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Calling all mechanical engineers - lever problem ... :-))

120.149.146.35

Posted on October 4, 2014 at 21:59:02
andyr
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I am hoping someone can tell me what the equation is to calculate the weight acting on each of 3 springs when the springs are different distances away from a (central) weight.

For instance, if I had 3 springs placed at the apexes of an equilateral triangle, then a 9lb weight placed in the middle of the springs (say 10" away from each spring) would result in an effective weight of 3 lbs on each spring.
NB: lets assume the 9lb weight is resting on a stiff, 3-pointed star which has no weight in itself. The ends of the star arms rest on the springs.

Now, what happens when we have different distances from each spring to the weight - say:
* 7.5"
* 8.25" and
* 10"?

Any takers? :-))


Thanks,

Andy

 

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Too much math!, posted on October 5, 2014 at 06:16:27
Wooden it be easier to use a board instead of a three pointed star? That way you don't have to worry about where the springs are located since you can move the weight around by hand until the board is level on the springs. Or conversely, place the springs in a wide pattern, then position the weight by hand until the board is absolutely level.

 

RE: Calling all mechanical engineers - lever problem ... :-)), posted on October 5, 2014 at 08:23:36
mark111
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I don't think there is any difference in load no matter the distance.There is no leverage involved. It would just be a third of the load on each point. 30.33333 lbs.
enjoy,
mark
BTW, I'm a machinist not an engineer.

 

RE: Calling all mechanical engineers - lever problem ... :-)), posted on October 5, 2014 at 08:30:40
You might consider asking Mark Kelly over on Vinyl Asylum.

 

RE: Calling all mechanical engineers - lever problem ... :-)), posted on October 5, 2014 at 08:31:20
Please posit one question per post.

Get back in line to posit another.

The answer to your first question is "yes, maybe".

:)

 

RE: Calling all mechanical engineers - lever problem ... :-)), posted on October 5, 2014 at 08:48:27
Rockethead26
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OP did ask one single question. What question are you answering?

 

RE: Calling all mechanical engineers - lever problem ... :-)), posted on October 5, 2014 at 09:48:00
It's the same. Although, due to the different distances, and therefore stress, one will break before another.

:)

 

RE: Calling all mechanical engineers - lever problem ... :-)), posted on October 5, 2014 at 11:43:57
Todd Krieger
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Now, what happens when we have different distances from each spring to the weight - say:
* 7.5"
* 8.25" and
* 10"?


Since it's no longer an equilateral triangle, the positions of the vertices are no longer fixed..... Too many unknowns to provide an answer.

 

Thanks, genungo ... :-)), posted on October 5, 2014 at 12:29:48
andyr
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Yes, I wouldn't be surprised if Mark could supply the answer. But as he's my favourite "go to person", I am embarassed to ask him yet another question - so was hoping to get an answer here.


Regards,

Andy

 

RE: Thanks, genungo ... :-)), posted on October 5, 2014 at 12:38:02
Posit the question from the perspective of turntable platform design and Kelly might not deem the question inappropriate.

 

RE: Too much math!, posted on October 5, 2014 at 12:39:58
andyr
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Yes, I may well have to do that (use a board instead of a three-pointed star). I was hoping the maths would be easy! :-))


Thanks,

Andy

 

That is true! :-)) nt, posted on October 5, 2014 at 12:41:50
andyr
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.

 

"Too many unknowns to provide an answer" ..., posted on October 5, 2014 at 13:20:15
andyr
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And if I fix the angle of the triangle apexes, relative to the central point carrying the weight?


Regards,

Andy

 

RE: "Too many unknowns to provide an answer" ..., posted on October 5, 2014 at 13:32:15
Todd Krieger
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"And if I fix the angle of the triangle apexes, relative to the central point carrying the weight?"

Then it would be solvable..... Some geometric calculations would be required.

 

RE: Calling all mechanical engineers - lever problem ... :-)), posted on October 5, 2014 at 14:02:40
Andy, "realize" your scenario. Yes, it's cool to say "a stiff, 3-pointed star which has no weight in itself.", but that has no practical value in your question.

What is your situation?

Any decent engineer can break it down to the pertinent components if you provide real world circumstances.

 

RE: Calling all mechanical engineers - lever problem ... :-)), posted on October 5, 2014 at 14:16:11
Steve O
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You need to specify both the length of the "arms" and the angles between them. With that info you can then set up three simultaneous equations and solve for the three unknowns: the force at the end of each arm that will result in static equilibrium of your star. Although you mentioned an equilateral triangle, you didn't specify what type of star you had.

 

"Some geometric calculations would be required." ..., posted on October 5, 2014 at 15:45:33
andyr
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And if I provide the exact angles and lengths, are you able to solve the equations, Todd?


Regards,

Andy

 

RE: "Some geometric calculations would be required." ..., posted on October 5, 2014 at 16:58:00
Andy, you are wasting people's time.

What do you need to do? Get to the point.

 

RE: "Some geometric calculations would be required." ..., posted on October 5, 2014 at 18:47:06
Todd Krieger
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"And if I provide the exact angles and lengths, are you able to solve the equations, Todd?"

Either the three angles or the three sides would suffice.... Of course, with the distances you gave, you cannot just come up with arbitrary lengths.... (For example, one of the sides cannot be longer than the sum of the two longest distances you stated.) And if you give angles, you must also specify which lengths to to which angle. You'd have to do some calculating yourself to make the geometry work.

 

"but that has no practical value in your question" ..., posted on October 5, 2014 at 18:57:26
andyr
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I'm not an engineer ... but my first degree was in applied mathematics, Inmate51.

As such, my approach is to move from the simplest case ... to more complex cases. You asked what is my situation:

* the simplest case (of my situation) is ... I'm interested in the apportioning of the weight of a bearing & platter which is located somewhere inside a right-angled triangle, on 3 springs which are fixed at the apexes of said triangle.

The "3-pointed star" comes from there being 3 (imaginary) arms of the 'star' which connects the central bearing location to the 3 springs.

The desired result is to get the weight of the bearing & platter apportioned equally, so that the 3 springs are equally compressed.

Hence my statement to ignore the weight of the (imaginary) 'star'.

* the more complex case is then to account for the weights of 2 arms which are located just off-board of 2 of the springs.


Regards,

Andy

 

Right angled triangle, posted on October 5, 2014 at 19:57:18
Bill the K
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What are the dimensions of this triangle?

Call this triangle ABC.

Tell us the distances of the load from A, B and C.

3 equations will solve the problem.

Cheers
Bill

 

Ding, ding, ding. We have a winner., posted on October 5, 2014 at 22:07:11
DragonEars
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.
Use Matrices to solve a system of equations.
In this case a matrix for three equations (possibly differential equations).

Not too bad if you are only taking into account the three weights/distances/angles.
It will start to get complicated if you bring in spring rate change with compression, angle changes with weight shifts and other factors.

Should be a piece of cake for a first or second year math major.
I don't know how badly you need this but your local University can hook you up with a math or engineering major. A cash starved student may jump at a few bucks to set you up.

Sorry Andy, more than three decades since I've had to do serious math. I can feel a tickle buried in my brain that wants/needs to do this but there is just not enough left. Math not brain :-).

 

RE: Ding, ding, ding. We have a winner., posted on October 6, 2014 at 00:14:58
andyr
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Haha - thank you for your input, DE.

I should have the maths to hand ... but I guess 45 years of alcohol since those days has depleted the brain cells. (Not including, in Dr Strangelove terms, the loss of essential bodily fluids! :-)) )


Regards,

Andy

 

Thank you Bill ..., posted on October 6, 2014 at 00:22:38
andyr
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Here is a sketch of the setup:





I thought it easier to draw the 3 lines connecting the weight to the position of the 3 springs (than the triangle). The weight (8.44 lbs) is located at the central cross.


Regards,

Andy

 

OK, Todd - here ya go: :-)) ..., posted on October 6, 2014 at 00:25:36
andyr
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Here is a sketch of the setup:





I thought it easier to draw the 3 lines connecting the weight to the position of the 3 springs (than the triangle). The weight (8.44 lbs) is located at the central cross.


And thanks very much.

Andy

 

I have the answer, posted on October 6, 2014 at 06:23:31
Bill the K
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andyr

Let us call the support 7.5" away as A, 8" away as B and 10" away as C. The triangle ABC is right angled at B and has side AB 15.1", BC 8 and AC 16.9" in length.

From my calculations I find Reaction at A as 4.3 lbs, at B as 2.61 lbs and at C as 1.53 lbs with the total load being 8.44 lbs.

Had to refer back to old textbooks like Timoshenko's Plates and Shells and the two other books he wrote with Young and Goodyear. I had to do some derivation to get the values for the triangular shape. It all turned into a simple equation of the ratio of the distance of the load to the opposite side of the triangle to the distance of the support to the same side. Thus it is a matter of 3 equations. Interestingly the sum of these 3 ratios add up to one in any triangle.

Thanks for the opportunity to exercise the old grey cells with some structural engineering.

Regards
Bill

 

RE: Calling all mechanical engineers - lever problem ... :-)), posted on October 6, 2014 at 06:37:44
Not the same, it would seem.

 

No you don't have the answer - I do., posted on October 6, 2014 at 08:28:23
mgbpuff
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  Since:
October 4, 2009
A + B + C MUST = 9 By definition I got 4.62# for A(7.5"); 2.81# for B(8"); and 1.57# for C (10")

 

Actually, posted on October 6, 2014 at 08:37:12
mgbpuff
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The problem I solved is the one framed for Krieger below and I thought the weight was 9# but I see he changed that to 8.44 #. Therefore, if I ratio my solution from 9# to 8.44#, I get the same results you did. You're correct - my bad!

 

RE: Actually, posted on October 6, 2014 at 09:45:03
Bill the K
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Hey thats wonderful. The answers match which means they are correct. Great.

Cheers
Bill

 

RE: "but that has no practical value in your question" ..., posted on October 6, 2014 at 10:19:50
I sent your initial post to my son, who is a fifth year mechanical engineering student. Haven't got a solution yet - he's probably ciphering the problem. Or, with any luck, he's spending his time on classes.

:)

 

Brilliant, Bill - thank you, thank you! :-)) ..., posted on October 6, 2014 at 13:41:39
andyr
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I intuited that the answer lay in the concept you said (the ratio of the distance of the load to the opposite side of the triangle to the distance of the support to the same side) but I didn't know how to express it, mathematically.

OK, so:
* 4.30 @ A
* 2.61 @ B, and
* 1.53 @ C.

That was the simple case! :-)) For the real-life situation, we need to get more complicated ... and allow for the added affect of 2 arm weights near B and C!! :-))

See my modified sketch:





It seems to me that if the arms were placed at apex B (arm weight = 0.99 lb) and apex C (arm weight = 1.65 lb) then the weight distribution would simply become:
* 4.30 @ A
* 3.60 @ B, and
* 3.18 @ C.

But because the arms are cantilevered out from B & C, there is a negative action on the apex which is at the other end of the particular triangle side?

So:
* the 0.99 lb arm outboard from B reduces the weight at C, and
* the 1.65 lb arm outboard from C reduces the weight at A?

But how to calculate the resulting apex totals, with these cantilevered weights? :-))

FYI, here is a pic of the 1st version of my 'SkeletaLinn' subchassis - which I completed in July, last year. This has the bearing positioned closer to A, causing an excessive compression of the spring at that apex (so I had to get a custom-made spring made up, which was 30% stiffer than the Linn springs).





Hence the weight analysis for v2! :-))

Thanks, Bill,

Andy

 

Wasn't that "Precious Essence"s? nt, posted on October 6, 2014 at 16:31:41
John Marks
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jm

 

RE: Wasn't that "Precious Essence"s?, posted on October 6, 2014 at 16:46:16
Ah, you mean the Communist conspiracy to sap and impurify all of our precious bodily fluids!

 

RE: Wasn't that "Precious Essence"s? nt, posted on October 6, 2014 at 16:51:37
andyr
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maybe, John,

It's been a long time since I saw the film. :-))


Regards,

Andy

 

RE: OK, Todd - here ya go: :-)) ..., posted on October 6, 2014 at 17:19:24
Todd Krieger
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The three points you selected come very close to forming a "right triangle"..... Presume the force is still 9 lb., as stated above.

Above image has some added geometric calculations..... (Using side-angle-side, side-side-side, etc.) These calculations were to enable distribution ratio calculations. Using the opposite two points as a "fulcrum" for each ratio calculation.

The force on the upper LH point can be approximated as 9 lb x (8" / 15.5") ... [didn't calculate exact distances, the ratio if roughly the same] Comes out to 4.64 lb.

The force on the lower RH point can be approximated as 9 lb x (1.43" / 7.95") [treated like a right triangle] ... Comes out to 1.62 lb.

The force on the upper RH point can be approximated as 9 lb x (2.19" / 7.17") .... Comes out to 2.75 lb.

Now if you add the three numbers, you get 9.01 lb total distribution on the corners.... Close enough due to lost precision. (If you did precise calculations, it would add up to exactly 9.00... )

 

Thanks for the diagram! :-)) nt, posted on October 6, 2014 at 18:09:07
andyr
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.

 

Yikes. Must be the Director's Cut Australian Hardcore edition :-) nt, posted on October 6, 2014 at 20:04:25
DragonEars
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.

 

Here's an updated diagram, Todd (slightly more complex!) :-)) ..., posted on October 7, 2014 at 01:44:56
andyr
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OK, 9 lb weight for the bearing/platters was an initial estimation; if you take a look at my responses to Bill K, you will see that I have put the weight at 8.44 lbs.

So this makes the weights you calculated:

* upper LH point (A): 4.35 lbs (instead of 4.64 lb)
* lower RH point (C): 1.52 lbs (instead of 1.62 lb)
* upper RH point (B): 2.58 lbs (instead of 2.75 lb)

... which agrees very well with Bill's calculations. Which were:
* (A): 4.30 lbs
* (C): 1.53 lbs
* (B): 2.61 lbs

So thank you, Todd. :-))

Now, here is a slightly more complex scenario - which is the actual situation ... include the weights of the arms which are cantilevered out from the 2 RHS apexes of the triangle.

Here is your diagram - suitably modified:





I would be interested in what your opinion is of the resulting weight balance. As I see it, because the arms are cantilevered out from B & C, there is a negative action on the apex which is at the other end of that particular triangle side?

So:
* the 0.99 lb arm outboard from B reduces the weight at C, and
* the 1.65 lb arm outboard from C reduces the weight at A?

But how to calculate the resulting apex total weights, with these cantilevered weights? :-))

Is it simply according to the relative distances? So:
* at C, weight is reduced by the fraction 2.0/7.95 * 0.99 lb = 0.25 lb
* at A, weight is reduced by the fraction 1.6/16.9 * 1.65 lb = 0.16 lb.

But should this decrease in weight be added to the weight at the fulcrum apex ... so that the total weight increase along each triangle side, sums to the weight of each cantilevered arm?


Regards,

Andy

 

RE: Calling all mechanical engineers - lever problem ... :-)), posted on October 9, 2014 at 00:48:05
stehno
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Finding the correct formula is moot.

You have 3 springs in the first place because the masses are convinced they can defy the laws of physics and isolate an object from all sources of vibration simultaneously. If you're only trying to isolate against one source of vibration, which is it? And if you succeed isolating against that source, how are you going to deal with the other sources of vibration. It also helps to know which source induces the most harm.

In other words, considering virtually everything in this universe is one form of vibration or another, the attempt to isolate anything is impossible to achieve, though many die trying. Even in a vacuum chamber the object must be levitating to achieve complete isolation. Short of that, once resonant energy enters into or makes contact with the object, it matters not whence it came. Because now the unwanted energy must escape before it induces its catastrophic harm.

About the only thing the springs are good for at this juncture is to ensure all vibrations already captured within remain trapped so that they cannot escape but must dissipate within. Oh, one more thing. The springs will also ensure that they and the platter they sustain will vibrate in sympathy with the already captured vibrations, hence taking an already poorly engineered situation and making it worse.

Don't believe me? Of course not. Try removing the springs for a few days and replace temporarily with some hardwood or better yet metal blocks and then place your massloading weight on it. If you do, I can all but guarantee you'll never go back to the springs.

Actually, the springs do serve one other purpose, the very purpose for which they were probably introduced. If your 400 lbs. spouse likes to 2-step on your suspended flooring system right in front of your turntable, the springs may prevent the stylus from jumping 14 grooves and instead jump only 5 grooves.

Anyway, good luck with the leveling formula.

 

RE: Calling all mechanical engineers - lever problem ... :-)), posted on October 9, 2014 at 03:12:21
In a vacuum chamber there is still the Earth's crust vibration to consider, that affects the object in the vacuum, whether it be optics or whatever. That is why isolation is necessary in a vacuum and also necessary outside a vacuum - to reduce seismic vibration. Your example of footfall affecting the stereo cartridge stylus demonstrates my point quite well, actually. No one is saying acoustic vibration should NOT also be addressed. Or the internal induced vibration, unless you believe motor noise and transformer noise are good things. If you are so enamored of vibration I suggest you consider putting the whole thing up on a shaker table.

 

RE: Calling all mechanical engineers - lever problem ... :-)), posted on October 9, 2014 at 19:06:26
stehno
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Geoff, as you've established so clearly to me at linkedin forums, IMO, you're little more than a fabulous speculator always pretending to speak from firsthand experience and knowledge that seems to require only a brief time between posts to realize your speculations this month contradict and void last month's statements of fact.

Just wish I would have recognized it sooner so I wouldn't have wasted so much time engaging in meaningful monologue.

BTW, contrary to popular folklore, the footfall as you call it from my 400 lbs. spouse 2-stepping example is referring to shock and impact which has nothing directly to do with proper vibration control. Though shock and impact will eventually result in subsequent vibrations, but leave it to you to speculate that the ensuing vibrations and not the shock and impact caused the needle to jump 14 grooves.

Your example of a shaker table is nearly as preposterous and gives clear evidence to me that as usual you lack discernment and your logic is unreasonable. Hence, your probable use of a sledgehammer and a Sherman Williams paint can shaker when performing all your "high-end" audio vibration controlling R&D as you seek to convert all that unwanted violent energy to harmless heat.

 

I see you are a mfr ..., posted on October 9, 2014 at 19:33:53
andyr
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of rack systems.

That might explain why you have displayed little knowledge of TTs? :-))

Linn (and others) use 3 springs not because they are convinced they can "defy the laws of physics and isolate an object from all sources of vibration simultaneously " but because they feel a sprung TT does not suffer from problems which solid-plinth TTs do.

Those mfrs who make non-sprung TTs rely on some kind of isolation devices to make their TT work well (eg, Vibraplanes, Minus Ks etc.) - not so much against footfalls caused by 400 lb spouses boogie-ing right in front of the turntable but from terrestrial vibrations, such as trains which are adjacent to the building structure. I would suspect that doing as you suggest (replace the springs with some hardwood or better yet metal blocks and then place your massloading weight on it) would produce such problems - however, it is easy enough for me to try this experiment. :-))

Linn (and the other suspended TTs) use springs - or in SME's case, rubber O-rings - to prevent these vibrations from getting to the stylus. AIUI, they only work down to (in Linn's case) about 5Hz - so there still may be a need for some other isolation device down to 1 or 2 Hz - depending on your surroundings.

Re. your comment "Finding the correct formula is moot" ... I find that very strange. :-)) It seems to me the springs will bounce better if the total weight of bearing/platter and 2 arms is shared equally across the 3 springs - rather than, say, the weight being 50% on one, 30% on another and 20% on the 3rd.

Thanks to the help I received here from Bill and Todd, I have been able to slightly adjust the position of the bearing which now - according to my calculations - gives me almost exactly equal loading, when the weight & position of the arms are also taken into account. I will post on this, soon.


Regards,

Andy

 

RE: I see you are a mfr ..., posted on October 10, 2014 at 00:41:01
stehno
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Hi, Andy. Yes, thank goodness I have very little experience with TT's. Knowing what I think I know about vibrations, their sources, and their behaviors, and most importantly how crippling they are to every last component's precision and accuracy, I wouldn't touch one with a 10 ft. pole. Now I do have a 35 - 40 year old Technics SL-1200 semi-auto buried in a closet that I haven't even viewed in perhaps 15 years just to prove that point.

When I think what gymnastics TT designers have gone through to compound isolation strategy upon isolation strategy upon isolation strategy, I can't help but think what a frickin' nightmare some of them have created. And most importantly how much they've crippled that source's musical potential.

That's why finding the right formula for your dilemma is moot to me.

A local friend who seems to swap out his equipment whenever the season changes had purchased some vintage turntable for which he sent out for a $5k makeover that he seemed proud of at the time and wanted me to listen.

As usual his system sounded like absolute crap even though it's always music to his ears. As he was showing off his custom turntable, I noticed how badly his speaker drivers were warbling, quite severe actually. He expressed his concern too and he only listens at maybe 86 - 88db tops.

Not knowing nor wanting to know anything about turntables but after studying his for a few minutes, I asked him for some paper towels. I created about 8 small wads of paper towels rolled up nicely and pinched them between his free-floating spring-loaded base and the outer chassis. The severity of the drivers warbling was now only about half as intense. Wadded up a bunch more and now he's wadding up a bunch and eventually made the spring-loaded free-floating base quite snug against the chassis and the warble ceased while the sound improved quite a bit.

It's been about a year and I've not inquired but if he still has the TT, I'll bet dollars to donuts that he's still got all those paper towel wads all around the perimeter.

Yes, TT's are a frickin' box of chocolates and the more exotic they are, usually the worse they are engineered from a proper vibration mgmt perspective.

Don't get me wrong, from an aesthetics and intimate engagement perspective I love TT's perhaps more than the next guy and they make some gorgeous ones. But performance-wise they simply cannot hold a candle to what is possible working with non-isolation-based digital sources.

When I exhibited at RMAF 2011, I spent an evening at a very high-end distributor's home listening on what I guess was about a $250k system (add at least another $200k for the gawdy room). He put on Boston on his highly rated TT. Then it was my turn and I pulled out my very best manners and said, "That sounds really nice." and he smiled and nodded with great pride. What I should have said was "Amazing." and left it at that. You know, as in, "Amazing how you could dedicate so much resources and time to a system and room and this is the best you can do?" and as in, "Amazing you are still in business with such untrained ears.", and as in, "Amazing that your customers must all have hearing abilities not too unlike your own."

During the show, because at that time he distributed my amplifier, that distributor was routinely sending his visitors and dealers up to my room, telling his customers, "if you really want to hear what this amp is capable of, go up to Dynamic Contrasts room." I recently downgraded from that joker's $8k amp to a $2k amp and because of a new methodology I employ, the new amp is perhaps 50% more musical than that $8 amp ever was. I think it also worth mentioning that his $6k power cable that he insisted I use was no better than my fully-cryo'ed DIY $100 custom power cable.

The funny thing is that my exhibiting system only retailed for maybe $40k and most important was that at a show my racking system operates at maybe 5% of its full potential (since it takes months to reach its full potential). Truth be told, since my rack does next to nothing at a show, what visitors were hearing mostly were my fabulous Foundation Research line conditioners which by themselves will put my system on an even keel with perhaps the very best of any other system provided they are not employing similar superior line conditioners.

I've certainly had other experiences listening to some fine TT's.

Think of it this way, a given component is really no different than any other sensitive instrument, e.g. an atomic force microscope, and it takes very little in the way of undercontrolled vibrations to absolutely saturate and hence cripple a component's precision and accuracy. That said, now consider the layer upon layer of vibration methodologies employed on any given TT including air bladders and air pumps, etc. Talk about battling the laws of physics? Without stripping the TT and completely reconstructing it by removing all the garbage, there simply is no hope for any major sonic improvements. It's impossible.

But yet we're all aware of experiencing a given TT's minor improvements. How can that be if what I say is true? I can only speculate that when a negative is compounded by another negative there is a possibility of achieving a little positive, e.g. -1 x -1 = -0.5. When it comes to high-end audio I never ever use the term isolation in a positive way because 1) it defies laws of physics, and 2) because of #1, it cannot be considered a valid methodology, but only an extremely gross and inferior version of the one true methodology, resonance energy transfer.

Think about it, the air-borne and internally-generated (motors, power supplies, etc) vibrational energy will always be there and is already trapped inside and if given a few milliseconds to bottleneck it's going to release that energy right then and there post haste. If it can't find an expedient exit path, it will attach itself to objects within and release its energy there. Only now matters are compounded because now you have innards vibrating while at the same time vibrating in sympathy with the vibrations captured.

Hence, if those most exciteable and sensitive internal objects are further isolated within, whoala, a little improvement and the designer pats himself on the back thinking he's done good. And if somebody's pride and joy toward sonic nirvana is a TT, then it's pretty much a given they ain't doing much of anything different at the other components to help control vibrations there either. I mean, if Walker uses air bladders, then so should I. So in the end, having springs in a TT and objects freefloating really isn't that catastrophic after all since the rest of the system is also severely compromised so the music's pretty much guaranteed to be pretty flat and lifeless anyway.

Of course, then there's the problem of friction of the stylus at the groove and there's no getting around that frictional energy.

But to come full circle, perhaps the most common cry of the vinyl lover is, "I hosted a party last night and some 400 lbs. woman dancing by the TT caused my stylus to jump 14 grooves. Please help me so this never happens again." Enter sandboxes, kitty litter, bungie cords, springs, tennis balls, Sorbathane, innertubes, cork, rubber, etc. All the beautiful "high-end" materials that instantaneously trap all crippling vibrations already captured at the TT.

Other than that, I love TT's.

 

Don't hold back on my account, posted on October 10, 2014 at 05:12:32
You might give some serious consideration to spending time in the library boning up on the pros and cons of mass on spring isoaltion systems. You might also bone up on seismic vibration and microseismic vibration while you're at it. It wasn't that long ago most of my customers had Rockport turntables, the one that's 60 grand, then dollars. Those are the ones that come with dedicated airspring isolation systems. My own nimbus is almost 20 years old and still the best performing airspring platform, by far.

Cheers, Geoff

 

RE: I see you are a mfr ..., posted on October 16, 2014 at 19:46:45
andyr
Manufacturer

Posts: 12548
Location: Melbourne
Joined: September 2, 2000
Hi Stehno,

Great post - thank you. :-)) However, it spurred me to comment on a few of the points you raised.

Firstly, I perceive the 'tone' of your discourse as similar to the arguments I see from some "engineering fundamentalists" who, because even good tube amps exhibit much higher levels of distortion than typical ss amps ... say they cannot possibly be as good to listen to! :-)) Whereas those of us with a wider-open mind understand that sound reproduction through tubes offers 'pluses' as well as 'minuses'!

Re. your comment that no TT can (performance-wise) compete with what is possible with non-isolation-based digital sources ... while I can see where you're coming from in terms of vibrations damaging the signal retrieval in a record-playing situation, you seem to have entirely forgotten the electrical aspect. IE. that (in the case of a CD, anyway) a brick wall filter introduced at 20KHz influences the sound lower down in the frequency range ... compared to a playback system which doesn't have this constraint.

Sure, if a digital system involves:
* a much higher sampling rate (say, 192K), and
* comes from memory, not from a computer disk

... then I would agree with you that it would sound much better than any TT. :-))

And re. "When I think what gymnastics TT designers have gone through to compound isolation strategy upon isolation strategy upon isolation strategy, I can't help but think what a frickin' nightmare some of them have created. And most importantly how much they've crippled that source's musical potential."

I bought an LP12 in about 1979 and have used it until mid last year, when I changed to v1 of my 'SkeletaLinn' (the current thread is because I am now building the chassis for v2). So I know a fair bit about LP12s - but I don't know much about how other mfrs of suspended TTs have gone about their suspensions. I also have friends with non-suspended TTs ... and have seen what they have needed to do, to to make them work well. (As you said "the layer upon layer of vibration methodologies employed on any given TT ...".

AIUI, the 3-point sprung suspension (AR, Thorens, Ariston, Linn - and preceeding all of these, an Australian TT called the 'Aurora') provides pretty good isolation down to about 5Hz ... so it can't handle terrestrial vibrations (like from a nearby railway track) but does a pretty good job of the rest. Providing, of course, the TT is on a rack or table which rests on a concrete slab; or you sit it on a wall-shelf which is attached to a masonry wall. If what you have is a springy wooden floor - then IMO, a TT is not for you ... particularly if you have a 400lb spouse who likes to boogie.

I like how my LP12 sounds and my current situation is merely the result of deciding I had to use a 12" arm that I fell in love with. :-)) My request for mechanical engineering assistance was simply to see if I could find out whether you could use calculations to find out where the cog would be, rather than finding this out heuristically.

And re. "TT's are a frickin' box of chocolates and the more exotic they are, usually the worse they are engineered from a proper vibration mgmt perspective." ... I'm not sure if you've come across the Aussie "Continuum" TT but vibration control was an essential part of the design - hence, I guess, the $160K price tag!

And re. "internally-generated (motors, power supplies, etc) vibrational energy will always be there and is already trapped inside and if given a few milliseconds to bottleneck it's going to release that energy right then and there post haste. If it can't find an expedient exit path, it will attach itself to objects within and release its energy there."

I agree - hence belt-driven TTs have a vibrational advantage, compared to DD TTs! But of course, some negatives come with belt drive. :-))

In an LP12, the motor is attached to the plinth ... but nominally, there is a pathway for vibrations to dissipate. In mine, I have chosen to make the motor external - so the only thing connecting it to the platter is the belt.

Then you said "Of course, then there's the problem of friction of the stylus at the groove and there's no getting around that frictional energy.". Absolutely! And it never ceases to amaze me how a vinyl playing system actually works at all! To say nothing of the LP manufacturing process!

I certainly do love TT's - 90% of my "serious listening" is done with my TT. But when a 192K system playing from memory appears on the market for, say, $8K ... I will probably switch over. :-))


Regards,

Andy

 

Current status ..., posted on October 17, 2014 at 23:45:38
andyr
Manufacturer

Posts: 12548
Location: Melbourne
Joined: September 2, 2000

Hi Todd,

Using your modified diagram that I included in my last post - and the lever calculations I suggested (to take the arm weights into account) - I adjusted the position of the centre of the bearing slightly.

Then I did the final work on the chassis and assembled it today, and mounted the bearing & platters, and used lead shot to mimic the weights of my 2 arms. See here:





Using the diagram you most kindly provided and adjusting for the arm weights, my calculations showed that I had got the weights almost equally divided between the 3 springs - the weight on the one nearest the platter was 4.8oz heavier than the other springs (8%). So not exactly equal - but within tolerance, IMO! :-))

Today's actual measurements backed this up; here's a side-on shot of the position of the chassis:





I measured the distance from the ply base to the underside of the chassis to be:
* 26mm at the LH spring
* 27mm at the top R spring, and
* 27mm at the bottom R spring

... which indicates that the weights are almost equal! :-))

So I will send the chassis off for powder-coating on Monday ... thank you very much for your help. :-))


Regards,

Andy

 

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