Talk:RECYCLING ROCKET EXHAUST - Revision history

Farred: Formatting difficulty addressed

2023-06-12T22:23:51Z

Formatting difficulty addressed

Farred: Formatting information addressed

2023-06-12T22:20:56Z

Formatting information addressed

Farred: talk

2023-02-24T16:27:08Z

talk

Farred: corrections

2023-02-19T18:46:02Z

corrections

Farred: some errors to be corrected

2023-02-19T18:34:53Z

some errors to be corrected

Farred: talk

2023-02-19T10:01:40Z

talk

Farred: add heading

2023-02-19T09:47:44Z

add heading

Farred: talk

2023-02-19T09:41:03Z

talk

Farred: talk

2023-02-18T00:56:52Z

talk

Farred: talk

2023-02-16T14:19:56Z

talk

@@ Line 85: / Line 85: @@
 ==Formatting==
 :::I came across what it was that I forgot about formatting.
-Those colons intent text.
+Those colons intent text.  [[User:Farred|Farred]] ([[User talk:Farred|talk]]) 23:23, 12 June 2023 (BST)

← Older revision		Revision as of 22:20, 12 June 2023
Line 81:		Line 81:

	One more mistake. I will just need to be extra careful in my old age, write out all of the steps and make sure the calculator did the calculation that I intended. Where above "8.72 lunar gravities" is written it should be 0.872 lunar gravities, a second change in this lunar gravities. I will make the changes. Perhaps I am the only one who reads this stuff. So, I am the only one who finds the mistakes. [[User:Farred\|Farred]] ([[User talk:Farred\|talk]]) 16:27, 24 February 2023 (GMT)		One more mistake. I will just need to be extra careful in my old age, write out all of the steps and make sure the calculator did the calculation that I intended. Where above "8.72 lunar gravities" is written it should be 0.872 lunar gravities, a second change in this lunar gravities. I will make the changes. Perhaps I am the only one who reads this stuff. So, I am the only one who finds the mistakes. [[User:Farred\|Farred]] ([[User talk:Farred\|talk]]) 16:27, 24 February 2023 (GMT)
		+
		+
		+	==Formatting==
		+	:::I came across what it was that I forgot about formatting.
		+	Those colons intent text.

@@ Line 67: / Line 67: @@
 ==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 per kilogram.  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 6.59295 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 8.72 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.  [[User:Farred|Farred]] ([[User talk:Farred|talk]]) 09:40, 19 February 2023 (GMT)
+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 per kilogram.  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 6.59295 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 0.872 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.  [[User:Farred|Farred]] ([[User talk:Farred|talk]]) 09:40, 19 February 2023 (GMT)
 Hi Michel,
@@ Line 77: / Line 77: @@
 *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.  [[User:Farred|Farred]] ([[User talk:Farred|talk]]) 18:34, 19 February 2023 (GMT)
 I am an old man.  If anyone can point out other errors in my statements I would appreciate it.  [[User:Farred|Farred]] ([[User talk:Farred|talk]]) 18:45, 19 February 2023 (GMT)

@@ Line 67: / Line 67: @@
 ==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.  [[User:Farred|Farred]] ([[User talk:Farred|talk]]) 09:40, 19 February 2023 (GMT)
+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 per kilogram.  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 6.59295 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 8.72 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.  [[User:Farred|Farred]] ([[User talk:Farred|talk]]) 09:40, 19 February 2023 (GMT)
 Hi Michel,
@@ Line 75: / Line 75: @@
 *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. [[User:Farred|Farred]] ([[User talk:Farred|talk]]) 18:34, 19 February 2023 (GMT)
+*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.  [[User:Farred|Farred]] ([[User talk:Farred|talk]]) 18:34, 19 February 2023 (GMT)

← Older revision		Revision as of 18:34, 19 February 2023
Line 71:		Line 71:
	Hi Michel,		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. [[User:Farred\|Farred]] ([[User talk:Farred\|talk]]) 10:01, 19 February 2023 (GMT)		*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. [[User:Farred\|Farred]] ([[User talk: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. [[User:Farred\|Farred]] ([[User talk:Farred\|talk]]) 18:34, 19 February 2023 (GMT)

← Older revision		Revision as of 10:01, 19 February 2023
Line 68:		Line 68:
	==Circular Mass Accelerator==		==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. [[User:Farred\|Farred]] ([[User talk:Farred\|talk]]) 09:40, 19 February 2023 (GMT)		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. [[User:Farred\|Farred]] ([[User talk: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. [[User:Farred\|Farred]] ([[User talk:Farred\|talk]]) 10:01, 19 February 2023 (GMT)

← Older revision		Revision as of 09:47, 19 February 2023
Line 66:		Line 66:
	* 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. [[User:Farred\|Farred]] ([[User talk:Farred\|talk]]) 00:56, 18 February 2023 (GMT)		* 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. [[User:Farred\|Farred]] ([[User talk: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. [[User:Farred\|Farred]] ([[User talk:Farred\|talk]]) 09:40, 19 February 2023 (GMT)		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. [[User:Farred\|Farred]] ([[User talk:Farred\|talk]]) 09:40, 19 February 2023 (GMT)

← Older revision		Revision as of 09:41, 19 February 2023
Line 65:		Line 65:
	* 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.		* 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. [[User:Farred\|Farred]] ([[User talk:Farred\|talk]]) 00:56, 18 February 2023 (GMT)		* 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. [[User:Farred\|Farred]] ([[User talk:Farred\|talk]]) 00:56, 18 February 2023 (GMT)
		+
		+	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. [[User:Farred\|Farred]] ([[User talk:Farred\|talk]]) 09:40, 19 February 2023 (GMT)

← Older revision		Revision as of 00:56, 18 February 2023
Line 59:		Line 59:

	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. [[User:Farred\|Farred]] ([[User talk:Farred\|talk]]) 14:19, 16 February 2023 (GMT)		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. [[User:Farred\|Farred]] ([[User talk: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. [[User:Farred\|Farred]] ([[User talk:Farred\|talk]]) 00:56, 18 February 2023 (GMT)

← Older revision		Revision as of 14:19, 16 February 2023
Line 49:		Line 49:
	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.		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.
	[[User:Farred\|Farred]] ([[User talk:Farred\|talk]]) 17:10, 12 January 2023 (GMT)		[[User:Farred\|Farred]] ([[User talk: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. [[User:Farred\|Farred]] ([[User talk:Farred\|talk]]) 14:19, 16 February 2023 (GMT)