION
« Reply #834 on: 12/01/2023 06:58 am »
2023-11-20 Raw data (Updated) - Time, Altitude, Speed (IFT-2).csv
We start with |vair| converted from km/h (speed, specifically magnitude of velocity vector in a rotating earth frame)
and sy in km (altitude, specifically y component of position in rotating earth frame. precise  [but not accurate] for S1, nearest km for S2)
time seems to increment at 30 Hz around hot-staging, but at other times is 49 Hz

At t = 163, using Stage 1 data
|vair| = 1,567 m/s
vy = 758 m/s  (from Δsy / Δt)

so the flight path elevation (in rotating earth frame)
γair = sin-1(758/1,567) = 28.9⁰

and horizontal velocity
vx (air) = √(1,567 - 758) = 1,371 m/s

The "air" is moving (in Earth's inertial frame) at 422 m/s (assuming lat=26⁰, az=90⁰, r=6371+sy)

I want to know what I expect |vair| and vy to do to a body in a ballistic trajectory. The cross-range effects of Coriolis are irrelevant, but I want to know the angular rate of where "down" is

ω = v/r = (|vx (air)| + 422)/(6371 + sy) = 0.000278 rad/s

so d/dt vy = Fy/m + 0.000278 * (vx (air) + 422) - μ/r
and d/dt vx (air) = Fx/m - 0.000278 * vy

which effect d/dt |vair| = d/dt vy * sin(γ) + d/dt vx (air) * cos(γ)
if F=0, d/dt |vair| = -9.11 * sin(28.9⁰) - 0.21 * cos(28.9⁰) = -4.59 m/s

Over the period from t = 160s to t = 166s, this value progresses from -4.62 to -4.54 m/s

If we assume the pitch angle is small, then add 4.6 m/s to get the axial acceleration in the ballistic inertial frame

P.S. I cut the corner in orange where the possible telemetry cut-out is. Similar conclusions to people in hot-staging thread, there's 4 - 8 m/s of (cumulative) deceleration, over an uncertain timeframe of 0.8 - 2.0 seconds. Booster pushed back a total distance of 3 or 4 meters (about 2 rings). No obvious signs of any re-collision with propellants though.
« Last Edit: 12/01/2023 10:30 am by Brigantine »

Offline CorvusCorax

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Re: SpaceX Starship IFT-2 : Starbase TX : 18 Nov 2023 DISCUSSION
« Reply #835 on: 12/01/2023 07:38 am »
ForF9 launches ( and I assume for Starship its the same) SpaceX displays velocity as absolute speed in a rotating earth fixed reference frame.

That means the displayed velocity standing on the pad is always zero.

And the velocity in geostationary orbit ( for the few launches we had telemetry overlay of an.upper stage in a stream that late )is also zero. That was especially weird when the stage did a circularisation burn - and the velocity kept dropping until.it was just a few hundred km/h - but makes sense in that frame

Offline InterestedEngineer

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Re: SpaceX Starship IFT-2 : Starbase TX : 18 Nov 2023 DISCUSSION
« Reply #836 on: 12/01/2023 08:43 pm »
1: so d/dt vy = Fy/m + 0.000278 * (vx (air) + 422) - μ/r
2: and d/dt vx (air) = Fx/m - 0.000278 * vy

3: which effect d/dt |vair| = d/dt vy * sin(γ) + d/dt vx (air) * cos(γ)
4: if F=0, d/dt |vair| = -9.11 * sin(28.9⁰) - 0.21 * cos(28.9⁰) = -4.59 m/s

Good stuff, but I got lost.  (1) makes sense but after that I'm confused.

Why is the 0.000278 * vy * vy component subtracted for finding d/dt vx (air) ?

similar question for lines 3:   Why do you add the sum of the vector components to get the magnitude instead of the sum of the squares of the components of the vectors?
« Last Edit: 12/01/2023 10:16 pm by InterestedEngineer »

Offline seb21051

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Re: SpaceX Starship IFT-2 : Starbase TX : 18 Nov 2023 DISCUSSION
« Reply #837 on: 12/01/2023 09:57 pm »
ForF9 launches ( and I assume for Starship its the same) SpaceX displays velocity as absolute speed in a rotating earth fixed reference frame.

That means the displayed velocity standing on the pad is always zero.

And the velocity in geostationary orbit ( for the few launches we had telemetry overlay of an.upper stage in a stream that late )is also zero. That was especially weird when the stage did a circularisation burn - and the velocity kept dropping until.it was just a few hundred km/h - but makes sense in that frame

So where is the point that it changes from rising to falling velocity, about halfway up to GEO?

Offline Brigantine

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Re: SpaceX Starship IFT-2 : Starbase TX : 18 Nov 2023 DISCUSSION
« Reply #838 on: 12/02/2023 02:34 am »
So where is the point that it changes from rising to falling velocity, about halfway up to GEO?
My best guess: (though I haven't seen the video and I took some shortcuts)

As soon as the engine cuts out on the GTO insertion burn.
Then during the coast it will change from falling to rising, as the ground track turns around and goes west
→ can calculate at what height from the max speed & equal swept area:
→ ALT = -6371 + √[1/(7.2921159 10−5) * speed*(6371 + altitude)] as at SECO 2
[EDIT: Added √, now I get 23,400 km altitude, but that's wrong since the vertical speed is changing so much too]

Then at apogee it would go from rising to falling, but the circularization burn makes it fall fast (ground track go west slower and slower)

1: so d/dt vy = Fy/m + 0.000278 * (vx (air) + 422) - μ/r
2: and d/dt vx (air) = Fx/m - 0.000278 * vy

3: which effect d/dt |vair| = d/dt vy * sin(γ) + d/dt vx (air) * cos(γ)
4: if F=0, d/dt |vair| = -9.11 * sin(28.9⁰) - 0.21 * cos(28.9⁰) = -4.59 m/s

Good stuff, but I got lost.  (1) makes sense but after that I'm confused.

Why is the 0.000278 * vy component subtracted for finding d/dt vx (air) ?

similar question for lines 3:   Why do you add the sum of the vector components to get the magnitude instead of the sum of the squares of the components of the vectors?
Thanks for engaging.

Why subtracted?
The Coriolis force pulls an upwards vy to the west.
Consider it this way - a 2D vector being rotated, as a matrix transformation, has that -sin term:

[vx (t+Δt)] = [cos(0.000278*Δt); -sin(0.000278*Δt)] [vx (t)]
[vy (t+Δt)]    [ sin(0.000278*Δt); cos(0.000278*Δt)   ] [vy (t)]

Or just consider a pure vertical vector up from Boca Chica, as seen from an observer in Florida - it looks like it's leaning away.

Why not sum of the squares?
There's a very subtle difference between the magnitude of (change in v), and change in (magnitude of v). Our best data is the latter.

Start with the sum of the squares, and then differentiate it:
|v| = vy + vx
2|v| * d|v|/dt = 2vy * dvy/dt + 2vx * dvx/dt

Where sin(γ) = vy/|v| and cos(γ) = vx/|v|

It isn't the total acceleration, just the component that changes the magnitude (instead of the direction) of vair.
You can think of it as a dot product between acceleration and a unit vector aligned with velocity i.e. a (v/|v|)
[ax]    [cos(γ)]
[ay] [ sin(γ)]
[az]    [   0    ] ← this zero is also why I ignored cross-range Coriolis effects

That "assume the pitch angle is small" is doing a lot of work here - since even after all this adjusting, we only have the component of acceleration in one direction (flight path). Luckily we only want to know the component of acceleration in one direction (booster longitudinal axis), and we hope the two are similar - otherwise they could have different signs, as they clearly do after t = 174s.


What if we assume no telemetry cut out and accept 2-3g deceleration of the rigid booster for <0.5s, and further assume(!) the Ship and Booster had equal thrust. What HSR pressure over 63m can accelerate a 200t rigid booster at 2g? 4 MN / 63m ~= 60 kPa. Significantly reducing over separation distances comparable to the HSR height.
It moves back 16m total and returns to zero around 171.3s, roughly where the acceleration decreases from 12-ish to 7-ish (but still no 'collision')
« Last Edit: 12/02/2023 06:47 am by Brigantine »

Offline Twark_Main

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Re: SpaceX Starship IFT-2 : Starbase TX : 18 Nov 2023 DISCUSSION
« Reply #839 on: 12/02/2023 06:28 am »
So where is the point that it changes from rising to falling velocity, about halfway up to GEO?
My best guess: (though I haven't seen the video and I took some shortcuts)

As soon as the engine cuts out on the GTO insertion burn.

...

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