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Assume we have a society that has sophisticated enough engineering to build a SSTO chassis that can survive the rigors of repeated launch (MaxQ) and re-entry (thermal stress). What thrust to weight ratio would they have to achieve from their propulsion systems in order to get the craft into orbit around an Earth-clone planet without the need for external boosters?

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  • $\begingroup$ https://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation DeltaV = Ve * ln(Mi/Mf) where DeltaV is the change in velocity -- about 7 km/s for earth. Ve is the effective exhaust velocity ln is the natural logarithm Mi is the initial mass with fuel Mf is the final dry weight. So DeltaV/Ve = ln (Mi/Mf) e^(DeltaV/Ve) = Mi/Mf $\endgroup$ – Sherwood Botsford Mar 4 '17 at 22:11
  • $\begingroup$ Are you aware of this? McDonnell Douglas DC-X - "en.wikipedia.org/wiki/McDonnell_Douglas_DC-X" - Screw it, if you follow this you find NASA Kills it. Shortly there after SpaceX and Blue Origin have some new employee's and exciting Reusable Rockets. $\endgroup$ – Enigma Maitreya Mar 4 '17 at 22:20
  • $\begingroup$ I've migrated this from Worldbuilding to Space Exploration as per the asker's request. I think it's okay for here, but if it's not, we'll take it back. $\endgroup$ – HDE 226868 Mar 4 '17 at 22:31
  • $\begingroup$ This is an interesting question with no easy answer. The SSTO needs TWR lower than 1 for flight on wings, but to reach any considerable fraction of orbital speed it must leave the atmosphere, as Ram Rise heating would obliterate it. For that it needs TWR approaching 1 - but not quite, because a part of its gravity drag is gone. Regardless, unless you have some magical propulsion of impossible delta-V, you'll want to make it in a good time to orbit to reduce gravity losses, so TWR>1 will be definitely welcome... $\endgroup$ – SF. Mar 4 '17 at 23:06
  • $\begingroup$ Do you talk some specific type of SSTO rocket style, jet style or in a general one? $\endgroup$ – MolbOrg Mar 4 '17 at 23:28
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What thrust to weight ratio would they have to achieve from their propulsion systems in order to get the craft into orbit

This wording indicates that you think there is a threshold. All designs with a TWR lower than this will fail to reach orbit, and designs with a higher TWR will succeed. This premise is incorrect. As SF says in his comment, a winged SSTO can get away with a TWR lower than 1. So depending on body design (large wings, small wings, lifting body or cylindrical rocket) you get a large range of possible TWR figures.

Other numbers are more critical than TWR:

  • You need sufficient delta-V to reach orbit.
  • You need a mass fraction that's good enough to bring some payload to orbit.

Those two combined (via the Rocket Equation) give a minimum specific impulse required to get that payload to orbit. TWR is a consideration via gravity losses: the longer you take to get to orbital speed, the more gravity losses you'll have and the more delta-V you'll need to account for.

So in order to be able to calculate anything, we need 2 parameters:

  • required payload
  • available specific impulse

Those two are enough to calculate the starting mass of the SSTO. Economics then decides if such an SSTO is worth building.

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  • $\begingroup$ I would respectfully point out your answer is Off Topic because it does not answer the question the OP asked. While I agree this will always be about delta-v that was not what was asked for. $\endgroup$ – Enigma Maitreya Mar 5 '17 at 17:03
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    $\begingroup$ Pointing out that a question has an incorrect premise is on topic. $\endgroup$ – Hobbes Mar 5 '17 at 18:13
  • $\begingroup$ With all due respect it would seem that is what messages are for else why this "This does not provide an answer to the question" as a reason for deleting an answer? $\endgroup$ – Enigma Maitreya Mar 5 '17 at 18:20
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    $\begingroup$ "The premise is incorrect but here is an answer to a closely related question that the correction of premise might reasonably lead to" is helpful enough to be worth making an exception. $\endgroup$ – Russell Borogove Mar 5 '17 at 18:29
  • $\begingroup$ @RussellBorogove - Thanks for that reply and regardless of how people may take things ... I prefer "helpful enough ..." but as I have read through a lot of questions and answers I get the impression that the value judgment is ... extremely broad indicating a possible preferential implementation. Regardless I accept your answer and thank you for confirming the Rules are not Literally the rules, rather guidelines to be used by people to do what they (subjectively) think is right. $\endgroup$ – Enigma Maitreya Mar 5 '17 at 19:01
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There is a chart and the explanation at wikipedia

Being limited by the performance of the existing chemical fuels and being limited by the sane take-off mass of less than 3K metric tonns (the weight of a Saturn V, the biggest rocket ever) then dry weight (everything but the fuel itself - the body, the payload itself, engines, re-entry fuel, safety margin fuel, landing gears, wings etc - everything) must fit into less than ~10% of the total mass, close to 5%, which is very challenging task to achieve.

Usage of better fuels like hydrogen improves the thrust to weight ratio but affects the dry weight because of the decreased density and insulating material for the cryogenic fuel tanks, so overall it does not make a big difference.

In order to SSTO to be technically possible we need the next generation of fuels / engines. Something like metallic hydrogen, thermonuclear fusion engine, photon engine etc.

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  • $\begingroup$ "In order to SSTO to be technically possible we need the next generation of fuels / engines. " only if you insist on having a payload. $\endgroup$ – Organic Marble Jun 13 at 17:11
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    $\begingroup$ @OrganicMarble Every problem we're having is because of this damn payload. No payload = no need for a rocket = no problem. $\endgroup$ – ilyakharlamov Jun 13 at 21:13
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    $\begingroup$ I'm with you! No more payloads. Just rockets. $\endgroup$ – Organic Marble Jun 13 at 21:15
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The question asked I am answering

What thrust to weight ratio would they have to achieve from their propulsion systems in order to get the craft into orbit around an Earth-clone planet without the need for external boosters?

I had wanted to use the Delta Clipper X but had a Eureka Moment that it was a scale testing unit and switched to this instead. To the best of my knowledge, there has not been a "Public" SSTO made or launched. Some may consider Lockheed a credible source.

https://en.wikipedia.org/wiki/VentureStar - VentureStar

While never built the work presented here seems to be consistent with a Full Scale Ship (SSTO)

Function    Manned Re-usable Spaceplane
Manufacturer    Lockheed Martin
Country of origin   United States
Size
Height  38.7 m[1] (127 ft)
Diameter    N/A
Mass    1,000,000 kg[1] (2,200,000 lb)
Stages  1
Capacity
Payload to LEO  20,412 kg[1] (45,000 lb)
Launch history
Status  Cancelled
Launch sites    Unknown
Total launches  0
First stage - VentureStar
Engines 7 RS2200 Linear Aerospikes[1]
Thrust  3,010,000 lb[1] (13.39 MN)
Fuel    LOX/LH2[1]

It gives both weight and thrust.

The following is based on the MASS because I have no clue if the Payload is included or not.

0.1643017177 lbs per Newton would be your answer in Newtons.

Or for my bud SF that likes to groan a lot :) 164.3017177 lbs per kN

Or 164,301.717699776 lbs per MN

Treat lbs of mass as lbf of weight, divide thrust in lbf by that and you have the clean dimensionless TWR

3,010,000 / 2,200,000 gives 1.36818118118182 - Just for you

For those that want a good and interesting read on Real Life SSTO's go here

https://en.wikipedia.org/wiki/McDonnell_Douglas_DC-X

The Delta Clipper, VentureStar, SpaceX, Blue Origin Are all linked together. The Delta Clipper was taken over by NASA killed in favor of their choice the VenturStar because of the landing being horizontal rather than vertical then canceled it for the Space Shuttle and guess were everyone went to work. Let us .... say the obvious, The Shuttle, Delta Clipper XA, VentureStar all ran separately and parallel to each other.

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    $\begingroup$ "lbs per kN" groan $\endgroup$ – SF. Mar 4 '17 at 22:59
  • $\begingroup$ The groaning presumably comes from mixing imperial (lbs) and SI (N) units, not imperial and kilo- or mega- units. $\endgroup$ – Nathan Tuggy Mar 4 '17 at 23:30
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    $\begingroup$ I like SF his taunts at me are cool and make me laugh (with both respect and all other good ways). Hopefully no one takes my response to him wrong. As in I would buy him a Beer any day of the week and have a good time. $\endgroup$ – Enigma Maitreya Mar 4 '17 at 23:32
  • $\begingroup$ @EnigmaMaitreya: Exactly. You've got both lbs and lbf, and kg and kN. What possessed you to mix the units? Treat lbs of mass as lbf of weight, divide thrust in lbf by that and you have the clean dimensionless TWR. $\endgroup$ – SF. Mar 5 '17 at 0:38
  • $\begingroup$ @SF - I would have thought that answer was obvious, being American and all and lbf vs kN = hum most scientist prefer kN in this scenario and lbs .... what can I say? Oh wait :) I am working a compromise here :) I note your not saying the mixing is wrong as in giving an error ... are you? To be clear as it seems people like to obfuscate things, I have dealt with the UK + Europe for longer than ... some posters and editors here have been alive. I knew exactly what you meant as it would not have been the first discussion of that nature .... not by a long shot. :) So are we good? $\endgroup$ – Enigma Maitreya Mar 5 '17 at 1:41

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