Downbearing

[Ric Brekne]

Hi Dave

Right you are.  Fortunatly my spread sheet had it right.. I just typed 
it in wrong.  I more or less followed your reasoning. But I didnt use 
your angle d.  Used the Cosinus longform ( c^2 = a^2 + b^2 - 2ab * Cos 
C) sentence to get the undeflected string length, then figured the front 
and back angles by way of Sin C / c = Sin A / a = Sin B / b.  That 
yeilds the rest... component angles of C and height. Which in turn yeild 
force vector values for the undeflected string, the deflected strings 
downwards force as a whole, and the componet values of this last for 
front and back lengths.

Sorry bout the lengthy numbers...  :)... just copied straight from a 
still very rough spreadsheet.

Cheers
RicB

---------
Ric,
I think the String Angle from front termination  =  0.66 deg
and the angle from the aliquot/hitchpin          =  1.33 deg

You got them swapped. Just by observation, the angle from the front 
should be smaller
than the back since the speaking length is larger than the back length.

(By the way, I was less bothered by the commas than by the 10 digits of 
precision.
Any more than 2 or 3 digits doesn't mean much.)

I have attached a diagram to show how I think you arrived at these 
numbers. The
red line represents the string. The angle at the bridge is c. The string 
displacement
is the length Z. a is the front angle, b is the back angle.
     Z = X sin(a)
     Z = Y sin(b)
     d = a + b      where d = (180 - c)
therefore
     X sin(a) = Y sin(b)
so
     sin a  =  (Y/X) sin( d - a )   =  (25/50) sin( 2deg - a )

solving for a (transcendental equation):
     a = arcsin (  (1/2) sin( 2 - a ) )

this gives a = 0.66 deg, b = 1.33 deg , Z = .58 mm

(Actually for such small angles, it's a pretty good approximation simply 
to say
     a = (1/2)b
     d = a + b
which gives you the same result to the 4th decimal place)

The force can be considered as vectors similar to the triangle of lengths,
where the "X" and "Y" contributions are added:

     force = 160 lbs ( sin(a) + sin(b) )  = 5.58 lbs

-- someone correct me if I'm wrong. --