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climbingfreak
May 25, 2004, 6:22 PM
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Whats the formula for falling, as in time you are falling. Just curious.
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l0wnsl0
May 25, 2004, 6:29 PM
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9.8 m/s^2
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overlord
May 25, 2004, 6:33 PM
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time equals square root (2distance/g).
if you use g=9.8 m/s2 you need to mput distance in meters to get time in seconds.
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studmuffin
May 25, 2004, 6:37 PM
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Time= the square root of twice your distance divided by acceleration.
If you're going to calculate that, make sure everythings in metric. Hope thats what you were looking for!
rock on
Justin
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jbell2355
May 25, 2004, 6:48 PM
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First of all, we have to assume that you come to a stop by hitting the ground. Otherwise we would have to calculate the amount of time it takes for the rope to stretch which is an unknown.
If you know the distance you've fallen in meters, divide it by 4.9 and then take the square root of the quotient. This will give you the number of meters fallen. For example: you fell 10 meters (about 30 feet).
10/4.9=2.04 sq. rt. of 2.04=1.23 seconds. In other words, you would have 1.23 seconds of free falling b4 you hit the ground.
Someone let me know if you see a flaw in my logic here.
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noshoesnoshirt
May 25, 2004, 7:34 PM
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square root 2.04 closer to 1.43
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organic
May 25, 2004, 7:57 PM
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first take some nice quartzite, a pair of mad rock shoes and some pumpy arms. Add a first time belayer(to keep you nervous) and sweaty palms, combined with an empty chalk bag give you the formula for falling.
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jbell2355
May 25, 2004, 8:15 PM
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noshoesnoshirt--you're right about the square root...i must have mis-entered a number. I was looking more for verification of my formula than the math, but thanks.
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opusxxii
May 25, 2004, 8:20 PM
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Simple kinematic equations will provide a close approximate.
x = v0t + (1/2)at^2
a = 2[(x - v0t)/t^2]
where v0 = zero because you are starting from a rest, right?
There are several ways to do this. If you wanted acceleration, that method will yield something very close to 9.8 m/s^2.
If you wanted velocity you might have to do both since this equation requires acceleration.
v^2 = v0^2 + 2ax
v = sqrt(v0^2 + 2ax)
a = 2[(x - v0t)/t^2]
v = sqrt(v0^2 + 2(2[(x - v0t)/t^2])x)
I think.. something like that. Start with the two base equations (x = blah, and v^2 = blah) and do simple algebra to yield whatever variable you want.
EDIT: This is what was said above basically.
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noshoesnoshirt
May 25, 2004, 8:22 PM
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yep jbell, your math is fine.
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j_ung
May 25, 2004, 8:32 PM
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This thread is so not what I thought it would be. :)
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bandycoot
May 25, 2004, 8:53 PM
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You can't really calculate it without it becoming really ugly due to the fact that there is wind resistance, rope stretch, maybe you're sliding down a slab or bounce off a ledge, etc.
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robgordon
May 25, 2004, 9:09 PM
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formula for falling:
you will need:
(1) pounds of .10c finger crack.
(7) metolius tcu's
(1) robgordon
(45) minutes
directions:
carefully mix rob and finger crack, adding in the tcu's gradually.
excessive falling will occur sometime after adding the third or fourth tcu.
Note: also produces precipitate cursing and bleeding, these can be scraped off and discarded if unwanted, but some feel that they enhance the falling.
Note: if falling does not occur after adding all tcu's, your finger crack is soft and you must add a similar amount of higher grade crack. you can substitute 5.11 bolted face climbing for the crack, but this produces a much higher volume of cursing.
good luck.
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tedc
May 25, 2004, 9:23 PM
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falling=failing
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jowanky
May 25, 2004, 9:24 PM
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X= Xo + Vo(t) + 1/2 (a) (t^2)
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bandycoot
May 25, 2004, 10:42 PM
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^ The problem with the above formula is what the hell is "a"?
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jgill
May 26, 2004, 1:57 AM
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Distance (in feet) equals 16 times the square of the time (in seconds). Or time (in seconds) equals 1/4 the square root of the distance (in feet). This simple formula ignores resistance due to air - a freely falling body with initial downward velocity equal to zero. 8^)
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pedro_burrito
May 27, 2004, 5:17 AM
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Yes, that is the calculation I remember when sky diving (just once). Using that calculation, you have 13.69 seconds before you bounce if the chute does not deploy. The big but is...the last five seconds do not count because the chute cannot deploy in time so you really have 8 seconds to deploy.
I was in the hangar at this skydiving place in DelMarVa hanging from the rafters practicing cut-aways when someone came on the speaker telling everyone to go inside the hangar. My instructor left me dangling as he ran outside. Someone had bounced.
My jumpmaster later told me that the jumpmaster for the person who bounced forgot to hook up her static line on her first jump. He said she did a perfect arch all the way down. That was 13 very quick or very long seconds.
A Washington, DC TV station did an investigative report on the sky diving facility. They killed more people during the prior 10 years than all other US facilities combined. They had an airtight liability waiver and never lost a lawsuit for negligence. They finally went out of business because the farmers surrounding the facility sued them for crop damages because of all the ambulances and coroner wagons driving around the fields.
I did not know how bad the place was before I jumped. They put us in a Cessna that could barely get off the ground and I was very happy to get the hell out of that plane. Thus ended my skydiving calling. I have turned to a much safer sport. Right?
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valeberga
May 27, 2004, 5:33 AM
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Well in binary it's f = 1 if you are falling :shock:
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overlord
May 27, 2004, 8:48 AM
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In reply to: Well in binary it's f = 1 if you are falling :shock:
:lol: :lol: :lol: :lol:
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