03 Jun 13
Originally posted by sonhouseSounds about right, I think I used 24 ft/s, an acceleration of 9.8 m/s² and then rounded the answer up to get 9 ft. (the formula is correct) In any case it only gives a lower bound since you neglect friction, which is a pretty poor approximation in this case.
So that would break down to H=V^2/2A? Using that I get 8.26 feet. Is that correct? Nice formula to remember if I have it right. (using 23 ft/second as velocity)
03 Jun 13
3 edits
Originally posted by KazetNagorraStill, it's a nice formula to add to my accel equations. The formula would work on the moon, when you use lunar accel of course. (1.6249 M/s^2 or in feet, about 5.33 Ft/s^2). I think that works out to about a drop of 50 feet on the moon for the same final velocity.
Sounds about right, I think I used 24 ft/s, an acceleration of 9.8 m/s² and then rounded the answer up to get 9 ft. (the formula is correct) In any case it only gives a lower bound since you neglect friction, which is a pretty poor approximation in this case.
03 Jun 13
Originally posted by sonhouseOn the moon the formula would work a lot better, considering the absence of significant friction. This is something for USArmyParatrooper to consider in an Iron Sky-scenario.
Still, it's a nice formula to add to my accel equations. The formula would work on the moon, when you use lunar accel of course. (1.6249 M/s^2 or in feet, about 5.33 Ft/s^2). I think that works out to about a drop of 50 feet on the moon for the same final velocity.
03 Jun 13
2 edits
Originally posted by sonhouseOkay. I was only calculating the distance that one would fall in one second with the parachute on. I admit I am wrong and you are right in this case. Does this make any sense? No. So what was I calculating?
I am baffled that you can't read. The question is how far do you fall to have a final velocity of 23 feet per second. You do need to figure the time. And like the man says, if you multiply 32 feet per second times .75 second^2 divide by two you get 9 feet.
So a final velocity of the parachute is the same as falling 9 feet not 11 as you think you calcula physics for a living? This is high school math. I am older than you and remembered that much.
The Instructor
04 Jun 13
So the answer is about 9 feet if in a vacuum? I presume it wouldn't be much different in normal atmosphere. How much of a difference can wind resistance make during a 9 foot fall? My guess is (if the calculation is correct), your final velocity would be pretty darn close if you do a standing jump off of a 9 foot wall.
You do land pretty hard, by the way. It's not just the vertical velocity you have to account for, the wind also causes drift. Some of that can be mitigated by pulling your riders into the wind, but not all of it. The higher the wind speeds, the greater the potential for injury.
04 Jun 13
Originally posted by USArmyParatrooperI wouldn't be able to say exactly how much difference friction will make here. As a first approximation, air friction tends to be proportional to velocity squared for small enough velocities, but the proportionality constant depends on atmosphere conditions, the size of the person and what they're wearing - someone in spandex is going to fall harder than someone wearing baggy jeans (or an army uniform).
So the answer is about 9 feet if in a vacuum? I presume it wouldn't be much different in normal atmosphere. How much of a difference can wind resistance make during a 9 foot fall? My guess is (if the calculation is correct), your final velocity would be pretty darn close if you do a standing jump off of a 9 foot wall.
You do land pretty hard, by ...[text shortened]... the wind, but not all of it. The higher the wind speeds, the greater the potential for injury.
05 Jun 13
2 edits
Originally posted by KazetNagorraThis begs an experiment, spandex Vs Baggis🙂 Quick, get out the stopwatches!
I wouldn't be able to say exactly how much difference friction will make here. As a first approximation, air friction tends to be proportional to velocity squared for small enough velocities, but the proportionality constant depends on atmosphere conditions, the size of the person and what they're wearing - someone in spandex is going to fall harder than someone wearing baggy jeans (or an army uniform).
You wouldn't even need a parachute, just get up on a nine foot ladder and jump onto a mattress. It would be complicated by having to figure out just how high you jumped up first. Maybe a hanging scaffod just sans rope around the neck....It could be a dual scaffold and you have laser detectors an inch above ground and then just have good timers see which one gets to the ground first.
06 Jun 13
Originally posted by RJHindsThe Parachute man fell on his head when he was a baby from a height of about two and a half feet with no elastic slowdown of the umbilical chord. What he wants to know is if on his next jump with his parachute, if he lands on his butt at the same velocity as he did on his head as a baby, will that cancel out his brain damage.
We are only concerned about the distance. So the first one is best.
Parachute landing - It is good to know how to fall.
http://www.youtube.com/watch?v=giKo8oKf4GU
The Instructor
06 Jun 13
Originally posted by joe beyserApparently this parachute stuff is not within my expertise of instruction, so I will defer to KazetNagorra or sunhouse.
The Parachute man fell on his head when he was a baby from a height of about two and a half feet with no elastic slowdown of the umbilical chord. What he wants to know is if on his next jump with his parachute, if he lands on his butt at the same velocity as he did on his head as a baby, will that cancel out his brain damage.
The instructor
07 Jun 13
Originally posted by USArmyParatrooperExcept it isn't 12 feet, it's like a 9 foot fall, from the top of the ceiling to the ground. In my house, we have 9 foot ceilings.
OK, so I brought up this discussion with a guy I work with and after overhearing the conversation, my Commander chimed in confidently, "12 feet. It's like jumping off the roof of a 1 story building."