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  1. Maglatron

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

    I made this file for those interested there are a couple of anomalies such as why the lifting of the block takes out 2.7 joules and I'd like to know how to change this value thanks I hope its correct!!
  2. Maglatron

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

    erm no was just going to use a fibonacci spiral
  3. Maglatron

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

    the actual distance to move is 1.256637061m thanks
  4. Maglatron

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

    is this correct To move the block upwards 1.256637061meters in 0.25 seconds, you need to counteract gravity's downward acceleration of 9.80665m/s² and provide an additional upward acceleration of 40.21238595m/s² to achieve the desired movement. Since both accelerations are in the...
  5. Maglatron

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

    its a casio fx CG50 definatly not binning it!!# do I not need the gravity constant 9.80665m/s^2? is 36.032 the acceleration solved? found the problem 1.256637061 * 2 is = 2.513274122(and not 2.252) and that / 0.0625 = 40.21238595 solved now need to know if I add the gravity in need to get it out...
  6. Maglatron

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

    2.513274122/0.0625 = 40.21238595
  7. Maglatron

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

    i assumed you needed to add it because thatl take care of gravity and even when I do the calculation 2.5132755 / 0.0625 = 40.213
  8. Maglatron

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

    so how to get the calculator to say 36.0212386m/s^2 it's really bugging me! and do you plus on to this the 9.80665?? thanks
  9. Maglatron

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

    I can't get my calculator to say 36.0212386 again I see so the cam need to account for the 0.05m so it's 0.05 + 0.126 = outer radius needs to be 0.176
  10. Maglatron

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

    The total acceleration required to move the block of 0.2753233775 kg upwards 1.256637061 meters in 0.25 seconds, overcoming gravity, is... what?? thanks I think 50.01903595m/^2
  11. Maglatron

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

    so then what should a be? thanks I've cross checked that calculation 2s/t^2 and get 40.21238595m/s^2 for a every time! okay if I add 40.21238595m/s^2 + 9.80665m/s^2 = 50.01903595m/s^2 or do I subtract 40.21238595m/s^2 - 9.80665m/s^2 = 30.40573595^2 because the gravity is in the opposite...
  12. Maglatron

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

    13.77140992 * 0.137 = 1.886683159N * max radius of cam 0.126 =0.237722078Nm correct??
  13. Maglatron

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

    sorry to be naive and probably will recognize when I see it how do you overcome gravity formula plaease, thanks wait a sec 40.21238595m/s^2 is correct chat gpt gave the same equation for acceleration a and also gave 40.21238595m/s^2 but you mean that the value for a is wrong because I have not...
  14. Maglatron

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

    whoops I redid the equation sum and the calculator says 36.0212386m/s^2 for a and the mass is 0.2753233775kg giving a force of 9.917489072N, is that correct now, thanks so did it again to make certain and I got 40.21238595m/s^2 again I'm certain this acceleration is correct!
  15. Maglatron

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

    ok cool could you help me modify the torque required to lif the weight up in 0.25sec is that (2 * 1.1256637061)/(0.25^2) = a = 40.21238595m/s^2 * mass 0.2753233775kg = force of 11.07140992N 11.07140992 * slope 0.137 = 1.516783159N * max radius of cam 0.1256637061m = torque of 0.1906045931Nm...
  16. Maglatron

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

    why thankyou Mr Delta Prime
  17. Maglatron

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

    what for?
  18. Maglatron

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

    oh nice one cool what have you learned?
  19. Maglatron

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

    so I'm reposting for the purpose that the question doesn't get lost in the thread I have a question if the force was worked out to be 3.05 newtons to lift the weight up in 1 second but the rotation of the cam took 0.25sec with a torque of 0.053Nm will that still lift the weight? thanks and whats...
  20. Maglatron

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

    ok that was NOT my intention the numbers are small because when you multiply numbers over 1 by numbers under 1 the resulting numbers get smaller and when you divide small numbers by large numbers the numbers get smaller and when you multiply numbers under 1 the numbers get even smaller I don't...
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