Pulling Firing Pin

[Dr.Gusta]

Okay so after some discussion on CD on another thread my team and I feel like we are pretty close to nailing down a design. The big issue we have right now is being able to pull the pin when there is large forces pulling on it. Let me. Try and explain what our plan is. We are going to have a Claw pick up and hold the ball. A surgical tubing "sling shot" will be pulled back by a pneumatic piston. We want to fire the "sling shot" by pulling the pin on the linkage between the surgical tubing and the piston. The issue is we are going to try and get 50-60lbs worth of force when the surgical tubing is pulled back. That is going to make it next to impossible. What my question is, is there any special linkage that can be easily pulled even though it is pulled on by large forces. Thank you for any help you can provide.

[Dr.Gusta]

One thing I was thinking about is you can claim a piston from the KOP that has a 2in bore but only a 1in stroke. At 60psi that is 120lbs of force. The just use the piston shaft as the actual pin.

[gurellia53]

We used a fence gate latch on our 10 point climber in 2013 and plan to use one again this year. Pulling on the "fence gate" part will not cause it to open unless it is unlatched. A small pneumatic cylinder can slide the latch open and closed.

[Adamz_]

Originally Posted by gurellia53

We used a fence gate latch on our 10 point climber in 2013 and plan to use one again this year. Pulling on the "fence gate" part will not cause it to open unless it is unlatched. A small pneumatic cylinder can slide the latch open and closed.

i just want to say that is one of the best ideas ive heard in a while

[Jared Russell]

Originally Posted by Dr.Gusta

One thing I was thinking about is you can claim a piston from the KOP that has a 2in bore but only a 1in stroke. At 60psi that is 120lbs of force.

How did you compute 120lbs of force?

[Jared]

Originally Posted by Jared341

How did you compute 120lbs of force?

A piston with a two inch bore has a radius of 1 inch, and an area of pi inches.
F=PA=(60 lbs/in^2)(pi)(1 in^2) = 188.49 lbs.

[Dr.Gusta]

That is a really good idea actually, super simple yet very effective. Thank you. As for the 120lbs of force 60psi of air pressure. The area of the 2in bore is a little less than 2 square inches. 60 pounds per square inch × 2 square inches = 120lbs of force. It is not exact but close enough.

[Dr.Gusta]

He beat me to it. I was typing my reply when he posted that. He did the math and all.

[gurellia53]

The fence gate latch is simple, effective, readily available, and $6.

A long bolt can be used instead of the bar piece for different mounting orientations.

Its not complicated at all, but I'll take a picture of our '13 mechanism on Monday.

[Kevin Ray]

We used that same latch in 2010 "Breakaway" to release our kicker. It was one of the most reliable parts of the robot.

[DonRotolo]

Originally Posted by Kevin Ray

We used that same latch in 2010 "Breakaway" to release our kicker. It was one of the most reliable parts of the robot.

Ditto.

One can modify the design slightly such that releasing a very large load needs only a small force. Commercially there are latches that can release several tons with a pull string.

[Brian C]

Originally Posted by Kevin Ray

We used that same latch in 2010 "Breakaway" to release our kicker. It was one of the most reliable parts of the robot.

X3 Cheap, easy and effective.

[roystur44]

Here is a nice design used in saling.

http://www.westmarine.com/webapp/wcs...7#.UtLpcfRDt8E

http://www.westmarine.com/webapp/wcs...7#.UtLqDvRDt8E