>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, Hopefully you spent four years doing something more enjoyable and ultimately more lucrative (oh, wait, you went into the piano business - well, hopefully you spent four years doing something more enjoyable). > 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. Sheesh. Someone's actually gonna check my numbers? Now I'm in trouble. >.. >>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. I'm sure that Ptolemy won't be happy to hear that. I don't know what you mean about the sin function getting less accurate as the angle increases. >>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? The force on the pin is a vector having magnitude 20.9 and direction along a line in the horizontal direction (to be more correct, which I'll now define as a horizontal line lying in the horizontal plane and perpendicular to the string speaking length - and to be more correct, being positive in the direction of the bridge pin - none of which is strictly correct, but more on that in a moment). I'm breaking down that vector into mutually perpendicular components (by using the sin and cos functions). The vector summation of those components is the square root of the sum of the squares [(20.9)^2 = ((20.2)^2 + (5.4)^2)^.5], if that makes sense. An easier way to think of it is that the force 20.9 is the hypotenuse of a right triangle, with the components 5.4 and 20.2 being the sides of the triangle. The square of the length of the hypotenuse is the sum of the squares of the side lengths. Clear as mud? Sorry, I'm not a math teacher, so I hope I haven't said something not strictly correct here. Caveat emptor. Please note that the 5.4 LB component is not strictly speaking a 'down force'. It is parallel to the bridge pin, but generally in the down direction (in other words, towards the bridge cap rather than away from it). As I mentioned before, defining the side force on the pin as being in the horizontal direction and perpendicular to the string speaking length is not strictly correct. The string bends around the pins (because of the side to side displacement of the string and also the portion of the string between the pins is probably not in the same plane as the speaking length). So the direction of the force will not be quite perpendicular to the speaking length of the string and will probably not be quite in the horizontal plane. But for talking purposes let's call it close enough. >>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. If the string is firm against the cap then you're correct. The point of my little exercise was to show that if the string was in fact above the cap that it might not want to slide down. So I was assuming that the string was not against the cap, in which case the down bearing will tend to pull the string down (and slightly away from the pin - depending on the pin angle. I think you can see that if the bridge pin angle was very severe and the string was up off the bridge that the down bearing would want to move the string down away from the bridge pin. The same thing is happening for very small pin angles, it's just that the component of the downbearing perpendicular to the pin is very small - as you see here, only 0.7 LB - I could have safely ignored it). >>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), It doesn't. The sidebearing and downbearing forces are two independent vectors. I'm doing a vector summation of those forces. I find it a little simpler or clearer to work with the components. Both of these vectors have components perpendicular to the bridge pin and parallel to the bridge pin. In this case the components parallel to the bridge pin are additive and the components perpendicular to the bridge pin are not. > and again, how can the sum of these two numbers exceed the total > downbearing force of 2.7 lb? The summation talked about above. >>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. I don't understand why you have a different number for moving down the pin and moving up the pin? Static friction is dependent on the normal force between the string and bridge pin and the coefficient of friction. The initial force to start an infinitesimal movement of the string should be the same whether it moves up the pin or down. > 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. Perhaps you can give me a little more detail on how you arrived at your 6.7 LB static resistance number. > I show a break even point of friction against down force at either a > 13.25 deg pin angle, I'll make my unknown angle a. Total Normal Force on Pin = N = 20.9 cos a - 2.7 sin a Static friction = FR = uN = 0.6(20.9 cos a - 2.7 sin a) Total Parallel Force = P = 20.9 sin a + 2.7 cos a String will move when P = FR, so 0.6(20.9 cos a - 2.7 sin a) = 20.9 sin a + 2.7 cos a Solving gives a = 23.6 degrees So, for this scenario you need a fairly steep pin angle for the string to want to slide down strictly under static load. > or a 0.666 (did you do that intentionally? <G>) friction coefficient. Guess who made me do it? <G> > Enlighten me, please. Notice that this is incompatible with my calcs, since I assumed a 0.6 coefficient of friction and showed that the string wouldn't move. So, if we're going to leave the pin angle alone and change the coefficient of friction so that the string will move it has to get lower, not higher. Assuming I'm leaving the 15 degree pin angle alone, then from the numbers above, and solving for Coefficient of Friction u: Friction = FR = uN = u (20.9 cos 15 - 2.7 sin 15) = 19.49 u Parallel Force from above P = 8.0 LB String will move when these are equal, so 19.49u = 8.0 u = 0.41 which is a fairly slippery interface. Happy calculating, Phil Ford
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