As those are the minimum angles the belts are at (thinking of horizontal and vertical tightropes given the corner attachments points).
And then trading off width vs height for a given material size you’re targeting so that both of the angles are as close to being the same as possible (hence maximising the minimal angle).
I think this is valuable info, and especially interesting to be able to see these angles when people have issues.
I wonder if the two angles is something you could both add to your frame visualiser @dlang / @bar?
It’s not difficult maths, but a pain to do manually each time.
I agree that sounds like a valuable way to think about it. I think that there might be some other factors too that we’d want to consider (especially in vertical orientation), but it might be nice to have an optimal aspect ratio to give folks for a 4x8 sheet
That’s roughly my inside X frame dimensions (more constraints based on my shed), but for outside horizontal X frame, I think 4500*4000, using 3m / 10ft ‘legs’ looks like a good, if big, config. When it eventually stops raining (yay, British weather!) that’s what i’m aiming to try for a full sheet frame.
since the sled can pivot, A and B don’t matter in themselves. It you look up in the (overly complex) settings for the page, you see a place to change angles, from the default as Bar designed it, to slightly winder numbers that I think cad measurements support, to what I think would work if you clipped the ears on the top clamp, to allowing you to set the limit entirely
there are three angles
how wide can the adjacent arms go as you near the top/bottom
how wide can the adjacent arms go as you near the sides
how narrow can the opposite arms go when you near the corners (which turns out to be a FAR bigger factor than anyone expected
I believe A and B matter immensely - I think the tightrope problem is what we need to solve in software to get the last bit of accuracy beyond flex in the device and flex in the frame.
Happy to have my maths corrected, but I believe A is a measure of the worst-case vertical component for the belts, and B is a worst-case horizontal component for the belts.
Assuming a vertical frame, sin(A) is the vertical component of the tension in the TR-belt when the Maslow is at the TL of the desired working area. That is to say, it’s what portion of the tension in the TR-belt goes to offset any downward force.
At A = 5 degrees, that is just under 0.1 - or to put it another way, the tension in the TR-belt has to be 10x the weight it’s supporting, ignoring any additional vertical forces from the bottom belt.
Given the Maslow is ~5kg, this is non-negligible.
Ultimately, I believe we will need to solve the series of…8? simultaneous equations (horizontal and vertical components for each belt) in software to understand how much belt stretch we need to account for, but for now, having some idea of how bad these angles are seems eminently useful…
I’m still not 100% sure that I understand it well enough to implement it, but if you make a really clear description of what you would want to see here: GitHub · Where software is built
I can have the AI give it a go. This seems like the kind of thing that it would be pretty good at.
That repo isn’t set up to make the process automatic the way some of our other repos are so I’ll have to manually assign the AI to implement your feature, but it should basically work the same way. It’s pretty magic.
I believe A and B matter immensely - I think the tightrope problem is what we need to solve in software to get the last bit of accuracy beyond flex in the device and flex in the frame.
Happy to have my maths corrected, but I believe A is a measure of the worst-case vertical component for the belts, and B is a worst-case horizontal component for the belts.
the highest tension on the top belts will be at the top center. the angle to the
opposite top belt gets worse as you go towards the corner, but more of the
weight of the sled is supported by the belt to the near top anchor
Assuming a vertical frame, sin(A) is the vertical component of the tension in
the TR-belt when the Maslow is at the TL of the desired working area. That is
to say, it’s what portion of the tension in the TR-belt goes to offset any
downward force.
At A = 5 degrees, that is just under 0.1 - or to put it another way, the
tension in the TR-belt has to be 10x the weight it’s supporting, ignoring any
additional vertical forces from the bottom belt.
Given the Maslow is ~5kg, this is non-negligible.
all this is correct but we don’t get anywhere close to 5 degrees, the closest an
arm can get to horizontal (if the sled does not rotate) is about 20 degrees
this is what the yellow bulge down from the top represents, the point where the
top belts are as far apart as they can get (~140 degrees)
Ultimately, I believe we will need to solve the series of…8? simultaneous
equations (horizontal and vertical components for each belt) in software to
understand how much belt stretch we need to account for, but for now, having
some idea of how bad these angles are seems eminently useful…
the problem with solving for belt stretch is that we don’t know how much tension
there is from the other belts, only from the weight (and we don’t know how much
the sled weighs, remember dust collection and cords)
Ahhhh balls, you’re right, i’m treating them as independent when they aren’t, well I did say check my maths This explanation post would have avoided some effort in the first place though
Quick maths with equal belts (T1 sin(angle1) + T2 sin(angle2) = w) tells me at 20 degrees each has a tension of 1.46 * W, which is not to be sniffed at. I wonder if doing the calculation for the mid point, or even a grid with a spacing of 1/8ths to look at the tensions might be interesting (I don’t know if others have done it before?).
I think I might bust out a spreadsheet.
Yeah, ultimately you want to solve all 8 simultaneous equations, and see what effect the lower attachments have as they are not running at zero tension unless the belts are actually slack.
I don’t buy the ‘we can’t measure tension’ generally - ultimately you could just mount the Maslow unit vertically on a wall, and for each arm in turn have it point down, hang a few different masses (1/2/3kg maybe) in turn, and let out a good length of belt (2/3/4m depending on frame size), and have it draw the weight up at a steady state at a few different speedand record the current change over time.
Yes it’s a massive faff, but it would give you a force/speed/belt position → current required map / look-up table for each arm similar to a car ECU map.
Ahhhh balls, you’re right, i’m treating them as independent when they aren’t, well I did say check my maths This explanation post would have avoided some effort in the first place though
modeling a maslow is surprisingly hard. I’ve been doing it for a while and while
I’ve found lots of stuff, I don’t for a minute belive I know all of the issues
Quick maths with equal belts (T1 sin(angle1) + T2 sin(angle2) = w) tells me at
20 degrees each has a tension of 1.46 * W, which is not to be sniffed at. I
wonder if doing the calculation for the mid point, or even a grid with a
spacing of 1/8ths to look at the tensions might be interesting (I don’t know
if others have done it before?).
I did a lot on the 2 chain maslow
I think I might bust out a spreadsheet.
Yeah, ultimately you want to solve all 8 simultaneous equations, and see what effect the lower attachments have as they are not running at zero tension unless the belts are actually slack.
I don’t buy the ‘we can’t measure tension’ generally - ultimately you could
just mount the Maslow unit vertically on a wall, and for each arm in turn have
it point down, hang a few different masses (1/2/3kg maybe) in turn, and let
out a good length of belt (2/3/4m depending on frame size), and have it draw
the weight up at a steady state at a few different speedand record the current
change over time.
the friction in the arms is what I think the big problem is. it will change
depending on so may factors (including how much tension there is), so the same
arm will take different current to apply the same force to the belt depending on
how much forst there is between the spool and the arm.
it will also change over time as things wear
Yes it’s a massive faff, but it would give you a force/speed/belt position →
current required map / look-up table for each arm similar to a car ECU map.
This explanation post would have avoided some effort in the first place though 