<!doctype html public "-//w3c//dtd html 4.0 transitional//en"> <html>   Ron Overs wrote: <blockquote TYPE=CITE> Phil Ford replied: <blockquote type="cite" cite>I'm not sure that I understand why you want to know this ratio.</blockquote>  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 (this latter figure is important in an action design for a sufficient speed of jack escapement during the let off).</blockquote> <blockquote TYPE=CITE>  In either case, the law of leverages dictates that the resulting ratio be the same regardless how you measure it. Any deviation from the actual ratio reveals an error in measurement, or that one is measureing something different to begin with.</blockquote> <blockquote TYPE=CITE>   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.</blockquote> <blockquote TYPE=CITE>Yes he did. And if you remember his method resulted in a figure closer to the Stanwood SW ratio then it did using the formula you give. Thats one of the things that still bothers me. But so far I have been mostly concerned with the SW ratio in itself.... what IT is in terms of lever arm lengths.</blockquote> <blockquote TYPE=CITE>  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. Given the basic formula for leverage  ie... D1 x W1 = D2 x W2 they actually are quite the same thing. And vectors dont come into this picture at all. A force vector will certainly influence the net input force on a lever, but the output is determined by the ratio of the lengths of the two levers alone. A 2:1 ratio will give a 2:1 result regardless of whatever input force vectors are involved. Bringing vectors into the picture only speaks to how the input is accepted and how the output is used... not the ratio of the lever itself.   Phil Ford continued: <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.</blockquote> </blockquote> None of this changes anything at all about the ratios. The whippen is a lever with a fixed ratio as well. Because  the input varies (as you correctly describe) so too will the output at the capstan.. and according to the ratio of the whippen in this case.  I say again... the variance of the amount of net input force does not alter the ratio of a lever. Nor does the output vector. Nor do any of these vectors change the ratio of the action as a whole. Only the variance of the length of an arm can change the ratio of a lever. <blockquote TYPE=CITE> <blockquote type="cite" cite>So first .... some distances.</blockquote> <blockquote type="cite" cite>-From center of hammer shank diameter to knuckle contact point 13 mm</blockquote> </blockquote> <blockquote TYPE=CITE> <blockquote type="cite" cite>A) From middle of center pin to middle of hammer molding straight down the shank -- 136 mm</blockquote> </blockquote> <blockquote TYPE=CITE> <blockquote type="cite" cite>B) From middle of center pin to center of gravity point on the hammer142 mm</blockquote> <blockquote type="cite" cite>C) From middle of center pin to tip of hammer 148 mm D) From middle of center pin to middle of knuckle molding -- 17.3 mm</blockquote> </blockquote> <blockquote TYPE=CITE> <blockquote type="cite" cite>E) From middle of center pin to knuckle contact point 21,64 calculated as root (17,3^2 + 13^2)</blockquote> </blockquote> <blockquote TYPE=CITE> <blockquote type="cite" cite>Now lets take a look at which convention most closely conforms to the already established ratio. A/D = 7.86  (given by Vincent RPT) B/D = 8.21  (discussed informally on the PTD list) B/E = 6.56  (I ran into this one in Stockholm) C/E = 6.84  (given by Overs) A/E = 6.28 (suggested informally in private correspondence) Its quite obvious which one of these comes out best.</blockquote> Yes quite obviously if you are interested in determining the weight ratio of the hammer assembly only. 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: Where angle q (36.92 degrees) is the included angle between a line from the knuckle contact point and the center pin and the horizontal. <center>(A/E)*(1/cos[q]) = 7.861</center> </blockquote> Yes... which is exactly the same thing as measureing both levers straight down the shank, use cos as you do simply takes the angle q out of the picture. You could just as easly multiply   A * E * cos(Q) = 7.861, or as I do... not bother with it in the first place since all of this is exactly the same as A/D to begin with.   <blockquote TYPE=CITE>  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.</blockquote> see above : I really think a better explanation for the disparagy here is  that the method you give for measuring action ratio is simply a different relationship then what the SW ratio is. If I were to attempt to ascertain the ratio you give in terms of  weights, I would have to look at how much weight at the top of the hammer is moved for a given input at the front of the key. The load in such a configuration is at the top of the hammer, instead of at the end of the shank. If we are to talk about what is remote from real piano action reality...... well :) But I dont think thats at issue here, this is not a matter of whats right and wrong per se... more a matter of whats what. As I have stated before I am not comfortable with the fact that a given method results in variance in total action ratio for bass and trebles, or any section with varying bore lenghts. But thats just a matter of taste I suppose. Any convention is ok as long as everyone is in aggreement at any given time which one we are talking about. I personally would prefer talking in terms of a ratio that aggrees with the SW ratio for obvious reasons. Tho perhaps if I was designing an action from scratch I might be interested in   <blockquote TYPE=CITE><img SRC="cid:part1.3E4BE9AE.5BD1201A@grieg.uib.no" height=269 width=490></blockquote> You do open for a nice translation between the two tho... simply useing the long form for cos on the angle formed by C and E  (which are the lengths you employ) and divide that into A/E, then solve as otherwise for the total action ratio. Cute.   <blockquote TYPE=CITE>   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 (of course this will vary somewhat throughout the hammer stroke).</blockquote> I dont aggree with this. I think its simply because you are measureing two different ratio relationships. The SW ratio does not "measure" the same arm lengths in the hammer shank as you do. That explains the difference. <blockquote TYPE=CITE>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. I hope that makes sense.</blockquote> Certainly.. that exact difference is at the heart of the whole thing issue. You are describing and entirely different lever then the lever described in the SW ratio. The other two levers appear to be dealt with in exactly the same manner for both systems.   <blockquote TYPE=CITE>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 (at the strike point). If you take the 5.8 ratio and multiply it by 130/138, you get 5.46.</blockquote> Where did you get these two figures ? Obviously from your action, but what are they ? Centerpin to molding / centerpin to hammertip ?? <blockquote TYPE=CITE>I believe this may explain the different figures arrived at via the two different measurement systems. 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. Ron O</blockquote> This one I have in anycase ruled out. Its describe above  (A/E) and results in the lowest hammershank ratio, and the one farthest away from what was ascertained by weighing.  I have figured these about 20 times now on that many actions and compared to the SW ratio. What I find is that A/D in the total ratio equation yeilds the result that best conforms to the SW ratio. <blockquote TYPE=CITE>  OVERS PIANOS - SYDNEY Grand Piano Manufacturers Web: <a href="http://overspianos.com.au">http://overspianos.com.au</a> <a href="mailto:info@overspianos.com.au">mailto:info@overspianos.com.au</a> _______________________</blockquote> Even MORE fun !! :) -- Richard Brekne RPT, N.P.T.F. UiB, Bergen, Norway <a href="mailto:rbrekne@broadpark.no">mailto:rbrekne@broadpark.no</a> <a href="http://home.broadpark.no/~rbrekne/ricmain.html">http://home.broadpark.no/~rbrekne/ricmain.html</a>  </html>