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

[Alec_t]

Re-thinking the maximum torque needed, it won't occur at the maximum radius of the cam as assumed so far; it will actually be at the minimum radius by my reckoning, for the following reason.
You want the weight to rise at constant velocity; but when the weight falls to its lowest point it instantly has zero velocity and then has to reverse to have a positive (upward) velocity. Let's assume that for 90% of the 0.5sec rise time the velocity v is constant. So v=(0.9 x 1.26m)/(0.9 x 0.5s) m/s = 2.52m/s. That means v has to reached from zero by an acceleration f in the first 10% of the 0.5sec, so f= v/(0.1 x 0.5) = 2.52/0.05 = 50.4m/s^2. That's about 5g which the cam has to provide at or near the minimum radius, in addition to the 1g to counteract gravity.

 

[Maglatron]

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

 

[Alec_t]

The more I look at your set-up the more I'm reminded of Roland Emmett's gizmos

 

[Maglatron]

ha

 

[Alec_t]

im thinking with the flywheel effect that is driving the cam it will "iron over" the "lump"

So the flywheel drives the cam, the cam drives the scissor mech, the mech moves the weight, the weight moves the gears, the gears rotate the flywheel, the flywheel drives the cam, ........ I hope you're not expecting perpetual motion?

 

[Delta Prime]

. I hope you're not expecting perpetual motion?

thanks to the fundamental physics of our universe, perpetual motion machines are impossible.We'd be wrong about simply everything, and nearly none of our observations would make any sense.
The first law of thermodynamics is the law of conservation of energy.









It states that energy is always conserved. It means that energy can be neither created nor destroyed. Instead, it simply changes from one form to another.
The second law of thermodynamics
I carry with me in my avatar.












Even today physicists are still considered to be "Not" cool.
But for me, thermodynamics entropy is getting cooler all the time.

 

[Maglatron]

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 input torque therefore -0.23526Nm make it negative because it's taking from the flywheel -0.23526Nm divide that by the inertia of the flywheel (10.8kgm^2) equal acceleration or rather deceleration -0.02178333333rad/s^2 then equation w2 = w1+alpha * time 3.362994729rad/s + - 0.02178333333rad/s^2 * 0.5 so w2 = 3.352103062
then 3.3629924729 - 3.352103062 = 0.01089166667
the intitial RKE = 3.362994729rad/s = 61.07256116J
3.352103062rad/s = 60.67761267J
61.07256116J - 60.67761267J = 0.3949484933J
block starts with PE 2.7 * 1.256637061 = 3.392920065J
3.392920065J : 0.3949484933J = approx 9 :1
9 steps forward 1 step back
I have purposly avoided that term "perpetual motion" however I do have the information that I was after now although it is interesting about the starting torque being alot higher at the start of the cam rotation

 

[Alec_t]

A problem I see with your gearing is that as the weight rises it will cause (or assist) the flywheel's anti-clockwise rotation, but that continued rotation will prevent the weight from subsequently descending.

 

[Maglatron]

So the flywheel drives the cam, the cam drives the scissor mech, the mech moves the weight, the weight moves the gears, the gears rotate the flywheel, the flywheel drives the cam, ........ I hope you're not expecting perpetual motion?

A problem I see with your gearing is that as the weight rises it will cause (or assist) the flywheel's anti-clockwise rotation, but that continued rotation will prevent the weight from subsequently descending.

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 lifting the backwheel of a pushbike and pushing down on the pedal once then lifting the padal back to it's start position, the wheel remains turning! and because it's on a sprag any anti-clockwise motion will not affect the clockwise motion of the flywheel because it's on a spag or (freewheel)

 

[Alec_t]

If the cam (not the gearing) provides the force to lift the weight and you want the weight to fall freely (with the gearing disengaged) then the gearing serves no useful purpose that I can see.

 

[Maglatron]

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 scissors mechanism as the weight lifts up it is true that the gears are in opposition to the motion of the wheel but the sprag is the for all intents and purposes - the spur gear on the flywheel
I think this is what you mean, thanks
gearing is engaged when the weight falls - that's what turns the flywheel!

 

[Maglatron]

Re-thinking the maximum torque needed, it won't occur at the maximum radius of the cam as assumed so far; it will actually be at the minimum radius by my reckoning, for the following reason.
You want the weight to rise at constant velocity; but when the weight falls to its lowest point it instantly has zero velocity and then has to reverse to have a positive (upward) velocity. Let's assume that for 90% of the 0.5sec rise time the velocity v is constant. So v=(0.9 x 1.26m)/(0.9 x 0.5s) m/s = 2.52m/s. That means v has to reached from zero by an acceleration f in the first 10% of the 0.5sec, so f= v/(0.1 x 0.5) = 2.52/0.05 = 50.4m/s^2. That's about 5g which the cam has to provide at or near the minimum radius, in addition to the 1g to counteract gravity.

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 accelerating for such a short period of time it won't affect the flywheel very much at all!!

 

[Alec_t]

if the velocity is reached in 1% of the time

That will require an acceleration of ~50g ! Somethings going to break!