Seating strings

[Ron Nossaman]

Phil,

I put a spreadsheet together about four years ago, for estimating 
these friction and force levels. I get somewhat different results. 
Having no formal engineering education, I did what I thought looked 
right and was supportable by what reference material I could find 
(and understand). It's subject to revision if better information is 
available, so I have some questions, if I may.


> OK, just for fun (or for the sake of argument, if you prefer) I'll take a 
> crack at this.  I thought I would throw a little math at this.

For fun, absolutely, and for education, which is also fun.


> My assumptions:
> 
> String tension T = 150 lbs.
> Side bearing angle 8 degrees
> Down bearing angle 1 degree
> Bridge pin angle (relative to cap surface) 15 degrees
> 
> Friction between string and bridge pin is given by static friction formula
> FR = u N
> 
> where:
> 
> u = coefficient of friction between string and bridge pin.  This will 
> depend on the material of the string and the material of the bridge pin and 
> on surface finishes of each.  For high polished steel on highly polished 
> brass it would be on the order of 0.2.  For rusty steel on rusty steel it 
> might be on the order of 1.0 or more.
> 
> N = normal force (force perpendicular to bridge pin) exerted by the string
> 
> The side bearing force is given by SB = T sin (8 deg) = 150 (.139) = 20.9 LB
> 
> The down bearing force is given by DB = T sin (1 deg) = 150 (.017) = 2.7 LB

The sin function gets less accurate as the angle increases, but yes, 
close enough.


> Each of these forces will have components normal to the bridge pin and 
> parallel to the bridge pin.
> 
> FOR THE SIDE BEARING:
> 
> Normal force N1 = 20.9 cos (15 deg) = 20.2 LB (note that this is towards or 
> into the bridge pin)
> 
> Parallel force P1 = 20.9 sin (15 deg) = 5.4 LB (note that this force is 
> down toward the bridge cap)

This puzzles me. I got 5.4 for the down force all right, but how can 
the total of 20.2 lb horizontal, and 5.4 lb vertical be more than 
the 20.9 lb the side bearing angle generates? Is this just an 
artifact of the inaccuracy of the sin and cos method at these big 
angles?


> FOR THE DOWN BEARING:
> 
> Normal force N2 = 2.7 sin (15 deg) = 0.7 LB (note that this is away from 
> the bridge pin and is counteracting the normal force from the side bearing)

This also puzzles me. The downbearing force should have nothing to 
do with the side bearing forces since it is acting directly on the 
bridge top and isn't bearing on the pin at all. I don't understand 
why this is here.


> Parallel force P2 = 2.7 cos (15 deg) = 2.6 LB (note that this is down 
> toward the bridge and is adding to the parallel force from the side bearing)

Double puzzlement. How can side bearing reduce downbearing force 
(absent friction), and again, how can the sum of these two numbers 
exceed the total downbearing force of 2.7 lb?


> TOTAL FORCES:
> 
> N = N1 + N2 = 20.2 - 0.7 = 19.5 LB
> 
> P = P1 + P2 = 5.4 + 2.6 = 8.0 LB
> 
> If the force parallel to the pin is higher than the friction generated by 
> the normal force then the string will want to move down the pin.  If the 
> friction force is higher than the force parallel to the pin then the string 
> will want to stay where it is (even if it's above the bridge cap).
> 
> FRICTION FORCE:
> 
> This is going to depend on the friction coefficient that you assume:  I 
> think a reasonable number might be 0.6.  For this assumption friction force is:
> 
> FR = u N = 0.6 (19.5) = 11.7 LB

I didn't break it out like this, but calculate a static resistance 
(with your quite reasonable 0.6 coefficient) moving down the pin at 
6.7lb, and up the pin at 17.5lb. By my reckoning, the 2.6lb of 
downbearing and the 5.4lb down force from the pin slant and side 
bearing exceeds the static resistance, and the string will seat on 
the bridge automatically. I show a break even point of friction 
against down force at either a 13.25° pin angle, or a 0.666 (did you 
do that intentionally? <G>) friction coefficient. Enlighten me, please.