
Post by Glom on Feb 16, 2012 19:09:20 GMT 4
The original cm3 of exhaust gas will now have accumulated 138 cm3 of exhaust gas compressed into a 1 cm3 area. You can't possibly believe how wrong this is. What's wrong Jay? Still working on the Ptolemaic belief in the Second Law of Thermodynamics. Piffle! Deepak Chopra proved conclusively that entropy needn't always increase provided you buy his fruit drink.



Post by Glom on Feb 16, 2012 19:10:11 GMT 4
Also dimensional consistency is for pussies!



Post by gwiz on Feb 17, 2012 12:04:07 GMT 4
After the exhaust gas initially contacts the lunar surface, a cm3 area of gas at the center will than move outward, say 1 cm, as it does a cm3 of gas must be combined into this cm3 area as a result of the continued flow of exhaust gas, this additional cm3 of gas must be compressed into the original cm3 area and so on until the original cm3 of gas reaches the perimeter of the plume, 137 cm from the center. The original cm3 of exhaust gas will now have accumulated 138 cm3 of exhaust gas compressed into a 1 cm3 area. Just to show that this argument doesn't make sense, try doing it in inches instead of cm. The radius of the plume becomes 54 inches, so by your argument the density increase is a factor of 55 instead of a factor 138. Now use feet, and radius becomes 4.5 ft so the density increase is only 5.5. Now use metres... An argument that gives an answer for a nondimensional factor based on the dimensional units used is obviously incorrect.



Post by profmunkin on Feb 20, 2012 12:02:07 GMT 4
From the altitude of 20 feet, 18 seconds elapsed before The Eagle touched down. Assuming the descent rate was fairly even, the descent rate was about 1.1 feet per second.
After The Eagle was on the lunar surface, the descent engine continued to operate for one second before being shut down.
I have not located specifications for The Eagle that detail the clearance between the exhaust cone and the ground, alternatively by taking measurements on NASA diagrams and photos of The Eagle, estimates indicate that the nozzle was about 18 inches from the landing pods. Because there is still a degree of uncertainty, I will use 24" as clearance distance.
Data for 24" clearance. Average distance per second, descent engine rocket nozzle clearance from lunar surface 1 second  rocket nozzle is at avg 5.85 feet 1 second  rocket nozzle is at avg 4.75 feet 1 second  rocket nozzle is at avg 3.65 feet 1 second  rocket nozzle is at avg 2.55 feet 1 second  rocket nozzle is at avg 2.00 feet
18.8 total feet / 5 seconds = average of 3.76 5 seconds at 3.76 feet or 114.15 cm from surface produces an area effected by exhaust to be 229 cm in diameter. An area of 41,187 cm2 Using equations provided in Reply #9 on Apr 5, 2011 179.35 x 1000 / 1.3 = 137,962 cm3 Depth = 137,962 / 41,187 = 3.34 cm (1.32 inch) of Lunar soil would be propelled at 1000 m / sec from an area 229 cm (90 inches) in diameter.


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Post by Bob B. on Feb 20, 2012 12:32:38 GMT 4
Depth = 137,962 / 41,187 = 3.34 cm (1.32 inch) of Lunar soil would be propelled at 1000 m / sec from an area 229 cm (90 inches) in diameter. Thanks for confirming my result. Even under the most conservative assumptions, no appreciable crater is formed.



Post by profmunkin on Feb 21, 2012 11:45:20 GMT 4
Correction to previous post Figure used for the bulking value of lunar regolith was 1.3 gms/cm3. This value had not been adjusted to lunar gravity, bulking value of regolith should be 0.22 gms/cm3
revised calculations 18.8 total feet / 5 seconds = average of 3.76 feet 5 seconds at 3.76 feet or 114.15 cm from surface produces an area effected by exhaust to be 229 cm in diameter. An area of 41,187 cm2 Using equations provided in Reply #9 on Apr 5, 2011 179.35 x 1000 / 0.22 = 815,227 cm3 Depth = 815,227 / 41,187 = 19.8 cm (7.82 inch) of Lunar soil would be propelled at 1000 m / sec from an area 229 cm (90 inches) in diameter.



Post by JayUtah on Feb 21, 2012 12:19:01 GMT 4
This value had not been adjusted to lunar gravity, bulking value of regolith should be 0.22 gms/cm3 Why are you adjusting a mass density for gravity?



Post by profmunkin on Feb 21, 2012 12:22:40 GMT 4
2nd Correction to previous post Because there is no signs of craters the error must be in how fast the regolith was propelled. The velocity must be closer to 3000 m/sec not 1000 m/s
2nd revised calculations 18.8 total feet / 5 seconds = average of 3.76 feet 5 seconds at 3.76 feet or 114.15 cm from surface produces an area effected by exhaust to be 229 cm in diameter. An area of 41,187 cm2 Using equations provided in Reply #9 on Apr 5, 2011 44.438 x 1000 / 0.22 = 90,581 cm3 Depth = 90,581 / 41,187 = 2.20 cm (.87 inch) of Lunar soil would be propelled at 3000 m / sec from an area 229 cm (90 inches) in diameter.



Post by profmunkin on Feb 21, 2012 12:24:11 GMT 4
Yes adjusting for lunar gravity is this not correct?



Post by JayUtah on Feb 21, 2012 12:25:00 GMT 4
No, it is not correct. Mass is mass, irrespective of gravity.



Post by profmunkin on Feb 21, 2012 12:35:49 GMT 4
The Eagle has a mass of 15,864+ pounds, it requires the thrust of 15,864 pounds to land safely, both of these figures are adjusted for lunar gravity to 2,644 pounds of thrust to support 2,644 pounds of lander. The lunar regolith on earth has 1.3 gms/cm3 and 0.22 gms/cm3 on the lunar surface.


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Post by Bob B. on Feb 21, 2012 12:38:01 GMT 4
2nd Correction to previous post Because there is no signs of craters the error must be in how fast the regolith was propelled. The velocity must be closer to 3000 m/sec not 1000 m/s A higher velocity means a smaller crater. We know the amount of kinetic with reasonably good confidence. Therefore, if the soil is moving faster, less mass is necessary to carry the same amount of kinetic energy. Increasing the velocity from 1000 m/s to 3000 m/s means that the amount of mass transported away is decreased by a factor of 9. Something else that you forgot to account for in your numbers is that the LM was moving laterally during the 18second period under consideration. You've assumed it was stationary over a single spot. 18 seconds prior to landing the LM's forward velocity was 4 ft/s.


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Post by Bob B. on Feb 21, 2012 12:42:47 GMT 4
The Eagle has a mass of 15,864+ pounds, it requires the thrust of 15,864 pounds to land safely, both of these figures are adjusted for lunar gravity to 2,644 pounds of thrust to support 2,644 pounds of lander. The lunar regolith on earth has 1.3 gms/cm3 and 0.22 gms/cm3 on the lunar surface. Thrust is counteracting the gravitational force acting on the LM, that is, the LM's weight. Density is mass per unit volume. Weight and mass are not the same thing. pounds = force grams = mass


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Post by Bob B. on Feb 21, 2012 12:54:25 GMT 4
Here’s another way of explaining it:
The LM had a mass of 15,864 poundsmass. In lunar gravity it had a weight of 2,644 poundsforce. To counteract this weight, a thrust of 2,644 poundsforce was needed.
(If you don’t know the difference between poundsmass and poundsforce then you need to take a time out and educate yourself before trying to work more problems.)
On the other hand, density is mass per unit volume. Mass is mass regardless of gravity. Therefore, 1.3 g/cm^{3} is 1.3 g/cm^{3} on Earth, the moon, or anyplace else.



Post by stutefish on Feb 21, 2012 13:20:03 GMT 4
On the other hand, density is mass per unit volume. Mass is mass regardless of gravity. Therefore, 1.3 g/cm ^{2} is 1.3 g/cm ^{2} on Earth, the moon, or anyplace else. It looks like your example gives mass per unit area...

