I’ll begin with the design of a modest-sized unit that is about half the size of the often-recommended 20-ton elevator carriage, which is imagined to be able to carry a 17-ton payload up in only a week. (Which I believe impossible)
First, in order to ascertain the starting speed to drive the elevator all the way up an accelerating path to the top while gravity effects diminish to zero, we need to accurately plot out this path of variable gravity. So, I designed a computer program to trace the path 10 miles at a time, with a new calculation of gravity pull and resultant speed in each of 2,175 steps from ground to orbit. A speed limit is provided, usually run at 130 MPH, since that has already been suggested and probably will be necessary to prevent traction and stability problems in the drive mechanism. In order to approach the desired transit time of one week, I ran the simulator with a starting speed of 65 MPH, and it got to 130 MPH at an altitude of 1,630 miles and went the rest of the way at that speed for a 7.2 day trip. This simulation only shows the ideal time with perfect power transmission and does not include any real-world interferences with power. So, here are the steps in the design:
- For a simple model, let’s start with 10 tons as a full-load limit and a speed of 65 MPH at ground level.
- That would take 3,467 HP to lift 20,000 lbs. straight up at 65 MPH from earth’s surface. We have to allow for at least an estimated 25% inefficiencies in the rollers and drive train, which brings the motors up to 4,622 HP to do the job. Allowing 15% inefficiency in the motors, it would take about 4 MW of electrical power out of the photovoltaic cells to drive them. The completed system may need more than that to overcome all the actual inefficiencies, but this will do for the present analysis.
- Now, with the total power known, the photovoltaic cell array can be developed from there, coordinating the voltage and current of the array with motor requirements. For a rough guide, the best practical solar cells I’ve seen reports on have an efficiency of about 42%. Here, I’m assuming laser energy to be equivalent to the sun on solar cells. Any improvement of laser energy over solar energy would upgrade this estimate in favor of a smaller array, of course.
On this basis, I would estimate that the area of the photovoltaic cell array would have to be about 9,500 square meters to provide the 4 MW of electrical power. That’s the area of a square that is 320 ft x 320 ft. Then I looked up commercial solar panels to get an approximation of weight per square meter. One typical panel was 1.64 meters by .992 meters that weighed 20 kg. So, I applied the math: (9,500 x 20 x 2.2) / (1.64 x .992) = 256,934 lbs. That monstrous array would weigh over 128 tons! WHOOPS–we lost control of the weight!
Not only are we more than twelve times overweight on the photovoltaic cell array alone, but that doesn’t even include the 320 ft x 320 ft framework that would hold all those panels together, nor the wiring to handle 4 MW, and we haven’t added that 4,622 HP worth of electric motors in yet. This study shows that it is a physical impossibility to power a 10-ton space elevator with laser power into photovoltaic cells and make it move up the tether at 65 MPH from ground level. It turns out that the two main power elements, the motors and the photovoltaic array, are also the predominant factors that control the weight of the elevator. Therefore, the limits on this design will affect every design at any size level.
Bad news, but all is not totally lost. We could just scale the whole project down to get the weight to come out at 10 tons. However, to continue this example, we simply have to gear the drive system down for extra torque and a slower start. Let’s also assume that diligent special design of the array and its support bracket constrain the weight and that the motors totaling 4,622 HP don’t add much more than 10 tons, for an estimated total weight with operator and supplies at 150 tons.
Now, we end up with an elevator that can start out at 10/150 x 65 MPH = 4.33 MPH. A run through the simulator, starting at 4.33 MPH, ends up with a total transit time of 4.3 weeks [30.4 days]. It finally attained its maximum speed of 130 MPH at an altitude of 12,500 miles, after 27.5 days had gone by. Then, it traveled for over 9,000 miles at that speed to reach orbit altitude.
Conclusion:
With current photovoltaic cells and electric motors, it is not feasible for an elevator driven by laser power to ascend all the way from earth to orbital altitude in as little as two weeks. The weight-limited starting speed under 5 MPH is the major bottleneck, and it will substantially impact all designs. This example shows a full month in transit, and little improvement can be made on this time with present-day materials. It must be kept in mind that this example is an ideal case with the best possible time, and does not include any of the hazardous interruptions of power cited in my original article. Obviously every one of those interferences would extend the transit time even further out.
– EngrGene
6 Responses to "Rapid Transit" Space Elevator Still Looks More Like Fiction
kert
October 16th, 2008 at 7:20 am
If you are looking at Power/weight of solar cells, you need to be looking at thin film, obviously. Yes, they are lower efficiency but power/weight is way better.
For a state of the art in that, you can probably check with the guys who FLY on solar power:
http://solar-flight.com/news/index.html
EngrGene
October 16th, 2008 at 2:29 pm
kert,
Your alternate arrays may be lighter, to be sure. However, I could not spend the time to research all photovoltaic cells for minimum weight/power ratios. Don’t forget, also, that they still need a frame, and your lower efficiency cells would have made a monstrously unwieldable size of a field of photocells to handle on the elevator.
In any case, the severity of overweight indicates that no photovoltaic cells would have fit anywhere near within the original ten tons, and the array is still a predominant factor of weight.
kert
October 16th, 2008 at 8:29 pm
we are not talking simply “lighter”, we are talking orders of magnitude lighter.
You calculated 128 tons for 4MW, state of the art thin film on Kapton substrate yields about 2000W/kg, i.e. 2 tons ..
jtkare
October 16th, 2008 at 11:38 pm
> For a rough guide, the best practical solar cells I’ve seen reports on have an efficiency of about 42%.
Photovoltaics illuminated with lasers can be more efficient; the ones we use are roughly 50%.
>Here, I’m assuming laser energy to be equivalent to the sun on solar cells.
Generally the laser flux is much higher; the limit is heating of the cells. 2x sunlight is easy, 10x is feasible but takes some effort in space — our cells run at 10X sunlight but are air-cooled. Assuming 8000 W/m^2 (~6x sunlight) incident and 50% efficiency, a 4 MW output array is 1000 m^2, not 9500.
Finally, the mass of PV arrays for high performance applications is much lower than your estimate. Current satellite solar arrays are around 5 kg/m^2. Advanced arrays such as the Entech stretched lens array are around 1 kg/m^2. That would make a 4 MW laser receiver mass around 1000 kg, or 11% of a 9000-kg climber. A full PV power system would mass a bit more, but on the other hand both PV cells and arrays are still improving.
There are certainly reasons to be skeptical about space elevators, but the laser power transmission system really isn’t one.
barakn
October 17th, 2008 at 1:34 am
…except I’m not sure the phrase “solar cell” is the correct one to use. Since the power source is a laser, the photovoltaic cells could be tuned to the laser’s wavelength and thus have a much higher efficiency than a typical solar cell.
reece
October 17th, 2008 at 6:55 pm
I’m not sure what your conclusion is: if it is that it is not currently feasible then I probably agree with you. If its that it will never be feasible or that it is not worth investigating then I disagree. If that is your argument then you should look at the best lab results (which will take 10 years or so to make to market) as the shortest timeframe for building a space elevator I have heard of is 10 years with most being 15-20 and todays technologies will be out of date by then.
I welcome a debate as the way we seem to advance is through the ‘head in the clouds’ people coming up with ideas, the ‘feet on the ground’ people shooting them down, and then the ones left over get developed. I feel however that it would be more worthwhile if you are going to engage in this kind of dialog to read the reference material that the current ideas are based on and try and find flaws in that (and when you have, do a quick internet search and see if others have come to the same conclusion and see the counter-arguments made for them).
I know thats not quite fair in that I’m interested/passionate about the subject and willing to spend time and money doing research and you are a skeptic and not inclined to spend precious resources on what you consider a useless pursuit but the total extent of my space elevator reference library is the internet plus 2 books:
The Space Elevator: A Revolutionary Earth-to-Space Transportation System (2002) by Brad Edwards and Eric Westling
Liftport: Opening Space to Everyone (2006) (of which half the book is sci-fi short stories with a space elevator in them)
Available from Amazon:
http://www.amazon.com/Space-Elevator-Earth-Space-Transportation/dp/0972604502
http://www.amazon.com/Liftport-Space-Elevator-Opening-Everyone/dp/1592221092
The reason for that semi-rant is simply that the numbers you state, as pointed out by jtkare, are nowhere near the correct ones and that similar ones to jtkare’s are used in a book published 6 years ago which is considered (by me at least) to be the best place to start for a technical description of lots of different parts of the space elevator. I like the way you think and I think you could make valuable contributions ‘to the cause’ if you were just primed with the right information. Buy the book and find flaws in that rather than risk being accused to constructing ’straw-man’ arguments. Heck, I’ll buy you a copy if you think you’ll actually read it.
Anyway, I thought I should do a short summary of the two sets of figures so far proposed and then look at what the figures in The Space Elevator book are:
From the original post:
4 MW total power for 10 ton climber
42% solar cell efficiency
9,500 square meters est.
1.62×0.992m 20kg = 12.293kg/m^2
= 256,934 lbs
From jtkare’s comments:
4 MW total power for 10 ton climber
50% efficient
6x sunlight = 8000W/m^2
1000m^2
1-5kg/m^2
= 1-5000kg = 2200-110000 lbs
From The Space Elevator (2002) book pg 63
“For our scenerio we will need 2.4MW of power delivered to our 20 ton climbers
…
For the longer term a 1×10^6kg climber would reuire 120MW of power delivered
…
If the Compower FEL system is selected, specifically designed GaAs phtovoltatic cells can be used with 59% conversion efficiency, 82% filling factor and a sizzling power density of 540kW/m^2 (D’Amato,1992, and Charlie Chu at Tecstar, private communication)
…
The smaller climbers [900kg mass - mainly used to build up the ribbon to take bigger climbers] with a 4m diameter base has 12m^2 of area and needs to produce 100kW of power. That only requires 8.3KW per square meter of PV cells, an easily achieved capacity given the generous margin from the 540kW/m^2 potential. These cells would also work well with a large laser diode array as proposed by Kwon, 1997.”
So we have:
2.4MW total power for 20 ton climber so we’ll say 1.2MW for a 10 ton climber
59% effiicient
lower bound of 8300W/m^2
1.75kg/m^2 (from the chapter before in the book the 12m^2 cells weigh 21kg of the 900kg climber)
= 253kg = 557 lbs
The key difference to the numbers from jtkare is simply the needed power which he just took as read. I’m not sure where the 2.4MW power needed for 20 ton climber (and the 120MW for the 1000 ton one) comes from in the book as I just opened it at the power beaming chapter but I’m sure if you’re interested you can find out :-)