Bridge reset

[JP Carlin]

So tomorrow is our fist day of actual competition at the Long Island Regional and our original bridge reset plan did not work. We are now reorienting the pneumatics that were going to lower the bridge to get them to direct more force on to the bridge hoping that that would bet us able to push it down more successfully, My question is that I know very little about pneumatics and all I really need to know is where I can get the specifications on the pneumatics im using to do the math and find out if im getting enough force out of it.

Dont know if I made complete sense so ill just add the main points of what I need to find here:
-where can I look up the specifications of my pneumatics and find out how much force they are delivering?
-What is the minimum amount of force needed to lower the bridge?

-And if anyone has tested both of the following which one have you had more success with: An arm lowering the bridge, or a wedge coming down and then driving into the bridge

[kevin.li.rit]

I'm told it is about 16 lbs.

The force your pneumatics delivers is dependent on the bore size.

The max working pressure on the robot is 60 PSI( Pounds per square inch). Calculate the area of the bore size to get the force.

[JustinRueb1891]

-And if anyone has tested both of the following which one have you had more success with: An arm lowering the bridge, or a wedge coming down and then driving into the bridge

We used a wedge, and it worked perfectly (until we bent the mechanical lock to keep it in place in elims :( ). IMO, the best way to manipulate the bridge is to lower a wedge and lock it into place, so the bridge won't fold your wedge right back into the robot. Good luck!

[Peck]

What is best to manipulate the bridge depends on what you have. Making a wedge that would work well requires a lot of things like:
space for the mechanism (wedge is often bigger then a pusher)
To be reliable, you have to have 2 wedges applying even force to both sides
it has to be joined together at farthest extent at all times (per the rules)
you have to be able to lower the entire system efficiently
greater attention to detail (a wedge is usually an all or nothing and if you get something wrong, you have to re-engineer the whole thing while with something pushing it down, if you don't go far enough, you cut a bigger whole, attach something to the end, or make a brace)
And the wedge design usually has more fail points though this depends on how u build it.

now, if u built a big wedge as part of your frame, that is a different story but right now is likely a little late to implement that.

[JP Carlin]

So i made the measurements and for what I intend on working with I have 2 pneumatics with a 1.5 inch bore and I believe an 10in stroke. I found a table giving the extending force of a pneumatic and my results are more then double what the table says so if someone could just double check my math and tell me if they have the same thing.
so force=area * psi
(3.14)(1.5in)*(60psi)=282.6lbs of force
table says that the force with this should be 106lbs

Here the link to the table I found
http://www.usfirst.org/sites/default/files/uploadedFiles/Robotics_Programs/FRC/Game_and_Season__Info/2012_Assets/2012%20FIRST%20Robotics%20Competition%20Pneumatics %20Manual.pdf

any idea on which force I should use

[kevin.li.rit]

So i made the measurements and for what I intend on working with I have 2 pneumatics with a 1.5 inch bore and I believe an 10in stroke. I found a table giving the extending force of a pneumatic and my results are more then double what the table says so if someone could just double check my math and tell me if they have the same thing.
so force=area * psi
(3.14)(1.5in)*(60psi)=282.6lbs of force
table says that the force with this should be 106lbs

Here the link to the table I found
http://www.usfirst.org/sites/default/files/uploadedFiles/Robotics_Programs/FRC/Game_and_Season__Info/2012_Assets/2012%20FIRST%20Robotics%20Competition%20Pneumatics %20Manual.pdf

any idea on which force I should use

106

It is pi*.75inch^2 * 60 PSI = 106lbs

This is because a 1.5 inch bore has a radius of .75 inches.

[Shaun Coon]

your math is correct

[DonRotolo]

PI R Squared is the area of a circle.
We know this because we are the PI-O-Neers...:rolleyes:

[JP Carlin]

PI R Squared is the area of a circle.
We know this because we are the PI-O-Neers...:rolleyes:

I see what you did there thanks for the corrections cant believe i mixed up diameter and radius but o well

[JP Carlin]

One last question and then i should have this pretty much down.

Ive only used these equations using metrics before so if
Torque=Force * Distance * sin of angle

Am i using distance as inches or feet when calculating this

[kevin.li.rit]

One last question and then i should have this pretty much down.

Ive only used these equations using metrics before so if
Torque=Force * Distance * sin of angle

Am i using distance as inches or feet when calculating this

Either one.

106lbs with a 1 ft lever arm would be 106Ft*lbs or 1272inch*lbs.

[slijin]

I'm told it is about 16 lbs.

Keep in mind that this is the vertical force required; if you're driving into the bridge with a wedge, that means the vertical component of the force has to be ~16+ lbf.

[JP Carlin]

Thanks for the help guys i was up manufacturing that part in the small shop I had in my garage all night and it worked out great. Doing very well so far in the competition