Because it won't leave me alone...
Jul. 13th, 2004 10:08 pm![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png){kind=small}
tanithryudo...and I have a sneaking suspicion that I might need it in the future when Jared tries to disprove the laws of physics or mathematics, again.
The equation for terminal velocity
Drag force: D = Cd * rho * v ^2 * A / 2
v = sqrt ( (2 * W) / (Cd * rho * A) )
Where Cd is the drag coefficient. (Let's assume 0.5 for simplicity.)
rho is the air density; let's use 1.2854 at -500 m.
Now, we know the terminal velocity of a skydiver without a parachute is about 55 m/s. Let's assume that's the end speed here.
Let's also assume that said skydiver is about 180 cm tall (almost 6') and 50 cm across (about 20") for a frontal area of 9000 cm^2 = 0.9 m^2. Let's also assume the guy has a mass of 80 kg (about 176 lbs).
So, since terminal velocity is the point at which weight = drag, my dilemma is just exactly when a person in free fall hits that point. More importantly, how that would be calculated...
The problem, of course, is that the acceleration isn't constant, which means the major equations of kinematics are useless.
I suppose one way would be to assume an *average* acceleration of 9.8/2 kg*m/s^2... in which case, starting from rest, we have v = at, or 55 = 9.8t/2, where t is about 11 s...
(tbc)
The equation for terminal velocity
Drag force: D = Cd * rho * v ^2 * A / 2
v = sqrt ( (2 * W) / (Cd * rho * A) )
Where Cd is the drag coefficient. (Let's assume 0.5 for simplicity.)
rho is the air density; let's use 1.2854 at -500 m.
Now, we know the terminal velocity of a skydiver without a parachute is about 55 m/s. Let's assume that's the end speed here.
Let's also assume that said skydiver is about 180 cm tall (almost 6') and 50 cm across (about 20") for a frontal area of 9000 cm^2 = 0.9 m^2. Let's also assume the guy has a mass of 80 kg (about 176 lbs).
So, since terminal velocity is the point at which weight = drag, my dilemma is just exactly when a person in free fall hits that point. More importantly, how that would be calculated...
The problem, of course, is that the acceleration isn't constant, which means the major equations of kinematics are useless.
I suppose one way would be to assume an *average* acceleration of 9.8/2 kg*m/s^2... in which case, starting from rest, we have v = at, or 55 = 9.8t/2, where t is about 11 s...
(tbc)
