James Campbell
@jcacampbell.bsky.social
Aerospace engineer and dog owner. All things aeronautics and space. Experience in modelling of structural crashworthiness and impact behaviour of satellites and aircraft.
The same applies to technical writing.
September 15, 2025 at 6:01 AM
The same applies to technical writing.
My current go at this is that the vessel with the feathers will have less weight. The mass of the feather/lead and vessel will be identical. However as keratin has a lower density than lead it requires more volume, so there is less air mass within it.
August 21, 2025 at 7:37 AM
My current go at this is that the vessel with the feathers will have less weight. The mass of the feather/lead and vessel will be identical. However as keratin has a lower density than lead it requires more volume, so there is less air mass within it.
Alternatively we place the lead/feathers in two identical sealed vessels, along with air at the same temperature and pressure. We then measure the weight of both vessels. Are they then the same?
August 21, 2025 at 5:40 AM
Alternatively we place the lead/feathers in two identical sealed vessels, along with air at the same temperature and pressure. We then measure the weight of both vessels. Are they then the same?
For mechanics problems the rule is ‘draw the free-body diagram’.
August 21, 2025 at 5:30 AM
For mechanics problems the rule is ‘draw the free-body diagram’.
The difference here is a measurement problem as you say. I have also assumed g is 9.81 m/s2, which is not strictly true as g varies with position on the Earth and height above/below sea level. But let’s assume we are measuring the lead and feathers at the same location.
August 21, 2025 at 5:30 AM
The difference here is a measurement problem as you say. I have also assumed g is 9.81 m/s2, which is not strictly true as g varies with position on the Earth and height above/below sea level. But let’s assume we are measuring the lead and feathers at the same location.