yes drop to minimum
did you read the pdf??
I want to know all of the maths concerning this scissor mechanism!
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so I think I've settled on the design I'm going to have 3 lobes and it will take 0.25sec to lift the load, 3x if turning 2pi rad/sec
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I skimmed through it, but it doesn't answer the questions I raised in post #139.
the scissor lifts a weight of 175g the cam turns clockwise without stopping it lifts the weight up and drops down again when the lobes go past each time
not at all! infact that would be good but why would it continue up and for how much further would it move when gravity takes over and the edge of the cam drops off at the end of the lobe I see this like accelerating a tennis ball up wardsat a certain accelerate the ball would leave the hand also the weight of the lever which I have not accounted for
The weight through the lever/ gear system has to accelerate against the inertia of the flywheel so it wouldn't be freefall if the flywheel is spun by the dropping weight through levers and gears and the initial rotational speed of the flywheel is 12pi rad/s 6 rotations/second or 360RPM and its inertia is 10.8kgm^2 and the acceleration is 0.05 rad/s^2 of the flywheel how to find the time the weight falls a distance of 1.26m and what will be the speed of the flywheel after the drop of the weight
this is important to me, thanks
this is important to me, thanks
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so imagine this: you're peddling down hill in first gear the bike gets too fast for your legs to keep up, its like this how fast can the wheel be going before its at the speed when the weight drops and it's no longer adding to the wheel because it requires the weight to drop faster than gravity pulls it down? for example 4pi rad/s, 6pi rad/s, 8pi rad/s and for the given cam the 0.9m diameter with the 3 lobes how do you work out the slope of the circular ramps it goes 2pi rad/s?
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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 components.
The weight and rotational inertia of the scissor mechanism itself, not considered yet, need to be taken into account in acceleration and deceleration calculations.
The maths is now above my pay grade; perhaps someone else would be willing to wade in.
I hope you can see by now why I objected to nine decimal places in your calculations
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The weight and rotational inertia of the scissor mechanism itself, not considered yet, need to be taken into account in acceleration and deceleration calculations.
The maths is now above my pay grade; perhaps someone else would be willing to wade in.
I hope you can see by now why I objected to nine decimal places in your calculations
.hmm just need to find the slopes first of the cam if you can shed any light on that, one step at a time! so I worked out that the wheight drops as 0.5068m/s free fall and that it takes 0.4995sec for the weight to drop if the wheel starts at 4pi rad/s so I would only 2 lobes on the cam rotating at 2pi rad/s if you can help me with the slopes as one last thing I'd be highly greatfull, thanks
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going to use this formula and where it says 2pi I will just use pi because the slopes are half rotation
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 d[imath]Θ[/imath], the cam moves along an arc of length r.d[imath]Θ[/imath] and moves vertically dr.
The slope is dr/r.d[imath]Θ[/imath]. So for a constant slope the increase in radius is proportional to both the rotation angle and the prevailing radius at that angle.
An alternative to a constant slope profile would be the Archimedian one, where r is proportional to [imath]Θ[/imath] only.
For an angle of rotation d[imath]Θ[/imath], the cam moves along an arc of length r.d[imath]Θ[/imath] and moves vertically dr.
The slope is dr/r.d[imath]Θ[/imath]. So for a constant slope the increase in radius is proportional to both the rotation angle and the prevailing radius at that angle.
An alternative to a constant slope profile would be the Archimedian one, where r is proportional to [imath]Θ[/imath] only.
this is for half the rotaion of the cam
Consider the spiral has an average radius Ravg half way between the minimum and maximum radii
Rmin, Rmax. (0.45 +0.324)/2 = 0.387m
The spiral length would then be π x Ravg. π * 0.387 = spiral length 1.215796357m
Over that length the cam-follower moves a vertical distance Rmax - Rmin. 0.45 – 0.324 = 0.126m
So, the average slope is (Rmax-Rmin)/(π x Ravg). (0.45 – 0.324)/(π * 0.387) =0.1036357769
The rotational force needed, ignoring friction, would be F = m * a (0.2753233775kg * 19.85974649m/s².) = 5.4678248N
then x slope
5.467852479N * 0.1036357769m =0.5666651397N acting at a radius of Ravg, which is a torque of 0.5666651397N x 0.378 Nm. = 0.214199422Nm
But we use the max radius of the cam
0.5666651397N * 0.45 = torque of 0.254999129Nm
Consider the spiral has an average radius Ravg half way between the minimum and maximum radii
Rmin, Rmax. (0.45 +0.324)/2 = 0.387m
The spiral length would then be π x Ravg. π * 0.387 = spiral length 1.215796357m
Over that length the cam-follower moves a vertical distance Rmax - Rmin. 0.45 – 0.324 = 0.126m
So, the average slope is (Rmax-Rmin)/(π x Ravg). (0.45 – 0.324)/(π * 0.387) =0.1036357769
The rotational force needed, ignoring friction, would be F = m * a (0.2753233775kg * 19.85974649m/s².) = 5.4678248N
then x slope
5.467852479N * 0.1036357769m =0.5666651397N acting at a radius of Ravg, which is a torque of 0.5666651397N x 0.378 Nm. = 0.214199422Nm
But we use the max radius of the cam
0.5666651397N * 0.45 = torque of 0.254999129Nm
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.
its not really a problem if it moves with constant velocity or not but I thought that if I put an unbalanced force that there would have to be acceleration also if the cam is moving 1 turn a second or 2pi then it would'nt be accelerating any way the slope would just lift up the weight at a constant with the slope all I know is that the cam needs to overcome the 2.7N to lift the weight and oh yeah it would take 0.5 second for each lift of the weight twice/revolution
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