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Calculating G-impulse of Drone Collisions with Ground.

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Patrick
Sun Jun 05 2016, 05:06AM Print
Patrick Registered Member #2431 Joined: Tue Oct 13 2009, 09:47PM
Location: Chico, CA. USA
Posts: 5639
I was told of this description in Steinberg's Vibration book for electronic equipment. This works if the ground being struck is stiff.

G=sqrt((2*h*k)/W)

Where,

h is drop height in inches
k is the stiffness of your structure in lb/in
W is the weight of the structure in lbs
G is the (a/g), where a is the acceleration and g is the acceleration of gravity.


but Id like to figure out a simple G force calculation.
Is It as simple as g=h/d
where:
h = is the height in meters.
d = distance of deceleration in meters.

so if my drones descend from 3 meters, and crumple (like a Volovo) to a stop within 5cm, then the g-load would be 60 g's ?
But theres a wrinkle. The three drone losses im investigating, all move toward a ridged surface at a constant 2 m/s despite falling from different heights. So the vertical speed is not a classical free fall.

So what value do i use for "h" ?

I choose the value of 5cm because the shedding of mass (crap breaking off) eases the fall, ive seen it at the center of mass during the crash, just for a fraction of a second. i wish i could measure it.

It feels like im missing a squared value or more some where.

Im interested in the quick g-load for evaluating the circuit boards and inertial components. semiconductor
survivability after the rest of the machine was destroyed is what im interested in.


1465104509 2431 FT1630 Crash 1



1465104509 2431 FT1630 Crash 2



1465104509 2431 FT1630 Crash 3



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Bored Chemist
Sun Jun 05 2016, 12:19PM
Bored Chemist Registered Member #193 Joined: Fri Feb 17 2006, 07:04AM
Location: sheffield
Posts: 1022
An answer to "So what value do i use for "h" ? " is the height it would have to fall from where, neglecting air resistance, it would hit the ground at 2 m/s.
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Patrick
Sun Jun 05 2016, 05:17PM
Patrick Registered Member #2431 Joined: Tue Oct 13 2009, 09:47PM
Location: Chico, CA. USA
Posts: 5639
Yes! thats what i was hoping some one would say. ok, so know to figure that out. Im not sure if im going crazy or not.

so:

t = V/a = (2 m/s) / (9.8 m/s^2) = t = 0.2 seconds

X = ( (a(t^2)) / 2 ) = (9.8 / 0.04) / 2 = X = 0.196 m

and then :

h/d =g is 0.2m / 0.05m = 4g's ?

Does this seem right ? 0.2 seconds does seem, by appearance to be the worst interval of the crash.


Are kiloponds equal to g's ?
one Newton equals one g right ?
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