? For math & physics guys

Originally posted by KazetNagorra
Sounds about right, I think I used 24 ft/s, an acceleration of 9.8 m/s² and then rounded the answer up to get 9 ft. (the formula is correct) In any case it only gives a lower bound since you neglect friction, which is a pretty poor approximation in this case.

For someone falling 9ft the correction is tiny. Sticking with imperial units, the acceleration is given by:

a = g - k v*v

where k = drag coefficient/mass. Using a = v * (dv/dx) and assuming a stationary start, I can integrate this to get:

v ^ 2 = (g/k) * (1 - exp(-2kx))

in the limit x -> infinity this gives v_t^2 = g/k. So I can write the above in terms of the terminal velocity to give:

v = v_t * sqrt((1 - exp(-2*g*x/(v_t^2)))

As a quick check, in the limit that v_t is very large the exponential expands to give:

v = v_t * sqrt ( 2*g*x / v_t^2)

=> v^2 = 2*g*x which is the vacuum result.

Wikipedia gives a terminal velocity for a human as 70 m/s, I'll approximate that as 200 ft/s. Putting this as well as your figure of 9 ft into the equation gives:

v = 200 * sqrt( 1 - exp (- 2 * 32 * 9/200*200))
= 200 * sqrt(0.0143) = 23.91, which rounds up to 24 ft/s

Which is the number you started with. I think the error due to air resistance is negligible over this distance. The effect of horizontal speed is going to be more important and could easily make the landing feel more like the 12 ft USArmyParatrooper's c/o quoted.

I could get a handle on the relative error in the speed. The formula for v^2 expanded to second order is:

v^2 ~ 2gx*(1 - gx/v_t^2)

so:

v ~ v_0 * sqrt(1 - gx/v_t^2) ~ v_0 (1 - gx/2v_t^2)

where v_0 = sqrt(2gx) is the speed when air resistance is neglected. The relative error in the speed is:

error = |v - v_0|/v_0 ~ gx/2v_t^2 = 32*9/2(200^2) = 0.36%

So the error from neglecting air-resistance is smaller than the error bound on the speed of the parachute (~4% ).

Originally posted by DeepThought
For someone falling 9ft the correction is tiny. Sticking with imperial units, the acceleration is given by:

a = g - k v*v

where k = drag coefficient/mass. Using a = v * (dv/dx) and assuming a stationary start, I can integrate this to get:

v ^ 2 = (g/k) * (1 - exp(-2kx))

in the limit x -> infinity this gives v_t^2 = g/k. So I can write the ant and could easily make the landing feel more like the 12 ft USArmyParatrooper's c/o quoted.

Yeah, looks like it is a better approximation than I guessed at first glance, although it should be noted that the terminal velocity would depend on what people are wearing (also, I'm not sure when the approximation of proportionality to velocity squared breaks down). Not sure what kind of clothing corresponds to the given terminal velocity, but I can imagine that it's quite a bit lower with very baggy clothing.

Originally posted by KazetNagorra
Yeah, looks like it is a better approximation than I guessed at first glance, although it should be noted that the terminal velocity would depend on what people are wearing (also, I'm not sure when the approximation of proportionality to velocity squared breaks down). Not sure what kind of clothing corresponds to the given terminal velocity, but I can imagine that it's quite a bit lower with very baggy clothing.

But still, it doesn't sound like the air resistance correction will be bigger than the +/- 4 % error of the chute.

Originally posted by sonhouse
But still, it doesn't sound like the air resistance correction will be bigger than the +/- 4 % error of the chute.

I think the figure from Wiki is probably based on people in free fall (i.e. parachutists before they opened their 'chutes); it matters a lot if you are feet first or splayed out and so catching the air. Fluid mechanics is distinctly non-trivial and one of the Millennium prizes is for proving the existence of solutions to the Navier-Stokes equation. My opinion is that the mass-gap problem is easier...

Originally posted by sonhouse
But still, it doesn't sound like the air resistance correction will be bigger than the +/- 4 % error of the chute.

Yeah, with DeepThroat's considerations in mind, even with baggy clothing and weird postures, the correction will probably be a few percent at worst.

Originally posted by KazetNagorra
Yeah, with DeepThroat's considerations in mind, even with baggy clothing and weird postures, the correction will probably be a few percent at worst.

I'm not crunching numbers, but intuitively I see air friction being a negligible factor at just 9 feet.

The 12 feet answer could account for most people crouching down when jumping off a 12' roof top. Or, it could be factoring in average drift (lateral speed) to give relative equivalence.

Originally posted by USArmyParatrooper
I'm not crunching numbers, but intuitively I see air friction being a negligible factor at just 9 feet.

The 12 feet answer could account for most people crouching down when jumping off a 12' roof top. Or, it could be factoring in average drift (lateral speed) to give relative equivalence.

Well, it is clear that it is less than 16 feet, because that is the distance one would free fall for a complete second in time to reach a speed of 32 feet/second. And you only need to reach a speed of 22 - 24 feet/second and that would require less than a second by free fall.

The Instructor

Originally posted by RJHinds
Well, it is clear that it is less than 16 feet, because that is the distance one would free fall for a complete second in time to reach a speed of 32 feet/second. And you only need to reach a speed of 22 - 24 feet/second and that would require less than a second by free fall.

The Instructor

Yes, and the distance from London to New York is less than a light year.

Originally posted by DeepThought
Yes, and the distance from London to New York is less than a light year.

😏

Originally posted by DeepThought
Yes, and the distance from London to New York is less than a light year.

Well, considering the Sun is a light 8 minutes away....

Originally posted by USArmyParatrooper
Well, considering the Sun is a light 8 minutes away....

The sun is the centre of our solar system.

Originally posted by USArmyParatrooper
Well, considering the Sun is a light 8 minutes away....

Right after AIT, I volunteered for airborne training because I considered myself in very good physical shape then and was looking for adventure. However, I was refused because I was too nearsighted they claimed. I was disappointed at the time, but now I think I was probably lucky.

The Instructor

Originally posted by RJHinds
Right after AIT, I volunteered for airborne training because I considered myself in very good physical shape then and was looking for adventure. However, I was refused because I was too nearsighted they claimed. I was disappointed at the time, but now I think I was probably lucky.

The Instructor

I was turned down for flight training because of the same issue. At least I didn't get shot down in Nam. I was in Thailand working on a search and rescue base fixing Troposcatter microwave communications links, a technology now superseded by satellite communications. I went to a military surplus shop one time and saw my old TRC 90 system there on sale for about 500 bucks🙂

Originally posted by sonhouse
Except it isn't 12 feet, it's like a 9 foot fall, from the top of the ceiling to the ground. In my house, we have 9 foot ceilings.

9 feet or 12 feet, my back and legs would be killing me either way!
Kelly