Arm resonance

NeverSatisfied

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#82
I never saw October sky,

But I did design some nozzles.
Jim, you ever do any military stuff? My dad designed the power supply for the guidance system in the Subroc missile. He told me stories of nozzle test failures he witnessed and some test equipment he designed being obliterated when a solid propellant rocket blew up the entire test cell and almost killed him.
 
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J!m

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#84
I did not design rocket nozzles for rockets. I designed them for “cold spray” guns.

Basically gas is accelerated and powdered (ductile) metal is injected into the stream and a deposit is formed by (predominately) kinetic energy.

It’s a little strange because you have to heat the gas (quite a bit) and it cools significantly after it expands in the nozzle. It’s not an efficient process at all but some coatings approach the mechanical properties of wrought or even forgings in some cases.

Typically done in atmosphere but I’m the only person (as far as I know) that has ever done cold spray in a vacuum chamber. That was in Switzerland over ten years ago. A potential advantage is the capture and reuse of the gas. Helium is the best but very expensive (and you use a lot). The chamber exhaust could be collected and filtered and then the helium recompressed for reuse.
 

J!m

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#85
I Grenaded the gun in the chamber I just remembered…(that was the end of that) the heat from the heated gas conducts through the gun body and coupled with the pressure cycles it fatigued to failure. I think my former boss still has it on the shelf in his office…

We hydrostatic tested to 2000psi. It was all torn open. Fun stuff.
 

J!m

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#86
The second design was trickier.

The original embodiment (KES-1) used radial injection of the powder just past the throat. The second embodiment (KES-2) is the one I invested serious time on. That used axial injection. The goal was to inject at the “sweet spot” where pressure is lowest. The area calculations at the throat get really tricky, so we got it close, and then I made a linear slide with a micrometer (sort of) scale to define the exac location of the injector where the pressure in the powder feed line was lowest.

With low (110psi) pressure, I was able to pull some vacuum there. I set it at lowest pressure and designed the rest around it.

It was based on an HVOF (high velocity oxy-fuel) gun that has an extended “nozzle” which was water cooled. What I did to try and increase efficiency was to take the “cold” gas and run it through the front in place of the water, and then through the big stainless tubing coil I had hooked to a welding power supply.

All I had to was regulate the pressure within a narrow window while simultaneously keeping the gas temperature in a narrow window. Three hands would have helped but after I melted a few coils down I got the hang of it.

It worked about as good as a regular gun did and needed probably two more generations of evolution to reach our goals. I don’t believe they ever happened after I left to do cylinder bore coatings for race engines in another part of the building.
 

AngrySailor

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#89
Made it yourself? I like! And...no, you are not measuring force with it, but comparing masses. It will show the exact same result on the Moon, whereby VTF will be only about 1/6 compared to the Earth, due to Moon's lower gravity. Measuring forces ain't an easy task.

My point is directed towards people claiming decimals at one end and speculating at the other. I don't mind, if somebody is anal about it, in fact I do appreciate it - just be anal the whole way! (buoyancy is only my over-the-top example):

1) VTF...if chasing the second decimal point, you must use some sort of a force gauge, not a weighing scale, or at least take into account the "agreed-upon average" gravitational acceleration on the face of the Earth is 9,81m/s²...the actual you don't know.
Hmmm, WTF seems the more appropriate term.
2) VTA
- do you know the cutting angle of your record(s)? If so, you surely set it for each different one? And why don't you?!
- record thickness...again, individual settings
- thicker/thinner mat...
- VTF/WTF...
4) azimuth...isn't there an expression about mirrors?...and smoke
5) anti-skating...eeehm...what?...where?...when?...why?
6) overhang, tracking angle...
7) matching an arm with a cartridge to get the desired resonant frequency...why is this so difficult?!:
- you know the stylus' compliance
- you know the arm's effective mass
- you know the masses of cartridge, shell, lifter, screws, shims and their effective masses
- you know the effective lenght
- now fill this data (don't forget the compliance conversion if stated at 100Hz) into one of the equations floating around, let it simmer for fifty years...et voilà!...an exact hit-and-miss ballpark aproximation of a checkee in a baskee.
8) and what not...but to start the anal-journey on a solid ground...first of all: Go linear; a radial arm is flawed by its nature!

Let common sense prevail, pretty please with sugar on top:

- weighin scale is fine...no need for a really expensive one
- set VTF according to manufacterer
- get a test record, if you fancy some setting-tweeking, or wanting to verify the stylus and/or cantilever are mounted within specs
- keep the records and stylus clean!
- if you just want to taste vinyl for the first time...get a new plug-and-play (cartridge mounted and pre-set) turntable.

Lean back and enjoy the music!!!...scrapped out of vinyl grooves with a stone - :eek: - I'll wear my precious records out by playing them...yada yada.

May the Hearing be with you!
Borut

Edit: BB, I see now you didn't make it yourself. Still like it.
Gravity will not effect VTF as it effects both side of the tone arm equally... look at it at torque/moments, g cancels out both sides of the equation so as long as g does not equal zero the VTF will be the same...
 

borchee

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#90
Hi @AngrySailor,

sure it will... force equals mass times acceleration. Since the gravitational accelaeration of/on the Moon is about six times smaller than that of/on Earth, the gravitational forces/weight/VTF are equally smaller. A kilogram weighs ~9.81N on the face of the Earth, but only ~1.6N on the face of the Moon. The balance-ratio will remain the same, though.

Hmm...my english may not be up to the task of an clear explanation, but this is physics 101.
 
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AngrySailor

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#91
Hi @AngrySailor,

sure it will... force equals mass times acceleration. Since the gravitational accelaeration of/on the Moon is about six times smaller than that of/on Earth, the gravitational forces/weight/VTF are equally smaller. A kilogram weighs ~9.81N on the face of the Earth, but only ~1.6N on the face of the Moon. The balance-ratio will remain the same, though.

Hmm...my english may not be up to the task of an clear explanation, but this is physics 101.
Hello fren, I think you must take into account that we are looking at a balance system, equilibrium. The counter weight negates the tone arm mass minus the desired VFT. As it’s a balance/equilibrium system, it’s strictly a mater of two forces reacting about a point. Those will be constant as gravity (g) acts equally on all...
 

vince666

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#92
yes, of course gravity will act the same on both the cart and the counterweight.
so, if the cart will appear lighter on the moon then also the counterweight will appear just as lighter.
 
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borchee

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#93
Ciao Enzo,

we both might be a victim of a language-barrier...the cartridge/counterweight/arm doesn't appear lighter, it just is!
 

Makymak

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#94
My trying to enlightening...

The system "cart - tonearm - pivot point - counterweight" isn't at equilibrium. It balances due to the needle touching the record, transferring the force that is on excess, to the vinyl. So, if you remove the vinyl, our system will continue to tilt until the needle (or any other part) touches somewhere else. This excess force is made by the difference in masses plus the distance each center of mass has from the pivot. Let's say that we have a cart setup at 1 gram. That means that the gravity of the earth pulls this "excess" of 1 gram of mass with a force of (0.001 x 9.81)N = 0.00981 Newton. If we take exactly the same setup at the moon, since no difference in masses or their center occurs, we will still have the same 1 gram difference (remember, the gram is a mass unit, not force), what will change now, is the force that the moon pulls that "excess" mass. And since the gravity on the surface of the moon is about the 1/6 of the gravity on the surface of the earth, then the force that the needle will put on the vinyl will be the 1/6 of the 0.00981 N.

I can't either explain it more simple.
 
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borchee

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#95
My trying to enlightening...
Yep!

...but to bring this across it needs even a simpler relate-to explanation/example, and I can't think/express of one (perhaps in Slovenian I can, although my English might suffice in a personal conversation) - am, obviously, a bad teacher...sorry, my fault! Still waiting to be proven wrong. :p
 

Makymak

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#96
The key phrase here is that the cart/tonearm/counterweight isn't balanced/equilibrium since an amount of force is put on the vinyl. This force is made out of mass. The mass produces force because the earth pulls it with a specific value (g). The moon will pull this exact mass with a specific value (1/6 of g). So, the force the needle will put on the vinyl on the moon, will be 1/6 of that on the earth.
 

Makymak

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#97
I found one simpler example!

Put on a double scale two weights: on one side a 6 grams and on the opposite side a 5 grams. Do we agree that the difference is 1 gram? If we glue a needle under the side of which we put the 6 grams, will this needle put on the vinyl this 1 gram difference?

Now, we put the 1/6 of each weight so now each side of the scale will have 1 gram and 0.83 gram respectively. The difference is just 0.17 gram and this is the weight our needle will put on the vinyl.

We have 1 gram x 1/6 (the difference of the gravity between earth and moon) = 0.17

Exactly the amount of our example!
 

AngrySailor

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#99
My trying to enlightening...

The system "cart - tonearm - pivot point - counterweight" isn't at equilibrium. It balances due to the needle touching the record, transferring the force that is on excess, to the vinyl. So, if you remove the vinyl, our system will continue to tilt until the needle (or any other part) touches somewhere else. This excess force is made by the difference in masses plus the distance each center of mass has from the pivot. Let's say that we have a cart setup at 1 gram. That means that the gravity of the earth pulls this "excess" of 1 gram of mass with a force of (0.001 x 9.81)N = 0.00981 Newton. If we take exactly the same setup at the moon, since no difference in masses or their center occurs, we will still have the same 1 gram difference (remember, the gram is a mass unit, not force), what will change now, is the force that the moon pulls that "excess" mass. And since the gravity on the surface of the moon is about the 1/6 of the gravity on the surface of the earth, then the force that the needle will put on the vinyl will be the 1/6 of the 0.00981 N.

I can't either explain it more simple.
The moons gravity also does not act on the counter weight either therefore the system is still in equilibrium + desired VTF.
 
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