Response to Derek, about the bow of the ribbon
Yesterday, for the first time in almost a year, I posted to our blog… more on why the delay, later. If you read yesterday’s post, you’ll see that a lot is coming.
I suspected that our blog was ‘dead’. Dead, in the sense that there was little or no traffic on it, considering the state of the project. Still, I am working to resurrect LiftPort v2.0 and so naturally, I posted a message about the work/school in Spain.
But I was surprised by a comment we got from Derek. Now, I don’t know Derek. But I thought he had some good thoughts. Rather than post them in the comments, I chose to write another blog, to answer in more detail. Here’s his text, in italics, with my response.
Michael,
I have often wondered why you have ignored the hard questions posed in the comments, but today you have explained this in your claims to stubbornness and strength, which can lead their owner into a myopic focus on their goals.
Two things. First, it’s not that I’ve been avoiding the ‘hard questions’. It’s that I’ve been avoiding ALL the questions. All of them. You read for yourself that I almost quit the whole project. The last thing I wanted to deal with was questions – hard or easy ones. So, don’t feel neglected, I wasn’t ignoring YOU, I was ignoring everyone. Face it, you were not my biggest priority. I came as close to declaring bankruptcy as a person can, without actually filing for it. This project cost me everything that I own, and left me deeply in the hole. Answering comments and questions from fans seemed liked a pointless waste of time, considering the other problems I was dealing with.
Second, I am stubborn. I am back in the game, and will not waiver again – you can count on it. I’ve been tested, and know what I can do. That stubborn streak of mine will be an asset (most of the time), Derek, but I admit it’s caused problems, too.
So, now that I’m back, I will start once again to respond to people. To give you perspective, there were over 31000 comments pending approval on the blog. Now, we both know that the vast majority were spam, but there is no way I am going to wade through to find the few nuggets of actual value. So, I’ve deleted all the comments of the past year, and am starting fresh.
If you sent a comment that you absolutely must have answered, please send it again. I’ll try and get it answered.
Myopia is a form of blinkers which are fine to ensure focus but only if someone has a firm grip of the reigns with clear sight of the whole situation.
Derek, you’ve nailed it. I certainly didn’t have a clear focus on either the current situation, nor the ongoing operation. I made a lot of mistakes that, in hindsight, seem obvious. I think I’ve cleared up this particular problem, and don’t think it will plague the project, or I, again.
In your case, the only people with clear view are your commenters,
Well, I disagree with this last point, but don’t feel the need to argue about it. Moving on.
so here goes again with two questions fundamental to the function of the SE.
First on the ascent, where do you propose to gain the increased angular momentum from to prevent the SE from being dragged backwards and downwards as the line is bent at the load point.
Good question. I wish I were exactly the right guy to answer it, but I’m not. I will do my best. However, please understand, with the near-failure of the team about 18 months ago, most of the technical people that were my staff have moved on to other jobs. That said, I still have access to some really great resources, and will do what I can to get your question answered.
In a VERY rough sense, there is a constant tension/tug of war between the gravity pulling the ribbon down, and the centripetal acceleration that is pulling the ribbon out and away from the earth. While often depicted as a ‘straight line’ the ribbon actually has a very significant bow in it, ‘down’ near the earth side of the system. That bow is thousands of miles long, but that is a direct effect of the ‘dragging backwards’ that you are asking about. The key is to maintain a semi-balanced system, where there is more ‘outward’ tension, than there is ‘downward’ pull. This is a factor of how big a counterweight, and how far out the weight is, in relation to the earth.
The center of mass of the system changes constantly, based on the load of the Lifter robots, as well as the position of the Lifter at any given time. See this lesson from one of our experiments, http://www.youtube.com/watch?v=QjcaQT_ssE4 as a great illustration. (make sure you read the “more info” section, it’s got useful details. It is important to remember that this is a dynamic system.
Second (which also comes into play with the first) how do you plan to shed the angular momentum of the payload when the load descends the SE, because if you do not shed the force, then it will push the SE forwards, lowering it and putting much more of the SE into a lower orbit and therefore traveling too fast for its orbit – it is a feedback loop which finishes with the cable wrapped forward around the earth.
Derek
Keep in mind that the ribbon is about 60,000 miles long. With that much ‘energy’ pulling the ribbon out, and away from the earth, it would take an enormous force to slow down the overall orbit. A lifter returning back to earth is not near enough to make a significant shift. It simply isn’t big enough to make a big, (there is a small) difference. It is the overall center of mass that is the key, as long as that remains at geosync ‘speeds’, then the lifter descending will have a minor effect.
I hope I’ve shed a little light on the problem. Sorry it’s not a more complete, technical answer, but it’s the best I can do tonight. Thanks for your interest in the project.
Take care. Mjl
www.twitter.com/mlaine
November 22nd, 2008 at 2:50 am
Hi Michael,
Thanks for the substantial reply. Sadly, there does not seem to be much traffic here yet, perhaps folks dropped the RSS feed as I nearly did on several occasions, so it might take a little time to build up the readership again and while it does, I guess I get you all to myself for a short while.
To the problem then. Yes, I appreciate that the SE is BIG both in length and in mass and that in comparison the elevator is small. However, this by itself does not negate the effect I described, because the effect is a positive feedback effect involving the whole of the SE including the elevator. I am not saying that the problem is not solvable, just that it has not been addressed in anything I have read to date. Of course, I might have missed something and be completely wrong, but until it is discussed I will assume the issue exists and in need of a solution.
To describe the problem a little better, let me simplify the SE by assuming it to be a non elastic ribbon, 60,000 miles long, tethered to the equator and standing in equilibrium straight out from the earth with the elevator at the GS point ca 22,000 miles out.
At this location, the elevator would have an orbital speed of ca 6,900 mph.
Now for the sake of clarity, let us move the elevator down the SE by say 2,000 miles. At this point the SE ribbon would have a stable orbital velocity of ca 6,400 mph giving us a net unresolved energy of the mass of the elevator times the square of 222 m/s (500 mph) – roughly 50 KJ for every kilo of elevator mass.
This will create a force pushing the SE ribbon FORWARD – i.e. in the direction of rotation. Now on a ribbon 60,000 miles long, even a few ounces of force is going to be able to move the ribbon due to the massive (almost infinite) mechanical advantage a lateral force has on a straight cord Sin (0) = 0 so you are dividing by zero to calculate the force vector needed to oppose the lateral force. The consequence is that even a small lateral force will push the SE ribbon forward until it has deflected the ribbon sufficiently for the sine of the angle created is sufficient to allow the tether tension to equal the lateral force.
But the SE is tethered to the earth, so by pushing it forwards, we have in fact rotated it about its tether point and in doing so, every point along the remaining 20,000 miles has moved fractionally closer to the earth and all of the tether beyond the elevator has also be pulled down by this same small amount.
So all of the elevator is now a little lower, and consequently it is moving just a little too fast for its new orbital height and so it ALL starts to move FORWARD relative to its equilibrium position. Now two little compounding effects check in.
First, because the SE is moving a little faster forward, it has a slightly less restraining ability on the elevator, which moves forward a little more and second, because the SE is now travelling slightly ahead of the earth and the earth is playing ‘catchup’ its force on the end of the cable is reduced, meaning its component of the Sine theta vector calculation is diminished and so the elevator moves forward yet a little further.
Nett effect is that the SE is now in a slightly lower orbit than its speed requires and so it is travelling slightly faster than needed for a perpendicular SE. But the whole SE is tethered to the earth, so it all rotates slightly and so it all moves down again slightly further compounding the forward speed anomaly. It is a positive feedback system and quickly escalates to a situation where the SE attempts to wrap itself forward around the earth.
This is a situation involving the whole SE, not just the elevator and it is the exact opposite of the elevator ascending problem where the elevator drags the SE backwards and this is compensated by drag at the anchor point to drive the necessary acceleration into the travelling elevator, but the descent of the elevator is the opposite with excess energy stored in the elevator and a positive feed back instead of a negative feedback.
I believe this is a real problem and if it is then I think I know how to solve it, but I would very much like to hear from you and your experts as to the actuality of my concerns.
Derek
November 24th, 2008 at 5:40 pm
Good to see you back, Michael. Like Derek, I had dropped my RSS feed at work, and was about to do the same at home, when I saw the *beep*
I will cogitate on your problem, Derek. I can’t claim to be an expert, though.
November 25th, 2008 at 7:00 pm
Welcome back Tony. I’ll try and post a lot more often. As the project is re-invigorated, I think there will be more activity here.
Derek, I can’t answer your question any better than I did the first time. I am certain you’ve got some errors in your assumptions, but as I said before, I am not the guy to tackle this problem. So if you will bear with me, I’ll hunt down someone that can give you the sort of answer you deserve. Clearly youve put some time and thought into this, and you’ve earned a solid reply.
Of course, you’ve also raised my curiosity by your last paragraph…
Where do you live? Maybe this is better as a conversation?
Take care.
Michael Laine
twitter.com/mlaine
November 26th, 2008 at 5:27 am
Hi Michael,
Oh I would love to take this up in conversation, but I am afraid that I live on the other side of the pond in Lincolnshire, England, so I will have to resort to pounding the keyboard for the time being until you are very wealthy and can afford to employ me on the team (Ho Hum).
One advantage of the keyboard route, is that the text is reasonably permanent, so more thought goes into its production and it remains in place to be reread as many times as needs be (and by as many folks as are interested).
Tony, good to see you here – and then we were three.
Re the solution, there is no point in posting what I believe might be the basis for a solution if there in actuality is no problem. I will wait to see if others think the forward momentum of a descending SE is a problem.
Derek
November 26th, 2008 at 8:29 am
An easy solution, given the simplistic design that Derek proposed, would be to place a crawling counterweight above GSO that moves up as the elevator crawler moves down, and vice versa. It wouldn’t have the same mass as the elevator car, so it would need to crawl at a different speed and to different heights in order to maintain the center of mass at GSO.
November 26th, 2008 at 12:18 pm
Derek – your numbers seem to be off, but more importantly, you seem to have the basic physics not quite internalized. The ribbon is under tension, with the bottom end tethered to the ground and the outer end (outside of geosynchronous) straining to break free and fly off, because of its high speed relative to orbital speed at that distance. That tension makes the system as a whole mechanically stable. When you perturb the ribbon, for example by moving the elevator up and down, both ends of the ribbon are pulling it back into an essentially straight up-and-down configuration. It may vibrate a bit, depending on its mechanical properties, but it won’t fall down without some really massive disturbance. The system is best understood in a rotating reference frame with competing gravitational and centrifugal forces – just as gravity is trying to pull the bottom half of the ribbon down to the ground, the centrifugal force of the rotating frame is pulling the top half of the ribbon outwards, and the two effects lead to a stable balance as long as the ribbon is sufficiently long (and strong enough).
November 26th, 2008 at 2:52 pm
Small world, Derek. I’m originally from Lincolnshire as well! Based in Melbourne these days. (Neither locations are particularly suited to SEs!)
Thinking about your problem:
Simply being placed below the stationary orbit imparts no forward force in itself (that is experienced as the elevator moves up or down: as it descends/ascends, the elevator will be pushed by the ribbon so as to make it slow down/speed up. The ribbon is in turn, sped up/slowed down due to the transfer of kinetic energy). However, elevator will then be experiencing a slight downwards force from uncompensated gravity. That could be handled by counterbalances moving out.
I think your point about flexing in the ribbon causing the whole thing to sink is true, but that neglects conservation of angular momentum. I think the net effect will be a perturbation into a slightly elliptical orbit rather than a death spiral.
(Take it from someone who hasn’t dabbled in orbital mechanics for thirty years! BTW, even if Michael’s been inactive of late, the forums on this site still appear to be moderately active)
November 28th, 2008 at 9:18 am
Derek: Arthur has it nailed (thanks Arthur!) Does this make more sense? The Ribbon is in the solar systems largest game to tug-of-war, and as long as the system remains in balance, it will stay up.
Tony: You are right, the forums have remained reasonably active, despite my disappearance. I am hopeful that they will become even more active, as my own activity ramps back up.
Arthur: Welcome back. I’m still plugging away, with a greatly revised business plan, new resources, and a strengthened resolve. It is gonna be hard, but it’s worth it. Thanks for chiming in and helping me answer the question above. (I guess that is the difference between having a PHD in physics vrs me just being an amateur hack!)
All, are you guys on twitter.com? Follow along if you want to get more involved.
Take care. mjl
President, LiftPort Group
http://twitter.com/mlaine
November 30th, 2008 at 3:50 pm
Arthur Smith;
Thanks Arthur for the correction, yes indeed I had left off the factor of 0.5 because the energy due to the unresolved velocity is equal to 0.5 mv^2 , so the unresolved energy is down to ca 25KJ per Kg mass of the elevator.
However, it really does not matter if the facto is 25, 2.5 or 0.25, the example chosen was meant only as a reference to establish the processes involved and the events.
The key elements are that the SE is tethered at its base and is held out taught by the end counter mass and that every single part of the whole SE has the same angular momentum which means that every single part having a unique distance from the centre of the earth has its own unique orbital velocity, and that each piece will attempt to maintain that velocity unless some force is involved to counter it.
Yes, the SE is under tension created by earth anchor and excess gravity on the part below GS orbit and excess centripetal force on the mass of the SE beyond the GS point (accepting that all parts of the SE are subject to various levels of gravitational force and centripetal force), with the part of the SE at GS orbit being under the greatest tension from these two opposing forces. Yes, those forces will be very large, but they will not be infinite and that is the force needed to prevent a lateral force from deflecting the SE in the direction of that force.
Reading the various posts, everyone seems to be accepting that the elevator at the new lower elevation will be travelling faster than the part of the SE it is attached to and that this will mean the elevator will attempt to force the SE forward against the tension present in the SE itself at that point (a very high tension).
At the heart of this exercise, is the force vector diagram which states that the force needed in the SE cable to restrain a lateral force is equal to the lateral force divided by the cosine of half the angle subtended at the point of applied force. If the elevator cable is to remain straight, then even a tiny one pound force would require a tension of 1/Cos(90) which is infinity. No matter how large the two opposing forces are, they will never be able to stop the elevator from pushing the cable forward by some amount until the angle subtended at the point the elevator is pressing is great enough to bring the SE tension to equate to the excess forces created by the excess orbital velocity of the elevator car.
At this point you face the nub of my argument. If the SE is non-elastic (an assumption I cited in the first post) then pushing the SE out of the straight line has the effect of lowering every single part and molecule of the SE, each and every part of which will have a slightly higher orbital velocity at this new reduced height than is required for it to remain in a relative stationary orbit. No matter how tiny the loss in height, no matter how tiny the increase in forward velocity over that needed for stable vertical orbit, it will move forward en mass and in doing so, because it is tethered, it will be rotating forward about its fixed point on the surface and so will be forced into even lower orbit – POSITIVE FEEDBACK.
At some point the forward motion dragging force of the whole of the SE will overcome the ground anchor ability and the SE will be ripped out of the anchor and any further rotation about the anchor point will stop, the SE will fly off into some new orbit like a conker and its string.
As for the idea of moving a counter weight out as the elevator car moves in, yes, that will work once. But now you have to get that counter weight back down the SE for the next car descent and now the counterweight suffers exactly the same positive feedback problem that the elevator car itself produced – as you bring the counterweight down, it pushes the SE forward and again descent and forward acceleration are created.
November 30th, 2008 at 4:00 pm
To solve this problem, you have to demonstrate a means of storing orbital momentum, passing it into the elevator car on its ascent, and then extracting it back from the car on its descent, while leaving the SE in a nett orbital momentum neutral state.
Derek
November 30th, 2008 at 4:02 pm
Michael,
No I do not Twitter. Can you imaging trying to handle this issue on Twitter. Even this medium is barely adequate lacking the facility to post diagrams.
Derek
December 3rd, 2008 at 10:26 am
Re-reading Derek’s analysis, he is correct. The sin(0) is, indeed, 0. The corrective force of an ~1000 tonne mass traveling at 73km/s will have no effect at correcting a perturbation in the curvature of the ribbon. That will be true for as long as the ribbon is straight.
As the perturbation increases, however, the angle also increases. At 4 * 10 ^ -8 degrees, the 5,000,000 newtons of centrifugal force will be acting with an angle sufficient to equal the ~.2 newtons of force a 20 tonne climber will have ascending or descending at 200 kph.
I’m sorry I don’t have to documenting ability that Tom had, but feel free to double check my math.
December 4th, 2008 at 1:13 am
Perhaps it would be better if I used an analogy to help explain the problem.
If a skater is rotating with arms outstretched and they draw their arms in, then the angular momentum of their hands and arms is passed into their torso and they spin faster. Unless some external work is done on the skater (like putting out a foot to brake against the ice) the skater does not slow down.
Or try this one. A child on the outer edge of a roundabout rotating slowly. The child pulls itself towards the centre of the roundabout taking with it its excess angular momentum. Because the child is going too fast for its new ‘orbit’ it is forced against the rails of the roundabout and presses forward on the roundabout. The roundabout reacts to this force in the only way it can, it absorbs the acceleration and because it is rigid, it distributes that acceleration throughout its whole self and comes to a new equilibrium where the child and roundabout are now rotating at some faster speed.
In a forward moving event, the SE cannot do work upon the ‘inward’ travelling elevator car, so the acceleration due to the new inner orbit has to be absorbed by the SE itself. The tension in the SE allows it to redistribute that acceleration along its length, but the nett result is that as the car descends the SE is accelerated forward, and as it is tethered. the earth end cannot move forward, so the SE is forced to rotate forwards and in doing so it moves en mass down towards the earth and so aquires an additional excess angular momentum which results in even greater forward speed anomaly.
Consider a simple experiment. A top spinning on an essentially frictionless point in a vacuum. Attached to the edge of the top is a thread, half way along the thread is a conker and at its outer end is a conker and they spin around with the top. Now we do work on this system, using a battery and a motor, we drag the inner conker towards the top.
The outer conker tries to keep the thread taught and the inner conker passes its excess angular momentum into this stiffened cord. The cord attempts to respond by absorbing the acceleration, but because it does not have a rigid connection to the top (like the platform and bars of the roundabout), it simply bends at the point the thread connects to the top. By bending and taking on a new V shape, the whole of the thread and double conker system has now moved inwards and so is now going faster than the top, so it continues to accelerate forwards.
As this happens, load is being applied to the thread / top junction, and if the strength of this junction is great enough, the additional angular momentum can be transferred into the top. But if the strength of this junction is not great enough, then it will break, releasing the cord and the conkers.
You cannot destroy angular momentum, you can only give it to something else, and in this case, that is the SE itself, which having been accelerated forward will in turn be required to travel faster than the earth and so must start to wind itself forward and hence down.
In case you are starting to think that this effect is indeed a real concern for the SE, then it is time to state a solution.
If the SE has too much forward angular velocity at its given height, then simply ‘pay out’ the SE until its new higher orbit matches its angular velocity. If it is going too slow, then ‘take in’ the SE until its new lower orbit again matches the earth rotation speed.
Using this method, the angular velocity stored in the SE can be utilised to take away and give back acceleration to the elevator car as required in both ascent and descent. Perhaps the great length of hte SE will need to be broken into ‘Stations’ so that orbital compensation can be applied to different regions of the SE as required.
DerekSmith
December 4th, 2008 at 1:22 am
Has anyone done the math to model the gravitational impact of the moon passing overhead on the centripetal balance forces, or the impact of the change in the effective CoG in the earth / ocean / moon group, or the tidal height impact on SE angular velocity?
December 4th, 2008 at 2:55 am
You knew of course that the moon does not rotate about the centre of the earth, they both precess around each other with the centre of this rotation within the earth but not at its centre, so the earth experiences a slight ‘slingshot’ effect on the side away from the moon and this is the cause of the second tide each day, effectively it is the ‘anti moon’ tide.
Now if the SE were to be positioned in space either geosynched between the moon and the earth or ‘anti moon’ then these gravity tides would not need to be considered, however, the solar gravity would need to be taken into account. This would need the SE to be free floating and some conventional means of accessing its lower (earth ) end would be required but would also mean that the lower end could be positioned above problematical weather systems. Such a system is effectively gravity tethered in the same spot, much as the tides always sit tin the same places relative to the moon earth pair. It also means that the SE would have the world rotate beneath it and so it would be accessible from many continents simply by ‘catching it’ as it passed overhead.
Has anyone calculated the control advantages of parking the SE in the essentially quite gravity well that exists between teh earth and the moon?
DerekSmith
December 5th, 2008 at 10:43 am
The scientific flaw in your hypothesis is your statement, “… he earth end cannot move forward …”. Indeed, the earth’s rotational speed will be effected by drawing from it’s inertial energy in the case of upward moving items, and adding to it’s inertial energy in the case of downward moving items.
I encourage you to visit our forums for more open discussion about this as it’s a better suited medium than blog comments are for discourse.
December 5th, 2008 at 11:59 am
Hmmm – a diagram here would help to understand what you’re getting at, Derek. But I think Joe Julian hits on at least one issue: your assumption about the Earth end of the elevator may be wrong. Joe – where is “your forum”? Or you mean the liftport forums here?
December 8th, 2008 at 12:36 am
Thanks Joe,
I had not known that there was a Forum but have located it now.
Time to read the back posts to see if the topic is already covered
DerekSmith
December 14th, 2008 at 5:10 am
Derek,
The ribbon would have to be much longer and stronger to be gravitationally locked between the earth and the moon and still have one end near the earth. L1 is 56,000km above the surface of the moon. http://en.wikipedia.org/wiki/Lunar_space_elevator
This would require a much greater taper ratio from the earth side to L1 than you would have by putting the center of weight at GSO.