Calculating terminal velocity

In summary, the conversation revolved around calculating terminal velocity for a skydive using the terminal velocity formula and making assumptions for variables such as density of air, coefficient of drag, and relative area. The calculated result of 22.14 m/s was significantly lower than the average terminal velocity for a skydiver, causing confusion and the question of why there is such a difference in Cd between a skydiver and a human in a wind tunnel. The importance of conducting experiments and obtaining real world data for accurate calculations was also mentioned.
  • #1
JAG.
3
0
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 I am 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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  • #2
One experiment is worth a hundred calculations, lots of "real world" data available for this question.
 
  • #3
Cd can vary for example with speed .Wiki gives the example of a sphere where Cd can vary between about 0.5 for laminar flow and 0.1 for turbulent flow.
 
  • #4
i've seen several sites with tables of figures for skydivers. they range from 50 - 56 m/s terminal velocity.

none explain their calculatoins or what Cd value was used, nor the mass or area used in the calculation. you'd think it would be easy to find these numbers online but I've had no luck so far.

still it strikes me odd that such a low Cd is used, considering a human in the spread-eagle position wearing gear isn't very 'streamlined'..
 

1. What is terminal velocity?

Terminal velocity is the maximum velocity that an object can reach when falling through a fluid (such as air or water). It occurs when the force of gravity pulling the object down is equal to the force of drag pushing against the object.

2. How do you calculate terminal velocity?

Terminal velocity can be calculated using the formula v = √(2mg/ρAC), where v is the terminal velocity, m is the mass of the falling object, g is the acceleration due to gravity, ρ is the density of the fluid, A is the cross-sectional area of the object, and C is the drag coefficient. This equation takes into account both the gravitational force and the drag force acting on the object.

3. What factors affect terminal velocity?

The factors that affect terminal velocity include the mass and shape of the falling object, the density and viscosity of the fluid, and the gravitational pull of the Earth. Objects with a larger mass or a greater surface area will have a higher terminal velocity, while denser fluids and stronger gravitational forces will increase the drag on the object and lower its terminal velocity.

4. Can terminal velocity be reached in a vacuum?

No, terminal velocity can only be reached in a fluid medium. In a vacuum, there is no air resistance to slow down the object, so it will continue to accelerate due to the force of gravity.

5. How is terminal velocity useful in real-world applications?

The concept of terminal velocity is used in various fields, including engineering, physics, and meteorology. It is important in designing parachutes and other air resistance devices, understanding the motion of objects in fluids, and predicting the behavior of falling objects (such as raindrops or debris) in the atmosphere.

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