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Robert Clark

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  • Location
    Philadelphia, PA USA
  • Interests
    Mathematics, Astronomy, Physics, Space Travel, Futurism, Science Fiction
  • Favorite Area of Science
    Space Travel
  • Biography
    I teach math at an east coast US university.
  • Occupation
    University Professor

Robert Clark's Achievements


Quark (2/13)



  1. Thanks for that. I had not heard of that before. Bob Clark
  2. Two instances were discussed in the film where unmanned cargo ships were mentioned to send up supplies to extend the time Whatney or the crew of the Hermes could survive. This brought back painful memories while watching the movie. It's such an obvious answer. In the space shuttle Columbia disaster NASA rejected a possible rescue because Atlantis could not be readied in time within the 16 days the Columbia's supplies would run out. So since "nothing could be done anyway" there was no need to do accurate imaging to even find out if the wing damage was survivable. Note in the movie they had to use China to do the resupply mission to the Hermes. If NASA had ordered the Columbia imaging, finding the damage unsurvivable, all the space-faring nations in the world, which are at least five, would have been working hard to send up a cargo mission to meet up with Columbia within the 16 day time frame. But they never were even given a chance to try. BTW, since this is in regards to "The Martian", for you chemistry heads out there, are there some foods, liquids, or common materials that might be on the shuttles that could have filtered out the CO2 in air other than the lithium canisters? For instance perhaps the solubility of CO2 in water is different than in O2 and there could have been a way to separate out the CO2 from the air that way. Bob Clark
  3. *Suggestions for Water Production for California.* California is still in the midst of a huge drought: Annie Sneed Science 03.23.15 7:00 am California’s About to Run Out of Water. We Have to Act Now. http://www.wired.com/2015/03/californias-run-water-act-now/ There appear to be however relatively simple methods of providing extra water. One is by deriving water from humidity in the air, the other by distilling the water from the ocean. Both would appear to have relatively low cost solutions. First the humidity solution. The amount of water in the form of water vapor is substantial, especially at high humidity. A key fact is air can store more water vapor at higher temperatures. But the point is the areas in California with the highest drought level are the areas with routine high temperatures, such as Los Angeles. So lets calculate the amount of water in air in Los Angeles. This page gives a graph of average relative humidity levels in LA: Average Weather For Los Angeles, California, USA. https://weatherspark.com/…/Los-Angeles-California-United-St… You see that the relative humidity commonly reaches the 80% range and above, especially during the warmer months. The relative humidity is the percentage of the maximum possible water vapor the air can hold based on that temperature, called the saturated vapor density. This page gives a calculator for the saturated vapor density based on temperature: Relative Humidity Calculation. http://hyperphysics.phy-astr.gsu.edu/hb…/kinetic/relhum.html For a temperature of 80 degrees F, the saturated vapor density is given as 25.4 gm/m^3. And for a relative humidity of 80%, this corresponds to an actual amount of water in the air of 20.3 gm/m^3, 0.0203 kg/m^3. *Residential Solutions.* Some ceiling fans can move quite large amounts of air at a time resulting in quite large amounts of water vapor inflow at a time, if placed for example in residential windows. For instance some fans on this page at their higher speed settings can move 5,000 cubic feet per minute (CFM) of air while using only 30 watts: ENERGY STAR Most Efficient 2015 — Ceiling Fans 52 inches and under. http://www.energystar.gov/index.cfm… Lower speeds give better energy efficiency, but at the cost of lowered air flow amounts. We’ll use the high speed numbers since we want large air flow to get high water vapor inflow. At this high amount of hot air inflow though we would probably want the fan in a basement or attic since we would also need a window for the hot air, once the water is removed, to exit. How much water would need to be produced? An average person uses about 50 gallons per day of water: City Utilities: Water Tips. http://www.cityutilities.net/resident/pgms/watertips.htm So say a residence for a family of 4 needed 200 gallons of water per day. A gallon is 3.785 liters, so this is 757 liters per day. Water weighs 1 kilo per liter so this is 757 kilos per day. Then how long would it take for a fan blowing in 5,000 cubic feet per minute to bring in this much water(vapor)? At 5,000 cfm this is 5,000/3.28^3 = 141.7 meters per minute. Each of these cubic meters of air would contain 0.0203 kg of water vapor. So this would amount to 2.88 kg per minute of water. This would take 262.8 minutes, about 4.4 hours. This amount of power, only 30 watts used for 4.4 hours, is also quite small in energy costs compared to what the county of Los Angeles charges for water. An additional question to be resolved however is how can we convert this water vapor to liquid water? Air conditioners are able to do this, by accident, by chilling the air. Air dehumidifiers also commonly work this way. This causes water to condense out like happens for example in cool morning temperatures with morning dew. On our relative humidity calculator page, at 80 degrees F and 80% relative humidity, the dew point is only 73.4 F. However, both air conditioners and air humidifiers use quite high power levels. We want to minimize additional power used. Some possibilities: 1.)If the water produced this way is an adjunct to the water received from the city, then we can use the cool water coming from the city water supply, typically around 50 degree F, to cool this air and get the water to condense out. 2.)It would be nice though, since the calculation showed this air-produced water alone is sufficient to supply the entire household water needs, to find a way that didn’t use the city water supply. a.)Another type of air dehumidifier uses desiccants to absorb water vapor. The desiccant material is then heated to release the water as liquid and the same desiccant is used over again. However, this material typically is a silica compound and you would not want remnants of this to be left in the water. This also uses additional power for the heating step. If a desiccant could be found that is a type of mineral you would normally see for example in spring water then this might work. You would though need to find a way to get the water to be released as a liquid. Heating as with air dehumidifiers would work. However it may be at the high temperature of southern California would be sufficient so this would happen naturally. b.) A similar possibility derives from the fact that rain droplets can frequently condense in the air out of water vapor at temperatures higher than they would normally do by having nucleation sites: Cloud condensation nuclei. http://en.wikipedia.org/wiki/Cloud_condensation_nuclei Then we could add nucleators into the air stream to get the air to condense. These nucleators though again would have to be a non-toxic if ingested. Ideal would again be some type of mineral commonly found in mineral water. c.)To get the water to condense we could also expand the air flow. Rapidly expanding the air would cause the temperature to drop thereby chilling the water. A problem here though is the air flow is so large it might require an unreasonable size of expansion needed to get the needed temperature drop. d.)Another possibility would be by increasing the pressure of the air. Just as increasing the pressure increases the temperature at which water makes the transition from liquid to gas, the boiling point, so also the temperature at which it makes the transitions from gas to liquid, the dew point, also increases. This page gives a calculator for how the pressure changes the dew point: Dew Point Conversion Calculator. http://www.howelllabs.com/…/dew-point-conversion-calculator/ Enter in 73.4 degrees F in the known dew point field for our 80 degree F and 80% relative humidity scenario. Enter in 0 for the “psig” field, which measures how far this is above standard pressure in psi. Then a psig of only 4 gives a dew point of 80.3 F. That is an increase in pressure of less than 30% results in the water condensing out. We might be able to generate this amount of extra pressure by circulating the air around in a circle by centrifugal force. A problem though is the size of the air stream coming from a large ceiling fan size diameter might make the circle size needed impractically large. We could constrict the air coming from the fan into a smaller pipe diameter, but by the Bernoulli principle this would reduce the pressure. It still may be possible though that some combination of restricted pipe diameter and circulation diameter size could provide the needed pressure change at a practical size. *Water Utility Solution.* For the water company using this method to produce extra water it becomes particularly simple. The large ceiling-fan type fans discussed in the residential solution only move air in the range of 3 to 4 miles per hour. But wind speeds commonly are above this speed especially in coastal areas. Then you would not even need to use fans for the water utility solution. You would just collect the air driven by the wind in large tubes for processing. Also, for California communities near the coast, using the ocean to supply the necessary cooling to condense the water vapor becomes especially simple. *Individual Water Production.* For a scenario where one is stranded out in the wild or in a life raft without water, the amount of power needed for the fan is so small that it could probably be generated by hand for a single individual just for enough water to sustain life. And considering the wind speed needed is only in the range of 3 to 4 mph a fan might not be needed at all. Also, at least for an 80 F and 80% relativity humidity scenario, the 30% extra pressure needed to allow water to condense could be easily supplied by hand. By the Ideal Gas Law PV = RT, to get a 30% increase in pressure we would need to make less than a 23% decrease in volume, assuming we did the compression slowly so as not to increase the temperature. For instance, a piston in a foot long cylinder would only need to make a 3 inch compression to get the needed pressure. The life boat case would also be very simple because the condensing could instead be done by using the cool temperatures of the ocean water. Bob Clark
  4. Deep Space Industries discusses the in space use of asteroid resources here: A valuable asteroid passing by this week February 13 2013 05:55:27 AM | by Clark Lindsey, Managing Editor https://www.newspacewatch.com/articles/a-valuable-asteroid-passing-by-this-week.html Bob Clark
  5. We agree costs are not linear with respect to delta-V. They are approximately linear with respect to the size of the rocket. Try using the rocket equation on a delta-V of 3.1 km/s and and Isp of 450 s. You'll find the mass ratio is about 2. This is how much bigger a rocket has to be to get the same size payload to escape velocity rather than just to LEO. Bob Clark SpaceX has said the cost for the Falcon Heavy will be in the range of $1,000 per pound, which is ca. $2,000 per kilo. You are aware of the fact that scaling a rocket up should make the costs cheaper, which is why there was support of the idea of a "big dumb booster" to cut the costs to space. I've seen some of your creative ideas for space on this and other forums. Suppose there were a big commercial push for space mining. Then we would need lightweight means of doing the processing of the precious metals. Then how would you do this task: separate out the expensive platinum group metals from the common metals such as iron, nickel, etc. and from the surrounding rock. Keep in mind there is little gravity on the small asteroids. How would you do it? Bob Clark
  6. Actually the cost estimate is taking into account the rocket equation. What Heinlein's statement was based on was that after you get to LEO to get to escape velocity you need an additional 3.1 km/s delta-V. By the rocket equation, using efficient hydrogen-fueled engines this about doubles the size of your rocket for the same size payload. So Heinlein was not saying the delta-V doubled, he was saying your rocket size doubled, which approximately doubles your cost. We agree his numbers aren't accurate in either direction. GIGO, as they say in computer science. A more relevant mining comparison to mining the asteroids would be like the large bucket excavators used in strip mining, since the valuable minerals are distributed uniformly throughout the asteroid. For the Mars rovers you were doing fine instrument measurements, not excavating large mass. Bob Clark
  7. A science-fiction film from 1969 also expressed the idea that asteroid mining could only be profitable by bringing the asteroid to the vicinity of the Earth, so I suppose this is a view that has long been expressed. The film was "Moon Zero Two": Moon Zero Two - Wikipedia, the free encyclopedia The film was a rather low budget endeavor, still it was enjoyable for a genre fan. It can be purchased on Amazon.com. However, you can see it for free on Youtube if you don't mind the Mystery Science Theater 3000 side comments: MST3k 111 - Moon Zero Two. BTW, the two "Deep Space Industries" videos I mentioned are at: Deep Space Industries. (sales video) Deep Space Industries Live Announcement. Bob Clark Yes, I noticed that too. Front-end loaders, dump trucks, etc. can last for decades with maintenance. This means many thousands or hundreds of thousands times greater mass can be mined than their mass. Bob Clark
  8. You would certainly need to have an undergraduate level understanding of math and physics. But if you major in those fields you should be able to understand it. Bob Clark
  9. At $1,500 per once, that's $24,000 per pound, $52,800 per kilo. Launch costs to LEO are about $10,000 per kilo now. Robert Heinlein famously said once you reach Earth orbit, you're half way to anywhere in the Solar System. So we can estimate getting to an asteroid as twice that cost. If you take the cost to get back as about the same then this total is still less than $52,800 per kilo. Also, SpaceX expects to offer launch to orbit at $2,000 per kilo on the Falcon Heavy, which improves the economics even further. Another company has announced asteroid mining intentions: Asteroid-Mining Project Aims for Deep-Space Colonies. by Mike Wall, SPACE.com Senior Writer Date: 22 January 2013 Time: 12:01 AM ET http://www.space.com/19368-asteroid-mining-deep-space-industries.html This is good news. My opinion is that space mining will prove to be the "killer app" that will make space flight routine. More companies entering the field will increase competition, and increase innovation. This will serve to advance the speed at which such a venture can come to fruition. Deep Space Industries. (sales video) Deep Space Industries Live Announcement. I like the optimistic approach taken in these videos. It's the idea espoused by Peter Diamandis that the upcoming times will be a period of abundance, not of need. Bob Clark Edit: The system won't allow youtube links. You can find the videos on youtube at the above titles.
  10. NASA administrator Charles Bolden told the NRC committee on human spaceflight that an asteroid mission didn't necessarily have to be a far trip: Bolden: Don't Have to Travel Far to Asteroid to Meet President's Goal. Marcia S. Smith Posted: 19-Dect-2012 http://www.spacepolicyonline.com/news/bolden-dont-have-to-travel-far-to-asteroid-to-meet-presidents-goal Perhaps he was referring to the Planetary Resources, Inc. proposal to bring a small asteroid to lunar orbit. But another possibility is a mission to near Earth asteroids that can be accomplished in about a month round trip travel time. See the table of NEO's here: Near-Earth Object Human Space Flight Accessible Targets Study (NHATS). http://neo.jpl.nasa.gov/cgi-bin/nhats Select max delta-v <= 12 km/s, visit time => 8 days, unlimited visual magnitude, the H parameter, and unlimited orbital uncertainty, the OCC parameter. Then there are several asteroids at 26, 34, and 42 day travel times, including stay times at or above 8 days. If you subtract off that stay time to make it only a day or so then the round trip travel time will be in the range of a month or so. This could serve as an intermediate step for BEO missions between the Apollo missions at max. 12 days and a Mars mission at 6 months one-way travel time. Bob Clark
  11. "Golden Spike" revealed their architecture for a commercial return to the Moon Dec. 6th: How Golden Spike's Moon Landing Plan Works (Infographic ) http://www.space.com/18805-golden-spike-private-moon-landing-graphic.html They estimated development costs in the $7 to $8 billion dollar range, less than 1/10 the cost of the Apollo or Constellation programs. However, even these numbers may be over inflated. The origin of the presented cost numbers were from NASA guys using NASA costing models. However, SpaceX has shown by following a commercial approach development costs can be cut by 1/5th to 1/10th that of NASA’s. So what I think Golden Spike should do is bring SpaceX on board. With the development costs reduced to this extent, then we would have the really exciting possibility of the flight costs being brought down perhaps to the $200 million range, especially if using the Falcon Heavy launcher. This clearly would have a major impact on the prospect of profitability. The only problem might be is that Elon appears to have no interest in the Moon, being focused on Mars as the ultimate goal. However the profitability motive may sway him. There is also the fact that these missions could serve to prove the capabilities of the Dragon even for BEO missions. It could also serve to prove the value of the Falcon Heavy for launching large payload at low cost, something Elon definitely wants for getting Air Force contracts. As I discussed here the importance of what SpaceX has accomplished is that it will make clear that manned space flight can be accomplished at a fraction of what was thought necessary, thus making manned space flight routine world-wide. Combining this with small, low cost approaches to BEO flight, suggests such missions can also happen on a regular basis. We are returning to the Moon, this time to stay. Bob Clark
  12. Just saw this: Exploration Alternatives: From Propellant Depots to Commercial Lunar Base. November 15th, 2012 by Chris Bergin http://www.nasaspace...ial-lunar-base/ I first thought the commercial plan was going to follow the Early Lunar Access (ELA) proposal because it mentioned landing two commercial passengers on the Moon. ELA was a lightweight architecture that used a small two-man capsule: Encyclopedia Astronautica. Early Lunar Access. http://www.astronaut...ft/earccess.htm But it is unlikely in the commercial plan they mean the passengers are to fly alone without one or more professional pilots. And also the article mentions the commercial plan is to use on orbit assembly. But by using the Falcon Heavy or the SLS you could launch the ELA architecture with a single launch. Still, using two launches of the Delta IV Heavy both at its maximum payload to orbit of 25 mT we could launch the ELA architecture. Even if the Delta IV Heavy is not man rated, we could use separate launchers to take the astronauts to orbit and transfer them to the Moon vehicle after it is assembled. For the NASA proposal, the article mentions the Lunar Surface Sortie (LSS) proposal. But this was still to use a 4 man capsule, which likely means the large, heavy Orion. It also would involve a separate lunar crew module, also at variance with the lightweight ELA architecture. This lunar lander of the LSS proposal would then likely be akin to the large, expensive Altair lunar lander. So this proposal would be similar to the Constellation program whose high expense caused it to be cancelled. Better would be if NASA went small following the ELA architecture to use a single, small capsule that would carry the astronauts all the way from LEO to the lunar surface and back again. This would allow a NASA return to the Moon with a proportionally small additional cost above that of the SLS itself, and in less than a decade. These commercial or NASA missions, if carried through, would allow a return to the Moon by the 50th anniversary of the Apollo missions if not of Apollo 11 itself. Bob Clark =================================================================== Just saw this article by legendary Apollo manager Chris Kraft mentioned on the NasaSpaceFlight.com forum: Space Launch System is a threat to JSC, Texas jobs By Chris Kraft and Tom Moser | April 20, 2012 | Updated: April 20, 2012 8:20pm Quote: We are wasting billions of dollars per year on SLS. There are cheaper and nearer term approaches for human space exploration that use existing launch vehicles. A multicenter NASA team has completed a study on how we can return humans to the surface of the moon in the next decade with existing launch vehicles and within the existing budget. This NASA plan, which NASA leadership is trying to hide, would save JSC and create thousands of jobs in Texas. http://www.chron.com/opinion/outlook...bs-3498836.php Since Kraft is opposed to the SLS and he says this plan uses existing launch vehicles, it can't use the SLS or the Falcon Heavy. It must then use something similar to the Early Lunar Access plan that uses orbital assembly, perhaps using two launches of the Delta IV Heavy. Like the suppressed report that suggested orbiting propellant depots could accomplish the goals of the SLS at lower cost, this report will eventually also come out. So whose got the inside scoop? Bob Clark
  13. Still, I haven't heard about that propellant you mentioned for your boosters. If they do provide better performance than the solids then you should propose their use to NASA. Remember another advantage of liquid fueled boosters is that those in the industry feel comfortable with them for manned flight since they can be turned off. Bob Clark
  14. Thanks for that. I looked at your proposal. I couldn't find there how much it requires to LEO. The Early Lunar Access(ELA) proposal only required 52 mT, which means it can be launched by the Falcon Heavy or SLS. Note also the cryogenic stages required for the ELA's space traverse and lunar landing already exist, which cuts majorly into development time and cost. About your launcher, it uses both new boosters and new core stage, quite expensive. But NASA does want to explore the possibility of using new liquid fueled boosters on the SLS and will be providing grant funds on this. Perhaps you could write a grant proposal for your new boosters for this advanced booster program for NASA. Bob Clark
  15. The argument for why this is doable is rather simple. The Early Lunar Access(ELA) proposal of the early 90's, which deserves to be better known actually, suggested that by using a lightweight 2-man capsule and all cryogenic in-space stages that a manned lunar lander mission could be mounted with only 52 mT required to LEO, half that previously thought necessary. The only technical complaint about its feasibility was that it required a crew capsule of only 3 mT empty weight. But the kicker is NASA is planning a Space Exploration Vehicle(SEV) at that same low 3 mT empty weight. So the SLS at a 70 mT payload capability will be able to launch such a mission using the SEV as crew capsule following the ELA architecture with plenty of margin. This would give the SLS a definite mission and in a short time-frame, less than a decade. No longer would it be referred to as a "rocket to nowhere". Bob Clark
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