Talk:RECYCLING ROCKET EXHAUST

     I need help with the style conventions of Lunarpedia.  I am out of practice in contributing for Lunarpedia and if someone could tell me what is needed to meet the ordinary style expected, I would appreciate that.  
     For the recycling of rocket exhaust from a rocket launched through a tube there would be some difficulty is testing the concept on Earth.  To prevent damage to the wall of the tube at the launch of the rocket, the rocket could be launched from a mobile launch platform accelerated in much the same way that fighter aircraft are accelerated.  This could give the rocket a velocity of 67 meters per second at the time of launch thus spreading the erosive effect of the exhaust out over a larger area of the tube wall.  This would also provide the ullage thrust necessary for the bubble in the fuel tank to be toward the front of the rocket and the fuel to be next to the fuel pump inlet.  Also it provides the first 4 percent of mission delta v.  However this does not work in reverse with the rocket decelerating to a stop and landing in a tube as would be necessary if one were performing a test on Earth and wanted to preserve the test rocket for another run.  
     The use of a larger tube for the tube landing area in the model test with T intersecting tube sections allowing the exhaust to dissipate into a larger area might serve the purpose of a model test or the purpose of an operational variant for which it is necessary to eject orbit bound cargo immediately after the rocket exits the acceleration tube with the rocket proceeding to a deceleration tube.  

Farred (talk) 01:14, 11 April 2022 (BST)

Some figures for a 1500 kilogram cargo rocket burning methane and oxygen, launched

4839 kilograms launch weight of tube launched rocket

2371 kilograms propellant (fuel & oxygen)

968 kilograms empty weight

1500 kilograms cargo

Vsub f = Vsub exh * (ln [Msub i/Msub f)

1680 meters per sec = 2500 meters per sec * (0.672)

exp (0.672) = 1.958 ############# ln (1.958) = 0.672

Msub f/Msub i = 51 percent.

51 percent final weight - 20 percent engines, structure and controls = 31 percent cargo

1500 kilograms cargo / 0.31 = 4839 kilograms initial mass

Farred (talk) 22:17, 28 June 2022 (BST)

Nice to see new work here! You need to divide the text into chapters and sub paragraphs. A long list of bullets is hard to read. Please keep consistent units, you switch from metric to US units. Just stick to metric.

If you are building a tube, why not just use magnetic acceleration? All fuel on the Moon needs to be manufactured. In a sense, fuel is just an energy storage medium. So when you use fuel to propel a rocket, instead of a magnetic launcher, you are replacing storing energy in a magnetic ring, or capacitors, with energy stored in a chemical form.

There are at least four problems with this: -The production of the fuel is not efficient compared to the direct storage of electrical energy. -The rocket is transporting the fuel, so energy is lost due to the rocket equation. You are accelerating fuel that will not be going anywhere. -The burned fuel, now in the form of water and CO2, will be hot. You need to cool it and that wastes energy. -It will take time to cool the exhaust and empty the tube to recycle it. this reduces the output significantly, compared to something like a magnetic launcher than can launch much faster. In theory.

Although I don't think the overall concept works (sorry about that) you would improve it by using oxygen and hydrogen. The only product is water and that is much more easy to condense than CO2.

I've added a few title and paragraphs as an example, please complete the page structure as you see fit!

best, Michel

Thanks for the discussion of the concept and the help with style. I have been busy with a number of personal concerns but I think I can finally put your advice to good use. Farred (talk) 17:10, 12 January 2023 (GMT)

• I looked up the weight disk alternators used to propel aircraft off of carriers. I got this: "High performance disk alternators" published by IEEE at (https://ieeexplore.ieee.org/document/22567)

The Institute of Electrical and Electronics Engineers: 100 kW/kg for one second pulses 15 kW/Kg continuously.

• The example rocket unfueled with cargo weighs 2468 kg. At 1680 meters per second that is 3480 mega joules. You can check my arithmetic. I come up with 34.8 thousand metric tons of high performance disk alternators to store all the power needed as compared with 2371 kilograms of propellant for the rocket. Your suggestion of an all electric launch would be appropriate for launching aircraft at 240 km/hr but at 6048 km/hr orbital velocity it is a "no go". It is not an alternative. I know of a way to launch electrically without using all of those alternators to store the power but that is for another day. For capacitor storage of power the situation is worse. ULTRACAPACITORS:

engadget at (https://www.engadget.com/2019-12-20-nawa-racer-e-bike-ultracapacitor.html#:~:text=Because%20they're%20built%20of,to%2065%20percent%20or%20so.) indicated that Nawa produced an electric bike with ultracapacitors weighing 10 kg and storing 0.1 kwh of power. That is 100 kg per kwh, 100 kg for 3.6 E6 joules, 36 kWseconds per kg.

As for cooling the exhaust to recycle it, that is not a problem, it is a process. Work must be done to accomplish anything. Recycling rocket exhaust is not easy but it is worthwhile.

Your difficulty seems to be that you know of things done at a normal earthly pace in an earthly way and just are not familiar with how large orbital velocities are and how the energy depends upon the square of the velocity. Farred (talk) 14:19, 16 February 2023 (GMT)

• I thought I might see some comment about my unfair treatment of "direct storage" of electrical power, taking the weight of disk alternators as fixed equipment for "direct storage" but not the weight of the radiator, the heat transport fluid or the tankage and piping for moving fuel and heat transport fluid around. If someone can add up the expense for all of the equipment in both cases, and they find electric launch is cheaper, I will not complain about having it. I am just not sure the electric launch is as close to ready to use technology as the fuel recycling for rocket launching, or as cheap. The equipment for recycling is technically simple and could be manufactured out of lunar materials. The disk alternators and ultra capacitors are high tech stuff that lunar industry probably could not handle for quite a while.
• Also, if people will ride in the electrically launched vehicle, 30 meters per second squared might be a reasonable acceleration, not the thousands of meters per second I have seen suggested for electromagnetic acceleration of cargo to L5. With a linear electric motor in the neighborhood of 48.28 kilometers long there might be a requirement for allowance for transverse and vertical motion of the moving portion of the motor more than the inches allowed for in magnetic levitated railroads.
• It is just sound policy to have more than one technology available in case problems arise with the one that people don't talk about because it is secret. Farred (talk) 00:56, 18 February 2023 (GMT)

Circular Mass Accelerator

Lets say it takes about 56 seconds to reach the desired launch speed of 1680 meters per second at 30 meters per second squared. Then the last second of electric acceleration requires about 50000 watt seconds of energy at a power of 50 kilowatts. I do not say that that is impossible for a linear electric motor to achieve, just undemonstrated and a point to keep in mind. For an electric launcher I would go with a circular track and not storing power in anything but the accelerated vehicle. The track could be 200 kilometers in radius with the center at 80 or 280 degrees East and 1.0493 degrees North or South so it is tangent to the equator. At 1680 meters per second for release speed that would be 1.41 meters per second squared centripetal force to deal with or about 8.71 lunar gravities. That would make a suitable launch to equatorial orbit. By varying the launch velocity between 2400 and 4800 meters per second and launching at an independently variable compass direction along the circular track and an independently variable universal standard time one could launch at from about 29 meters per second squared to about 115 meters per second squared centripetal force into orbits that terminate at just about any point on the lunar or Earthly surface or depart the Earth/moon system in just about any direction. That would be 2.9 to 11.8 Earth gravities centripetal force which would be hard for people to take on the high end of the scale but suitable for many types of cargo. Farred (talk) 09:40, 19 February 2023 (GMT)

Hi Michel,

• Thanks again for your advice and discussion. I hope I have answered your question about "why not use magnetic acceleration." It is mainly about money. Magnetic accelerators are expensive and the copper for the best equipment does not come from the moon. Farred (talk) 10:01, 19 February 2023 (GMT)

Some errors concerning a circular mass accelerator

• Where I wrote: "50000 watt seconds of energy at a power of 50 kilowatts." that should be "50,000 watt seconds of energy at a power of 50 kilowatts per kilogram." So, I will make the change.

• Where I wrote: "would be 1.41 meters per second squared centripetal force to deal with or about 8.71 lunar gravities." the 200 kilometers is measured at the surface should equal 6.59295 degrees so the euclidean radius would be 1.9989 E2 km. Velocity squared, 1680 *1680, divided radius is 1.4120 meters per second squared or about 8.72 lunar gravities. So, I will make the changes. Farred (talk) 18:34, 19 February 2023 (GMT)