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Author
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Topic: Return compensator - how does it work?
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Hans Holzherr
Member From: Ostermundigen, Switzerland
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posted 16 May 2006 09:45 AM
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I know WHAT a return compensator is supposed to, and I know how to install one (thanks to Carl Dixon) - but HOW is it supposed to do what it does? Doesn't it merely change the idle (rest) position of the lower changer finger? Doesn't tuning the compensator cause the C pedal F# on the 4th string to be flat? Doesn't it cause the 4th string E itself to be flat after pedal C has been engaged and released?Hans |
Leon Roberts
Member From: Tallahassee,FL USA
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posted 16 May 2006 02:29 PM
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Hans, Once the return compensator has been adjusted, the raise and lowers must be tuned again. When I had my ZumSteel, once the return compensators were adjusted, they never required tweaking. Even if I changed strings the returns were always right on the money. As for why a return compensator is needed on some guitars, I haven't got a clue. It is my opinion that the older Sho-Buds didn't need one because the tuning head allowed a straighter pull past the nut roller allowing the string to return to proper pitch. The Newer Sho-Buds didn't allow this. At least that's been my experience. It goes without saying the the nut rollers on any guitar must be lubed and free to turn. That's about all I know about the subject of return compensators.Leon |
Bobby Lee
Sysop From: Cloverdale, North California, USA
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posted 17 May 2006 03:59 PM
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Bump.
I've never had them. A mechanical description of how they work would be great. Can anyone here describe how to add return compensators to a regular push-pull guitar? |
Brint Hannay
Member From: Maryland, USA
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posted 17 May 2006 08:13 PM
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Bump.
Seems a lot of us are curious about this. I hope someone can enlighten us! |
Jerry Roller
Member From: Van Buren, Arkansas USA
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posted 17 May 2006 09:13 PM
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I will try a rather brief explanation on a return compensator. It is a rod from a lowering finger connected to a fixed point. This rod has a rubber "O" ring between the finger and the tuning hex nut. This "O" ring serves as a cushion. The normal tendency is for a string which is raised to return slightly flat and a string which is lowered to return slightly sharp.
When properly adjusted the "O" ring expands when the string is lowered because the lowering finger is pulled away from the "O" ring allowing it to expand so when the lowering finger returns against the "O" ring it is cushioned causing a slight lowering action. When the raise finger is pulled then released it compresses the "O" ring back to its compressed state so the slight lowering is cancelled out and the raise returns correctly. This is my basic understanding of it.
Jerry |
Jim Sliff
Member From: Hermosa Beach California, USA
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posted 17 May 2006 11:54 PM
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"The normal tendency is for a string which is raised to return slightly flat and a string which is lowered to return slightly sharp."I have to ask - "why"? Honestly, that "tendancy" makes no sense from a mechanical or physics standpoint. |
Hans Holzherr
Member From: Ostermundigen, Switzerland
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posted 18 May 2006 01:45 AM
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Jerry, the question remains, why, on the lowering-release action, the lower return spring is not strong enough to compress the ring, but on the raise-release action, it is?Jim, this has been discussed elsewhere. Due to friction on the roller, the tension in a stretched/lowered and released string will not be immediately and absolutely equal in both string parts left and right of the roller. As an alternative explanation, the state of the molecular grid is not totally preserved, and/or needs some time to reach its initial state. Hans |
Curt Langston
Member From: ***In the shadows of Tulsa at Bixby, USA***
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posted 18 May 2006 05:32 AM
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quote:
"The normal tendency is for a string which is raised to return slightly flat and a string which is lowered to return slightly sharp."
I have to ask - "why"? Honestly, that "tendancy" makes no sense from a mechanical or physics standpoint.
Jim, quite simple. The return compensator has to "compensate" for the nut not allowing the string to return to pitch. This is a hallmark problem with keyed guitars. The longer the keyhead is, the more prone the guitar will be to hysteresis or not returning to pitch. IMHO, that alone is enough for me to avoid a keyed guitar. |
b0b
Sysop From: Cloverdale, California, USA
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posted 18 May 2006 07:36 AM
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quote:
I have to ask - "why"? Honestly, that "tendancy" makes no sense from a mechanical or physics standpoint.
"Why" is a different topic. People have theories. Let's keep this topic on track, so that it doesn't dissolve into a debate about those theories.
The stated topic is "Return compensator - how does it work?". Please start another topic if you need to ask "why" return compensators are needed.------------------
 Bobby Lee
-b0b- quasar@b0b.com
System Administrator
My Blog |
Charlie McDonald
Member From: Lubbock, Texas, USA
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posted 19 May 2006 08:25 AM
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Well, 'why' might work if I can be indulged.
Remember, I could be wrong.You have two string segments, short and long, and they're at equal tension at rest.
Pull the string up in pitch; let it drop, and it will take a short period of time for the tension in the two segments to re-equalize.
Only in a perfect system, where there is zero friction at the nut, would the two segments re-equalize tension instantaneously. |
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