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

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