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

Maglatron

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so if the snail shell cam max radius is 0.12m and the radius of the thinest part of the cams radius is 0.05m spiraling outward then I would need 10 levels of scissor linkage, because the furthest radius on the cams axis is 0.12m one rotation of the cam would lift the 0.275kg weight up to the top that is 1.26m high
 
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Maglatron

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F = m g
gravitaional force, weight mass m (0.2753233775 kg)
g acceleration due to gravity 9.80665m/sec^2

Fgrav = 2.7N

force for upward acceleration F = m a
m = (0.2753233775 kg)
a = 1m/sec^2 (arbitrary acceleration)

Facc = is just 0.2753233775N

total force required

Ftot = Fgrav+Facc
2.975323377N

or 2.98N rounded off
modify the force for upward acceleration so that it accelerates at 1.26m/sec^2
Ftot = Fgrav + Facc
3.05N
so the weight would lift in 1 second with this force applied! so because the cam moves the pivot on the 10th level it need to push with 10 times the force = 30.5N now not too sure how to work out the next bit to do with the transfer from the wheel to the cam
 
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Maglatron

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can anyone point me to a page or pdf that explains the math of the snail cam mechanism ie torques, radial force, power, time etc the way I see it is a rotational ramp incline
 

Maglatron

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It might not even need acceleration the motion upwards could be a constant velocity with the force being the same as the downward load of 2.7N
yep I need a nudge in the right direction with this, thanks
 

Alec_t

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and by the mechanical advantage if I put a unit of force into the level 4 then the it would get divided by 4 too
I didn't say that at all. I said there is no mechanical advantage, meaning the applied upward force = 1 x the downward load. To get a mechanical advantage you need to apply a horizontal force between the left and right pivot points at the bottom leve, such as is done with a screw mechanism in a vehicle scissor jack.
 

Maglatron

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ok I misunderstood
I didn't say that at all. I said there is no mechanical advantage, meaning the applied upward force = 1 x the downward load. To get a mechanical advantage you need to apply a horizontal force between the left and right pivot points at the bottom leve, such as is done with a screw mechanism in a vehicle scissor jack.
nonetheless if I moved an upward force up at that pivot, a distance dy, then at stage 4 then it would have moved 4*dy. And is it true when I said if the cam's max distance from the axis of rotation is 0.126 the and has 10 levels then the top of the scissor will have moved 1.26m = 10 * dy
 

Alec_t

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if I moved an upward force up at that pivot, a distance dy, then at stage 4 then it would have moved 4*dy. And is it true when I said if the cam's max distance from the axis of rotation is 0.126 the and has 10 levels then the top of the scissor will have moved 1.26m = 10 * dy
Yes and yes.
 

Maglatron

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I didn't say that at all. I said there is no mechanical advantage, meaning the applied upward force = 1 x the downward load. To get a mechanical advantage you need to apply a horizontal force between the left and right pivot points at the bottom leve, such as is done with a screw mechanism in a vehicle scissor jack.
ok such as is done with a screw mechanism in a vehicle scissor jack. I understand now if it was not from the horizontal the the weight of the car is going down and that would be the weight force of the car downward and would be like lifting it directly
if the load is 2.7N (2753233775kg / 9.80665n/m^2) then I need an upwards force of 2.7N from the cam? thanks is that correct?
 

Maglatron

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ok cheers mate got a book on mechanisms and related math on monday too1716642805830.jpeg
 

Maglatron

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this might sound stupid but if the forces are in equilibreum (downwars force 2.7N and upward force from the edge of the cam 2.7N) how/why will the scissor/cam mechanism move the weight upwards?
 

Maglatron

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thanks Mr bertus! my internet does not allow for that site
 
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Maglatron

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I understand that something will stay at rest or in uniform motion unless acted upon by an outside force and the forces are balanced but a little confused!
this might sound stupid but if the forces are in equilibreum (downwards force 2.7N and upward force from the edge of the cam 2.7N) how/why will the scissor/cam mechanism move the weight upwards?
 

bertus

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Hello,
my internet does not allow for that site
Why does your internet block this open website?
Can you change the DNS settings?
If so, you could try OpenDNS:

Bertus
 

Maglatron

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Hello,

Why does your internet block this open website?
Can you change the DNS settings?
If so, you could try OpenDNS:

Bertus
1716646466025.png
my internet blocks allsorts
 

bertus

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Hello,

It is a security setting in Firefox.
Can you use Brave as internet browser?
 
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