Forums: Skydiving: Incidents:
8 May 2013 Fatality Skydive Deland

 

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format  (B 15348)

May 22, 2013, 10:48 PM
Post #151 of 161 (741 views)
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Re: [Ron] 8 May 2013 Fatality Skydive Deland [In reply to] Can't Post

In reply to:
How long would you estimate I took from chop to pull?
1.84 sec from canopy distortion (chopped) to first sign of red (reserve exiting freebag).
Can't see actual pull so I'll say it took you less then 1.5 and more than 1 sec.


unkulunkulu  (C License)

May 23, 2013, 12:54 PM
Post #152 of 161 (633 views)
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Re: [format] 8 May 2013 Fatality Skydive Deland [In reply to] Can't Post

format wrote:
I don't see connection between wing or jumper's mass with it but if there is one, someone will explain.
Sorry, but if you don't see a connection between "center of mass" and rotational movement, maybe you should not share your computations of g forces and other things about physics? I mean, really, make an experiment: take a pencil and rotate it by quickly knocking it on its end, it will probably rotate around it's center. Then glue something to the other end of the pencil and make the same knock. It will rotate around a point much closer to the glued mass. It's called the "center of mass", very important thing for rotational movement and other things :)


(This post was edited by unkulunkulu on May 23, 2013, 12:57 PM)


format  (B 15348)

May 23, 2013, 1:30 PM
Post #153 of 161 (609 views)
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Re: [unkulunkulu] 8 May 2013 Fatality Skydive Deland [In reply to] Can't Post

unkulunkulu wrote:
format wrote:
I don't see connection between wing or jumper's mass with it but if there is one, someone will explain.
Sorry, but if you don't see a connection between "center of mass" and rotational movement, maybe you should not share your computations of g forces and other things about physics? I mean, really, make an experiment: take a pencil and rotate it by quickly knocking it on its end, it will probably rotate around it's center. Then glue something to the other end of the pencil and make the same knock. It will rotate around a point much closer to the glued mass. It's called the "center of mass", very important thing for rotational movement and other things :)
Thank you for contributing to better understanding of g-forces involved. It is more clear to me now and you are right.
Please continue, eventually a simple way will arise for easy G-calculation that no one would question.




ChrisD  (No License)

May 23, 2013, 8:39 PM
Post #155 of 161 (528 views)
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Re: [Festus] 8 May 2013 Fatality Skydive Deland [In reply to] Can't Post

thanks festus!

My favortie actor from Ripcord!!!

Angelic


SethInMI  (A 47765)

May 24, 2013, 5:08 AM
Post #156 of 161 (457 views)
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Re: [unkulunkulu] 8 May 2013 Fatality Skydive Deland [In reply to] Can't Post

unkulunkulu wrote:
if you don't see a connection between "center of mass" and rotational movement, maybe you should not share your computations of g forces and other things about physics?

Don't get carried away unk. The center of mass of the system is important, but keep in mind we are not in a vacuum, and the aerodynamic forces acting on the parachute are more significant in finding the rotation point of the system. The rotation point of a diving canopy is usually far above the top skin. It is only for situations where the system is rotating scary fast that the rotation point moves below the top skin of the canopy, and it NEVER gets all the way to the center of mass of the system, somewhere in the torso of the jumper.

Seth


Premier billvon  (D 16479)
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May 24, 2013, 12:58 PM
Post #157 of 161 (368 views)
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Re: [unkulunkulu] 8 May 2013 Fatality Skydive Deland [In reply to] Can't Post

>I mean, really, make an experiment: take a pencil and rotate it by quickly knocking it
>on its end, it will probably rotate around it's center. Then glue something to the other
>end of the pencil and make the same knock. It will rotate around a point much closer
>to the glued mass.

Now put an airfoil on the other end and repeat, this time spinning very fast. Its center of rotation will change, and it will depend on speed.


chuckakers  (D 10855)

May 24, 2013, 2:02 PM
Post #158 of 161 (353 views)
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Re: [format] 8 May 2013 Fatality Skydive Deland [In reply to] Can't Post

Don't bother, guys. Dood is clearly a genius. After all, only such a brilliant person could calculate a 7-second reserve ride like he did. He apparently even knew exactly how long his reserve would take to open. Wink

Fools die young - and damage the sport.


Premier wmw999  (D 6296)

May 24, 2013, 2:21 PM
Post #159 of 161 (341 views)
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Re: [chuckakers] 8 May 2013 Fatality Skydive Deland [In reply to] Can't Post

I've met Ron. He's neither a fool, nor (particularly) young. OTOH, it does sound calculated a whole lot more carefully than I would in the press of the moment.

Wendy P.


chuckakers  (D 10855)

May 24, 2013, 2:53 PM
Post #160 of 161 (326 views)
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Re: [wmw999] 8 May 2013 Fatality Skydive Deland [In reply to] Can't Post

wmw999 wrote:
I've met Ron. He's neither a fool, nor (particularly) young. OTOH, it does sound calculated a whole lot more carefully than I would in the press of the moment.

Wendy P.

Calculated carefully? Listen to his description of the incident. It's ludicrous.

He can do all the calculating he wants. The only difference between a 7-second reserve ride and death is nothing more than deployment time, and THAT can't be calculated with any regularity or accuracy by anyone.

THAT is foolish no matter what other positive attributes a jumper may demonstrate, and it's a ridiculous and dangerous message to send to other jumpers - especially the young sponges that take what more experienced jumpers say to heart.

Can you honestly say someone who believes they can calculate a 7-second reserve ride is anything other than a fool?


(This post was edited by chuckakers on May 24, 2013, 2:54 PM)


unkulunkulu  (C License)

May 24, 2013, 4:49 PM
Post #161 of 161 (290 views)
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Re: [billvon] 8 May 2013 Fatality Skydive Deland [In reply to] Can't Post

Right, the lift of the canopy will change things. But we have to calculate the center, not just say "if we assume it at the end of lines, we get 10G", right? I don't know how to account for that, but I'm thinking :)


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