Ric and list, The downbearing calculations shouldn't be too terrible= .  The triangle idea is on the right track.  It is called vector = summation.  The hypotenuse of the right-triangle is the string tension= .  The small side is the downbearing force (assuming small downbearing= angle).  The angle used should be the angle between the speaking length of the strin= g and the backscale length. Example: Downbearing angle = 2 degrees= , string tension = 160 pounds: String Tension  x  Sin(do= wnbearing angle) = Downbearing Force i.e.   160 x Sin(2 degrees) = 5.58 pounds  (3.49% of string= tension) This would be real simple to add to a spreadsheet like the= one Ron Overs posted on Jan. 21st. Also, be careful with spreadsheet calculations in EXCEL.  My vers= ion uses angular units in "radians" not "degrees".&nbsp= ; 180 degrees are equal to "pi" radians (pi = ~3.1416), so one = radian = ~57.3 degrees.  The formula above in EXCEL could be "=160*sin(2/57.3)". = This advice has nothing to do with how much downbearing is desirable, only = how it could be calculated. Having just said that, it appears that i= f the desired total downbearing is only in the ballpark of 0.5% to 1.0% of total string tension, then the average downbearing angle wo= uld be in the range of 0.29 to 0.57 degrees.   Hmm...  = Have fun! Best Regards, Steve Fujan <a href="http:/= /www.fujanproducts.com"> www.fujanproducts.com</a>    <div>On 2/19/06, Ric Brekne <<a hr= ef="mailto:ricbrek@broadpark.no" target="_blank" onclick="return top.= js.OpenExtLink(window,event,this)"> ricbrek@broadpark.no</a>> wrote:<blockquote class="gmail_quote"= style="border-left: 1px solid rgb(204, 204, 204); margin: 0pt 0pt 0pt 0.= 8ex; padding-left: 1ex;"> Please correct if this is entirely wrong... but I thought that since the<br= >string was being measured in terms of its tension (pounds)  one = could simply the problem  as a like sided triangle with half t= he pounds on each leg. Since the measurement is taken in the deflected condition... y= ou have basically the hypotenus and all angels of a right angle triangle= available to figure the amound of deflection.. pounds in this case.&nbs= p; So 160 pounds with a 2 degree deflection at the bridge  y= ields Sin 1 x 80  = 1.396192515  lbs downbearing,&n= bsp; which is 1.745 % of the string tension. er... yes ??<br= > RicB ------------- >   So knowing all of t= he above, what is the equation that will calculate > an approximate string bearing load under the conditions I describ= e? Beats me. I use the SIN(RADIANS(degree measurement))*tension p= er unison, and add them up in my spreadsheet. __________________________= _____________________ Pianotech list info: <a href="http://www.ptg.org/mailman/listinfo/pia= notech" target="_blank" onclick="return top.js.OpenExtLink(window,event= ,this)">https://www.moypiano.com/resources/#archives</a> </blockquote><= /div>