How much do winds really push you in freefall

[tdog 0]

So... Lets say you get out of the plane and a few thousand feet below the winds are blowing at 40 knots, for the sake of argument, perpendicular to the direction that they are blowing above...

Logically, you are going to get blown (or ride with) the winds in the direction the winds are going...

But, there are a few variables...

Since the winds changed direction, your mass has to accelerate to the new direction of the winds, while your momentum carries you in the direction the winds used to be going. The winds are not an inelastic hook pulling you, instead they accelerate you by the friction of air as it blows across your body... That friction is not 100% efficient in transferring the energy to your body - and I believe (but have no proof) that as the delta between your speed and the wind speed approaches zero, the acceleration changes, thus this is not a linear formula.

Another example of this dynamic... Hot air balloons... They find equilibrium with the winds eventually, and once they do the wind across the basket is zero, even when the ground speed is hot... But when the balloon crosses into air going a different direction or speed, you momentarily feel a breeze in the basket until the new equilibrium is found and the balloon changes speed or direction to match the new wind...

Anyone have some real data on how quickly a human body in freefall finds equilibrium with the winds in which the body is flying in??? Anyone know of any papers on the subject? I am trying to build a computer model to help disprove/prove theories I hear around...

[Andy9o8 2]

Quote

Logically, you are going to get blown

Get your mind out of the gutter.

[JohnMitchell 16]

When you jump out into a wind, the plane from which you jumped was already accellerated by the winds aloft. In othe words, the effects of winds aloft start when you leave the plane. Your scenario of no/low winds at jumprun altitude and high winds below are very unlikely unless your are jumping from 45-50 thousand feet.

[billvon 3,107]

>and I believe (but have no proof) that as the delta between your speed
>and the wind speed approaches zero, the acceleration changes, thus this is
>not a linear formula.

It changes by the cube root of total airspeed. The amount of force changing your speed laterally would be the component of that drag force given by your angle to the relative wind, which of course is composed of both vertical and horizontal components.

In a nutshell, you come VERY close to the windspeed very quickly, then very slowly approach zero relative difference.

To get scientific about it:

If V is your airspeed, then the force (F) on you is (V^3) * X. X is composed of your frontal area, drag coefficient etc but since those don't change much during a jump we can treat them as a constant. Your lateral acceleration is then given by F * arctan(theta) where theta is the angle away from vertical the wind is coming from. That angle in turn is a vector sum of your vertical speed and the transient upsetting lateral wind.

[jas8472 0]

I've done jumps before with 50 knot uppers, the plane is barely moveing across the ground and had over a mile of free fall drift. Get out deeeeeep!!

[mdrejhon 8]

Quote

I've done jumps before with 50 knot uppers, the plane is barely moveing across the ground and had over a mile of free fall drift. Get out deeeeeep!!

I'm happy I spot myself from a Cessna well enough to know that!

[IanHarrop 42]

Check out John Kallend's site, there are some free fall modelling tools there that will let you play with the winds and see. They're really designed to show what happens when you don't leave enough time between jumpers exiting the plane but you'll get the idea as you change value for either upper or lower winds what happens to the jumper's trajectory.

http://www.iit.edu/~kallend/skydive/

"Where troubles melt like lemon drops, away above the chimney tops, that's where you'll find me" Dorothy

[chuteless 1]

When I left the aircraft at 36,916 ft, the winds were 117 MPH. I drifted four miles east and two miles south of the exit point before opening after 2 1/2 minutes of freefall.

Bill Cole D-41

[tso-d_chris 0]

Quote

I'm happy I spot myself from a Cessna well enough to know that!

Good on you for knowing how to spot themselves. I fear that for too many newer jumpers, spotting means locating the green light.

[tdog 0]

edit: Deleted my post, I forgot something that I wanted to calculate into what I was saying.... I will repost corrected version when I have time.

[kallend 2,146]

Quote

Anyone have some real data on how quickly a human body in freefall finds equilibrium with the winds in which the body is flying in??? Anyone know of any papers on the subject? I am trying to build a computer model to help disprove/prove theories I hear around...

If you believe Newton's Laws are real, then yes, I do.

...

The only sure way to survive a canopy collision is not to have one.

[kallend 2,146]

Quote

>

In a nutshell, you come VERY close to the windspeed very quickly, then very slowly approach zero relative difference.

.

When I ran the numbers it took around 6 seconds to close around 95% of the initial difference in relative velocity, IIRC.

...

The only sure way to survive a canopy collision is not to have one.

[firstime 0]

the way I was taught was 1) wind direction from 13K
to 3K in 4 incredments and divide it by 4 eg:
13k 180, 9k 160, 6k 160 and 3k 180. now average
direction would be 170. now use the wind speed in the same equation. Wind speed @ 13k is 20 mph,
9k is 20mph, 6k is 20 mph and and 3k is 10mph.
If my math is correct, you will drift approx 1/3 of a
mile during your 60 second freefall at 170 degrees.

Please correct me if I am wrong

[sixtysevengt5oo 0]

the idea of avg. the direction and wind speed to find your spot is for a simple estimate, so that you have an idea of where to get out, its not meant to be exact. All this high tech math is slightly pintless unless you plan on eventless dive, and opening a round main that you cant direct. In both freefall and canopy you can control where you are going to a certain extent, find a general spot, say 1/2 mile in diameter (or radius if you have enough wind), and jump control yourself the rest of the way.

[diverdriver 7]

Quote

All this high tech math is slightly pintless unless you plan on eventless dive, and opening a round main that you cant direct. In both freefall and canopy you can control where you are going to a certain extent, find a general spot, say 1/2 mile in diameter (or radius if you have enough wind), and jump control yourself the rest of the way.

Unless you're spotting a world record attempt with 300+ jumpers.


Then it becomes REALLY important to be precise.

Chris Schindler
www.diverdriver.com
ATP/D-19012
FB #4125

[firstime 0]

***its not meant to be exact.

I don't get out with GPS on my chest strap. My post was in fact a general estimate on the spot. Did my
post indicate something different????

[diverdriver 7]

Quote

***its not meant to be exact.

I don't get out with GPS on my chest strap. My post was in fact a general estimate on the spot. Did my
post indicate something different????

If the winds aloft average 60 knots and you are in freefall for 60 seconds then you will drift exactly 1 nautical mile (6,000+ feet).

If the average is 30 knots then you will drift .5 mile or 3,000 feet (about the size of many DZ runways).

Is that easier?

Chris Schindler
www.diverdriver.com
ATP/D-19012
FB #4125

[Erroll 80]

Quote

Quote

All this high tech math is slightly pintless

Unless you're spotting a world record attempt with 300+ jumpers.

Yip. you can have as many pints as you like afterwards.