this post was submitted on 09 Jun 2024
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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.