ampage Tube Amps / Music Electronics 
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6/4/2005 6:44 PM  
Dr. Strangelove 
Re: Iteration is not required to find turns count Joe Gwinn wrote:
[RE: beveled blade edges]
You need to remember that these pickups were sold on the merit of being refined to Danny Gatton's taste over a 10 year period. If taped edges had been acceptable to Mr. Gatton, then Barden would have taped them. The low fill factor suggests low winding tension used to avoid cutting the wire. drh  

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6/5/2005 7:58 AM  
Joe Gwinn 
On 6/5/2005 12:44 AM, Dr. Strangelove said:
I would guess that the classic pickup design where the wire is wound directly on the six 5mm magnets or slug poles was the model, carried over more or less directly to blade pickups, even though blades are 1.6mm, far thinner than 5mm. And it did work at first. While the pickups tended to die young, this would have been well after Mr. Gatton had moved on to another experimental pickup.


6/5/2005 8:55 PM  
Joe Gwinn 
Actually, I now see a simple way to show that the only the mean turn length is required, without calculus. Draw the following diagram: The Y axis (vertical) is turn length in inches (or centimeters). The X axis is the height of the turn above the core, in the same length units as for the Y axis. From the geometry, we know that the length of a turn varies linearly as a function of height above the core. At zero height, right on the core, the length is Lmin. At the top of the winding, where height equals winding depth, the length in Lmax. Draw a line on the diagram from Lmin at zero height to Lmax to Lmax at height=depth. Halfway between Lmin and Lmax is Lmean. Specifically, Lmean= (Lmin+Lmax)/2 is at height=depth/2. Mark this point on the diagram. It's right in the middle. Draw a light horizontal line through Lmean, parallel to the X axis. Now seen are two identical triangles, one to the left of and below Lmean, the other to the right and above. No matter what the values of Lmin and Lmax, these two triangles are identical, so the total area under the line from Lmin to Lmax is Lmean times Depth. This is obvious when looking at the diagram, even if the words here don't quite do it. 

6/7/2005 12:14 AM  
Dr. Strangelove 
Joe Gwinn wrote:[QUOTE]Actually, I now see a simple way to show that the only the mean turn length is required, without calculus... This is obvious when looking at the diagram, even if the words here don't quite do it.[/QUOTE] Can you draw a crude diagram with a paint program and post a link? drh  

6/7/2005 6:38 AM  
Joe Gwinn 
On 6/7/2005 6:14 AM, Dr. Strangelove said: [QUOTE]Joe Gwinn wrote: "Actually, I now see a simple way to show that the only the mean turn length is required, without calculus... This is obvious when looking at the diagram, even if the words here don't quite do it." Can you draw a crude diagram with a paint program and post a link?[/QUOTE]Yes. Actually, I should combine the drawing with a short description of the algorithm. This will take a few days. 

6/10/2005 2:37 PM  
Joe Gwinn 
On 6/7/2005 12:38 PM, Joe Gwinn said: [QUOTE]On 6/7/2005 6:14 AM, Dr. Strangelove said: "Joe Gwinn wrote: "Actually, I now see a simple way to show that the only the mean turn length is required, without calculus..." Can you draw a crude diagram with a paint program and post a link?" Yes. Actually, I should combine the drawing with a short description of the algorithm. This will take a few days.[/QUOTE]I just posted on my website (http://home.comcast.net/~joegwinn/) a threepage document with drawings, explanation, stepbystep algorithm, and a worked example. 

6/10/2005 4:23 PM  
Dr. Strangelove 
Looks good, Joe. Thanks. drh  

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