Thank you both! I'm pretty sure now that I was not using the online calculator I had found correctly. strangequark, that page is exactly what I really needed, thank you! A 1.6% difference in density makes a lot more sense and that must be well within the tolerances of the NK-33 engine. For future reference, anyone discussing the merits of LOX densification, etc. LOX at -184 C = 1141.7 kg/m3 (just below boiling point)LOX at -192 C = 1185.4 kg/m3LOX at -196 C = 1204.3 kg/m3LOX at -210 C = 1268.4 kg/m3LOX at -217 C = 1299.0 kg/m3 (just above freezing point)

Quote from: whitelancer64 on 12/08/2015 03:09 PMThank you both! I'm pretty sure now that I was not using the online calculator I had found correctly. strangequark, that page is exactly what I really needed, thank you! A 1.6% difference in density makes a lot more sense and that must be well within the tolerances of the NK-33 engine. For future reference, anyone discussing the merits of LOX densification, etc. LOX at -184 C = 1141.7 kg/m3 (just below boiling point)LOX at -192 C = 1185.4 kg/m3LOX at -196 C = 1204.3 kg/m3LOX at -210 C = 1268.4 kg/m3LOX at -217 C = 1299.0 kg/m3 (just above freezing point)Boiling & freezing points at what pressure though?

12/03/13 Falcon 9 v1.1 F9-7 SES 8 3.183 CC 40 295x80000x20.8 GTO+01/06/14 Falcon 9 v1.1 F9-8 Thaicom 6 3.016 CC 40 295x90000x22.5 GTO+03/02/15 Falcon 9 v1.1 F9-16 Eutelsat 115WB/ABS 3A 4.159 CC 40 400x63300x24.8 GTO+

Here is a question. Falcon9 has done three GTO missions, where the end orbit was super-synchronous. 12/03/13 Falcon 9 v1.1 F9-7 SES 8 3.183 CC 40 295x80000x20.8 GTO+01/06/14 Falcon 9 v1.1 F9-8 Thaicom 6 3.016 CC 40 295x90000x22.5 GTO+03/02/15 Falcon 9 v1.1 F9-16 Eutelsat 115WB/ABS 3A 4.159 CC 40 400x63300x24.8 GTO+If I understand correctly, the reason for that is that it makes circularizing to GEO for the payload easier. I want to know how I can convert the above stats to a - X m/s equivalent. In other words, I want to find out how much dv the payloads had to use to get to their end orbits. How can I do that?

12/03/13 Falcon 9 v1.1 F9-7 SES 8 3.183 CC 40 295x80000x20.8 GTO+ -> 1,511m/s01/06/14 Falcon 9 v1.1 F9-8 Thaicom 6 3.016 CC 40 295x90000x22.5 GTO+ -> 1,504m/s03/02/15 Falcon 9 v1.1 F9-16 Eutelsat 115WB/ABS 3A 4.159 CC 40 400x63300x24.8 GTO+ -> 1,594m/s

Quote from: Dante80 on 01/04/2016 03:51 PMHere is a question. Falcon9 has done three GTO missions, where the end orbit was super-synchronous. 12/03/13 Falcon 9 v1.1 F9-7 SES 8 3.183 CC 40 295x80000x20.8 GTO+01/06/14 Falcon 9 v1.1 F9-8 Thaicom 6 3.016 CC 40 295x90000x22.5 GTO+03/02/15 Falcon 9 v1.1 F9-16 Eutelsat 115WB/ABS 3A 4.159 CC 40 400x63300x24.8 GTO+If I understand correctly, the reason for that is that it makes circularizing to GEO for the payload easier. I want to know how I can convert the above stats to a - X m/s equivalent. In other words, I want to find out how much dv the payloads had to use to get to their end orbits. How can I do that?If you assume two burns from a super synchronous GTO, you can use my spreadsheet (it is done in LibreOffice). But I guess you could use it for sub synchronous, too. Normal GTO usually have a single burn and thus I'm not sure it will work. It should since it takes the delta-v requirements of two burns which should be the same. But I'm not sure.BTW:12/03/13 Falcon 9 v1.1 F9-7 SES 8 3.183 CC 40 295x80000x20.8 GTO+ -> 1,511m/s01/06/14 Falcon 9 v1.1 F9-8 Thaicom 6 3.016 CC 40 295x90000x22.5 GTO+ -> 1,504m/s03/02/15 Falcon 9 v1.1 F9-16 Eutelsat 115WB/ABS 3A 4.159 CC 40 400x63300x24.8 GTO+ -> 1,594m/s

1. Can you drop the class2. The second part seems to be describing the pertubation from the ideal orbit with the given parameters. So it seems like you need to set the rocket equation for deltaV equal to the amount of deltaV caused by the pertubation. I may help you to know that in the rocket equation, exhaust velocity can be replaced by ISP x Gravity ( 4500 x 9.807 m/s^2) ln(mo/m1) and then solve for m1

Astronautics: The Physics of Space Flight - Ulrich WalterRocket and Spacecraft Propulsion - Martin J. L. Turner

I have a really boring question. Obviously liquid propellant rockets have valves. Valves for the pre-burner and for the main combustion chamber and a load of others.I don't think I've ever seen a picture of a main oxidizer or fuel propellant - or perhaps I didn't recognize it.I mean, take the F-1 - that's got to have some seriously meaty propellant valves. With bloody enormous actuators.Or perhaps more interestingly, take the original Atlas. Those booster engines are gonna get dropped off, so you have a separation plane too.So what kind of valves are they - obviously they need to be lightweight, but have to take a very high flow rate, etc..Just my weird curiosity.. If this is covered elsewhere, apologies, I haven't noticed anything appropriate..

how big would your pressure vessel need to be [...] before you have to unload to external structure to stay mass-efficient?

I have a really boring question. (valves)

With bloody enormous actuators.

Does it matter if that structure is inside or outside the pressure vessel wall?

A completely unrelated question of my own: does the "residence time" requirement (in the sense of the characteristic length) for a combustion chamber change dramatically when the propellants are being injected as gases, e.g. in a full-flow staged combustion design?