This is the thread I found that quote above in: HERE
And here’s the link to the video I’m going to be talking about: THE_VIDEO
So I was going around the Internet looking for information about 6 bar lifts. I found a great video explaining some ratios between the lengths of some of the bars. Then I became interested in making a formula for finding the height at which a 6 bar goes, based on only the length of the horizontal middle bar and the length of the veritcal middle bar. The lower horizontal bar would be just half of the middle horizontal bar, and the uppermost horizontal bar would also be just half of the middle horizontal bar. The outward most vertical bar would just be half of the middle vertical bar. It’s great I can find the lengths of the other bars by just knowing the lengths of two bars, but then my progress stopped. I didn’t know how to continue making the formula. So I looked online and found a thread about some 6 bar lift math. I checked it out and the only possible answer was the quote above. @Tim had replied with two links. If you try it yourself, the links don’t actually work. So now I’m even more stuck. So I have two questions: 1) Can someone help me with my formula; 2) Can someone explain where those links could have gone. Thank you in advance.
I don’t know a formula but I build models in fusion 360 to experiment with new configurations. Here is a pic that might help with your search. That note is a general guide and not a formula.
A big question is how far apart you can space your bars. If all the bars need to lie in two planes (verticals in one, non-verticals in other) because you have to keep it really narrow, then the width of the bars will limit angles first. If you can use more planes, you can get higher with the connectors (presumably spacer-diameter) limiting the angles. But as you start moving them apart, things will get unstable. Not knowing the specifics (pieces used for arms, etc.), I could probably throw something together for MS Excel tomorrow morning.
I know this formula doesn’t calculate the height of a 6 bar but its still important. At the beginning of the season I did some research for an educational video online challenge. I did research on finding the torque output of a lift and then finding the amount of weight the lift could lift. Its a very important thing to know before building and for designing for obvious reasons. After researching torque equations I applied them to robotics and came up with a new equation to find the torque output of lift. Here it is
(number of motors x each motors torque output x gear ratio motor:lift) / the length of the lifts middle bar(the distance from the lifting point to the lifts center of rototion) = the amount of weight a lift can lift.
the torque out of each motor can of course be found on the vex product description of motors
Here the video i made explaining the equation, i apologize ahead of time for the poor quality of the video since it was a more of a last two days to enter entry.
Thanks for responding. So the reason why I’m trying to make a formula is because I wanted to know the length of the bars to make the lift go up to 4 feet. So if I did find a formula, then all I needed to do is plug in 4 and solve, and so forth and so forth. So @pietrofesar, your note, is that like an estimation of what could have been the height of the lift? So @callen, I’m lost st what you said about planes. Like 2 dimensional planes? And finally @RunTheCode, thank you for that video, it really did inform me about torque which is actually what I was going to do next.
Be really careful using that. It is problematic in a couple ways. First, if you’re using the maximum torque the motors can supply, then you’re looking at the load that will stall them. You really want to consider that we’re generally trying to accelerate a load and so want to take into account rotational performance. Second, you have no reduction for friction, instead using an ideal value. Third, you’ve completely left out the weights of the arms. If you’re lifting something heavy, that may be insignificant. But if you’re working with long arms and a light load light the cones in this year’s competition, then the calculation will be way, way off, even for the point when it stalls.
If you look end-on, are all the vertical pieces lined up perfectly (in one 2-dimensional plane)? Are all the non-vertical pieces also lines up (in a different 2-dimensional plane)? Or have you offset them a little sideways so that that is not the case? Unless your vertical bars are really short, the difference in vertical height it can reach won’t be huge, while trying to work with more planes may make it less stable.
I plotted the angle to axle spacing relationship with the help of Fusion and sheets. Using this plot you can calculate the max height of a four bar parallelogram, limited by the structural contact when fully extended. Use the spacing of your tower axles to determine theta(the max angle of extension). Then use some trig to find the height of from the top pivot point to the max height. Add this to the current height of your top pivot point and you will have the theoretical max lift of a four bar. You can derive 6 bar, DR4B, and DR6B calculations using this method-unless I missed something. This doesn’t take into account the hardware involved that will eventually limit your actual height(bearings, motors, other pieces of structure), it’s theoretical.
So is the horizontal bar from the axle to the first pivot point with the middle vertical bar or the entire thing? Also how did you get 10 spaces to 78 degrees?
The graph is a reference chart to determine theta based on the axle spacing. I measured the spacing and angle relationships over a series of intervals using modeling software, and then plotted it using Google Sheets. That chart is necessary because the relationship isn’t linear. I couldn’t find a function that fits the curve. It looks like a square root function but I couldn’t get it work a(x)^.5 = y.
Regardless you can trust the reference chart. The following is a diagram to help understand how the procedure works for a variety of common configurations. Again if you break any configuration down to a series of 4 bars the theoretical height can be determined.
The problem fitting the function is because Google Sheets’ graphing is weak. I love Google Docs, but I can’t even use Google Sheets’ graphing for beginning science classes.
What you’re looking at is an inverse cosine graph, specifically:
theta = arccos( width / axle distance )
Here width is the average width of the two arms and axle distance is as you used it. I could probably scan an image showing why this is so later.
Oh okay, i get it @pietrofesar. Theta is actually the angle of an imaginary triangle that is created when you put a dashed line through it to represent one of the sides. I was a little confused when you said in your last image that 10 spaces was 78 degrees. I get it now.
Here is the scan of the theoretical calculation: https://s6.postimg.org/n5d1kcudd/4barcalculation.jpg
Here is the resulting graph, formatted to look like yours: https://s6.postimg.org/acoxkfirl/4barangle.jpg
Interestingly, I also get 78 degrees for 10 holes, but your graph shows 77 degrees, which is what I get for 9 holes. And further to the right (and some to the left) we also seem to be off by 1 hole on the graph. At first I figured it was because of the reality of the bars not matching their stated widths. But with your 78-degree statement alongside the 77-degree bit on your graph, I’m not sure. Regardless, they’re quite similar.
@callen Thank you, this explains exactly what I was missing. I’m not sure about the hole discrepancy; I measured 77.1 at 10 holes using the CAD model.
I used your diagram to summarize the entire process. I hope you don’t mind. I haven’t had to do trig simplification for many years so I’m not sure if it can be simplified further. If it can by all means chime in.
I get 77.16 degrees at 9 holes. I wonder if we’ve defined the differently. For me N holes means they’re N/2 inches apart, measured center of axis to center of axis. So N=10 would be 5" apart. That would be when there are 9 holes between them, though, and when the bars occupy 11 holes (the 2 they’re in and the 9 between).
The whole sin(acos(w/d) can be written as sqrt(d^2-w^2)/d, if you want to avoid inverse-trig and trig computation.
@callen that is the problem. I defined 10 as 1 being the starting point, and 10 being the ending point, leaving 8 holes in between. I will simplify and define everything later; unless you would like to? Thanks for all the help, this will be a great unit for my students.
I would shy away from your definition because you’ve labeled it a distance and are using values that, although they map one-to-one with distance, are not distances. You could also halve my values to have it in inches so the distances correspond to what could be quickly measured.
Here is a graph that I made on Desmos. It shows the heights of the different lifts at a given angle. The different lifts are a 4 bar, 6 bar, DR4B, and DR6B.
The x axis is the angle of the lift (in degrees) and the y axis is the height (in inches.)
This is assuming that the angle is the angle of the bar on the top of the tower that it is supported on. Also, the distance between the bars on the tower that the bars are supported on is 6 inches, and the top bar that is supported on the tower is always 20 inches long.
This graph shows how the 4 bars and 6 bars both go up at about the same rate, which is also the same as the DR4B and DR6B. However, what this graph doesn’t show is that each 4 bar and 6 bar each have different starting angles.
But, what we can conclude is that DR4B’s and DR6B’s in fact go up faster than regular 4 bars and 6 bars.
Hope this helps explain the math a bit more.
Red line: 4 bar
Blue line: 6 bar
Green line: DR4B
Orange/yellow line: DR6B
