13
(There is one exception: notice the plateau section of
reversible sleeve 1T type curves). In the example the effective
area of a #22, at 4.0 inches using the 80 psi curve, is:
7,180 lbs.
80 Ibs/in
2
= 89.8 in
2
at 9.0 inches in height it is:
4,670 lbs.
80 Ibs/in
2
= 58.4 in
2
An air cylinder with 89.8 in
2
of area would have an 80 psi curve
as shown by dotted line [10].
The volume curve [3] may also be of importance:
a. If one needs to know the amount of free air (then com-
pressed by the compressor) to perform a desired operation.
b. If the actuation must be completed quickly and calculations
of flow through the air inlet (orifice) are required.
In each case above, the change in internal volume is required.
Read up from the two heights involved to the intersecting point
with the volume curve. Then move to the left and read from the
volume scale. In the example at 4.0 a #22 (notice most volume
curves are at 100 psig) has an internal volume of 349 in
3
[11]
and at 9.0 the volume is 752 in
3
[12]. The change in volume is
then 752 in
3
– 349 in
3
, or 403 in
3
. The volume at minimum
height (349 in
3
) would not be subtracted if exhausting the air
spring to atmospheric pressure.
Notice the shaded area [13]. We do not recommend that an
air spring be used at heights extending into this section. The
“beginning of the shaded area” for a #22 is at 101 inches [5].
SEE PAGE
15
FOR A MORE DETAILED DISCUSSION OF
ACTUATION.
AIRMOUNT ISOLATION
Because of lateral stability considerations (see page 23 for
more details) we recommend that each air spring be used at a
specific height when used as an isolator. This specific height is
called the “Airmount design height” [6]. The vertical line running
through this height [7] is darkened so that it is easy to see
where it intersects the static curves for load readings.
EXAMPLE: Support a 4,100 pound load with an air spring.
Would a #22 be appropriate, and if so, at what height? The
height isn’t much of a problem, as this part SHOULD BE
USED AT 9.5 INCHES. Simply move up the darkened line to
where it intersects 4,100 Ibs [14]. That point falls between the
80 and 60 psig curves. Exactly what pressure would be
required? Use the formula:
Effective Area =
Load (Ibs.)
Pressure (Ibs/in
2
)
Determine the effective area at 9.5 inches (using the 80 psig
curve, since 80 psig would be closer to our exact pressure
than 60 psig), or:
Effective Area =
4,280 Ibs. [15]
= 53.5 in
2
80 Ibs/in
2
Then divide the actual load by the effective area:
4,100 Ibs.
= 76.6 in
2
53.5 in
2
The pressure required to support 4,100 Ibs. with a #22 at a
design height of 9.5 inches is therefore 76.6 PSIG.
Please note that the static data can be converted to dynamic
data (the air spring is in motion) by applying the formulas that
are presented in the Airmount isolation section on page 22.
SEE PAGE 21 FOR A MORE DETAILED DISCUSSION OF
VIBRATION ISOLATION.
INTERNAL RUBBER BUMPERS
Some parts are available with internal rubber bumpers. Where
a bumper is available, it is shown as a dotted line in the cross
sectional view of the air spring. Additionally, please note that:
1. the minimum height is increased to the “bumper contact”
point [16] (this reduces the total available stroke somewhat,
by 4.2 – 3.0 = 12 inches in our #22 example), and
2. the order block contains the proper ordering numbers for
parts with bumpers.
