Moonshot Artemis

Ishmael

52,348
12,187
Fuctifino
Apologies for disrupting the spacecraft maintenance conversation, but given the 1.54 degree obliquity to the ecliptic, and the lunar diameter of 3476 kilometers, that polar peak only needs to be about 630 meters above the spherical surface to be in constant sunlight (other than during a lunar eclipse.)

It's not a perfect sphere, it's an oblate spheroid.
 

BeSafe

Super Anarchist
7,980
1,290
Given that the majority of the thrust for the first couple of minutes is largely coming from the solid rocket boosters, the Space Shuttle launches aren't a bad proxy.

The answer to your question ( http://www.aerospaceweb.org/question/aerodynamics/q0025.shtml). From the graph, the first '5 miles' up (26,000 feet or so) takes about 48 seconds, during which, the solid rocket boosters are consuming about 11,000 lbs of fuel per second for the two of them (https://www.nasa.gov/sites/default/files/113069main_shuttle_trivia.pdf) or about 48*11,000 = 528,000 lbs of fuel. There's some additional fuel being burned by the main engine but that's the lion's share at that altitude. The problem with that math is that while you have only climbed 5 miles, you've also accelerated to ~810 mph, so things are more complicated.

The biggest issue is that your final destination is much more important than your initial condition. You're final goal is a velocity > 7 mps, starting from around 0.27 mps. The only real benefit you get from higher altitude is lower wind resistance some improvement in engine efficiency, neither of which are particularly large at 5 miles up vs sea level.

Good luck!


maxq.png
 

βhyde

Super Anarchist
8,204
1,791
Beside Myself
Given that the majority of the thrust for the first couple of minutes is largely coming from the solid rocket boosters, the Space Shuttle launches aren't a bad proxy.

The answer to your question ( http://www.aerospaceweb.org/question/aerodynamics/q0025.shtml). From the graph, the first '5 miles' up (26,000 feet or so) takes about 48 seconds, during which, the solid rocket boosters are consuming about 11,000 lbs of fuel per second for the two of them (https://www.nasa.gov/sites/default/files/113069main_shuttle_trivia.pdf) or about 48*11,000 = 528,000 lbs of fuel. There's some additional fuel being burned by the main engine but that's the lion's share at that altitude. The problem with that math is that while you have only climbed 5 miles, you've also accelerated to ~810 mph, so things are more complicated.

The biggest issue is that your final destination is much more important than your initial condition. You're final goal is a velocity > 7 mps, starting from around 0.27 mps. The only real benefit you get from higher altitude is lower wind resistance some improvement in engine efficiency, neither of which are particularly large at 5 miles up vs sea level.

Good luck!


View attachment 537673
Fcking atmosphere! We need to build a big open top cylinder with the bottom firmly anchored to the ground and the top sticking into space. Suck all the air out of it and that Max-Q just goes away. Hell, you could save weight by getting rid of the windows and all that nosecone stuff. Would also make for a much smoother ride. Problem solved.
 

Stingray~

Super Anarchist
11,618
3,098
PNW
This is being advertised as a long-term stepping stone to eventual human exploration of Mars and maybe it has some small worth in that direction. Mars has better gravity the Moon for humans but both would need terra forming including atmospheres to be even barely habitable and so maybe the Moon is the best place to start on that?
 

βhyde

Super Anarchist
8,204
1,791
Beside Myself
Then *you* do the math. In my universe, *everything* is perfect.
You'll be happy to know I did a quick calculation of the lateral surface area of the Shacketon crater. Using a conical frustum as an approximation, I get roughly 191,043,966m2. I'll check Amazon and see if they have enough panels. If only half the panels are in sunlight, we get 95521983m2 * 1366w/m2 = 130Megawatts @ 100% efficiency (hopelessly optimistic). That's great new. The bad news is it looks like that many panels is going to weigh in at about 6 million kg. That's going to take about 300 Artemis flights if my cocktail napkin is correct (it isn't). Totally doable!
 

Steam Flyer

Sophisticated Yet Humble
44,208
9,586
Eastern NC
You'll be happy to know I did a quick calculation of the lateral surface area of the Shacketon crater. Using a conical frustum as an approximation, I get roughly 191,043,966m2. I'll check Amazon and see if they have enough panels. If only half the panels are in sunlight, we get 95521983m2 * 1366w/m2 = 130Megawatts @ 100% efficiency (hopelessly optimistic). That's great new. The bad news is it looks like that many panels is going to weigh in at about 6 million kg. That's going to take about 300 Artemis flights if my cocktail napkin is correct (it isn't). Totally doable!
We can cut that in half by launching from the top of Mt Everest, right?
 

kent_island_sailor

Super Anarchist
27,222
5,136
Kent Island!
Launching from a mountain in FLORIDA would help, farther from the equator not so much, you gain in one area and lose in another.
EDIT - Small rockets are sometimes launched from airplanes, so the general idea of saving fuel by launching at altitude is a thing ;)
 

Steam Flyer

Sophisticated Yet Humble
44,208
9,586
Eastern NC
Launching from a mountain in FLORIDA would help, farther from the equator not so much, you gain in one area and lose in another.
EDIT - Small rockets are sometimes launched from airplanes, so the general idea of saving fuel by launching at altitude is a thing ;)

Florida is not that close to the equator. How about a mountain in Key West? There's a railroad that goes there, cut down transport costs.......
 

MR.CLEAN

Moderator
46,273
4,420
Not here
Launching from a mountain in FLORIDA would help, farther from the equator not so much, you gain in one area and lose in another.
EDIT - Small rockets are sometimes launched from airplanes, so the general idea of saving fuel by launching at altitude is a thing ;)
If all the bullshit in florida were shoveled into one pile it'd be quite a bit taller than everest. Problem solved.
 

Snaggletooth

SA's Morrelle Compasse
33,811
5,460
If all the bullshit in florida were shoveled into one pile it'd be quite a bit taller than everest. Problem solved.
That woude meane addittionalle tourisime dolleres and a niew locatione foire Sherpas to settelle! Winne winne!
 

Mark_K

Super Anarchist
Given that the majority of the thrust for the first couple of minutes is largely coming from the solid rocket boosters, the Space Shuttle launches aren't a bad proxy.

The answer to your question ( http://www.aerospaceweb.org/question/aerodynamics/q0025.shtml). From the graph, the first '5 miles' up (26,000 feet or so) takes about 48 seconds, during which, the solid rocket boosters are consuming about 11,000 lbs of fuel per second for the two of them (https://www.nasa.gov/sites/default/files/113069main_shuttle_trivia.pdf) or about 48*11,000 = 528,000 lbs of fuel. There's some additional fuel being burned by the main engine but that's the lion's share at that altitude. The problem with that math is that while you have only climbed 5 miles, you've also accelerated to ~810 mph, so things are more complicated.

The biggest issue is that your final destination is much more important than your initial condition. You're final goal is a velocity > 7 mps, starting from around 0.27 mps. The only real benefit you get from higher altitude is lower wind resistance some improvement in engine efficiency, neither of which are particularly large at 5 miles up vs sea level.

Good luck!


View attachment 537673
Obviously it's not a simple as subtracting all the fuel used in the first five vertical miles due to the velocity at that altitude in the comparison. However we are talking about only 3-5 tons of payload. If only 10% of that fuel was saved....
 




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