David,
Your posts are coming in faster than I can type<G>
To other readers, this post includes a long length of antecedent posts,
but I retained them for reference. And to Jude --- thanks for the
compliment. RE the Boston Seminar, I've begun saying that my primary
motivation for accepting the invitation was to visit your shop, but you
attended a wedding instead and bummed me out<G>. Thanks for your posts
and insights.
The following addresses only the concept of the relationship of W / S to
Action Leverages.
The problem you are having with W / S is that the process of action
design or modification does not begin here (although it does include the
factor of a 10 mm dip + or - as a limiting factor). The whole point of
the EQ
EQ 1: W / S = X| RAs / X| EAs
Note that the X| means the "product of" the Resistance Arms (RA) divided
by the "product of" the Effort Arms (EA). It might be simpler to say
that w / s = action leverage ratio as geometrically laid out with the
action parts.
EQ 2: W / S = Action Leverages Ratio
Now the whole point here is to put the cart before the horse. EQ 1 and 2
are meant to show a relationship of equality. What is on the left side
of the = sign must equate to the right side. Since a 10 mm dip (+ or -)
has been a standard for decades, and since the W factor, which is blow
distance minus letoff, is also a limiting factor built in to case design
and action parameters of most pianos, etc. that leaves the S factor,
that unknown portion of the dip which exists prior to aftertouch (called
AT, a spongy and slippery character). Now don't go away.
If we consider the right side of the EQs only, we note that the action
leverages must interact to produce a 5.50 AR (or thereabout)
relationship and considering that there are six "levers" to consider
(two on the key, two on the whippen, two on the hammer shank), this 5.50
AR can be achieved in an infinite number of ways, and most of them could
be "wrong". Why wrong? Because although given that the right side
leverages might configure to a 5.50 AR, the parts might not fit together
well, or require extra long keys, or weird looking whippens, etc.
So let's assume that a limited but somewhat broad choice of parts
dimensions and leverage layouts will produce a satisfactory 5.50 AR
Before continuing, it is important to understand AT from a design
standpoint and not confuse it with a regulation parameter, although a
relationship exits. Consider:
EQ 3: S = (V x Ra x N x W) / (H x Rs x K)
(See emails below for factor designations, which come from Pfeiffer's
books)
This EQ solves for S ("engaged dip"). The difference between engaged dip
and full dip I refer to as the escapement dip. We must keep in mind
exactly where this escapement dip (or theoretical AT) is occurring; it
will not be under the key and above the front rail punching. It can only
exist at the front-most part of the key, its very front edge, assuming
the key stick is measured this way. Same is true for total dip. This
might be a little tricky to test, but forgetting about front rail
punchings, the allotment of key dip prior to letoff vs. what is left for
AT must fit into a total of 10 mm, and still allow for a practical
regulation AT.
Thus, with a total dip of 10 mm which is allocated as 8 mm of "engaged
dip" with the remaining 2 mm as "escapement dip", we are good to go. As
you can see, we really don't even have a unified language on here.
Again, theoretical AT in light of these discussions must be understood
to take place the moment the jack tender makes contact with the letoff
button, and continues to the bottom of a 10 mm drop, located at the
front most part of the key stick.
Let's take two actions for example:
1) Where
W = 46 mm - 2 mm = 44 mm)
S = (10 - aftertouch)
Note that of the four factors given here at the outset, only the
aftertouch is unknown.
H Rear Key = 126 V Front Key = 245
Rs Resistance whip = 94 Ra Effort Whip = 67
K Resistance Shank = 141 N Effort Shank = 18.50 (Jack
to knuckle contact taken at half stroke)
The Action Ratio works out to 5.50, engaged dip = 8 mm, and escapement
dip = 2 mm. This action will regulate OK and without any weight issues.
2) Where
W = 46 mm - 2 mm = 44 mm)
S = (10 - aftertouch)
H = 135 V = 250
Rs = 93 Ra = 69.5
K = 141 N = 18.50 (Jack to knuckle contact taken at
half stroke)
With these parameters, the Action Ratio works out essentially the same
with a 5.51 AR, engaged dip = 7.99 mm, and escaped dip = 2.01 mm. This
action will regulate OK even though the parts dimensions and leverages
layout is different from Action #1; also no weight issues.
An indirect proportion exists as to Action Ration (AR) relative to Dip
parameters, and that is the whole point of EQ 2: W / S = Action Ratio
1) As AR goes up (mechanical advantage lessens as distances increase),
the engaged dip decreases and escapement dip increases, all fitting into
the total dip of 10 mm.
2) As the AR goes down (mechanical advantage increases as distances
decrease), the engaged dip increases and escapement dip decreases.
For example: a 5.50 AR should yield an 8 mm engaged dip, followed by a 2
mm escaped dip
A 5.00 AR should yield an 8.80 mm engaged dip,
followed by a 1.20 mm escaped dip.
My personal preference is the 5.50 AR
Finally, and although this post has been about theory and not action
modification (which is what we are always dealing with as rebuilders)
remember that from a practical standpoint, any modifications made to
increase the action's mechanical advantage (make it lighter to the
touch) must be paid for by decreasing the distances that the parts move
upward. Eventually we run out of headroom. It is not uncommon to hear of
instances where the total dip needed to be increased to 11 mm and more.
Steinway's specs for dip have been officially stated for models S, M, L
at approximately 0.420" (10.63 mm), and the B, D at approximately 0.390
(9.867 mm).
Unofficially the specs are stated as approximately anything that works.
David, I will look again at your other posts.
Nick Gravagne, RPT
Piano Technicians Guild
Member Society Manufacturing Engineers
Voice Mail 928-476-4143
_____
From: pianotech-bounces at ptg.org [mailto:pianotech-bounces at ptg.org] On
Behalf Of David Love
Sent: Wednesday, January 06, 2010 10:06 PM
To: pianotech at ptg.org
Subject: Re: [pianotech] Action Ratios
OK, I get it but what bothers me is that the "choice" of defining the
aftertouch is just that, a choice in order to make the numbers work.
It's a bit too loosely defined to my liking. With Nick's side of the
equation 44 mm makes sense because, in fact, 46 mm blow - 2 mm let-off
is about how we set it. But where does 2.6 mm aftertouch come from. It
appears as if it's an arbitrary number that must be assigned to an
overall key travel of 10.5 mm in order to make the w/s ratio conform to
the effort/resistance ratio. Similarly on yours, the 1.5 mm makes more
sense, I suppose (though it's really more like .75 mm in practice if we
use the blow, let-off numbers as a reality check) but then your blow
distance is artificial in that it must deviate from the actual blow
distance in order to make your w/s ratio conform to the effort
resistance ratio. So what does the w/s then really tell us? That we
can play with the numbers to make things work but it doesn't apparently
have any real world application. It's arbitrary. So the question
still stands as far as I'm concerned. Suppose we want 10 mm dip, 46
blow distance where do we set the effort/resistance ratio and who can we
calculate that (if we can)? Or do we resign ourselves to "experience
has shown."
David Love
www.davidlovepianos.com
----- Original Message -----
From: David <mailto:davidlovepianos at comcast.net> 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
_____
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