tag:blogger.com,1999:blog-3596550435682943926.post4585374987359781208..comments2024-08-22T06:13:55.957-07:00Comments on Hop's Blog: Beanstalks, Elevators, Clarke TowersHop Davidhttp://www.blogger.com/profile/12923433894475072056noreply@blogger.comBlogger13125tag:blogger.com,1999:blog-3596550435682943926.post-63349380005653754552015-07-27T12:53:04.873-07:002015-07-27T12:53:04.873-07:00Symeon, I don't mind if you delete your commen...Symeon, I don't mind if you delete your comment. In any case, it's your comment, you don't need my permission.<br /><br />I would like to maintain contact with you, though. As I said I like the numeric methods in your spreadsheet more than what I use. Would it be possible to send me the spreadsheet as a Microsoft Excel document? My email is hopd@cunews.info<br /><br />As I mentioned, I want to do some blog posts looking at tether mass to payload mass ratio. Would you mind if I mention you? With your permission I want to use your spreadsheet as my tool for looking at various scenarios. Although there are a few changes I'd like to make to the sheet. Hop Davidhttps://www.blogger.com/profile/12923433894475072056noreply@blogger.comtag:blogger.com,1999:blog-3596550435682943926.post-52330110567151102862015-07-27T09:34:17.366-07:002015-07-27T09:34:17.366-07:00Thank you so much; that was one number I didn'...Thank you so much; that was one number I didn't have much confidence in.<br />I updated with your value, or rather calculated the same value from the orbital period of Phobos. I get a taper ratio of 59.89, so it looks like our approaches are pretty close (within 2%). The 'up' tether values are less amazing now but still feasible.<br />It turns out that most rockets use a safety factor of 1.25 to 1.4; that might be reasonable to use for a prototype or experiment but I'd still like to see at least 2 for a structure that will catch or launch humans; I've set the starting safety factor to 2 for all tabs.<br /><br />I corrected errors that resulted from the bad value I used and also added the gravity of Phobos. My idea of a tether launch to the gas giants turned out to be unreasonable, so I've removed that tab. I also broke up the three main forces (Mars gravity, Phobos gravity, centripetal force) into their own columns for clarity.<br /><br />Would you mind if I deleted my comment on the 24th? It is based on bad data and could be misleading.<br /><br />You and anyone else are welcome to use the sheet, formulas, ideas, etc. with or without attribution. I declare it to be in the public domain and absolve myself of any responsibility for damage or loss resulting from its use.<br />It could be adapted to other bodies with some effort. It's not perfectly accurate but I think it stands as a good validation of the formula used by your sources.Chris Wolfehttps://www.blogger.com/profile/11247630943891521469noreply@blogger.comtag:blogger.com,1999:blog-3596550435682943926.post-71053902321565649832015-07-25T06:02:26.733-07:002015-07-25T06:02:26.733-07:00Symeon, I've taken a look at your spreadsheet....Symeon, I've taken a look at your spreadsheet. I like that you've set tether thickness of a step by looking at total newtons from step lengths below. A more direct approach than the expression I use for thickness: e^(complicated exponent)*e^(horribly complicated exponent).<br />Going over it I agree with most quantities except cell B7. B7 is supposed to be Phobos' angular velocity in radians per second? You have 2.3192e-5 where I get 2.28e-4. <br />Except for cell B7, I like your spreadsheet much more than my awkward effort. I am hoping you'll change that cell and see if your numbers agree with mine.<br />My spreadsheet didn't have any safety factor. Pearson and Aravind use a quantity h they call characteristic height. To introduce a safety factor of 3, I've divided h by 3. With a safety factor of 3, 5.8e9 pascal tensile strength, 1560 kg/meter^3 density, foot radius - 3.697e6 meters, balance point radius 9.379e6 meters, I'm getting a taper ratio of about 61.Hop Davidhttps://www.blogger.com/profile/12923433894475072056noreply@blogger.comtag:blogger.com,1999:blog-3596550435682943926.post-22377045334280223452015-07-24T11:35:46.230-07:002015-07-24T11:35:46.230-07:00Payload tension is highest when the payload is jus... Payload tension is highest when the payload is just leaving Phobos. I adjusted the sheet to test for max payload tension at each step, so each step of the tether is sized for the worst-case tension. The 'up' tether is still very achievable.<br /><br />In early stages the tether would only be needed during arrival windows from Earth or launch windows from Phobos, so it could be retracted for the two-year idle period to extend its lifespan. A '4-year' tether could last over 40 years by doing this.<br /><br /> Using Zylon fiber and a safety factor of 3, the taper ratio is 11.34 and the mass ratio is 3.22 for the 7,980km tether and can launch payloads with a Vinf up to 3.27km/s. A 10.6t tether system could catch multiple 2.5t payloads from Earth as long as they arrive one orbit (~7h40m) apart and could return payloads at the same rate. A single 3.3t payload every three days could also be sent or captured. A tether of that mass could (in theory) be launched to Phobos on a single Falcon 9H with 2.5t of margin for deployment equipment. Costs for such a mission could be under $300 million and could return tens to hundreds of tons of Phobos samples over the next few launch windows. Three payloads per day, 6t (excludes 500kg climber allowance), times the launch window (0-60 days depending on orbits); this system wouldn't be able to return samples until 2026 but that year has a launch window two months wide for this Vinf. That one window could return 360 tons of mass from Phobos to Earth orbit, roughly a 6.8:1 mass advantage over the 53t to LEO from the single launch.<br /><br />Other windows where sample return to Earth isn't feasible would still allow the tether to catch a series of exploration vehicles and then relay them to the asteroid belt at the appropriate times. Four or five vehicles of 2.5t could be sent in a single Falcon 9H launch and then deployed throughout the belt; with the right instruments this fleet could form a VLBI array around 5.5AU in diameter for incredibly high resolution radio astronomy. This leg of the program could be done for perhaps $700 million with immensely useful science return including asteroid and comet orbit determination, spectroscopy, parallax distance measurements of stars and many others.<br /><br /> Using the same parameters with a 10,000km tether gets you a taper ratio of 14 and mass ratio of 3.75, with a maximum Vinf of 4.0km/s. That would widen the 2026 return window to almost 100 days / 600 tons but would require an SLS mission or multiple commercial launches. It would also open up launches to main belt objects during bad alignments, overall increasing its usefulness. Let's call it about $600 million, or $1,000 per kg of Phobos samples. Even if half the 'sample' mass is heat shield and containment that's still $2,000 per kg, cheaper than launching that mass from Earth.<br /><br /> That's all well and good, but what if we get really ambitious? A tether 31,330km long earns you a Vinf of 9.27km/s, which is enough to inject directly into a trans-Neptune Hohmann transfer orbit. This powerful tether would handle the same 2.5t / 3.3t payloads as the other two for less than 18t with a taper ratio of ~480 and mass ratio of ~5.4. This is only two Falcon 9H launches or one SLS launch; call it about $700 million for the tether system and the same $700 million for four spacecraft and you could send an orbiter to all four gas giants for around $1.4 billion, with follow-on craft every 2 years for $175 million each. We would have to be careful of Deimos. Launch windows vary from 1.9 years for Neptune to 2.2 years for Jupiter.Chris Wolfehttps://www.blogger.com/profile/11247630943891521469noreply@blogger.comtag:blogger.com,1999:blog-3596550435682943926.post-5672073963136571422015-07-23T14:39:45.678-07:002015-07-23T14:39:45.678-07:00Thanks for the reply. I appreciate all the effort ...Thanks for the reply. I appreciate all the effort you've put into collecting references and spreading information, both here and in the many space-related forums where you post. Between this site, Wikipedia and Project Rho an enormous depth of knowledge can be obtained.<br /><br />I added a tab for the 'up' tether and also columns for V, Vesc and Vinf at each step.<br />It probably needs some additional thought, since making the tether longer seems to reduce the tension. Perhaps the max tension for that tether occurs when the payload is at the surface, rather than at the end of the tether.<br />One thing to consider: not all trips from Mars to Earth require the same Vinf, so it may be worth making the tether another 5,000km longer (~22,000km or almost to Deimos orbit) pushing the maximum Vinf over 4km/s in order to widen the launch and capture windows to and from Earth.<br />I will revisit my sheet when time allows, but for now consider the 'up' tab preliminary.Chris Wolfehttps://www.blogger.com/profile/11247630943891521469noreply@blogger.comtag:blogger.com,1999:blog-3596550435682943926.post-5219256212064896592015-07-23T13:53:34.960-07:002015-07-23T13:53:34.960-07:00Symeon, The expressions for tether width I get fro...Symeon, The expressions for tether width I get from Aravind's and Pearson's PDFs. If I remember right, those PDFs are linked to in the above blog post.<br /><br />They use many steps and I have to admit I'm no wiz at integral calculus.<br /><br />Just a couple of weeks ago I constructed a spread sheet where I take an approach similar to yours (Breaking the tether into small lengths and summing them). I'm pressed for time and energy (as always). I hope to look at your spreadsheet soon. It will probably be helpful when comparing to my own efforts.<br /><br />In a future blog I hope to look at tether mass to payload mass ratios.Hop Davidhttps://www.blogger.com/profile/12923433894475072056noreply@blogger.comtag:blogger.com,1999:blog-3596550435682943926.post-89941860374659901832015-07-23T10:10:30.130-07:002015-07-23T10:10:30.130-07:00Hi Hop,
I'm having a lot of trouble followin...Hi Hop,<br /><br /> I'm having a lot of trouble following the math for tether taper ratios and masses for the Phobos tethers. Some references seem to be assuming that the force of gravity is uniform along the length of the tether and others simply say they arrived at their result through numerical integration without explanation.<br /><br />Can you tell me where I'm going wrong, or provide your procedure for arriving at a taper ratio and (hopefully) a tether mass based on payload?<br /><br /><br />I have a <a href="https://docs.google.com/spreadsheets/d/12Lfb8cHC782L6OmhotOoHFTPEYFWO1SEPL87De_e4xI/edit?usp=sharing" rel="nofollow">spreadsheet</a> demonstrating my method down to 10km steps using Zylon, but my results are dramatically different than most references. The taper ratio and mass ratio should be static regardless of payload and tether mass; they are after accounting for the station mass, but I still get a taper ratio of 276:1 and a mass ratio of 1,233:1 tether to payload. If I change the safety factor to 2 I get a taper ratio of 43:1, still several times larger than expected.<br /><br />This is what I'm trying to do to get a numerical approximation that will fit in a spreadsheet. I'll walk through the process in words:<br /><br />For the Phobos tether, assume the maximum tension is at the surface of Phobos and that the gravity of Phobos can be ignored. This is inaccurate but only by a very small amount. We also neglect any force due to the atmosphere of Mars. Minimum tension is at the foot, where force due to the cable mass is 0.<br />Step 1: Force due to any capture hardware at the foot is (gravity at foot m/s² - centrifugal acceleration at foot m/s²) * foot station mass kg = force N (Newtons or kg*m/s²); this force is static.<br />Step 2: Force due to the maximum payload mass at the foot is equal to (gravity at foot m/s² - centrifugal acceleration at foot m/s²) * payload mass kg = force N; this force is also assumed to be static.<br />Step 3: These forces must be resisted by the tether, whose required cross-section is: force N / material strength Pa (which is N/m² or kg/ms²) = cross sectional area m². Most sources want at least a factor of 2 for safety, but let's assume a factor of 3 to allow for strength loss in splices / joints, etc. Short form is now f / (strength / 3) or 3f/strength.<br />Step 4: Since this is a numerical approach, I need to find the mass of a short length of the cable based on the cross section found in the previous step. The mass of this length of cable is cross section m² * step length m * density kg/m³ = mass kg. Add this to the total tether mass for tracking purposes.<br />Step 5: The force produced by this section of tether is (gravity at section m/s² - centrifugal acceleration at section m/s²) * step mass kg = step force N. Add this to the total force for the next cycle.<br />I repeat steps 3, 4, 5 using a new radius r = old radius + step length until the new radius reaches the surface of Phobos.<br /><br />Please feel free to copy the sheet and play with the numbers. Hopefully it is useful, but based on my results so far I also hope I've made a mistake somewhere and the real-world values will be better.Chris Wolfehttps://www.blogger.com/profile/11247630943891521469noreply@blogger.comtag:blogger.com,1999:blog-3596550435682943926.post-27391473543458318522015-01-12T21:02:38.973-08:002015-01-12T21:02:38.973-08:00Hi David,
Re pumping fluids up a Ceres elevator:
...Hi David, <br />Re pumping fluids up a Ceres elevator:<br />Yes, you could not vacuum the export water from orbit, it would have to be pumped up the elevator from the surface. I have a column of water at Clarke orbit as having a base pressure of 76MPa ~ 10,000 psi. That is high pressure, but less than the Mariana Trench and well within off the shelf industrial design. You can buy water pumps online that can pump up to 36,000 psi. The pipe would have to be insulated and heated, but that technology was already done with the Alaska oil pipeline.<br />CheersJoyhttps://www.blogger.com/profile/07024219782322596271noreply@blogger.comtag:blogger.com,1999:blog-3596550435682943926.post-86496305065636581012015-01-04T15:27:50.185-08:002015-01-04T15:27:50.185-08:00Thanks, Joy. A Phobos elevator could release a pay...Thanks, Joy. A Phobos elevator could release a payload into mars atmosphere at a speed of .6 km/s wrt to mars surface. Someone from Mars surface could rendezvous with the tether foot with a small suborbital hop. A mass driver from Ceres couldn't do this. Power requirements for the mass driver would be huge and require a massive power source. An elevator on the other hand could provide delta V by borrowing from Phobos' orbital momentum.<br /><br />You are correct a Ceres elevator has low enough stress you could have electric cables along its length. Thus the power source for the climbers becomes much less of an issue.<br /><br />I believe gases and fluids would be a major export for Phobos but I'm wondering how a pipeline along the elevator would work. I don't think you could use a vacuum pump to lift the fluids and gases to orbit.<br /><br />Quite a few asteroids have a healthy angular velocity. I believe elevators may be helpful for getting around the Main BeltHop Davidhttps://www.blogger.com/profile/12923433894475072056noreply@blogger.comtag:blogger.com,1999:blog-3596550435682943926.post-91628235241934956732015-01-03T20:06:52.064-08:002015-01-03T20:06:52.064-08:00Thanks for the link on Centauri Dreams. Good work...Thanks for the link on Centauri Dreams. Good work! I agree that Luna, NEAs, and Phobos are more accessible near term than Ceres. But I still think Ceres would be the sweetest and lowest hanging fruit for a Clarke elevator. The Phobos tether is pretty cool for tossing things around, but building a mass driver on Phobos to do the same thing might be easier.<br /><br />Tensile strength not being a problem at Ceres would also allow bells and whistles such as a (mostly) nonstructural HV power cable from a solar array in Clarke orbit to the surface. (actually steel conductors wrapped in kevlar could be structural at Ceres) Climbers could draw power from a cable as well. Finally, a pipeline to the surface could expedite import/export of gases and fluids.<br /><br />Cheers, JoyJoyhttps://www.blogger.com/profile/07024219782322596271noreply@blogger.comtag:blogger.com,1999:blog-3596550435682943926.post-47948812250772391422014-11-21T07:51:07.605-08:002014-11-21T07:51:07.605-08:00There are ways ot deal with coriolis induced oscil...There are ways ot deal with coriolis induced oscillations. An elevator car traveling up will push the tether in a retrograde direction, a car traveling down will push prograde.<br /><br />By timing ascent or descent, oscillations can be dampened.<br /><br />And a tether end swinging like a pendulum isn't necessarily bad. At mid swing the end is traveling fastest with regard to its anchor mass. Releasing a payload at midswing could be a way to enjoy some extra delta V. Also releasing a payload mid swing would take some of the momentum out an oscillation.Hop Davidhttps://www.blogger.com/profile/12923433894475072056noreply@blogger.comtag:blogger.com,1999:blog-3596550435682943926.post-51391998706952532552013-11-17T19:34:39.762-08:002013-11-17T19:34:39.762-08:00Alan, I'm embarrassed to admit I don't kno...Alan, I'm embarrassed to admit I don't know much about coriolis force. Googling it seems that it is indeed a concern for gravity gradient stabilized tethers. I need to read up on this. Thank you for your comments.Hop Davidhttps://www.blogger.com/profile/12923433894475072056noreply@blogger.comtag:blogger.com,1999:blog-3596550435682943926.post-40771365690759332572013-11-17T16:52:06.639-08:002013-11-17T16:52:06.639-08:00I mentioned this post here:
http://imgur.com/gall...I mentioned this post here:<br /><br />http://imgur.com/gallery/6cdch<br /><br />In short, the Phobos throw seemed dicey to me. It's not instantly obvious how you could dangle something on that tether without it swinging around everywhere. But with certain assumptions, it's clear that it CAN be done, even if there will be significant deflection of the tether. Logistically it would still be difficult, or at least rather involved.AlanSEhttps://www.blogger.com/profile/08119889774421336682noreply@blogger.com