OK I understand that. So my next question is what would you consider the
minimum escapement dip that you can live with and what do you consider a
reasonable ceiling for AR overall in terms of conforming hammer weight (or
strike weight if you prefer).
From: pianotech-bounces at ptg.org [mailto:pianotech-bounces at ptg.org] On Behalf
Of Nick Gravagne
Sent: Thursday, January 07, 2010 11:07 AM
To: pianotech at ptg.org
Subject: Re: [pianotech] Action Ratios
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 Love <mailto:davidlovepianos at comcast.net>
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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