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#660
by
Marine_Mustang
on 27 Jun, 2019 21:37
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Here, I think.
Looks like they removed the legs from the booster still near the upper pad and laid them down on the pad.
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#661
by
scr00chy
on 28 Jun, 2019 00:03
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How the heck did they move the LZ-2 booster all the way there? With a mobile crane?
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#662
by
russianhalo117
on 28 Jun, 2019 00:58
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How the heck did they move the LZ-2 booster all the way there? With a mobile crane?
Wonder if they have Land Roomba's.
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#663
by
russianhalo117
on 28 Jun, 2019 00:59
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Here, I think.
Looks like they removed the legs from the booster still near the upper pad and laid them down on the pad.
NOTE: the legs looks white on the booster hugging side so they are probably from FH-01.
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#664
by
wannamoonbase
on 28 Jun, 2019 05:31
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How the heck did they move the LZ-2 booster all the way there? With a mobile crane?
Wonder if they have Land Roomba's.
A crane moving with a load under hook is not unusual. They just need to make sure the crawling surface is suitable as a crane pad.
I worked on a slurry wall construction project with rebar cages that were longer, wider and heavier than an F9 core. It’s not a problem.
Why they needed to ,over it is a more interesting question.
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#665
by
EspenU
on 28 Jun, 2019 06:48
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During the booster boostback burn there was something "flying around" with a gaseous "tail" in the plume interactions. It can be seen in the IR view from T+02:45 to 02:50 (from 27:43 in the Youtube video). Does anyone know what it was?
I'm not suggesting that anything went wrong, I'm just curious.
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#666
by
Steven Pietrobon
on 28 Jun, 2019 08:25
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So with these tweaks, we should use an effective g of 9.8 - 0.5 - 1.58 = 7.72 m/s. Knowing the 370 second coast, the 123 km start altitude, and the 60 km end altitude, then we solve for 123000 + v*t - 1/2*g*t = 60000, to get vertical v = 1257 m/s. This is within 1% of the correct value of 1265 m/s, as determined by the physics based computation below (in Python). The estimated peak altitude is 226.6 km as opposed to the correct 225.6 km. So the few tweaks get a result that is likely more accurate than the input data we are using.
Python makes my head spin! I redid the code in Pascal. Here's my output, which finds the maximum altitude. If I were to do the code though, I would use fourth-order Runga-Kutta to solve the differential equations, plus add in the drag from the thin upper atmosphere. :-)
vx = 3214 m/s. Earth rotation added 407 m/s.
t= 0.0 s, g=9.4950 m/s^2, dr= 0.000 km, y=6479.000 km, alt= 123.000 km
t=163.1 s, g=9.2011 m/s^2, dr= 437.223 km, y=6560.991 km, alt= 225.638 km
t=370.0 s, g=9.6823 m/s^2, dr= 996.027 km, y=6311.894 km, alt= 60.003 km
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#667
by
ugordan
on 28 Jun, 2019 08:43
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During the booster boostback burn there was something "flying around" with a gaseous "tail" in the plume interactions. It can be seen in the IR view from T+02:45 to 02:50 (from 27:43 in the Youtube video). Does anyone know what it was?
I'm not suggesting that anything went wrong, I'm just curious.
It's basically the high altitude version of a Mach diamond. It's normal and can be seen with many kerolox rockets, here's an example on a
Saturn V Recent F9 launch seen pretty much from sideways, starting at around 30 seconds in:
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#668
by
Vettedrmr
on 28 Jun, 2019 10:52
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During the booster boostback burn there was something "flying around" with a gaseous "tail" in the plume interactions. It can be seen in the IR view from T+02:45 to 02:50 (from 27:43 in the Youtube video). Does anyone know what it was?
I'm not suggesting that anything went wrong, I'm just curious.
I was curious about that as well. My first guess was that it's the cross-brace that ties the 3 boosters together right above the grid fins, but then I remembered that retracts forward against the booster body.
It sure doesn't look like any kind of plume interaction. So I guess I have no idea what it is.
Have a good one,
Mike
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#669
by
cscott
on 28 Jun, 2019 13:07
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Why they needed to ,over it is a more interesting question.
I believe that's where the fixed stand is that they attach the booster to in order to fold/remove the legs.
I think both boosters are currently on their fixed stands, having been brought from their landing spots (grey smudges on concrete) by mobile crane.
Don't know why they brought the legs over and laid them out on the pad, but I'd wager that's not where the legs were removed.
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#670
by
LouScheffer
on 28 Jun, 2019 13:17
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So with these tweaks, we should use an effective g of 9.8 - 0.5 - 1.58 = 7.72 m/s. Knowing the 370 second coast, the 123 km start altitude, and the 60 km end altitude, then we solve for 123000 + v*t - 1/2*g*t = 60000, to get vertical v = 1257 m/s. This is within 1% of the correct value of 1265 m/s, as determined by the physics based computation below (in Python). The estimated peak altitude is 226.6 km as opposed to the correct 225.6 km. So the few tweaks get a result that is likely more accurate than the input data we are using.
Python makes my head spin! I redid the code in Pascal. Here's my output, which finds the maximum altitude. If I were to do the code though, I would use fourth-order Runga-Kutta to solve the differential equations, plus add in the drag from the thin upper atmosphere. :-)
vx = 3214 m/s. Earth rotation added 407 m/s.
t= 0.0 s, g=9.4950 m/s^2, dr= 0.000 km, y=6479.000 km, alt= 123.000 km
t=163.1 s, g=9.2011 m/s^2, dr= 437.223 km, y=6560.991 km, alt= 225.638 km
t=370.0 s, g=9.6823 m/s^2, dr= 996.027 km, y=6311.894 km, alt= 60.003 km
You are certainly correct that that any number of better integration schemes could be used. Here I used forward Euler, the fancy name for extrapolation, just to make the code simple. This in turn requires a small time step (if you increase dt to 1 sec, you will give km-size errors) and risks roundoff errors (here not a problem since we are computing only a fraction of a smooth orbit.)
But at least the same algorithm (in Pascal or Python) gives the same answers:
vx = 3213.996 Earth rotation added 406.860 m/s
t=163.1 g=9.2011 m/s^2 dr= 437.223 km y=6560.991 alt= 225.638 km
t=370.0 g=9.6819 m/s^2 dr= 996.027 km y=6311.894 alt= 60.003 km
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#671
by
abaddon
on 28 Jun, 2019 13:56
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Maybe I'm seeing things... but am I the only one who counts four (!) sets of legs (one set still attached)?
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#672
by
Vettedrmr
on 28 Jun, 2019 14:02
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I think the legs are off the left booster, still attached on the right booster.
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#673
by
stcks
on 28 Jun, 2019 14:03
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Maybe I'm seeing things... but am I the only one who counts four (!) sets of legs (one set still attached)?
Definitely 3 sets.. one on the concrete at LZ-1, one in the grass/dirt between the LZ and the hangar, one on the booster near the hangar... but wheres the fourth?
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#674
by
abaddon
on 28 Jun, 2019 14:06
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Maybe I'm seeing things... but am I the only one who counts four (!) sets of legs (one set still attached)?
Definitely 3 sets.. one on the concrete at LZ-1, one in the grass/dirt between the LZ and the hangar, one on the booster near the hangar... but wheres the fourth?
I'm seeing what look a lot like legs, but they are a darker grey color, not white. Immediately to the right of the small white building in the center.
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#675
by
stcks
on 28 Jun, 2019 14:10
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Maybe I'm seeing things... but am I the only one who counts four (!) sets of legs (one set still attached)?
Definitely 3 sets.. one on the concrete at LZ-1, one in the grass/dirt between the LZ and the hangar, one on the booster near the hangar... but wheres the fourth?
I'm seeing what look a lot like legs, but they are a darker grey color, not white. Immediately to the right of the small white building in the center.
Yes, I think those are legs. I'm counting those as the ones between LZ and hangar above. Still not seeing 4 sets though
Edit: I think we're in the wrong thread too... updates vs discussion
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#676
by
pb2000
on 28 Jun, 2019 15:02
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#677
by
abaddon
on 28 Jun, 2019 15:04
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Maybe I'm seeing things... but am I the only one who counts four (!) sets of legs (one set still attached)?
Definitely 3 sets.. one on the concrete at LZ-1, one in the grass/dirt between the LZ and the hangar, one on the booster near the hangar... but wheres the fourth?
I'm seeing what look a lot like legs, but they are a darker grey color, not white. Immediately to the right of the small white building in the center.
Yes, I think those are legs. I'm counting those as the ones between LZ and hangar above. Still not seeing 4 sets though
Edit: I think we're in the wrong thread too... updates vs discussion
Continuing from Updates...
Four on the left pad, four on the booster on the bottom right, the four grey ones immediately to the right of the building in the center of the image, and then four down a little bit and to the left of those.
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#678
by
Robotbeat
on 28 Jun, 2019 15:15
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Guys, you cannot just plug in earth-surface kinematic equations and expect accuracy.
For one example, the value of "g" at 250km is a good 7% less than at the surface.
Rather than try to compensate for each of the oddities, just approach the problem using *orbital* equations?
It should be quite easy to determine the exact shape of an orbit, given known velocity , direction and altitude at a point on this orbit. Just calculate the Apogee of that orbit to know how high the stage will be at max.
The reason you might want to compensate for each of the oddities is that then you can do the computation in one line on a calculator. The pure physics solution you suggest is called Lambert's problem and has no closed form solution. All known solutions require iterations and mathematicians have been arguing about the best way to solve it for centuries.
Furthermore, the three main tweaks get very close to the right answer. They are:
(a) Gravity is weaker higher up, by about 0.5 m/s at 150 km
(b) The forward velocity generates an additional acceleration of v^2/r in the Earth frame. For this you need the inertial velocity, not the rotating frame velocity shown in the SpaceX webcast (I made this mistake above). This means the horizontal velocity is about 400 m/s faster than the SpaceX numbers. Plugging in 3200 m/s horizontal and Earth's radius gives 1.58 m/s^2.
(c) The Earth is not flat. This effect is small (the landing location is about 2 km below a plane defined by the start of coast location). Since we are guessing the 60 km altitude for start of entry burn anyway, this can be ignored.
So with these tweaks, we should use an effective g of 9.8 - 0.5 - 1.58 = 7.72 m/s. Knowing the 370 second coast, the 123 km start altitude, and the 60 km end altitude, then we solve for 123000 + v*t - 1/2*g*t = 60000, to get vertical v = 1257 m/s. This is within 1% of the correct value of 1265 m/s, as determined by the physics based computation below (in Python). The estimated peak altitude is 226.6 km as opposed to the correct 225.6 km. So the few tweaks get a result that is likely more accurate than the input data we are using.
import math
Re = 6356000 # Radius of Earth in meters
mu = 3.985744e14 # Gravitational constant of Earth
Erot = 40000000/86400.0*math.cos(28.5/180.0*math.pi) # Rotation speed of earth at 28.5 degrees north
x = 0;
y = Re+123000 # Initial altitude is 123 km
vy = 1265.0 # Initial vertical velocity (tweak this to get right result)
vx = math.sqrt(3079**2 - vy**2) # Initial velocity in Earth-rotating frame is 3079 m/s, find X component
vx += Erot # Add Earth's rotation to convert to velocity in inertial frame
print("vx = ", vx, "Earth rotation added", Erot, "m/s")
r = math.sqrt(x**2+y**2)
t = 0
dt = 0.1
while t < 370-dt/2:
a = mu/r**2 # Find the acceleration magnitude
ax = a * (x/r) # Find x and y components
ay = a * (y/r)
vx = vx - ax*dt # Update velocity = integral of acceleration
vy = vy - ay*dt
x = x + vx*dt # Update position, integral of velocity (in inertial space)
y = y + vy*dt
t += dt
r = math.sqrt(x**2+y**2)
theta = math.pi/2 - math.atan(y/x) # angle from vertical
dr = Re*theta - Erot*t # distance downrange, compensation for Earth rotation
print("t={0:5.1f} g={1:6.4f} m/s^2 dr={2:8.3f} km y={3:8.3f} alt={4:8.3f}".format(t,a, dr/1000.0, y/1000.0, (r - Re)/1000.0))
Here, I pasted it into repl.it for ya.
Anyone should be able to run and modify it:
https://repl.it/repls/BurlyCylindricalEngineers (name was auto-generated lol)
Side note:
this kind of analysis is our forum at its best. Hats off to y'all. Additionally, Euler's method is terrible but I love it for its absolute simplicity and portability...
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#679
by
FlokiViking
on 28 Jun, 2019 18:10
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Did they bring home just the one side, or both sides of the fairing? Haven't really been able to tell from the pictures...