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[Jeremy McCreary]

NO, Jeremy ! None of these questions is meant for you, not before you show us Miss Skimpy in Strings!

Chains, ortwin, chains! The only thing better than a string bikini on a top? A skimpy chain bikini, of course.

The answer to your energy question is a 5-minute diversion:

D = I₃ s² / (2 M g e),

where M is the mass dropped, D is the vertical distance it dropped, I₃ is the top's AMI, s is the target spin rate in rad/s, g is the acceleration of gravity, and e is the energy conversion efficiency.

If the only thing being dropped is the top itself, then

D = J₃ s² / (2 g e),

where J₃ is the top's specific AMI.

Consider yourself doing well at e > 30%.You can take it from there.

[ta0]

I tried these push-screw tops without the pushing: just let the tops fall under gravity.



The larger ones don't spin fast enough to reach critical speed. The red and white stays up for about 2 seconds. The flat one for about 5 seconds.
If the screws were 50% to 100% longer, I think they would work fine.

Ortwin: there is a nice analogy between the equations of linear motion and angular motion. If you know that kinetic energy is 1/2 m v² then you automatically know that rotational energy is 1/2 I ω². Equating it to the potential energy, m g h, you get what Jeremy wrote.

[Jeremy McCreary]

Top kabobs, anyone?

[the Earl of Whirl]

Those tops are cool!  I love them!!!

[ortwin]

At first I did not expect this topic to have that much POTENTIAL.  
 

[ortwin]

I tried these push-screw tops without the pushing: just let the tops fall under gravity.
...

I tried a few things with that "lead screw" you can see in the video a few weeks ago. It did not work well at the time.

But now that ta0 reported some results on something very similar I tried once more.

Adding a tip and MORE AMI helped.

I am taking a base for a spin.

 



While playing like that, I realized that it should also be possible just to hold the brass nut in hand and let the screw with the tip gain rotational speed while it falls through. In that case one has to ad AMI to the lower end of the screw of course. Maybe I try one day.

[ortwin]

Chains, ortwin, chains! The only thing better than a string bikini on a top? A skimpy chain bikini, of course.

I guess I am a visual person, since you STILL never showed us Miss Skimpy in any way, I might have fantasies about her that do not match with reality.

...

Consider yourself doing well at e > 30%....

 

No way! Efficiency needs to be >90%!!!Those Maxwell wheels are used in Physics classes to show the conservation (and conversion) of mechanical energy. The efficiency is surely above 90 %.

...
D = J₃ s² / (2 g e),
...

 

...

Ortwin: there is a nice analogy between the equations of linear motion and angular motion. If you know that kinetic energy is 1/2 m v² then you automatically know that rotational energy is 1/2 I ω². Equating it to the potential energy, m g h, you get what Jeremy wrote.

Thank you Jeremy and ta0, I should have made it clearer, I did not ask for the formula, I am fine with that end. I wanted to know some examples
to what height a starting speed would convert if we assume 100% efficiency.
You were talking before about a plan for a data base with curves and all for different types of spinning tops. This data that I mean would be a picture in my brain I can hold on to. And maybe we find something interesting if we put a lot of different data points into that picture.
I start by calculating those numbers for one of Iacopo's tops, and of KBB (Kindergarten Brass Band).
I will edit this post and put the results here.

Iacopo's Nr.29 : h = 52 cm
KBB                : h = 1.7 cm
I strongly doubt my results!!
I used the following data:Nr.29: 0.119 kg ; AMI 0.0000635 ; 1320 RPM at startKBB  : 0.125 kg ; AMI 0.0000151 ;  500  RPM at start
The data for Nr. 29 I found in posts and videos here in the forum. The AMI for KBB is calculated with a formula from a book (ignoring the everything but the brass ring).
h= 1/2 *1/g * AMI/m * (RPM*2pi / 60 )²  was what I used to calculate h from that data. The large difference in the AMI of the two tops must be wrong somehow.

[ta0]

Wow! Using the base as top was thinking out of the box! 
I would have thought that the pitch of that screw was too small to be used, but I was wrong.
You crushed the gravity-powered spins of my real tops with your improvised top! 

[ortwin]

... Using the base as top was thinking out of the box!...

But not really new here: remember, that versatile aluminum flywheel of my "easy listening" top was once also part of a base.

...
I would have thought that the pitch of that screw was to small to be used, but I was wrong.
...

Since it is a triple start thread the pitch looks three times smaller then it actually is.


Fun fact: that screw was originally bought to play a role in an adjustable base for a levitron. But for that role the pitch was far too big. I had fallen into the same trap as you. I was not aware of multti start threads at the time either. Here that screw became the stem of a  top  the flywheel of which is also a base!

...
You crushed the gravity-powered spins of my real tops with your improvised top! 

But there was luck involved: by accident it turned out to be a top with recessed tip giving low critical speed.  That surely helped!
Lucky also that the jar working as dedicated base was at hand and fitted perfectly.

[the Earl of Whirl]

Fun video of the lead screw top!

[Jeremy McCreary]

Well done, ortwin! A low-drag, low-CM version could spin a long time.

[ta0]

Since it is a triple start thread the pitch looks three times smaller then it actually is.

Ah! A 3X lead makes it more plausible.

I strongly doubt my results!!
I used the following data:Nr.29: 0.119 kg ; AMI 0.0000635 ; 1320 RPM at startKBB  : 0.125 kg ; AMI 0.0000151 ;  500  RPM at start
. . . The large difference in the AMI of the two tops must be wrong somehow.

Inner diameter ~77 mm, outer diameter ~93 mm. Height ~8 mm. 

That should give you an AMI of about 0.00009 kg m²

[ortwin]

... A low-drag, low-CM version could spin a long time.

Thank you, but you know, spin time is not everything! If it proves that something works in principle I can be happy with one minute of spin time. Especially if it would not be "legit" by any rules for a hand (or finger or arm or foot ) spun endurance top. ;-)
Edit: This IS a low-CM version! It is so low that we are dealing with a recessed tip!

[ortwin]

..
That should give you an AMI of about 0.00009 kg m²

Whatever, we are in the range of a few centimeters.   
Edit: I found that the formula  for the AMI of the torus that I used from the book is off by factor of 16 from other formulas I find on that on the net. Or maybe I misinterpreted what they meant in the book.
So if I multiply the AMI of by KBB by 16,  I am at 0.000242 kg m², resulting in h = 27.2 cm 




Now how about the Giant Top of Joe Mauk? To calculate the height here, we do not even need the AMI or the RPMs it reaches.
Only the mass of the top, the mass of the bumblebee and the height the bumblebee goes down.

For the top I found something like 800 kg in the thread, may be around 1000 kg now with the tarpaulin.
For the bumblebee I have not seen its weight anywhere, do you know it ? And how many meters is it going down?
There should be bigger lead screws available than the one used, ....hm.... 
On wikipedea there is this on "lead screw":


I guess there are lead screws that can support 800+ kg! But how long does it need to be for the Giant Top 2.0?  Is one meter enough?

[ta0]

For a thin ring like yours you don't need to use the torus equation to get a reasonable approximation. Just use the basic definition of mass times radius square.

You may want to look at this thread Re-thinking spin dynamics for world's largest top