Pneumatic Cylinder to open Hatch

AI Thread Summary
The discussion focuses on designing a pneumatic system to open a 500lb hatch using a pneumatic cylinder and bell crank mechanism. The user seeks assistance in calculating the necessary force to size the cylinder appropriately, aiming for the hatch to open 110 degrees in 10 seconds. A suggested approach includes calculating acceleration using the equation of motion and determining torque based on the inertia of the hatch. From the torque, the required force can be derived using the formula F=Tr, where r is the radial distance of rotation. Diagrams and a clear layout of the hatch's constraints are recommended for better understanding.
jrobinson
Messages
1
Reaction score
0
Hey everyone,

Having a little trouble with something a little out of my element. I've been tasked to design a pneumatic system to open a ~500lb hatch. I think I'm leaning towards a pneumatic cylinder to act on a bell crank. I need the hatch to open to 110 degrees from zero in about 10 seconds.

Where I'm getting stuck is on the force necessary to perform this action so I can size my cylinder and then the rest of the system. Any help would be greatly appreciated.
 
Engineering news on Phys.org
What is the layout like? You need to describe how the hatch is constrained and what it looks like. Diagrams are really helpful in these cases.
 
jrobinson said:
Hey everyone,

Having a little trouble with something a little out of my element. I've been tasked to design a pneumatic system to open a ~500lb hatch. I think I'm leaning towards a pneumatic cylinder to act on a bell crank. I need the hatch to open to 110 degrees from zero in about 10 seconds.

Where I'm getting stuck is on the force necessary to perform this action so I can size my cylinder and then the rest of the system. Any help would be greatly appreciated.

1st of all calculate the acceleration

acceleration can calculated by using equation of motion,

aceeleration= 2ө/ (tXt)
where ө= total angle travelled, (in your case it is 110 degrees)
t= time taken for traveling the angle ө(in your case it is 10 sec.)

torque, T= (inertia of the rotating object) X acceleration


From th torque u can calculate the force required...

Force F=Tr, where r is the radial distance @ which u r rotating the object...
 
Posted June 2024 - 15 years after starting this class. I have learned a whole lot. To get to the short course on making your stock car, late model, hobby stock E-mod handle, look at the index below. Read all posts on Roll Center, Jacking effect and Why does car drive straight to the wall when I gas it? Also read You really have two race cars. This will cover 90% of problems you have. Simply put, the car pushes going in and is loose coming out. You do not have enuff downforce on the right...
Thread 'Physics of Stretch: What pressure does a band apply on a cylinder?'
Scenario 1 (figure 1) A continuous loop of elastic material is stretched around two metal bars. The top bar is attached to a load cell that reads force. The lower bar can be moved downwards to stretch the elastic material. The lower bar is moved downwards until the two bars are 1190mm apart, stretching the elastic material. The bars are 5mm thick, so the total internal loop length is 1200mm (1190mm + 5mm + 5mm). At this level of stretch, the load cell reads 45N tensile force. Key numbers...
I'm trying to decide what size and type of galvanized steel I need for 2 cantilever extensions. The cantilever is 5 ft. The space between the two cantilever arms is a 17 ft Gap the center 7 ft of the 17 ft Gap we'll need to Bear approximately 17,000 lb spread evenly from the front of the cantilever to the back of the cantilever over 5 ft. I will put support beams across these cantilever arms to support the load evenly
Back
Top