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Bill Stanley
Joined: 15 Apr 2004
Posts: 315
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 Posted: Thu May 06, 2004 12:22 am Post subject: Arcs with a zero sweep angle |
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Arcs with a zero sweep angle
It appears that while fixing ‘flat arc picking’ I unintentionally did something that makes the software more sensitive to arcs with a zero sweep angle. You might want to select the menu item Clean Up Drawing and remove the arcs with a sweep angle too small to be seen. One might also question whey objects are in the drawing that cannot be seen but we will leave that for another day. The algorithm that calculates the pick point relative to the arc is simply more sensitive. Additional bulletproofing will be added.
Sorry. Thanks.
Bill Stanley |
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Jeff H
Joined: 14 Apr 2004
Posts: 29
Location: Naples, FL
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| Posted: Tue May 18, 2004 6:35 pm Post subject: |
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I work on translated ACAD files from civil engineers frequently. It is quite common to find arc segment of only a few minutes with very long radii. At scale the arcs are more than a few pixels in length. A seemingly small sweep angle does not necessarily make a "small object" that should be targeted for "clean up".
Jeff |
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Matt
Joined: 13 Apr 2004
Posts: 391
Location: Sterling, Virginia
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| Posted: Wed May 19, 2004 6:37 pm Post subject: |
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Philosophical digression:
let's examine an arc segment of a few minutes in length and with a very long radiius. One that one of these surveyors might draw.
For fun, let's pretend that they are drawing at full scale, 1:1, at actual size, not at 1"=40' or 1:500 or something.
and let's say that the arc subtends 6 minutes (or 6/60 of 1/60 of 1/60 of a circle) and that the radius is, oh, 200 feet. that's big but not too big.
so how long is that line, and how much bend is there in it?
well, the circumference of a circle is 2*pi*radius which is 1256.637061435917 feet or 15079.644737231 inches.
so the length of our arc is that /60 /60 /10 or 0.4188790204786389 inches, just about 7/16" in the real world.
how much bend is there in that line?
well, the triangle in question has a 200' hypotenuse, an angle of 0 degrees, 0 minutes, and 3 seconds. if we can determine the length of the base of the triangle, we'll know the bend in the arc.
The base is equal to the cosine of the angle times the hypotenuse (which is 2400 inches).
the cosine of the angle is 0.9999999999999706; consequently the base of the triangle is 2399.999999999929 inches long.
this means that the 7/16" line has a deflection of .00000000008 inches in it's length.
an iodine atom is approximately .00000000004 inches across, meaning that this hypothetical 7/16" line deflects around two iodine atoms side-by-side - if you could get them adjacent, I couldn't find how to do that in my haste tonight.
(remember, we're drawing at full size here, not at a shrunken down scale.)
What is the thickness of the thinnest line you use in your drawings? I have used .005, however I find that it does not photocopy reliably, but let's use that as our reference..
the arc your ACAD friend is drawing has a deflection of approximately 1/62,500,000th of the thickness of that thin line. At full scale. that's the maximum difference between the arc he sent you and a straight line.
for reference, it's about 93,000,000 miles to the sun.
Now if I were in this dilemma, I would probably give my AutoCAD buddy a call and suggest drawings of this type are probably costing him a trifle more effort to create then they are worth, and he might want to consider bringing his standards down, oh, maybe a few orders of magnitude.
but that's just me. and I don't know, maybe the jurisdiction you're practicing in requires eleven or twelve significant digits, maybe your buddy has uses angstrometers to do his survey work.
I'd hate to be the judge who had to adjudicate the property line dispute over those two iodine atoms.
in the end, I'd probably give one to each property owner.
-------------
but never fear, in the meanwhile, I don't think PowerCADD would delete any of these objects without your consent. |
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Derek
Joined: 13 Apr 2004
Posts: 568
Location: Melbourne, Australia
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| Posted: Wed May 19, 2004 6:56 pm Post subject: |
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Does anyone else here think Matt has too much time on his hands?  |
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Matt
Joined: 13 Apr 2004
Posts: 391
Location: Sterling, Virginia
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| Posted: Thu May 20, 2004 7:49 am Post subject: |
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| not enough time apparently, because those calculations were done based on an arc of six seconds rather then what I said: six minutes. ALWAYS CHECK YOUR WORK.. dang! |
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JohnMorse
Joined: 14 Apr 2004
Posts: 294
Location: Birmingham, AL
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| Posted: Thu May 20, 2004 7:54 am Post subject: |
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| Matt wrote: | | ...let's say that the arc subtends 6 minutes (or 6/60 of 1/60 of 1/60 of a circle)... |
360 degrees in a circle, 60 minutes in a degree, 60 seconds in a minute.
Therefore 6 minutes is 6/60ths of 1/360th of a circumference. 6 seconds would be 6/60ths of 1/60th of 1/360th of a circumference.
As for the rest, it's too far over my head to check - but the point is well taken, anyway. |
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Matt
Joined: 13 Apr 2004
Posts: 391
Location: Sterling, Virginia
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| Posted: Thu May 20, 2004 8:10 am Post subject: |
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right. So an arc of 6 minutes with a radius of 200 feet is 4.188790204786389 inches long (say 4 1/8") and the bend in the arc is actually .000000007050857675494626 inches, or 176 iodine atom-diameters.
sorry for the confusion. |
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JohnMorse
Joined: 14 Apr 2004
Posts: 294
Location: Birmingham, AL
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| Posted: Thu May 20, 2004 9:25 am Post subject: |
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Don't apologize to me Matt, think of the 174 iodine atoms. javascript:emoticon(' ') |
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Matt
Joined: 13 Apr 2004
Posts: 391
Location: Sterling, Virginia
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| Posted: Thu May 20, 2004 12:27 pm Post subject: |
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| I know! I know.. |
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