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A Math Condundrum

I'm drawing on a simple system my dad had on a 3 wheeled crane we used to lift wall panels & set them when building a house. It involves a inexpensive winch some 3/8" aircraft cable some pullies & a 1/2 " drill motor to drive the winch. Dad removed the drill handle & remounted it so there was a cord between it & the drill motor a walk around corded remote with trigger & reverse switch, a single speed motor will work fine. I think a key lock out switch would be wise to keep kids from playing with it would be wise. a forward & reverse switch top & bottom & a current on switch top & bottom & a track for the platform to follow. I would make the platform able to fold up against the unit when not in use. In all actual reality you would only need the control at the top of the run. You could run it down when you leave to go shopping & load it upon return then go up & run it up just using the drill motor as manufactured. So the winch & the drill motor are both at the top of the run, maybe inside a protective locked box.
 
My method used before is mildly inaccurate but didn't involve to much thought on my part to start with (and gets a "close enough" answer). The correct solution is actually a bit easier, but I had to go look it up :D.

This link pretty nicely describes the difference between a class 1 (what you're thinking of: fulcrum in the middle) and class 3 (what I was thinking of: fulcrum on the end opposite the load) lever and the relative distance advantages between them clearer than I can. The short version is that putting the fulcrum on the end of the lever (class 3) gives you significantly more lift for a given length of lever arm than a class 1.

https://www.school-for-champions.com/machines/levers_increase_distance.htm#.WonEfIKIbBI

So taking their formula:
DO/dO = DI/dI

We have multiple "knowns"

D0=36 - this is the lifting height, you have a constant goal of 36
D1=24 - the capacity of your jack which also constant at 24"
d0=48 - some assumed maximum length of the lever arm (we have to make something up for this)

So we can plug that in and get:
36/48 = 24/? which is 0.75 = 24/? which is 0.75 * 24 which is 18.0"

pre-doing the algebra that's d1 = D0/d0 * D1
which is: 36/?load arm length? * 24 so you can just plug in different load arm lengths and get an answer for the offset with that.

Thus for a class 1 lever you would need a total lever length of 66" (load arm of 48" plus effort arm of 18") with the fulcrum 18" from the jack to lift the load end 36". I could re-jigger the formula to get the total lever length but just plugging in a set of possible numbers is fast enough here.

For a class 3 lever you would need a total lever length of 48" (load arm of 48") with the jack 18" in from the fulcrum end to lift the load end 36".

The advantages of a 3rd class lever for this use case become reasonably apparent here. This isn't accounting for any problems of practicality in having angles that steep, that will depend on what you have for pivots, etc.. Note also that the ends of the load and effort arms describe an arc, so that will have to be accounted for in the design (the easiest way is probably to allow the fulcrum to pivot somewhat as well as both ends of the jack - that way you can have the lift go more or less straight up and down which will be easier to stabilize on a track or something).

Visually diagramming the two example cases for a class 1 and class 3 lever.
class1class3.jpg

Pre-calculating for some set of possible load arm lengths.. Remember that for a class 1 lever the total lever length is the load arm plus the effort arm but for a class 3 the effort arm is "inside" the load arm (so the load arm is the total lever length). Some of these (especially towards the short and long ends..) are likely infeasible due to the angles involved.

load arm effort arm
36 24.0
37 23.35
38 22.74
39 22.15
40 21.6
41 21.07
42 20.57
43 20.09
44 19.64
45 19.2
46 18.78
47 18.38
48 18.0
49 17.63
50 17.28
51 16.94
52 16.62
53 16.3
54 16.0
55 15.71
56 15.43
57 15.16
58 14.9
59 14.64
60 14.4
61 14.16
62 13.94
63 13.71
64 13.5
65 13.29
66 13.09
67 12.9
68 12.71
69 12.52
70 12.34
71 12.17
72 12.0
73 11.84
74 11.68
75 11.52
76 11.37
77 11.22
78 11.08
79 10.94
80 10.8
81 10.67
82 10.54
83 10.41
84 10.29
85 10.16
86 10.05
87 9.93
88 9.82
89 9.71
90 9.6
91 9.49
92 9.39
93 9.29
94 9.19
95 9.09
 
I am thinking so, Rennie. A linear actuator of the design I have is very much like a pneumatic or hydraulic cylinder. On this case, a screw drive via a small electrical motor.
 
How about something like this for the last twelve inches. Pull it up with a rope, add a stop at the top plus center mechanism like a lounge chair keeper to keep it from sliding back down.

carol.jpg
 
Carol Reed said:
I am thinking so, Rennie. A linear actuator of the design I have is very much like a pneumatic or hydraulic cylinder. On this case, a screw drive via a small electrical motor.
Click to expand...

Someone else will have to do the math, but it seems to me that, depending on the design of the scissor, you could get that 2' actuator capability to give you the 3' of vertical travel you need.
 
http://www.econolift.net/econolift_pdf/lifts_tilts/dble_scissor_lift_tables.pdf

The picture in this link will give you a conceptual view of what you want. Obviously a dual scissor gives you twice the lift. I would suggest getting some popsicle sticks and making a model, then you can scale your model up from there. Most of the ones I made were for assembly tables with adjustable height for individual preferences. I used 1 X 2 X 1/8 wall steel tubing which gave about 750 lb. capacity for industrial applications. A lighter material can be substituted. I'm certain plywood strips could easily be used for a table with a much lighter capacity. Keep in mind the linear orientation of the grain. Popsicle sticks and #6 nuts and bolts are a cheap way to figure out what you want. Hope this helps.
 
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Rennie Heuer said:
Someone else will have to do the math, but it seems to me that, depending on the design of the scissor, you could get that 2' actuator capability to give you the 3' of vertical travel you need.
Click to expand...

Yes, but its more than a little complicated. You might need to go to a two stage lift. Turns out designing scissor lifts is harder than designing simple levers by a lot (who'd have thunk.. its actually much worse than I'd have thought).
https://www.engineersedge.com/engineering-forum/showthread.php/1789-Scissor-Lift-Design-Help

The short version is that the initial angle of the lifts arms DRAMATICALLY impacts its load capacity, if you're starting near flat you're going to have a really bad time. The optimal starting angle is right about 45 degrees, at which point the load on the cylinder is about equal to the weight being lifted (its a non-linear increase from there). Figuring the load and distance requirements requires a lot of math that I forgot over 20 years ago... but we'll give it a shot stealing liberally from other peoples work.

Found a couple of calculators for figuring load depending on the actuator position:
https://www.engineersedge.com/calculators/scissor-lift-center-loaded.htm
https://www.engineersedge.com/calculators/scissor-lift-bottom-loaded.htm

In either case to lift 100lbs with a starting angle of 45d takes right about 100lbs of force, but if you drop the starting angle to 10degrees it takes closer to 600lbs of force. My understanding is that the force is roughly multiplied by the number of stages, so a 2 stage system lifting 100lbs at 10 degrees would require about 1200lbs of force.

As for the height, its Pythagoras to the rescue again if you know the length of you desired arm ("c") and the total width of the platform you need to solve for the other side twice then take the difference.

So if we had a 55" arm ("c") and started with it 10" up in the air ("a") then the base is calculated as square_root(55*55 - 10*10) = a 54" base. Push the base in 24" and we now have a triangle with a 30" base ("b") and a 55" arm ("c") so the height is square_root(55*55 - 30*30) or about 46" - and 46" (finish height) minus the 10" initial height gives us 36".

This also happens to work out to about a 10 degree starting angle ("A") so using the other calculator we can figure that we'll need around 600lbs of force to get 100 lbs moving or about 300lbs of force for 50lbs of weight. More than I'd have naively expected, I guess that's why they use hydraulics for these things.

I played around with some other numbers but it starts getting unattractive size wise pretty quickly with a single stage system. You can get 36" with a single stage with a 70" arm starting 20" off of the ground (giving a 16degree starting angle) which needs 350+lbs of force to move. The difference after that starts to smooth out.

For a two stage system if you had 30" arms with a 26" base (which gives a combined initial height of 36" - 15" for each arm) you would get 36" of lift, an initial angle of 30 degrees which is about 350lbs of force to lift 100lbs (175lbs to lift 50). This is probably approaching the "best" set of trade offs.. not sure how high you need the initial platform to be.
 
There might also be some interference problems with the sizes I figured.. a scale model like Roger suggested would help find those... (someone who's a much better engineer than I will ever pretend to be could undoubtedly determine that easily enough).
 
Just reading all the suggestions and solutions and I find them all good, but maybe too sophisticated, but if what you want to lift is basically groceries or similar weights, let's say about 10Kg.max (22 lbs) why not just use a pulley and an electric screwdriver attached to a reel? you would have enough force ( I guess) plus the possibility or reverse direction. If the screwdriver wasn't strong enough, you could use a pulley belt to demultiply the turns and get more power. Conecting it to a 12V DC transformer would get rid of the hassle of recharging bateries. You could have a commuted switch on top of the stairs and another one at the bottom to operate it from up stairs or down stairs.
A blinder motor would also do an you could mount it so that the rope would wound onto it, and those are powerful.
Just thinking...
 
Thanks, Toni. What is a blinder motor?

The heaviest thing are the three gallon water bottles. Don't think the screwdriver will lift them, even one at a time though.

Still noodling through this. Learning all sorts of things with ropes and pulleys. Now working on levers and their relative position to the fulcrum. Not to mention forces, mass, loads, distances, speed, etc. This really did turn into a math conundrum.
 
Carol Reed said:
Thanks, Toni. What is a blinder motor?
O0
The heaviest thing are the three gallon water bottles. Don't think the screwdriver will lift them, even one at a time though.

Still noodling through this. Learning all sorts of things with ropes and pulleys. Now working on levers and their relative position to the fulcrum. Not to mention forces, mass, loads, distances, speed, etc. This really did turn into a math conundrum.
Click to expand...

Sorry Carol, I meant a blind motor, or a motor for blinds I think that in US you call them shutters. They come in many different forces depending on the width and weight of your shutter, they are fast enough and silent, and also powerful.
A simple box or plastic bucket on a pulley would do the job, or even a cloth bag hung from the rope with a hook. If you don’t want the load to turn while going up, you could use one or two tight ropes or wires from ceiling to floor acting as rails if you had a box with a metal loop on eac side and the rail rope going throug them. That would be a quick an easy installation and those motors are not expensive I changed one at home and it cost me about 50bucks. So I guess in US they are even cheaper.
I hope I explained the idea.
 
I think a scissor type lift is the way to go. Actually you don't need to start from the ground do you? For instance, if the table is beside the van/car/truck, to decrease back issues with lifting, I find it easier to move something across and aisle way at the same height easier than setting it down on the floor. So possibly you are starting your table height at 2' but for your actuator it could be zero. a2 + b2 = c2 is what was said in one video of actuator length to lift. I am not a mathematically working person, I have to design the lift bars and then place the cylinder so it gives me the desired action. Here is a video I found interesting and then it's follow up came on and I found it amusing so am sending you two videos. I tried reading through this entire thread, I think my brain exploded somewhere in the middle of it. Basically, what is the height you would be starting from? What is the height you want to achieve? Are there space restrictions? The length of your cylinder retracted? Length of cylinder extended?
 
Though I already have a scissor lift table that I built some years ago, I am intrigued by the mini dual cross table. The single cross leaves one end of the table in the uppermost position hanging precariously out there. I can just see one of the full three gallon water bottles out there hanging ten. Scary. There is also the side to side movement due to all the joints. One does not need to have it tip over sideways. I also saw another video where a motor was added to the mini table. Having seen the physics of that, I can envision the linear actuator replacing the threaded rod and any cranking mechanism.

The actuator I have has a dynamic load of 50 pounds.

So, Ryan. Does that dynamic load translate into the torque of the threaded rod? If so, I am golden. If not, ????
 
I'm going to claim no, but the whole dynamic versus static load bits make my head hurt worse than usual :huh:

The mini dual cross table is pretty cool. I don't ?think? it changes the load calculations any but it would certainly help stability.

With a 50 lb actuator, you should be able to lift right about 50 lbs if the angle between the two arms starts at 45 degrees (I was misreading that before, I thought it was the angle between the arm and the base so its half as bad as I thought). If you listen near the beginning of the first video you can head the drill moaning a bit about the load when the table is lower then easing out some as it gets higher. You can hear the same thing in the last video around 1:29 in versus the noise when he jumps to a higher level at 1:36. I was mildly surprised at the apparent ease the wooden rig was lifted with in the second video after having looked at this more. But then working the math he said it was around 40kg (88lbs) so maybe half that (or less) was in the top (40lbs) and figuring a starting angle of 10 (250lbs of force) to 20 (120 lbs of force) once you factor in the screw giving you a 15x advantage he only had to put 8-16lbs of torgue on the handle... I'm not going to try to figure the leverage the length of the handle gave him (and better than even odds I'd get it wrong anyway) but it was definitely non zero.

A 3 gallon water bottle comes in just a smidge north of 25lbs (pints a pound-ish the world round, except in merry old england where it weights more) but its about 8.35lbs per US gallon anyway. So if that's the heaviest thing you're putting on there.. you'd have a wee bit of room for the rest of the assembly. I hate to cut things like this to tight though, engineer for a bulldozer - use like a butterfly :).
 
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