[CAUT] Sperrhake Harpsichord wire

Fred Sturm fssturm at unm.edu
Fri Dec 4 11:48:58 MST 2009


On Dec 4, 2009, at 11:06 AM, Ron Nossaman wrote:

>>
>>    Part of my assumption is based on the fact that bridges are  
>> curved to a scale, and it is not a logarithmic one that doubles  
>> lengths for every octave. A given string at the same tension should  
>> give an octave lower at double the length, two octaves lower at 4  
>> times the length, etc. Scales foreshorten these distances. Hence, I  
>> reason that the given size wire would need to be at a lower tension  
>> if the octave lower length were shorter. Is this not true?
>
> It's certainly irrelevant to whether replacing a given string with a  
> different size changes break%, which has, I think, been the question.


	Let me make another stab at explaining this reasoning. We'll make a  
one octave piano (for simplicity) with one gauge. We'll make the  
bottom string length two times the top string length, and do a  
logarithmic scale between. All strings will be at precisely the same  
tension, and percent of breaking point, correct?
	Now we foreshorten the scale, still logarithmic, but the lowest  
string is less than twice the top string. Tension will reduce evenly  
as we go down the scale, correct? Hence we will be at a lower percent  
of breaking point as we go down the scale, correct?
	All pianos and harpsichords have foreshortened scales. It doesn't  
take a spreadsheet to see that if you strung a piano with 13 wire  
throughout, the tension would decrease as you went down the scale, and  
the percent of breaking point would also decrease. And that the same  
thing would be true if you started at the point where any other gauge  
began and strung the lower portion of the piano with that gauge.
	Or is there some fallacy I am missing in this reasoning?
Regards,
Fred Sturm
University of New Mexico
fssturm at unm.edu







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