# Calculate the minimum gas pressure inside the gas shocks when full extended and full retracted

By Ernie Parker, CFPAI, CFPSD, CFPS, CFPMM, CFPMT, CFPMIP, CFPMMH, CFPMIH, CFPE

### Given:

Hatch door on a car may weigh 889.6 N (200 lb.) uniformly loaded.

2 cylinders support the weight of the hatch back door.

The gas shocks have a 19.05 mm (3/4″) inside diameter bore & 304.8 mm (12″) stroke.

The gas shocks retract 152.4 mm (6 inches).

Calculate the minimum gas pressure inside the gas shocks when fully extended and fully retracted.

Calculate the minimum gas pressure inside the gas shocks when full extended and full retracted.

##### See Solution

Air Teaser Sept 2023 Metric Solution:

The lift cylinders have a 2:1 mechanical disadvantage. The weight pushing straight up on center of door will be 1779.2 N. Since there are 2 cylinders supporting the load, there will be an 889.6 N load on each side pushing straight down.

Step 1:

Determine the angle that the cylinders are pushing at. A triangle has 180°.

180°- 83°- 83°= 14°

Step 2:

889.6 N / sine of 14°= 3677.22 N minimum push per cylinder to hold the door up

Step 3:

Calculate the area of a 19.05 mm bore cylinder.

A = D² × .7854

19.05²× .7854

A = 285.02 mm²

Step 4:

F = p×A

3677.22 / 285.02

p = 12.9 MPa for extension

Step 5:

Calculate the pressure when the door is closed (retracted 152.4 mm)

p1× V1= p₂ × V₂

Step 5a:

Calculate the volume of the extended cylinder V₁ = A × Stroke

V₁ = 285.02 mm²× 304.8 mm = 86,874.1 mm³

Step 5b:

Calculate the volume of the retracted cylinder V₂ = A × Stroke

V285.02 mm² × (304.8-152.4 mm) = 43,437.05 mm³

Solving for p₂ = (p1 × V1 )/V₂

p₂ = (12.9 × 86,874.1) / 43,437.05 = 25.8 MPa for retraction

Proof

12.9 MPa × 145 = 1870.5 psi extension

25.8 MPa × 145 = 3741 psi retraction

Air Teaser Sept 2023 US Customary Solution:

The lift cylinders have a 2:1 mechanical disadvantage.  The weight pushing straight up on center of door will be 400 lbs. Since there are 2 cylinders supporting the load, there will be a 200 lb. load on each side pushing straight down.

Step 1:

Determine the angle that the cylinders are pushing at.

A triangle has 180°.

180°- 83°- 83°= 14°

Step 2:

200 lbs. / sine of 14°= 826.71 minimum pounds of push per cylinder to hold the door up.

Step 3:

Calculate the area of a ¾” bore cylinder.

A = D² × .7854

.75²× .7854

A = .4418 sq. inches

Step 4:

F = p×A   826.71 / .4418

p= 1871.23 psi for extension

Step 5:

Calculate the pressure when the door is closed (retracted 6”)

p1 × V1= p₂ × V₂

Step 5a:

Calculate the volume of the extended cylinder V₁ = A × Stroke

V₁ = .4418 × 12” = 5.3 in³

Step 5b:

Calculate the volume of the retracted cylinder V₂ = A × Stroke

V₂= .4418 × (12-6)” = 2.65 in³

Solving for p₂ = (1871.23 × 5.3) / 2.65 = 3742.46 psi for retraction

Note: To extend the life of your gas shocks in the winter, close your hatch door slowly.