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  1. Maglatron

    I want to know all of the maths concerning this scissor mechanism!

    ok my argument is that when the scissor drops onto the cam the slope starts immediatly and then keeps the weight going up at constant speed/velocity and will be gradual hence the radius of the cam therefore the accelerating period will be near instant, I definatly don't think it would be...
  2. Maglatron

    I want to know all of the maths concerning this scissor mechanism!

    har har :)
  3. Maglatron

    I want to know all of the maths concerning this scissor mechanism!

    I'm back!! so I did a bit of maths going for dinner now but will explain my findings soon!
  4. Maglatron

    I want to know all of the maths concerning this scissor mechanism!

    and if the velocity is reached in 1% of the time because the slope occurs the instant the the scissor drops down on to the cam it will be going back up the moment the scissor hits the cam because of the slope starts at pi rad from lowest point to the highest point 0.45m and because it's...
  5. Maglatron

    I want to know all of the maths concerning this scissor mechanism!

    the gearing might not be necessary BUT wanted to have the weight to be light the weight falls engaging the flywheel to turn clockwise though the compound gear and levers, once the weight has reached the bottom of the drop the flywheel connects to the cam and pushes the weight back up through the...
  6. Maglatron

    I want to know all of the maths concerning this scissor mechanism!

    Nope!!!!! the flywheel is allowed to freewheel after each successive drop!! through use of a sprag clutch or something similar!! I've not figured out all of the mechanisms yet, just getting the bare bones of the machine at the moment, but the freewheel on the flywheel is essential. It like...
  7. Maglatron

    I want to know all of the maths concerning this scissor mechanism!

    fixed the laptop, took out the socket in the laptop and snipped the connector off the end of the cable and soldered it in and dowsed in hot glue. the required torque to move the cam is 0.1259174689Nm output speed(cam) 2pi output torque at cam 0.1259174689Nm input speed of flywheel = 3.362994729...
  8. Maglatron

    I want to know all of the maths concerning this scissor mechanism!

    ha:)
  9. Maglatron

    I want to know all of the maths concerning this scissor mechanism!

    but its momentary im thinking with the flywheel effect that is driving the cam it will "iron over" the "lump"
  10. Maglatron

    I want to know all of the maths concerning this scissor mechanism!

    mgh in Physics: Refers to the potential energy (PE) of an object in the context of gravitational potential energy. The formula is: PE=mgh where: m is the mass of the object, g is the acceleration due to gravity, h is the height above the reference point. in the context of the gravitational...
  11. Maglatron

    I want to know all of the maths concerning this scissor mechanism!

    hmm this is why I like decimals without the g factor it equals 0.3459815599J of PE and I argue that you do need the 9.80665 because mgh is potential energy mass in kg, g 9.80665m/s^2, h 1.2566m
  12. Maglatron

    I want to know all of the maths concerning this scissor mechanism!

    so my laptop is fully f***** and won't charge so when the battery dies probably won't be on for at least 3 weeks. Two weeks before I can order and a a few days for postage. So if you can answer this before it does die I'd be well happy :)
  13. Maglatron

    I want to know all of the maths concerning this scissor mechanism!

    so this is an important question the block at height 1.26m has potential energy of (1.256637061m * 9.80665m/s^2 * 0.2753233775kg) = 3.392920065J but the torque * radian (the rotational equivalent to F * d) eg work 0.1259174689Nm * π = 0.3958147464J how is this?
  14. Maglatron

    I want to know all of the maths concerning this scissor mechanism!

    oh shizer ok average radius of (0.324m + 0.45m)/2 = 0.387 slope is (Rmax-Rmin)/(π x Ravg). (0.45 - 0.324)/(π * 0.387) = 0.1036357769 weight * 9.80665 = 2.7N lateral force = 2.7N * 0.1036357769 = 0.2798165976N MAX Torque = 0.2798165976N * 0.45m = 0.1259174689Nm correct?? I used the correct...
  15. Maglatron

    I want to know all of the maths concerning this scissor mechanism!

    erm ok can you explain please, thanks
  16. Maglatron

    I want to know all of the maths concerning this scissor mechanism!

    :D
  17. Maglatron

    I want to know all of the maths concerning this scissor mechanism!

    oh ok average radius of (0.324m + 0.45m)/2 = 0.387 slope is (Rmax-Rmin)/(π x Ravg). (0.45 - 0.324)/(π * 0.387) = 0.1036357769 weight * 9.80665 = 2.7N lateral force = 2.7N * 0.1036357769 = 0.2798165976 MAX Torque = 0.2343918253N * 0.45m = 0.1259174689Nm correct??
  18. Maglatron

    I want to know all of the maths concerning this scissor mechanism!

    ok average radius of (0.324m + 0.45m)/2 = 0.396 slope is (Rmax-Rmin)/(π x Ravg). (0.45 - 0.342)/(π * 0.396) = 0.08681178714 weight * 9.80665 = 2.7N lateral force = 2.7N * 0.08681178714 = 0.2343918253N MAX Torque = 0.2343918253N * 0.45m = 0.1054763214Nm correct??
  19. Maglatron

    I want to know all of the maths concerning this scissor mechanism!

    average radius of (0.324m + 0.45m)/2 = 0.342 slope is (Rmax-Rmin)/(π x Ravg). (0.45 - 0.342)/(π * 0.342) = 0.1005189114 weight = 0.2753233775kg lateral force = 0.2753233775kg * 0.1005189114 = 0.0276752062N MAX Torque = 0.0276752062N * 0.45m = 0.01245384279Nm correct??
  20. Maglatron

    I want to know all of the maths concerning this scissor mechanism!

    any chance you could use the numbers provided to explain what you mean please Mr Alec_t could I use the mean slope at max radius to give the max torque needed?
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