Duh -- huh??? • m*******e@a****.***u.edu 03/08/1996 00:00:000 UTC I don't know what you're getting at with this stuff about linear absorption of radiation. The point is that the load bearing ability of the support structures is what allows internal open spaces to exist in the ship without collapsing under the stress of accelerations. Since Star Wars & similar shows have ships with obvious engines at the back of the ship, this consideration applies to the Death Star & such. It would be intersting to look up the maximum theoretical strength of various alloys and compare them to the torque on Voyager's nacelles & saucer. I still think that they'd break off. I don't know why the crew can't just beam down like in earlier series. The ship is obviously not designed for planetary landings. Yeah, you can talk about additional support from force fields & such but since it is well established that Federation equipment doesn't work in the presence of electromagnetic radiation I feel free to ignore them. See for example the crappy DS9 episode (there are so many!) where Sisko & O'Brien (of the Inner Party -- just kidding) are stuck on the planet of the Luddites (who, note, have invented many primitive devices of their own and are thus hypocrites). Just as bad is this nonsense about not sensing ships over a planet's magnetic pole. Why not station people at windows with telescopes? If only they hired REAL SF writers like in the first series they'd have an infinitely better show. Oh, yeah -- I do have Space: 1889, which is quite amusing. I'll have you know that I personally invented a better ether propellor that Thomas Edison's! I am currently playing a game of Macho Women With Guns and trying to figure out some of the subtleties in the rules. --- Brian ------------------------------------------------------------------------------- Direct all criticism to m*******e@***.com This is an equal opportunity net. We do not discriminate on the basis of race, gender, or sexual orientation. Please list your: Race _____________ Gender ___________ Sexual orientation ______________ ------------------------------------------------------------------------------- Internet autograph service 1. Insert book into laser printer, opened to the title page 2. Click on the button below This week's guest writer: Cordwainer Bird [Start] • j********o@***.com 04/08/1996 00:00:000 UTC whoops! g(r) = g(R)*r/R! (was thinking in terms of g(depth), I suppose) Also, I think I integrated wrong! (Ed.: also, uncertain of bearing area, but I'll correct old math first.) so we integrate (r^3/R) * (4 Pi Rho g(R)), getting the much simpler form of Pi Rho g(R) (R^4 - r^4)/R If we divide by A(r) = 4 Pi r^2, and rearrange, we get P = Weight/Area = g(R) Rho (R^4 - r^4)/(4 R r^2), if my math has no further errors. We can reformulate in terms of Rho, g(R) = 4 Pi G Rho r/3: P = (Pi G /3) Rho^2 (R^4 - r^4)/Rr^2, _If_ all my math is correct. Pi G/3 = 7 E-11 (MKS) With contemporary materials, max load >~= 200 Kpsi = ~ 130 ATM >= 10 E10 N/m^2 Assuming Rho = .1 kg/l (100 kg/m^3) at 1/10 the density of water and if the central void (or bottom deck) has a radius/height of 10 km, the max R we can have with contemporary materials is about 1100 km (1010 km). Coolness. Now this all floats on a raft of assumptions (that I have the right definition of strength, that this is the right area over which the pressure acts (perhaps the cross section of the bottom deck is more apt?)), and I haven't checked the math, but it's inneresin, anyway. If the deck cross section is the correct area (which, come to think of it, seems more sensible), things are different. The weight remains (Pi Rho g(R)) (R^4 - r^4)/R, and expressing here in terms of Rho this time, F (Weight) = 4 (Pi Rho)^2 (R^4 - r^4)/3. Approximating A as > 2 Pi r*t (t = thickness of bulkhead), call A ~= 7 r t: P <= .2 (Pi Rho)^2 (R^4 - r^4)/(r*t) Pi^2*G = 6.6 E-10, so P <= (1.3 E-10) Rho^2 (R^4 - r^4)/(r*t). Using our same density and a bottom deck thickness of 1 km, I get a max R of ~560 km for contemporary materials, which is what I was assuming about the Death Star. Still cool, assuming there aren't any further errors (this is hot off the presses, so to speak). I'm still considering what the correct area is... maybe the first idea was right..., since the deck can't squish up due to the symmetry involved... Gorno "No, I haven't got a chance: There's no silver lining... I'm far beyond the point of doing it right. But I can still strike the match, That will start the canyons shining: Might see something new in that light. I'll do this, I'll do that, I'll be burning canyons for you. I'll do this, I'll do that, I can't wait for Heaven, I won't be there." • j********o@***.com 04/08/1996 00:00:000 UTC In article <4tutlu$***@g*****.***u.edu>, m*******e@a****.***u.edu (Brian McGuinness) writes: >I don't know what you're getting at with this stuff about linear absorption >of radiation. If you actually wanted to do this indirect support with light, say, you could have a stack of translucent decks. The thing is, each deck would absorb some fraction of whatever light passed through it, so upper decks would get less light, so would take up less momentum, and not keep up. (The light extinction is exponential.) If you make the upper decks increasingly more opaque, they will absorb the same amount of light out of a dimmer beam. (The light extinction is linear.) (As an amusing aside, this harkens back to Planetary Atmospheres, which I think we all took at the same time! Hmm, maybe Brain was the TA...) Using super-science, you might or might not have to do something similar. >The point is that the load bearing ability of the support structures is what >allows internal open spaces to exist in the ship without collapsing under >the stress of accelerations. Direct load-bearing needs to be transmitted over a surface area, but mass is proportional to volume, hence the runaway superstructure requirements. Using a suitable indirect method to transmit weight or impulse seems to get around this problem. >Federation equipment doesn't work in the presence >of electromagnetic radiation I feel free to ignore them. True enough, especially since we see their ships get the crap beaten out of them, after losing power, without crumpling. Naturally, if your Death Star is held up by neutrinos, and the neutrino generator can and does give out, it's BAD. (Another idea along these lines is a reflective ring or shell under each deck, where huge numbers of photons bounce along the inner surface (only), transmitting momentum from one side of the surface to the opposite, resulting in a self-supporting shell. Assuming this whimsy is sound, photons would appear to offer a means of applying pressure only to the shell or ring above, not to that below. There might be some analogous way of using superconducters in a similar fashion.) But considering the stresses on a large naval vessel, I don't think the materials strengths needed for Star Trek, even "Blunderer," are really that large. Consider a sub: the modern ones really can rear up out of the water, and slap down without rupturing. Ahh, found my book! The example given is for Copper. The ordinary (if I read this correct) weakest crystaline axis has a shear strength of 140 psi. A refined calculation gives a theoretical strength of 1 million psi. Steel be at least five times this, based on shear strength. The (1974) max strength measured for tiny (.001 inch diameter) whiskers, single crystals nearly devoid of dislocations, is near 400,000 psi. Some (non-crystaline) steels in here appear to have ultimate strengths greater than 200,000 psi; iron whiskers are listed at up to 2,000,000 psi. I'm too lazy to actually figure out the (purely self-gravitational) stress in a Death Star, and I can't quite see how to deal with the center geometry (hmm: maybe the question is, what is the size of the void space at (or _nearest_ ) the center: its outer surface, proportional to r squared, is that over which the pressure must be supported. And likewise for each level above (above any solid core). If the Death Star is hollow and thin, there's only a minimal "root space" problem; if it's solid, there's no surface to support; so the question is always dependent on the specific lay out of the decks concerned.) Ok, maybe I'll do the math some time after all. We want to integrate g(r) * dM from outer to inner surface to get the weight on the latter. (hmm, Rho*gh? I may be overlooking a term) Assuming uniform density and a surface g of .01 (from earlier), g(r) = .01*(1 - r/R). dM = rho * dV = Rho*4*Pi*r^2*dr: so we integrate the whole beasty, which is an r^2 - r^3, resulting in an r^3 - r^4, but from larger to smaller r (note, the g(r) term is decreasing, so the subtracted term is never larger than the positive, i.e., size won't levitate it for us!) This gives the weight borne by any particular shell's inner surface area. So now, do the work, I'm too tired (I assumed R = .1 R Earth (+), Rho = .1 Rho (+): other values for Rho can easily be introduced, modifying g proportionately). There's surely an equation directly relating radius and deck height to necessary strength, and maximum available materials strength and deck height to maximum radius of Death Star. >See for example the >crappy DS9 episode (there are so many!) where Sisko & O'Brien (of the Inner >Party -- just kidding) are stuck on the planet of the Luddites Actually, I liked that one: it exposed the natural extension of their PC mentality. The end was what sucked: the mob would have wanted to lynch that bitch, and if it had been Old Trek, Kirk would have had to step in: "You're civilized people again; citizens of the United Federation of Planets, not her little tribe of barbarians. Don't sink back to the level she trapped you in!" And off he would have carted her, and her pansy son, to a high tech reeducation center where they can eat synthetic food for the rest of their lives, and watch all the TV they want! "I'm sorry, prisoners are not allowed real books. Have some more synth-ale." Gorno • j********o@***.com 05/08/1996 00:00:000 UTC Hadn't considered the possibility that each shell can support itself (which is the cross sectional bearing case I think). The spherical surface is the case where beams are used to support upper levels (haven't yet doped out whether there's a middle ground... maybe there, each shell need only support its own weight that is not above the pylons... need to think further: the cross-sectional issue may have to do with individual decks only, that is, the assumed structure within a deck). Brian's probably more concerned with acceleration than self-gravity, but this is a problem that has always tempted me, and I've finally bothered to look into it. One might also spin one's Death Star, to reduce stress and render the situation into a more conventional building problem, perhaps... Gorno Mustn't sleep! They get you in your dreams! Cthulhu Faughn Ryleh! Gentle Gods of Earth, I am falling into the sky! Yiiieahhhh!