- #1

- 8

- 0

## Main Question or Discussion Point

How do I determine how much drag is required to orient a spinning spherical projectile with a small parachute just before apogee? The projectile is going straight up. The parachute will not be deployed until around 6.5 seconds into flight when the object is traveling at 14.2 m/s. This isn't a requirement, just an arbitrary choice. I would think 1 second would allow for orientation/stabilization. The only requirement is to orient the object so that the bridle attachment point of the parachute is straight down when apogee is reached. The changed apogee would need to be calculated so that an event can occur before the object rolls/becomes unstable. Essentially, just the new apogee height and time need to be known. I have attempted to calculate how much drag a 14" parachute would create, but need to determine how much is needed.

Object size: 12" diameter sphere. 11.3 kg

Initial velocity: 78 m/s.

Apogee without parachute: 310m at 7.5 seconds.

Parachute deployment 6.5 seconds at 300m when the object is traveling at 14.2 m/s.

For a 14" parachute at 14.2 m/s:

Drag force:

Fdrag= ½⋅(ρ⋅A⋅Cd⋅V²)

p - air density = 1.2041kg/m³ @ 20c

A - surface area = 0.11m²

Cd - coefficient of drag = 0.9

V - velocity = 14.2m/s

2.7 lbf of initial drag would be generated by this parachute from what I understand.

How do I determine how much drag is required for orientation as described above, and how is the new apogee height and time determined?

Object size: 12" diameter sphere. 11.3 kg

Initial velocity: 78 m/s.

Apogee without parachute: 310m at 7.5 seconds.

Parachute deployment 6.5 seconds at 300m when the object is traveling at 14.2 m/s.

For a 14" parachute at 14.2 m/s:

Drag force:

Fdrag= ½⋅(ρ⋅A⋅Cd⋅V²)

p - air density = 1.2041kg/m³ @ 20c

A - surface area = 0.11m²

Cd - coefficient of drag = 0.9

V - velocity = 14.2m/s

2.7 lbf of initial drag would be generated by this parachute from what I understand.

How do I determine how much drag is required for orientation as described above, and how is the new apogee height and time determined?