<!doctype html public "-//W3C//DTD W3 HTML//EN"> <html><head><style type="text/css"></style><title>Re: Hammer Shank Ratio</title></head><body> <blockquote type="cite" cite>Richard Brekne wrote:</blockquote> <blockquote type="cite" cite><br></blockquote> <blockquote type="cite" cite>>Folks<br> ><br> >I keep being bothered by the differing conventions for measuring the</blockquote> <blockquote type="cite" cite>>ratio of the hammer shank. . .>Below are a couple pictures showing a method of checking the hammer<br> >shank ratio by measuring the weights involved. First is the SW and</blockquote> <blockquote type="cite" cite>>second is a difficult (but doable)  quantity to measure, the actual<br> >weight of the hammer assembly at the knuckle. The flange is replaced by<br> >an adapter to the Stanwood kit so as to achieve as solid and friction<br> >free a measurement as possible. And the wood dowel used to contact the<br> >knuckle is glued in place to tray so as to remove any influence from it<br> >moving around. About 50 samples were taken, and compared to a 60 second<br> >sampling under vibration.... the variation of the samples was at highest<br> >80,4 grams and at lowest 73.1 grams. About 80 % of the samples were<br> >right around 76 grams, and the average weight under vibration was at<br> >about 75 grams, though as soon as the vibration was shut of the scale</blockquote> <blockquote type="cite" cite>>settled at 76.5... which ended up being the picture I took. . .>  All in all</blockquote> <blockquote type="cite" cite>>it seems pretty reasonable that in this case the established ratio is</blockquote> <blockquote type="cite" cite>>76.5 / 10.1 which gives a ratio of  7.57 for this hammer shank assembly.</blockquote> <div><br></div> <div>Phil Ford replied:</div> <blockquote type="cite" cite>I'm not sure that I understand why you want to know this ratio.</blockquote> <div><br></div> <div>Me either. The real figures which matter are the average hammer movement with respect to that of the key front, and the average jack tip movement with respect to the key front<font color="#0000FF"> (this latter figure is important in an action design for a sufficient speed of jack escapement during the let off)</font>.</div> <div><br></div> <div>Terry Farrell came up with a very good practical method for measuring hammer/key leverage ratio in one of his recent posts. He suggested measuring the key travel and resultant hammer travel to derive the ratio. Now this idea would seem to have considerable merit if you have a dial guage on hand, since you could depress the key exactly 2 mm for example, and measure the resultant hammer movement quite easily with a ruler. Makes a lot of sense to me Terry, the idea is practically foolproof and you would get an answer in a very short time.</div> <div><br></div> <div>Richard B, you are measuring a weight ratio, not a leverage ratio. They are not quite the same thing unless the vectors are in the same direction, and in the case of the experiment you conducted they are not.</div> <div><br></div> <div>Phil Ford continued:</div> <div><br></div> <blockquote type="cite" cite>Given the parameters that you've established I agree that this particular ratio is 7.57.  I question whether this is useful for anything.  When the shank is parallel to the scale platform and the support under the knuckle is vertical then this ratio is 7.57.  This doesn't represent a configuration of any real piano action so in what way is this useful?  In a real action the support for the knuckle is being applied perpendicular to a line between the whippen center and the whippen/knuckle contact point. So, when the hammer shank is horizontal, the force applied by the whippen at the contact point will be less than 76.5g since the moment arm, or lever arm, is 21.64 mm - see D below, assuming that the contact point is on the magic line at this moment.  This number seems more relevant to what is happening in a real action.<br> <br> <br> >  Now the interesting part of all this comes when you compare the<br> >different conventions for finding the ratio by measuring lever arm<br> >distances. Remember that whatever method is chosen simply must conform</blockquote> <blockquote type="cite" cite>>reasonably to the ratio established above.</blockquote> <div><br></div> <div>Phil Ford continued:</div> <blockquote type="cite" cite>Why?<br> <br> >  So first .... some distances.<br> ><br> >--From center of hammer shank diameter to knuckle contact point 13 mm</blockquote> <blockquote type="cite" cite>>A) From middle of center pin to middle of hammer molding straight down</blockquote> <blockquote type="cite" cite>>the shank -- 136 mm</blockquote> <blockquote type="cite" cite>>B) From middle of center pin to center of gravity point on the hammer</blockquote> <blockquote type="cite" cite>>142 mm</blockquote> <blockquote type="cite" cite><br></blockquote> <blockquote type="cite" cite>How did you establish this?<br> <br> >C) From middle of center pin to tip of hammer 148 mm<br> >D) From middle of center pin to middle of knuckle molding -- 17.3 mm</blockquote> <blockquote type="cite" cite>>E) From middle of center pin to knuckle contact point 21,64 calculated</blockquote> <blockquote type="cite" cite>>as root (17,3^2 + 13^2)</blockquote> <blockquote type="cite" cite>></blockquote> <blockquote type="cite" cite>>Now lets take a look at which convention most closely conforms to the<br> >already established ratio.<br> ><br> >A/D = 7.86  (given by Vincent RPT)<br> >B/D = 8.21  (discussed informally on the PTD list)<br> >B/E = 6.56  (I ran into this one in Stockholm last year in informal<br> >discussions)<br> >C/E = 6.84  (given by Overs)<br> >A/D = 6.28 (suggested by a technical editor informally in private<br> >correspondence)<br> <br> <br> I believe this last one was supposed to be A/E = 6.28.<br> <br> <br> >Its quite obvious which one of these comes out best.</blockquote> <div><br></div> <div>Yes quite obviously if you are interested in determining the weight ratio of the hammer assembly only.</div> <div><br></div> <div>If you are particularly interested in the weight ratio with respect to the hammer only, and you would like to derive a formula to reflect the fact that you are deriving both weights with the hammer shank positioned horizontally, then an appropriate formula might be:</div> <div><br></div> <div>Where angle<font face="Symbol" color="#000000"> q</font> (36.92 degrees) is the included angle between a line from the knuckle contact point and the center pin and the horizontal.</div> <div align="center">(A/E)*(1/cos[<font face="Symbol" color="#000000">q</font>]) = 7.861</div> <div align="center"><br></div> <div>However, as Phil Ford mentioned in his post, the vector force of the jack tip on the roller in a real piano action will not be the same as your setup.</div> <div align="center"><br></div> <div align="center"><img src="cid:a05100302ba714ce1e53e@[61.8.27.1].1.0"></div> <div><br></div> <div>I have noticed that when calculating the hammer/key ratio by measuring the lever lengths you will always get a larger figure than that derived by weight. Only recently did I realise that this is because the weight of the hammer head is bearing down on the end of an almost horizontal hammer shank. Therefore the hammer weight is bearing down on the hammer shank lever at approximately the distance from the hammer center to the center of the hammer moulding<font color="#0000FF"> (of course this will vary somewhat throughout the hammer stroke)</font>. When calculating the hammer leverage ratio using the (hypotenuse) distance from the hammer center pin the hammer strike point, we will necessarily get a larger figure since the tip of the hammer will travel further relative to the key front when compared to the end of the hammer shank.</div> <div><br></div> <div>I hope that makes sense. When our piano was exhibited at Reno, David Stanwood made a calculation of the leverage ratio of the action using his weighing method. He got a figure of 5.5:1. Now I designed the action for this piano on CAD to have a ratio of 5.8<font color="#0000FF"> (at the strike point)</font>. If you take the 5.8 ratio and multiply it by 130/138, you get 5.46. I believe this may explain the different figures arrived at via the two different measurement systems.</div> <div><br></div> <div>So if you wish to arrive at a measured-lever-lengths figure for the hammer/key ratio which agrees reasonably closely to the figures derived by the Stanwood method, you could substitute the length from the hammer center pin to the strike point with the length from the hammer center pin to the center of the hammer moulding.</div> <div><br></div> <div>Ron O.</div> <x-sigsep><pre>-- </pre></x-sigsep> <div>_______________________<br> <br> OVERS PIANOS - SYDNEY<br> Grand Piano Manufacturers<br> <br> Web: http://overspianos.com.au<br> mailto:info@overspianos.com.au<br> _______________________</div> </body> </html>