5
$\begingroup$

Dropping the external boosters and also the entire 75-100 ton orbiter from the picture, would the tank all by itself have made orbit? This would require bolting at least 5, better 6, perhaps even 9 SSMEs to the bottom of the tank. I am assuming 7 below for parity with the Shuttle's ~half-G initial acceleration.

I am grossly simplifying here obviously: 26 tons empty tank plus 7 x 3.5 tons of engines is 50.5 tons together against 735 tons of fuel mass. Sea level exhaust velocity is 3.6 km/s, and vacuum 4.4, so I'll assume 4 (optimistically, pessimistically?) to plug into the Wolfram Alpha calculator linked from the Wikipedia page on the Tsiolkovsky rocket equation which says that the delta-v to reach orbit including gravity and air drag is 9.7 km/s.

What I come out with is that I have at least 20 tons left to play with for things like a harness around to the bottom of the tank to attach the engines to, control gear and payload. Which would apparently leave at least 10 tons of payload allowance?

What the equations I used completely ignore is initial thrust to gross launch mass which'd surely affect gravity and air drag? With 6 SSMEs at ~21 tons and ~1116 tons aggregate sea level thrust I'd have a little less than the Shuttle's initial acceleration, about 1.38, with 9 at ~32 tons it would be about twice the Shuttle's. What is the assumption behind the 9.7 km/s delta-v on the Wikipedia page as to air/gravity drag fraction and initial launch acceleration?

How far off am I with this sloppy math?

EDIT: Can someone derive the optimum number of SSMEs that yields the largest payload, and share their logic? It's gravity drag vs dry-mass fraction vs air drag: More engines means less gravity drag, but worse mass fraction and also air drag since you'd get faster early on in the lower dense atmosphere, and vice versa.

Improve this question
$\endgroup$
4
  • 1
    $\begingroup$ One thing to note is that the SSME isn't capable of being restarted, this complicates circularizing the orbit. $\endgroup$
    – Lex
    Mar 29, 2019 at 23:40
  • $\begingroup$ Could someone with sufficient reputation create the tags gravity-drag and air-drag if they see this? It seems an omission not to have those. $\endgroup$
    – Prototypist
    Mar 30, 2019 at 9:03
  • 1
    $\begingroup$ atmospheric-drag exists, I didn't see a synonym for gravity-drag though. $\endgroup$
    – Lex
    Mar 30, 2019 at 15:56
  • $\begingroup$ Ah thanks, didn't see that. Turns out I can't attach more than 5 tags anyway. $\endgroup$
    – Prototypist
    Mar 30, 2019 at 19:50

1 Answer 1

Reset to default
4
$\begingroup$

What the equations I used completely ignore is initial thrust to gross launch mass which'd surely affect gravity drag? ... What is the assumption behind the 9.7 km/s delta-v on the Wikipedia page as to gravity drag fraction and initial launch acceleration?

9.7km/s is towards the high end of delta-v to orbit requirements. It varies with both the aerodynamics (dominated by mass-to-cross-section ratio, thus favoring larger launchers) and with the initial acceleration and specific impulse (because of acceleration curve and gravity losses).

Counter-intuitively, for the same initial mass and thrust, a lower specific impulse yields lower delta-v requirement because the rocket throws mass away faster and thus accelerates more rapidly (an explanation of this phenomenon is here on yarchive.net; thanks to user Arris for finding it).

So your ∆v cost to orbit might be slightly higher for your all-hydrolox design than it would be for the shuttle with its low-specific-impulse SRBs, but 9.7km/s is still probably a conservative estimate.

Your analysis seems reasonable given your 6 SSME edit. Some of the payload mass would be eaten by thrust structure and payload adapter, but you'd be in the ballpark of 20 tons payload with a 9.6km/s ∆v target. You'd probably want to shut down some of the engines as you ascended; at 68% throttle you'd be pulling over 10g at burnout with all 6 firing.

Bear in mind this design throws away an entire shuttle fleet's worth of SSMEs (modern cost around \$50-\$60 million per SSME) every two launches.

Improve this answer
$\endgroup$
12
  • 2
    $\begingroup$ We don't do SSTO because there's no particular advantage to doing expendable SSTO. It's cheaper and more straightforward to do 2STO. If you intend to recover the single stage, you need a thermal solution and a landing solution, which eats up your payload. $\endgroup$
    – Russell Borogove
    Mar 28, 2019 at 19:49
  • 1
    $\begingroup$ The usual figure tossed around for recovery of the Falcon 9 1st stage is that it costs 15%-30% of its "performance" (very vague!) relative to a comparable expendable -- fuel reserve for braking, flyback, and landing, weight of grid fins and landing legs. And that's for a reentry from ~2km/s rather than orbital velocity. If you intend to recover an orbital SSTO you're going to pay a very large penalty. $\endgroup$
    – Russell Borogove
    Mar 28, 2019 at 20:05
  • 1
    $\begingroup$ @qqjkztd Not sure I understand what/why you're asking, but it's relatively straightforward to symmetrically add two boosters to an F9 and leave the second stage as-is; a "Falcon 18" would need a new interstage and new upper stage design and would be less capable than a Falcon Heavy. $\endgroup$
    – Russell Borogove
    Mar 28, 2019 at 22:33
  • 1
    $\begingroup$ The simplicity in part count of an SSTO is attractive, true, but hydrolox in general has a lot of disadvantages (some discussed here ), and other propellants need a significantly better mass fraction to do SSTO because of their lower specific impulse. 2STO lets you scale your hydrolox problems down to a comparatively small second stage, and use cheaper (solids or kerolox) solutions for the first stage. $\endgroup$
    – Russell Borogove
    Mar 29, 2019 at 2:00
  • 1
    $\begingroup$ @Prototypist, You might be interested in aquarius. It was an attempt to maximally reduce the complexity of a rockets design in the way you are suggesting. $\endgroup$
    – Lex
    Mar 29, 2019 at 23:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.