Author Topic: Quickest (fantasy) journey to Mars with chemical rockets?  (Read 23349 times)

Offline RanulfC

  • Senior Member
  • *****
  • Posts: 4595
  • Heus tu Omnis! Vigilate Hoc!
  • Liked: 900
  • Likes Given: 32
Re: Quickest (fantasy) journey to Mars with chemical rockets?
« Reply #20 on: 07/26/2011 07:34 pm »
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.pdf

You could probably aerobreak away the excess speed...

...and "litho" break (note I copied YOUR spelling which I suspect was intentional? :) ) away the rest....

Randy
From The Amazing Catstronaut on the Black Arrow LV:
British physics, old chap. It's undignified to belch flames and effluvia all over the pad, what. A true gentlemen's orbital conveyance lifts itself into the air unostentatiously, with the minimum of spectacle and a modicum of grace. Not like our American cousins' launch vehicles, eh?

Offline Arthur

  • Full Member
  • ***
  • Posts: 351
  • Liked: 0
  • Likes Given: 5
Re: Quickest (fantasy) journey to Mars with chemical rockets?
« Reply #21 on: 07/26/2011 08:17 pm »
I don't know if it would help, but P&W had a design called Triton for a nuclear engine with a LOX augmentation that might provide a fast start (like a pure Chemical rocket) with a high ISP continuous thrust flight.

[sort of pushing the limits of 'existing chemical', but not pure fantasy either.]

Refueling in Mars orbit for the return trip would help too.

Offline ANTIcarrot

  • Member
  • Full Member
  • *
  • Posts: 115
  • Liked: 10
  • Likes Given: 30
Re: Quickest (fantasy) journey to Mars with chemical rockets?
« Reply #22 on: 07/27/2011 03:10 am »
4146 / 39^1.242 = ~44 km/sec
Can this be done chemically with a single stage?

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.

You *can* get MRs in the 1000+ range if you're using nuclear thermal engines and storing your reaction mass as solid ice (no tank weight ) or in the 1,000,000+ range if your engine 'floats' in a large sphere of high surface tension reaction mass. But not with chemicals. Not any way I ever heard of or thought of at any rate.

Long and short - use a nuke, or solar power - or laser power beamed from earth. Or a rotational mechanical slingshot; I forget the technical name. Chemical rockets can get you from LEO to LMO with moderate reliablity, safety, and low cost - but not quickly.
« Last Edit: 07/27/2011 03:15 am by ANTIcarrot »

Offline Warren Platts

So if we wanted to go to Mars 1-way in 39 days then:

4146 / 39^1.242 = ~44 km/sec
Can 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.

Huh? I was talking specifically about a mission staged from L2 that would do an Oberth slingshot to get to Mars. Furthermore, I answered my own question as "No", but not because of the mass fraction, but because an Oberth bonus factor of 4 isn't realistic for such high delta v's. Here is the equation (thanks to Isaac Kuo):

((vesc + Δv)2 - vesc2)1/2 = vinf

For a 1-way mission there would be two burns; just splitting it up, one gets:

Earth Burn = 7.5 km/sec
((11.1 + 7.5)2 - 11.12)1/2 = 14.9 km/sec

14.9 / 7.5 = Oberth Bonus Factor of 2.0

Mars Burn = 3.5 km/sec
((5.03 + 3.5)2 - 5.032)1/2 = 6.89

6.89 / 3.5 = ~2.0 OBF

So staging from L2, one can 22 km/sec out of RL-10 motors with a single stage and yet have a decent mass fraction.

As I wrote a couple of posts back:

Δt (days) = 796.3 / Δv0.802 = 796.3 / 220.802 = ~67 days.

I think 67 days is "fast" by any reasonable definition of the word. 6 months in weightless space is done routinely at ISS. With 67 day transits, 6-month Mars missions would be enabled with 40+ days allowed on the surface.

All done with conventional--even reusable--RL-10 powered spacecraft.
« Last Edit: 07/30/2011 07:28 am by Warren Platts »
"When once you have tasted flight, you will forever walk the earth with your eyes turned skyward, for there you have been, and there you will always long to return."--Leonardo Da Vinci

Offline douglas100

  • Senior Member
  • *****
  • Posts: 2177
  • Liked: 227
  • Likes Given: 105
Re: Quickest (fantasy) journey to Mars with chemical rockets?
« Reply #24 on: 07/27/2011 01:21 pm »
Then there is the business of stopping once you arrive at Mars...
Douglas Clark

Offline kch

  • Full Member
  • ****
  • Posts: 1758
  • Liked: 496
  • Likes Given: 8807
Re: Quickest (fantasy) journey to Mars with chemical rockets?
« Reply #25 on: 07/27/2011 01:37 pm »
Then 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 ... ;)

Offline Warren Platts

Then 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 ... ;)

Amateurs.... Instead of resorting to the giggle factor fallacy, you could have pointed out an obvious flaw in my reasoning above: departing Earth with a 12.3 km velocity would require 25.5 km/sec of braking--to be done fully propulsively--not with lithobraking. Can't be done with single-stage chemical, no matter how abundant.

So to refine the equation: 

((vesc + Δv)2 - vesc2)1/2

Two burns:

Earth Burn = 2 km/sec
((11.1 + 2)2 - 11.12)1/2 = 7.0 km/sec

7 / 2 = Oberth Bonus Factor of 3.5

Mars Burn = 9 km/sec
((5.03 + 9)2 - 5.032)1/2 = 13.1 km/sec

13.1 / 9 = ~1.5 OBF

total delta v = ~20 km/sec, average OBF = 1.8

To rephrase my earlier post:

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.

As I wrote a couple of posts back:

Δt (days) = 796.3 / Δv0.802 = 796.3 / 200.802 = ~72 days.

I still think 72 days is "fast" by any reasonable definition of the word. 6 months in weightless space is done routinely at ISS. With 72 day transits, 6-month Mars missions would be enabled with 30+ days allowed on the surface.

Or just make it a 7 or 8 month mission....

All done with conventional, fully reusable, RL-10 powered spacecraft.

« Last Edit: 07/27/2011 03:39 pm by Warren Platts »
"When once you have tasted flight, you will forever walk the earth with your eyes turned skyward, for there you have been, and there you will always long to return."--Leonardo Da Vinci

Offline IsaacKuo

  • Full Member
  • ****
  • Posts: 435
  • Liked: 2
  • Likes Given: 1
Re: Quickest (fantasy) journey to Mars with chemical rockets?
« Reply #27 on: 07/27/2011 04:02 pm »
Two burns:

Earth Burn = 2 km/sec
((11.1 + 2)2 - 11.12)1/2 = 7.0 km/sec

7 / 2 = Oberth Bonus Factor of 3.5

Mars Burn = 9 km/sec
((5.03 + 9)2 - 5.032)1/2 = 13.1 km/sec

13.1 / 9 = ~1.5 OBF

total delta v = ~20 km/sec, average OBF = 1.8

You 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:

Quote
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.

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.

Offline Warren Platts

Two burns:

Earth Burn = 2 km/sec
((11.1 + 2)2 - 11.12)1/2 = 7.0 km/sec

7 / 2 = Oberth Bonus Factor of 3.5

Mars Burn = 9 km/sec
((5.03 + 9)2 - 5.032)1/2 = 13.1 km/sec

13.1 / 9 = ~1.5 OBF

total delta v = ~20 km/sec, average OBF = 1.8

You 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.

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. The goal is to calculate delta t--not to calculate the tons of propellant we need, because that's already a given. The ULA  MTV would carry about 726 tonnes of LH2/LO2. I don't think the way I'm using the word is improper English.

Quote from: Isaac
Why is your confused language a problem?  Because it results in confused thoughts:

Quote
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.

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.

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. 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 Δv, 72 days, 13.4 Δv 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.)

Thus, using Nordley's figures, you can do a 72 day 1-way transit to Mars chemically if you have 20 km/sec delta v. This cannot be done chemically with a single stage--unless, however, you stage from high orbit and take full advantage of the Oberth effect. In that case, you can still do a 72-day transit even though the spacecraft (the ULA MTV in this case) only has an 11 km/sec nominal delta v.--and do it fully propulsively--no fancy heat shielding required.



Quote from: Isaic
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.

This is interesting. I never thought of that.

Quote
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.

True, but if you have abundant chemical propellant, the lander could refuel in LMO, and then land fully propulsively. In which case, it would be nice to a high temperature resistant titanium skin capable of handling 1W/meter of heating

Quote
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.

If you can get to LMO, one would only need to carry ~4 km/sec worth or propellant for a fully propulsive landing.
"When once you have tasted flight, you will forever walk the earth with your eyes turned skyward, for there you have been, and there you will always long to return."--Leonardo Da Vinci

Offline IsaacKuo

  • Full Member
  • ****
  • Posts: 435
  • Liked: 2
  • Likes Given: 1
Re: Quickest (fantasy) journey to Mars with chemical rockets?
« Reply #29 on: 07/27/2011 09:06 pm »
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.

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.

Quote
Quote from: Isaac
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.

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?

Quote
When I said "get to Mars" above, I meant LMO.

This is a very important distinction.  Anyway, the proven way to get to LMO is to propulsively insert into a highly elliptical orbit and then use aerobraking to circularize the orbit (lower the apoapsis).

Quote
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.)

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.

Quote
If 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.

Offline gbaikie

  • Full Member
  • ****
  • Posts: 1592
  • Liked: 49
  • Likes Given: 5
Re: Quickest (fantasy) journey to Mars with chemical rockets?
« Reply #30 on: 07/28/2011 02:00 am »
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.

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.

So, if starting from 1500 km circular orbit it is a bad idea:)
But what if you not starting from LEO, but are from L-points, so one is essentially starting from an earth escape trajectory. Do you simply use  11 instead of 7.8 or 7.0 km/sec.
Or a similar type question, what if orbit isn't circular. Let's take your 1500 km circular orbit, lower it's perigee to say 200 km, so it's 1500 by 200 km orbit. You use the velocity at perigee?


Offline Warren Platts

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.

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.

OK. This is a good point. Delta-v and v_inf aren't the same thing, and I was conflating the two. Maybe it would help if I knew what the "inf" stood for. Perhaps you could enlighten us on that one.

Quote
Quote
Quote from: Isaac
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.

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?

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.

Quote from: Isaac
Quote
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.)

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.

OK, I see what you're saying about the v_inf thing. I'll review my numbers yet again. 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.

Quote
Quote
If 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.

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....
« Last Edit: 07/28/2011 02:57 am by Warren Platts »
"When once you have tasted flight, you will forever walk the earth with your eyes turned skyward, for there you have been, and there you will always long to return."--Leonardo Da Vinci

Offline Andrew_W

  • Full Member
  • ****
  • Posts: 754
  • Rotorua, New Zealand
    • Profiles of our future in space
  • Liked: 17
  • Likes Given: 12
Re: Quickest (fantasy) journey to Mars with chemical rockets?
« Reply #32 on: 07/28/2011 04:28 am »
Quote
Maybe it would help if I knew what the "inf" stood for. Perhaps you could enlighten us on that one.
infinity, v inf is I think the theoretical velocity the vehicle has relative to the body its performed the Oberth maneuver around when the gravitational pull of that body on the vehicle has declined to zero with distance.
I confess that in 1901 I said to my brother Orville that man would not fly for fifty years.
Wilbur Wright

Offline IsaacKuo

  • Full Member
  • ****
  • Posts: 435
  • Liked: 2
  • Likes Given: 1
Re: Quickest (fantasy) journey to Mars with chemical rockets?
« Reply #33 on: 07/28/2011 04:31 am »
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.

Offline gbaikie

  • Full Member
  • ****
  • Posts: 1592
  • Liked: 49
  • Likes Given: 5
Re: Quickest (fantasy) journey to Mars with chemical rockets?
« Reply #34 on: 07/28/2011 10:36 am »
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.

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?

Quote
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.

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?

And that if instead the spacecraft was near Earth and applied it's rocket thrust near earth that it's possible to gain much in terms of efficiency- so much so, that instead of 16 km/sec, it would be around 1/2 of this or around 8 km/sec of delta-v [from starting point of  Earth/Sun L-1] giving as equally fast trajectory to Mars?

[I picked Earth/Sun L-1 because it's always in same position- when Earth/Moon L-1 or 2 was between Sun and Earth it would actually be a more likely location to start the journey to Mars.]

I don't see how you get the same trajectory from LEO as one could get from high earth- or is this incorrect?

Offline IsaacKuo

  • Full Member
  • ****
  • Posts: 435
  • Liked: 2
  • Likes Given: 1
Re: Quickest (fantasy) journey to Mars with chemical rockets?
« Reply #35 on: 07/28/2011 02:02 pm »
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.

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.

Quote
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].

This can make things simpler to calculate, but it will make all of your calculations invalid for an actual mission.

Quote
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?

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.

Quote
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?

Due to the Oberth effect, the v_inf would be up to sqrt(11^2+16^2) = 19.42km/s.

Quote
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?

No, you should not assume.  You should calculate.

The orbit you describe never crosses Mars orbital distance.

At 1AU, the orbital speed is 33.8km/s = 1.135*29.8km/s.

At 1.5AU, orbital speed would be 29.8*sqrt(1.135^2-2*1^2+2*(1/1.5)^2) = 12.5km/s.  This is calculated by conservation of energy--adding specific energy at 1AU and subtracting specific energy at 1.25AU.

At 1.5AU, transverse speed would be 29.8*1/1.5 = 19.9km/s.  This is calculated by conservation of angular momentum--starting with the initial transverse speed, multiplying by the initial moment arm and dividing by the latter moment arm.

Oops!  12.5km/s is less than 19.9km/s!  This is not a possible situation!  Orbital speed can't be less than transverse speed.  This means the orbit never reaches 1.5AU.

In an elliptical orbit, orbital speed always exceeds transverse speed except at apoapsis and periapsis.  At those two points, orbital speed is equal to transverse speed.

Offline Warren Platts

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.

Thanks for the clarification. It's still a little confusing, because although one may have a positive v_inf WRT to Earth, I guess, because of the Sun, it's still technically in an elliptical orbit, it's just around the Sun instead of the Earth.
"When once you have tasted flight, you will forever walk the earth with your eyes turned skyward, for there you have been, and there you will always long to return."--Leonardo Da Vinci

Offline IsaacKuo

  • Full Member
  • ****
  • Posts: 435
  • Liked: 2
  • Likes Given: 1
Re: Quickest (fantasy) journey to Mars with chemical rockets?
« Reply #37 on: 07/28/2011 04:30 pm »
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?

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.

The time spent during aerobraking depends on how timid you want to be, our increasing knowledge of Mars's upper atmosphere, and our increasing experience with Mars aerobraking.

But if you really want to trim the days to a minimum, there's no contest--the best approach is direct atmospheric entry.  Unlike aerocapture, we've actually done it this way (Apollo).  It doesn't require fine tuning the amount of braking.  It just requires a robust heat shield.

Quote from: Isaac
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.

Well, what's the delta-v budget?  11km/s?  I would burn it all at once, at perigee.  11km/s requires a mass ratio of 12.1 for the disposable MTV--but the awesome heat shield might boost the mass ratio up to 18+ for the landing capsule.

In this scenario, where we care about speed above cost, the MTV is simply disposed of.  Maybe it is broken up into several capsule-sized pieces and sent ahead at slightly different altitudes.  Tracking the pieces as they enter Mars's atmosphere is used to determine the optimal entry angle.

No big loss, though--in this case the MTV is mostly just supplies for the relatively short journey.  Most of the life support hardware is on the landing capsule.  A second MTV is needed for the return journey, but it can take a slower path (even within the same launch window).  But let's get back to the first MTV:

Assuming that we start off with escape velocity, the v_inf would be sqrt((11+11)^2-11^2) = 19km/s.

Honestly, I don't know the ideal hyperbolic escape angle.  I'll just guess a slant of 30 degrees outward.  This gives a transverse velocity of about 30+19*cos(30) = 46.5km/s and an outward velocity of 19*sin(30) = 9.5km/s.

The orbital speed at 1AU is sqrt(46.5^2+9.5^2) = 47.5km/s.

The orbital speed at 1.5AU will be sqrt(47.5^2+2*24^2-2*30^2) = 40.1km/s.  Transverse speed will be 46.5*1/1.5 = 31.0km/s, so outward speed is sqrt(40.1^2-31.0^2) = 25.4km/s.

I'll go with a rough estimate of average outward speed of (25.4+9.5)/2 = 17.45km/s, so it takes roughly 50 days to travel outward by .5AU.  This is a very rough estimate.  The actual outward speed varies according to a curve.  Also, the optimal trajectory is likely different from my guess of 30 degrees.

Anyway, the v_inf at Mars is sqrt((31.0-24)^2+25.4^2) = 26.4km/s, so the relative velocity upon atmospheric entry is sqrt(26.4^2+5^2) = 27km/s.  This is how much velocity must be bled off by the heat shield.  Pretty awesome, but still only a third as energetic as the Galileo atmospheric entry probe.

The error ellipse on this atmospheric entry is going to be something crazy, so there's no hope of landing on top of a prepared base.  Instead, the main supplies and launch rocket should wait in orbit--they should drop after the landing capsule since they will have far smaller error ellipses.

So, what about the second MTV?  If we assume the same delta-v budget of 11km/s from Earth escape, it takes roughly 1km/s to Hohmann transfer to park it co-orbital near Mars.  The crew capsule docks with this MTV before periapsis (this is a highly nontrivial part of the mission involving dropping a launch rocket with near escape delta-v).  The MTV then performas a periapsis burn of 10km/s to get a v_inf of sqrt((10+5)^2-5^2) = 14km/s.

As before, I don't know the ideal hyperbolic escape angle.  As before, I'll guess a 30 degree slant from the orbital path.  This gives a transverse speed of 11.9km/s and an inward speed of 7km/s.  At 1.5AU, the orbital speed is 13.8km/s, so orbital speed at 1AU is sqrt(13.8^2+2*30^2-2*24^2) = 29km/s.  The transverse speed at 1AU is 11.9*1.5/1 = 17.85km/s so inward speed is sqrt(29^2-17.85^2) = 23km/s.

This gives a very rough average inward speed of (7+23)/2 = 15km/s.  So, very roughly, the return journey takes around 60 days.

Like the first MTV, the second MTV is disposed of while the crew capsule makes an awesome direct reentry to Earth.  This time, the v_inf is sqrt((30-17.85)^2+23^2) = 26km/s.  The reentry speed is sqrt(26^2+11^2) = 28km/s.

These are some really good trip times (albeit very very rough estimates).  The big enabler of these fast speeds is the assumption of awesome atmospheric entry.  I mean awesome.  It may be necessary to immerse the crew in water to provide an acceptable safety margin on the gee forces.

Quote
Quote
Quote
If 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....

A manned MTV doesn't need to be hauling propellant for landers if the lander doesn't use propellant.

Clearly you are doing your thinking backwards--you have already decided upon the conclusion that "fully propulsive" is the best solution without actually doing correct calculations to support this conclusion.

Anyway, assuming a "fully propulsive" solution, the fuel still has to get their somehow, and we're restricting ourselves to chemical rockets only.  If this is sent ahead of time, you need another vehicle with its own engines and tanks, which adds to costs and mass.  Since this vehicle is taking a slow Hohmann transfer, it needs to be launched in the previous launch window--meaning boiloff losses for two years.

And then there's the question of time needed to rendezvous.  Normally I wouldn't say this is a big deal.  In this case, though, the objective is quickest trip time.

Offline gbaikie

  • Full Member
  • ****
  • Posts: 1592
  • Liked: 49
  • Likes Given: 5
Re: Quickest (fantasy) journey to Mars with chemical rockets?
« Reply #38 on: 07/28/2011 06:11 pm »
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.

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.

Quote
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].

This can make things simpler to calculate, but it will make all of your calculations invalid for an actual mission.

Quote
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?

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.

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?

 
Quote
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?

Quote
Due to the Oberth effect, the v_inf would be up to sqrt(11^2+16^2) = 19.42km/s.
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.


Offline IsaacKuo

  • Full Member
  • ****
  • Posts: 435
  • Liked: 2
  • Likes Given: 1
Re: Quickest (fantasy) journey to Mars with chemical rockets?
« Reply #39 on: 07/28/2011 06:26 pm »
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.

You're just going to have to bite the bullet and learn something about orbital mechanics if you want to understand things and calculate them out.  Wikipedia is not a bad place to start.  Browse through the equations related to elliptical orbits and hyperbolic escape trajectories.
 
Quote
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?

This is not a helpful way to calculate things, because the gravity and velocity angles aren't constant.  Instead, it's easier to use conservation of energy and angular momentum to calculate things.

Quote
Due to the Oberth effect, the v_inf would be up to sqrt(11^2+16^2) = 19.42km/s.

I'm sorry, I made a mistake and used a bogus equation.  The correct equation was v_inf = sqrt((11+16)^2-11^2) = 24.7km/s.  That's actually a pretty decent boost.

Quote
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.

If you boost at an altitude where escape velocity is 10km/s, the v_inf is sqrt((10+16)^2-10^2) = 24km/s.  The speed is only 26km/s at perigee.  As it escapes Earth, Earth's gravity slows it down.

Tags:
 

Advertisement NovaTech
Advertisement Northrop Grumman
Advertisement
Advertisement Margaritaville Beach Resort South Padre Island
Advertisement Brady Kenniston
Advertisement NextSpaceflight
Advertisement Nathan Barker Photography
0