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good day all, out of sheer curiosity i tried to calculate what could be my terminal velocity in a skydive, since i will be diving in soon.

using the terminal velocity formula: Vt = sqrt[ (2mg) / (pACd)

where:

Vt = terminal velocity

m= mass

g = acceleration due to gravity

p = density of fluid [air]

A = relative area

Cd = coefficient of drag

m and g are easy to fill, but for the remaining variables i made somewhat reasonable assumptions.

the density of air [considering it changes between higher and lower altitudes i took to be 1.22 kg/m3 @ 15 degrees C.

for Cd, i looked up charts of drag coeffs for typical shapes and objects. i could not find anything specific on a skydiver but i did find the Cd of a human standing upright. i assumed this meant a human standing in a stream of air as in a wind tunnel, the human standing vertically and the air moving aginst him horizontally.

so if you rotate that entire system 90* so that the human is falling vertically thru air, i figured similar Cd values could be used. in this case, the Cd range is between 1.0 and 1.3 so i used the mean of 1.15.

it seemed reasonable but i later found this is not the case.

for the area, i generalized it into a 2x1 meter rectangle just for simplicity's sake.

so crunching all those numbers i get a result of terminal velocity Vt = 22.14 m/s

now, the situation is that several references i've seen all say that the average terminal velocity for a skydiver is about 55 m/s. which means im under half that of that!

i figured some of my assumptions may be off so i worked backwards using the Vt = 55m/s to find the other variables.

m and g really can't be argued with, the relative area is simple but reasonable. density of air could affect it, but i doubt significantly.. that only leaves drag coefficient Cd.

solving for that i get Cd = .172 !!

according to a table on wikipedia (http://en.wikipedia.org/wiki/Drag_coefficient), this is closer to a 'streamlined half body'.

and according to the table at the bottom of this page (http://www.bookrags.com/wiki/Drag_coefficient [Broken]) it is closer to a smooth sphere, which i am not.

what am i missing here.. why is there such a difference fin Cd or a human skydiver than a human in a wind tunnel??

it's probably a very simple and obvious answer but it's beyond me at the moment. can anyone explain this to me.. i'm stumped.

using the terminal velocity formula: Vt = sqrt[ (2mg) / (pACd)

where:

Vt = terminal velocity

m= mass

g = acceleration due to gravity

p = density of fluid [air]

A = relative area

Cd = coefficient of drag

m and g are easy to fill, but for the remaining variables i made somewhat reasonable assumptions.

the density of air [considering it changes between higher and lower altitudes i took to be 1.22 kg/m3 @ 15 degrees C.

for Cd, i looked up charts of drag coeffs for typical shapes and objects. i could not find anything specific on a skydiver but i did find the Cd of a human standing upright. i assumed this meant a human standing in a stream of air as in a wind tunnel, the human standing vertically and the air moving aginst him horizontally.

so if you rotate that entire system 90* so that the human is falling vertically thru air, i figured similar Cd values could be used. in this case, the Cd range is between 1.0 and 1.3 so i used the mean of 1.15.

it seemed reasonable but i later found this is not the case.

for the area, i generalized it into a 2x1 meter rectangle just for simplicity's sake.

so crunching all those numbers i get a result of terminal velocity Vt = 22.14 m/s

now, the situation is that several references i've seen all say that the average terminal velocity for a skydiver is about 55 m/s. which means im under half that of that!

i figured some of my assumptions may be off so i worked backwards using the Vt = 55m/s to find the other variables.

m and g really can't be argued with, the relative area is simple but reasonable. density of air could affect it, but i doubt significantly.. that only leaves drag coefficient Cd.

solving for that i get Cd = .172 !!

according to a table on wikipedia (http://en.wikipedia.org/wiki/Drag_coefficient), this is closer to a 'streamlined half body'.

and according to the table at the bottom of this page (http://www.bookrags.com/wiki/Drag_coefficient [Broken]) it is closer to a smooth sphere, which i am not.

what am i missing here.. why is there such a difference fin Cd or a human skydiver than a human in a wind tunnel??

it's probably a very simple and obvious answer but it's beyond me at the moment. can anyone explain this to me.. i'm stumped.

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