Physics of Terminal Velocity question

[dterrick 0]

To get my B CoP (CSPA) I needed to demonstrate swooping to a formation (Coach 2) with a 1-2 second delay. This I have done and (as you all know) have found a 2 second delay swoop takes a long time to dock. Once, though, I broke at the waist and went virtually headdown but it ATE the altitude up like nothing.

Can anyone quantify (in English) the physics of swooping before we reach TV? I understand it's a drag vs. "1G" gravity thing but I'm wondering about the ramp-up effect while accelerating to "normal" TV120. Does it still take 12 seconds and 1483 ft (for eg) to reach TVheaddown? At 6 seconds of headdown are you at 1/2 TVhd? Would you also be at 1/2(1483 ft) as well?

In a C-182 DZ we only get 9,000 ft to play with. Altitude awareness (and gear issues) notwithstanding I'd also like an idea what kind of fun (or trouble) I could get into with freestyle/freefly and I think some physics would help me.

Anyone feeling Brainy?

the Dave

Life is very short and there's no time for fussing and fighting my friend (Lennon/McCartney)

[Chief 0]

I believe, in English, you are falling at 32 feet per second - squared; per second falling until you reach terminal. Hence in 12 seconds you fall about 1483 feet. If you have access to a pro track read out, it displays you fall graphically, although not 100 per cent accurately.

[bmcd308 0]

x = 1/2 a t^2

x=distance covered
a=acceleration
t=time

a is determined by the difference between gravity and drag. If gravity = drag, then a=0 (TV). When you change your body position, you change your coefficient of drag, which changes your acceleration.

EDITED TO ADD:

Your velocity at a point in time is determined by the product of acceleration and time. So to answer your question in two parts:

1. In a vacuum, yes. After 6 seconds of freefall, your velocity will be half ow what it would be after 12 seconds of freefall.

2. As a practical matter, no. The above only holds true in a vacuum. Drag is a function of velocity, so as your speed picks up, your drag picks up also, such that you actually approach terminal velocity slowly. In the equation above, think of a going down over time slowly until it reaches 0 as you reach terminal. Drag increases with the square of velocity, so you are accelerating much more (your vertical speed is changing more per unit time) early in your freefall.

BMcD

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www.jumpelvis.com

[kallend 2,106]

Quote

To get my B CoP (CSPA) I needed to demonstrate swooping to a formation (Coach 2) with a 1-2 second delay. This I have done and (as you all know) have found a 2 second delay swoop takes a long time to dock. Once, though, I broke at the waist and went virtually headdown but it ATE the altitude up like nothing.

Can anyone quantify (in English) the physics of swooping before we reach TV? I understand it's a drag vs. "1G" gravity thing but I'm wondering about the ramp-up effect while accelerating to "normal" TV120. Does it still take 12 seconds and 1483 ft (for eg) to reach TVheaddown? At 6 seconds of headdown are you at 1/2 TVhd? Would you also be at 1/2(1483 ft) as well?

In a C-182 DZ we only get 9,000 ft to play with. Altitude awareness (and gear issues) notwithstanding I'd also like an idea what kind of fun (or trouble) I could get into with freestyle/freefly and I think some physics would help me.

Anyone feeling Brainy?

the Dave

Go to

www.iit.edu/~kallend/skydive/

and download the freefall separation program. Although meant to illustrate separation, it also models the entire freefall including acceleration, TV, etc.

[Staso 0]

for speed skydiving body position (head down, arms tight next to the body, legs straigt,
toes pointed) which could be considered as ultimate swoop :), there is no terminal velocity
for regular altitude exits

stan.

--
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