iTopSpin
Current Posts => Collecting, Modding, Turning and Spin Science => Topic started by: Aerobie on July 19, 2016, 01:34:18 AM
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I picked up a bit of forehead oil on my finger and wiped it onto a concave mirror. Then I wiped it thinner, with a clean finger three times (pressing hard) to get a super-thin haze of oil.
My home made 208 gram tungsten top (44mm OD) had substantially lower decay rate on this oil haze. This top spins on a 1/4" tungsten-carbide ball tip.
RPM decay on oil was 28% lower than on the same mirror clean. Measurements were made below 1,000 RPM, where aerodynamic drag is low and tip friction dominates. I made spot measurement at 30 second intervals.
I developed this lube method with Euler discs. It made any even greater difference there. I've also tried various oils, including watchmaker's oil. Some were as good as my forehead, but some were worse.
It's essential to wipe the mirror multiple times, pressing hard, to reduce the thickness of the oil film to a few molecules. Too much oil will slow the top.
Best regards,
Alan
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Great tip, Alan! No cheaper lube on the planet, and I always remember to bring my forehead along (though not always the brain behind it).
I'm also going to try forehead oil on my favorite surface for LEGO tops -- polished fine-grained black granite. (This stuff is generally even better than glass for my ABS plastic tips.)
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very interesting ???
ive used wiping my forehead and running my fingers through my hair as a means of getting friction for snapstarts when i have dry hands which is the exact opposite lol!!! ;D
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Aerobie, could I know where did you get the tungsten for your top ?
For comparing spin times, I would clock more spins and calculate an avarage spin time, because spin time is variable, also if the top starts always from the same RPM, and the spinning surface and the lube are always the same.
For example, the following numbers are the RPM of my top Nr. 15 after five minutes of spinning, starting always from 450 RPM: the tip of the Nr. 15 is a conical, sharp, carbide tip, and the base is made of tungsten carbide, very smooth. Weight of the top 186 grams.
Without oil:
330 288 311 316 331
Motor oil:
330 331 328 332 329 324 325
Multicut G, oil for cutting tools for the lathe:
327 328 326
WD40:
334 335 322 329 334 330 331 331 332
WD40, very thin layer:
344 338 331
I didn't experiment more because these data were enough for me to decide to use a super thin layer of WD40 for my record, and it worked fine, I did the over 50 minutes spin in this way.
But I never tried skin oil...
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Hi Iacopo,
I bought my tungsten carbide balls from mcmaster.com, in California. They are relatively inexpensive.
I think spot checks are better for this than total spin time. Spot checks permit focusing on low speeds, where tip friction dominates. My tachometer saves the current reading when I release the button. So as I approach the x:00 or x:30 on my stopwatch, I begin reading, then I release the button at :00 or :30 and record the RPM. I take RPM1/RPM2 as the decay over that period.
Alan
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Hi Alan,
so I did, only my spot checks in this oil test were 5 minutes long, instead of 30 seconds.
They have been made on low speeds, from 450 RPM to 300-330 RPM.
I thought it is the flywheel of your top to be made of tungsten because you have written "208 gram tungsten top". I would like to try a tungsten flywheel but I don't know where I could find it. For the tip, I have found a high quality tungsten carbide (there are various kinds of it) which works well for my needs. Thank you anyway for the address.
Now I am going to test skin oil.. :)
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Hi Iacopo,
I bought some tungsten years ago from a scrap metal dealer. I'll search my records. I've also seen tungsten metal offered for sale on ebay.
Surprisingly, my tungsten is very easy to machine.
The specific gravity is 16!
Best,
Alan
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Hi Alan,
Maybe you have an alloy of tungsten and copper, some time ago someone suggested me to look for this alloy because it is relatively easy to machine and at the same time still quite heavy.
But I can't find it. I looked on E-bay, there are mainly little objects made of tungsten, but I can't find something I could use.
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Hi Alan,
Maybe you have an alloy of tungsten and copper, some time ago someone suggested me to look for this alloy because it is relatively easy to machine and at the same time still quite heavy.
But I can't find it. I looked on E-bay, there are mainly little objects made of tungsten, but I can't find something I could use.
You probably won't be as lucky as Jim who found some on the street: what is it? (http://www.ta0.com/forum/index.php/topic,4437.0.html) But I am sure Jim would like to make an exchange with you :)
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You probably won't be as lucky as Jim who found some on the street: what is it? (http://www.ta0.com/forum/index.php/topic,4437.0.html) But I am sure Jim would like to make an exchange with you :)
Since the letters are written mirrored, probably it is a stamp. But 28 mm diameter is still too little..
Anyway, thank you for the hint.
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You probably won't be as lucky as Jim who found some on the street: what is it? (http://www.ta0.com/forum/index.php/topic,4437.0.html) But I am sure Jim would like to make an exchange with you :)
Since the letters are written mirrored, probably it is a stamp. But 28 mm diameter is still too little..
Anyway, thank you for the hint.
Hi Iacopo,
I'm still searching for sources of tungsten. What diameter and length to you seek?
Alan
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I'm still searching for sources of tungsten. What diameter and length to you seek?
I have read your e-mail. Thank you very much for the info.
Regards
Iacopo
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I'm thinking of standardizing on brass and abandoning tungsten alloy. I created a spreadsheet with various flywheels of equal diameter and varied height for constant weight.
Tungsten alloy (SG=18) only reduced effective surface area (and thus aerodynamic drag) by about 18% compared to brass. Aerodynamic drag has little effect on finger twirls, and perhaps a one third effect with string launches. So an 18% reduction in aero drag might extend total spin time by a few percent for one-pass finger twirls, and about 6% for string launches. That's not much, considering that tungsten costs $100 per pound.
I still have some small pieces of tungsten and will probably used them. But after that I may stick with brass.
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Alan: Do you have any photos of your tops to share -- especially the tungsten and brass tops described in your last post about abandoning tungsten? I'd love to see them and may have some insight into the behavior you observed depending on rotor shape.
What's the SG of the brass you're using?
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Hi Jeremy,
I'll get some pics together in the next few days.
My prior post was about hypothetical flywheels which I did not build. They were all 2" diameter cylinders and each material had a height which resulted in the same weight as a 0.5" high brass wheel. For surface area, I assumed that the top and bottom face area was 70% of actual, because area near the center is moving as lower surface speed than the perimeter. I used 100% of the perimeter area, which is pi*diameter*ht.
Because all have the same diameter and weight, the only variable is surface area.
Best,
Alan
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Below is a pic of seven tops. The two Delrin tops are my latest. They were designed for base regeneration. Both are pretty similar but the 52g spins on a 1/2" machined aluminum hemisphere and the 64g has a 5/8" steel ball. Both work quite nicely, but the 64g is best.
Stems are drilled to accept a rod to stabilize them for string spins. The 190g and 208g tungsten spin about 10-12 minutes with a finger twirl and over 30 minutes with string launch. The 47g tungsten spins about the same time with a twirl but I haven't explored string launches yet, it's exceptionally easy to twirl for a 2,000 RPM start. Tungsten diameters are 1", 1.5" and 1.73".
Update: I replaced the .5" diameter aluminum hemisphere with a .5" diameter steel ball set in a brass "cup". That increased the weight to 68g. And lowered the Cg slightly. Performance didn't change much. The 64g top with .625" steel ball is still best for regeneration. I'll try larger balls in about a week.
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if you want longer spins, use a lighter material in the middle and put a heavier metal on the outer edge.. the RPM will be lower but it will maintain it longer.
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Yes, I agree completely. The Megatop is hollow. So it has air in the center, which is pretty light.
The 275g brass top in the upper row has an aluminum center top plate, and air below that.
Best,
Alan
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Hi Alan,
I know tungsten will not make miracles. When I used a lead flywheel instead of a brass one, spin times improved by only a few minutes, so this is what I can expect with tungsten instead of lead, no more than another few minutes improvement. Nevertheless, I would like to make at least one top with a tungsten flywheel. 100 $ per pound would be still pretty much affordable, but I'm afraid you intended 100 $ per ounce. I am waiting for an answer from the dealer, if he has something available and not excessively expensive, I will buy it.
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Prices quoted on ebay vary, but several are about $0.227 per gram, which is $103 per pound.
Alloys are cheaper, but lower density.
Of course, what you really want for your large tops is a heavy ring. But the melting point of tungsten is so high that you are faced with machining away the center of a cylinder, and discarding several $100 worth of tungsten. An alternative would be to machine a brass ring, then drill an array of holes in the ring to accept small tungsten plugs. However the effective density is probably about equal to lead.
Adding an array of tungsten plugs to a lead ring would definitely increase its weight. Straight sides would not be as pretty as your toroid, but would be easier to make, heavier, and (in my estimation) perform well.
Alan
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Iacopo,
We discussed knurling of small diameter stems a few weeks ago.
The 275g brass top in my photo has a tapered stem with straight grooves. I cut the taper with the "compound" on my lathe set at that angle. Then with the motor off, I scraped the straight grooves by cranking the compound along the stem with the cutter about 1/4 mm deep into the aluminum.
Alan
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.... Stems are drilled to accept a rod to stabilize them for string spins. The 190g and 208g tungsten spin about 10-12 minutes with a finger twirl and over 30 minutes with string launch....
Beautiful tops with impressive spin times, Alan! Especially like the stark, almost stealth look of the Delrin tops. Love the idea of the guide rod for string starts.
As I'm sure you know, string-based spinners of various kinds have been around almost as long as tops, and that's saying a lot. I recently made this "string theory" spinner for my LEGO tops. Different from your string-based method but similar to many ancient designs.
https://www.youtube.com/watch?v=hgMu-vAEr20
Surprisingly high torque capacity. Works well with all but the lightest tops.
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Alan,
I asked the dealer a tungsten cylinder with diameter 50 mm and height 10 mm.
With these dimensions the weight is 379 grams. If I can have it for let's say $ 100, I will take it.
About half of that weight will be lost machining it, but, as you say, there isn't much I can do about it.
I thought also to use littler pieces of tungsten in another metal, also there is wire of tungsten, or sand, that I could use, but I am afraid that, apart from a more complicated making, I am not sure if I can make the flywheel sufficiently balanced as needed. It could be balanced in a second moment, but all becomes more complicated, for a density that in the end wouldn't be so high in any case.
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The 275g brass top in my photo has a tapered stem with straight grooves. I cut the taper with the "compound" on my lathe set at that angle. Then with the motor off, I scraped the straight grooves by cranking the compound along the stem with the cutter about 1/4 mm deep into the aluminum.
It is the same way I do. The knurling tools I have seen (I never bought, nor used them), seem to work only for cylindrical surfaces, but I want my stems to be slightly tapered, so I can't use them.
My cuts too are about 1/4 mm deep. But I make many grooves, (their sides overlap in the narrower part, so they are littler, upside), you have made many less.
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If any of you ever buy a knurling tool, I strongly recommend the scissors type.
Alan
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If any of you ever buy a knurling tool, I strongly recommend the scissors type.
Alan
Thank you for the tip, I will remember if I will buy one.
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You can find companies to make you tungsten rings using wire EDM machines, I've been quoted yo-yo sized rings (about 2.25" outside diameters) in almost pure tungsten for around $20-40 each in the past. With wire EDM there isn't really any wasted material.
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You can find companies to make you tungsten rings using wire EDM machines, I've been quoted yo-yo sized rings (about 2.25" outside diameters) in almost pure tungsten for around $20-40 each in the past. With wire EDM there isn't really any wasted material.
This also is interesting, thank you.
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I'm thinking of standardizing on brass and abandoning tungsten alloy. I created a spreadsheet with various flywheels of equal diameter and varied height for constant weight.
Tungsten alloy (SG=18) only reduced effective surface area (and thus aerodynamic drag) by about 18% compared to brass. Aerodynamic drag has little effect on finger twirls, and perhaps a one third effect with string launches. So an 18% reduction in aero drag might extend total spin time by a few percent for one-pass finger twirls, and about 6% for string launches. That's not much, considering that tungsten costs $100 per pound.
Took the liberty of extending your spreadsheet analysis to estimate the total drag on (i) a tungsten top very much like your 47 g beauty, and (ii) a brass top identical in every way other than rotor density and axial rotor length. The tungsten rotor's aerodynamic advantage turn out to be slight -- indeed, even less than the ratio of rotor lengths would suggest. Its greatest advantage could well be the reduced minimum speed for stable sleeping brought about by its lower center of mass (CM) and lesser transverse moment of inertia (TMI).
Assumptions:
(i) Stems and tips are negligible in comparison to top rotors WRT both top motions and aerodynamics.
(ii) One top has a rotor of brass (subscript "b"), and the other, a rotor of tungsten (subscript "t") with respective densities db = 8.5e3 and dt = 18e3 kg/m^3.
(iii) Both tops resemble your 47 g tungsten top in rotor profile and have identical rotor masses m = 0.047 kg, identical rotor radii R = 0.0127 m, and identical release speeds w0 = 209 rad/sec (2,000 RPM).
(iv) Each rotor is modeled as a solid cylinder with flat "faces" (subscript "f") above and below and a vertical "edge" (subscript "e") in between.
(v) Rotors differ in axial length Li so as to keep rotor mass m constant. Hence, Lb = Lt (dt / db) > Lt.
(vi) Both tops are subject to the same air density dair = 1.225 kg/m^3 and kinematic viscosity v = 1.5e-5 m^/s.
The initial Reynolds number Re == w0 R2 / v = 2,252 is the same for both tops. Since 60 < Re < 300,000, I made 2 more assumptions:
(vii) Swirling laminar von Karman flows prevail over the rotor faces, but a dominantly circumferential turbulent flow prevails around the rotor edges.
(viii) The face and edge flows don't interact.
I'm betting that these last 2 assumptions are close enough to allow useful estimates. The drag equations relevant to assumption (vii) come from White's Viscous Fluid Flow and the Childs chapter that ta0 recently linked for us. Happy to show my work, but for now, I'll just give the results.
Results: Let Qb be the total aerodynamic braking torque on the brass rotor due to its edge and both faces, let Qt be the same for the tungsten top, and let Af = 5.1e-4 m^2 be the area of a rotor face. The tungsten/brass drag ratio is then
Qt / Qb = (a + b / dt) / (a + b / db) = 0.89
where
a == Cf R = (3.87 / sqrt(Re) R = 1.04e-3 m, and b == Ce m / Af = 2.22 kg/m^2.
Comments: The result Qt / Qb = 0.89 confirms that the tungsten rotor encounters less total drag by virtue of the reduced edge torque that comes with a shorter axial length (5.2 vs. 10.9 mm). The advantage is much smaller than the axial length ratio Lt / Lb = 0.47 would suggest because both rotors encounter the same face torques, and the face torques are several times larger than the edge torques in both cases.
In the definitions of a and b above, Cf == 3.87 / sqrt(Re) is the dimensionless laminar torque coefficient for both faces together, and Ce is the dimensionless turbulent torque coefficient for the edge. There's no simple formula for Ce, but it can be calculated iteratively, as I've done here.
Cf and Ce depend only on sqrt(Re) but in different ways. This has several ramifications:
(i) Each coefficient is the same for both tops.
(ii) Each depends weakly on speed in its own way.
(iii) The 0.89 figure for the tungsten/brass drag ratio applies only to the 2000 RPM release speed assumed above. My sense is that tungsten's aerodynamic advantage wouldn't grow much during spin-down and might even decay. Seeing how that goes would require a lot more spreadsheet work, but I may get around to it.
Since the tops are assumed to have identical stems and tips and identical weights, they should see identical braking torques due to tip friction. And since both tops have the same axial moment of inertia, they'll also have the same resistance to deceleration by their respective total braking torques.
Under these circumstances, the tungsten top would spin longer by virtue of reduced total drag alone. However, the tungsten top has the added advantage of a reduced minimum speed for stable sleeping for 2 reasons: (i) Its CM is signifcantly closer to its tip. (ii) Its TMI about the tip is correspondingly smaller.
Estimating relative sleep times is out, as we have no way to quantify the frictional braking torques, but I'm guessing that tungsten's multiple advantages would result in a sleep time that's noticeably longer but not spectacularly so.
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I'm away from my computer for a few days, limping along on my tablet.
Effective face area is reduced because surface velocity is lower towards the center.
With aero drag proportional to V squared, face drag is reduced substantially.
Thus the relative importance of perimeter drag is greater than its area only implies.
I reviewed my decay measurements on this top and aero drag dominated above 1,000 RPM. Below that, tip drag dominated. I presume that inertia/tip drag is independent of material. But agree that the lower Cg of tungsten is significant.
Alan
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Effective face area is reduced because surface velocity is lower towards the center.
My approach took all of that into account.
With aero drag proportional to V squared, face drag is reduced substantially.
If by "drag", you mean the braking torque exerted on a top by the air flow it induces, I have 3 excellent sources in fluid dynamics that show quite clearly that drag is not proportional to the speed squared in a purely rotational context by any definition of "speed" -- not over the faces, and not over the edge. That's not just theory. There's a great deal of experimental evidence to back it up.
Moreover, the highly exponential spin decays evident in the data reported by various forum members are not consistent with drag that varies with the square of speed, and the same goes for the slower decays observed at low speeds.
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Hi Jeremy,
I'm not doubting or debating you, but please cite your 3 sources. If the drag exponent is not 2, what's an approximate number that we can use as the exponent for surface drag of a smooth spinning wheel?
I've encountered situations where some smaller exponent works fairly well. At extremely low Reynolds numbers 1.0 works. However my experimental data says it's greater than 1.0 for spinning tops.
Sincerely Alan
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Hi Alan,
Here you go...
Childs, PRN, 2011, Rotating Flows, Chap 6 (laminar and turbulent cylinder edge flows), easily found online. (The earlier chapter on cylinder face flows doesn't seem to be online.)
Coran, D, Childs, PRN, Long, CA, 2009, Windage sources in smooth-walled rotating disc systems, J. Mechanical Engineering Science, v. 223, p. 1-16.
Diab et al., 2004, "Windage losses in high-speed gears -- preliminary experimental and theoretical results", Journal of Mechanical Design, v. 126, p. 903-908. (Analysis starts with toothless cylinder. Data suggests that laminar edge flow could be realistic for our tops.)
Miclavcic, M, Wang, CY, 2004, The flow due to a rough rotating disk, Z. agnew Math. Phys. v. 54, p. 1-12.
Vrancik, JE, 1968, Prediction of windage power loss in alternators, NASA Technical Note D-4849.
White, FM, 2006, Viscous Fluid Flow, 3rd ed., p. 155-160 for laminar face flow (very realistic for our tops), p. 103-105 for laminar edge flow (may or may not be realistic).
I've encountered situations where some smaller exponent works fairly well. At extremely low Reynolds numbers 1.0 works. However my experimental data says it's greater than 1.0 for spinning tops.
As you'll see in the sources above, exposing the full dependence of drag torque on angular speed from literature equations is complicated by the fact that aerodynamic braking torques are often expressed as products of (i) a speed-dependent dimensionless torque or moment coefficient (typically a function of Re) and (ii) a speed-dependent reference torque Qref. The latter is typically
Qref = ½ d w2 R5
for face flows, and
Qref = ½ d w2 R4 L
for edge flows, where d is air density, w is rotor angular speed, R is rotor radius, and L is rotor axial length.
When the full speed dependencies of the torques are laid out, the exponents usually found for angular speed are as follows:
For laminar face flows, 1.5
For laminar edge flows, 1.0
For turbulent edge flows, hard to tell due to the iterative nature of many of the formulas.
Mathematically, truly exponential spin decay demands an angular speed exponent of exactly 1.
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Thank you Jeremy. I may not tackle the references until next week when I am reunited with my computer. But in the meantime do we agree on the following?
1. At high RPM the decay slope is steeper and curved due to aero drag. (I call this exponential, but the exponent isn't necessarily 2, nor constant).
2. The aero drag of the faces per square cm is less than the aero drag of the perimeter per square cm, because the inner regions of the faces are moving slower.
3. The decay slope at low RPM (where aero drag is insignificant) is pretty much independent of material density for tops of identical geometry because tip friction and polar inertia are both proportional to weight.
Best,
Alan
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Thank you Jeremy. I may not tackle the references until next week when I am reunited with my computer. But in the meantime do we agree on the following?
1. At high RPM the decay slope is steeper and curved due to aero drag. (I call this exponential, but the exponent isn't necessarily 2, nor constant).
2. The aero drag of the faces per square cm is less than the aero drag of the perimeter per square cm, because the inner regions of the faces are moving slower.
3. The decay slope at low RPM (where aero drag is insignificant) is pretty much independent of material density for tops of identical geometry because tip friction and polar inertia are both proportional to weight.
Best,
Alan
Alan,
Agree on No.1.
Disagree on No.2 -- at least for tops like your 47 g tungsten and its simulated brass counterpart. The face torques exceed the edge torques in both cases, and the difference is greater than would be expected from the surface area difference alone. The same has been true in similar calculations done on my own tops. This pattern reflects the higher speed and radius exponents and the lower viscosity exponent in the face case. You'll see what I mean when you check the references.
Of course, these are all estimates based on simplifying assumptions that apply to only a small class of finger tops with rotors that approximate cylinders. They certainly don't hold for conventionally shaped tops or even for most of your tops.
I need to think about No. 3.
In the data I've seen on this forum, the decay slopes and curvatures at the lowest speeds vary considerably, and some of the low-speed tails are quite inconsistent with speed-independent tip friction. The underlying processes could be complicated. Heck, physicists can't even agree on the kind of friction that should prevail at the tip. IMO, the only way forward is to amass and share more data on the spin decays of real tops and to analyze that data in a consistent way from top to top.
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Here's a restatement regarding faces.
Face torque divided by total face area is much lower than perimeter torque divided by perimeter area because:
a. The inner portions of the faces see lower air velocity.
b. The inner portions of the faces are closer to the center, and thus have a shorter lever arm to convert surface drag to torque.
If we assume surface drag is proportional to V^2 then a face factor is the definite integral of r^4, which equals 0.2.
If surface drag is proportional to V then the integral is r^3, which equals 0.25.
Alan
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Agree on points (a) and (b), but I think we should continue this after you've looked at the equations.
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Until today I've calculated decay rate as follows:
(RPM1/RPM2-1)/seconds
But this method fails to show the change from aero dominated decay to friction dominated decay.
So I've changed to linear decay (revs per second)
(RPM1-RPM2)/seconds
This computes as -7.6 revs/sec at about 4,000 RPM and decays exponentially over 20 minutes, then levels at about -0.8 revs/sec (at 700 RPM) and ends only slightly lower at -0.7 revs/sec at 220 RPM at t=30 minutes.
Alan
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Hi Jeremy,
Back home, I searched for your references. I didn't find them all, but found some interesting items. One was the statement,
"In general aerodynamic power is proportional to w^2.8 and R^4.6"
This is power, which is proportional to torque*w. So torque is proportional to w^1.8, which is in the neighborhood of w^2, but the exponent is less than 2 because drag coefficients diminish with Reynolds number. I observe exponential decay at high w (RPM), but perhaps the exponent is only about 1.8.
An interesting item too is R^4.6. Radius of gyration is proportional to R and polar inertia is proportional to mass*radius of gyration. But mass is proportional to R^3, assuming constant R/Ht. So polar inertia is proportional to R*R^3 or R^4. At a fixed speed we have Inertia / drag proportional to R^4 /R^4.6 which simplifies to 1/R^0.6 and suggests that smaller tops spin longer in the aerodynamic dominated range (for constant material and constant proportion of R/ht).
I haven't seen this. But I have noted that just scaling up a top doesn't do much. Perhaps it's closer to size independent because the benefits of Reynolds number are greater than the approximation of R^4.6.
What I really want to learn is the optimum proportion of R/ht and the optimum shape. I recently made a cylinder top which had twice the decay rate of an approximately similar cone top. I think this was probably due to a couple of factors:
1. The bottom face of the cylinder was close to the base and there was more aero-shear between the bottom face and the base than with the cone. Some of the references discuss this effect because it's an issue with computer disks spinning near their enclosure walls.
2. The cylinder was 1.5" high and had a lot more circumferential area than the cone, which was only 0.25" high at the circumference.
At very low speeds, where there was hardly any aerodynamic drag, both tops have about the same decay rate.
Best,
Alan
Update: I machined the cylinder into a cone. Cylinder weighed 374g, cone weighs 280g.
The minimum RPM before topple went from 300 to 359 due to the higher Cg of cone.
The decay rate at 1,500 RPM is 5% lower as a cone. The decay rate at 1,000 RPM is 17% lower as a cone as the benefit of lower tip weight becomes more apparent. The decay rate at 500 RPM was 39% lower for the cone. At this speed aero drag is pretty much gone. The total spin time, for the same launch is very similar because the higher topple speed negated the lower decay rate.
Ah yes, there's no free lunch. Improve decay rate and give it back due to higher Cg.
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There have been some posts here about top materials, tungsten and brass. I started a thread recently about tip materials and will post something tonight about top body materials. Although not a tip, it's at least material and thus closer to "on-thread" than this thread.
Alan
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"In general aerodynamic power is proportional to w^2.8 and R^4.6"
This is power, which is proportional to torque*w. So torque is proportional to w^1.8, which is in the neighborhood of w^2, but the exponent is less than 2 because drag coefficients diminish with Reynolds number. I observe exponential decay at high w (RPM), but perhaps the exponent is only about 1.8.
What were the assumptions/geometries/conditions underlying those exponents on angular speed? Whether they apply to our tops or portions thereof depends very much on the details. They look a lot like the exponents given in empirical expressions for the drag due to turbulent swirling flow over the face of a thin spinning disk. Such expressions typically apply only at Re > 106 or 107.
Since our tops are lucky to hit Re = 104, any disk faces they might present to the surrounding air would operate entirely in the laminar regime at Re < 3x105. The flow regime matters because the exponent on angular speed in laminar drag is usually less than that in the turbulent case. And the exponents for laminar and turbulent drag on the edge of a disk tend to be lower still.
An interesting item too is R^4.6. Radius of gyration is proportional to R and polar inertia is proportional to mass*radius of gyration. But mass is proportional to R^3, assuming constant R/Ht. So polar inertia is proportional to R*R^3 or R^4. At a fixed speed we have Inertia / drag proportional to R^4 /R^4.6 which simplifies to 1/R^0.6 and suggests that smaller tops spin longer in the aerodynamic dominated range (for constant material and constant proportion of R/ht).
I haven't seen this. But I have noted that just scaling up a top doesn't do much. Perhaps it's closer to size independent because the benefits of Reynolds number are greater than the approximation of R^4.6.
I think you meant to say that polar inertia (what I've been calling axial moment of inertia, or AMI) is proportional to the square of the radius of gyration. (Check Wikipedia.) If we're talking about a cylinder of constant radius/length ratio (R/ht in your notation) with negligible tip friction, the angular deceleration then ends up proportional to R-0.4, which is close to your R-0.6.
In the laminar case applicable to disk faces on our tops, however, the drag is proportional to R5, and the effect of radius on deceleration rate vanishes. If tip friction and the drag on the top's flank were both negligible, that would match your observation that scaling up doesn't have much of an effect -- at least not on spin decay.
However, tip friction isn't always negligible and grows with top weight by almost any theory of tip friction. As you pointed out, rotor weight grows as R3 at constant density and radius/length ratio. Hence, tip friction in such a top would grow very rapidly with radius.
The effect of radius on spin time is even more complicated, as spin time also depends very strongly on the minimum speed for stable sleeping or steady precession. That speed is unaffected by drag and tip friction, depends on size and shape in very different ways than spin decay does, and is definitely lower for greater maximum rotor radius for any reasonably shaped body spinning about an axis of rotational symmetry, regardless of mass. (See my post at http://www.ta0.com/forum/index.php/topic,1496.msg46536.html#msg46536 (http://www.ta0.com/forum/index.php/topic,1496.msg46536.html#msg46536) for details.)
So as you noted farther down, the ultimate impact of radius on spin time is hard to predict. Ditto for rotor shape.
What did you mean by "benefits of Reynolds number"?
What I really want to learn is the optimum proportion of R/ht and the optimum shape.
We're definitely on the same page there.
I recently made a cylinder top which had twice the decay rate of an approximately similar cone top. I think this was probably due to a couple of factors:
1. The bottom face of the cylinder was close to the base and there was more aero-shear between the bottom face and the base than with the cone. Some of the references discuss this effect because it's an issue with computer disks spinning near their enclosure walls.
2. The cylinder was 1.5" high and had a lot more circumferential area than the cone, which was only 0.25" high at the circumference.
At very low speeds, where there was hardly any aerodynamic drag, both tops have about the same decay rate.
....
Update: I machined the cylinder into a cone. Cylinder weighed 374g, cone weighs 280g.
The minimum RPM before topple went from 300 to 359 due to the higher Cg of cone.
The decay rate at 1,500 RPM is 5% lower as a cone. The decay rate at 1,000 RPM is 17% lower as a cone as the benefit of lower tip weight becomes more apparent. The decay rate at 500 RPM declined 39% for the cone. At this speed aero drag is pretty much gone. The total spin time, for the same launch is very similar because the higher topple speed negated the lower decay rate.
Ah yes, there's no free lunch. Improve decay rate and give it back due to higher Cg.
Very interesting observations! I'll definitely give them more thought. How are you measuring your decay rates?
Totally agree, though, no free lunch with tops. Every design parameter has its trade-offs, and how best to play them is often unclear without exhaustive testing. In the end, we'll need a very consistent (and very tedious) testing campaign with well-controlled variables to sort out some of the trade-offs we've been discussing for real tops.
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Although I expressed an interest in linear decay rate because it shows when the decay shifts from aero dominated to tip friction dominated, I'm now using this exponential formula because it is relatively constant at all speeds and more convenient for comparing tops.
Decay = (RPM1/RPM2)^(1/seconds)
seconds = time between measurements of the two RPM's
my better tops have a decay ~ 1.0010
When grading decay, I ignore the leading "1", so I view 1.0020 as twice the decay rate of 1.001
Alan
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Jeremy,
Thank you for noting my error. Yes, with rotational inertia proportional to R^5, then spin time (with constant material and aspect ratio) is proportional to R^0.4. This seems about right. Larger tops spin slightly longer.
Best,
Alan
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Jeremy wrote, "What did you mean by "benefits of Reynolds number"?"
I mean that (in the low speed regime) as low Re increases, the drag coefficients diminish. Thus the effective exponent can be viewed less than 2. Example: drag ~ V^1.8.
Alan
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Flat Mirror:
With a 43 gram twirler, spinning on a 0.156" diameter silicon nitride ball.
I compared the decay rate (at low speed, where tip drag dominates) of the same top on a flat mirror and a 3x magnifying mirror. The decay on the flat mirror was about 55% of the decay on the magnifying mirror.
I visualize the flat mirror contacting a smaller diameter patch at the tip of the ball. Thus I expect it to have lower decay. I've also posted previously that I prefer low magnification mirrors, like 2x and 3x, over higher magnification such as 10x, because they are closer to flat.
Alan
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ForeverSpin DLC Mirror Disappointing:
I tested my 47g tungsten and 43g bronze tops on this mirror. Decay times, at low speed where tip drag dominates, were about equal for the 47g and slightly worse for the 43g, compared to my 3x glass mirror with a very thin haze of forehead oil.
The DLC mirror is deeply concave, perhaps about 10x. As I've mentioned above, high magnification mirrors are inferior to lower magnification mirrors.
Alan
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Predicting minimum RPM (topple speed):
I think the topple speed is affected by balance, but I haven't attempted to measure balance. Ignoring balance, this formula works pretty well for me most of the time.
minimum RPM = (500/top diameter)*(Hcg/Ball diam)^0.5
(Ball is the tip ball)
All dimensions are in inches. Give it a try and let us know how it does for you.
Alan
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minimum RPM = (500/top diameter)*(Hcg/Ball diam)^0.5
Sorry for the Foreverspin mirror, I really thought it would have performed better than so. Maybe with skin oil ?
Usually I use spiked tips for my tops, but I have data for two tops for which I used a diameter mm 4 sapphire tip:
Top Nr. 8
minimum RPM: 320
diameter: 1.73
Hcg: 0.55
Ball diam: 0.16
Top Nr. 6
minimum rpm: 190
diameter: 2.95
Hcg: 0.63
Ball diam: 0.16
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ForeverSpin DLC Mirror Disappointing:
I tested my 47g tungsten and 43g bronze tops on this mirror. Decay times, at low speed where tip drag dominates, were about equal for the 47g and slightly worse for the 43g, compared to my 3x glass mirror with a very thin haze of forehead oil.
The DLC mirror is deeply concave, perhaps about 10x. As I've mentioned above, high magnification mirrors are inferior to lower magnification mirrors.
Alan
I don't have any hard data, but my sense is that LEGO tops behave the same way on concave surfaces of all kinds: The greater the surface curvature at the contact patch (magnification in the mirror's case), the shorter the spin time. The tips involved have all been of ABS plastic with hemispherical ends and radii of curvature of 1.5, 3.0, and 5.0 mm. The only dynamically significant variable in such tests would seem to be tip friction.
I should test flat-ended tips as well, as the contact patch area would be a different function of mirror and tip curvature.
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Iacopo and Alan: What constitutes a "toppling" event WRT your data?
I ask because the formulas generally given for the minimum angular speed wmin (rad/sec) for stable sleeping or steady precession refer to the speed at which wobbling becomes possible, not the speed at which the top ultimately falls to the ground. (See the last 1/3 of my post at http://www.ta0.com/forum/index.php/topic,1496.msg46533.html#msg46533 (http://www.ta0.com/forum/index.php/topic,1496.msg46533.html#msg46533) for details.)
These formulas assume no perturbations due to imbalance, air currents, or tip or surface irregularities. To first order, however, they're independent of the dissipation involved, including tip friction. A top spinning without wobble can continue to do so below wmin in a metastable state. However, it doesn't take much of a perturbation to trigger wobbling below wmin.
These formulas can't predict the speed at which the top ultimately falls to the ground, and I have yet to see an analytical (as opposed to numerical) treatment that even tries to do so. Below wmin, the math becomes intractable, and too many unknowables enter the problem.
The upshot: To compare observed minimum speeds to each other and to well-established formulas in a meaningful way, I suggest that we report only the speeds at which major wobbling begins. Of course, identification of the onset of "major wobbling" will be somewhat subjective, and we'll be assuming that metastable states don't last long. But that's probably the best we can do in the real world.
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minimum RPM = (500/top diameter)*(Hcg/Ball diam)^0.5
I have data for two tops for which I used a diameter mm 4 sapphire tip:
Top Nr. 8
minimum RPM: 320
diameter: 1.73
Hcg: 0.55
Ball diam: 0.16
Top Nr. 6
minimum rpm: 190
diameter: 2.95
Hcg: 0.63
Ball diam: 0.16
My formula gives 535 and 336 RPM for these tops, which is way too high. Ditto for two tops which I made with 0.156" diameter tips. I could change the constant (now 500) for small diameter tips, but what I really want is a formula which works for a wide range of tops. Well, it's a start.
Alan
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Iacopo and Alan: What constitutes a "toppling" event WRT your data?
I ask because the formulas generally given for the minimum angular speed wmin (rad/sec) for stable sleeping or steady precession refer to the speed at which wobbling becomes possible, not the speed at which the top ultimately falls to the ground. (See the last 1/3 of my post at http://www.ta0.com/forum/index.php/topic,1496.msg46533.html#msg46533 (http://www.ta0.com/forum/index.php/topic,1496.msg46533.html#msg46533) for details.)
These formulas assume no perturbations due to imbalance, air currents, or tip or surface irregularities. To first order, however, they're independent of the dissipation involved, including tip friction. A top spinning without wobble can continue to do so below wmin in a metastable state. However, it doesn't take much of a perturbation to trigger wobbling below wmin.
These formulas can't predict the speed at which the top ultimately falls to the ground, and I have yet to see an analytical (as opposed to numerical) treatment that even tries to do so. Below wmin, the math becomes intractable, and too many unknowables enter the problem.
The upshot: To compare observed minimum speeds to each other and to well-established formulas in a meaningful way, I suggest that we report only the speeds at which major wobbling begins. Of course, identification of the onset of "major wobbling" will be somewhat subjective, and we'll be assuming that metastable states don't last long. But that's probably the best we can do in the real world.
My formula is an attempt to fit math to experimental observations.
All but one of my tops go from major wobble to falling within a second. Only one top precesses at about 30 degrees lean for about 5 seconds.
I'm calling minimum RPM the last reading, within a second of falling. Sometimes I extrapolate this point by extending the last RPM reading by using the last decay reading. But I'm only doing this to intervals of about ten seconds.
A nice low minimum RPM might contribute to total spin time. Because at the end of spin, my better tops are loosing less than one RPM per second. My new 3" best is loosing about 1/4 RPM per second near the end. So reducing min RPM by 50 (by lowering Hcg) might add 200 seconds at the end-of-spin.
But moving the body of the top closer to the tip increases the aero drag shear on the bottom surface, which would cause end-of-spin to occur sooner. So, as I've written before, "There's no free lunch".
Alan
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Iacopo and Alan: What constitutes a "toppling" event WRT your data?
In my case I mean RPMs when the top topples down.
Wobbling appears and builds up very gradually, during the last 2 - 5 minutes of the spin.
About spinning surfaces curvatures: I have conflicting data so I am not sure how much the curvature of the spinning surface influences spin time. My best spin (57 minutes) has been obtained on a spinning surface with a very little radius of curvature (about mm 4.5). The tip is spiked, but the contact point of the tip is a flattened area with diameter about 0.1 mm. More precisely, not perfectly flat, but slightly rounded, like the section of the surface of a sphere, with a diameter of maybe mm 2, or something so.
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Hello Iocopo,
It appears to me that your tip shapes work better with curved mirrors than balls do. I would guess that there is little increase in contact area due to base curvature.
Alan
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How Much Oil is Best With my Heavy Top?
I've previously written that a very thin "haze" of oil has been best. I re-visited this with my newest 440 gram top, thinking perhaps this heavyweight liked more oil.
But the thin haze was still best.
My method is to measure decay at low speed, where tip drag dominates.
Regards,
Alan
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The ball bearing example supports what you are saying. Serious yo-yo and spin top players all clean their bearings and play them DRY. Ball bearings at high speed need oil to reduce the heat from friction. We play at speeds not high enough for friction to be a problem.
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Iacopo and Alan: Thought I'd share some of the things I just learned about the surfaces of concave mirrors. The goal is to be able to compare the radius of curvature of the mirror to that of the tip. The formulas apply to both spherical and parabolic mirrors, but only near their centers, where tops usually end up.
Consider a concave mirror with the parameters below and assume that the glass thickness is a small fraction of the mirror's radius of curvature (reasonable for a typical cosmetic mirror).
d == rim diameter, taken through optical axis
h == rim height at center
f == focal length
R == radius of curvature at center
s == distance from center to subject (e.g., eyebrows in a cosmetic mirror)
m == magnification of subject seen at distance s from mirror
Parameters d and h are easily measured, and the rest are easily calculated from them. Parameters h, f, R, and s are all distances from the mirror surface along the optical axis perpendicular to its center. Near the axis, we have
f = d2 / 16 h
m = s / (f - s)
R = 2 f = d2 / 8 h = 2 s (1 + 1 / m)
Most of the practical points one can draw from these equations are already familiar, at least qualitatively:
(i) The radius of curvature R is just twice the focal length f.
(ii) The more deeply dished the mirror (i.e., the greater the central depth h at fixed mirror diameter d), the shorter the focal length f, and the smaller the radius of curvature R.
(iii) The magnification m depends on both focal length f and subject distance s in nonlinear ways. Calling a mirror "5x" means that m = 5 at some unspecified subject distance s.
(iv) The greater the magnification m at fixed subject distance s, the shorter the focal length f, and the smaller the radius of curvature R.
(v) The image is upright when the magnification m is positive (s < f) and inverted when m is negative (s > f). Hence, the image flips at s ~ f. This fact can be used to estimate f and therefore R without having to do the math.
Now consider a top spinning at the center of the mirror. Assume a tip that approximates a spherical cap with radius of curvature r < R. Increasing magnification closes the gap between R and r. When R = r, we would expect full tip-mirror contact and maximum friction.
All of the cosmetic concave mirrors at my house have conspicuous dimples and rigdes when viewed from a distance. These irregularities represent significant departures from ideal geometry, but the mirrors are probably closer to spherical than parabolic, as the former are cheaper to make.
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A nice low minimum RPM might contribute to total spin time.
Couldn't agree more! Each of my tops seems to have a maximum attainable release speed for each spin-up method (hand, string, etc.). Hence, best spin times are strongly limited by the minimum speed wmin for stable sleeping or steady precession. And the strongest single influence on wmin is CM height, followed closely by axial moment of inertia.
This isn't just theoretical. Many of my tops have readily adjustable CM heights. The correlation between CM height and spin time is obvious.
(http://images.mocpages.com/user_images/99234/1469834963m_SPLASH.jpg)
But moving the body of the top closer to the tip increases the aero drag shear on the bottom surface, which would cause end-of-spin to occur sooner. So, as I've written before, "There's no free lunch".
I may have some good news here: As long as the ground clearance c beneath the rotor exceeds d, the boundary layer thickness over the lower rotor face, the supporting surface won't affect the drag on the rotor.
This effect is well known to engineers who worry about windage on rotors in tightly fitting shrouds. Taking advantage of it in top design calls for an estimate of d. For a top with laminar swirling (von Karman) flow over a flat lower face of radius R,
d = 5.4 sqrt(v / w) = 5.4 R / sqrt(Re),
where d and R are in meters, v ~ 1.5e-5 is the kinematic viscosity of room temperature air in m^2/s, w is the angular speed in rad/sec, and Re is the dimensionless rotational Reynolds number given by
Re == w R2 / v
This formula for d is valid for Re < 300,000. Counter-intuitively, the boundary layer thickens as spin-down proceeds.
For your 47-gram tungsten top spinning upright at 2,000 rpm,
R = 0.50 in = 0.0127 m
w = 209 rad/s
Re = 2,252 (well within the laminar regime)
d = 0.0014 m = 1.4 mm
Upshot: You can reduce CM height until c ~ d without incurring a drag penalty, but twirlability will suffer if c gets too small.
One measure of twirlability is the "grounding angle" a between the stem and the vertical when the top is leaning on its rotor:
a = tan-1(c / R)
The smaller the grounding angle, the less wiggle room the twirler has to avoid a rotor strike. For your 47 g tungsten, a is only 6° when c = d = 1.4 mm. That's pretty tight for a twirl by hand but easily doable with your string-launch system.
A simple spin-up aid like the one below really helps with tops with small grounding angles, as it lets one hand control the stem angle while the other applies the spin-up torque. This division of labor tends to benefit both spin time and the likelihood of getting a sleeper from the get-go.
https://www.youtube.com/watch?v=SL9KBh2z6uQ
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I tell you about a little experiment with oil, which I made months ago.
I made the top spin in a base with a lot of motor oil, (level of the oil about mm 3).
So the tip was fully submerged in the oil. Not only the tip, but also the support of the tip was in contact with the oil.
The diameter of the support, cylindrical, was about mm 10, (I modified the support and I don't remember the exact diameter).
How much drag from all this oil had the top ?
The weight of the top is 329 grams, radius of gyration mm 30.2.
In normal conditions, with only a very thin layer of oil, this top could spin for up to about 27 minutes.
With mm 3 of oil, the spin time has been only 9 minutes. Which is a dramatic reduction of the spin time !
So the level of the oil is important as for the spin time, because oil adds viscous friction at the sides of the tip, the larger the quantity of oil, the worse.
On the other hand, the oil between the contact points lowers their friction.
As Alan, I too had my best spin times with a very thin layer of oil.
Without any oil at all, my spin times generally result slightly shortened.
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In normal conditions, with only a very thin layer of oil, this top could spin for up to about 27 minutes.
With mm 3 of oil, the spin time has been only 9 minutes. Which is a dramatic reduction of the spin time !
So the level of the oil is important as for the spin time, because oil adds viscous friction at the sides of the tip, the larger the quantity of oil, the worse.
That's a quite an effect, Iacopo! Clearly, dissipative forces don't need large contact areas or long lever arms to generate significant braking torques on spinning tops.
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I've been experimenting with a method which doubles the length of string I can pull.
Instead of pulling on the string, I loop the string through a pulley and tie the end to an anchor (such as my left wrist). Then I pull on the pulley and each increment of motion of the pulley produces twice that movement on the string which is wrapped around the top. So I can use twice the string length.
So far I'm learning the art of this pull. But I think it has potential.
Alan
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That's a quite an effect, Iacopo! Clearly, dissipative forces don't need large contact areas or long lever arms to generate significant braking torques on spinning tops.
It surprised me when I saw it, because I didn't expect this strong effect.
Then, after some thinking, I considered that, after all, oil has a density and a viscosity that are really much higher than those of air. If air influences the spin time, let's figure out oil. About 60 - 100 mm^2 of surface in contact of oil at the center of the top, (low leverage), reduced the spin time to one third, from 27 to 9 minutes.
Using mm 1 of oil in the base, I guess it would still slow down this top, but much less dramatically, (since only the tip would be in contact with the oil), with spin times of maybe 26 minutes instead of 27. In the past I was used to use about 1/2 to 1 mm of oil in the base. After this test, I became acquainted to use less oil in the base, using often just a thin layer of it.
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I've been experimenting with a method which doubles the length of string I can pull.
Instead of pulling on the string, I loop the string through a pulley and tie the end to an anchor (such as my left wrist). Then I pull on the pulley and each increment of motion of the pulley produces twice that movement on the string which is wrapped around the top. So I can use twice the string length.
So far I'm learning the art of this pull. But I think it has potential.
Alan
I believe it was Jeffs who first posted trying this, but it was on the previous forum. I did find the photos I posted when I tried it myself (dated March 2008).
With a Mexican top that has a launcher (my left hand holds both the launcher and the string):
(http://www.ta0.com/Button-n-String/pulley_Mexican.jpg)
With a top that has dual bearing tips:
(http://www.ta0.com/Button-n-String/pulley_bulldog1.jpg)
(http://www.ta0.com/Button-n-String/pulley_bulldog2.jpg)
I believe I just held the tips between my fingers and pulled but on the photos it looks like I may have thrown it. I don't recall! :-\
Joe Mauk is also using this system on the current version of the world's largest top.
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To date, I've not beaten the starting RPM of conventional launches. But, as I wrote, I think it has potential. I may make a custom pulley which has deep narrow groove for string.
How did it work for you?
Alan
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How did it work for you?
I did not do any careful measurements or comparisons. It was just a proof of principle so if need arose in the future it could be in my bag of tricks.
The case presented itself with Joe's humongous top where the length of the pull is restricted by the length the ramp: it worked great to get efficient energy coupling.
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Just received this "optical glass" biconcave lens from Amazon (https://www.amazon.com/SEOH-Double-Concave-2000mm-Length/dp/B0088AQXHI/ref=sr_1_3?ie=UTF8&qid=1471825462&sr=8-3&keywords=seoh+concave+lens (https://www.amazon.com/SEOH-Double-Concave-2000mm-Length/dp/B0088AQXHI/ref=sr_1_3?ie=UTF8&qid=1471825462&sr=8-3&keywords=seoh+concave+lens)) to test as a spinning surface. It was $10.86 before shipping.
(http://images.mocpages.com/user_images/99234/1471825422m_SPLASH.jpg)
(http://images.mocpages.com/user_images/99234/1471825426m_SPLASH.jpg)
d == rim diameter = 0.075 m (given)
f == focal length = 2.00 m (given)
h == rim height at center = 0.00031 m (calculated, looks right)
R == lens radius of curvature at center = 2.20 m (calculated)
r == tip radius of curvature = 0.0015 m (measured)
The minimal surface curvature is enough to attract test tops to the center in still air, albeit very slowly. However, it's not enough to hold test tops on the lens in a gentle breeze. Hence, you'll need a deeper one for outdoor demonstrations.
With r << R, I don't expect much of a tip friction penalty due to lens curvature, but my ABS plastic tips see more tip friction with some kinds of glass than others. No idea yet where this glass stands in that regard.
This is the least curved biconcave lens I could find on Amazon at a reasonable price. I'll be trying lenses with greater curvatures over the next few weeks.
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Oops! I confess to fuzzy thinking. After several pulley launches which were about equal in RPM to conventional string launches, I analyzed the pulley launch.
Although the string lengthens twice as much as the motion of the pulley, the lengthening is divided equally between the portion wound around the top and the portion which is anchored. So it's equal to a conventional pull at the top.
Alan
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Mm, I think you were right the first time. Although the length is divided equally between the two sides, it still all has to come out of the winding of the top (if it remains stationary).
Perhaps you are pulling slower because of the extra resistance. The real goal is to have the correct gear ratio for your arm power (what an EE would call impedance matching).
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I wanted to measure the focal point of the Foreverspin "diamond" base by focusing the sun on a piece of paper (and then doubling the value to get the radius of curvature). Unfortunately, it has been raining almost continuously for more than a week. So I used a laser pointer instead. But with a small spot, the determination of best focusing is not very precise. The value I got was about 23 cm for the Foreverspin base, while a nominal 5X bathroom mirror gave me 30 cm. Doubling these values, the radios of curvature for the base and the mirror are respectively 46 cm and 60 cm: the base is more curved than the 5x mirror. As discussed above, this is probably not good for long spins.
I did check the results by imaging an object. I moved a pen between me and the mirrors until I saw an upside-down image of the pen with exactly the same width as the original (magnification equal to 1). This happens when the object is at the center of the mirror.
By the way, if you look at your own image and move the mirror, the image will invert close to the center of the mirror, not the focal point.
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Mm, I think you were right the first time. Although the length is divided equally between the two sides, it still all has to come out of the winding of the top (if it remains stationary).
Perhaps you are pulling slower because of the extra resistance. The real goal is to have the correct gear ratio for your arm power (what an EE would call impedance matching).
Your point is well taken. I'll noodle on this a bit more.
I tried another approach to lengthening my string. I tied it to a foot of PVC pipe, making my arm about a foot longer. My first two attempts (with a 210g top) were short of my prior best. But the third try was about 5% faster. That's only enough to add about 30 seconds to the run.
Alan
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Quicken Lubricant:
I bought some Quicken brand lubricant. They make great claims for this lube, which has "novel spherical NanoDiamonds" suspended in the lube.
I tried various amounts of it, spread thin on my mirror. Decay rates were identical to forehead oil or ordinary watch/clock oil.
Alan
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Quicken Lubricant:
I bought some Quicken brand lubricant.... I tried various amounts of it, spread thin on my mirror. Decay rates were identical to forehead oil or ordinary watch/clock oil.
It tickles me no end that fancy oils can't beat forehead oil, but I have to wonder: Could there be better oils from other body parts? And do some people have better oils than others? If so, I bet the oils found on Donald Trump are just FANTASTIC! ;^}
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Mm, I think you were right the first time. Although the length is divided equally between the two sides, it still all has to come out of the winding of the top (if it remains stationary).
Perhaps you are pulling slower because of the extra resistance. The real goal is to have the correct gear ratio for your arm power (what an EE would call impedance matching).
Try this line of thought:
The pulley moves one inch away from the top and the anchor. We know that the anchored strand got one inch longer, this only leaves one more inch for the strand wound around the top. So there is no multiplication.
Alan
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Try this line of thought:
The pulley moves one inch away from the top and the anchor. We know that the anchored strand got one inch longer, this only leaves one more inch for the strand wound around the top. So there is no multiplication.
ta0's right, and so were you the first time around. For every inch of outward pulley travel, the top (being the only source of string in the system) has to reel out 2 inches of line. Hence, double the top speed for the same hand translation.
Sailors call this block and tackle arrangement a "runner". If the load is attached to the block (pulley), the sailor has to pull or let out 2 m of rope for every meter that he moves the load. His mechanical advantage over the load is 2:1, but he has to process twice as much rope to get the job done. In our case, the sailor (corresponding to the top) is paying out line.