There is a ton of ways to solve this, but I believe what might help you the most is a tool that allows you to a) solve these kinds of problems repeatedly, with variations in the setup and b) helps you re-learn the basic math.
Therefore, I suggest you setup your problem in Geogebra:
You can introduce an interactive slider to vary one or more parameters (e. g. extension of the linear rod) and immediately read all other values, like lengths and distances.
It’s a triangle, why not just use Pythagorean theorem? I don’t know if you can get the full 90 degree rotation from a liner actuator like that but it could get pretty close as long as A and B are long in relation to c’s minimum distance.
It's only a right triangle at horizontal, so I'm not sure how I'd use that to figure out the lengths that give me anything for the vertical position. It's easy enough for the fully retracted situation, but after that I'm lost figuring out how to get the angle to 0 at 2C.
Edit: OK, so you're saying if the pole is 5' long, pick an LA that's close to half that length and just wing it. Which I could do, but I'd hoped to maybe figure it out approximately ahead of buying an expensive LA if I could get a smaller one for less.
Ok I understand better what you’re trying to do, this is a pretty basic trig problem there are a ton of triangle calculators out there that will give you a good idea of the lengths you want. This site even has some explanations of how sine and cosine can be used to find missing sides and angles. Also just a note, any triangle can be split to create 2 right angle triangles, it’s not the ideal way to solve these problems but can help simplify some concepts to make things easier to understand.
OK, so there's some way to use the idea that at retracted, b^2 = a^2 + c^2 and at extended, 2C = A + B.
Since I have a 78" long panel I was going to hinge about 1/3 of the way from the top, it seems like a 60" tall post would be a reasonable height to work from. Just plugging random numbers in, if I have an LA of 24" and randomly select A=18, at horizontal B=30 and at vertical those add up to 2C=48.
It seems like if I make A=.75C and figure out B with pythagorean, then the second equation works out as well, ie: C=18 and A=13.5, then B=22.5 and it comes out right at 2C=36. Same with C=16 and A=12, B=20 and 12+20=32.
No clue how I make the above equations prove that, but seems to work.
Thanks for reviving that part of my brain. Now I can go back to killing those brain cells with alcohol.
Thanks for all your responses! I should have mentioned that pneumatics and hydraulics are off the table at this time.
Right now we're thinking of manufacturing a flexible seal that can travel with the rod while maintaining a seal with the housing. That might be over engineering a bit, but we'll be doing some testing this week to see how it goes!
If you need a shaft to slide through the wall of your housing, Google "shaft seal" or "oil seal." The seal usually presses into a hole in your housing, so you will want the housing to be metal rather than plastic.
A dynamic seal like this is going to be difficult to implement. The shaft will need to be very smooth and carefully aligned to the housing. Some small amount of water is likely to get past the seal.
I'm with @Longpork3. Consider other ways to actuate your claw that don't require moving a part through the housing wall. I'm addition to pneumatic or hydraulic actuators outaide your housing, here's an idea using magnets and screws:
Would depend on how deep underwater and the pressure and so forth, but maybe look at how steam engine piston rods are sealed - an o-ring or a wiper seal would probably do the job
If you have an air line to the surface, then running the enclosure at positive pressure to negate the force of water pressure at depth might be a starting point.
I would also reconsider whether a linear actuator is the right approach. A hydraulic piston external to the housing where your motor resides reduces the complexity of any waterproofing, and might be a better approach if you're dealing with significant depths/pressures.
There was a rumor today that rumblings had been heard at Zulof - a town only 30 km from my own. Worried whispers could be heard as I went to the market for supplies - how long? how deep? - but no one knew the answers. The automatic tunneling machines had been turned loose long ago, with nuclear engines and a directive to never stop, to keep connecting the cities of the world.
Too late the Engineers realized their inertial guidance systems were drifting, as swathes of cities collapsed into tunnels cut too shallow. The tunneling machines could not be stopped or reprogrammed, to prevent nation-states from turning them to their own whim, but now they acted as agents of terror by their own accord.
Those 30 km could be cut in a day or a week, depending on the hardness of the rock and the condition of the machine. It would score through the town, wreaking havoc as it went. My plan was to camp in the mountains for a week or two, if the market had enough supplies. Chances were high that others had the same idea.
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