One of the quickest way to get to mars with curent technology would involve something like three shuttle external tanks cross feed to a central engine. That's about elaborate as you'd want to get with chemical engines, and because of fun-happy parallel staging you get an effective mass ratio of 25, and a final velocity of 14.2kmps.Based on option D in this document, I believe that would translate to a minimum trip time of roughly 70 days:www.smad.com/analysis/rirtme.pdfYou could probably aerobreak away the excess speed...
4146 / 39^1.242 = ~44 km/secCan this be done chemically with a single stage?
Quote from: Warren Platts on 07/24/2011 03:47 pmSo if we wanted to go to Mars 1-way in 39 days then: 4146 / 39^1.242 = ~44 km/secCan this be done chemically with a single stage? I think so: Oberth effect slingshots around Earth from L2 give a delta v bonus factor ranging from 2 to 7 IIRC. DeltaV = Exhaust Velocity * Design Multiplier.(An engineer's view of the rocket equation)In this case we want a DM of 11 for LOH/LH.For a single stage this translates to a mass ratio of ~22,000.That would be a 'no' then.
So if we wanted to go to Mars 1-way in 39 days then: 4146 / 39^1.242 = ~44 km/secCan this be done chemically with a single stage? I think so: Oberth effect slingshots around Earth from L2 give a delta v bonus factor ranging from 2 to 7 IIRC.
Then there is the business of stopping once you arrive at Mars...
Quote from: douglas100 on 07/27/2011 01:21 pmThen there is the business of stopping once you arrive at Mars...Stopping is not a problem. Mars is a large, massive object -- your spaceship isn't going to damage it very much (if all else fails, lithobraking *will* do the job). Surviving the stop is another matter entirely ...
Two burns:Earth Burn = 2 km/sec((11.1 + 2)2 - 11.12)1/2 = 7.0 km/sec7 / 2 = Oberth Bonus Factor of 3.5Mars Burn = 9 km/sec((5.03 + 9)2 - 5.032)1/2 = 13.1 km/sec13.1 / 9 = ~1.5 OBFtotal delta v = ~20 km/sec, average OBF = 1.8
So staging from L2, one can get 20 km/sec out of RL-10 motors with a single stage and yet have a decent mass fraction--and still get to Mars fully propulsively.
Quote from: Warren Platts on 07/27/2011 03:37 pmTwo burns:Earth Burn = 2 km/sec((11.1 + 2)2 - 11.12)1/2 = 7.0 km/sec7 / 2 = Oberth Bonus Factor of 3.5Mars Burn = 9 km/sec((5.03 + 9)2 - 5.032)1/2 = 13.1 km/sec13.1 / 9 = ~1.5 OBFtotal delta v = ~20 km/sec, average OBF = 1.8You really need to stop saying things this way. In no sense is delta-v equal to ~20km/sec. Delta-v is 2km/s + 9km/s, or 11km/s. The fact that v_inf at Earth and v_inf at Mars total ~20km/s is an interesting tidbit, but it isn't the same thing as delta-v.
Why is your confused language a problem? Because it results in confused thoughts:QuoteSo staging from L2, one can get 20 km/sec out of RL-10 motors with a single stage and yet have a decent mass fraction--and still get to Mars fully propulsively.You don't get to Mars fully propulsively. You get to a parabolic escape trajectory from Mars. You can certainly use aerocapture/aerobraking at periapsis to get from here to an orbit around Mars, but it's no longer "fully propulsively".Assuming you don't want to do aerocapture, you can spend slightly more delta-v to insert into a highly elliptical atmosphere grazing orbit, and then use aerobraking to circularize the orbit.
Personally, I'd leave the main spacecraft in the parabolic escape trajectory. This parks the main interplanetary vehicle in a nearby solar orbit--co-orbital with Mars. This co-orbit can be arranged to encounter Mars again just before the return leg.
The landing module can detach from the main spacecraft just after the periapsis burn and use its own rocket to propulsively brake into a highly elliptical orbit. At apoapsis, it does a slight burn to lower periapsis to the upper atmosphere. Thus, the landing module can use aerobraking to circularize its orbit and then perform atmospheric entry.
There are numerous variations on this theme. You could save mass by detaching the landing module before the Mars flyby so it can aerocapture or even directly land. (This requires a heavier heat shield.)But none of these variations are "fully propulsive" unless you include another 5+ km/s of delta-v to get from Mars escape velocity to Mars.
I see what you're saying. Yes, there's the "nominal" delta v that a space craft has if it would burn all its propellant in a gravity free vacuum. Then there's the Oberth bonus one can obtain with chemical rockets that low thrust rockets cannot. They are two different things. Yet it's also the case that velocity is velocity. One could start in LEO and do a burn 7 km/sec burn and have a v_inf of 7 km/sec, or one could start at L2, do a 2 km/sec burn and still have a 7 km/sec v_inf. You still have a total change in velocity of 7 km/sec. That's delta v.
Quote from: IsaacAssuming you don't want to do aerocapture, you can spend slightly more delta-v to insert into a highly elliptical atmosphere grazing orbit, and then use aerobraking to circularize the orbit.No no! Absolutely not! Fully propulsive is what we want--and it's all you're going to get if you visit a place like Ceres or Mercury. The whole point of the "abundant chemical" paradigm is to avoid complications like aerocapture by simply throwing tons of propellant at the problem.
Assuming you don't want to do aerocapture, you can spend slightly more delta-v to insert into a highly elliptical atmosphere grazing orbit, and then use aerobraking to circularize the orbit.
When I said "get to Mars" above, I meant LMO.
The figures I used came from Nordley (2006). Well, in truth I had to extrapolate a little from his Table 1: it only goes to 6.5 km/sec departure from Earth (delta t = 76 days; delta v to LMO = 12.163 km/sec; compared to my 7 km/s Earth departure ?v, 72 days, 13.4 ?v to LMO--OK I rounded down, but I could have gone 6.9 and 13.1. The delta t is the same within rounding errors.)
If you can get to LMO, one would only need to carry ~4 km/sec worth or propellant for a fully propulsive landing.
Quote from: Warren Platts on 07/27/2011 05:12 pmI see what you're saying. Yes, there's the "nominal" delta v that a space craft has if it would burn all its propellant in a gravity free vacuum. Then there's the Oberth bonus one can obtain with chemical rockets that low thrust rockets cannot. They are two different things. Yet it's also the case that velocity is velocity. One could start in LEO and do a burn 7 km/sec burn and have a v_inf of 7 km/sec, or one could start at L2, do a 2 km/sec burn and still have a 7 km/sec v_inf. You still have a total change in velocity of 7 km/sec. That's delta v.If you start in LEO and do a 7km/s burn, you have a v_inf of 9.9km/s. This depends on your starting LEO orbit; I'm assuming a circular orbit with a 7.8km/s orbital speed. v_inf = sqrt( (7.8+7)^2 - 2*7.8^2) = 9.87km/s. If your initial LEO orbit is at an altitude where the orbital speed is 7km/s (an altitude of about 1500km), then you do get a v_inf of 7km/s after a 7km/s burn.
Quote from: Warren Platts on 07/27/2011 05:12 pmI see what you're saying. Yes, there's the "nominal" delta v that a space craft has if it would burn all its propellant in a gravity free vacuum. Then there's the Oberth bonus one can obtain with chemical rockets that low thrust rockets cannot. They are two different things. Yet it's also the case that velocity is velocity. One could start in LEO and do a burn 7 km/sec burn and have a v_inf of 7 km/sec, or one could start at L2, do a 2 km/sec burn and still have a 7 km/sec v_inf. You still have a total change in velocity of 7 km/sec. That's delta v.If you start in LEO and do a 7km/s burn, you have a v_inf of 9.9km/s. This depends on your starting LEO orbit; I'm assuming a circular orbit with a 7.8km/s orbital speed. v_inf = sqrt( (7.8+7)^2 - 2*7.8^2) = 9.87km/s. If your initial LEO orbit is at an altitude where the orbital speed is 7km/s (an altitude of about 1500km), then you do get a v_inf of 7km/s after a 7km/s burn.Anyway, the total change in velocity isn't 7km/s. In neither case did the vehicle start off with a zero velocity relative to Earth. The resulting v_inf is 7km/s relative to Earth, but the starting velocity wasn't 0km/s relative to Earth.Because the starting velocity was actually a changing variable as the spacecraft revolved around Earth, it's just not a useful way to determine delta-v.Anyway, you really should stop confusing v_inf and delta-v this way. It's just confusing to others and yourself. For example, you have confused v_inf and delta-v from a 400km LEO orbit. They're simply not the same thing.
QuoteQuote from: IsaacAssuming you don't want to do aerocapture, you can spend slightly more delta-v to insert into a highly elliptical atmosphere grazing orbit, and then use aerobraking to circularize the orbit.No no! Absolutely not! Fully propulsive is what we want--and it's all you're going to get if you visit a place like Ceres or Mercury. The whole point of the "abundant chemical" paradigm is to avoid complications like aerocapture by simply throwing tons of propellant at the problem.I can understand avoiding aerocapture in some cases because it requires a heat shield. But aerobraking doesn't require a heat shield. I see no logic in avoiding aerobraking when an atmosphere is available. We already use it at Earth and at Mars. It's proven and it's cheap. Why avoid it?
QuoteThe figures I used came from Nordley (2006). Well, in truth I had to extrapolate a little from his Table 1: it only goes to 6.5 km/sec departure from Earth (delta t = 76 days; delta v to LMO = 12.163 km/sec; compared to my 7 km/s Earth departure ?v, 72 days, 13.4 ?v to LMO--OK I rounded down, but I could have gone 6.9 and 13.1. The delta t is the same within rounding errors.)You should calculate this from basic orbital mechanics equations rather than try and extrapolate. Particularly since you aren't entirely clear on what counts as "delta-v" and you seem to be making mistakes interpreting the table.For example, you say you want a v_inf of 7km/s. But you claim that it's necessary to extrapolate outside of the table. But it is in the table! A v_inf of 7.02km/s is the 6th entry from the bottom. This corresponds to a 400km LEO delta-v of 5.250km/s and a trip time of 92.9 days.On the other hand, you want a 72 day transit, which is outside of the table. The table tops off at a v_inf of 9.12km/s and a trip time of 76.0 days. To get a trip time of 72 days, you would need a v_inf greater than 9.12km/s.Since you're making mistakes here, I would recommend you start again and this time don't try any fancy extrapolation. Just stick with the data on the chart--a 76 day transit and a v_inf of 9.12--and see where that gets you.Assuming we start with Earth escape velocity, we can do an Earth grazing perigee burn of 3.29km/s to get a v_inf of 9.12 (sqrt(11^2+9.12^2)-11). To perform a fully propulsive Mars grazing burn to LMO would be another 12.163km/s, but it's only 10.600km/s if we accept aerobraking.So, the delta-v requirements from Earth escape to fully propulsive LMO insertion are 3.29+12.163 = 15.45km/s.The delta-v requirements from Earth escape to a propulsive insertion to an elliptical Mars orbit are only 3.29+10.600 = 13.89km/s.This doesn't sound like a big difference, but if we assume the a specific impulse of 450s it translates to a whopping 42% increase in mass ratio. The mass ratio for 15.45km/s is 33:1; the mass ratio for 13.89km/s is 23:1. For each ton of dry mass, this is a difference of ten tons.And for what? Aerobraking costs almost nothing.
QuoteIf you can get to LMO, one would only need to carry ~4 km/sec worth or propellant for a fully propulsive landing.4km/s on top of 15.45km/s would be 19.45km/s. With a specific impulse of 450s, this translates to a mass ratio of 82:1. If you use aerobraking, the mass ratio is reduced to 58:1--but this is still ridiculously high.In contrast, a suitable heat shield might have a mass on the order of 15% of the lander mass. This reduces the mass ratio down to 27:1. Which in my opinion is still needlessly too high. Since we already have a heat shield, we can use that heat shield for a modest amount of aerocapture. If we shave off just 3km/s, the mass ratio is reduced from 27:1 down to 14:1.I'd say the advantages of using Mars's atmosphere are quite compelling. These reduced mass ratios don't merely translate into reduced propellant cost. They reduce the number of stages needed, they reduce thruster mass and cost, they reduce gravity losses (increased wet mass translates to decreased acceleration which translates to longer burn times which leads to lower Oberth effect benefits as more of the burn takes place further away from periapsis).Simply throwing more propellant mass at the problem just doesn't get you very far due to the exponential nature of the rocket equation.
Maybe it would help if I knew what the "inf" stood for. Perhaps you could enlighten us on that one.
I'll consider your other questions with the attention to detail they deserve later when I have more time. In the meantime, real quick:v_inf means "velocity at infinity". It refers to the orbital speed that an escape trajectory would have at an infinite distance away. It doesn't apply to elliptical orbits because they can't reach an infinite distance.
In reality, every planet and star has some Hill sphere beyond which their gravitational pull is no longer dominant. The velocity at this distance is close enough to v_inf to be good enough for BOE calculations. If you're actually flying a mission, of course, it's worth conducting more accurate (and far more complex) numerical simulations.
Quote from: IsaacKuo on 07/28/2011 04:31 amv_inf means "velocity at infinity". It refers to the orbital speed that an escape trajectory would have at an infinite distance away. It doesn't apply to elliptical orbits because they can't reach an infinite distance.I don't understand the difference in this regard of elliptical orbit and circular orbit- neither will reach velocity at infinity without adding velocity. Are you saying elliptical orbits are excluded whereas circular are included?
v_inf means "velocity at infinity". It refers to the orbital speed that an escape trajectory would have at an infinite distance away. It doesn't apply to elliptical orbits because they can't reach an infinite distance.
For purposes of making it simpler, let's remove the Earth's gravity well, and assume a spacecraft is at earth orbital distance following same orbital path [but tens of millions of kilometer from Earth, or say least 60 million kilometer from Earth].
Say, this spacecraft has a total delta-v of 16 km/sec and it's burn is in a direction of 90 degree away from it's orbital path [with it's back to Sun]. Wouldn't the trajectory result from addition it's orbital velocity [a vector 29.8 km/sec at 180 degrees] to the vector from rocket [16 km/sec at 90 degrees]. With the resultant vector of 33.8 km/sec at 151.7 degrees?
And if we do the same thing but this time being in Earth gravity well and starting from Earth/Sun L-1, then passing close to Earth [200 km] and burning the 16 km/sec when near earth, one could expect to get a significantly faster trajectory to Mars? A steeper angle and/or faster velocity?
Would I be correct to assume that in first example where spacecraft is not anywhere near earth, that the trajectory would cross Mars orbital distance in a time period of somewhere around 55 days?
I'll consider your other questions with the attention to detail they deserve later when I have more time. In the meantime, real quick:v_inf means "velocity at infinity". It refers to the orbital speed that an escape trajectory would have at an infinite distance away. It doesn't apply to elliptical orbits because they can't reach an infinite distance.In reality, every planet and star has some Hill sphere beyond which their gravitational pull is no longer dominant. The velocity at this distance is close enough to v_inf to be good enough for BOE calculations. If you're actually flying a mission, of course, it's worth conducting more accurate (and far more complex) numerical simulations.
Quote from: IsaacKuo on 07/27/2011 09:06 pmI can understand avoiding aerocapture in some cases because it requires a heat shield. But aerobraking doesn't require a heat shield. I see no logic in avoiding aerobraking when an atmosphere is available. We already use it at Earth and at Mars. It's proven and it's cheap. Why avoid it?Well that sounds all right, but IIRC, some of these Mars probes spent several months trying to circularize their orbits. The title of the thread is "Quickest Journey"--not most fuel economical, cheapest journey where time is no object.
I can understand avoiding aerocapture in some cases because it requires a heat shield. But aerobraking doesn't require a heat shield. I see no logic in avoiding aerobraking when an atmosphere is available. We already use it at Earth and at Mars. It's proven and it's cheap. Why avoid it?
But why don't you help us out? The thread is about how fast it would take to "get to Mars" using [abundant] chemical. The ULA MTV is a realistic baseline to work with. Zegler and Kutter say it's got a nominal delta v of 11 km/sec. You're the mission planner. You have permission to burn it all on a 1-way transit, because it will get refueled on the back end. How fast can you get us there? You can stage from L2 with as much Lunar propellant as you can fit into 6 ACES-121 modules.
QuoteQuoteIf you can get to LMO, one would only need to carry ~4 km/sec worth or propellant for a fully propulsive landing.4km/s on top of 15.45km/s would be 19.45km/s. With a specific impulse of 450s, this translates to a mass ratio of 82:1. If you use aerobraking, the mass ratio is reduced to 58:1--but this is still ridiculously high.Straw man. The two functions will be separated. Rendezvous with a lander in LMO. The Lander will be topped off in orbit at LMO depot. The propellant will either come from the Moon, Phobos, or Mars itself eventually. A manned MTV definitely wouldn't be hauling prop for landers....
QuoteIf you can get to LMO, one would only need to carry ~4 km/sec worth or propellant for a fully propulsive landing.4km/s on top of 15.45km/s would be 19.45km/s. With a specific impulse of 450s, this translates to a mass ratio of 82:1. If you use aerobraking, the mass ratio is reduced to 58:1--but this is still ridiculously high.
Quote from: gbaikie on 07/28/2011 10:36 amQuote from: IsaacKuo on 07/28/2011 04:31 amv_inf means "velocity at infinity". It refers to the orbital speed that an escape trajectory would have at an infinite distance away. It doesn't apply to elliptical orbits because they can't reach an infinite distance.I don't understand the difference in this regard of elliptical orbit and circular orbit- neither will reach velocity at infinity without adding velocity. Are you saying elliptical orbits are excluded whereas circular are included?Huh? Of course not. Obviously circular orbits don't reach an infinite distance. I don't see how you could possibly interpret my words as saying that circular orbits have a v_inf.QuoteFor purposes of making it simpler, let's remove the Earth's gravity well, and assume a spacecraft is at earth orbital distance following same orbital path [but tens of millions of kilometer from Earth, or say least 60 million kilometer from Earth].This can make things simpler to calculate, but it will make all of your calculations invalid for an actual mission.QuoteSay, this spacecraft has a total delta-v of 16 km/sec and it's burn is in a direction of 90 degree away from it's orbital path [with it's back to Sun]. Wouldn't the trajectory result from addition it's orbital velocity [a vector 29.8 km/sec at 180 degrees] to the vector from rocket [16 km/sec at 90 degrees]. With the resultant vector of 33.8 km/sec at 151.7 degrees?First off, 16km/s is exceedingly high for chemical rockets. Your vector math is accurate. Remember that this velocity is only momentary--the Sun's gravity affects it.
Due to the Oberth effect, the v_inf would be up to sqrt(11^2+16^2) = 19.42km/s.
Yes, 16 km/sec is about as much velocity as has ever been achieve with chemical rockets. How much the sun gravity affects it, is my fundamental question. If "somehow" the resultant vector could be 90 degree instead of 151.7 degrees it would be easier.
The sun' gravity at earth distant is .0059 m/s/s and at 90 degrees, I assume it loses this much velocity per sec and some fraction of this if instead it's at the 151.7 degrees?
When spacecraft reaches 200 km from earth it should going around 10 km/sec relative to earth and then one would add 16 km/sec from rocket- giving around 26 km/sec.