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Nov 30, 2020 - 2:28:21 PM
1572 posts since 4/13/2017

I have a physics question. I tighten my banjo heads to 6.5 inch pounds. All 24 hooks are torqued to 6.5.

What is the total force coming down on the tone ring?

Nov 30, 2020 - 3:56:48 PM
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1448 posts since 4/13/2009

156

Nov 30, 2020 - 4:56:04 PM

1 posts since 11/30/2020

What I believe you are asking is what tension should the head be set at. So, you're gonna want a drum dial. Typically, the tension is set at about a 90.  You will need to test the tension at multiple points around the head - you will not need to adjust everything.  (typically)

Edited by - Buckeye101 on 11/30/2020 16:57:24

Nov 30, 2020 - 5:04:06 PM

1572 posts since 4/13/2017

quote:
Originally posted by Buckeye101

What I believe you are asking is what tension should the head be set at. So, you're gonna want a drum dial. Typically, the tension is set at about a 90.  You will need to test the tension at multiple points around the head - you will not need to adjust everything.  (typically)

 


What I am actually asking is how much pressure is being put on the tone ring. Like, if all 24 hooks are pulling X pounds on the head, then the tone ring has Y pounds being applied to it. 

I believe Deestexas answered my question.

Nov 30, 2020 - 5:19:49 PM

481 posts since 1/28/2011
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The torque to tighten the nuts is not the same thing as pull down pressure on the head.

Nov 30, 2020 - 5:27:30 PM

1572 posts since 4/13/2017

quote:
Originally posted by latigo1

The torque to tighten the nuts is not the same thing as pull down pressure on the head.


Oh...so how would I determine that?

Nov 30, 2020 - 5:37:48 PM

461 posts since 2/6/2011

42!
Latigo1 is correct. In fact, placing 6.5 inch-lbs of torque on one nut does not even mean that it is pulling down with equal force as any of the other nuts. Consider if one hook is cross-threaded, or rusty, and another has a little oil on the threads transferred from your fingers. If the torque placed on each was equal, it would take more revolutions to reach that value in the latter case.

Nov 30, 2020 - 6:01:53 PM

784 posts since 12/30/2008

Yes, I agree as well with Dave Bird here. The 6.5 inch pounds isn't the tension or 'pull-force' on the threaded hook. It's the torsional force of the nut. An inch pound is the torque of one pound applied to a 1 inch lever (as the distance measured from the pivot axis). Likewise a foot pound is the torque of one pound applied to a 1 ft. lever.

The torque would likely have some sort of correlation to head tension, but it wouldn't be linear and as already pointed out, would also be affected by variable friction due to different factors such as lubrication, thread condition, thread pitch, etc.

Edited by - championofnorthhuron on 11/30/2020 18:03:12

Nov 30, 2020 - 6:17:31 PM
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RB3

USA

889 posts since 4/12/2004

I believe that what you're looking for is the axial force on the tension hook that results from the torque applied to the nut. Once you determine that axial force, you can multiply it by the number of tension hooks and you will have the total force applied to the tone ring.

Below is a link to a web site that has a calculator that allows you to input the parameters associated with your situation and it will do the calculation for you. You enter the screw major diameter, the applied torque and an estimate of the coefficient of friction for the material and condition you're using and it will calculate an estimate of the resulting axial force.

Axial Force Vs. Applied Torque

Nov 30, 2020 - 6:23:27 PM

481 posts since 1/28/2011
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quote:
Originally posted by Lemon Banjos
quote:
Originally posted by latigo1

The torque to tighten the nuts is not the same thing as pull down pressure on the head.


Oh...so how would I determine that?


If you had an allen wrench with one end chucked in a vise and the other end sticking out horizontaly and then hung a one pound weight exactly one inch from the center of the chucked up part you would be applying one inch pound of torque to the end sticking out.  I have no idea how you would relate that to pull down pressure on a threaded part.  banjukebox is correct that the condition of the threads will vary.  Also the threads per inch would make a difference as well as how tight or loose the threads are from the manufacturer.  Even a drum dial wont tell you head pressure.  All it measures is thousandths of an inch of head defection.  Neither method is going to answer your question.  

Nov 30, 2020 - 6:30:52 PM

1572 posts since 4/13/2017

quote:
Originally posted by RB3

I believe that what you're looking for is the axial force on the tension hook that results from the torque applied to the nut. Once you determine that axial force, you can multiply it by the number of tension hooks and you will have the total force applied to the tone ring.

Below is a link to a web site that has a calculator that allows you to input the parameters associated with your situation and it will do the calculation for you. You enter the screw major diameter, the applied torque and an estimate of the coefficient of friction for the material and condition you're using and it will calculate an estimate of the resulting axial force.

Axial Force Vs. Applied Torque


Do I combine the coefficient of friction of steel with that of mylar?

Nov 30, 2020 - 8:12:52 PM

184 posts since 4/3/2009

After reviewing the online calculator, and the coefficient of friction link, I'd say that the CoF would apply to the fastener materials themselves. That would provide the axial bolt "clamp" force that is applied to tightening the mylar head.

If, for example, the friction was brass on steel (the link doesn't list brass on brass), the CoF is 0.5 for dry contact and 0.19 for lubricated contact.  The former would result in about 63 lbs. per 10-24 or 10-32 hook, and about 166 lbs. for the latter.

So unless I'm missing something:

  • There is in general a whole lot of tension on a 24 hook banjo.  It no longer surprises me that flange damage can occur on a highly tightened banjo head.
  • Head materials like mylar must be pretty tough stuff!
  • The variation between "dry contact" (CoF ~0.5 to 1.0 depending on metal) and "lubricated contact" (CoF ~ 0.1 to 0.2) is so great, with so little control, that any correlation of hook and nut torque to axial force, and thus head tension, would be wildly inconsistent.  This agrees with a number of posters.

And therein lies the rub (pun intended):  One could make a calculation of the downforce on the tone ring, and the result could be off by up to a factor of 10.  Would that be accurate enough?

Nov 30, 2020 - 8:36:11 PM

1572 posts since 4/13/2017

So if one said that each hook was pulling 60lbs, that's 1440lbs on the tone ring!

Nov 30, 2020 - 8:36:37 PM

nakigreengrass

New Zealand

5502 posts since 5/16/2012
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I reckon the cof could be as much as .75 for brass on brass ....which would give you about 50 lbs of force on each hook @ 6.5 " lbs of torque...which means about 1080 lbs all up...which sounds a huge load...but.... considering that the circumference of the head is 34.5 "...the pull towards the center is only about 31 lbs per inch of circumference...which sounds about right.  edit.....Even 1440 lbs total sounds ok.

Edited by - nakigreengrass on 11/30/2020 20:38:21

Nov 30, 2020 - 10:01 PM

1053 posts since 1/9/2012

I'd go with the DrumDial idea.  Use that to estimate the head tension and multiply by the circumference.  I wrote to them once, and they kindly sent me a calibration curve.  Their numbers are the dots in figure 18 on page 25 in the JUNE 2018 entry on http://www.its.caltech.edu/~politzer .  (I've no idea how they got those numbers, but they're reasonably consistent with other things I know.)  For a typical mylar head on a resonator banjo, I get something around 1000 to 1100 lbf ("pounds of force" -- not a unit used by people in my business).  I'm also never very good with arithmetic.  So it's worth doing for yourself.

Dec 1, 2020 - 3:45:20 AM
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4559 posts since 11/20/2004

I don't know the answer, but multiplying the number of hooks seems inaccurate. A head tightened to a deflection of 90 with 24 hooks will not have 50% more pressure if tightened to 90 by 36 hooks.

Dec 1, 2020 - 6:02:42 AM

784 posts since 12/30/2008

quote:
Originally posted by lightgauge

I don't know the answer, but multiplying the number of hooks seems inaccurate. A head tightened to a deflection of 90 with 24 hooks will not have 50% more pressure if tightened to 90 by 36 hooks.


I had been pondering the same. In your scenario I think you are correct that the head wouldn't have 50% more pressure exerted on the tone ring. Rather, each hook would require 33.33% less tension to maintain the (constant) deflection of 90. 

But 36 hooks each tightened to say, 6 inch pounds, would create more tension on the head than 24 hooks tightened to 6 inch pounds. Think of each hook instead as being a 50# weight hanging from a cord tied to the perimeter of the head. Each additional 50# weight that is hung does equal more pulling force. While the drawing force of each weight (hook) is somewhat localized, the counterbalancing force of hanging more 50# weights is additive across the entire surface of the membrane.

Edited by - championofnorthhuron on 12/01/2020 06:16:11

Dec 1, 2020 - 7:18:14 AM

RB3

USA

889 posts since 4/12/2004

The question posed in the original post was how much force is applied to the tone ring. No question was posed about any forces associated with the head.

The question I would ask Mr. Lemon is why would you want to know how much force is being applied to the tone ring and if you had that information, what would you do with it?

A good way to begin an analysis of a situation like this is to create a free body diagram of the system.  A free body diagram allows you to visualize and identify the forces acting on a system.  Below is a link to a Wikipedia explanation of free body diagrams.

Free Body Diagram


 

Dec 1, 2020 - 8:02:44 AM

rmcdow

USA

883 posts since 11/8/2014

This was also discussed in this archived post. It might be worth a read by you

banjohangout.org/archive/360410

Dec 1, 2020 - 8:18:43 AM

816 posts since 2/19/2012

quote:
Originally posted by RB3



The question I would ask Mr. Lemon is why would you want to know how much force is being applied to the tone ring and if you had that information, what would you do with it?
 


Let me guess.....

Hunter, you're wanting to recreate a ball bearing head system, aren't you?

Dec 1, 2020 - 8:22:10 AM

1572 posts since 4/13/2017

quote:
Originally posted by Parker135
quote:
Originally posted by RB3



The question I would ask Mr. Lemon is why would you want to know how much force is being applied to the tone ring and if you had that information, what would you do with it?
 


Let me guess.....

Hunter, you're wanting to recreate a ball bearing head system, aren't you?


Not at all. I would prefer to keep my motives to myself for wanting this information.

I have found all the information I would possibly need if I wanted to recreate the ball bearing system. I could do so if I had the willpower to do it. And the money haha.

Dec 1, 2020 - 9:15:29 AM

RB3

USA

889 posts since 4/12/2004

I suspect that Mr. Lemon is conjuring up some new system to apply the force necessary to tighten a banjo head. Perhaps he's going to supplant the balls with tapered roller bearings. Inquiring minds want to know. But, as Bill Monroe once said, it must be "right smart of a secret".

Dec 1, 2020 - 11:15:18 AM

rmcdow

USA

883 posts since 11/8/2014

If you want to take a shot at the math to figure out the pressure on the tone ring, the Nechville tensioning ring is 11.7" in diameter with a 12 tpi pitch, turned by a 12 tooth pinion .585" in diameter. I don't know how much pressure this exerts on the tone ring sitting on the ring, but based on the mechanical advantage that the pinion turning inside the 11.7" diameter body, driven by the 12 tooth pinion, it is a lot. I'm pretty careful tensioning the head of the Nechville, as it would be pretty easy to over tension it if it was not for the obvious force that it takes to turn the wrench with the pinion on it.

Dec 1, 2020 - 11:25:49 AM

ChunoTheDog

Canada

509 posts since 8/9/2019

You could always build some sort of flange/rim jig with a large diameter digital scale where the tone ring would be and get accurate readouts of the pressure being exerted onto the rim/ring/scale at various drumdial settings...or something similar, no?

Dec 1, 2020 - 1:06:18 PM

1572 posts since 4/13/2017

One also may be able to hang a rim and flange upside down, put all 24 hooks and nuts on, then "weave a basket" between all 24 hooks with some sort of strong string material, then progressively add weight to it until the hooks match the torque wrench.

Dec 1, 2020 - 4:06:50 PM

481 posts since 1/28/2011
Online Now

quote:
Originally posted by Lemon Banjos

One also may be able to hang a rim and flange upside down, put all 24 hooks and nuts on, then "weave a basket" between all 24 hooks with some sort of strong string material, then progressively add weight to it until the hooks match the torque wrench.


You are still confused about the difference between POUNDS of TORGUE, and POUNDS of PULL DOWN WEIGHT on the tension hoop.  They are like apples and oranges.  If you had two hooks with different threads one very loose and one very tight, it would require more inch pounds of torque to get the one with tight threads to have the same pull down weight as the one with loose threads. If you had some hooks with fine threads and some with coarse threads, it would take more torque on the ones with coarse threads to reach the same tension as the ones with fine threads because the fine threads have more mechanical advantage than coarse threads ( like the difference between a long crowbar and a short one)..  Torque is measured in pounds but torque is not weight.

Edited by - latigo1 on 12/01/2020 16:25:39

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