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surface calculus


francoisdenis
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The large square areas can be calculated with straight measurements, then the remaining "odd" areas can be measured with a planimeter.  Accuracy is advertised as 0.2%  Several measurements should be made, and the total (added-up) result should be divided by 3 in order to get best accuracy.Planimeter.thumb.jpg.d9468f929b053e2f84d709194cb11e7f.jpg

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13 hours ago, David Beard said:

I just did a quick version of this in a graphics program, using a Bros A example.

You can ballpark the surface as 3/4  the framing rectangle.  My actual rough calc gave .76x.

So:

Surface ~ (width x height) x .75

Or ~   (35.4 × 20) x .75 = 541 sqr cm

Interesting,  so the approximative formula would be 

Surface ~ (width x height) x .75

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Cut it out on good 100g/sq.m paper. The weigh the paper on a precise scale. 1 gram is 1/100 square meter or one square dm. 0.01 gram is one square centimeter, which is about the limit if resolution on this.

Before modern computerized nethods, mineral distributions in thin sections of rocks was determined in a similar way, using tin foil instead of paper. When I was a young geology student I printed photos on thick photo paper, cut out all the grain shapes and sorted them according to mineral, and weighed the heaps. I only did this once, to prove that my alternative computerized version gave the same result, but it still works for a one-off - at least on simple shapes. Come to think of it, good tin foil would be better...

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15 hours ago, Felefar said:

Cut it out on good 100g/sq.m paper. The weigh the paper on a precise scale. 1 gram is 1/100 square meter or one square dm. 0.01 gram is one square centimeter, which is about the limit if resolution on this.

Before modern computerized nethods, mineral distributions in thin sections of rocks was determined in a similar way, using tin foil instead of paper. When I was a young geology student I printed photos on thick photo paper, cut out all the grain shapes and sorted them according to mineral, and weighed the heaps. I only did this once, to prove that my alternative computerized version gave the same result, but it still works for a one-off - at least on simple shapes. Come to think of it, good tin foil would be better...

Ok, probably the simple way, I will do that and return the result

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I use both the weighing cut-out paper method and the computer tracing methods.

Attached is an example of the areas calculated in my computer drawing program Microspot MacDraft.  I traced the upper portion of the outline of a CT scan of the rib section of a DG violin with a free form drawing tool.  I traced the plate where it wasn't glued and free to vibrate.

The program shows the resultant area.  I did the same for the bottom portion.

I had made the upper and lower portion split right on top of the sound post.  It appears that the sound post location splits the vibrating plate area into equal parts.

Area drawings.jpg

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25 minutes ago, Marty Kasprzyk said:

I use both the weighing cut-out paper method and the computer tracing methods.

Attached is an example of the areas calculated in my computer drawing program Microspot MacDraft.  I traced the upper portion of the outline of a CT scan of the rib section of a DG violin with a free form drawing tool.  I traced the plate where it wasn't glued and free to vibrate.

The program shows the resultant area.  I did the same for the bottom portion.

I had made the upper and lower portion split right on top of the sound post.  It appears that the sound post location splits the vibrating plate area into equal parts.

Area drawings.jpg

Thank you for posting this result.  I find this extremely interesting.  :)

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Here,s a possible method ,,,, if one were to for instance cover the surface with some material, close fitting like a cloth or foil maybe a paper dampened, , then to cut an accurate outline ...for holes what not , one could simply weigh the materials and compare to a known weight for x sq inches , might take a bit of trials , and testing, but in theory, should work.

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9 hours ago, Marty Kasprzyk said:

I use both the weighing cut-out paper method and the computer tracing methods.

Attached is an example of the areas calculated in my computer drawing program Microspot MacDraft.  I traced the upper portion of the outline of a CT scan of the rib section of a DG violin with a free form drawing tool.  I traced the plate where it wasn't glued and free to vibrate.

The program shows the resultant area.  I did the same for the bottom portion.

I had made the upper and lower portion split right on top of the sound post.  It appears that the sound post location splits the vibrating plate area into equal parts.

Area drawings.jpg

My bad for not reading the whole thread, .... 

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On 9/1/2019 at 4:24 PM, Felefar said:

Cut it out on good 100g/sq.m paper. The weigh the paper on a precise scale. 1 gram is 1/100 square meter or one square dm. 0.01 gram is one square centimeter, which is about the limit if resolution on this.

Before modern computerized nethods, mineral distributions in thin sections of rocks was determined in a similar way, using tin foil instead of paper. When I was a young geology student I printed photos on thick photo paper, cut out all the grain shapes and sorted them according to mineral, and weighed the heaps. I only did this once, to prove that my alternative computerized version gave the same result, but it still works for a one-off - at least on simple shapes. Come to think of it, good tin foil would be better...

IIRC, Galileo used this method for measuring the area covered by a curtate cycloid. However, in this era of computers, I second the recommendation by  @Don Noon to use a CAD program. Google around for one that does this calculation.

I see that @Marty Kasprzyk has the answer. I too need to read the entire thread before typing.

I thought that the center of mass with the f-hole cutouts was at the nicks, but I never examined it closely. The sounpost location would make better sense to drive the plate uniformly. However, I never tested that idea too.

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8 hours ago, David Beard said:

Note that because of soundholes, the center of mass and the dividing line of area will differ some.

I had originally thought that the bridge feet should be at the center of the area but I proved myself to be wrong because it was the sound post not the bridge location at the center.

I then thought it wasn't the line that divided the upper and lower bouts to be be in equal in area but rather they should be at the balance point (first moment of area) which is more difficult to calculate. Then I thought the line should be at the center point of polar moment of inertia (second moment of area) important for vibration dynamics which can only be calculated with much more expensive engineering software programs. 

I then thought that this could turn out to be a waste of time and money.

 

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