Patrick1000
Banned
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- Jul 22, 2011
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Correction to the compact car thing
Much thanks to Jay at Clavius for his section debunking the CT blast crater objection. http://www.clavius.org/techcrater.html
In that segment, one learns the mass of a lunar lander without fuel is 7,184 kg. I had done my calculation above assuming the mass of the lander to be 15,000 kg roughly, so twice what the number should have been.
Running my numbers again 7,184 X 9.8 X 12.20 all divided by 6 gives 143,153 joules. So a 40 foot or 12.2 meter drop for an unloaded lander would find the lander moving at square root of 2gh or about 6.3 meters per second, or 20.7 feet per second or 14 miles per hour.
For the compact car with the lander's legs attached one finds for a terrestrial solution;
square root of 2 X energy divided by the mass or square root of 143,153 X 2 divided by 1705 or roughly 13 meters per second or 29 miles per hour.
I think that is the first answer Border Reiver got, though I believe he used 300,000 joules for the energy. Anyway, my apologies regardless Border Reiver. I suspect your errors cancelled. You used twice the correct energy given we both did the calculation assuming the lander's fully loaded weight. Thanks to the Clavius details, we now know that the mass of the lander is half of its fueled up mass. I suspect your error was with your "2" in your calculation above and this cancelled your having used twice the correct energy.
Anyhoo, we agree now......
Go back and get a refund for your physics and maths degrees Pat - they taught you the wrong formula for deriving kinetic energy.
Kinetic energy is derived from the formula e=1/2m*v2 where m is the mass in kg and v is the velocity in metres per second.
Using your velocity for the drop of the lunar lander, we get
e = 7600* 40.0689 = 304,523.64 joules of energy
Now your compact car, the speed of impact is a lot different from what you've derived:
In this case we're solving for v, so your formula is
v2 = e/(1/2 m)
or
v = square root of the energy divided by one half the mass
using your mass for the car, I derive a velocity of 13.366 m/s or 29.83178 miles per hour.
All this of course is an exercise in remembering my high school physics from 1986 and pointless unless I knew what is the level of energy that the legs of the lander were designed to withstand?
Much thanks to Jay at Clavius for his section debunking the CT blast crater objection. http://www.clavius.org/techcrater.html
In that segment, one learns the mass of a lunar lander without fuel is 7,184 kg. I had done my calculation above assuming the mass of the lander to be 15,000 kg roughly, so twice what the number should have been.
Running my numbers again 7,184 X 9.8 X 12.20 all divided by 6 gives 143,153 joules. So a 40 foot or 12.2 meter drop for an unloaded lander would find the lander moving at square root of 2gh or about 6.3 meters per second, or 20.7 feet per second or 14 miles per hour.
For the compact car with the lander's legs attached one finds for a terrestrial solution;
square root of 2 X energy divided by the mass or square root of 143,153 X 2 divided by 1705 or roughly 13 meters per second or 29 miles per hour.
I think that is the first answer Border Reiver got, though I believe he used 300,000 joules for the energy. Anyway, my apologies regardless Border Reiver. I suspect your errors cancelled. You used twice the correct energy given we both did the calculation assuming the lander's fully loaded weight. Thanks to the Clavius details, we now know that the mass of the lander is half of its fueled up mass. I suspect your error was with your "2" in your calculation above and this cancelled your having used twice the correct energy.
Anyhoo, we agree now......