[pianotech] Action Ratios

Jude Reveley (Absolute Piano) juderev at verizon.net
Wed Jan 6 21:31:13 MST 2010


Great posts...David, you ask the best questions and Nick, you're a treasure to the industry and I'm still bummed that I missed you in Boston. 

There seems to be some chicken and egg going on so here's my two cents FWIW.

  1.. Replace (blow distance - letoff) in the equation with (overall hammer travel-let-off).
  2.. The overall hammer travel equation that I use (thanks to Bruce Clark)  requires a specification for aftertouch. However, it allows us to calculate the the complete distance the hammer travels along its arc. The attached spreadsheet demonstrates how an AT can be selected (cell B6 highlighted) that will also produce the AT as calculated according to Nick Gravagne's EQ. Column E is Nick's example and Column F demonstartes my suggestion. I came up with 1.5mm to balance the two approaches, which seems more realistic to me.
  3.. As for the key input lever, it seems to me that the input should begin at the top of the key (perhaps even closer to where a finger touches down) rather than at the bottom as shown in the diagram that Chris Robinson and Rick Baldassin produced (or perhaps originated by Pfeiffer).
  4.. The action ratio equation that David posted still balances out but it bothers me that is doesn't jibe with the overall hammer travel/ keydip.

Jude Reveley, RPT
Absolute Piano Restoration, LLC
www.absolute-piano.com


  ----- Original Message ----- 
  From: David Love 
  To: pianotech at ptg.org 
  Sent: Wednesday, January 06, 2010 8:17 PM
  Subject: Re: [pianotech] Action Ratios


  You wrote the following so I used that as a guide in my previous post:

   

  .S = (V x Ra x N x W) / (H x Rs x K) = (245 x 67 x 18.25 x 44) / (126 x 94 x 141) = 7.89 mm which is the theoretical key dip minus aftertouch. Thus aftertouch = 2.61 mm. Said another way; dip prior to aftertouch is 75% of the key stroke, while aftertouch accounts for the remaining 25%...

   

  So let me restate the question.  

   

  The products of RAs/EAs can be figured by physical measurement of the three levers and will produce some number known as the action ratio. 

   

  The right side of the equation which you refer to as W/S is not arbitrary however if you assign a number to the S then you can calculate W based on the number that R/S must produce from your products of RA/EA.  Since you implied that aftertouch accounts for about 25% of the total dip, I took 10mm as the target dip and subtracted 2.5 mm (25%) and assigned the value S as 7.5.  If S = 7.5 then if the product of RAs/EAs = 5.5, W must equal 7.5 x 5.5.  But, as you can see, the theoretical key dip minus aftertouch with such an action does not produce a W number that makes much sense.  

   

  So,

   

  Let's change the question and assume that the action is in compliance, i.e. the elevations and convergence lines are within reason.  Let's say I want an action that regulates with 10 mm total did and 46 mm blow distance.  What action ratio by the products of RAs/ EAs will produce that result.  Another way to ask this is how much total hammer travel is necessary as a percentage of the target blow distance.  For example, with an action whose ratio by products is 5.0, 10 mm of dip will produce a total of 50 mm of hammer travel (if the hammer travel were unimpeded by let off).  If the target blow distance is 46 mm then the total potential hammer travel equals 109% of the desired blow distance (50/46 expressed as a percentage).  Is that enough?  Can that number be used as a rule of thumb?  If so, then by setting up an equality whose left side is the product of RAs/EAs, you can calculate the blow distance/key dip by choosing an arbitrary target key dip and the calculate the total amount of potential hammer travel and see whether or not it meets the requirements of 109% of the target blow distance.  

   

  Let's say that the action ratio by products is 5.0 (keep it simple here) and the target is 46 blow and 10 mm dip.  We can calculate that 10 mm dip will produce 50 mm of hammer travel and we can see that 50/46 = 109% and we are in business.  But let's say we need 48 mm of hammer travel for some reason but we still want 10 mm dip. Then we see that 10 mm dip produces hammer travel of only 104% of targeted blow distance-not enough.  So we have to increase the key dip (or raise the hammer line).  How much?  We can figure out that 10.5 mm dip will produce 52.5 mm hammer travel and see that 52.5/48 in fact equals 109% .  So if we need 48 mm blow distance we will need to increase the key dip to 10.5 mm.

   

  So now the question is, can we use the percentage produced by total hammer travel/blow distance as a guide for the right side of the equation to determine whether or not a particular action ratio will produce the desired regulation requirements.  If so, what should that percentage be?  109%, 105%, 115%?   Or what is the range?  Or is this a valid way to do this at all?

   

  That's my question.  

   

   

  David Love

  www.davidlovepianos.com

   

  From: pianotech-bounces at ptg.org [mailto:pianotech-bounces at ptg.org] On Behalf Of Nick Gravagne
  Sent: Wednesday, January 06, 2010 10:04 AM
  To: pianotech at ptg.org
  Subject: Re: [pianotech] Action Ratios

   

   

  David,

   

  I think I see where you are trying to go with this, and I think I can see you frustration, but I am a bit uncertain. You begin with:

   

  >Let's assume that the left side of the equation produces an action ratio of 5.5-a fairly standard target.  

   

  What EQ? Is it Pfeiffer's W / S = the products of the RAs / EAs? I assume are you referring to: 

  (keyout/key in) x (wippen out/wippen in) x (shank out/shank in) = (blow distance - letoff) / (key dip-aftertouch)

   

  >Then if we look at the right side of the equation and target a key dip of 10mm, say.  By your analysis the denominator would be 7.5 (.75 x 10) making the numerator 39.7mm representing blow distance minus let-off???  

   

  I assume now we are referring to W / S = the products of the RAs / EAs. How did you arrive at 7.5?

   

  In order for the ratio of W / S to = your 5.5, and given that S = your 7.5, then W (i.e., hammer blow - AT) would have to = 41.25 since 41.25 / 7.5 = 5.5. 

   

  Given your 10 mm total dip, aftertouch (the moment that the jack tender makes contact with the let-off button) would be 2.5 mm. Still, if hammer blow is actually 46, then let off would have to occur at 4.75 mm. Things are not adding up here and I will address it later.

   

  For now, and given your 10 mm total dip, and a 5.5 Action ratio*:

   

  W = 44 (not 39.7 or even 41.25)

  S = 8 (not 7.5)

  At = 2

   

  *NOTE all of the Kawai action values below are retained, but in order to yield David's required 5.5 ratio rather than the 5.58 Kawai original, the short shank lever arm was increased by 0.25mm.

   

  >That doesn't seem to bear any resemblance to what one would expect from an action ratio of 5.5 in practice. 

   

  The EQ W / S = the products of the RAs / EAs resolves to a perfect balance of the EQs in that W is to S as the RAs are to the EAs, which is the point if one wishes to approach the subject via these EQs.

   

  >Clearly the dip-aftertouch number is at issue but the assignment of the AT number seems somewhat arbitrary in order to make the formula work.  

   

  The AT number is neither arbitrary nor assigned; it is required to make the relationships balance. Note that the W value relates to all the RAs on the right side the EQ, and that the S value relates to all the EAs. 

   

  >What, then, is the point of the right side of the equation at all?

   

  Again, if the right side of the EQ is according to Pfeiffer: "The ratio W / S.is of much less importance for us than the right side of the equation, which permits us to survey at a glance not only the relationship between our lever arms, but also the effect which changing one of the lever arms has on the others, or on the stroke of the key or hammer blow distance." Page 110 - 111

   

  If the right side of the EQ is (blow distance - letoff) / (key dip-aftertouch), then I see your frustration in trying to assign a total key dip.

   

  More later.

   

  Nick

   

  From: pianotech-bounces at ptg.org [mailto:pianotech-bounces at ptg.org] On Behalf Of Nick Gravagne
  Sent: Tuesday, January 05, 2010 8:24 PM
  To: pianotech at ptg.org; joegarrett at earthlink.net
  Subject: Re: [pianotech] Action Ratios

   

  Any confusion with the formulas as shown in emails below exists with the (key dip - aftertouch). You cannot arbitrarily assign a number of your choosing to this value, although from a practical regulating standpoint this is often done. The idea of the overall transmission (or action) ratio is that the component ratio of (hammer travel - let off) to (key dip - aftertouch) is that of the product of the (output levers) to that of the (input levers). My studies assign input levers as effort arms and output levers as resistance arms.

   

  So, if we designate: 

   

  W = (hammer travel - let off)

  S = (key dip - aftertouch)

   

  H = rear key lever arm resistance (key out)

  Rs = whippen lever arm resistance (whippen out)

  K = hammer lever long arm resistance (hammer out)

   

  V = front key lever effort (key in)

  Ra = whippen lever arm effort (whippen in)

  N = hammer lever short arm effort (hammer in)

   

  The relationship, then, of W / S should be that of the product of Resistance Arms / Effort Arms. But note that S (key dip - aftertouch) implies a calculated value, not an arbitrary one. In order to isolate S the formula works out thus:

   

  S = (product of Effort Arms times W) / (product of Resistance Arms)

   

  Once the theoretical value of S is isolated; the Action ratio can be calculated.

   

  So, using some values from a Kawai action model:

   

  W = 46 mm - 2 mm = 44 mm)

  S = (10.5 - aftertouch)

   

  H = 126             V = 245

  Rs = 94             Ra = 67

  K = 141             N = 18.25 (Jack to knuckle contact taken at half stroke)

   

  S = (V x Ra x N x W) / (H x Rs x K) = (245 x 67 x 18.25 x 44) / (126 x 94 x 141) = 7.89 mm which is the theoretical key dip minus aftertouch. Thus aftertouch = 2.61 mm. Said another way; dip prior to aftertouch is 75% of the key stroke, while aftertouch accounts for the remaining 25%.

   

  Given this, the ratio of W to S is equal to the ratio of lever arms thus: W / S = 44 / 7.89 = 5.58 and the ratio of the products of the RAs / EAs = 5.575. The ratios not only agree, but they define the Action Ratio at half stroke.

   

  Now, the so-called aftertouch value of 2.61 mm seems odd, but it is important to realize that the measurement for this (if we can call it that) begins the exact moment that the jack tender makes contact with the let-off button in a well regulated action, and continues to the a solid bottom at full key dip. In addition, if the aftertouch value is far off from the theoretical remember that 

   

  In any case, measurements aside, this is how the ratios interact. For more, see Pfeiffer's The Piano Hammer pages 110 and 111. It is necessary to work the formulas and read a few things between the lines as some of the info references his other book The Piano Key and Whippen.

   

  Nick Gravagne, RPT

  Piano Technicians Guild

  Member Society Manufacturing Engineers

  Voice Mail 928-476-4143

   


------------------------------------------------------------------------------

   
-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://ptg.org/pipermail/pianotech.php/attachments/20100106/f2aecd25/attachment-0001.htm>
-------------- next part --------------
A non-text attachment was scrubbed...
Name: Action ratio experiment.xls
Type: application/vnd.ms-excel
Size: 15360 bytes
Desc: not available
URL: <http://ptg.org/pipermail/pianotech.php/attachments/20100106/f2aecd25/attachment-0001.xls>


More information about the pianotech mailing list

This PTG archive page provided courtesy of Moy Piano Service, LLC