Design Suggestion

[Abbreve]

I see pneumatic pistons have been tested for use in prosthetics in the
past. I'm sure this idea is swirling around out there somewhere, but
just to satisfy myself that it doesn't get overlooked...

A simple design would seem to be using the body of the prosthetic as a
pressure vessel, and having pneumatic pistons, bladders, etc as
actuators. Pressure charge would be aquired through an electrical
pump, gas charged cyclinder, hand/foot pump, or even charged simply by
walking (bladder pumps under the heels of the foot.)

Control would be another matter, although it could be achieved
pneumatically, electrical control valves would seem to be a more
elegant solution.

Using bladders in the hand could give a more life-like grip, while
using them on the sole of a foot could lend itself to balance and a
more natural gait.

This technology could be used to make simple rugged devices, or more
complex/lifelike ones.

The existing prosthetics are marvels to me, showing the true beauty of
the KISS principle. Considering the speed of technological
advancement of the past 30 years though, I can only assume that a
people capable of producing the Large Hadron Collider, are ready for
the next step in prosthetics.

Perhaps you can help,

Abbreve

[Daniel]

Abbreve:

As an ME I am attracted to your vision. Pneumatics would offer a nice
advantage because the energy can be stored gradually and used in
bursts. It makes me smile that you appreciate the KISS principle in
prosthetics.

Here is a very quick and dirty analysis (sorry I don't think in
metric):

For simplification, lets pretend that your compressor can make 100 psi
and your pneumatic is pulling on a cable that is powering a typical
hook.

A hook with 8 rubber bands requires about 100 lbs of cable force. So
using P=F/A the piston area would need to be 100lbs/100psi = 1 in^2.
If your rod diameter is .375 inch then the rod area would be 3.14*(.
375/2)^2 = .110 in^2. That means the piston OD would have to be 2*((1
- .11)/3.14)^.5 = 1.065 inch = 27.05mm. Also a hook requires about
1.5in (38.1mm)excursion. 1.07OD X 1.5LNG (28mm X 38mm) doesn't seem to
bad. I think that would be reasonable.

Now that I think about it, I think getting the 100 psi there in the
first place might be the harder challange. I can think of three
options:

1) Have an low power electric compressor built into the arm that is
constantly running
2) The user has a compressor at their home and work place and
periodically recharges the arm throughout the day. Possibly they have
a portable compressor they the can take with them, or they have a
second non-pneumatic arm for situations where compressed air is
unavailable
3) The user has some kind of device in there shoe that steals energy
from their gait.

Does anybody have any suggestions for how to get a tank, valve, and
piston?

[Adam M Whiton]

Daniel, Abbreve,

I've actually been tinkering with a pneumatic design that uses a micro
air pump
to charge/re-charge a small air tank but at a very low pressure around 30psi.
the initial goal was not for powering a gripper but that had crossed my mind.

From a design perspective, electric micro/mini air pumps that are small enough
to fit in an arm or comfortably on a person won't provide high pressures. the
higher the pressure the bigger and heavier the pump, at least thats what i've
found. plus as they reach their max pressure their current draw increases
dramatically reducing battery life.

a compressor at home can provide higher pressures and their are even
hand pumps
that can get up to 3000psi so something like paintball compressed air tank
could be a good start, they come in smaller sizes and can get up to 4500psi.
that said, tanks at those pressures can be very dangerous! also the pressure
will fluctuate based on ambient temperature as well.

i've thought that for developing nations a pneumatic prosthetic that could be
charged with bicycle tire pumps might be viable. bicycle technology is fairly
ubiquitous even in the poorest of countries. although i believe bicycle pumps
max at around 150psi. so depending on tank volume and 150psi how many gripper
actions could be achieved?

adam

[Daniel]

"how many gripper actions could be achieved? "

The formula just came to me. You have to differentiate the ideal gas
law with respect to time for the tank. V1 is the volume of the tank:

pV1 = mRT ->

dp/dt*V1 = dm/dt*R*T

This means that pressure change times volume of the tank equals mass
flow rate out of the tank times the constant R times the temperature.

now we should use m = rho*V to get:

dp/dt*V1 = rho*dV2/dt*R*T

Now dV2/dt is the volumetric flow rate out of the tank. Now we can
use an alternate form of the ideal gas law:

P = rho*R*T -> rho = P/(R*T)

Now combining this we get:

dp/dt*V1 = P*dV2/dt

Now lets multiply by dt and divide by P:

dp/P*V1 = dV2

After integrating we get:

ln(pf/pi)*V1 = -V2 (the negative is because gas is leaving)

This means the the natural log of the final tank pressure divided by
the initial tank pressure times the tank volume is equal to the volume
of gas that left the tank. The volume of gas that left the tank would
be equal to the working volume of the pnuematic cylinder. It will be
convenient to rearrange the formula to this:

Pf = Pi*e^(-V2/V1).

Lets put numbers to this. Lets say that our tank is 2 liters = 2000
cc. The working volume of the cylinder I previously suggested would
be 1*2.54*2.54*1.5*2.54= 24.6 cc. So if you start at 100 psi and use
your gripper 100 times then the final pressure would be Pf=
100*e^(-24.6*100/2000) = 29.2 psi. Or if you had 1000 psi you would
have 292 psi left over.