Tell me how this is not the case.A 92 lb. person, accelerating themselves to 3x their weight (by jumping up, and grabbing on) should be able to do this.
Tell me how this is not the case.A 92 lb. person, accelerating themselves to 3x their weight (by jumping up, and grabbing on) should be able to do this.
All these numbers make my old head hurt.Just glad that Turtlelady had figured all that out.
Most of us end up poorer here but richer for being here. Olddog, Fulltime hammocker, 365 nights a year.
Well sure, if you want to do something silly that will obviously have a high probability of damaging the ridgepole (it's not a trapeze).Originally Posted by skree
If that's the way we're going to look at it, then we better start calculating jumping into our hammocks from a platform 10' above them and stopping our cars at the trailhead by driving into a tree.
gmcttr: If you're going to retask the long pipe, I vote trebuchet.
olddog: I'm pretty sure those numbers I gave earlier, are wrong.
Oops.
. . .
For a different (and somewhat irrelevant) set of starting conditions--and only to show how the math can be worked out, I give ya this one:
m = 136 kg (300 lb.)from www.askthephysicist.com
(classical mechanics)
How many Newtons are exerted when a 300 lb. man falls 3 ft.?
ANSWER:
What matters is how long it takes the falling guy to stop.
g = 9.8 m/s^2
v = 4.5 m/sIf he falls 3 ft (about 1 m) he will be going about 4.5 m/s.
t = 1/4 (0.25 seconds) (wild guess)
F ≈ m(g+(v/t)) = 136(9.8+(4.5/0.25)) = 3783 Newtons
F ≈ (3783/9.8) * 2.205 = 851 lb.
Note: numbers slightly rounded. Corrections welcomed.
Last edited by skree; 09-10-2013 at 17:14.
Umm t doesn't need to be a wild guess. He has 1m to fall and the rate of his fall is controlled by gravity, assuming nothing to impede the fall. So he will hit the ground in:
1 sec = 9.8m travel for the first sec, more distance than needed.
1sec/9.8m = time(t)/1m So t=1/9.8 or 0.102 about 1/10th of a sec
That makes your calc:
F ≈ m(g+(v/t)) = 136(9.8+(4.5/0.102)) = 7332.8
F ≈ (7332.8/9.8) * 2.205= 1649.88 lbs
Most of my falls in a hammock were from a foot or foot and a half and I didn't even have time to think, Oh ####. So it must be pretty fast.
Most of us end up poorer here but richer for being here. Olddog, Fulltime hammocker, 365 nights a year.
I can't see myself hanging in a hammock 3 feet up from trees or TDS. It's usually about 1 1/2 feet- 2 feet. That makes the calcs wrong, but who's counting.
Enjoy and have fun with your family,before they have fun without you
Well, this pretty well explains a bruise on the butt for a sudden and total stand failure.
Next would be to add the calculations for the mass and speed of the ridge pole as it will follow by falling from 72" to 24" and impact the skull.
P.S. It's a good thing Happy added the disclaimer to the directions:>)
Enjoying the simple things in life -
Own less, live more.
Nift:
Free Fall Calculator
A person in free-fall, with a constant acceleration of g (9.8 m/sec) in a time interval of 0.102 seconds (with an initial velocity of zero) achieves a velocity of about 1 meter per second, and travels a distance of about 0.05 meters (1.96 inches) during the 0.102 seconds you allotted.
Similarly, if he was at rest initially, and falls through 1 meter, he falls for 0.45 seconds, and achieves a velocity of 4.43 meters per second (4.5 m/sec given in the earlier example).
A drop of the ridgepole through 49" accelerates the pole to a velocity of 11 mph, and it takes somewhere near a half of a second to drop that far.
Energy of falling object has a nice calculator with diagrams, showing a bit more where delta t comes from (but does not give a means to calculate delta-t).
Last edited by skree; 09-11-2013 at 20:54.
Whatever else, but my only experience with a TDS falling was to realize that the top rail coming down was more hurtful than the ground coming up. YMMV.
Posting Permissions
Reply With Quote
Bookmarks