Page 69 - Parker - Rodless Air Cylinders

Catalog #AU03-0928/NA Guided Cylinders
Engineering Data HBBY Series

A074
Kinetic Energy

Steps to sizing a guided cylinder with shocks:

1) Determine the “Moving Weight”, W. Table 18
Use Table 18 to determine the “Kinetic Energy Weight” of Base Stroke
a given slide. This value should be added to the weight of Base Stroke Weight*, Adder*,
Model Weight Adder Oversized Oversized
the load the slide will be carrying.
(lb) (lb per in) (lb) (lb per in)
Moving Weight (lbs) = HBBY15 7.93 0.09 7.93 0.09
Kinetic Energy Weight (lbs) + Weight of Load (lbs) HBBY20 13.94 0.22 13.94 0.22
2) Determine the velocity of the load, V (ft/second) HBBY25 25.03 0.42 25.03 0.42
* Support rods do not move with the carriage, so kinetic energy
3) Determine the cylinder force output at the operating
weights are the same for standard and oversized support rods.
pressure, F cylinder (lbs)
4) Detemine the Kinetic Energy of the load:
2
KE = 0.2 × W × V (lb-in)
5) Determine the Energy per Cycle, E cycle (lb-in): Table 19
E cycle = KE + F cylinder × Shock Stroke Total Energy Total Energy Effective Velocity
(unless stroke adjusters are used, 1 inch is standard) per Cycle per Hour Weight Range
Size (lb-in) (lb-in) (lb) (in/sec)
This value should be less than the value listed in Table 19.
15 600 600,000 20 - 3000 6 - 144
6) Determine the Energy per Hour: E hour (in-lbs) 20 900 800,000 30 - 4500 6 - 144
25 1500 670,000 28 - 3800 6 - 120
E hour = 2 × E cycle × # of cycles in one hour
(a cycle is defined as the extension and retraction
of the slide)
This value should be less than the value listed in Table 19.
D
7) Determine the Effective Weight of the load
E cycle
W effective =
0.2 × V 2
This value should be between the values listed in Table 19.

Example:

An HBBY20-10DH1-B with standard support rods and shock
absorbers will be carrying a load of 40 lbs at a velocity of
17 in/second (cycling 15 times per hour) while operating at
80psi. Is this unit properly sized?

1) Moving Weight = [13.94 + (10 × 0.22)] + 40 lbs = 56.14 lbs
2) V = 17 in/second = 1.4 ft/second
3) F cylinder = 251 lbs
4) KE = 0.2 × 56.14 × 1.4 = 22.01 lb-in
2
5) E cycle = 22.01 + 251 x 1 = 273.01 lb-in
6) E hour = 2 × 273.01 × 15 = 8190 lb-in

273.01
7) W effective = = 696 lbs
0.2 × (1.4) 2

The shock will dissipate the energy of the load.

Parker Hannifin Corporation
61 Actuator Division
Wadsworth, OH 44281