- #1
- 2
- 0
My first question, so be kind (!)
Long time since I was at school, and can only remember the basics of beams and fulcrums. In essence, that a 1kg weight 5mtrs from a fulcrum / pivot needs 5kg at 1mtr to counterbalance. But what about the beam itself? Do the two sides cancel each other out?
For example, I plan to create a "XXX" scissor / accordian arrangement. The metalwork is in ten "X" sections, each approx 1kg, and extends to 5mtrs. One kg weight on the end. The counterweight side has 1.5 "X" sections. All things equal, that gives a c/w of 3kg ... but doesn't account for the 10kg of beam metalwork. So how to calculate the counterweight required. Any ideas how to proceed?
Next (related) question. If the "X" were vertical, (they're not: At most 45 degrees) How much 'force' is needed to push the pivots in the middle of the "XX" apart. (My current thinking is a 8mm lead screw that has to rotate 40 times to cover the 300mm travel). If that's motorised, what sort of torque would that motor need? ... if it's even practical. Since the outer edges of the X are extending the next section, they must (in theory) be adding to the friction forces, (they will be bolts acting as pivots, not ball bearings)
Now whilst this question may appear that I have a smattering of knowledge on the subject, it's exactly that: basics. All the online formulaes that deal with "Newton/Meters", gravity, force, torque and all the other variables leave me out in the cold. So if poss, can you give any answer in "schoolboy" basics that someone of 12 ~ 14 could comprehend, (alas, I'm physically 4x that age range) Or simply say: You need XYZ for counterweight / ABC size for the motor
Long time since I was at school, and can only remember the basics of beams and fulcrums. In essence, that a 1kg weight 5mtrs from a fulcrum / pivot needs 5kg at 1mtr to counterbalance. But what about the beam itself? Do the two sides cancel each other out?
For example, I plan to create a "XXX" scissor / accordian arrangement. The metalwork is in ten "X" sections, each approx 1kg, and extends to 5mtrs. One kg weight on the end. The counterweight side has 1.5 "X" sections. All things equal, that gives a c/w of 3kg ... but doesn't account for the 10kg of beam metalwork. So how to calculate the counterweight required. Any ideas how to proceed?
Next (related) question. If the "X" were vertical, (they're not: At most 45 degrees) How much 'force' is needed to push the pivots in the middle of the "XX" apart. (My current thinking is a 8mm lead screw that has to rotate 40 times to cover the 300mm travel). If that's motorised, what sort of torque would that motor need? ... if it's even practical. Since the outer edges of the X are extending the next section, they must (in theory) be adding to the friction forces, (they will be bolts acting as pivots, not ball bearings)
Now whilst this question may appear that I have a smattering of knowledge on the subject, it's exactly that: basics. All the online formulaes that deal with "Newton/Meters", gravity, force, torque and all the other variables leave me out in the cold. So if poss, can you give any answer in "schoolboy" basics that someone of 12 ~ 14 could comprehend, (alas, I'm physically 4x that age range) Or simply say: You need XYZ for counterweight / ABC size for the motor
!