Regarding calculations of ft-lbs of kinetic energy

[nikku-kun]

I’ve seen some comments about the need for parachutes to reach threshold kinetic energy levels for new FAA rules. High K/E calculations and parachute retrofits are seemingly impractical to impossible for any commercial drones ever made or purchased so far.

Perhaps the calculation the government could have intended was not how everyone else has interpreted it. It could be referring to potential danger of injury linked to ballistics related kinetic energy calculations.

Excerpt example from FAA:
Category 2 eligible small unmanned aircraft must not cause injury to a human being that is equivalent to or greater than the severity of injury caused by a transfer of 11 foot-pounds of kinetic energy upon impact from a rigid object

Two links:
Kinetic Energy Calculator
Colorado State explanation of ballistics and how it is calculated.

If so, according to the calculator below, a 595g drone falling at 20 m/s would be less than 8.5 ft/lbs kinetic energy.

 

[sar104]

I'm not sure how you calculated that, but your answer is incorrect by an order of magnitude:

Kinetic energy is given by E = ½mv², where E is the energy in Joules, m is mass in kg, and v is velocity in meters per second.

In your example, m = 0.595 kg and v = 20 m/s, giving a kinetic energy, E, of 119 J.

The unit ft-lbs (not ft/lbs) is an alternative unit of energy, where 1 J = 0.738 ft-lbs, and so in this case, 119 J = 88 ft-lbs, not 8.5 ft-lbs.

 

[brett8883]

I'm not sure how you calculated that, but your answer is incorrect by an order of magnitude:

Kinetic energy is given by E = ½mv², where E is the energy in Joules, m is mass in kg, and v is velocity in meters per second.

In your example, m = 0.595 kg and v = 20 m/s, giving a kinetic energy, E, of 119 J.

The unit ft-lbs (not ft/lbs) is an alternative unit of energy, where 1 J = 0.738 ft-lbs, and so in this case, 119 J = 88 ft-lbs, not 8.5 ft-lbs.

Yea I don’t really see how category 2 could be accomplished without a parachute do you?