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#80 by docmordrid on 22 Jul, 2018 07:20
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WOW!!
TomCross @_TomCross_
Tonights Falcon 9 max-Q was super awesome! #spacex #falcon9 @Teslarati #TelStar19V
2:13 AM - Jul 22, 2018 · Florida, USA
https://twitter.com/_TomCross_/status/1020914676371271680?s=19
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#81 by DigitalMan on 22 Jul, 2018 07:23
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That was quite a long trail of green as S2 lit up. I guess I hadn't noticed that before. I didn't see a jellyfish plume on S2 this time, perhaps it needs light reflecting off it from the sun or moon.
Great launch.
You'll only see a jellyfish if the launch is just before dawn or after sunset. The 2nd stage plume needs to be illuminated by the sun to create that effect, and in this case the 2nd stage did not get into sunlight until a couple of minutes after engine cutoff.
Yea, too far away by then for me to see it with my binoculars.
Binoculars would have made no difference :-), the stage was way over the horizon and practically over Africa before it hit sunlight.
Yea, I was just pointing out that I had been watching it with my binoculars until it went out of view earlier. -
#82 by JimO on 22 Jul, 2018 08:06
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Burn-2 over Africa seems to have been against a post-sunrise sky so we won't get UFO reports this time, like we did for the Falcon-Heavy trans-Mars insertion burn.
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#83 by tp1024 on 22 Jul, 2018 08:39
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Looked very sub-synchronous. Telemetry froze for a while, but when it resumed (at 30:43) it showed 32835 km/hr = 9120 m/s. Add in the 402 m/s from Earth's rotation, to get 9522 m/s. That's about a 14000 km apogee (assuming no inclination reduction), and another 600 m/s to go to get to a nominal GTO. So GEO - 2400 m/s.
The variable we won't know until we get TLE's is the inclination. There could be a bunch of inclination reduction in that burn as well that won't show up in the final velocity.
Unless you have another look at worldmap with the trajectory on the livestream. You'll see it shifting north from crossing Lake Bangweulu to crossing above Lake Mweru - or roughly half the north-south extend of Lake Malawi. Which is a difference of about 300km at a distance of about 3000km from the equator. So it lost on the order of 6 degrees of inclination.
So ths reduces the delta-v required for plane change to 245 m/s, saves 75m/s and makes it approximately a GEO -2325m/s trajectory. [Goof corrected.] -
#84 by Ultrafamicom on 22 Jul, 2018 10:29
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Can you provide the source? I remember she had said that, but forgot whenSpaceflight Now is reporting this sat's mass is 7075kg, the heaviest GTO sat ever. Link to article: https://spaceflightnow.com/2018/07/21/record-setting-commercial-satellite-awaits-blastoff-from-cape-canaveral/
At last the mythical 7-ton sat Gwynne mentioned shows up
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#85 by RLA on 22 Jul, 2018 10:37
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Be I the only one who thinks the second stage burned 10 seconds less then what was planned for the second burn?
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#86 by tp1024 on 22 Jul, 2018 10:46
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Looked very sub-synchronous. Telemetry froze for a while, but when it resumed (at 30:43) it showed 32835 km/hr = 9120 m/s. Add in the 402 m/s from Earth's rotation, to get 9522 m/s. That's about a 14000 km apogee (assuming no inclination reduction), and another 600 m/s to go to get to a nominal GTO. So GEO - 2400 m/s.
The variable we won't know until we get TLE's is the inclination. There could be a bunch of inclination reduction in that burn as well that won't show up in the final velocity.
Unless you have another look at worldmap with the trajectory on the livestream. You'll see it shifting north from crossing Lake Bangweulu to crossing above Lake Mweru - or roughly half the north-south extend of Lake Malawi. Which is a difference of about 300km at a distance of about 3000km from the equator. So it lost on the order of 6 degrees of inclination.
So ths reduces the delta-v required for plane change to 245 m/s, saves 75m/s and makes it approximately a GEO -2325m/s trajectory. [Goof corrected.]
After doing more math (-> https://ocw.mit.edu/courses/aeronautics-and-astronautics/16-512-rocket-propulsion-fall-2005/lecture-notes/lecture_34.pdf) and noticing that a 6 degree change in inclination is a fairly stupid and about 2-3 degrees is the way to go (6 degrees correction while accelerating from LEO costs about 100m/s delta V extra compared to either a smaller change or doing it a apogee) ...
I also noticed at least some of the apparent shift in the trajectory is because the trajectory is higher above the ground after raising it (duh).
The reason why it makes sense to change inclination a little is a neat bit of trigonometry during the acceleration burn. A pure plane-change of 2 degree in LEO would normally cost 270m/s of delta V, in a burn at a right angle to the trajectory - such that your new velocity vector is no longer 7800 m/s to the right, but turned up by 2 degree. And you can't do anything to improve that.
But if you want to accelerate from 7800m/s to 9500m/s and change the plane by 2 degree at the same time, your new velocity vector is 1700m/s longer to the right and 330m/s up (because you are faster after the acceleration, the delta-v between planes is also larger).
Now you have three options.
1) Do plane-change first, accelerate later -> 270m/s + 1700m/s = 1970m/s
2) Accelerate first do plane-change later -> 1700m/s + 330m/s = 2030m/s
3) Do both at once. The real acceleration vector is (roughly) the hypotenuse of the triangle you would get with option 2. And old man Pythagoras tells us that this this is 1732 m/s long - compared to 1700m/s without any plane-change at all.
Which is a real bargain, because the same maneuver in GTO at 36000km apogee would cost 56m/s. (Which leads me to believe that the real change was a bit larger than 2 degree, because the Merlin MVac is more efficient than the satellite engines and it makes more sense to do some more work with that.) -
#87 by CorvusCorax on 22 Jul, 2018 12:18
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Also the Falcon upper stage is expendable. So its best to squeeze as much delta/v out of it as possible and have the sat, which needs to spent its life with a finite amount of storeable propellant, have to do as little work as possible.
This can be done by pushing the sat into a supersynchronous orbit (or at least make it less subsynchronous) or make some extra inclination change even though it would be more efficient at apogee (but paid with finite hypergolics instead of Kerolox that's going to reenter anyway) or both. Whichever gives the sat the best possible overall deltav budget. -
#88 by Alexphysics on 22 Jul, 2018 12:47
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This sat has a lot of fuel, that's why it is so heavy. The sat has been injected into a lower orbit and it will raise its orbit over the next few weeks. Once in GEO, station-keeping maneuvers will be done via ion thrusters so the amount of hydrazine fuel left at GEO insertion is irrelevant in this case because it uses different propellants, so if it saves hydrazine or not, that won't matter.
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#89 by theonlyspace on 22 Jul, 2018 12:49
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Were the fairings halves recovered?
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#90 by seruriermarshal on 22 Jul, 2018 12:57
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Congratulations !
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#91 by deruch on 22 Jul, 2018 13:08
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Were the fairings halves recovered?
If you mean were they caught by a ship with the net on it? No. If the fairings had the recovery hardware installed, they may get picked up from the ocean surface. But SpaceX doesn't currently have a fairing catching ship on the east coast. Their only one Mr. Steven, which operates out of Los Angeles and attempts catching fairing halves from launches out of VAFB. Next attempt will be on the upcoming Iridium 7 launch. -
#92 by edkyle99 on 22 Jul, 2018 13:33
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And now we see TLEs showing 243 x 17,863 km x 27 deg for 7,075 kg Telestar 19V. We missed the maximum velocity at engine cutoff due to the data dropout (screen freeze), which led us to our lower initial apogee estimate.Looked very sub-synchronous. Telemetry froze for a while, but when it resumed (at 30:43) it showed 32835 km/hr = 9120 m/s. Add in the 402 m/s from Earth's rotation, to get 9522 m/s. That's about a 14000 km apogee (assuming no inclination reduction), and another 600 m/s to go to get to a nominal GTO. So GEO - 2400 m/s.
That's what I'm getting too, something like a 250 x 14,000 km (or less) orbit. I think we expected subsynchronous, but I wonder about the "very" part.
- Ed Kyle
HISPASAT 30W-6 (with its smaller mass of 6,092 kg) was launched by Falcon 9 Block 4 to 184 x 22,261 km, 26.97°
They had planned to attempt OSCILY landing for the Hispasat booster, but sea states prevented an attempt.
As I see it, VA-189/TerraStar 1 still holds the record for GTO mass (6,910 kg to 250 x 35,786 km x 6 deg), since Telestar 19V went subsynchronous.
- Ed Kyle -
#93 by ZachF on 22 Jul, 2018 13:56
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In the updates thread Soltasto calculated this as GTO-2065m/s.
Doing a little math then puts the GTO-1800 capability of Block 5 with ASDS at ~6,150kg. -
#94 by ZachF on 22 Jul, 2018 14:00
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And now we see TLEs showing 243 x 17,863 km x 27 deg for 7,075 kg Telestar 19V. We missed the maximum velocity at engine cutoff due to the data dropout (screen freeze), which led us to our lower initial apogee estimate.Looked very sub-synchronous. Telemetry froze for a while, but when it resumed (at 30:43) it showed 32835 km/hr = 9120 m/s. Add in the 402 m/s from Earth's rotation, to get 9522 m/s. That's about a 14000 km apogee (assuming no inclination reduction), and another 600 m/s to go to get to a nominal GTO. So GEO - 2400 m/s.
That's what I'm getting too, something like a 250 x 14,000 km (or less) orbit. I think we expected subsynchronous, but I wonder about the "very" part.
- Ed Kyle
HISPASAT 30W-6 (with its smaller mass of 6,092 kg) was launched by Falcon 9 Block 4 to 184 x 22,261 km, 26.97°
They had planned to attempt OSCILY landing for the Hispasat booster, but sea states prevented an attempt.
As I see it, VA-189/TerraStar 1 still holds the record for GTO mass (6,910 kg to 250 x 35,786 km x 6 deg), since Telestar 19V went subsynchronous.
- Ed Kyle
It would be interesting to know the final mass of both when they reach their final orbits. If I had to guess I would probably tip the hat towards TerraStar. -
#95 by eriblo on 22 Jul, 2018 14:38
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2018-059A/43562 (243 km x 17863 km x 27.00°) has a delta v to GTO of 2064.5751 m/s or it is in GTO-2065
What parameters did you use? I get 2277 m/s using that approach (which seems to agree with this online implementation)...? BTW, doing a third of a degree of the inclination change in the first apogee raising burn saves ~4 m/s.
2018-059B/43563 (242 km x 17860 km x 27.00°) has a delta v to GTO of 2064.7341 m/s or it is in GTO-2065
according to my C/C++ program based on this. -
#96 by soltasto on 22 Jul, 2018 14:48
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2018-059A/43562 (243 km x 17863 km x 27.00°) has a delta v to GTO of 2064.5751 m/s or it is in GTO-2065
What parameters did you use? I get 2277 m/s using that approach (which seems to agree with this online implementation)...? BTW, doing a third of a degree of the inclination change in the first apogee raising burn saves ~4 m/s.
2018-059B/43563 (242 km x 17860 km x 27.00°) has a delta v to GTO of 2064.7341 m/s or it is in GTO-2065
according to my C/C++ program based on this.
I used the TLE data along with the program written by LouScheffer. I literally "translated" that program.
This is the program, it should compile with GCC using C++11:
#include <iostream>
#include <iomanip>
#include <cmath>
using namespace std;
int main() {
int stay;
do {
stay=0;
double perigee, apogee, inclination;
cout << "Enter perigee in km, apogee in km, inclination in degrees." << endl;
cin >> perigee >> apogee >> inclination;
cout << setprecision(4) << fixed << "Perigee: " << perigee << " km" << endl << "Apogee: " << apogee << " km" << endl;
double sma;
sma = ((perigee + apogee) / 2) + 6371; //semi-minor axis, 6371 is Earth radius in km
cout << setprecision(4) << fixed << "Semi-major axis: " << sma << " km" << endl;
double const MU = 3.986005e14;
double vp, va;
vp = sqrt((MU * (apogee + 6371) * 1000) / ((perigee + 6371) * sma * 1e6));
va = sqrt((MU * (perigee + 6371) * 1000) / ((apogee + 6371) * sma * 1e6));
cout << setprecision(4) << fixed << "Speed at perigee: " << vp << " m/s" << endl << "Speed at apogee: " << va << " m/s" << endl;
double rpd = 8681663.653 / pow(sma, 3 / 2); //revolutions per day
cout << setprecision(4) << fixed << "Revolutions per day: " << rpd << endl << "Days per revolution: " << 1 / rpd << endl;
double sync = 35786;
sma = (sync + apogee) / 2 + 6371;
perigee = sync;
double nvp, nva;
nvp = sqrt((MU * (apogee + 6371) * 1000) / ((perigee + 6371) * sma * 1e6));
nva = sqrt((MU * (perigee + 6371) * 1000) / ((apogee + 6371) * sma * 1e6));
cout << setprecision(4) << fixed << "Speed at perigee: " << nvp << " m/s" << endl << "Speed at apogee: " << nva << " m/s" << endl;
double cross, along, need_along, dv_top, dv_bot, total;
double const GEO_V = 3075;
cross = va * sin((inclination / 180) * M_PI);
along = va * cos((inclination / 180) * M_PI);
cout << setprecision(4) << fixed << "Cross v at apogee: " << cross << " m/s" << endl << "Along track: " << along << " m/s" << endl;
need_along = nva - along;
dv_top = sqrt(pow(cross, 2) + pow(need_along, 2));
cout << setprecision(4) << fixed << "Delta-v at top: " << dv_top << endl;
dv_bot = nvp - GEO_V;
cout << setprecision(4) << fixed << "Delta-v at bot: " << dv_bot << endl;
total = dv_top + dv_bot;
cout << setprecision(4) << fixed << "Delta-v (total): " << total << endl;
cout << "Enter 1 to restart or any key to close." << endl;
cin >> stay;
}while(stay==1);
return 0;
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#97 by gongora on 22 Jul, 2018 15:07
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This is the interview where Gwynne mentioned having 7-ton satellites on the manifest:
http://interactive.satellitetoday.com/via/april-2017/shotwell-ambitious-targets-achievable-this-year/ -
#98 by LouScheffer on 22 Jul, 2018 15:30
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2018-059A/43562 (243 km x 17863 km x 27.00°) has a delta v to GTO of 2064.5751 m/s or it is in GTO-2065
What parameters did you use? I get 2277 m/s using that approach (which seems to agree with this online implementation)...? BTW, doing a third of a degree of the inclination change in the first apogee raising burn saves ~4 m/s.
2018-059B/43563 (242 km x 17860 km x 27.00°) has a delta v to GTO of 2064.7341 m/s or it is in GTO-2065
according to my C/C++ program based on this.
I used the TLE data along with the program written by LouScheffer. I literally "translated" that program.
This is the program, it should compile with GCC using C++11:
...
Unfortunately, that program is only correct for synchronous or greater. It does not account for the perigee burn to raise the apogee, gets the wrong sign for circularization if the apogee is below GEO, and will do the inclination reduction at the wrong burn if sub-sync. So the numbers from this program will be wrong for the sub-sync case, which this is. Sorry, it should check for that and either do it right, or at least give a message, rather than silently doing it wrong. -
#99 by Avron on 22 Jul, 2018 15:42
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No word yet on T19V health - should have had something by now ? I may have missed it - I just don't see any news on health
