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[ortwin]

It would be a surprise if none of you would show a picture of an really existing  top of this type, but I was not able to find any such thing, so here we go:
The general appearance of the top should be like this:



But the flywheel should be made of two different materials, like this:




The brown part should be some lightweight material, for example wood.
The golden part, some high density metal (brass, copper, tungsten).
The center of mass will move away from the geometric center (that is the whole point actually).
How far off center will CM be for a given pair of densities? Let's say 1g/cm³ and 10 g/cm³ ?
Where do I need to draw that interface line to maximize the offset?
The "beer can Ansatz" we discussed earlier will hold here as well. That is why I tried to put the black circle, symbolizing the stem, on the line that separates the two materials.  The top would have a nice smooth movement that one would not expect at first glance with that offset stem.
That paper Jeremy was pointing me to, for a different reason, the one with the spinning elephant, gave me this idea.
So although there might already be such a top, I think two of you will enjoy the mathematical challenge (that I was not willing to take on).

[ta0]

I don't think I have such a top in my collection. Perhaps I should 3-D print one 

Damn. Even using the Ansatz shortcut I get pretty cumbersome expressions for the position of the line.

r₀= 2/3 m/(M+m) sin³φ /(φ - sinφ cosφ) r
with:
M = π r² ρ₁
m = (φ-sinφ cosφ) r² (ρ₂-ρ₁)

where r₀ is the distance from the center to the line, r is the circle radius, φ is the half-angle of the circle sector and ρ₁ and ρ₂ the densities of the light and heavy sides.

I'm not going to try take the derivative of that thing. I might plot it.

[ortwin]

I don't think I have such a top in my collection. Perhaps I should 3-D print one 

...
I'm not going to try take the derivative of that thing. I might plot it.

I expect you need a big density difference in the two materials to have an impressive offset. Not so easy with 3-D print?
You could simulate a different density by making one side thinner or by leaving the inner part hollow.


If you plot it, please tell us where the minimum maximum  r₀ lies for a desity ratio of two, ratio of 10 and a ratio of 20.
Things become very easy if you go from a circle to a square. A specific solution:




The ratio of the width of the different rectangles is just the inverse of the ratio of the densities.

Edit: very wrong! See reply #9
 
But how to get to the circular disk from this? "Piece of pi(e)!"- I hear Jeremy shout. Or did he say tau?

[ta0]

I was missing a 2 in the equation. I fixed that and realized there is no derivative needed (that was taken care by the Ansatz shortcut). I only needed to equate r₀ =  r cosφ.
I get this equation:

cosφ [π / (ρ2/ρ1 - 1) + φ - sinφ cosφ] / sin³φ = 2/3

Now, it's a matter of finding φ that verifies this identity for a given ratio of densities, what is easy to do with Excel.

[Jeremy McCreary]

@ta0: How about a Simonelli-esque density contrast like wood and brass -- say, 8,500 vs. 850 kg/m³, or 10:1?

BTW, I see that you're also using Greek letters in your equations now. Pretty handy. I'm using Excel's "insert symbol" function to put them in a cell. From there, I just copy them to the forum text box. Where are you getting yours?

In addition to the usual conventions like using ρ for mass density, I really like Greek letters for ratios and proportions. For example, in my top calculations, I usually scale all top dimensions by the top's max radius R. If the top's axial length is L, then its transverse aspect ratio becomes λ = L / R.

This approach makes crystal clear which results depend only on top proportions and which also have some dependence on absolute size. There are more of the former than you might think.

[ta0]

I got these values for the position of the line:

ρ2/ρ1		φ		r0/r
2		81.61		0.146
10		63.15		0.452
20		55.99		0.559

Note: the angle on the table is in degrees but on the equation in radians.

@Jeremy: I'm just copying the Greek letters from the Wikipedia English article of the Greek alphabet. 

[Jeremy McCreary]

@Jeremy: I'm just copying the Greek letters from the Wikipedia English article of the Greek alphabet.

Great idea! I then copied the alphabet from there to a Keep Notes page on my Android phone so they're always handy.

[ortwin]

So to get this quite impressive offset for the circle, we need a factor ~10 in the difference of densities where the difference needs to be only a factor three for the square. 



 


In a top like this setscrews in the  flywheel would definitely make sense for getting it balanced.
A dove tail connection between the two materials would be nice.

[ta0]

The ratio of the width of the different rectangles is just the inverse of the ratio of the densities.

Don't forget that both the levers and the masses depend on the widths of the rectangles. So the ratio of the densities should be equal to the square of the widths.
For a 3:1 ratio of the widths, the ratio of the densities should be 9.

[ortwin]

I see, so even in this simple case of the square I made a big mistake in reply #3. 

[ta0]

I first made the same mistake. I only caught it when I tried to apply the method I used for the circle.
Jeremy didn't catch it either 

[ortwin]

Jeremy is concentrating on filming "Miss Skimpy in Chains". He has no time  for theoretical tops at the moment.

[Jeremy McCreary]

ta0 nicely tackled the offset top with a circular rotor. Here's a general solution for the not-so-simple square rotor case. (Sorry, ortwin, have to shoot Miss Skimpy in Chains and her fluorescent friends at night and just haven't had the energy for all the takes that will involve.)

Setup: Consider a square top rotor of side S. In map view, a straight "boundary" a distance B < ½ S from the closer parallel side divides the rotor into 2 parallel rectangles — a smaller "Rectangle 1" with short side B (white below) and a larger "Rectangle 2" with short side S - B (dark gray).



Each rectangle i of mass M_i, area A_i, mass density ρ_i, and thickness L_i has its own areal density

α_i = M_i / A_i = ρ_i L_i

Areal density ratio: Rectangle 1 gets the greater areal density in this problem — whether by thickening it, densifying it, or both. Hence, the areal density ratio given by

δ = α₂ / α₁ = ρ₂ L₂ / ρ₁ L₁ < 1

In a rotor of uniform thickness where L₁ = L₂, this density ratio reduces to δ = ρ₂ / ρ₁. Likewise, the ratio reduces to δ = L₂ / L₁ when ρ₁ = ρ₂. The wobble-free LEGO offset top above and below models the latter scenario with δ = L₂ / L₁ = 1/3. The stem and tip accurately mark a rotor CM well off the boundary.



Rotor CM: The respective rectangle CMs are located at the following distances, not from the boundary, but from the parallel rotor edge on Rectangle 1:

y_CM,1 = ½ B
y_CM,2 = ½ (B + S)

The distance to the rotor's overall CM is then

y_CM = (M₁ y_CM,1 + M₂ y_CM,2) / M = ½ S (β + M₂ / M) = ½ S (β² + q) / (β + q),

where the proportion β = B / S, M = M₁ + M₂ is the rotor's total mass, and q = δ / (1 - δ) = α₂ / (α₁ - α₂). The LEGO model above confirms this expression for β = 1/4 and δ = 1/3, which makes q = 1/2 and y_CM = 3/8 S.

CM on the boundary: Putting the rotor's CM smack on the boundary means that B = y_CM, or equivalently, that β = y_CM / S. To generalize the solutions forced by this restriction to a square rotor of any size, we'll focus now on β as a function of δ via q. The problem then reduces to the quadratic equation below. With this 1:1 relationship between β and q when the CM's on the boundary, you can set either one and easily solve for the other.

β² + 2 q β - q = 0

Problem 1: For a desired areal density ratio δ, find the β needed to put the CM on the boundary using

q = δ / (1 - δ)
β = q [√(1+ 1/q) - 1]

Problem 2: For a desired β, find the δ needed to put the CM on the boundary using

q = β² / (1 - 2 β)
δ = q / (1 + q)

Does it work? You bet! In the uniform density model below, δ = 1/5 and q = 1/4. The calculations give B = 4.94 on this rotor with S = 16, but couldn't get the boundary, stem, and tip any closer than B = 5.00. Still, the wobble's not bad.




Golden ratio: Can you put the CM on the boundary with δ = β? Yes! Just set them both to 2 - φ ~ 0.382, where φ = ½ (1 + √5) ~ 1.618033 is the golden ratio. Unfortunately, can't (yet) see a way to make a smoothly spinning golden LEGO offset top with δ = β = 0.382.

[ortwin]

... not-so-simple square rotor case...

Admittedly, only the specific case that made use of the "beer can Ansatz" is so simple.

But it seems you have the general solution now in its full glory!How would you judge the play value and the surprise/astonishment effect for the "non top expert" of the versions that you built? Are they keepers for you? Would you say (as I feel) that circular versions would be better in these regards?

[Jeremy McCreary]

How would you judge the play value and the surprise/astonishment effect for the "non top expert" of the versions that you built? Are they keepers for you? Would you say (as I feel) that circular versions would be better in these regards?

Definitely keepers, thank you. Play value for me 8/10 (very good). Astonishment value for the general public probably 9/10 for both shapes. The main advantage of the circular rotor in my mind would be greater play value via greater spin time via reduced air resistance.

Remember, most people aren't used to seeing square tops. As long as the top has a conspicuous asymmetry but spins smoothly anyway, I foresee a decent surprise factor even when the stem and tip aren't on the border but still way off center.

You've hit on an important driver in my own top designs. Before the pandemic, my LEGO club would put on many public displays a year at events with attendances often in the thousands.

So my tops have played to 3 main audiences over the years: Me, this forum, and the general public. And with the last, they've generally been quite popular. They also really like my starters.

Astonishment is by far my favorite public response to a top (or any of my other LEGO gizmos on display). Really tickles me when I hear "Whoa, that's just not right!" or see it in their faces. Unexpected visual effects at speed are generally the most potent triggers. And tops are great canvasses in that regard, as they concentrate a lot of speed across the visual field in a very small space.

Admittedly, only the specific case that made use of the "beer can Ansatz" is so simple.

Couldn't figure out how to turn the Ansatz into a practical design tool in this case, so I went to the general solution for its predictive value. That way, I could start with a LEGO-friendly δ or β value and see if the other was also doable without a lot of guess and check.