Reconsidering DR4B Rubber Banding

VEX V5 General Discussion Technical Discussion
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I haven’t worked with one in a while, but I was thinking about the standard advice on applying rubber bands to a DR4B. I think it’s a little bit off. The whole triangle thing is clever, and I’m not saying to avoid that. I’m saying the targets seem to be incorrect.

Why? Maintaining uniform force on the lift is not desired, nor is uniform torque. Aiming for this is incorrect to begin with. What we really want is to provide the most torque when the bars are horizontal and less as the bars are more vertical (toward the top or bottom). And, unfortunately, we always end up with the most force from the elastics at the bottom since all the elastics are trying to lift the mechanism the whole time.

My thought? It’s a pain to try to get something to really fit those sine curves well, both mathematically and because we would use lots of connection points. Using the triangles accomplishes what we want pretty well and more simply. But we could still target a maximum force when the bars are horizontal. Set up triangles of elastics on the upper and lower bars. On one set of bars do them normally. The key to my idea is the other set of bars. Instead of doing the same triangles, we aim for something different. We want to force the bars away from somewhere a little under horizontal, both up and down. This way, when the lift is low the elastics counteract each other so there could actually be less lift at the bottom and more in the middle; this would allow for more strength assisting the motors without the problem of the lift not staying down to start. Also this way, when the lift is above horizontal and the regular triangle of elastics is getting weaker, it gets augmented. So the elastics on this part would need to be the most stretched when the lift is horizontal.

Mathematically, really roughly speaking, we’re dealing with the left half of a concave up parabola for the potential energy right now (positive x being upward). Then add a gentler concave down parabola with its vertex a little bit to the left of whatever x is when the bars are horizontal. Ideally this could flatten out the curve some when x is smaller than this vertex while making a bigger drop-off afterward. As the negative of the slope is the force, that would give less positive force early on and boost it a bit around when the bars are horizontal.

Is it possible? The first issue is finding a good way to pull away from horizontal. I think this could be done with a triangle again, just making sure the smallest-perimeter triangle is formed when the bars are horizontal. The second issue is if this can actually be matched well against elastics trying to lift the whole thing. Varying elastics could certainly help, but I’m more worried about things skewing than about trying to find the right relative forces. The third is I’m not sure how well this would really create the most torque when the bars are horizontal, but I would think it would help move things in that direction. Fourth, there might accidentally be a bit too much boost when the lift gets really high, which would need to be avoided as best possible.

Anyway, just thought I’d post this to bounce ideas off of others.

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You are right - most teams are doing the triangle wrongly.

This uniform tension rubber band method is a lot more than just spamming a triangle at the sides.
If one is to look closely at how those pioneers teams did it, they will notice that the different lengths of the triangle matter.

Maybe the name uniform tension is misleading, but the main purpose is not to provide the same tension throughout, but rather, to make the lift goes up and down uniformly throughout, so that it will be easier to control the lift.

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Do you have any specific examples of teams that do this well? I believed that the goal with triangle tension was to create a triangle with similar SA from down to up.

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+1 I’d like to take a look at these teams for future reference. No way to really easily tell the difference.

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I actually disagree with this. A lift requires the most force when accelerating. Since the lift accelerates up when it’s fully down and down when it’s fully up, and since it doesn’t take much force to maintain the current velocity, or position if your velocity is 0, I think you actually want the most force upwards when you’re fully down and vice versa. When the lift is all the way down, it needs some extra help accelerating up, and when it’s returning to a collapsed position from all the way up, it needs some extra force to help it brake. Here’s a video of my lift last year (ft. my messy bedroom.)

With that robot, I intentionally chose not to use triangle bracing because I wanted the lift to accelerate as fast as possible. You can see how quickly the lift gets up to full speed despite the pretty high overall lift speed. Granted there were 4 motors and my passive intake was lighter than an active intake, but I still got much better acceleration than I would have with bracing like Callen is describing.

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There are actually many teams that did it well.

this is the original idea.

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Think you misunderstood the point.

What I am saying is that this uniform tension rubber-band system (UTRS or triangle method to most ppl) is misleading, this is because what we are trying to achieve is not uniform tension (or force), but rather uniform acceleration (as much as possible).

What all of us can agree is that the lift will definitely need more force when it is going up as compared to going down (with gravity acting on the lift as well). But what we are looking at is to get an as uniform resultant force as possible, so that the lift will always be accelerating at about the same rate at every point of its motion.
And what @callen (correct me if i am wrong) is saying is that we should play around with the triangle to get an uniform acceleration, rather than to aim for uniform force.

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You’re not understanding what I’m saying. I’m not saying you should have less force at the bottom than you had. I’m saying you could adjust so that you have the same force near the bottom while not lacking so much as it moves up. This would provide the same acceleration you want near the beginning, and slightly more for the first bit as it moves up. Meanwhile, at the top, gravity will still tend to dominate, just not by as much, but if you’re carrying extra weight it will still dominate that much more anyway. Overall is that the acceleration will be more uniform.

Yes, @meng understands what I’m looking at.

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Oof, sorry for being unclear. Let me try to clarify: uniform acceleration isn’t really a realistic goal because the lift accelerates from different points. Uniform acceleration from the bottom will require more rubber band force when the lift is at the bottom, while uniform acceleration up from the middle will require more rubber band force when the lift is centered. I disagree that you want uniform force all the way up the lift. This makes the lift not require as much force as without any rubber bands, but extra force at the bottom still helps to accelerate it up, and extra force at the top still helps to accelerate it down. In other words, the non linear nature of the rubber bands is an advantage, not a disadvantage.

Ok I think I did misunderstand what you said, but I think you also misunderstood what I said. :slight_smile: The truth is somewhere in the middle. My lift intentionally had less rubber band force at the top (gravity > bands) so that the lift could accelerate down more easily, and it had more rubber band force at the top (bands > gravity) so it could accelerate up. When you say I could have the same force at the bottom and more at the top, triangle banding would have accomplished that. The point is I actually wanted less force at the top to help me accelerate downwards.

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Yes… realistically it will be almost impossible to achieve uniform acceleration throughout, and that’s why I said “as much as possible “.

And no… I didn’t say I want uniform force. I want uniform resultant force, which is different from uniform force.
Eg. When the lift is going up, resultant force = (force from motor) + (tension from rubber band) - (weight)

Actually I think we are all trying to achieve the same result, just that you seem to misunderstood my point on resultant force.
You are right to say that there are certain moments that the lift will require more or less force from the rubber band (and that’s when we make use of the different length or sides of the triangle), but the resultant force should be as uniform or constant as possible, so that we can achieve as uniform acceleration as possible.
And if done properly, there should be tension from the rubber band to help to accelerate the lift downward as well.

Edit: but of course if there is no intention to achieve uniform acceleration, then there is totally no need to use the triangle method.

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Actually, I understood you really well. But some of it was that I recognized a physics error you were making and just didn’t have the time to also address that. But since that’s what’s getting in the way now, I’ll address it. I’m going to try to drastically simply things to show this one issue more clearly. The way I’m going to do it is to estimate all force changes as linear and to use the “resultant force” approach @meng is explaining to you, so a lot of what I’m saying should be addressed with torques and I’m avoiding it to highlight this one issue.

First, let’s assume you have no extra load at all and that you have no special catch mechanism to hold the lift down to keep it legal before the game starts, just that the elastics are just shy of pushing it up on their own, perfectly balancing gravity at the bottom. We’re also dropping every bit of friction.

Now also recognize that the motor doesn’t care which direction it’s going. It can create the same force in either direction.

So your approach is to set the elastic force as Fe=Fg at the bottom and decrease toward some fraction of Fg at the top. Let’s let it lose n*Fg at the top, leaving it flexible to go to 0.6Fg (n=0.4), 0.2Fg (n=0.8), etc. My and @meng 's explanation of “uniform resultant force” would have Fe=Fg everywhere. So yours becomes ours if your force drops to Fg (no drop) at the top, if n=0. So the question is if n>0 helps or not. Let’s say the motor can produce force Fm. Meanwhile, we’re accelerating the same mass M upward. So let’s see which method can get the mass some fixed distance h in the least time. Let’s call x=0 at the bottom and measure x upward.

Your upward acceleration would be (Fm - (x/h)nFg)/M for the first portion, and (-Fm - (x/h)nFg)/M for the second portion. Our method would be Fm for the first portion and -Fm for the second portion. Note that rise h will not be split into the same sized portions between the two methods. Our method will split it equally, while yours will not. Our method will take a time of t=2sqrt(hM/Fm) (Lots of calculus and algebra later…) Your method’s lowest time will be accelerating upward for a distance of x=(0.5 + Fm/(nFg) - sqrt(Fm^2/(nFg)^2 - 1/4))h, which is a bit higher than h/2; after that it accelerates downward. This will take t=2sqrt(h/(ng))(arcsin(sqrt(nFgx/(2hFm))) + arcsinh(sqrt(nFg*(h-x)/(2hFm)))), using x from the last sentence. Yes, while the first is inverse sine, the second is the inverse of hyperbolic sine. The inverse of hyperbolic sine, where sinhø=y, is ø=ln(y+sqrt(y^2+1)).

Now, visually comparing those equations isn’t beautiful. We can simplify. Let’s say h=1, M=1, and g=1 to establish our special theoretical units (h sets the units for length, M for mass, and g for time by way of h). That makes Fg=1. All we really need to choose is value for n and for Fm, the strength of the motor, with 1 being where it could just barely lift the lift on its own because it could only just match gravity. Our method yields t=2/sqrt(Fm). For your method, x=0.5+Fm/n-sqrt(Fm^2/n^2-1/4) and t=2/sqrt(n)(arcsin(sqrt(nx/(2Fm))) + arcsinh(sqrt(n(1-x)/(2*Fm)))).

I’ve produced three graphs for you, with Fm=0.5, Fm=1, and Fm=1.5 The red line shows you the time using our method. To get to the top most quickly, you want it to take the least time, so you want your method to go lower on the graph than the red line.



As you can see, your method never does better. Sure, you have it slowing down faster at the top. But this never makes up for it rising more slowly at the bottom than our method. Keeping the two accelerations balanced always works out to be faster.

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