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

#### Maglatron

Jul 12, 2023
1,798
because my intuition says somethings up, bassically I don't know, it would be harder to lift up that way, through the scissor than lifting it strait because you're decreasing the distance moved by a factor of 10 (the point at which the cam touches) in the same time it takes to move the weight 10x further how can the force be the same?!?

#### Maglatron

Jul 12, 2023
1,798
Consider the energy used. Work = force x distance.
I think that that is my point 2.7N * 1.26 = 3.39J but if you take the 2.7N at the bottom of the scissor with force 2.7N and it moves 0.126m that equals 0.3402J!!

#### bertus

Moderator
Nov 8, 2019
3,412
Hello,

Most times the drive point is at an other location as yours.
Have a look at the attached pdf.

Bertus

#### Attachments

• scissors lift.pdf
1.1 MB · Views: 1

#### Maglatron

Jul 12, 2023
1,798
Hello,

Most times the drive point is at an other location as yours.
Have a look at the attached pdf.

Bertus
so I have seen this pdf before and stated a while ago that the point of application of the force is not included in the document there is some that come close and I bet someone on here can utilise the equations in that document and modify for my application but my maths is not up to scratch, thanks if you could move over point P to central and point Q to the intersection up to the right these might be helpfull!!! see what you guys can come up with

#### Maglatron

Jul 12, 2023
1,798
so I worked out that 3.392920066J / 0.126 = 26.92793703N at the cam

#### Maglatron

Jul 12, 2023
1,798
so I worked out that 3.392920066J / 0.126 = 26.92793703N at the cam
does that sound right? because energy over distance equal force?

Jul 7, 2015
3,694

#### Maglatron

Jul 12, 2023
1,798
don't mean to confuse but I changes somethings to make the theoretical machine makeable ( the first design was not plausible because the gear on the shaft of the flywheels radius was 1/5 of a millimeter - now the radius is 35mm) the height the block moves is now 1.32m and the weight is now 0.2622kg
so the cam is still 0.45m at the highest points of the lobes but 0.318m for the lowest points on the cam
it still has two lobes
Radius of cam avg = 0.45 + 0.318 / 2 = 0.384m
spiral length = pi * 0.384 = 1.206371579m
slope = Rmax - Rmin / pi * Ravg (0.45 - 0.318) / (pi * 0.384) = 0.01613449151

v = (0.95 * 1.32) / (0.95 * 4) = 0.33m/s
F = v / 0.05 *4 = 1.65N
F + g = 1.65 + 9.80665 = 11.45665N
lowest point on cam 0.318 * by force 11.45665 = torque 3.6432147Nm

so this need altering but I don't know how!!

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#### Alec_t

Jul 7, 2015
3,694
don't mean to confuse
Well, you're confusing me I'm afraid, with all the changes you keep making. Sorry, but I can't spare the time to go through all the maths again.

#### Maglatron

Jul 12, 2023
1,798
ok it's only a minor change to the cam (0.318m lowest part of cam, largest 0.45m) rotates at 2pi in 4sec and so lifts the weight through the scissor every 2 seconds, and weight 0.2622kg and distance it moves 1.32m I really do appreciate your time if you could have a go one last time before writing me off please mr Alec_t

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#### Maglatron

Jul 12, 2023
1,798
I get it - I've got to go to college where they are paid to help and that you can only expect so much of people who are just helping where they can! thanks anyhow

#### Alec_t

Jul 7, 2015
3,694
Which bit of the calculation set are you having a problem with?
I don't agree with the slope value or v value in post #268, btw.
When are you going to start making allowances for friction and scissor weight? Without those parameters all your calculation results so far could be way out from practical values.

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#### Maglatron

Jul 12, 2023
1,798
okay so the initial force needs to be calculated that accelerates the weight (in the first 10% of the 2 second cycle) to the velocity that it stays at (for the remaining 90% of the 2 second cycle) - the force to keep it at that velocity till it reaches the top of the 1.32m, then along with the slope, what's the torque required given the cam has two lobes and at the widest radius of the cam is 0.45 and the lowest point is 0.318m and will make the scissor and weight move up in 2 seconds (the cam makes a full revolution in 4 seconds then weight is 0.2622kg and it moves up 1.32m, then when I have all the calculations made in a perfect system I will remove the losses due to friction inertia of the cam and the weight of the scissor, thanks Mr Alec_t

#### Maglatron

Jul 12, 2023
1,798
slope = Rmax - Rmin / pi * Ravg (0.45 - 0.318) / (pi * 0.384) = 0.1094190234
v = (0.9 * 1.32) / (0.9 * 2) = 0.66m/s

I'm planning on using lightweight plastic for the scissor mechanism with a guide so that it doesn't bend or cripple (its only got to lift 262grams) and for the friction I was just going to go for 90% of the starting energy to compensate for the friction through the levers and gear meshes and bearings and I think thats conservative to be fair

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#### Alec_t

Jul 7, 2015
3,694
Let us see a pic of the completed project in due course. Good luck with the build.

Jul 12, 2023
1,798
thanks

#### Maglatron

Jul 12, 2023
1,798
if you do happen to have a change of heart and get a minute to share your expertise on the problem I would be seriously overjoyed:
okay so the initial force needs to be calculated that accelerates the weight (in the first 10% of the 2 second cycle) to the velocity that it stays at (for the remaining 90% of the 2 second cycle) - the force to keep it at that velocity till it reaches the top of the 1.32m, then along with the slope, what's the torque required given the cam has two lobes and at the widest radius of the cam is 0.45 and the lowest point is 0.318m and will make the scissor and weight move up in 2 seconds (the cam makes a full revolution in 4 seconds then weight is 0.2622kg and it moves up 1.32m, then when I have all the calculations made in a perfect system I will remove the losses due to friction inertia of the cam and the weight of the scissor, thanks Mr Alec_t

#### Maglatron

Jul 12, 2023
1,798
is there anybody else that might have some input to my querie??

#### Maglatron

Jul 12, 2023
1,798
Which bit of the calculation set are you having a problem with?
I'm specifically having problems with the force required to lift the weight at the point where the cam touches the scissor mechanism

#### Alec_t

Jul 7, 2015
3,694
Force = mass x acceleration (not forgetting g-force). Acceleration is controlled by cam profile and cam rotation rate.
In view of all the changes you keep making, why not put all the formulae you need into a spreadsheet so that changes to parameters result in any dependent parameters being updated automatically?

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