Creating a Drag Shield for a free fall object

[mackwithamac]

Would anyone know how to determine the area that a drag shield would have to be to allow an object above not to experience no more then .001 Gs?
The drag shield that I was thinking about is just a flat sheet attached to the bottom of the top object by a string. Would this even be possible?

 

[Travis_King]

Flat surfaces are generally very bad for stopping objects from falling. They don't catch any air.

http://en.wikipedia.org/wiki/Drag_equation

 

[mackwithamac]

Is it possible to be done?

 

[Travis_King]

Sure I guess. But its not the right way to do it and you won't have any stability and you will likely wind up with a dart (when the flat plate tilts into the wind and becomes a blade)

 

[Kevin_Axion]

I'm not sure if this makes sense or not but it's like the shape of a jet engine with an airfoil on the outside: http://i.imgur.com/jV82y.png. The top is solid.

 

[DaveC426913]

You might want to look into whether there is any practical application of a .001g deceleration.

For one, at that rate of deceleration, the object will still be going pretty much full speed after virtually any distance of fall.

On the other hand, at virtually anything but the slowest velocity, air resistance alone will create a deceleration of greater than .001g's, which means in order to keep it lower than that, you may need to actually apply thrust forward, otherwise the object will naturally decelerate faster than that.

I suspect that you are looking for a solution to a problem that can't realistically exist.

 

[mackwithamac]

thank you for all the feedback. and I'm sorry to confuse you but its .01Gs I was wrong when I initially gave the info.
let me give you a little more info and maybe there is a simple solution to my problem:
I have an experiment that will be dropping from a 10ft 5in tower and I need to be able to reuse the experiment after the 1st initial drop and have the experiment only experience .01gs. The reason that the drag shield was brought to my attention is because as it was explained to me, the drag shield will help me reduce the amount of gs on the experiment.

 

[DaveC426913]

thank you for all the feedback. and I'm sorry to confuse you but its .01Gs I was wrong when I initially gave the info.
let me give you a little more info and maybe there is a simple solution to my problem:
I have an experiment that will be dropping from a 10ft 5in tower and I need to be able to reuse the experiment after the 1st initial drop and have the experiment only experience .01gs. The reason that the drag shield was brought to my attention is because as it was explained to me, the drag shield will help me reduce the amount of gs on the experiment.

.01g. That's 9.8cm/s^2. That's tiiiiiiiny.

The only way for your device can not exceed .01g and still arrive at the ground at zero velocity is if it never gets up any speed in the first place.

Maybe attach a giant balloon to it.

How heavy is your device?

 

[Travis_King]

Even then, you'll have a tough time grabbing enough air to slow you down to .01g's from 10 feet up. Could you be more specific with the application and why you require such little deceleration?

Is it a question of impact?

 

[Bob S]

What you need to do is match the downward force F_down = mg with the upward drag force F = ½CρAv², where ρ is the density of air, C is the drag coefficient, and A is the cross sectional area of the plate or parachute.

So

mg = ½CρAv²

Solve for v to get the terminal velocity. Once an object reaches the terminal velocity, the velocity is constant, and the acceleration is zero.

Bob S