iTopSpin

Please login or register.

Login with username, password and session length
Advanced search  

News:

Author Topic: offset top  (Read 23391 times)

ortwin

  • ITSA
  • Hyperhero member
  • ********
  • Posts: 1493
Re: offset top
« Reply #45 on: May 21, 2021, 09:18:40 AM »

I got these values for the position of the line:

ρ21    φ    r0/r
2    81.61    0.146
10    63.15    0.452
20    55.99    0.559

...
@tao: those were the values you gave us for the "Ansatz" case of the circular offset top.
May I ask you to give us the position of center of mass also for density ratios of 2 and 3 respectively? But this time the interface should stay exactly
in the middle. That could be easily built with 6 or 8 of these, respectively:






@Jeremy: are the tip positions ta0 provides here LEGO friendly? For radius of 12 studs? Or even for a radius of 4 studs, which also seems available (as semicircles)?



 
« Last Edit: May 21, 2021, 09:39:44 AM by ortwin »
Logged

In the broader world of tops, nothing's everything!  —  Jeremy McCreary

Jeremy McCreary

  • ITSA
  • Demigod member
  • **********
  • Posts: 3786
    • MOCpages
Re: offset top
« Reply #46 on: May 21, 2021, 10:22:37 AM »

ρ21    φ    r0/r
2    81.61    0.146
10    63.15    0.452
20    55.99    0.559
@Jeremy: are the tip positions ta0 provides here LEGO friendly? For radius of 12 studs? Or even for a radius of 4 studs, which also seems available (as semicircles)?

For the ρ21 = 2 Ansatz case, r0 = 12 studs x 0.146 = 1.75 studs from center.

No easy way to secure a stem and tip there in LEGO. You should aim for integer or half-integer CM locations.
Logged
Art is how we decorate space, music is how we decorate time ... and with spinning tops, we decorate both.
—after Jean-Michel Basquiat, 1960-1988

Everything in the world is strange and marvelous to well-open eyes.
—Jose Ortega y Gasset, 1883-1955

ortwin

  • ITSA
  • Hyperhero member
  • ********
  • Posts: 1493
Re: offset top
« Reply #47 on: May 21, 2021, 11:57:50 AM »

...
For the ρ21 = 2 Ansatz case, r0 = 12 studs x 0.146 = 1.75 studs from center.

No easy way to secure a stem and tip there in LEGO. You should aim for integer or half-integer CM locations.


I am not sure we are talking about the same case. I meant NOT the Ansatz case where  CM is at the interface. That is not possible with LEGO. So I chose a disk that is easy to build with Lego, the  question is then where the CM is lying.
I Suggested four different  disks (radius four, radius 12, weight per area ratio 2 and 3, interface in the middle for all of them.) With some luck and a bit of additional balancing with single studs there could be a chance I thought.
Logged

Jeremy McCreary

  • ITSA
  • Demigod member
  • **********
  • Posts: 3786
    • MOCpages
Re: offset top
« Reply #48 on: May 21, 2021, 12:07:30 PM »

OK, misunderstood "interface". Look up the centoid of a semicircle and the lever rule.
Logged

ta0

  • Administrator
  • Olympus member
  • *****
  • Posts: 14366
    • www.ta0.com
Re: offset top
« Reply #49 on: May 21, 2021, 04:34:19 PM »

I Suggested four different  disks (radius four, radius 12, weight per area ratio 2 and 3, interface in the middle for all of them.) With some luck and a bit of additional balancing with single studs there could be a chance I thought.

As Jeremy said, using the center of mass of a semicircle and equating the levers from them to the position of the center of mass, x, one gets a general solution for a radius R and a ratio of densities ρ2/ρ1:

x = 4/(3π) (ρ2/ρ1-1)/(ρ2/ρ1+1) R
Logged

Jeremy McCreary

  • ITSA
  • Demigod member
  • **********
  • Posts: 3786
    • MOCpages
Re: offset top
« Reply #50 on: May 22, 2021, 03:34:47 PM »

NB: Sorry, this post has been rewritten. Talk about burying the lead! Facts and results unchanged, so need to read again if you got them the first time.

...using the center of mass of a semicircle and equating the levers from them to the position of the center of mass, x, one gets a general solution for a radius R and a ratio of densities ρ2/ρ1:

x = 4/(3π) (ρ2/ρ1-1)/(ρ2/ρ1+1) R

Turns out that ta0's solution for a circular "half-and-half" rotor generalizes nicely to a wide range of rotor "shapes" as seen in plan view.

General formula: You have a half-and-half rotor like the one ortwin envisioned in Reply #45. The rotor has a certain "shape" in plan view with a sharp density contrast along a straight "boundary" passing through the shape's center and dividing it into equal "halves". For shapes like circles, ellipses, rectangles, rhombi, squares, reqular hexagons, regular octagons, and many others sharing certain key properties, the perpendicular distance x from the boundary to the rotor's CM is then

x = (γ - 1) xcentroid / (γ + 1),

where γ > 1 is the ratio of areal densities across the boundary, and xcentroid is the perpendicular distance from the boundary to the centroid of the chosen half-shape. Wikipedia's list of centroids is a good resource here.

Octagonal example: In the half-and-half top below, the rotor shape is an octagon of side S = 32 mm = 4 studs. The chosen boundary between halves is clear. The centroid location comes out to xcentroid = 17.2 mm = 0.543 S on both halves. (You don't wanna see the math.) With γ = 5 in this example, the overall CM is then on the thickened half at distance x = 11.6 mm = 1.45 studs from the boundary.




Unfortunately, wobble is barely tolerable. The yellow stem and tip mounts at xactual = 1.50 studs were the closest I could get to x = 1.45 studs, and the difference matters.

Which shapes qualify? Applicable rotor shapes must have a center of symmetry crossed by at least one pair of perpendicular "mirrors" (lines of mirror symmetry at right angles to each other). All of the shapes mentioned above fit that description. But stars are out because they're not convex. The regular pentagon is out because it has no perpendicular mirror pairs. And the isoceles triangle is out because it has only one mirror and no center of symmetry.

If the rotor shape has more than one perpendicular mirror pair through its center (e.g., a regular hexagon), pick a pair, call one of its mirrors the "boundary", and divide the shape into halves accordingly. For either half, xcentroid is the perpendicular distance from the boundary to that half's centroid.

For example, a rectange with sides a > b has a single perpendicular mirror pair through its center of symmetry. If you pick the mirror parallel to b as the boundary, then xcentroid = a / 4. A circle of radius R, on the other hand, has an infinite number perpendicular mirror pairs. Just pick a pair and make one of its mirrors the boundary. Result: 2 semicircles, each with xcentroid = 4 R / 3 π.

Why areal densities? Because they're more general than mass densities measured in kg/m³. Recall that if half-rotor i has uniform uniform thickness Li and mass density ρi in kg/m³, then its areal density is αi = ρi Li in kg/m². As long as you watch out for couple unbalance, doesn't matter then whether you change ρi or Li to adjust αi. All that counts here is the final value.

In these terms, γ = α1 / α2 > 1, where the subscript "1" points to the denser half.
« Last Edit: May 22, 2021, 04:44:27 PM by Jeremy McCreary »
Logged

Jeremy McCreary

  • ITSA
  • Demigod member
  • **********
  • Posts: 3786
    • MOCpages
Re: offset top
« Reply #51 on: May 22, 2021, 03:47:10 PM »

Aside for @ortwin: The octagonal half-and-half top above isn't a keeper. But no end to the fun (non-offset) LEGO tops you can build on these octagonal plates. Great spinning platforms with plenty of room and AMI for all kinds of additions, as they're 80 mm = 10 studs across and plenty rigid — especially when stacked. When I want to boost specific AMI, I just stack some of the hollowed out versions (yellow) above or below the full octagonal plates.



If your kids don't have any of these octagonal plates, buy some on BrickLink.

Logged

ortwin

  • ITSA
  • Hyperhero member
  • ********
  • Posts: 1493
Re: offset top
« Reply #52 on: May 24, 2021, 04:35:08 PM »

...
x = 4/(3π) (ρ2/ρ1-1)/(ρ2/ρ1+1) R
Thank you!
From this formula and the LEGO parts I thought were in reach, the best results I got, are for the semicircle with a four stud radius. When I choose the areal density ratio to be three, I get an offset of 0.85 studs. Is that close enough to an integer? I guess that depends on how much wobble you can accept or how much additional balancing with small parts you can tolerate.
The upper side looks like this:



 Since I found only three of those semicircles in my children's LEGO where I actually would have needed four, the bottom side looks very improvised:
 


@JEREMY: I think we have to exchange a few videos to determine what each of us thinks that tolerable wobble is and what not.

https://youtu.be/uHkp29SSxeM
 
Logged

Jeremy McCreary

  • ITSA
  • Demigod member
  • **********
  • Posts: 3786
    • MOCpages
Re: offset top
« Reply #53 on: May 24, 2021, 06:37:48 PM »

Interesting result! When I lack the parts to give a promising idea its due, I just go to BrickLink.

I guess that depends on how much wobble you can accept or how much additional balancing with small parts you can tolerate.

Exactly.

@JEREMY: I think we have to exchange a few videos to determine what each of us thinks that tolerable wobble is and what not.
https://youtu.be/uHkp29SSxeM

For me, that much wobble would be fine for just fooling around but no go for a keeper.

Yes, it would be interesting to compare wobble tolerances. Mine's generally pretty low, but if a top has other redeeming quailities, I might put up with more.
Logged

ta0

  • Administrator
  • Olympus member
  • *****
  • Posts: 14366
    • www.ta0.com
Re: offset top
« Reply #54 on: May 24, 2021, 11:30:05 PM »

To me the wobble looks tolerable, but I'm known for playing with crooked tips and to never balance my tops (I hope Iacopo is not reading this).  ::)
Logged

Jeremy McCreary

  • ITSA
  • Demigod member
  • **********
  • Posts: 3786
    • MOCpages
Re: offset top
« Reply #55 on: May 26, 2021, 01:07:50 AM »

Rectangular offset tops

Turns out that the general solution for square offset tops in Reply #42 also applies to their rectangular cousins. For the Ansatz case with the top's CM on the boundary between the small and large sub-rectangles,

(S - B)² / B² = γ,

where S is the total length of the side of the rectangular rotor perpendicular to the boundary, B < ½ S is the small sub-rectangle's short side, and γ > 1 is the areal density ratio (small / large sub-rectangle). In terms of the small sub-rectangle's area share β = B / S, this becomes

(1 - β)² / β² = γ,

The 12x16-stud Ansatz top with β = 3/8 and γ = 3 below has negligible wobble at high speed and acceptable wobble at low speed, as did its square counterpart in Reply #28...



If I'd had contrasting 6x12 plates on hand to thicken its small sub-rectangle, some of the small black and white balancing weights on the large sub-rectangle might have been unnecessary. Close to what you see at speed in person...



In the non-Ansatz case, the distance to the CM from the free edge of the small sub-rectangle parallel to the boundary is

xCM = ½ S [(γ - 1) β² + 1] / [(γ - 1) β + 1]

If released carefully, this 12x16-stud non-Ansatz model with β = 1/4 and γ = 3 spins with minor wobble at all speeds using only the single white balancing weight shown at right. In contrast, its square counterpart in Reply #28 needed no balancing weights and spun with no visible wobble at any speed. Why the difference? Beats me!



But no combination of balancing weights can tame the wild wobble in this much narrower 6x16-stud non-Ansatz model, also with β = 1/4 and γ = 3...



Perhaps this last case is starting to butt up against the intermediate axis theorem for triaxial bodies with 3 different central moments I3 > I2 > I1. Per Bächer et al. (2014), critical speed is then given by

ωC² = M g H / (I3 - I2),

where M is the top's mass, g is the acceleration of gravity, and H is the CM-contact distance. As the rectangular rotor narrows, the moment difference in the denominator decreases, and critical speed increases. Perhaps this top's critical speed is close to the release speed can attain by hand.
« Last Edit: May 26, 2021, 01:11:21 AM by Jeremy McCreary »
Logged

ortwin

  • ITSA
  • Hyperhero member
  • ********
  • Posts: 1493
Re: offset top
« Reply #56 on: May 26, 2021, 02:33:06 AM »

Rectangular offset tops

... negligible wobble at high speed and acceptable wobble at low speed

..

... spins with minor wobble at all speeds using only the single white balancing weight shown at right. In contrast, its square counterpart in Reply #28 needed no balancing weights and spun with no visible wobble at any speed. ...

...
You have built and calculated an impressive range of offset tops by now Jeremy!
 
Maybe you could take the chance with these rectangular tops to show us what you mean exactly by some terminology you are using?
- negligible wobble
- no visible wobble
- high speed
- low speed
-minor wobble

Since those are qualities that are not so easy to quantize and can be very relative, the best would be a simple video. Just so we get a better feel of what you mean.

Another question: How careful do you have to be when choosing different LEGO parts? I mean stuff like edge effects on the areal weight: If you use say two 4 x 8  stud plates instead of a 8 x 8  plate, do you have to think about the orientation of the junction? Do the two smaller plates weigh exactly the same as the larger one? The weight distribution is definitely a bit different.





 
Logged

Jeremy McCreary

  • ITSA
  • Demigod member
  • **********
  • Posts: 3786
    • MOCpages
Re: offset top
« Reply #57 on: May 26, 2021, 03:28:44 AM »

Another question: How careful do you have to be when choosing different LEGO parts? I mean stuff like edge effects on the areal weight: If you use say two 4 x 8  stud plates instead of a 8 x 8  plate, do you have to think about the orientation of the junction? Do the two smaller plates weigh exactly the same as the larger one? The weight distribution is definitely a bit different.

Yes, the edge effects are a significant issue in this context. Definitely best to thicken the small sub-rectangles on these square and rectangular offset tops with single plates above and below. And I've tried to do that here whenever possible.

Since those are qualities that are not so easy to quantize and can be very relative, the best would be a simple video. Just so we get a better feel of what you mean.

Useful, yes. But I'm afraid that isn't a video I'm all that motivated to make. Videos that I'm willing to post to my YouTube channel in order to show here are much more work and bother than writing posts illustrated with a few quick photos.
Logged

ortwin

  • ITSA
  • Hyperhero member
  • ********
  • Posts: 1493
Re: offset top
« Reply #58 on: May 26, 2021, 04:25:01 AM »

...Videos that I'm willing to post to my YouTube channel in order to show here are much more work and bother than writing posts illustrated with a few quick photos.
As I understand it, you are concerned with the overall quality of your YouTube channel and don't want to compromise that by "quick and dirty" videos?
Is that about right?
I myself do not use the YouTube channel as a stand alone platform, I mainly use it these days for videos  (quick and dirty for sure!) that I post here.
Nevertheless I use in the "visibility" settings for those videos "not listed" (I am not sure YouTube is using those words in English as I translated them from my German version).
Doing it this way, should make this videos not appear in my channel, only with the link I post in the forum they can be watched, correct?
Would this not be a way to go for you, Jeremy?


« Last Edit: May 26, 2021, 10:56:57 AM by ortwin »
Logged

ta0

  • Administrator
  • Olympus member
  • *****
  • Posts: 14366
    • www.ta0.com
Re: offset top
« Reply #59 on: May 26, 2021, 10:34:54 AM »

What ortwin suggests is what I do. More than half of my youtube videos are unlisted.
Logged