Ratios across keystroke, 15 vs 17 knuckle
Posted on Wed 4th March 2015 12.42PM in category Qtrs1thru3_2015

I modeled up two actions in CAD, one with a 15 mm knuckle, the other 17 mm.  They both have the same average "static" ratio across the pre-let-off stroke (5.97).  David Love had asked me if an action were configured in two different ways (one with a 15 knuckle, the other 17), while keeping the AR the same, would there be a difference in inertia between the two cases?  I admitted I didn't know, but highly suspected that the ratios (both "static" and gear) would vary more across the stroke, for the 15 mm knuckle location.  I also detailed the ramifications such a move would have on let-off forces (they go up for the 15 mm case)!  To maintain the same average ratio between the two cases, David suggested simply sliding the key and balance rail fore/aft, thus changing only one of those two key-to-wippen levers.

See Figures


An image of the CAD model for a midstroke position of the 17 mm case is given as Figure 1.  Figure 2 is the same, but for the 15 mm knuckle configuration.

From the model, I extracted two types of "static" ratio:  one based on vertical movement of the tip of the hammerhead, the other based on vertical movement of the approximate center of mass (c.m.) of the hammerhead.  For either configuration, the AR based on the tip varies much more across the stroke than the AR based on the c.m.!   David had indicated that some technicians believe there may be a different AR emanating from mass-based (or force based?) methods, as opposed to displacement (movement)-based methods.   I haven't been able to find sufficient information out there to determine what they actually mean by mass-based or force-based AR's.  However, my next post will show that an AR calculated from vertical force components at finger and hammerhead is identical to an AR calculated from vertical movements at those same locations.  It should be noted that most displacement-based methods seem to focus on movement of the tip/crown of the hammer.  Force-based methods, on the other hand, almost inherently use the approximate c.m. of the hammerhead.   Certainly this is the case for my own method of determining an action ratio, and would appear to be the case with David Stanwood's Strike Weight Ratio as well.  It may be that this attempted comparison of apples (tips) to oranges (c.m.'s) has led some to believe there are unique "force" and "displacement" AR's.  As far as a "mass" based AR, there is no "static" (AR) ratio related to mass.  When it comes to mass, the Gear Ratio (GR) is king, and must be known to calculate hammer-related Inertia at the Key.

Figure 3 shows two kinds of displacement-determined "static" ratios, for the design having the 17 mm knuckle location.  It also shows the gear ratio [GR].  All three are instantaneous values, calculated from the CAD model every 1/3rd of a millimeter across the keystroke, up to a typical start of let-off (7.3 mm).  Recall that the gear ratio is simply the instantaneous angular movement
of the shank/hammer, divided by the instantaneous angular movement of
the keystick.  Figure 4 is the same basic thing, but for the 15 mm knuckle location configuration.  I took great pains to ensure that the average static ratio (using the tip) was identical for both designs.  As the average values in the graph show, this was done accurately.  As you can see, the other static ratio (using the c.m.) also remains identical between the two designs.   Forcing the average static ratios to be equal also causes the average gear ratios to be equal as well!  This allows me to answer David Love's question.  Since the "local" hammer assembly inertia (about its own axis) is not varying, and the GR is also not varying between the two designs, the hammer inertia reflected to the keystick cannot vary either.  The local value multiplied by the square of GR is the value at the key.  Since neither are changing, the hammer inertia - at the key - cannot change.  Now, the inertia will be a little higher early in the stroke - and lower late in the stroke - for the 15 mm knuckle design.  But the average pre-let-off inertia will be exactly the same between the two designs.

A very glaring fact can be seen from Figures 3 and 4:  the AR based on the center of mass varies much less across the stroke than the AR based on the hammerhead tip/crown.  The hammerhead used here had a tip height of 51 mm above the shank centerline.  The pattern would seem to be that the taller the hammer is, the more the "tip AR" will vary across the stroke.  Of course, both ratios definitely decrease with increasing key travel.  The next thing seen is that the average "c.m. AR" is about 3.5% smaller than the average tip-based AR.  Of course, this will vary somewhat as the hammerhead height and c.m. location vary.  But 3.5% is a good starting point for discussion.

Finally, you can see that the instantaneous gear ratio (MR) also decreases quite continuously with increasing key depression.  On a percentage basis, it varies less than the tip-based AR, and approximately the same as the c.m.-based AR.  Remember, the gear/moment ratio is the ratio of instantaneous shank rotation to instantaneous keystick rotation.  As with the static ratios, averaging all these instantaneous values across the pre-let-off stroke yields the average values shown.

Astute observers can see from all this that I have the tools to go ahead and calculate theoretical force-based AR's at all points along the stroke.  I have the horizontal lever arm (not shown in drawings) of the tip and center of mass.  Plus, I have the gear ratio at that same point in the stroke.  The torque created by a force at the hammer is multiplied by the instantaneous gear ratio to get the torque (and thus force at the finger/AP) at the keystick.  It will be interesting to do this analysis, and see how the two force-based static ratios compare to the two displacement-based ratios discussed in this post!  I hope to make a post about those results next week.

Rick Voit

Comments:
Nibby
Fri 13th May 2016 6.48AM
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Dave
Thu 2nd July 2015 10.48AM
Figure 3 shows two kinds of displacement-determined "static" ratios, for the design having the 17 mm knuckle location. It also shows the gear ratio [GR]. All three are instantaneous values, calculated from the CAD model every 1/3rd of a millimeter across the keystroke, up to a typical start of let-off (7.3 mm). Recall that the gear ratio is simply the instantaneous angular movement
of the shank/hammer, divided by the instantaneous angular movement of
the keystick. Figure 4 is the same basic thing, but for the 15 mm knuckle location configuration. I took great pains to ensure that the average static ratio (using the tip) was identical for both designs. As the average values in the graph show, this was done accurately. As you can see, the other static ratio (using the c.m.) also remains identical between the two designs. Forcing the average static ratios to be equal also causes the average gear ratios to be equal as well! This allows me to answer David Love's question. Since the "local" hammer assembly inertia (about its own axis) is not varying, and the GR is also not varying between the two designs, the hammer inertia reflected to the keystick cannot vary either. The local value multiplied by the square of GR is the value at the key. Since neither are changing, the hammer inertia - at the key - cannot change. Now, the inertia will be a little higher early in the stroke - and lower late in the stroke - for the 15 mm knuckle design. But the average pre-let-off inertia will be exactly the same between the two designs.
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