SpaceX updates Mars Colony rendering with new BFR

The new version of the BFS 100-person spaceship is longer than the old design. It is 55 meters instead of 48 meters.

50 thoughts on “SpaceX updates Mars Colony rendering with new BFR”

  1. In the different-angle closeups in the other posts it looks more like 8 per row, so 48 quartes total. If they can fit two people per quarter, then that works out to ~100.

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  2. Assuming fairly cramped 2m long quarters near the outer wall, oriented radially, 60 cm wide on the inner end (just barely wide enough for a door): 9 – 2*2 = 5 m inner hallway diameter. Circumference = 5 * pi = 15.7 m 15.7 / 0.6 m = 26 doors per row. So 100 quarters would fit in 4 full-circle rows or 8 half-circle rows. They could probably arrange it better, but at least the math works out on the number of people. The rendering shows 6 rows, which would need 100/6 = 17 quarters per row (102 total). The rendering looks more like 12, and 10-11 for the two rows closest to the nose. Which would give 4*12 + 11 + 10 = 69 quarters. Probably the rendering still isn’t accurate.

    Reply
  3. In the different-angle closeups in the other posts it looks more like 8 per row so 48 quartes total. If they can fit two people per quarter then that works out to ~100.

    Reply
  4. Assuming fairly cramped 2m long quarters near the outer wall oriented radially 60 cm wide on the inner end (just barely wide enough for a door):9 – 2*2 = 5 m inner hallway diameter.Circumference = 5 * pi = 15.7 m15.7 / 0.6 m = 26 doors per row.So 100 quarters would fit in 4 full-circle rows or 8 half-circle rows. They could probably arrange it better but at least the math works out on the number of people.The rendering shows 6 rows which would need 100/6 = 17 quarters per row (102 total). The rendering looks more like 12 and 10-11 for the two rows closest to the nose. Which would give 4*12 + 11 + 10 = 69 quarters. Probably the rendering still isn’t accurate.

    Reply
  5. Minus the dome, maybe. Though I suppose it’s possible you might use a very low pressure dome as the entrance to an underground colony, just to avoid dust getting in.

    Reply
  6. Minus the dome maybe. Though I suppose it’s possible you might use a very low pressure dome as the entrance to an underground colony just to avoid dust getting in.

    Reply
  7. Phobos is really tiny, though, more of a captured asteroid than an actual moon. Its larges dimension is only 14 miles. You wouldn’t notice any difference in illumination if Phobos were white. Maybe if it were covered entirely with mirrors pointed at the colony, though even that would be questionable unless they were optically perfect.

    Reply
  8. Phobos is really tiny though more of a captured asteroid than an actual moon. Its larges dimension is only 14 miles.You wouldn’t notice any difference in illumination if Phobos were white. Maybe if it were covered entirely with mirrors pointed at the colony though even that would be questionable unless they were optically perfect.

    Reply
  9. If you’ve done the math to show it’s “more than enough”, then let’s see it. Otherwise, go do the math (and then post it here). If you can’t do the math, then don’t just assume it will work. Most people have very poor understanding of orders-of-magnitude, and tend to oversimplify and over/under-estimate things by large factors. By my own quick look into this: A) Apparent size: Phobos distance = ~9400 km orbital radius minus ~3400 km Mars radius = ~6000 km. (It’s so close that Mars’ radius isn’t negligible – an important point for optics calculations.) Sun distance = ~210 million km near perihelion (closest point to the Sun, where the Sun appears largest.) Distance ratio = 210e6 / 6000 = 35000. Phobos radius = ~11 km Sun radius = ~700000 km Radius ratio = 700000 / 11 = ~64000 => Phobos should appear about half the size of the Sun from Mars’ surface. That alone sounds significant, but then you need to consider: B) Amount of light available: Mars radius = ~3400 km Phobos radius = ~11 km Cross section area ratio = (11/3400)^2 = 1e-5 = 0.001% Let that sink in: *total* amount of light that Phobos receives and can reflect is just *0.001%* of the amount of light that Mars receives directly. On a planetary scale, that’s completely negligible. If you focused all the reflected light at the base, maybe that’d make a difference for just the base. But you’d need some pretty impressive optics to hit a few-hundered-meter-radius target from 6000 km away while accounting for relative motion.

    Reply
  10. If you’ve done the math to show it’s more than enough””” then let’s see it. Otherwise go do the math (and then post it here). If you can’t do the math then don’t just assume it will work. Most people have very poor understanding of orders-of-magnitude and tend to oversimplify and over/under-estimate things by large factors.By my own quick look into this:A) Apparent size:Phobos distance = ~9400 km orbital radius minus ~3400 km Mars radius = ~6000 km.(It’s so close that Mars’ radius isn’t negligible – an important point for optics calculations.)Sun distance = ~210 million km near perihelion (closest point to the Sun where the Sun appears largest.)Distance ratio = 210e6 / 6000 = 35000.Phobos radius = ~11 kmSun radius = ~700000 kmRadius ratio = 700000 / 11 = ~64000=> Phobos should appear about half the size of the Sun from Mars’ surface.That alone sounds significant but then you need to consider:B) Amount of light available:Mars radius = ~3400 kmPhobos radius = ~11 kmCross section area ratio = (11/3400)^2 = 1e-5 = 0.001{22800fc54956079738b58e74e4dcd846757aa319aad70fcf90c97a58f3119a12}Let that sink in: *total* amount of light that Phobos receives and can reflect is just *0.001{22800fc54956079738b58e74e4dcd846757aa319aad70fcf90c97a58f3119a12}* of the amount of light that Mars receives directly. On a planetary scale that’s completely negligible.If you focused all the reflected light at the base”” maybe that’d make a difference for just the base. But you’d need some pretty impressive optics to hit a few-hundered-meter-radius target from 6000 km away while accounting for relative motion.”””

    Reply
  11. If you’ve done the math to show it’s “more than enough”, then let’s see it. Otherwise, go do the math (and then post it here). If you can’t do the math, then don’t just assume it will work. Most people have very poor understanding of orders-of-magnitude, and tend to oversimplify and over/under-estimate things by large factors. By my own quick look into this: A) Apparent size: Phobos distance = ~9400 km orbital radius minus ~3400 km Mars radius = ~6000 km. (It’s so close that Mars’ radius isn’t negligible – an important point for optics calculations.) Sun distance = ~210 million km near perihelion (closest point to the Sun, where the Sun appears largest.) Distance ratio = 210e6 / 6000 = 35000. Phobos radius = ~11 km Sun radius = ~700000 km Radius ratio = 700000 / 11 = ~64000 => Phobos should appear about half the size of the Sun from Mars’ surface. That alone sounds significant, but then you need to consider: B) Amount of light available: Mars radius = ~3400 km Phobos radius = ~11 km Cross section area ratio = (11/3400)^2 = 1e-5 = 0.001% Let that sink in: *total* amount of light that Phobos receives and can reflect is just *0.001%* of the amount of light that Mars receives directly. On a planetary scale, that’s completely negligible. If you focused all the reflected light at the base, maybe that’d make a difference for just the base. But you’d need some pretty impressive optics to hit a few-hundered-meter-radius target from 6000 km away while accounting for relative motion.

    Reply
  12. If you’ve done the math to show it’s more than enough””” then let’s see it. Otherwise go do the math (and then post it here). If you can’t do the math then don’t just assume it will work. Most people have very poor understanding of orders-of-magnitude and tend to oversimplify and over/under-estimate things by large factors.By my own quick look into this:A) Apparent size:Phobos distance = ~9400 km orbital radius minus ~3400 km Mars radius = ~6000 km.(It’s so close that Mars’ radius isn’t negligible – an important point for optics calculations.)Sun distance = ~210 million km near perihelion (closest point to the Sun where the Sun appears largest.)Distance ratio = 210e6 / 6000 = 35000.Phobos radius = ~11 kmSun radius = ~700000 kmRadius ratio = 700000 / 11 = ~64000=> Phobos should appear about half the size of the Sun from Mars’ surface.That alone sounds significant but then you need to consider:B) Amount of light available:Mars radius = ~3400 kmPhobos radius = ~11 kmCross section area ratio = (11/3400)^2 = 1e-5 = 0.001{22800fc54956079738b58e74e4dcd846757aa319aad70fcf90c97a58f3119a12}Let that sink in: *total* amount of light that Phobos receives and can reflect is just *0.001{22800fc54956079738b58e74e4dcd846757aa319aad70fcf90c97a58f3119a12}* of the amount of light that Mars receives directly. On a planetary scale that’s completely negligible.If you focused all the reflected light at the base”” maybe that’d make a difference for just the base. But you’d need some pretty impressive optics to hit a few-hundered-meter-radius target from 6000 km away while accounting for relative motion.”””

    Reply
  13. If you’ve done the math to show it’s “more than enough”, then let’s see it. Otherwise, go do the math (and then post it here). If you can’t do the math, then don’t just assume it will work. Most people have very poor understanding of orders-of-magnitude, and tend to oversimplify and over/under-estimate things by large factors.

    By my own quick look into this:
    A) Apparent size:

    Phobos distance = ~9400 km orbital radius minus ~3400 km Mars radius = ~6000 km.
    (It’s so close that Mars’ radius isn’t negligible – an important point for optics calculations.)
    Sun distance = ~210 million km near perihelion (closest point to the Sun, where the Sun appears largest.)
    Distance ratio = 210e6 / 6000 = 35000.

    Phobos radius = ~11 km
    Sun radius = ~700000 km
    Radius ratio = 700000 / 11 = ~64000

    => Phobos should appear about half the size of the Sun from Mars’ surface.

    That alone sounds significant, but then you need to consider:
    B) Amount of light available:

    Mars radius = ~3400 km
    Phobos radius = ~11 km
    Cross section area ratio = (11/3400)^2 = 1e-5 = 0.001%

    Let that sink in: *total* amount of light that Phobos receives and can reflect is just *0.001%* of the amount of light that Mars receives directly. On a planetary scale, that’s completely negligible.

    If you focused all the reflected light at the base, maybe that’d make a difference for just the base. But you’d need some pretty impressive optics to hit a few-hundered-meter-radius target from 6000 km away while accounting for relative motion.

    Reply
  14. Phobos is really tiny, though, more of a captured asteroid than an actual moon. Its larges dimension is only 14 miles. You wouldn’t notice any difference in illumination if Phobos were white. Maybe if it were covered entirely with mirrors pointed at the colony, though even that would be questionable unless they were optically perfect.

    Reply
  15. Phobos is really tiny though more of a captured asteroid than an actual moon. Its larges dimension is only 14 miles.You wouldn’t notice any difference in illumination if Phobos were white. Maybe if it were covered entirely with mirrors pointed at the colony though even that would be questionable unless they were optically perfect.

    Reply
  16. Minus the dome, maybe. Though I suppose it’s possible you might use a very low pressure dome as the entrance to an underground colony, just to avoid dust getting in.

    Reply
  17. Minus the dome maybe. Though I suppose it’s possible you might use a very low pressure dome as the entrance to an underground colony just to avoid dust getting in.

    Reply
  18. In the different-angle closeups in the other posts it looks more like 8 per row, so 48 quartes total. If they can fit two people per quarter, then that works out to ~100.

    Reply
  19. In the different-angle closeups in the other posts it looks more like 8 per row so 48 quartes total. If they can fit two people per quarter then that works out to ~100.

    Reply
  20. Assuming fairly cramped 2m long quarters near the outer wall, oriented radially, 60 cm wide on the inner end (just barely wide enough for a door): 9 – 2*2 = 5 m inner hallway diameter. Circumference = 5 * pi = 15.7 m 15.7 / 0.6 m = 26 doors per row. So 100 quarters would fit in 4 full-circle rows or 8 half-circle rows. They could probably arrange it better, but at least the math works out on the number of people. The rendering shows 6 rows, which would need 100/6 = 17 quarters per row (102 total). The rendering looks more like 12, and 10-11 for the two rows closest to the nose. Which would give 4*12 + 11 + 10 = 69 quarters. Probably the rendering still isn’t accurate.

    Reply
  21. Assuming fairly cramped 2m long quarters near the outer wall oriented radially 60 cm wide on the inner end (just barely wide enough for a door):9 – 2*2 = 5 m inner hallway diameter.Circumference = 5 * pi = 15.7 m15.7 / 0.6 m = 26 doors per row.So 100 quarters would fit in 4 full-circle rows or 8 half-circle rows. They could probably arrange it better but at least the math works out on the number of people.The rendering shows 6 rows which would need 100/6 = 17 quarters per row (102 total). The rendering looks more like 12 and 10-11 for the two rows closest to the nose. Which would give 4*12 + 11 + 10 = 69 quarters. Probably the rendering still isn’t accurate.

    Reply
  22. Phobos is really tiny, though, more of a captured asteroid than an actual moon. Its larges dimension is only 14 miles.

    You wouldn’t notice any difference in illumination if Phobos were white. Maybe if it were covered entirely with mirrors pointed at the colony, though even that would be questionable unless they were optically perfect.

    Reply
  23. Assuming fairly cramped 2m long quarters near the outer wall, oriented radially, 60 cm wide on the inner end (just barely wide enough for a door):

    9 – 2*2 = 5 m inner hallway diameter.
    Circumference = 5 * pi = 15.7 m
    15.7 / 0.6 m = 26 doors per row.

    So 100 quarters would fit in 4 full-circle rows or 8 half-circle rows. They could probably arrange it better, but at least the math works out on the number of people.

    The rendering shows 6 rows, which would need 100/6 = 17 quarters per row (102 total). The rendering looks more like 12, and 10-11 for the two rows closest to the nose. Which would give 4*12 + 11 + 10 = 69 quarters. Probably the rendering still isn’t accurate.

    Reply

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