First, I would like to thank everybody on this thread for their input. I think we can put the Earth landing to rest as being impractical. However, there are 3 more logical targets out there - the moon, Mars, and Earth orbit where the nickel and iron would be very valuable. Someone is already...
Assuming we can land it without burning it up, it is still an immense undertaking. I think the
following steps would be required:
- getting the stuff for the 'asteroid mission' up there at a minimum cost. Perhaps an
air breathing, tail first reusable craft to get the stuff into low Earth...
The asteroid in question is not an ELE sized rock but a carefully selected nickel/iron
asteroid as big as a house. Since it is starting from rest and not going 17,000 mph,
the use of steerable parachutes is feasible. Remember, Antarctica is a continent so
there is lots of places to land. A...
ok - so the asteroid free falls to its terminal velocity - remember, it is not in orbit but starting from rest. Could a system of parachutes slow it down enough to let it crash land (remember, we don't care if it is damaged.) 100 meters/sec would not do much damage on the south pole.
can a small nickel/iron asteroid brought near Earth (matching its solar orbital velocity) and soft landed at the south pole where Earth's rotation is not an issue?
Thanks, that did it
Thanks - you got me off the dime and I have my answer - is amazing what
the squared term does to flywheel energy. We has a 60 pound flywheel
on the cankshaft and it wasn't getting the job done. A 40 pound flywheel
rotating at 20 revs/second gave us more torque than we...
Google calculator.
Put "degrees to radians" into Google search field. Also read more about Googles built in calculator. Also, put "calc98" to get location of free engineering calculator that has an astonishing number of unit calculations from torque
to velocity and much more.
Our air motor has 2 cylinders and is a low rpm device. It therefore
needs a flywheel to store enough torque to get the crankshaft
past the dead spots at 10 degrees before and after top dead center and
10 degrees before and after bottom dead center.
My question is how to calculate the...