A Dragon derived general purpose Mars lander?

[docmordrid]

- As we know Dragon needs to prove a number of its [yet] theoretical capabilities, but my personal impression about the Nasa personnel doing the presentations is that they are non as skeptical as some over here ...

Interesting, so Dragon can manage a re-entry through a CO2 atmosphere? I wonder how cost effective it is compared to standard entry options such as MSL.

MSL's is PICA and SpaceX's is PICA-X, an evolved version they worked on with NASA Ames and which they make in-house.

http://www.airspacemag.com/space-exploration/Visionary-Launchers-Employees.html?c=y&page=3

According to Mueller, SpaceX’s material, called PICA-X, is 10 times less  expensive than the original, “and the stuff we made actually was better.”

[douglas100]

Interesting, so Dragon can manage a re-entry through a CO2 atmosphere? I wonder how cost effective it is compared to standard entry options such as MSL. Pity that it's limited to only 1mT, but it's a stepping stone to a large manned Mars lander.

Thank you for your comments on my points about a Dragon derived lander.

Slightly OT:

Your point about N2 as opposed to CO2: I have never heard this idea before, that the Russians lost the 71 landers by believing that the Martian atmosphere was nitrogen. Do you have any references? Remember that they claim that the Mars 3 lander reached the surface and survived a short time.

It's not clear to me why the molecular makeup of the atmosphere would have much affect on entry. I would have thought that the temperature and density profiles would be much more important. Remember that CO2 is about 50 percent denser than nitrogen under the same conditions. If the Russians had designed for a nitrogen atmosphere you would think that the fact it was actually CO2 would have made landing easier.

[go4mars]

Are we even confident that the heat shield shape would be the same for a red dragon? 

Would the superdraco plume effects increase drag by concentrating atmosphere toward the heatshield, subduing the noted historical shape effect through active compensation? 

[charliem]

The info disclosed these days during the Concepts and Approaches meeting is inspiring. It also gave us some more bits of data with which nourish speculations (don't know whether that's good or bad, but fun, yes that is ).

If Red Dragon can really land 1 mT of payload on Mars, I wonder how hard could be to expand it somewhat.

Red Dragon is said to start the propulsive deceleration 800 m from the surface, flying at an angle of, maybe, 10-15 deg, and doing about mach 2.2 (~540 m/s), burning a maximum of 1,900 kg of propellants.

From some BOE calcs seems that increasing the fuel to 3,400 kg max payload should reach 2 mT, although the speed at which the engines had to be started would also grow, to mach 3.1 (~765 m/s).

One of the presenters, from Nasa Langley I think, commented that they were not aware of any test for supersonic retro-propulsion, but that they thought it was the most promising EDL technology.

Another one from Nasa Ames talked a bit about a possible future sample recovery mission using also a Dragon.

Well, I suspect but don't know for sure, that a SuperDraco can do its job against a mach 2.2 flow (or 3.1). Anyhow it seems that we are not going to have to wait too much for experimental proof (or disproof). There's a lot of people interested.

And this time the idea to take two MER like rovers and a MAV rocket to Mars within a Dragon wasn't mine ...

[Robotbeat]

BTW, for the sort of manned Mars mission concepts which come out of NASA, the payloads require at least: 1) supersonic retropropulsion or 2) a deployable heatshield. This is true even with the really big fairings. The SuperDracos are so powerful that they could be strong enough to be put on even a manned-class lander for supersonic retropropulsion (at the expense of higher gravity losses).

[MikeAtkinson]

So the Dragon capsule has to start its SuperDraco engines at ~550 m/s, 800 m before crashing, and decelerate at 7 g ... what a ride!!

Those numbers can't be right. I've rechecked them against the presentation and this time it's not my mistake.

In one of the slides we can read: "* Performs a direct transition to powered flight at Mach 2.24 800 m above ground utilizing SuperDraco thrusters".

That speed can't be vertical, because then a simple v=sqr(2.h.a) says that the minimum distance to brake from 550 m/s to zero at 7 gees is 2200 meters, so the only possibility that remains is a very very shallow descent.

+1, there's an inconsistency somewhere. Forgetting about gravity, that works out to be about a 70 degree angle.

Could they have meant 800m above the datum, which would mean landing 1400m below the datum.

Add in some constant velocity descent (which was mentioned) and we have a landing at about 1500m below the datum.

[charliem]

Could they have meant 800m above the datum, which would mean landing 1400m below the datum.

Add in some constant velocity descent (which was mentioned) and we have a landing at about 1500m below the datum.

The written paragraph is clear: "800 m above ground" (sic).

And it's not necessary after all.

I've done some more math. It seems that indeed the flight path of a Dragon would be really flat, under 20 degrees while crossing the 0 altitude MOLA reference.

You can see it in this graphic for the entry path: http://forum.nasaspaceflight.com/index.php?action=dlattach;topic=28668.0;attach=392917

Vertical and horizontal scales are very different. Once you compensate for that it's evident how shallow the whole trajectory is. Quite different that an earth reentry.

Following that path 2,200 meters ahead on the final approach mean only about 500 or 600 of descent.

[modemeagle]

Could they have meant 800m above the datum, which would mean landing 1400m below the datum.

Add in some constant velocity descent (which was mentioned) and we have a landing at about 1500m below the datum.

The written paragraph is clear: "800 m above ground" (sic).

And it's not necessary after all.

I've done some more math. It seems that indeed the flight path of a Dragon would be really flat, under 20 degrees while crossing the 0 altitude MOLA reference.

You can see it in this graphic for the entry path: http://forum.nasaspaceflight.com/index.php?action=dlattach;topic=28668.0;attach=392917

Vertical and horizontal scales are very different. Once you compensate for that it's evident how shallow the whole trajectory is. Quite different that an earth reentry.

Following that path 2,200 meters ahead on the final approach mean only about 500 or 600 of descent.

Finally, a chance to post on this topic!!