If you can't get an opponent OOTA with 10 flips, it will become an endurance fight.
Can your machine drive upside down?
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If you can't get an opponent OOTA with 10 flips, it will become an endurance fight.
Can your machine drive upside down?
Better to use spring(bungee) return, that saves gas and gets the flipper in default closed.
The thing is I'll be using a spring as the main force to deploy the flipper the pneumatics to cock the spring. It's still at design stage but before starting the actual build I want to make sure it's within the weight limits
Attachment 6321
Is there a way to calculate what throw would this set up be able to produce?
The pneumatics would run on 55bar.
How free flowing the gas supply is has as much influence as the geometry. Small tank, pipe, regulator and valve will restrict gas flow and you wont get full pressure. The flipper arm mass has a small effect too, as does where the opponent is on the arm. You could get from 30cm to 3m depending on the system efficiency.
Assuming a very light arm and perfect pnuematics:
H = P*(R^2)*T/10/M
H=flip height
P = pressure in pascals
R = cylinder radius in metres
T= cylinder throw in metres
M = opponent mass in kg.
but that will only tell you the upper limit. The pressure drop will be significant.
Let's leave the efficiency of the pneumatics out of the equation for now, what I'm after is the geometry. With a ram stroke of 38mm and a force of approx 1000kg will the geometry work?
I'm wondering if I put too much negative mechanical advantage, and can't figure out how to get approximate figures.
@Mark, your formula seems to do without leverage ratios
The flipperarm will push the extended rod down and lock it, making retracting difficult or even impossible.
It will be pneumatic retract ram Mario so should not be an issue, what do you think in terms of flip height tho?
Going from real life examples with rams in that size range you get this kind of flips.
Starting on 18 seconds.
https://www.youtube.com/watch?v=HM349ORrbq4