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

[Alec_t]

torque needed to get 3.05N out of the cam at the furthest from the axis please

That's where the maths starts getting tricky. Think of the cam-follower climbing a slope presented by the cam spiral. The precise shape and length of the spiral will affect the relation between torque and vertical force at any rotational angle of the cam. The formula for the length of the spiral is not trivial (according to ChatGPT). Additionally there will be friction between cam and cam-follower.
As an approximation you could consider the spiral has an average radius Ravg half way between the minimum and maximum radii Rmin, Rmax. The spiral length would then be 2 x pi x Ravg. Over that length the cam-follower moves a vertical distance Rmax - Rmin. So, the average slope is (Rmax-Rmin)/(2 x pi x Ravg). The rotational force needed, ignoring friction, would be 3.05N x slope acting at a radius of Ravg, which is a torque of 3.05 x slope x Ravg Nm.

 

[Maglatron]

just for my understanding can you eplain wih arbitrary numbers, or even the numbers im working with Rmin=0.05m Rmax=0.126m also if it was designed for using Rmax instead of ravg then that would over compensate and therefore be enough if that makes any sense, thanks
I think this is a question calculus was invented for

 

[Maglatron]

I have a snail shell drop cam who's larger radius is 0.126m and minimum radius 0.05m I want it to rotate and push up on a follower with 3.05Newtons the cam turns through angle 2pi rad before resetting I need to know the torque I need to produce this motion in 0.25seconds

Step 1: Calculate the Angular Velocity​

calculate angular velocity
Given that the cam completes one full rotation (2π radians) in 0.25 seconds, we can calculate the angular velocity (ω) as follows:


ω= dtheta/dt 2pi rad/0.25seconds= 8pi rad/sec

Step 2: Determine the Force Exerted by the Cam​

The force exerted by the cam on the follower is given as 3.05 Newtons.

Step 3: Average Radius of the Cam​

radius Ravg=(Rmax+Rmin)/2=0.088m

Step 4: Calculate the Torque​

Torque (τ) is the product of the force (F) and the radius (r) at which the force is applied:
3.05 * 0.088 = 0.2684Nm of torque

Summary​

The torque needed to rotate the cam and push up on the follower with a force of 3.05 Newtons, completing a full rotation in 0.25 seconds, is 0.2684 Nm0.2684Nm
and using calculus brings the exact same number!! so it seems this approach is sufficient!

 

[Maglatron]

my result is different to yours Mr Alec_t

 

[Alec_t]

... because you are not taking the slope factor into account.
Unless you know the cam profile (i.e what is the relationship between radius and rotation angle) and allow for friction you can't expect an accurate answer for the torque.

 

[Maglatron]

ok how do you do that, could you do the calculation so I can see where I went wrong please? can you clarify that the answer is 0.0369Nm
so I used the chatgpt and entered the variables and it confirms the torque of 0.03696Nm

 

[Alec_t]

Average slope = (0.126-0.05)/(2 x 3.14 x 0.088) = 0.137.
Since this mythical frictionless slope is linear and constant, the horizontal force to move the cam-follower up it will be constant (the same at Rmin and Rmax) at ~ 3.05N x 0.137 = 0.42N. So the torque at Rmax would be ~ 0.42 N x 0.126 m = ~ 0.053 Nm.
This calculation is for a quasi-static situation; not for accelerating the load at a particular value.

 

[Maglatron]

consider the spiral has an average radius Ravg half way between the minimum and maximum radii Rmin, Rmax. The spiral length would then be 2 x pi x Ravg. Over that length the cam-follower moves a vertical distance Rmax - Rmin. So, the average slope is (Rmax-Rmin)/(2 x pi x Ravg). The rotational force needed, ignoring friction, would be 3.05N x slope acting at a radius of Ravg, which is a torque of 3.05 x slope x Ravg Nm.

from what you said Ravg halfway between 0.126 and 0.05 = 0.088
spiral length = 2pi * 0.088
average slope = (0.126 - 0.05)/(2pi * 0.088) = 0.13745
rotational force needed, ignoring friction, would be 3.05N x slope acting at a radius of Ravg
3.05 x slope x Ravg Nm = 3.05 * 0.13745 * 0.088 = 0.0369Nm
a little confused!

 

[Maglatron]

Average slope = (0.126-0.05)/(2 x 3.14 x 0.088) = 0.137.
Since this mythical frictionless slope is linear and constant, the horizontal force to move the cam-follower up it will be constant (the same at Rmin and Rmax) at ~ 3.05N x 0.137 = 0.42N. So the torque at Rmax would be ~ 0.42 N x 0.126 m = ~ 0.053 Nm.
This calculation is for a quasi-static situation; not for accelerating the load at a particular value.

so I think I know what you're saying the torque at the tip of the cam before it drops needs a slight increase in torque and if I made that the torque on the input then it would be sufficient to turn through the entire angle 2pi rad

 

[bertus]

Hello,

I have corrected the quote.
About the time failure I found this:
https://answers.microsoft.com/en-us...n-failed/efdd97ba-b362-4b84-a51f-b3d83d3e792e

Bertus

 

[Maglatron]

so I want to relate the speed of the wheel which is 3.3873422 rad/sec to that of the cam, I want the cam to rotate once in 0.25 seconds which is 8 rad/sec if the gear on the flywheel is 0.01m then velocity equals 0.033873422m/s tangientially add compound gear so that the smaller is meshed with the gear on the wheel and make the larger on the compound drive the cam perhaps through another gear, now we have the correct angular velocity for the cam we then work out the force back to the wheel through the gear system and then the new torque and therefore work out decceleration rough guide

 

[Maglatron]

so I've found that if the speed of the input gear is 3.3874322rad/sec and the output speed is 8pi and the output torque is 0.053Nm then the input torque is 0.3932Nm and the Gr is 1 : 7.4196 now I need to design a compound gear
-0.3932Nm/10.8(the inertia)= negative acceleration (-0.03640740741/sec^2) this is alpha
the equation ω2 = ω1+ alpha * t
ω1 equal to 3.3873422rad/sec
time = 0.25sec
what is ω2 of the flywheel
okay alpha * t = -0.009101852852
3.3873422+-0.009101851852 should equal ω2 and I get3.378340348rad/sec
this is really good !!!
because on the last drop of the weight initial velocity was 3.293291461rad/sec and it accelerated to 3.3873422rad/sec in 1.881014774sec
now it lifts the wheight to the top height of 1.26m in 0.25sec and only looses 0.009101851852rad/sec the inital RKE was 61.96007076Joules
and the new RKE is 61.62754238Joules difference of 0.3325283762Joules and seeing as the block at full height is mgh
(0.2753233775kg *m9.80665m/sec^2 * 1.26m) the stored potential was 3.392920066Joules
now its a matter of using the gear ratio and making a suitalbe compound gear and design the cam and scissor but proof of concept has been finalised (in my head anyway)

 

[Maglatron]

now need to add in losses for realism

 

[Martaine2005]

losses for realism

That’s the best comment you’ve made since you started.
Your math has drifted into pico and micro. Why?.
Tools to measure this costs millions of pounds. Just my 2 cents.

 

[Maglatron]

started the working from the beggining theres a couple of little creases to iron out but it's going well I'm going to put together a PDF file explaining from beggining to end, thanks

 

[Maglatron]

That’s the best comment you’ve made since you started.
Your math has drifted into pico and micro. Why?.
Tools to measure this costs millions of pounds. Just my 2 cents.

Your math has drifted into pico and micro. Why?.

It looks like you're looking at the size of the number and not following the math of how they were derived or what they mean and stand for! thanks

 

[Delta Prime]

not following the math of how they were derived or what they mean and stand for!

Pooh. Gaslighting. you're a bad boy.

It looks like you're looking at the size of the number

1 2 3 4 5 6

 

[Maglatron]

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 gaslighting I need to lift the weight in 0.25 seconds does this affect the torque or can I use the 0.053Nm still I want the cam to rotate at 8pi rad/s thanks

 

[Delta Prime]

thanks and whats gaslighting

You must have been born on a highway because that's where most accidents happen!
That's gaslighting.
Your question is gibberish because you're not following the math and do not know how it's derived. I am smarter than you.
That's gaslighting.
And I will make condescending remarks as to why you're even on this site.
That's gaslighting. That's being a bad boy.

 

[Maglatron]

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 know what else to say wouldn't be too bothered by that to be fair