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

[Delta Prime]

I think you mean trouser pockets

I'm happy we are separated by an ocean. You guys just weird...

 

[Maglatron]

F off did you get chatgpt to draw that for you

 

[Delta Prime]

I just googled difference between trousers and pants... Weird is good I'm not judging.

still don't know what the scissor has to do with windmills

That's good! That was a psyc- test
You're not supposed to find a correlation between windmills and scissors.
But if you decide to charge it in full armor
Riding horseback you would tell us right?

 

[Delta Prime]

quid skrillar moola wonga p's coinage doe paper etc

Gotcha." Wonga " had to look that one up.

 

[Maglatron]

Yes.

If you do happen to have any other questions fire away, I'm ready!

 

[Maglatron]

trying to work out the force at gear 5 (input) can't get an answer from chatgpt, it's giving a different answer every time

 

[Maglatron]

anyone?

 

[bertus]

Hello,

Here are some pages about gearing forces:

Gear Forces | KHK Gears

Gear Forces is a page to study the forces in various types of gears in details. This is a part of KHK's Gear Technical Reference for machine designers.

khkgears.net

Force Analysis

Gear forces in spur and helical gears.

drivetrainhub.com

I hope you can get the wanted info from that pages.

Bertus

 

[Maglatron]

thanks Mr Bertus! that info is good but was kind of asking for the simplistic answer or formula to work out tangential force on the input fifth gear if the out put on gear 1 is 5.142857143 with this coumpound gear train the gear 5 turns 1/4 of the output!

 

[Maglatron]

can anyone have a crack? I'd prefer a formula though because I want to experiment with different sizes of gear 5! thanks

 

[Alec_t]

You have 4:1 gearing, so the torque at gear 4 shaft = 4 x the torque at gear 1 shaft. The tangential force on the 20-tooth gear is the gear 4 shaft torque divided by the radius of the 20-tooth gear.

 

[Maglatron]

ok cool so its 31.16883117N cool thanks

 

[Maglatron]

so whatever I do the energy to move the 10.8km^2 inertia flywheel at 0.05rad/s^2 that weights approx 40kg in the ring format the energy is 3.392920066J but if I have (1 newton then that weighs approx 102grams) J/N = meter (high) then the height it needs to lift up is 3.392920066m and the lever arm is s = r theta 3.392920066/0.3490658504 equals 9.72m lever arm the flywheel moves at 2pi rad/s initially the wheel is about 392 times the mass and it's moving I think the problem is starting me in the face but common sence says that I should be able to use the mass of the moving wheel to shoot the 102gram wheight up easily

 

[Maglatron]

if the force downawrd is 0.25N (25gram 0.02549290532kg) it needs to lift 13.57168026 meters It just makes sence that the moving mass of 40kg 10.8kgm^2 can lift the 25 gram weight 13.57168026m easily given the right mechanism what do you guys think if it hits with impact to "shoot" it up and not lose 3.392920066J in doing it, I think it's about power which is J/sec so if I move it up slower than it falls eg 1 sec to fall equals 3.392920066watts but if it lifts up in 2 seconds then it will use 1.69460033watts to lift

 

[Maglatron]

weather the distance of the arc of the lever is lifted 1.319468915m and 2.571428573N downwards or arc of 13.57168026m and 0.25N the energy is = to 3.392920066J fo acceleration of 0.05rad/s^2 and inertia of 10.8kgm^2

 

[Alec_t]

what do you guys think

I think you're confusing yourself and us by using too many decimal places and not enough punctuation in your posts.

 

[Maglatron]

sorry

so whatever I do, the energy to move the 10.8km^2 inertia flywheel at 0.05rad/s^2, that weights approx 40kg - in the ring format, the energy is 3.3929J. But if I have (1 newton then that weighs approx 102grams) then J/N = 3.3929J / 1N =meter (in this case the distance moves up, high) then the height it needs to lift up is 3.3929m, and the lever arm is r, and s = r theta 3.3929/0.34906 equals 9.72m = lever arm, the flywheel moves at 2pi rad/s initially, the wheel is about 392 times the mass, and it's moving. I think the problem is starting me in the face, but common sence says that I should be able to use the mass of the moving wheel to shoot the 102gram wheight up easily

if the force downawrd is 0.25N (25gram 0.02549kg), it needs to lift 13.5716 meters, upward. It just makes sence that the moving mass of 40kg 10.8kgm^2; can lift the 25 gram weight 13.5716m easily - given the right mechanism, what do you guys think? if it hits with impact to "shoot" it up - and not lose 3.3929J in doing it, I think it's about power which is J/sec, so if I move it up slower than it falls eg 1 sec to fall equals 3.39292watts, but if it lifts up in 2 seconds then it will use 1.6946watts to lift

weather the distance of the arc of the lever is lifted 1.3194m and 2.5714N downwards, or arc of 13.5716m and 0.25N, the energy is = to 3.3929J for acceleration of 0.05rad/s^2 and inertia of 10.8kgm^2

now I'm going the other way heavier weight, less height, so less stage of scissor and for a cam profile of 5% slope!!
hope that you can read it now!!

 

[Maglatron]

so now I'm thinking 3.3929J / 0.3m (height) = 11.3097N approx 1.1528kg, and have a cam that varies with 1 lobe 10cm drop off (need to design the cam so that the slope is 5%) and have the scissor have only 3 levels, this way Fv will be 3*f and then so Fv = 3 * 11.3097 = 33.9292, and the Hf will be = 5% of Fv, 33.9292 * 0.05 = Hf = 1.6964N

From energy considerations, for a n-stage scissor mechanism the vertical force Fv the cam needs to provide will be n x the vertical force fv needed to accelerate the load at the top. But if the cam has a s% slope then the horizontal force Fh on the cam follower will be s% of Fv,
If n=10 and s=10% (which I think is the arrangement you presently have) then Fh will just happen to equal the force f.
It is Fh which determines the cam torque needed.

need to design a cam that varies over it's slope 10cm and must be 5% any ideas people, 1 lobe! max radius min radius or a formula - must tick 5% slope box and 10cm from max to min box?? ideas, thanks does not need to be 10 cm the important part is it needs to be 5% slope, thanks but would be good if the solution ticked both boxes for the design of the cam

 

[Maglatron]

slope = Rmax - Rmin / 2pi * Ravg

can I rearrange this formula for 5% slope??
average radius Ravg half way between the minimum and maximum radii Rmin, Rmax.
the smaller the lower radius the better because lower radius = lowers required torque!

 

[Harald Kapp]

can I rearrange this formula for 5% slope??

Obviously you can't, why else would you ask

average radius Ravg half way between the minimum and maximum radii Rmin, Rmax.

makes for Ravg = 1/2 × (Rmax+Rmin).
Insert into
slope = (Rmax - Rmin) / (2×Pi × 1/2×(Rmax+Rmin)) = (Rmax - Rmin) / (Pi × (Rmax+Rmin)) = 0.05.
You now have an equation with two variables: Rmax and Rmin. You need to define one of the two, e.g. Rmin, then solve the equation for Rmax. Or vice versa.