Search results

  1. Alec_t

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

    Just spotted another error. In post #187, lateral force is right but max torque uses the wrong lateral force.
  2. Alec_t

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

    Ok.:D
  3. Alec_t

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

    Still wrong. You used the value 0.342 instead of 0.324.
  4. Alec_t

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

    The average radius value is wrong, so are the slope and lateral force values as a result. You also forgot the 9.81 factor if you want the lateral force in Newtons.
  5. Alec_t

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

    Wt (weight) is the force exerted by gravity acting on the mass. Weight = mass x gravity = 0.275 Kg. Yes, since without complicating things further you will only get an approximation anyway, given all the gearing and unknowns (scissor weight, friction etc). Mean slope = (Rmax-Rmin)/(Rmean x pi)...
  6. Alec_t

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

    If your cam profile and constant rotation rate cause the weight to move vertically at a constant speed then the upward force needed = the weight Wt. The lateral force, Lf, to follow the profile = Wt x slope, so the torque T = Lf x radius. Taking the mean slope at the mean radius should give an...
  7. Alec_t

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

    For a near-constant rate the acceleration a will be >1g initially then near 1g. Force=effective-mass*a. Don't forget to take into account the scissor mass and the effect of the inertia of the flywheel (via the gearing) when calculating the effective mass. Force will also be affected by friction...
  8. Alec_t

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

    If you want the load to start from stationary and reach a certain height in 0.25 sec it must accelerate. The acceleration will be constant if the force moving the load is constant, or will vary if the force varies. If you want the cam-follower to stay in contact with the cam then the cam...
  9. Alec_t

    A Little Help for Brainstorming Ideas

    Since the car must hit the wall you could simply count wheel revolutions and use the wall impact to reverse the motor.
  10. Alec_t

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

    So how do they get to that constant speed? Any thing which is initially static and then starts moving MUST accelerate. The acceleration may not be noticeable to a human, or may start off at a finite value and then quickly reduce to zero.
  11. Alec_t

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

    "can I make the cam movement like this so that it moves the scissor up with ease but with no acceleration,?" The laws of physics dictate that you have to accelerate the load if you want it to move from rest to a desired height in a desired time.
  12. Alec_t

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

    Don't see why not, except that would result in the load rising at a constant velocity, whereas you want it to rise at an accelerating rate under constant force unless you accelerate the cam rotation rate.
  13. Alec_t

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

    Given all the unknowns and approximations, I don't think the actual slope is that critical in practice for a prototype. Try using the approximate average slope as per post #155. For an angle of rotation dΘ, the cam moves along an arc of length r.dΘ and moves vertically dr. The slope is...
  14. Alec_t

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

    Adding the spring above the load, to limit the upward movement , will result in the load's initial downward acceleration having a component which depends on the spring's unknown free length and compressibility. The flywheel's inertia, via the gearing, friction and gravity are the other...
  15. Alec_t

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

    If the load has a constant upward acceleration for the whole 1.26m of its travel, it won't stop at 1.26m if you are relying solely on gravity to cause it to descend. Will that be an issue?
  16. Alec_t

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

    I skimmed through it, but it doesn't answer the questions I raised in post #139.
  17. Alec_t

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

    Without knowing what is supposed to happen after 1.26m vertical movement its hard to advise on the cam profile. For example, if the cam suddenly stops, the load (and cam-follower) will carry on moving up until slowed to a halt by gravity and friction. Is that the intention? Or is the cam...
  18. Alec_t

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

    That would help to give a constant slope, if that is indeed what you want. However, as the load is accelerating you will also then need to accelerate the rotation speed of the cam to keep the cam follower in contact with it. Have you considered that? A further point. Up to now you have not said...
  19. Alec_t

    What is the value of this resistor (or is it possibly not a resistor)?

    .... which suggests the fault is in something downstream of that resistor drawing excess current..
  20. Alec_t

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

    The calculations so far have assumed the cam presents a constant slope to the cam follower. In reality that won't be the case (at least for the usual spiral forms), so your refinements need to take actual slope into account in calculating force, torque and acceleration. Sorry, but I don't have...
View older results
Top