So how high would the lunar version be? How high is a geostationary (if that's the right term) orbit for the moon?
The aim is to put the counter weight 263,980 km above the lunar surface.L1 is 56,000 km above the lunar surface.
Quote from: A_M_Swallow on 08/28/2012 10:00 pmThe aim is to put the counter weight 263,980 km above the lunar surface.L1 is 56,000 km above the lunar surface.263,980 km?I was going to ask if this counterweight was going to ride in a boat, but it couldn't be, because your sense of humor is dry.
Quote from: mto on 08/28/2012 06:04 pmSo how high would the lunar version be? How high is a geostationary (if that's the right term) orbit for the moon?Well, if the Moon were all by herself (rotating at the usual once per month), then it'd be ~90000km or so. But the Moon is orbiting the Earth, and close enough to the Earth such that the Hill Sphere radius of the Moon is less than 90000km. So, instead, the lunar elevator would go to Earth-Moon Lagrange 2, which is roughly a Hill Sphere radius from the Moon, or about ~60000km. So the answer is about 60000km (plus whatever is needed for the counterweight).
{snip}Deploying it would be tricky, in my opinion. The EML1 part of the tether needs to be held in place while the ends are heading earthward and moonward. A small nudge from EML1 can send a mass far from EML1. Once the tether's anchored on the moon and the tether foot is well earthward, the tether is gravity gradient stabilized. But until the tether ends reach their destination, it's a tricky balancing act:{snip}
Quote from: Hop_David on 08/29/2012 07:04 pm{snip}Deploying it would be tricky, in my opinion. The EML1 part of the tether needs to be held in place while the ends are heading earthward and moonward. A small nudge from EML1 can send a mass far from EML1. Once the tether's anchored on the moon and the tether foot is well earthward, the tether is gravity gradient stabilized. But until the tether ends reach their destination, it's a tricky balancing act:{snip}Surprisingly that is the easy bit. They use gravity and inertia to make the tether self stabilizing.
During construction the tether is a satellite whose centre of gravity (or mass or what ever it is called) is at EML-1. At the start of the deployment the counter weight is placed at EML-1. Liftport then feed out cable from the counter weight, this causes the counter weight to get further from the Moon and the foot to get nearer to the Moon. Conservation of momentum then takes care of the ratios.
What's the smallest lunar "elevator" that could be built?Could it bootstrap?This LADDER concept looks interesting, is it fleshed out anywhere or is that it?
Minimum length is 600000km, other dimensions are up to you! (Needs to have enough surface for the climber to grab on to.)
The LADDER mission would erect a 264,000 km space elevator from theLunar surface, past the L1 Lagrange point, to a counterweight deep in cislunar space.
Quote from: Robotbeat on 08/29/2012 11:48 pmMinimum length is 600000km, other dimensions are up to you! (Needs to have enough surface for the climber to grab on to.)hmmm..QuoteThe LADDER mission would erect a 264,000 km space elevator from theLunar surface, past the L1 Lagrange point, to a counterweight deep in cislunar space.I guess it's the mythical counterweight that allows them to claim a shorter distance.
Quote from: Comga on 08/28/2012 10:42 pmQuote from: A_M_Swallow on 08/28/2012 10:00 pmThe aim is to put the counter weight 263,980 km above the lunar surface.L1 is 56,000 km above the lunar surface.263,980 km?I was going to ask if this counterweight was going to ride in a boat, but it couldn't be, because your sense of humor is dry. Distance between the Earth and Moon is 384,405 km - 6,378 km - 1,738 km = 376,289 kmhttp://en.wikipedia.org/wiki/MoonSo on average there is a gap of 376,289 - 263,980 = 113,000 km between the Earth and counter weight.p.s. So the counter weight will be a dry rock. If you want a siren I sure that we can transmit a recording of a woman singing.
Actually, no, I added one too many zeros. The minimum distance is 60000km, not 600,000km.
Quote from: A_M_Swallow on 08/29/2012 05:49 pmQuote from: Comga on 08/28/2012 10:42 pmQuote from: A_M_Swallow on 08/28/2012 10:00 pmThe aim is to put the counter weight 263,980 km above the lunar surface.L1 is 56,000 km above the lunar surface.263,980 km?I was going to ask if this counterweight was going to ride in a boat, but it couldn't be, because your sense of humor is dry. Distance between the Earth and Moon is 384,405 km - 6,378 km - 1,738 km = 376,289 kmhttp://en.wikipedia.org/wiki/MoonSo on average there is a gap of 376,289 - 263,980 = 113,000 km between the Earth and counter weight.p.s. So the counter weight will be a dry rock. If you want a siren I sure that we can transmit a recording of a woman singing. Oops. Got that metric vs Imperial issue again. 250,000 miles 376,000 km.Once upon a time I did a numerical evaluation of the minimum mass of an Earth space tower. It was an astonishingly large number. For some useful amount of force at the bottom, what is the mass of this 326,000 km long lunar space tower, even with some magic material like mass-produced graphene cable?P.S Re-transmit someone singing from a system far out in space? Who would do such a silly thing? (On, yeah. JPL just did that with MSL, and were so very proud of it.)
Quote from: Comga on 08/30/2012 03:58 amQuote from: A_M_Swallow on 08/29/2012 05:49 pmQuote from: Comga on 08/28/2012 10:42 pmQuote from: A_M_Swallow on 08/28/2012 10:00 pmThe aim is to put the counter weight 263,980 km above the lunar surface.L1 is 56,000 km above the lunar surface.263,980 km?I was going to ask if this counterweight was going to ride in a boat, but it couldn't be, because your sense of humor is dry. Distance between the Earth and Moon is 384,405 km - 6,378 km - 1,738 km = 376,289 kmhttp://en.wikipedia.org/wiki/MoonSo on average there is a gap of 376,289 - 263,980 = 113,000 km between the Earth and counter weight.p.s. So the counter weight will be a dry rock. If you want a siren I sure that we can transmit a recording of a woman singing. Oops. Got that metric vs Imperial issue again. 250,000 miles 376,000 km.Once upon a time I did a numerical evaluation of the minimum mass of an Earth space tower. It was an astonishingly large number. For some useful amount of force at the bottom, what is the mass of this 326,000 km long lunar space tower, even with some magic material like mass-produced graphene cable?P.S Re-transmit someone singing from a system far out in space? Who would do such a silly thing? (On, yeah. JPL just did that with MSL, and were so very proud of it.)Too bad the MSL doesn't actually even have an audio codec. All it did was send back a string of (to it) ones and zeroes. As for the cable mass, it states right there in the paper: approximately 11 tonnes, of which most is presumably cable.A simple BOTE calculation tells me that, for an *average* area of 1mm^2 and a density of 1gm/cc (for simplicity's sake), I get a cable that weighs 264 tonnes. So, given Zylon's density of 1.5gm/cc we probaly get a thin cable of about 0.0025 mm^2 or about a twentieth of a mm across.
Quote from: Lampyridae on 08/30/2012 08:14 amAs for the cable mass, it states right there in the paper: approximately 11 tonnes, of which most is presumably cable.A simple BOTE calculation tells me that, for an *average* area of 1mm^2 and a density of 1gm/cc (for simplicity's sake), I get a cable that weighs 264 tonnes. So, given Zylon's density of 1.5gm/cc we probaly get a thin cable of about 0.0025 mm^2 or about a twentieth of a mm across.You are off by a factor of 100. A 1mm^2 cross sectional area cable that has a weight of 1.5gm/cc would weigh 1.5gm for a length of 1 meter. And for 60,000km
As for the cable mass, it states right there in the paper: approximately 11 tonnes, of which most is presumably cable.A simple BOTE calculation tells me that, for an *average* area of 1mm^2 and a density of 1gm/cc (for simplicity's sake), I get a cable that weighs 264 tonnes. So, given Zylon's density of 1.5gm/cc we probaly get a thin cable of about 0.0025 mm^2 or about a twentieth of a mm across.
would weigh 90mt.
So a cross section of .2 mm^2 would give you less than 20mt weight for the cable.
...sqrt(.027) = ~.167. So the tether cross section would average a sixth of a millimeter by a sixth of a millimeter.
Quote from: Hop_David on 09/01/2012 05:18 pm...sqrt(.027) = ~.167. So the tether cross section would average a sixth of a millimeter by a sixth of a millimeter.In other words, almost impossible to climb...
The Liftport paper says 11 tonnes of tether. So we'd need to shrink Lampyridae's cross section by a factor of 11/396. That's .027. So the cross section should be .027 mm2. Lampyridae got an extra zero in there somehow.sqrt(.027) = ~.167. So the tether cross section would average a sixth of a millimeter by a sixth of a millimeter.
Quote from: Hop_David on 09/01/2012 05:18 pmThe Liftport paper says 11 tonnes of tether. So we'd need to shrink Lampyridae's cross section by a factor of 11/396. That's .027. So the cross section should be .027 mm2. Lampyridae got an extra zero in there somehow.sqrt(.027) = ~.167. So the tether cross section would average a sixth of a millimeter by a sixth of a millimeter.So for Zylon with a strength of 5.8 GPa, an .027 mm2 cable has a limit of 157 N. So if the factor of safety is set down at 2, a single crawler, payload, power supply, motor, etc. of under 50 kg can be hung on the elevator at any time. And that's assuming that there is zero force on the cable base when the elevator is just above the surface.This speaks to the throughput issue.Did I get those calculations correct?
Let's see, a 50 kg elevator car just above the lunar surface would give a tug of about 80 newtons. This 80 newtons would be felt throughout, including at EML1. At EML1, the tether is feeling the 80 newtons of the elevator car as well as the weight of all the tether between EML1 and the moon's surface. As well as the weight of all the tether between EML1 and earth's surface.
Quote from: Hop_David on 09/07/2012 10:18 pmLet's see, a 50 kg elevator car just above the lunar surface would give a tug of about 80 newtons. This 80 newtons would be felt throughout, including at EML1. At EML1, the tether is feeling the 80 newtons of the elevator car as well as the weight of all the tether between EML1 and the moon's surface. As well as the weight of all the tether between EML1 and earth's surface.If the elevator car is moving then its engine and brakes will also be supplying a force on the cable.
Yes, Hop David. 80 N is about half of 157 N, which is the safety factor of two. But this is just the maximum tension that could be supported by that size Zylon cable, which would be felt at the ground anchor.
Yes, Hop David. 80 N is about half of 157 N, which is the safety factor of two. But this is just the maximum tension that could be supported by that size Zylon cable, which would be felt at the ground anchor.No, A_M_Swallow. An elevator moving at constant speed imparts no additional along-cable force beyond its weight. (There would be lateral Coriolis force, but these would be tiny with a once-per-month rotation rate.) Once the elevator climbs up a ways where its weight is decreased, it can accelerate, adding force, but staying within the limit. It might help to push off the ground with a spring, coming or going, to not add the force of the initial acceleration or final deceleration to the cable tension.The point remains: The notional plan is for more than 11 tons of hardware starting at EML-1 and then capable of infrequently lofting or lowering 50 kg elevators. This does not make practical sense.
Couldn't a 50kg elevator keep elevating new threads?
Quote from: baldusi on 09/20/2012 06:17 pmCouldn't a 50kg elevator keep elevating new threads?On Earth definitely. It is a little harder on the Moon.
{snip}The tether's about 3 km/s from LEO and 1 km/s from GEO.Delta V to LEO is about 9 km/s, so it's take about 12 km/s to get Zylon to the tether. Then the elevator would need to haul the material 160,000 km to EML1, the region of maximum stress.
Quote from: Hop_David on 09/21/2012 05:29 pm{snip}The tether's about 3 km/s from LEO and 1 km/s from GEO.Delta V to LEO is about 9 km/s, so it's take about 12 km/s to get Zylon to the tether. Then the elevator would need to haul the material 160,000 km to EML1, the region of maximum stress.Also the delta-v at EML-1 is about 0 km/s, it is a docking.
A robotic arm with a special grip should be able to berth to the cable. Delta-v LEO to EML-1 is 3.77 km/s.
Alternate thickening the cable above EML-1 and below EML-1. The ratio is about 3:1 with modifications for tapering.
Quote from: A_M_Swallow on 09/21/2012 11:15 pmQuote from: Hop_David on 09/21/2012 05:29 pm{snip}The tether's about 3 km/s from LEO and 1 km/s from GEO.Delta V to LEO is about 9 km/s, so it's take about 12 km/s to get Zylon to the tether. Then the elevator would need to haul the material 160,000 km to EML1, the region of maximum stress.Also the delta-v at EML-1 is about 0 km/s, it is a docking.This is like saying it's 42 kilometers to Tampa. From Where?? Unless you say where you're coming from, your statement is meaningless.
Quote from: A_M_Swallow on 09/21/2012 11:15 pmA robotic arm with a special grip should be able to berth to the cable. Delta-v LEO to EML-1 is 3.77 km/s.True. But Baldusi was asking "Couldn't the elevator keep elevating new threads?" I was showing the elevator could, but you'd need at least 3 km/s from LEO and that would deliver the new threads to a point well below EML1.
Quote from: A_M_Swallow on 09/21/2012 11:15 pmAlternate thickening the cable above EML-1 and below EML-1. The ratio is about 3:1 with modifications for tapering.It was easy for me to find papers for tapers to synchronous orbit. But I still haven't been able to put together a spreadsheet for tapers to L1 or L2. Where'd you get that 3:1 taper? A cite would be helpful.Pearson, Levin, Oldson and Wykes write "For the Moon, we can build a non-tapered lunar ribbon if the characteristic height is 275 km or more." bottom page 11 of this pdf
Quote from: Hop_David on 09/22/2012 12:10 amQuote from: A_M_Swallow on 09/21/2012 11:15 pmQuote from: Hop_David on 09/21/2012 05:29 pm{snip}The tether's about 3 km/s from LEO and 1 km/s from GEO.Delta V to LEO is about 9 km/s, so it's take about 12 km/s to get Zylon to the tether. Then the elevator would need to haul the material 160,000 km to EML1, the region of maximum stress.Also the delta-v at EML-1 is about 0 km/s, it is a docking.This is like saying it's 42 kilometers to Tampa. From Where?? Unless you say where you're coming from, your statement is meaningless.EML-1 to cable at EML-1.
QuoteQuote from: A_M_Swallow on 09/21/2012 11:15 pmA robotic arm with a special grip should be able to berth to the cable. Delta-v LEO to EML-1 is 3.77 km/s.True. But Baldusi was asking "Couldn't the elevator keep elevating new threads?" I was showing the elevator could, but you'd need at least 3 km/s from LEO and that would deliver the new threads to a point well below EML1.Quote from: A_M_Swallow on 09/22/2012 12:37 amYou can thicken a cable by adding at the 'middle' instead of the ends. Due to weird gravity and orbital effects the centre of mass is at EML-1.
You can thicken a cable by adding at the 'middle' instead of the ends. Due to weird gravity and orbital effects the centre of mass is at EML-1.
Quote from: A_M_Swallow on 09/22/2012 12:37 amQuote from: Hop_David on 09/22/2012 12:10 amQuote from: A_M_Swallow on 09/21/2012 11:15 pmQuote from: Hop_David on 09/21/2012 05:29 pm{snip}The tether's about 3 km/s from LEO and 1 km/s from GEO.Delta V to LEO is about 9 km/s, so it's take about 12 km/s to get Zylon to the tether. Then the elevator would need to haul the material 160,000 km to EML1, the region of maximum stress.Also the delta-v at EML-1 is about 0 km/s, it is a docking.This is like saying it's 42 kilometers to Tampa. From Where?? Unless you say where you're coming from, your statement is meaningless.EML-1 to cable at EML-1.EML1 would be part of the elevator. This is a little like saying it's 0 kilometers from Hawaii to the earth.
Quote from: A_M_Swallow on 09/22/2012 12:37 amQuoteQuote from: A_M_Swallow on 09/21/2012 11:15 pmA robotic arm with a special grip should be able to berth to the cable. Delta-v LEO to EML-1 is 3.77 km/s.True. But Baldusi was asking "Couldn't the elevator keep elevating new threads?" I was showing the elevator could, but you'd need at least 3 km/s from LEO and that would deliver the new threads to a point well below EML1.That point had better be the lunar surface, any where else and you are doing an aircraft carrier landing. The cable would be acting as both deck and catch cable. The cable is weak so if the speed difference between the point on the cable and a fast moving spacecraft is more than a couple of miles per hour the cable will snap. This is an instantaneous docking between two moving items.Quote from: A_M_Swallow on 09/22/2012 12:37 amYou can thicken a cable by adding at the 'middle' instead of the ends. Due to weird gravity and orbital effects the centre of mass is at EML-1.The region of maximum stress is at EML1. If it needs to be thickened, EML1 is where it'd need it the most. If you added to the elevator you would start at EML1 and build earthward and moonward simultaneously.Baldusi was asking if the elevator could deliver more threads to the elevator. The answer is yes. 3 km/s from LEO could deliver material to a point 160,000 km below EML1. Then the elevator could lift the materials 160,000 km to EML1.Of course the same materials could be delivered directly to EML1 from LEO and it would take 3.77 km/s as you say. But the question was whether materials to add to the elevator could be lifted by the elevator.Because of asymmetry of acceleration gradient, EML1 isn't in the middle. The strands dangling earthward from EML1 are longer. If the strand were of uniform thickness throughout, it would need to be a total of 290,000 kilometers instead of the 264,000 kilometers Liftport suggests. But I believe liftport plans include a counterweight at the end of that 264,000 kilometers.
QuoteQuote from: A_M_Swallow on 09/21/2012 11:15 pmA robotic arm with a special grip should be able to berth to the cable. Delta-v LEO to EML-1 is 3.77 km/s.True. But Baldusi was asking "Couldn't the elevator keep elevating new threads?" I was showing the elevator could, but you'd need at least 3 km/s from LEO and that would deliver the new threads to a point well below EML1.That point had better be the lunar surface, any where else and you are doing an aircraft carrier landing. The cable would be acting as both deck and catch cable. The cable is weak so if the speed difference between the point on the cable and a fast moving spacecraft is more than a couple of miles per hour the cable will snap. This is an instantaneous docking between two moving items.Quote from: A_M_Swallow on 09/22/2012 12:37 amYou can thicken a cable by adding at the 'middle' instead of the ends. Due to weird gravity and orbital effects the centre of mass is at EML-1.
To dock the two items have to be at the same speed and same velocity, anything else causes a crash (or a miss).The lunar elevator has zero relative velocity to an item next to it at only 2 places - the lunar surface and EML-1.
The spacecraft and cable laying machines will make a good addition to the counterweight.