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[ortwin]

Can't you just have the results presented as formula where that slope is still a parameter?

Sure. Is your triangle slanted out (taller on the outside) or slanted in (taller close to the axis)?

Taller outside.

[ta0]

Taller outside.

By integration I get that the moment of inertia increases linearly with the height of the triangle, if you you keep the the other side and the distance to the axis constant.

The equation is:

I = ρ 2 π h/L [(R+L)⁴ (L/5-R/20) + R⁵/20]

where ρ is the density, R the inner radius, L the base of the triangle and h the height.

Edit: Changed the last - to +. Thanks Jeremy for pointing out the error.
I also realized that it could be calculated without any integration. Just add the moment of inertia of a cylinder and a small cone and subtract a large cone.
In general the moment of inertia of any toroid with triangular cross section can be calculated by the sum and subtraction of the moments of inertia of cones and perhaps a cylinder.

[ortwin]

...
The equation is:

I = ρ 2 π h/L [(R+L)⁴ (L/5-R/20) - R⁵/20]

where ρ is the density, R the inner radius, L the base of the triangle and h the height.

Thank you very much, but how to compare this now to the results Jeremy presented? For example to the triangle with three 60 degree angels that you suggested?

[Jeremy McCreary]

The equation is:
I = ρ 2 π h/L [(R+L)⁴ (L/5-R/20) - R⁵/20]
where ρ is the density, R the inner radius, L the base of the triangle and h the height.

Thank you very much, but how to compare this now to the results Jeremy presented? For example to the triangle with three 60 degree angels that you suggested?

To bring a new generator into my Pappus- and Diaz-based framework, I need to parameterize its orientation, size, and exact shape (mass distribution), including the coordinates of its centroid.

Specifying "equilateral triangle" or "isoceles right triangle" (as ortwin did earlier) allows all that in a way that "generic right triangle" does not.

I can move ahead with the isocelese case with one side facing outward. And I'll see how much I can generalize. But in the end, I'll need a specific right triangle in a specific orientation to compare the resulting toroid to the ones already evaluated.

Remember, we want to end up with toroid mass, critical speed, and total surface area at constant AMI. And for the critical speed, we need TMI as well.

[ortwin]

Once evolution took a few further steps in the right direction and the flywheel provides good results ( spin times), I might get T-shirt with something corresponding to this one:

Please order an extra for me!

My wife sometimes thinks I'm an evolutionary throwback. Now that I see that last step in posture, she could be right. Just this once.

So I take it you and your wife are already wearing these? :

[Jeremy McCreary]

@otrwin: Great T-shirt, but I don't think she views my continuing de-evolution as an improvement.

[Jeremy McCreary]

Will need a specific slope (e.g., rise over run) to put numbers to it.

Can't you just have the results presented as formula where that slope is still a parameter?

Figured out a way to parameterize a right triangle generator with one leg (non-hypotenuse) facing outward in terms of its inradius and the angle pointing inward. Will explore 2 toroids with angles of 30° and 60° and go from there.

[Jeremy McCreary]

Examined 3 more toroids using right triangle generators with one leg parallel to the plane of the toroid and inner angles of 30°, 45°, and 60°. Computing the moments of inertia was no fun.

Highlights...
1. The square's toroid still had the lowest critical speed at 410 RPM, but the 30° right triangle's was a close 2nd at 416 RPM. The 60° right triangle had the highest at 495 RPM -- mainly due to having the largest central TMI and CM-contact distance.
2. The square's toroid had 13% more surface area than the circle's, while the 30° and 60° right triangles' had ~43% more. All other toroids were in between.
3. All toroids had maximum radii of 38±1 mm.
4. As before, the circle's toroid had largest central opening. The 30° right triangle came in with the smallest.
5. As before, the square's toroid had the shortest axial length. And as you might guess, the 60° right triangle came in with the tallest.
6. Toroid masses varied by <1%. Hence, when mounted on the same core, the resulting tops would have about the same tip resistance.

Actual relative air resistances remain unknown.

[ortwin]

Examined 3 more toroids using right triangle generators with one leg parallel to the plane of the toroid and inner angles of 30°, 45°, and 60°. Computing the moments of inertia was no fun.

No fun is not good! Thanks for doing it anyway! What I take home from your highlights is, that the right angle triangle that I proposed is  probably not an evolutionary step forward from the square.   But I am quite sure slanting it like on the left side of the stem would provide a little improvement.

That is by reasoning:

My reasoning for slanting it like this, is to bring the CM further down compared to the contour on the right side while keeping the scraping angle and AMI the same.

The surface area is a bit larger, but my feeling is that the lower CM is more important. 
This  optimization step in construction should work for most generators:
- imagine your flywheel is cut from a roll of aluminum sheet rolled around the stem
- draw your scraping line
- let all your thin aluminum tubes gravitate towards that scraping line. 
 -> CM came down, AMI stayed constant, surface increased a bit
Ah, and Jeremy, you probably missed my request below:

Could you show me the formula for the critical speed? If I am not too scared by the first look of it, I just might try to understand some of it and draw conclusions.

 

[Jeremy McCreary]

Could you show me the formula for the critical speed? If I am not too scared by the first look of it, I just might try to understand some of it and draw conclusions.

Not missed, just deferred, as the formula deserves some explanation.

For a top in steady precession (no wobble) at tilt angle θ, the formula you usually see is for the square of the critical angular speed ω_C in rad/s:

ω_C² = 4 M g H I_1tip cos(θ) / I₃²

where M is the mass in kg, g = 9.81 m/s² is the acceleration of gravity, H is the CM-contact distance in m, I_1tip is the transverse moment of inertia (TMI) about the tip (contact point) in kg m², and I₃ is the axial moment of inertia (AMI) in kg m². When you have a sleeper with θ = 0, the cos(θ) factor goes away.

Technically, ω_C includes both the pure spin rate and a component of the precession rate. But for design purposes, you can interpret ω_C as the critical spin rate to a good approximation, as the precession rate involved is usually much smaller.

Making that which is hidden seen
The problem with using I_1tip is that it buries a hugely important dependency on H. To get the latter out in the open where it belongs, I prefer the "central TMI" I_1cm -- i.e., the TMI about the CM. An added advantage is that I_1cm is the TMI you see in tables of formulas like the one on Wikipedia. The parallel axis theorem shows how the two TMIs are related:

I_1tip = I_1cm + M H²

With H exposed in all its glory, the critical speed is then

ω_C² = 4 M g H (I_1cm + M H²) cos(θ) / I₃²

So your plan to minimize H and maximize I₃ for a given scrape angle is well founded, as H is the most influential parameter here, with I₃ not far behind. You can also see why we sometimes talk about the TMI/AMI ratio.

Forget the mass
But the mass M turns out to be just confusing clutter, as both moments are also proportional to M. To give M the boot it deserves, I prefer to use the "specific moments" J_i (aka moments per unit mass), like so...

J₃ = I₃ / M and J₁ = I₁ / M

For example, in the formula I₃ = 1/2 M R² for the AMI of a thin disk of radius R , the specific AMI is just the 1/2 R² part.

The critical spin rate then reduces to

ω_C² = 4 g H (J_1cm + H²) cos(θ) / J₃²

In words, critical speed has nothing to do with mass. To minimize it, you need to focus on maximizing J₃ and minimizing both J_1cm and H. As an added bonus, you can then get all the AMI you need with the least weight on the tip.

Practical applications
Unlike the regular moments, the specific moments are purely geometric in nature. For example, the specific AMI J₃ measures the efficiency of a top's mass distribution about the spin axis without regard to how much mass there actually is.

Suppose you have two lumps of clay, A and B, of the same mass M. You mold A into a solid disk and B into a hollow toroid of the same maximum radius R. Since B will get more AMI out of the same mass, it will have the greater specific AMI.

I find it useful to think in terms of specific moments when I'm optimizing top designs -- especially J₃. But you also have to play a delicate trade-off with the absolute AMI I₃, as AMI is the inertia that opposes any change in spin rate -- during both spin-up and spin-down. Too much AMI, and you'll limit the release speed your fingers can attain in the brief time available during a single twirl. Too little, and air and tip resistances will chew up your hard-won release speed that much faster. Test, test test!

[ortwin]

You explained that formula  very nicely Jeremy!  I just need to find an explanation of how this formula comes about. It seems only then I am fully prepared to believe I can too draw conclusions from it. Don't worry, if you don't have an explanation ready, I try to find something in the books. "Critical angular speed" is the term I should start looking for?

[Jeremy McCreary]

Hope it helps your design process. Look for Jerry Ginsberg's excellent textbook Advanced Engineering Dynamics, free online.

His formulas differ a bit in that he gives the critcal spin rate (not the total angular speed) directly -- but with the complication of a non-physical result in the unlikely event that TMI about the tip is less than the AMI. I like his derivation and stability analysis, though. See chapter 10.

The excellent article by Provatidis, "The spinning top revisited" gives the formula I used explicitly. It's also less intimidating but is no longer free online.

Another good online freebie is Chloe Elliot's "The spinning top" -- maybe part of a thesis. Nice explanations of many things, but she takes a lot longer to get around to critical speed.

The above are all for "symmetric tops" with only 2 distinct principal moments of inertia. Most real tops are of that kind but need not be.

For a fascinating engineering approach to asymmetric tops, see the superb free online article "Spin-it: optimizing moment of inertia for spinnable objects" by Bacher et al. of Disney Research. They give a critical speed formula unique to such tops but don't derive it. I found the nomenclature in their source for it utterly impenetrable.

[Jeremy McCreary]

@ortwin: Since we both seem to like space stations and tops, one of my favorite tops...





An older version below. Of course, no need to worry about air or tip resistance in space.



One of the best musical themes in all of TV if you ask me.

[ta0]

That wormhole sink makes me laugh quite hard! 

The excellent article by Provatidis, "The spinning top revisited" gives the formula I used explicitly. It's also less intimidating but is no longer free online.

I could still download it for free and I'm attaching it to the post just in case it's unavailable in the future. I don't think I would call it less intimidating  Also, I didn't see the equation for the critical spin when I browsed it.

[Jeremy McCreary]

It's kinda scary having a wormhole in your sink.

You know how they say never to put your hand down a Dispose-All? Well, that goes about a thousand times for a wormhole when the rest of you won't fit.

And you wouldn't believe the things that come out of it! My kitchen's that Star Wars bar scene in miniature every day!