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wiring
how many jaguars are legal to have on a robot?
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Re: wiring
As long as you don't go over size, weight, or cost constraints, as many as you like.
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Just make sure they're all somewhere where they won't be easily filled with aluminum shavings, or guard them with a bag while working above them.
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Re: wiring
120 pounds worth
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Re: wiring
As many as can be plugged into the power panel/digital sidecar.
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Re: wiring
There is no limit other than the practical one (weight, size, volume, I/O ports on 2 digital sidecars, etc). You ARE limited in which motors you can use though. Realistically few teams use less than 4 (4 CIM drive motors) and more than 10.
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Those teams do tend to use IFI Victor speed controllers, though. |
Re: wiring
motors, Eric, Don's talking about motors.
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That would be TOO cool...
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Re: wiring
How would one recommend keeping up with the demands of a pneumatic engine? What bore and stroke should we use (assuming stroke = 2 times the crank pin's offset from the crankshaft axis), What type of solenoid should we use to handle the high-ish switching frequency (and what control method should we use)? :rolleyes:
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Re: wiring
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Edit: I just ran some calculations, and this actually kind of looks easy :D . According to the motor curves, the maximum CIM torque is 340 oz-in. If we use 3/4" bore pistons, our force at 60 psi is about 25 Ibs. Moving along, let's say we're using 4" stroke pistons, with a 2" offset from the drive shaft. 25 Ibs * (1/6' [=2"]) = 4.167 ft-Ibs = 800 oz-in. If we end up using two pistons (mounted 90 degrees with respect to each other, to prevent the engine from catching in place/going the wrong direction) that's 1600 oz-in. That's more than enough to replace two CIMs on a toughbox, a max of 680 oz-in, or anything else for that matter. One of these 2-piston engines on both sides, and you're more than set. |
Re: wiring
All right, so we load up the robot with some of those plastic air tanks that seem to be becoming quite popular, put the fastest legal compressor we can find on there, fill the tanks (using one battery) in the pits, come out to the field, and start driving around.
Now, how do we steer? Servos are legal, so let's make life easy on ourselves. 3-wheeled robot, one pivoting wheel controlled by a bunch of ganged servos. Directional control front/back: Easy, just reverse the up-down cylinder without warning. Cantilevered shafts would be a must, just about, or some really funny axles, but that's doable, I think. Now, the manipulator: Springs, servos, gravity, and whatever air is left over from the drivetrain. Anyone want to give this year's game a shot so far? |
Re: wiring
Here are some more calculations:
Continuing with the same theoretical engines from my previous post: To rotate one of the engines once, both pistons would have to cycle back-and-forth once. Let's find the total volume of air required to do that: 1.767 cubic inches (fills one chamber of a 4" stroke 3/4" bore piston) x 2 (motions per piston) x 2 (pistons) = 7.068 cubic inches. Multiply that by two again, and you get driving both sides of the drivetrain forward one turn: 14.136 cubic inches. Remembering that that's in 60 psi, we can half it (this is an educated guess) to get the equivalent in 120 psi, our store pressure: 7.068 cubic inches. If we are using those fancy white plastic tanks (28 cubic inches), that's 3.962 turns of both sides of the drivetrain per tankful. We need some gearing. Since 4 rotations to a tank is sort of ridiculuously low, lets see how this needs to compare the the 2010 KOP drivetrain (continuing the toughbox thing): In a conservative estimate, the CIM's will be moving at 2000 rpm. With an overall gear reduction of 8.693:1, that works out to the wheels on the robot turning at 230 rpm. With a 2.5 min match, that's 575 wheel rotations in a maximized scenario, 1150 for both wheels. Now let's figure out how to fit 1150 wheel rotations into, say, 64 tanks. We know that one tank will give us 3.962 engine rotations on both sides per tank, so 64 tanks will give us 254 engine rotations. The gear ratio necessary for each side is easy now: 1:4.53. Now, once again, back to torque: with the 2010 KOP gear reduction, the wheel torque of those two maxed-out CIM's on one toughbox will be 680 oz-in x 8.693 = 5911 oz-in. With one engine, we have 1600 oz-in. / 4.53 = 353 oz-in. - pretty weak. You could get this to work out, eventually, but it would require an absolutely ludicrous amount of air tanks. With the 64 tanks in these calculations, you would have to make the robot rather light to get ideal preformance, not to mention that you would spend the entire morning of competition filling those tanks up to 120 psi for one match, at least with anything close to a KOP compressor. Conclusion: it won't work :rolleyes: |
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