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Chockstone Forum - Gear Lust / Lost & Found

Rave About Your Rack Please do not post retail SPAM.

Topic Date User
Ripped daisy bartack Tuesday, 17 April 2012 At 11:10:58 AM Olbert
Message
On 17/04/2012 Miguel75 wrote:
>. Forgive my simple physics, and
>please correct me if I'm wrong;
>
>F=Ma
>
>M=110kg
>a=~3.26 m/s-2 (assuming I've remembered it correctly, a= ~9.8 m/s-2, and
>assuming I statically 'slumped' 30cm onto the daisy)
>
>Therefore force = ~359N, or less then the rated strength.

Sigh - I like you M75, you're so nice. That is why this is so hard to say:

Wrong wrong wrong. Wrong!

First point, F = Ma - correct.

Second M = 110Kg - no idea, you probably know your own weight better than me.

Third a = -3.26m/s2 - incorrect.

The acceleration due to gravity is ~9.8m/s2 but that doesn't mean you just multiply it by the amount of metres you fell to get your acceleration. You can't even use that for getting the velocity, for that you use v2 = u2 + 2as. v = final velocity (unknown), u = initial velocity (0m/s), a = acceleration (9.8m/s2), s = distance (0.3m). Using this, your velocity is ~2.4m/s in the downwards direction.

The acceleration that you undergo to go from 2.4m/s to rest is calculated with the same equation: v2 = u2 + 2as where v is the final velocity (0m/s - you are at rest), u is the initial velocity (-2.4m/s - you are falling down, hence the negative), a is the acceleration that will cause you to go from -2.4m/s to 0m/s and s is the distance you do that in. There are two unknowns, the acceleration and the distance. The distance you can estimate as this is the stretch in the daizy (this is a simplified calculation with average acceleration, not accounting for the 'screamer' action of the daisy). I would estimate the distance (s) to be anywhere from 1cm to 5cm - somebody with more knowledge of falls and nylon daisy chains might be able to provide a better estimation. Now the acceleration you undergo can now be calculated.

v = 0, u = -2.4, s = (0.01 <-> 0.05) a = ?

a due to you stopping is therefore in the range (-57.6 <-> -288)m/s2.

The force the daizy chain must impart to stop you can be calculated using F=ma.
F is therefore in the range of ~(-31.7 <-> -6.3)KN (the negative signifies it's downward direction).

The total force on the daizy chain is this force plus the regular force it must impart to keep you from falling due to gravity, so add another -9.8m/s2 x 110Kg = -1.1KN.

So the total force on the daizy chain is something between (7.4 and 32.8) KN. (The negative has been taken away because now the calculation is done who cares what direction the force is in.)

Now that I have performed the relevant physics calculations for you I will explain why they are irrelevant and that the daizy undergoes nothing like that amount of force, even if it were not to rip in the manner of a screamer***. This calculation assumes that you (the 110Kg mass) are a completely rigid body but you are not. Your body is a bit squishy and elastic, all your 110Kg mass will not decelerate at the same rate. At such slow speeds (~2.4m/s) you will act as the 'screamer' in this case and the peak force on the daizy chain will be greatly reduced.

I would be surprised if the daizy chain was pulling anything near 3KN.

Heh heh heh


***On a side note, 'screamers' do not absorb the force. It reduces the peak force by spreading the load over time. Using a different formula: a = (v-u)/t which is basically saying acceleration is the difference in your velocity divided by the time. In a situation where you fall on a screamer the time it takes for you to accelerate from your falling speed to rest is increased. As a result of t increasing, a is smaller. At no point in time is force absorbed!!!

Sorry about this last note, I just get pissed off when people bullshit on about things they know nothing about! (and no that wasn't directed at you M75)

Heh heh heh


Also, who am I kidding, I enjoyed that.

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