Yet Another Weird Idea for Space Transportation

[rogers_buck]

Moon Chucker. MC is a tower built at one of the poles of the moon. At the top is a giant motor with a boom attached and a variable counter weight. There is a sturdy cable that deploys from the end of the boom and the boom and counter weight are rotated by the monster motor. So, if you want to go to mars a crane loads your space ship onto a track on the boom and you connect to the cable. The motor starts to rotate the aparatus and when it achieves enough RPMs that the parabolic sag as the force of lunar gravity is balanced by the centrifugal force the spaceship drives off the track and starts to get winched out by the cable. Around and around the whole thing goes as the cable plays out for a few kilometers and the rotation rate is steadilly increased by the electric motor. Then, when the desired linear velocity is acheved, the spaceship is released from the cable and goes soaring off to mars. A similar rig on Phobos will send it back to earth when it is time to go.<br /><br />This would be a cheap way to send billots or other cargos back to earth and could also be adapted to capture payloads from earth.<br /><br />To capture a payload, the above scenario is reversed.<br />

 

[vogon13]

If it would work, it would work most anywhere on the moon. Carefully adjusting the time and speed of release would send your cargo most anywhere you want to go. You could even fling toward Venus or Jupiter to utilize a gravitional boost to further open up places to reach with this.<br /><br />Note the material strength of the cable is going to be a big challenge.<br /><br /><br /> <div class="Discussion_UserSignature"> <p><font color="#ff0000"><strong>TPTB went to Dallas and all I got was Plucked !!</strong></font></p><p><font color="#339966"><strong>So many people, so few recipes !!</strong></font></p><p><font color="#0000ff"><strong>Let's clean up this stinkhole !!</strong></font> </p> </div>

 

[nexium]

A variation I saw, had the giant motor at the top of the highest polar mountain. this allowed the ribbon = cable to turn at lower RPM without hitting other polar peaks. This has the advantage that the motor does not need to be slowed as much, nor the ribbon retracted between pay loads. The pay load can just slide along the cable until it reaches the release point. The slide speed is added vectorally to the cable speed, reducing stress on the ribbon and the giant motor, plus alowing more frequent pay load tosses. Neil

 

[publiusr]

How about we just support heavy-lift and chunk Rube Goldberg nonsense?

 

[josh_simonson]

A lunar space elevator would work better, and it would drop you off at LL1, from which it is easy to leave the earth/moon system.

 

[rogers_buck]

And here I thought I had come up with an original idea... Oh well, probably read it here years ago and forget about it.<br /><br />These are some pretty simple calculations to determine the gee force loading on the vehicle as a function of tether length and tangential velocity. I won't run the numbers here because I think it would be a good exercize for some of our younger members who are keen on space.<br /><br />As a side calculation, they might want to consider the problem of an astronaut with a bola. Ignoring for a moment the need to move the arm faster than the arm can move, what would be the terminal velocity of a released bola weighing 40 grams assuming the astronaut could hold on to 50kg force in his gloved hands?<br />

 

[rogers_buck]

Rockets can't realize the kinds of economies needed to import/export from the moon. Even on the earth, damned few modes of transportaion use rockets. So in essence, the rocket is the Rube Goldberg contrivance...<br /><br />A similar setup could be made using the natural topography of the moon and mag-lev technology. Imagine a large crater several 10s of kilometers in diameter. A mag-lev rail is laid down inside the rim of the crater, supported by the crater wall. A payload is circulated around the crater rim track to build up its linear velocity. Sort of a storage ring as is used in particle accelerators. When the desired velocity is attained the payload + mag-lev carrier are switched out of the crater to a linear launch track. The payload is then release from the mag-lev carrier (which is then braked) and goes flying off into space.<br />

 

[rogers_buck]

Humm, I was thinking about yet another variation that has an extremely interesting aspect - crack the whip...<br /><br />Imagine the giant motor rotating an extremely long and magnetic cable. The cable is a few km long and is rotating at a high RPM rate in the vacuum. The payload is initially sitting at the axis of the motor-cable and is attached at the leading edge of the cable. Ok, the crew is throwing up violently, so let's say its just billots.<br /><br />When the cable system achieves the desired RPM the payload activates a mag-lev and imparts an initial motion outward along the cable. The a correolis force will impart and acceleration to the payload fueled by the angular momentum of the entire cable system integrated over the entire length of the cable. The whip will be cracked and the cable will only require sufficient tensile strength to hold together its own mass plus a small margin.<br /><br />For anyone who isn't familiar with the crack of a bull whip, the crack is caused by the tip of the whip breaking the sound barrier. This scheme has great POTENTIAL (pun intended).<br /><br />A perfectly crafted system like this would result in the rotating cable being stopped in its tracks with the entire energy being transfered to the surfer payload. It's huge!<br /><br />Is this the nexus of the variation you were speaking of?<br /><br />

 

[rogers_buck]

Y'know, this technique would work in Earth orbit too. The conductive cable could be set rotating with respect to a counter-mass perhaps using a solar-electric manipulation of an electric field on the long cable - using the earth's magnetic field as part of the motor. When the cable was spun up to the desired total energy the mag-lev payload would surf the cable and steal it's energy.<br />

 

[scipt]

Maglev launches from the moon of any sort are a very sensible idea. No air resistance, low G plenty of solar power, or build a nuclear reactor. <div class="Discussion_UserSignature"> </div>

 

[nyarlathotep]

>"Imagine a large crater several 10s of kilometers in diameter. A mag-lev rail is laid down inside the rim of the crater, supported by the crater wall. A payload is circulated around the crater rim track to build up its linear velocity."<br /><br />Wouldn't it be easier just to bore the track though the sides of the crater? Your payload processing facility and most of the track is going to have to buried in regolith for micrometeoroid protection anyway, why not make the track a bit longer and dead straight?

 

[vogelbek]

I like the lunar maglev launch system concept. Its extremely elegant, requires no propellants, and would be (relatively) easilly serviced by any population living in the nearby vicinity.<br /><br />I'm guessing that the acceleration forces required for a reasonably short track would be too high for live payloads though.

 

[scottb50]

One problem I see with the crack the whip idea is the speed of sound has no bearing on the moon, or in Space. A better idea would be a simple rotating cable built on a lunar mountain top, with no air resistance it should look similar to a propeller. Attach a payload at the hub, spin it up to speed, release the payload down the cable and centrifical force would fling the payload off the end at the appropriate time and in the right direction.<br /><br />Definitely not Rube Goldberg, most of his stuff would probably work. I think it closer to British engineering. I used to fly Jetstreams and the nose gear had what should have been a simple squat switch that did about thirteen things and was so convoluted in design that it probably weighed half what the entire assembly did. One good thing is it weighed so much they had to embrass modern technology and use chemical milling of the skins to reduce the overall weight to keep it under 12,500 pounds. Eventually they gave up completely, when they could get bigger engines and the 3100 and 3200 worked pretty well, if you discount the brakes, definitely a Rube Goldberg design.<br /><br /> <div class="Discussion_UserSignature"> </div>

 

[rogers_buck]

Well Soctty, I won't have you as engineer on my star ship. The "crack" reference to speed of sound is merely an earthly mundane example of the phenomena.<br /><br />What your proposing is the same thing. When the payload slides down the cable coriolis force pushes it against the cable. This creates a torsion wave which builds in amplitude as the payload accelerates. When the payload leaves the tip it has robbed most of the angular momentum from the whip. That is the crack.<br /><br />Think about a balerina with long skinny arms doing a perroette (sp). When she extends her arms she slows down here spin. Same dynamics only in our scenario here arms were already outstretched and her shoulder bracelot slips loose riding on the kink in her arm as it snaps bone and goes flying off and kills a judge. 1-1-2-1-0<- dead judge's don't rank.<br /><br />Where the speed of sound does come into play is that it is a factor in calculating the translation velocity of the torsion wave. Faster than sound the bone not only breaks in the balerina's arm, it gets hot.<br /><br /><br />

 

[mdodson]

With a lunar escape velocity of 2.37 km/sec http://home.tiac.net/~cri/1999/moon.html , my math [d = ve**2 / a ] shows that a mag lev track using a leisurely 2g acceleration would be 140 km long. For a 4km long track, the acceleration would be 70g's. Billets only, please.

 

[barrykirk]

aha, but to get one half of lunar escape velocity would at say 4g's would require.<br /><br />d = v^2/a<br /><br />v = 2.37km/sec for escape velocity, but 1.66 km/sec for<br />lunar orbital velocity.<br /><br />For a = 4 gs<br /><br />So, now d = ( 1660 m / sec ) ^ 2 / ( 9.8 m / sec * 4 )<br /><br />d = 70 km.... That's better but not good enough.<br /><br />If an oribtal tether with a tip velocity of .75 km / sec is used to provide a boost ( this should be relatively easy to acheive ).<br /><br />Now initial velocity required is 1.66 km /sec - 0.75 km/sec or 0.91 km / sec.<br /><br />Plugging back in <br /><br />d = (991 m /sec )* ( 991 m /sec ) / ( 39.2 m /sec )<br /><br />d = 25 km.... That might just be do able.

 

[rogers_buck]

Huzzah! Someone on this forum actually did a calculation! By so doing they nailed down one end of free floating speculation with the tether of reality. Well done MDobson.<br /><br />Now let's talk about the physics of your answer MDobson. <br /><br />distance = velocity * time<br /><br />taking the second derrivative with respect to time<br /><br />distance = (1/2) * acceleration * time^2 [1]<br /><br />We want chimpanzees to be able to not just survive this ride but to have cocktails, so let's cap acceleration at just one Earth g's (9.8m/s^2).<br /><br />We also know (thanks to MDobson's research) that we want to be going 2.37km/sec when we come off the track.<br /><br />velocity = acceleration * time<br /><br />2.37km/sec = 9.8m/sec^2 * time<br /><br />time = velocity/acceleration [2]<br />time = 2370m/sec/(9.8m/s^2) = ~242sec<br /><br />distance = (1/2) * 9.8m/s^2 * (242s)^2 = 287km track<br /><br />and at MDobson's 2g acceleration ~ 140km as indicated.<br /><br />Combining [1] and [2]<br />distance = (1/2) * acceleration * (velocity/acceleration)^2<br /><br />gives<br /><br />distance = (1/2) velocity^2/acceleration [3]<br /><br />So even though MDobson did the calculation correctly in his head, he attempted to mislead us by giving us a formula that was off by a factor of two! As a result many have perished by crashing into the far side of the moon. (-;<br /><br />Let's see how large a circular track would be required to produce the velocity MDobson so kindly researched for us lazy people.<br /><br />First, we know there will be three types of g forces the chimps in our maglev will have to tollerate in a circular track. 1) Fg_lunar; 2) Fg_accelerator; 3) Fg_centrifugal. The moon fixes 1), we can stick with our abbove calculation 1g limits for 2), then we can solve for 3).<br /><br />r = c/(2pi) = 140km/2*pi ~ 22km.<br /><br />The maximum centrifugal force an obese lunarian would be subjected to would then be<br /><br />Fc = mv^2/r [4]<br />Fc = (1kg*2.37km/sec^2)/22km ~ 256gm/sec^2<br />g_equiv = (256g

 

[mdodson]

Yep, I screwed up the transposition of that formula. Thanks for catching my error!<br /><br />Let's try another one: For centripetal acceleration, I show a = omega**2 / r as opposed to your v**2 / r. Setting omega = v / (2 * pi * r) and using the moon's escape velocity, I end up with a = .101 / r in km**2 / sec**2. So I end up with a 10 km radius crater to keep a primate happy. And the 2.26 rpm might not mess his inner ear up too much.<br />I'm launching off of here to see if I can find a way to show formulas on the web. (laughing)

 

[rogers_buck]

I must confess, your answer feels more "right" than mine, but...<br /><br />Centripetal acceleration is given by:<br /><br />a = omega^2 * r [1] - buzzz, units cm/s**2<br /><br />cycletime = (2pi*r)/v [2]<br /><br />omega = 2pi / cycletime [3] in radians/second<br /><br />omega = v/r [4] combining [2] and [3],<br /><br />a = (v^2/r^2)*r = v^2/r combining [1] and [4]...<br /><br />You killed a few more monkeys today, there is a nice plaque for them on the rim of MDOBSON Crator. (-;<br /><br />I still think there is something wrong with my result, but that ain't it...<br /><br />

 

[mdodson]

But at least on the moon, the monkeys I killed won't stink. I expect letters from several schools rescinding my passing grades for my previous reply.<br />I don't see anyway to include equations in their traditional form in messages, and it's darned hard to put them on a web page without a specialized package.

 

[rogers_buck]

Don't feel bad. You stepped up and ventured a calculation which added tonnes of reality to this wildly speculative thread. For what we do here orders of magnitude are sufficient to determine if something is loony or possible. What usually happens here is that speculation is met with more speculation about engineering details rather than the basic physics. That's kind of fun too, but once the a handle is gotten on the basic physics a wild idea becomes even more coherent in the minds eye.<br /><br />Now let's thicken the plot. We have what appears to be a correct result (even though I still don't like it) for a maglev on a circular track. This same result also applies to a payload on the end of a cable being slung around. The difference is only in where the centripetal force comes from. From the crator wall for the track and the cable for the latter case...<br /><br />So here's a question to ponder. Given the monkeys can survive a 1km radius rotation and that one earth gee would make the hyperbolic cable slack clear the surface of the moon, what would the tensile strength for the cable need to be for a 10,000kg payload?<br />

 

[semiliterate]

Ummm ... monkeys would not be able to survive a 1km radius rotation.<br /><br />Centrifugal acceleration is given by<br /><br />a = v^2/r<br /><br />Escape velocity for the moon is: v = 2,369 m/s<br /><br />and g = 9.81 m/s^2<br /><br />Therefore the radius required for 1 gee centrifugal force is:<br /><br />r = v^2/a<br />r = (2,369 m/s)^2/(9.81m/s)<br />r = 572,086 m<br />r = 572 km <br /><br />For a 1km radius rotation, you would get<br /><br />a = v^2/r <br />a = (2,369m/s)^2/(1000 m)<br />a = 5,612 m/s^2<br />a = 572 gees

 

[rogers_buck]

Ah, good job. My turn to kill some monkeys. But I think my way of squishing them was much more sinister... No crator, no plaques.<br /><br />Looks like I missed the k in 2.37 km/sec. I knew my answer was wrong but I couldn't spot it.<br /><br /><br />I also like your way of calculating the radius for the gee force. Much cleaner than my train of thought method.<br />

 

[rogers_buck]

Hey guys, were not done here. Now we have to do a sanity check on material strength or they won't fund us. Ok, they won't fund us anyway, but at least we'll know how off the mark these ideas are.<br /><br />1) We now know thanks to several memebers contributions that a circular maglev track with a radius of a few hundred kms can accelerate to lunar escape velocity while serving drinks.<br /><br />We know that the same kinetics apply to a payload on the end of a cable which was the original topic of this thread. The question that begs answering is how thick a cable would it take to fullfill this fantasy.<br /><br />The next tool we need to develop is the means of calculating the total centripetal force the cable itself would have to withstand.<br /><br />We know that the mass of the cable is rotating at constant angular velocity in radians/second, but the cable itself is at distances from 0 to the length of the cable. We need to integrate these tiny steps to derrive the definite solution for a cable of arbitrary lenght, radius, and density.<br /><br />m(x) = pi*r^2*x<br />Ac(x) = omega^2*x<br /><br />So who wants to go first?<br />

 

[barrykirk]

This idea certainly does have promise in that a cable would certainly be cheaper, lighter, and require less raw materials than a maglev track.<br /><br />You would need a short distance of track to get things started. Kind of like an airplane needs a runway to land and takeoff, but not in between.