There is a more detailed explanation of the control software for the lift in this thread. Improved lift control software

I captured some data for the lift moving full speed between lower and upper limits. The graph shows motor speed, motor drive value and motor current against time. The motor current calculation is pulled from the smartMotor library. This is for the lift going up, manual control was used, joystick was pushed full forwards. Maximum lift speed is limited to 80rpm. The left vertical axis is speed and motor control (either rpm or control value). The right hand vertical axis is current. Horizontal axis is time in seconds.

The motors accelerate (there are two motors on the lift) until they reach 80rpm which is programmed as the maximum speed, the motor control value needed to maintain this speed is a little below the maximum value of 127. Current spikes during acceleration and then settles at about 1.2A, this corresponds quite well with the theoretical value of 1.12A shown in the graph here.

When the lift reaches the full height the motor is sent a holding value of 15 to maintain position, current is about 0.25A with this value and will not trip the internal motor PTC which has a nominal rating of 0.9A (1.8A is trip current, there are lots of forum threads on PTC performance, you need 1.8A for several seconds to trip the PTC )

The time to move the lift is about 2.5 seconds, during this time the encoder changes by 1710 counts (not shown on the graph) which would be 2.73 rotations of the motor (1710/627.2 counts per rev). Each rotation of the motor on a 12 tooth sprocket will move the chain 4.63 inches (vex spec is 54 inches for 140 chain links) so total chain distance moved is 12.64 inches. Due to the way this lift works this means the total lift displacement is 25.28 inches (ie. double the distance). The final encoder error was 4 counts.

Here is the graph of the lift going down.

Notice that lift speed is also about 80rpm (not quite as stable) but motor drive is much less at about -40. Motor current is very low, about 0.12A, gravity is driving the motor which now acts as a generator and reduces current. As the lift comes to a stop drive is reversed and holding power applied, this creates a brief (100mS) current spike of about 3A as the motor is now braking. The final encoder error was 5 counts.

Now I need some help from all the mech-eng students out there.

I retested the lift earlier today with more a substantial game object, a 44oz weight. As expected, lift speed was reduced to about 68rpm, this corresponds to a motor torque of about 4.8 in-lbs from my graph. I have two motors so total motor torque was 9.6 in-lbs. The 12 tooth sprocket Iām using has en effective diameter of 0.75 inches (based on 12 chain links wrapped around it being 4.63 inches). So in theory the motors could pull 9.6/0.75 = 12.8 lb. We can guestimate that there will be losses due to friction etc. and perhaps we have 80% efficiency, this would reduce that number to about 10lb.

So I had a 44oz weight, my third stage weights about 14oz and my second 24 oz. Total weight of all that is 82oz or 5.1lb, way less than the motors were apparently moving.

To understand what was happening I drew a free body diagram, and this is where I need some help. What I have here is more commonly called a cascade elevator, when you simplify the chain connections it would look like this.

S3 means weight of stage 3
S2 means weight of stage 2
W is the weight of the game object

The math here says my motor would see an equivalent weight of (2x44) + (2x14) + 24 = 140oz (or 8.75lb). Still not the theoretical 10lb but much closer.

Adding yet another stage would make this situation even worse.

This is why in my opinion I stray away from elevator lifts that have any more than like 3 stages, because of the friction from the seperate stages. 1103ās lift worked well but thats because it was only 2 stage, or double extending.

Despite this one of the teams at my school is going with an elevator lift, for the full 60" too. However the way he is building it is trying to prevent the way elevators lean in when fully extended (and thus creating more friction). He is building two sets of elevators, each one only one linear slide thick per stage. He then attached them to the chassis at a 45 degree angle, so that the stages dont lean in, because the two sets of elevators will be supporting each other. Although with your lift the elevators are in the center so unless you rebuild it this is not an easy fix.

Also, if the same lift were to be built without the stages moving together, it would be slower. So with the lift with the continiual loop of chain you could just gear up to lift more, and still have the same torque to speed ratio as a cascading lift(i believe thats what its called).

This is a familiar problem that my teams faced when building for Singapore Vex. Our lifts were all well above 50".

The elevator lift will lean in.
You can have the best quality of construction (which I believe my teams do), but once you have added in the intake and the cubes (especially if you are talking about 2 or more cubes), the elevator will lean inward.

This leaning in will definitely create a lot of unnecessary friction, which means the motors will need to work a lot more harder (than in theory).

We had tried counter-weight, but it was only successful to a certain extent.

I am having problem visualising how this looks like.
Mounted 45 degree tilting backward? So that with the weight of the intake and cubes it will be vertical?

If this is the case, wouldnāt it be creating additional friction when it is not carrying any cubes (since it will be leaning backward by default)?

I agree, last year we dabbled with a counter weight it is not very successful. We are working on our own version that uses two set of lift to counter act each other.

Your missing the point, this is not my concern, this design (admittedly only two moving stages) does not lean significantly. I think it would be ok with a third moving stage. The problem is that the required motor torque effectively doubles as each stage is added.

Some more numbers

Lets assume stage 2 is 30 oz, stage 3 is 20 oz and the intake plus game object is 50 oz.

So motors on the lift as designed would need
(2x50) + (2x20) + 30 = 170 in-oz (10.6 in-lb) of torque (which would be about 4 in-lb on each motor with the 12 tooth sprocket and ignoring friction etc.)

If we add another stage the equation becomes (where S4 is weight of stage 4 etc.)
4(W+S4) + 2(S3) + S2
assuming S2 and S3 are the same (both 30oz) we have
(4x50) + (4x20) + (2x30) + 30 = 370 in-oz (23.1 in-lb) or torque, more than double.

We now need at least 4 motors (everything else being equal).

Adding yet another stage would more than double yet again, 8 motors needed or they would have to be geared down and run the lift at half speed.

So I see what your saying now, by adding a fifth stage I came up with 770oz, or about 48.1lbs. That would be 64.16 for your motors. So to make that work you would need at least 6 motors keeping the torque for each at 10 lbs and not overload them.

Gearing them might not be too bad of a deal because of the exponential lift speeds correct?

Would you also be reaching the limits of the chain? It is rated at 50lbs.

maybe i can make it more clear
imagine holding up an elevator with your hand, and teh bottom is resting on the table. Rotate the elevator to the left 45 degrees, not tilting it back or forward, but left and right

We tried using cascade during Singapore Vex. And we faced exactly the same problems raised over here - we switched from 4 motor to 6 motor lift. And even then, it has to be painfully slow.
And not only that, the chain kept snapping, esp when we added the cubes in.

So in the end, we used continuous instead of cascade.
But we are back in experimenting with cascade currently.

The elevator you have built uses multiple loops of chain as to of opposed to 1103ās lift. Also it is much faster than the single loop of chain. When you add more stages, the torque requirement increases, but so does the speed of the lift.

In turn you just have to gear it up, because right now it is really fast. A continious loop chain with teh same sprocket would be slower than this multiple loop elevator.

the 1103 lift is stronger than the lift in this thread, and also slower

if both lifts were to have same number of motors and same sprocket size, the 1103 lift would lift more wieght. (thus lifting requiring less torque to lift the equivelant weight on the lift in this thread).

I know. I was asking a rhetorical question yesterday just to get the discussion going.

So my lift is a variation on the cascade lift (a variation because Iām using loops of chain, there are other ways of designing it).

The 1103 lift is called (unsurprisingly) a continuous lift. The force needed to raise the continuous lift will increase as each stage is lifted, the worse case is when it is near full extension and has to raise the weight of the game object and all moving stages combined. So using my example from post #28, the motor on a three stage lift would see a force of 100oz (as opposed to 170oz). A four stage would only see the additional force needed to raise the extra stage, 130oz (as opposed to 370oz).

The downside is that the lift will be slower. My three stage lift needs to move 12.5 inches of chain to raise the lift 25 inches. A four stage variant would still only need to move 12.5 inches of chain but the lift would raise 37.5 inches. A continuous four stage lift would need to move 37.5 inches of chain to move that same distance, ie. 1/3 the speed with 1/3 the force. I could gear my lift motors to have a 1:3 ratio and then the apparent force needed would drop to 123oz and I could run at the same speed.

I threw this out there yesterday to try and partly explain whyā¦

A much simpler, layman way of understanding will be this - 1103 lift (which is continuous elevator) will requires less torque. You can imagine it as lifting stage by stage.

As for cascade elevator lift, you can imagine it as lifting all the stages at the same time. Much faster, but definitely requires a lot more torque.

More physics thinking - perhaps a different way of measuring āgear ratioā might help when comparing different lift styles:

distance lift raised / motor revolution

eg1 for a single stage linear slider lift, powered using a rack and 12t pinion directly off a motor, the ratio would be:

pi*0.5 inches raised per motor revolution ā 1.57"/revolution

(A 12t pinion has a pitch diameter of 0.5", and C = pi*D)

eg2 for a chain powered single stage or continuous lift driven by a 12t sprocket directly off a motor, the ratio would be

12t*0.386 inches = 4.63"/revolution

(0.386 inches is the chain pitch or the length of one link)

eg3 A 2-stage cascade lift like jpearmanās would be twice that speed = 9.26"/revolution

This ratio can be used for comparing different types of lifts easilyā¦ and can even compare to non-linear lifts to an extent (e.g. think about the point where an arm is horizontal, which is where you get maximum vertical lifting per motor revolution)

The weight of the mechanism is a separate question of course.

A purist might prefer using radians to revolutions, at which point you should probably change to SI units too!

I re-worked your free body diagram from scratch and came up with the same result. I donāt think thereās any forces being left out. (I say this because thatās what Iām always worried about when I draw a free body diagramā¦ )

Unfortunately, there are only two major ideas I have where error could creep in:

Is your chain nearly vertical? I canāt tell from the pictures in the OP. If it was angled, I would expect that to increase the overall required force from the motor. (I havenāt worked it out, but I think that the larger force would be proportional to cos(theta) or 1/(cos(theta)), where (theta) is the angle (from vertical) of the chain. This factor is ~1 when (theta) is small, so it shouldnāt make a very large difference, but itās still something.) We assumed it was vertical in both of our free body diagrams.

If we decrease W to a smaller weight (or take it away entirely), does the lift perform closer to what the free body diagram predicts? If so, it could be that the larger W ābendsā the lift more and increases friction in the lift. You said the lift didnāt bend significantly, though, so this probably wouldnāt be the only cause.

These are both āsmallā things, but perhaps the combination of the two could be significant enough. Sorry this isnāt more help.

What wasnāt on the free body diagram are the slider points.

When you add those in, you also have to account for the torque applied from the center of gravity to the connection point. That is the āleaningā source as you add weight and distance. The older slides had a longer contact surface that could be managed.

A static view of the system gets you the chainās forces, but adding in the touch points upon sliding, gets you the friction that needs to be overcome before you move for dynamic motion to start. (as evidenced by the nice current mountain required to get that thing moving)

The new versions have this plastic with a lower coefficient of friction but a lower contact surface area. Good in very linear applications but when you get a bit of twist in there, it does not yield that nice low friction any more.