Tony Szamboti
Illuminator
- Joined
- Jun 2, 2007
- Messages
- 4,976
If you examine this part of your newly edited word salad, you will find that the same object is both decelerating at 1g, and have zero acceleration, both at the same time.
You have to remember that a static load is the weight of the building when stationary, it does not disappear just because it is moving. So when something is moving downwards and decelerating the load will be the usual 1g plus the g of the deceleration.
Decelerating a drop at 0.2g would be mass times (1g+0.2g) for a total of 1.2 times static load.
Let's describe it in terms of the velocity change. To get a load from an impact of a freefalling object which is equal to 1.2 times the static load you need to have a deceleration of 32.2 ft/sec/sec x 1.2 = 38.64 ft/sec/sec, in other words a 1.2g deceleration.
That is equal to the static load plus an amplification of 0.2.
It seems there is confusion because I am speaking of deceleration occuring relative to full gravitational acceleration. The static load is being decelerated by 1g to begin with relative to a full gravitational acceleration state.
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